Exercises

Heat transfer modes and steady conduction

Insulation around a spherical vessel

A spherical vessel has insulation extending from 𝑟1 =0.50 m$r_1=0.50\ \mathrm{m}$ to 𝑟2 =0.55 m$r_2=0.55\ \mathrm{m}$. Take the insulation conductivity as 𝑘 =0.040 W m−1 K−1$k=0.040\ \mathrm{W\,m^{-1}\,K^{-1}}$. Its inner and outer insulation surface temperatures are 80 °C$80\,\degree\mathrm{C}$ and 20 °C$20\,\degree\mathrm{C}$. Assume steady radial conduction and neglect heat flow through supports and pipe connections. What is the heat loss through the insulation?

Worked solution

The given temperatures are at the insulation surfaces, so the required resistance is the insulation resistance alone. From Eq. 7,

𝑅sph=10.50−10.554𝜋(0.040)≃0.362 K W−1$$R_{\mathrm{sph}}=\frac{\frac{1}{0.50}-\frac{1}{0.55}}{4\pi(0.040)} \simeq0.362\ \mathrm{K\,W^{-1}}$$

The outward heat loss is

˙𝑄𝑟=80−20𝑅sph≃166 W$$\dot Q_r=\frac{80-20}{R_{\mathrm{sph}}} \simeq166\ \mathrm{W}$$

If the given temperatures were fluid temperatures, the fluid-film resistances would also be needed. For a vessel containing a cold fluid, the same resistance method gives the inward heat gain.

Heat transfer modes and steady conduction

Glass window: heat loss and inside surface temperature

Consider a 0.8-m-high and 1.5-m-wide glass window with a thickness of 8 mm and a thermal conductivity of 𝑘 =0.78 W/(m⋅°C)$k=0.78\ \mathrm{W/(m\cdot\degree C)}$. Determine the steady rate of heat transfer through this glass window and the temperature of its inner surface for a day during which the room is maintained at 20 °C$20\ \degree\mathrm{C}$ while the temperature of the outdoors is −10 °C$-10\ \degree\mathrm{C}$. Take the heat transfer coefficients on the inner and outer surfaces of the window to be ℎ1 =10 W/(m2⋅°C)$h_1=10\ \mathrm{W/(m^2\cdot\degree C)}$ and ℎ2 =40 W/(m2⋅°C)$h_2=40\ \mathrm{W/(m^2\cdot\degree C)}$, which includes the effects of radiation.

Worked solution

Heat passes from the room air to the inner glass surface, through the glass, and then from the outer glass surface to the surroundings. These three steps form a series resistance network.

Figure 8: Heat flows from room air at 20 degrees Celsius through the inside film, glass and outside film to air at minus 10 degrees Celsius. The two glass surface temperatures are the unknown intermediate temperatures.

Data

Quantity	Symbol	Value
Window height	𝐻$H$	0.8 m$0.8\ \mathrm{m}$
Window width	𝑊$W$	1.5 m$1.5\ \mathrm{m}$
Glass thickness	𝐿$L$	8 mm =0.008 m$8\ \mathrm{mm}=0.008\ \mathrm{m}$
Glass thermal conductivity	𝑘$k$	0.78 W m−1 K−1$0.78\ \mathrm{W\,m^{-1}\,K^{-1}}$
Room air temperature	𝑇∞,1$T_{\infty,1}$	20 °C$20\,\degree\mathrm{C}$
Outside air temperature	𝑇∞,2$T_{\infty,2}$	−10 °C$-10\,\degree\mathrm{C}$
Inside heat transfer coefficient	ℎ1$h_1$	10 W m−2 K−1$10\ \mathrm{W\,m^{-2}\,K^{-1}}$
Outside heat transfer coefficient, including radiation	ℎ2$h_2$	40 W m−2 K−1$40\ \mathrm{W\,m^{-2}\,K^{-1}}$

A temperature difference of 1 °C$1\,\degree\mathrm{C}$ equals 1 K$1\ \mathrm{K}$. We can therefore express the supplied conductivity and heat transfer coefficients using kelvin-based units in the calculation.

We need the heat loss ˙𝑄$\dot Q$ and the inside glass surface temperature 𝑇𝑠,in$T_{s,\mathrm{in}}$.

Assumptions

• Steady state.
• Unidirectional heat flow, normal to the glass.
• Heat transfer through the edges and frame is neglected.
• No heat generation within the glass.
• Constant thermal conductivity and heat transfer coefficients.
• Uniform temperature on each glass surface.

The supplied outside heat transfer coefficient includes radiation, so no separate radiation resistance is added.

Calculations

First calculate the area normal to heat flow:

𝐴=𝐻𝑊$$A=HW$$

Substituting the window dimensions gives

𝐴=0.8×1.5=1.2 m2$$A=0.8\times1.5=1.2\ \mathrm{m^2}$$

The inside film resistance is

𝑅in=1ℎ1𝐴$$R_{\mathrm{in}}=\frac{1}{h_1A}$$

Using the inside coefficient and window area gives

𝑅in=110×1.2=0.083333 K W−1$$R_{\mathrm{in}}=\frac{1}{10\times1.2} =0.083333\ \mathrm{K\,W^{-1}}$$

For conduction through the glass,

𝑅𝑔=𝐿𝑘𝐴$$R_g=\frac{L}{kA}$$

Substituting the thickness, conductivity and area gives

𝑅𝑔=0.0080.78×1.2=0.008547 K W−1$$R_g=\frac{0.008}{0.78\times1.2} =0.008547\ \mathrm{K\,W^{-1}}$$

The outside film resistance is

𝑅out=1ℎ2𝐴$$R_{\mathrm{out}}=\frac{1}{h_2A}$$

Substituting the outside coefficient gives

𝑅out=140×1.2=0.020833 K W−1$$R_{\mathrm{out}}=\frac{1}{40\times1.2} =0.020833\ \mathrm{K\,W^{-1}}$$

The same heat rate passes through all three resistances, so they add in series:

𝑅total=𝑅in+𝑅𝑔+𝑅out$$R_{\mathrm{total}}=R_{\mathrm{in}}+R_g+R_{\mathrm{out}}$$

Substituting the three resistances gives

𝑅total≃0.112714 K W−1$$R_{\mathrm{total}}\simeq0.112714\ \mathrm{K\,W^{-1}}$$

The temperature difference between the air on either side is

Δ𝑇=𝑇∞,1−𝑇∞,2$$\Delta T=T_{\infty,1}-T_{\infty,2}$$

Substituting the air temperatures gives

Δ𝑇=20−(−10)=30 K$$\Delta T=20-(-10)=30\ \mathrm{K}$$

The heat loss can now be calculated using Eq. 2. Writing that relation for this window gives

˙𝑄=𝑇∞,1−𝑇∞,2𝑅total$$\dot Q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{\mathrm{total}}}$$

Substituting the temperature difference and total resistance gives

˙𝑄=300.112714≃266.16 W$$\dot Q=\frac{30}{0.112714}\simeq266.16\ \mathrm{W}$$

Heat flows from the warmer room to the colder outdoors. To find the inside glass temperature, consider the temperature drop across the inside film:

𝑇∞,1−𝑇𝑠,in=˙𝑄𝑅in$$T_{\infty,1}-T_{s,\mathrm{in}}=\dot QR_{\mathrm{in}}$$

The calculated heat rate gives an inside-film temperature drop of

Δ𝑇in=266.16×0.083333≃22.180 K$$\Delta T_{\mathrm{in}}=266.16\times0.083333 \simeq22.180\ \mathrm{K}$$

Rearranging the inside-film relation gives

𝑇𝑠,in=𝑇∞,1−˙𝑄𝑅in$$T_{s,\mathrm{in}}=T_{\infty,1}-\dot QR_{\mathrm{in}}$$

Subtracting the temperature drop from the room temperature gives

𝑇𝑠,in=20−22.180=−2.180 °C$$T_{s,\mathrm{in}}=20-22.180=-2.180\,\degree\mathrm{C}$$

We can check this result by working from the outside air towards the room. The outer glass surface must be warmer than the outside air by ˙𝑄𝑅out$\dot QR_{\mathrm{out}}$:

𝑇𝑠,out=𝑇∞,2+˙𝑄𝑅out$$T_{s,\mathrm{out}}=T_{\infty,2}+\dot QR_{\mathrm{out}}$$

Substitution gives

𝑇𝑠,out=−10+266.16×0.020833≃−4.455 °C$$T_{s,\mathrm{out}}=-10+266.16\times0.020833 \simeq-4.455\,\degree\mathrm{C}$$

The temperature drop through the glass is therefore approximately

−2.180−(−4.455)=2.275 K.$$-2.180-(-4.455)=2.275\ \mathrm{K}.$$

Dividing this drop by 𝑅𝑔$R_g$ recovers the heat rate of approximately 266 W$266\ \mathrm{W}$. The surface temperatures also have the expected order: temperature falls from the room air to the inner glass surface, then to the outer surface and finally to the outside air.

Answer

The heat loss from the room through the window is

˙𝑄≃266 W$$\dot Q\simeq266\ \mathrm{W}$$

The inside glass surface temperature is

𝑇𝑠,in≃−2.18 °C$$T_{s,\mathrm{in}}\simeq-2.18\,\degree\mathrm{C}$$

Most of the resistance is in the two fluid films, with the largest drop across the inside film. This explains why the glass surface is much colder than the room air. Adding a trapped air layer, as in double glazing, increases the resistance and reduces heat loss at the same air temperatures. The calculated inside surface is below freezing, so frost may form if sufficient moisture reaches it.

Heat transfer modes and steady conduction

Temperature and conservation in a heated wire

A wire has radius 𝑅 =0.002 m$R=0.002\ \mathrm{m}$, length 𝐿 =0.50 m$L=0.50\ \mathrm{m}$, thermal conductivity 𝑘 =15 W m−1 K−1$k=15\ \mathrm{W\,m^{-1}\,K^{-1}}$ and electrical conductivity 𝑘𝑒 =106 S m−1$k_e=10^6\ \mathrm{S\,m^{-1}}$. Its current density is 𝐼 =106 A m−2$I=10^6\ \mathrm{A\,m^{-2}}$, and its surface is held at 𝑇0 =25 °C$T_0=25\ \degree\mathrm{C}$. Use the wire model above to calculate the centre temperature, mean temperature and heat loss through the curved surface. Verify that the heat loss equals the total heat generation.

Calculation and answer

The electrical current produces heat at the volumetric rate 𝑆𝑒 =𝐼2/𝑘𝑒 =106 W m−3$S_e=I^2/k_e=10^6\ \mathrm{W\,m^{-3}}$.

From Eq. 11, the centre-to-surface temperature difference is

106(0.002)24×15=0.066667 K.$$\frac{10^6(0.002)^2}{4\times15}=0.066667\ \mathrm{K}.$$

The mean rise from Eq. 12 is half this value, 0.033333 K$0.033333\ \mathrm{K}$. Adding these rises to 𝑇0$T_0$ gives 𝑇max⁡ =25.0667 °C$T_{\max}=25.0667\ \degree\mathrm{C}$ and ¯¯¯¯𝑇 =25.0333 °C$\overline T=25.0333\ \degree\mathrm{C}$.

The surface flux is

𝑞𝑟(𝑅)=106×0.0022=1000 W m−2.$$q_r(R)=\frac{10^6\times0.002}{2}=1000\ \mathrm{W\,m^{-2}}.$$

Multiplying by the curved area

2𝜋𝑅𝐿=0.00628319 m2$$2\pi RL=0.00628319\ \mathrm{m^2}$$

gives ˙𝑄𝑅 =6.28319 W$\dot Q_R=6.28319\ \mathrm{W}$.

Independently, the total heat generation is 𝑆𝑒𝜋𝑅2𝐿 =6.28319 W$S_e\pi R^2L=6.28319\ \mathrm{W}$. This equals the surface heat loss, confirming the energy balance. The wire loses approximately 6.28 W$6.28\ \mathrm{W}$ at steady state.

Heat transfer modes and steady conduction

Temperature drops through a furnace wall

A furnace wall separates hot gas at 600 °C$600\ \degree\mathrm{C}$ from air at 25 °C$25\ \degree\mathrm{C}$. From the hot side, the layers are 0.12 m$0.12\ \mathrm{m}$ of firebrick, 0.060 m$0.060\ \mathrm{m}$ of insulation and 0.0020 m$0.0020\ \mathrm{m}$ of steel. Their thermal conductivities are 1.2$1.2$, 0.050$0.050$ and 16 W m−1 K−1$16\ \mathrm{W\,m^{-1}\,K^{-1}}$, respectively. The hot-side and cold-side heat transfer coefficients are 25$25$ and 10 W m−2 K−1$10\ \mathrm{W\,m^{-2}\,K^{-1}}$, respectively.

Assume steady, one-dimensional heat transfer, constant properties, perfect contact and negligible radiation. Calculate the heat flux and all surface and interface temperatures. Determine which solid layer has the largest temperature drop. Then calculate the heat flux if the insulation thickness is doubled, with all other conditions unchanged.

Worked solution

Heat passes through the hot-side film, firebrick, insulation, steel and cold-side film in series. We need the heat flux and the temperature at every solid interface.

Figure 10: The furnace wall has three solid layers between hot gas and outside air. A common heat flux passes through all five resistance elements.

Data

Quantity	Symbol	Value
Hot-gas and cold-air temperatures	𝑇ℎ,𝑇𝑐$T_h,T_c$	600,25 °C$600,25\,\degree\mathrm C$
Firebrick thickness and conductivity	𝐿𝑓,𝑘𝑓$L_f,k_f$	0.12 m,1.2 W m−1 K−1$0.12\ \mathrm m,1.2\ \mathrm{W\,m^{-1}\,K^{-1}}$
Insulation thickness and conductivity	𝐿𝑖,𝑘𝑖$L_i,k_i$	0.060 m,0.050 W m−1 K−1$0.060\ \mathrm m,0.050\ \mathrm{W\,m^{-1}\,K^{-1}}$
Steel thickness and conductivity	𝐿𝑠,𝑘𝑠$L_s,k_s$	0.0020 m,16 W m−1 K−1$0.0020\ \mathrm m,16\ \mathrm{W\,m^{-1}\,K^{-1}}$
Hot-side and cold-side coefficients	ℎℎ,ℎ𝑐$h_h,h_c$	25,10 W m−2 K−1$25,10\ \mathrm{W\,m^{-2}\,K^{-1}}$

Assumptions

• Steady one-dimensional heat flow with constant properties.
• Perfect contact between the layers.
• No generation, edge loss or separate radiative transfer.

Calculations

We can calculate the heat flux directly by using 𝑅′′ =𝐴𝑅$R''=AR$, with units of m2 K W−1$\mathrm{m^2\,K\,W^{-1}}$. The temperature drop is then Δ𝑇 =𝑞′′𝑅′′$\Delta T=q''R''$. For each solid layer, 𝑅′′𝑖 =𝐿𝑖/𝑘𝑖$R_i''=L_i/k_i$; for each film, 𝑅′′ℎ =1/ℎ$R_h''=1/h$. In order from hot gas to cold air, these values are

Element	Resistance per unit area, m2 K W−1$\mathrm{m^2\,K\,W^{-1}}$
Hot gas film, 1/ℎℎ$1/h_h$	0.040$0.040$
Firebrick, 𝐿𝑓/𝑘𝑓$L_f/k_f$	0.100$0.100$
Insulation, 𝐿𝑖/𝑘𝑖$L_i/k_i$	1.200$1.200$
Steel, 𝐿𝑠/𝑘𝑠$L_s/k_s$	0.000125$0.000125$
Cold air film, 1/ℎ𝑐$1/h_c$	0.100$0.100$

The total is 𝑅′′total =1.440125 m2 K W−1$R_{\mathrm{total}}''=1.440125\ \mathrm{m^2\,K\,W^{-1}}$. Dividing the gas-to-air temperature difference by this resistance gives

𝑞′′=600−251.440125=399.3 W m−2$$q''=\frac{600-25}{1.440125}=399.3\ \mathrm{W\,m^{-2}}$$

The temperature drop across each element is 𝑞′′𝑅′′𝑖$q''R_i''$. Subtracting these drops successively from the hot-gas temperature gives

• Hot gas: 600 °C$600\ \degree\mathrm{C}$.
• Hot firebrick surface: 584.0 °C$584.0\ \degree\mathrm{C}$.
• Firebrick-insulation interface: 544.1 °C$544.1\ \degree\mathrm{C}$.
• Insulation-steel interface: 64.98 °C$64.98\ \degree\mathrm{C}$.
• Outer steel surface: 64.93 °C$64.93\ \degree\mathrm{C}$.
• Cold air: 25 °C$25\ \degree\mathrm{C}$.

The insulation has the largest temperature drop, 479.1 K$479.1\ \mathrm{K}$. The drops through the firebrick and steel are 39.93 K$39.93\ \mathrm{K}$ and 0.0499 K$0.0499\ \mathrm{K}$, respectively. Since the heat flux is the same through all three layers, these differences are determined by 𝐿𝑖/𝑘𝑖$L_i/k_i$. The insulation has the largest resistance because of its combination of thickness and low conductivity.

Doubling the insulation thickness adds another 1.200 m2 K W−1$1.200\ \mathrm{m^2\,K\,W^{-1}}$ to the total resistance. The new heat flux is

𝑞′′𝑛𝑒𝑤=5751.440125+1.200=217.8 W m−2$$q''_{new}=\frac{575}{1.440125+1.200} =217.8\ \mathrm{W\,m^{-2}}$$

The heat loss decreases by 45.5%$45.5\%$. It does not halve because the films, firebrick and steel still contribute the same resistances. For the original wall, adding all five temperature drops recovers the full 575 K$575\ \mathrm{K}$ difference between the gas and air.

Answer

The outward heat flux is 399 W m−2$399\ \mathrm{W\,m^{-2}}$. The solid temperatures from the hot side are 584.0$584.0$, 544.1$544.1$, 64.98$64.98$ and 64.93 °C$64.93\,\degree\mathrm C$. Doubling the insulation thickness reduces the flux to 218 W m−2$218\ \mathrm{W\,m^{-2}}$, a reduction of 45.5 percent.

The two cold-side interface temperatures retain guard digits to show the small contact-temperature drop.

Heat transfer modes and steady conduction

Insulation and boil-off from a spherical vessel

A thin spherical vessel of radius 0.50 m$0.50\ \mathrm{m}$ holds boiling liquid nitrogen at 77 K$77\ \mathrm{K}$. Its shell is at the liquid temperature, and the surrounding air is at 293 K$293\ \mathrm{K}$. Insulation with conductivity 0.040 W m−1 K−1$0.040\ \mathrm{W\,m^{-1}\,K^{-1}}$ extends from the vessel surface to an outer radius of 0.54 m$0.54\ \mathrm{m}$.

Take the outside heat transfer coefficient as 10 W m−2 K−1$10\ \mathrm{W\,m^{-2}\,K^{-1}}$ for both insulated and bare vessels. Neglect shell resistance, radiation and heat transfer through supports. The liquid remains at its boiling temperature.

Calculate the heat leak into the insulated vessel, the nitrogen boil-off rate and the outer insulation temperature. Compare the heat leak and boil-off rate with those of the bare vessel. The latent heat is 198 kJ kg−1$198\ \mathrm{kJ\,kg^{-1}}$. Use the supplied outside coefficient; no convection correlation is required.

Worked solution

Warm air supplies heat through the outside film and spherical insulation. That heat evaporates nitrogen at its maintained boiling temperature.

Figure 11: The cold spherical vessel has inner insulation radius ri and outer radius ro. Heat flows inward from ambient air through the outside film and insulation.

Data

Quantity	Symbol	Value
Vessel and outer insulation radii	𝑟𝑖,𝑟𝑜$r_i,r_o$	0.50,0.54 m$0.50,0.54\ \mathrm m$
Liquid and air temperatures	𝑇𝑁,𝑇∞$T_N,T_\infty$	77,293 K$77,293\ \mathrm K$
Insulation conductivity	𝑘$k$	0.040 W m−1 K−1$0.040\ \mathrm{W\,m^{-1}\,K^{-1}}$
Outside coefficient	ℎ$h$	10 W m−2 K−1$10\ \mathrm{W\,m^{-2}\,K^{-1}}$
Nitrogen latent heat	ℎ𝑓𝑔$h_{fg}$	198000 J kg−1$198000\ \mathrm{J\,kg^{-1}}$

Assumptions

• Steady heat transfer with constant properties.
• Negligible shell resistance, support losses and radiation.
• The liquid remains at its boiling temperature.
• The same supplied outside coefficient is used for the insulated and bare comparisons.

Calculations

Heat flows from the warmer air through the outside film and then through the insulation to the nitrogen. The film and spherical conduction resistances are in series. Using Eq. 7 gives

𝑅𝑖𝑛𝑠=10.50−10.544𝜋(0.040)=0.2947 K W−1$$R_{ins}=\frac{\frac{1}{0.50}-\frac{1}{0.54}}{4\pi(0.040)} =0.2947\ \mathrm{K\,W^{-1}}$$𝑅𝑓𝑖𝑙𝑚=110(4𝜋)(0.54)2=0.02729 K W−1$$R_{film}=\frac{1}{10(4\pi)(0.54)^2} =0.02729\ \mathrm{K\,W^{-1}}$$

Adding these resistances gives the heat leak:

˙𝑄𝑖𝑛=293−77𝑅𝑖𝑛𝑠+𝑅𝑓𝑖𝑙𝑚=670.8 W$$\dot Q_{in}=\frac{293-77}{R_{ins}+R_{film}} =670.8\ \mathrm W$$

The outside film uses the area at 𝑟 =0.54 m$r=0.54\ \mathrm{m}$. The insulation resistance uses both radii to account for the changing area through the layer. A plane-wall expression with a single area would only approximate the spherical conduction resistance.

The heat entering the vessel evaporates nitrogen while its temperature remains constant. Dividing the heat rate by the latent heat gives

˙𝑚𝑏𝑜𝑖𝑙=˙𝑄𝑖𝑛ℎ𝑓𝑔=670.8198000=0.003388 kg s−1=12.20 kg h−1$$\dot m_{boil}=\frac{\dot Q_{in}}{h_{fg}}=\frac{670.8}{198000} =0.003388\ \mathrm{kg\,s^{-1}} =12.20\ \mathrm{kg\,h^{-1}}$$

The drop from the air to the outer insulation surface is ˙𝑄𝑖𝑛𝑅𝑓𝑖𝑙𝑚$\dot Q_{in}R_{film}$. The surface temperature is therefore

𝑇𝑜=293−˙𝑄𝑖𝑛𝑅𝑓𝑖𝑙𝑚=274.7 K,$$T_o=293-\dot Q_{in}R_{film}=274.7\ \mathrm{K},$$

between the air and liquid temperatures as expected for inward heat flow.

For the bare vessel, there is no insulation resistance. The outside film now acts over the spherical area at 𝑟 =0.50 m$r=0.50\ \mathrm{m}$, giving

˙𝑄𝑏𝑎𝑟𝑒=10(4𝜋)(0.50)2(293−77)=6786 W$$\dot Q_{bare}=10(4\pi)(0.50)^2(293-77) =6786\ \mathrm W$$

The resulting boil-off rate is

˙𝑚𝑏𝑎𝑟𝑒=123.4 kg h−1$$\dot m_{bare}=123.4\ \mathrm{kg\,h^{-1}}$$

The insulation reduces both heat leak and boil-off by 90.1%$90.1\%$ for the specified conditions. This comparison uses the same ℎ$h$ in both cases. If ℎ$h$ were calculated from a convection correlation, changes in diameter and surface temperature could also affect its value.

Answer

With insulation, the heat leak is 671 W$671\ \mathrm W$, nitrogen boil-off is 12.2 kg h−1$12.2\ \mathrm{kg\,h^{-1}}$ and the outside insulation surface is at 274.7 K$274.7\ \mathrm K$. The bare vessel receives 6.79 ×103 W$6.79\times10^3\ \mathrm W$ and boils off 123 kg h−1$123\ \mathrm{kg\,h^{-1}}$.

Heat transfer modes and steady conduction

A heat-generating cylinder with an unknown surface temperature

A long solid cylinder has radius 0.010 m$0.010\ \mathrm{m}$, length 1.0 m$1.0\ \mathrm{m}$ and thermal conductivity 5.0 W m−1 K−1$5.0\ \mathrm{W\,m^{-1}\,K^{-1}}$. Uniform heat generation occurs at 𝑆𝑒 =2.0 ×106 W m−3$S_e=2.0\times10^6\ \mathrm{W\,m^{-3}}$. Its curved surface is cooled by fluid at 300 K$300\ \mathrm{K}$, with a heat transfer coefficient ℎ =200 W m−2 K−1$h=200\ \mathrm{W\,m^{-2}\,K^{-1}}$.

Assume steady radial conduction, constant properties and negligible heat loss through the ends. Derive the temperature profile and calculate the surface and centre temperatures. Check the overall heat balance.

Compare two modifications: doubling ℎ$h$, and replacing the cylinder by four cylinders of half the radius. Each smaller cylinder has the same length, conductivity, heat generation per unit volume and original heat transfer coefficient. The coolant temperature is unchanged.

Worked solution

The radial geometry is the same as the wire model in Figure 9. Here the surface temperature must be found from convection before the centre temperature can be evaluated.

Data

Quantity	Symbol	Value
Cylinder radius and length	𝑅,𝐿$R,L$	0.010,1.0 m$0.010,1.0\ \mathrm m$
Thermal conductivity	𝑘$k$	5.0 W m−1 K−1$5.0\ \mathrm{W\,m^{-1}\,K^{-1}}$
Volumetric heat generation	𝑆𝑒$S_e$	2.0 ×106 W m−3$2.0\times10^6\ \mathrm{W\,m^{-3}}$
Coolant temperature	𝑇∞$T_\infty$	300 K$300\ \mathrm K$
Surface coefficient	ℎ$h$	200 W m−2 K−1$200\ \mathrm{W\,m^{-2}\,K^{-1}}$

Assumptions

• Steady radial conduction with uniform heat generation.
• Constant conductivity and convection coefficient.
• Negligible heat flow through the ends.
• Each comparison cylinder sees the same specified coolant temperature.

Calculations

The governing equation is the same as for the heated wire. At the axis, symmetry requires a zero radial temperature gradient:

1𝑟𝑑𝑑𝑟(𝑟𝑑𝑇𝑑𝑟)+𝑆𝑒𝑘=0[𝑑𝑇𝑑𝑟]𝑟=0=0$$\begin{gathered} \frac1r\frac{d}{dr}\left(r\frac{dT}{dr}\right) +\frac{S_e}{k}=0\\ \left[\frac{dT}{dr}\right]_{r=0}=0 \end{gathered}$$

Integration with the symmetry condition gives

𝑑𝑇𝑑𝑟=−𝑆𝑒𝑟2𝑘.$$\frac{dT}{dr}=-\frac{S_er}{2k}.$$

The surface temperature is now unknown, so we need a convection boundary condition at 𝑟 =𝑅$r=R$. At steady state, the conductive flux reaching the surface equals the convective flux leaving it:

−𝑘[𝑑𝑇𝑑𝑟]𝑟=𝑅=𝑆𝑒𝑅2=ℎ(𝑇𝑠−𝑇∞)$$-k\left[\frac{dT}{dr}\right]_{r=R} =\frac{S_eR}{2}=h(T_s-T_\infty)$$

Rearranging gives

𝑇𝑠−𝑇∞=𝑆𝑒𝑅2ℎ.$$T_s-T_\infty=\frac{S_eR}{2h}.$$

Integrating the temperature gradient and using this surface temperature gives

𝑇(𝑟)=𝑇∞+𝑆𝑒𝑅2ℎ+𝑆𝑒4𝑘(𝑅2−𝑟2)$$T(r)=T_\infty+\frac{S_eR}{2h} +\frac{S_e}{4k}(R^2-r^2)$$

The two temperature rises come from different parts of the model. The surface rise follows from the overall generation-to-convection balance:

𝑇𝑠−𝑇∞=𝑆𝑒𝑅2ℎ=2.0×106(0.010)2(200)=50.0 K$$T_s-T_\infty=\frac{S_eR}{2h} =\frac{2.0\times10^6(0.010)}{2(200)}=50.0\ \mathrm K$$

The conduction profile then gives the additional centre rise:

𝑇𝑚𝑎𝑥−𝑇𝑠=𝑆𝑒𝑅24𝑘=2.0×106(0.010)24(5.0)=10.0 K$$T_{max}-T_s=\frac{S_eR^2}{4k} =\frac{2.0\times10^6(0.010)^2}{4(5.0)}=10.0\ \mathrm K$$

The surface-to-coolant temperature difference is 50.0 K$50.0\ \mathrm{K}$. The centre-to-surface difference is a further 10.0 K$10.0\ \mathrm{K}$. The temperatures are therefore 𝑇𝑠 =350.0 K$T_s=350.0\ \mathrm{K}$ and 𝑇max⁡ =𝑇(0) =360.0 K$T_{\max}=T(0)=360.0\ \mathrm{K}$.

The total heat generated is 𝑆𝑒𝜋𝑅2𝐿 =628.3 W$S_e\pi R^2L=628.3\ \mathrm{W}$. The surface heat loss is

ℎ(2𝜋𝑅𝐿)(𝑇𝑠−𝑇∞)=628.3 W,$$h(2\pi RL)(T_s-T_\infty)=628.3\ \mathrm{W},$$

so the overall energy balance is satisfied. This check also confirms the surface boundary condition used to obtain the temperature profile.

Doubling ℎ$h$ halves the surface-to-coolant difference to 25.0 K$25.0\ \mathrm{K}$. The centre-to-surface difference stays at 10.0 K$10.0\ \mathrm{K}$ because 𝑆𝑒$S_e$, 𝑅$R$ and 𝑘$k$ are unchanged. The new centre temperature is

300+25+10=335.0 K.$$300+25+10=335.0\ \mathrm{K}.$$

Four cylinders of half the radius have the same total volume and heat generation as the original cylinder, but twice its total curved surface area. In each smaller cylinder, the surface-to-coolant difference is 25.0 K$25.0\ \mathrm{K}$ and the centre-to-surface difference is 2.50 K$2.50\ \mathrm{K}$. The centre temperature is therefore

300+25+2.50=327.5 K.$$300+25+2.50=327.5\ \mathrm{K}.$$

The smaller radius reduces both temperature differences. At fixed 𝑆𝑒$S_e$, ℎ$h$ and 𝑘$k$, the surface-to-coolant difference varies as 𝑅$R$, whereas the centre-to-surface difference varies as 𝑅2$R^2$. These results assume that every smaller cylinder is exposed to the same coolant temperature and specified heat transfer coefficient.

Answer

The original cylinder has 𝑇𝑠 =350.0 K$T_s=350.0\ \mathrm K$, 𝑇𝑚𝑎𝑥 =360.0 K$T_{max}=360.0\ \mathrm K$ and total heat loss 628 W$628\ \mathrm W$. Doubling ℎ$h$ lowers the centre to 335.0 K$335.0\ \mathrm K$. Four cylinders of half the radius lower each centre to 327.5 K$327.5\ \mathrm K$ while preserving the original total heat generation.

Transient heat conduction

Cooling a steel sphere

A hot spherical particle made out of carbon steel (1.25 cm diameter, 500 °C$500\ \degree\mathrm{C}$, 𝑘 =40 W/(m⋅K)$k=40\ \mathrm{W/(m\cdot K)}$) is cooled by a fan, with convection coefficient of 110 W/(m2⋅K)$110\ \mathrm{W/(m^2\cdot K)}$ and air temperature of 25 °C$25\ \degree\mathrm{C}$. How long would it take for the sphere to reach a temperature of 100 °C$100\ \degree\mathrm{C}$?

Worked solution

Figure 1: The steel sphere cools in air. Its radius is half its diameter. After checking the Biot number, the lumped model treats its centre and surface as having the same temperature.

Data

Quantity	Value
Diameter	𝐷 =0.0125 m$D=0.0125\ \mathrm m$
Initial temperature	𝑇𝑖 =500 °C$T_i=500\ \degree\mathrm C$
Target temperature	𝑇 =100 °C$T=100\ \degree\mathrm C$
Air temperature	𝑇∞ =25 °C$T_\infty=25\ \degree\mathrm C$
Heat transfer coefficient	ℎ =110 W m−2 K−1$h=110\ \mathrm{W\,m^{-2}\,K^{-1}}$
Steel conductivity	𝑘 =40 W m−1 K−1$k=40\ \mathrm{W\,m^{-1}\,K^{-1}}$
Steel density	𝜌 =8131 kg m−3$\rho=8131\ \mathrm{kg\,m^{-3}}$
Steel specific heat capacity	𝐶𝑝 =434 J kg−1 K−1$C_p=434\ \mathrm{J\,kg^{-1}\,K^{-1}}$

The original slide does not supply density or specific heat capacity. These two values come from the lecturer’s notes.

Assumptions

• Uniform initial temperature.
• Constant material properties and heat transfer coefficient.
• Constant air temperature.
• Radiation and heat transfer through supports are neglected.
• No internal heat generation.

Calculations

First test whether conduction within the sphere is fast enough to keep its internal temperature nearly uniform. For a sphere,

𝐿𝑐=𝑉𝐴𝑠=𝐷6=0.01256=0.0020833 m$$L_c=\frac{V}{A_s}=\frac{D}{6} =\frac{0.0125}{6}=0.0020833\ \mathrm{m}$$

The Biot number compares internal conduction resistance with the surface convection resistance:

Bi𝑐=ℎ𝐿𝑐𝑘=110(0.0020833)40=0.00573$$\mathrm{Bi}_c=\frac{hL_c}{k} =\frac{110(0.0020833)}{40}=0.00573$$

Since Bi𝑐 <0.1$\mathrm{Bi}_c<0.1$, the lumped model is appropriate. The energy balance is

𝜌𝑉𝐶𝑝 𝑑𝑇𝑑𝑡=−ℎ𝐴𝑠(𝑇−𝑇∞).$$\rho VC_p\,\frac{dT}{dt}=-hA_s(T-T_\infty).$$

Its time constant is

𝜏=𝜌𝑉𝐶𝑝ℎ𝐴𝑠=𝜌𝐶𝑝𝐿𝑐ℎ=8131(434)(0.0020833)110=66.84 s$$\tau=\frac{\rho VC_p}{hA_s}=\frac{\rho C_pL_c}{h} =\frac{8131(434)(0.0020833)}{110}=66.84\ \mathrm{s}$$

Integrating the balance, as in Eq. 3, gives the time to the requested temperature:

𝑡=𝜏ln⁡𝑇𝑖−𝑇∞𝑇−𝑇∞$$t=\tau\ln\frac{T_i-T_\infty}{T-T_\infty}$$

Substituting the temperatures and time constant gives

𝑡=66.84ln⁡500−25100−25=123.4 s$$t=66.84\ln\frac{500-25}{100-25}=123.4\ \mathrm{s}$$(9)

The required excess-temperature ratio is 75/475 =0.1579$75/475=0.1579$, between 𝑒−1$e^{-1}$ and 𝑒−2$e^{-2}$. A time between one and two time constants is therefore consistent with the result. Radiation would increase heat loss, particularly at the initial high temperature, so the result applies to convection alone.

Answer

The sphere reaches 100 °C$100\,\degree\mathrm{C}$ after approximately 123 s$123\ \mathrm{s}$, or 2.06 min$2.06\ \mathrm{min}$, in the convection-only model.

Transient heat conduction

Quenching brass spheres in water

In a manufacturing facility, 5 cm diameter brass spheres (𝑘 =111 W/(m⋅K)$k=111\ \mathrm{W/(m\cdot K)}$, 𝜌 =8500 kg/m3$\rho=8500\ \mathrm{kg/m^3}$, 𝐶𝑝 =385.2 J/(kg⋅K)$C_p=385.2\ \mathrm{J/(kg\cdot K)}$) initially at 121 °C$121\ \degree\mathrm{C}$ are quenched in a water bath at 20 °C$20\ \degree\mathrm{C}$ for a period of 2 minutes at a rate of 120 spheres per minute. If the convection heat transfer coefficient 240 W/(m2⋅K)$240\ \mathrm{W/(m^2\cdot K)}$, determine:

• The temperature of the spheres after quenching
• The rate at which heat needs to be removed from the water in order to keep its temperature constant at 20 °C$20\ \degree\mathrm{C}$.

Worked solution

Figure 2: Radial section of a sphere in a fluid bath, showing radius, centre temperature, surface temperature and the external convection condition. In the lumped model the centre and surface temperatures are approximately equal.

Data

Quantity	Value
Sphere diameter	𝐷 =0.050 m$D=0.050\ \mathrm m$
Brass conductivity	𝑘 =111 W m−1 K−1$k=111\ \mathrm{W\,m^{-1}\,K^{-1}}$
Brass density	𝜌 =8500 kg m−3$\rho=8500\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =385.2 J kg−1 K−1$C_p=385.2\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Initial temperature	𝑇𝑖 =121 °C$T_i=121\ \degree\mathrm C$
Bath temperature	𝑇∞ =20 °C$T_\infty=20\ \degree\mathrm C$
Residence time	𝑡 =120 s$t=120\ \mathrm s$
Heat transfer coefficient	ℎ =240 W m−2 K−1$h=240\ \mathrm{W\,m^{-2}\,K^{-1}}$
Sphere throughput	˙𝑁 =120/60 =2 s−1$\dot N=120/60=2\ \mathrm{s^{-1}}$

Assumptions

• Uniform initial temperature in each sphere.
• Constant material properties and heat transfer coefficient.
• Uniform bath temperature, maintained at 20 °C$20\,\degree\mathrm{C}$.
• No internal heat generation.
• Heat transfer from the bath to other surroundings is neglected.
• Steady throughput for the bath heat-duty calculation.

Calculations

The characteristic length for a sphere is

𝐿𝑐=𝐷6=0.0506=0.0083333 m$$L_c=\frac{D}{6}=\frac{0.050}{6}=0.0083333\ \mathrm{m}$$

Test the internal temperature gradients using

Bi𝑐=ℎ𝐿𝑐𝑘=240(0.0083333)111=0.01802$$\mathrm{Bi}_c=\frac{hL_c}{k} =\frac{240(0.0083333)}{111}=0.01802$$

This is below 0.1, so each sphere can be treated as a lumped thermal capacitance. Its time constant is

𝜏=𝜌𝐶𝑝𝐿𝑐ℎ=8500(385.2)(0.0083333)240=113.69 s$$\tau=\frac{\rho C_pL_c}{h} =\frac{8500(385.2)(0.0083333)}{240}=113.69\ \mathrm{s}$$

The integrated energy balance gives

𝑇(𝑡)=𝑇∞+(𝑇𝑖−𝑇∞)𝑒−𝑡𝜏$$T(t)=T_\infty+(T_i-T_\infty)e^{-\frac{t}{\tau}}$$

After two minutes,

𝑇(120)=20+(121−20)𝑒−120113.69=55.15 °C$$T(120)=20+(121-20)e^{-\frac{120}{113.69}} =55.15\,\degree\mathrm{C}$$

Now calculate the heat released by one sphere. Its volume and mass are

𝑉=𝜋𝐷36=𝜋(0.050)36=6.5450×10−5 m3$$V=\frac{\pi D^3}{6}=\frac{\pi(0.050)^3}{6} =6.5450\times10^{-5}\ \mathrm{m^3}$$𝑚=𝜌𝑉=8500(6.5450×10−5)=0.5563 kg$$m=\rho V=8500(6.5450\times10^{-5})=0.5563\ \mathrm{kg}$$

The decrease in its stored thermal energy is

𝐸1=𝑚𝐶𝑝[𝑇𝑖−𝑇(120)]=0.5563(385.2)(121−55.15)=14.11 kJ$$E_1=mC_p[T_i-T(120)] =0.5563(385.2)(121-55.15)=14.11\ \mathrm{kJ}$$

At steady throughput, the cooling equipment must remove heat at the same average rate as the spheres release it to the bath:

˙𝑄bath=˙𝑁𝐸1=12060(14.11)=28.22 kW$$\dot Q_{\mathrm{bath}}=\dot N E_1 =\frac{120}{60}(14.11)=28.22\ \mathrm{kW}$$(10)

The exit temperature lies between the bath temperature and the initial sphere temperature, as required. There are 240 spheres in the bath at steady operation, but heat duty depends on the throughput of two spheres per second. Inventory determines how many spheres are cooling at once.

Answer

The spheres leave at approximately 55.2 °C$55.2\,\degree\mathrm{C}$. The required heat-removal rate is 28.2 kW$28.2\ \mathrm{kW}$.

Transient heat conduction

Cooling an insulated brick wall

A flat wall of fire-clay brick (0.5 m thick) initially has a uniform temperature of 200 K. One of its faces is suddenly exposed to hot gas at 1200 K. If the heat transfer coefficient on the hot side is 7.38 W/(m2⋅K)$7.38\ \mathrm{W/(m^2\cdot K)}$ and the other face of the wall is insulated, determine:

• The time it takes for the centre of the wall to reach 600 K
• The temperature of the insulated wall at the time evaluated in (a)

Properties of fire-clay brick:

• 𝑘 =1.125 W/(m K)$k=1.125\ \mathrm{W/(m\,K)}$
• 𝐶𝑝 =919 J/(kg K)$C_p=919\ \mathrm{J/(kg\,K)}$
• 𝜌 =2310 kg/m3$\rho=2310\ \mathrm{kg/m^3}$

Worked solution

The insulated face has zero normal temperature gradient. Reflecting the brick about that face produces a symmetric wall heated from both sides.

Figure 5: A brick wall heated by gas on one face and insulated on the other. Reflection makes the insulated face the symmetry plane of a wall twice as thick.

Data

Quantity	Value
Real brick thickness	𝑏 =0.50 m$b=0.50\ \mathrm m$
Initial temperature	𝑇𝑖 =200 K$T_i=200\ \mathrm K$
Gas temperature	𝑇∞ =1200 K$T_\infty=1200\ \mathrm K$
Heat transfer coefficient	ℎ =7.38 W m−2 K−1$h=7.38\ \mathrm{W\,m^{-2}\,K^{-1}}$
Brick conductivity	𝑘 =1.125 W m−1 K−1$k=1.125\ \mathrm{W\,m^{-1}\,K^{-1}}$
Brick density	𝜌 =2310 kg m−3$\rho=2310\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =919 J kg−1 K−1$C_p=919\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Reflected-wall half-thickness	𝐿 =0.50 m$L=0.50\ \mathrm m$
Physical brick-centre position	𝑥 =0.25 m$x=0.25\ \mathrm m$, 𝑋 =𝑥/𝐿 =0.5$X=x/L=0.5$
Target temperature at the brick centre	𝑇 =600 K$T=600\ \mathrm K$

Measure position from the insulated face. Reflection makes that face the midplane of the equivalent wall.

Assumptions

• Uniform initial temperature.
• Unidirectional conduction normal to the wall.
• Perfect insulation at the rear face.
• Constant material properties, gas temperature and heat transfer coefficient.
• No internal heat generation.
• Radiation is neglected.

Calculations

The insulated face obeys ∂𝑇/∂𝑥 =0$\partial T/\partial x=0$, which is the symmetry condition of the finite-wall solution. First calculate the diffusivity:

𝛼=𝑘𝜌𝐶𝑝=1.1252310(919)=5.2992×10−7 m2 s−1$$\alpha=\frac{k}{\rho C_p} =\frac{1.125}{2310(919)}=5.2992\times10^{-7}\ \mathrm{m^2\,s^{-1}}$$

The wall Biot number is

Bi=ℎ𝐿𝑘=7.38(0.50)1.125=3.28$$\mathrm{Bi}=\frac{hL}{k}=\frac{7.38(0.50)}{1.125}=3.28$$

For this one-sided wall, 𝑉/𝐴𝑠 =0.50 m$V/A_s=0.50\ \mathrm{m}$ also. The Biot number is well above 0.1, so a spatial temperature profile is needed. The one-term wall solution from Eq. 18 is

Θ(𝑋,𝑡)=𝐴1𝑒−𝜆21Focos⁡(𝜆1𝑋),Fo=𝛼𝑡𝐿2$$\Theta(X,t)=A_1e^{-\lambda_1^2\mathrm{Fo}}\cos(\lambda_1X), \qquad \mathrm{Fo}=\frac{\alpha t}{L^2}$$

The eigenvalue relation Eq. 15, 𝜆1tan⁡𝜆1 =Bi$\lambda_1\tan\lambda_1=\mathrm{Bi}$, gives 𝜆1 =1.21582$\lambda_1=1.21582$. Substituting that root into the coefficient relation gives

𝐴1=4sin⁡𝜆12𝜆1+sin⁡(2𝜆1)=4sin⁡(1.21582)2(1.21582)+sin⁡(2.43164)=1.21638$$A_1=\frac{4\sin\lambda_1}{2\lambda_1+\sin(2\lambda_1)} =\frac{4\sin(1.21582)}{2(1.21582)+\sin(2.43164)}=1.21638$$

At the real brick centre, the requested temperature corresponds to

Θ(0.5,𝑡)=600−1200200−1200=0.600$$\Theta(0.5,t)=\frac{600-1200}{200-1200}=0.600$$

The value 0.600$0.600$ applies at the physical centre of the brick, 𝑥 =0.25 m$x=0.25\ \mathrm m$ or 𝑋 =0.5$X=0.5$. The reference position for the symmetric wall solution is 𝑥 =0$x=0$, at the insulated face. This is the centre of the equivalent reflected wall, so it is a different location from the physical centre of the brick.

At 𝑋 =0$X=0$, the cosine factor is one and the one-term solution gives Θ0 =Θ(0,𝑡) =𝐴1𝑒−𝜆21Fo$\Theta_0=\Theta(0,t)=A_1e^{-\lambda_1^2\mathrm{Fo}}$. At the physical brick centre, it therefore gives Θ(0.5,𝑡) =Θ0cos⁡(𝜆1/2)$\Theta(0.5,t)=\Theta_0\cos(\lambda_1/2)$. Divide by this cosine factor to find the dimensionless temperature at the insulated face:

Θ0=Θ(0.5,𝑡)cos⁡(𝜆12)=0.600cos⁡(1.215822)=0.73096$$\Theta_0=\frac{\Theta(0.5,t)}{\cos\left(\frac{\lambda_1}{2}\right)} =\frac{0.600}{\cos\left(\frac{1.21582}{2}\right)}=0.73096$$

Thus 0.600$0.600$ and 0.73096$0.73096$ are dimensionless temperatures at different positions at the same time. The value 0.73096$0.73096$ is the insulated-face reference value needed to find the elapsed time.

Use

Θ0=𝐴1𝑒−𝜆21Fo$$\Theta_0=A_1e^{-\lambda_1^2\mathrm{Fo}}$$

to find the elapsed time:

Fo=ln⁡(𝐴1Θ0)𝜆21=ln⁡(1.216380.73096)1.215822=0.34452$$\mathrm{Fo}=\frac{\ln\left(\frac{A_1}{\Theta_0}\right)}{\lambda_1^2} =\frac{\ln\left(\frac{1.21638}{0.73096}\right)}{1.21582^2}=0.34452$$(28)

Since Fo >0.2$\mathrm{Fo}>0.2$, the one-term approximation is appropriate. Convert this dimensionless time to seconds:

𝑡=Fo𝐿2𝛼=0.34452(0.50)25.2992×10−7=1.6253×105 s=45.1 h$$t=\frac{\mathrm{Fo}L^2}{\alpha} =\frac{0.34452(0.50)^2}{5.2992\times10^{-7}} =1.6253\times10^5\ \mathrm{s}=45.1\ \mathrm{h}$$

At the insulated face, 𝑋 =0$X=0$ and the cosine factor is one. Its temperature is

𝑇0=𝑇∞+Θ0(𝑇𝑖−𝑇∞)=1200+0.73096(200−1200)=469.0 K$$T_0=T_\infty+\Theta_0(T_i-T_\infty) =1200+0.73096(200-1200)=469.0\ \mathrm{K}$$

The insulated face is colder than the brick centre because heat enters from the opposite face. The ordering 200 <𝑇0 <600 <1200 K$200<T_0<600<1200\ \mathrm{K}$ is consistent with one-sided heating. Constant properties over this temperature range are an idealisation of the classroom model.

Answer

The brick centre reaches 600 K$600\ \mathrm{K}$ after approximately 45.1 h$45.1\ \mathrm{h}$. At that time, the insulated face is at 469 K$469\ \mathrm{K}$.

Transient heat conduction

Cooling a short steel cylinder

A steel cylinder (𝐷 =30.5 cm$D=30.5\ \mathrm{cm}$, 𝐿 =61 cm$L=61\ \mathrm{cm}$) is initially at 645 K. It is immersed in an oil bath with a constant temperature of 310K. If the convection heat transfer coefficient at the steel surface is 34 W/(m2⋅K)$34\ \mathrm{W/(m^2\cdot K)}$, determine the temperatures at the following locations:

• at the centre of the cylinder after 1 hour?
• at the centre of the top surface of the cylinder?

Properties of steel:

• 𝑘 =42.9 W/(m K)$k=42.9\ \mathrm{W/(m\,K)}$
• 𝐶𝑝 =473 J/(kg K)$C_p=473\ \mathrm{J/(kg\,K)}$
• 𝜌 =7820 kg/m3$\rho=7820\ \mathrm{kg/m^3}$

Worked solution

An axial section identifies the two requested points and the two directions of conduction.

Data

Quantity	Value
Cylinder diameter	𝐷 =0.305 m$D=0.305\ \mathrm m$
Full cylinder length	𝐿 =0.610 m$L=0.610\ \mathrm m$
Half-length for the plane-wall factor	𝐿𝑤 =0.305 m$L_w=0.305\ \mathrm m$
Cylinder radius	𝑟𝑜 =0.1525 m$r_o=0.1525\ \mathrm m$
Initial temperature	𝑇𝑖 =645 K$T_i=645\ \mathrm K$
Oil temperature	𝑇∞ =310 K$T_\infty=310\ \mathrm K$
Heat transfer coefficient	ℎ =34 W m−2 K−1$h=34\ \mathrm{W\,m^{-2}\,K^{-1}}$
Steel conductivity	𝑘 =42.9 W m−1 K−1$k=42.9\ \mathrm{W\,m^{-1}\,K^{-1}}$
Steel density	𝜌 =7820 kg m−3$\rho=7820\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =473 J kg−1 K−1$C_p=473\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Time for both requested temperatures	𝑡 =3600 s$t=3600\ \mathrm s$

Assumptions

• Uniform initial temperature.
• Axisymmetric conduction in both radial and axial directions.
• Constant material properties.
• Constant oil temperature and uniform heat transfer coefficient on all surfaces.
• No internal heat generation.
• Radiation is neglected.

Calculations

The ends cannot be neglected because the length is only twice the diameter. For a constant-property body with these separable boundary conditions, the short-cylinder excess-temperature ratio is the product of a wall factor and a long-cylinder factor:

Θ(𝑥,𝑟,𝑡)=Θ𝑤(𝑥,𝑡)Θ𝑐(𝑟,𝑡)$$\Theta(x,r,t)=\Theta_w(x,t)\Theta_c(r,t)$$(29)

Here 𝑥 =0$x=0$ is the axial midplane and 𝑟 =0$r=0$ is the axis. The product solves

∂𝑇∂𝑡=𝛼[∂2𝑇∂𝑥2+1𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)],$$\frac{\partial T}{\partial t}=\alpha\left[\frac{\partial^2 T}{\partial x^2}+ \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)\right],$$

because each factor solves its own one-dimensional diffusion equation. The diffusivity is

𝛼=𝑘𝜌𝐶𝑝=42.97820(473)=1.1598×10−5 m2 s−1$$\alpha=\frac{k}{\rho C_p} =\frac{42.9}{7820(473)}=1.1598\times10^{-5}\ \mathrm{m^2\,s^{-1}}$$

Each factor needs its own Biot and Fourier numbers. For the wall factor,

Bi𝑤=ℎ𝐿𝑤𝑘=34(0.305)42.9=0.24172$$\mathrm{Bi}_w=\frac{hL_w}{k} =\frac{34(0.305)}{42.9}=0.24172$$Fo𝑤=𝛼𝑡𝐿2𝑤=1.1598×10−5(3600)0.3052=0.44884$$\mathrm{Fo}_w=\frac{\alpha t}{L_w^2} =\frac{1.1598\times10^{-5}(3600)}{0.305^2}=0.44884$$

For the cylindrical factor,

Bicyl=ℎ𝑟𝑜𝑘=34(0.1525)42.9=0.12086$$\mathrm{Bi}_{\mathrm{cyl}}=\frac{hr_o}{k} =\frac{34(0.1525)}{42.9}=0.12086$$Focyl=𝛼𝑡𝑟2𝑜=1.1598×10−5(3600)0.15252=1.7954$$\mathrm{Fo}_{\mathrm{cyl}}=\frac{\alpha t}{r_o^2} =\frac{1.1598\times10^{-5}(3600)}{0.1525^2}=1.7954$$

Both Fourier numbers exceed 0.2. Solving 𝜆𝑤tan⁡𝜆𝑤 =0.24172$\lambda_w\tan\lambda_w=0.24172$ and

𝜆𝑐𝐽1(𝜆𝑐)=0.12086𝐽0(𝜆𝑐)$$\lambda_cJ_1(\lambda_c)=0.12086J_0(\lambda_c)$$

gives 𝜆𝑤 =0.472701$\lambda_w=0.472701$ and 𝜆𝑐 =0.484321$\lambda_c=0.484321$. The corresponding coefficient relations, Eq. 17 and Eq. 21, give 𝐴𝑤 =1.03703$A_w=1.03703$ and 𝐴𝑐 =1.02960$A_c=1.02960$.

The wall centre factor is

Θ0,𝑤=𝐴𝑤𝑒−𝜆2𝑤Fo𝑤=1.03703𝑒−0.4727012(0.44884)=0.93807$$\Theta_{0,w}=A_we^{-\lambda_w^2\mathrm{Fo}_w} =1.03703e^{-0.472701^2(0.44884)}=0.93807$$

The cylinder centre factor is

Θ0,𝑐=𝐴𝑐𝑒−𝜆2𝑐Focyl=1.02960𝑒−0.4843212(1.7954)=0.67573$$\Theta_{0,c}=A_ce^{-\lambda_c^2\mathrm{Fo}_{\mathrm{cyl}}} =1.02960e^{-0.484321^2(1.7954)}=0.67573$$

At the body centre, both position factors are one. The product and resulting temperature are

Θ𝑂=Θ0,𝑤Θ0,𝑐=0.93807(0.67573)=0.63388$$\Theta_O=\Theta_{0,w}\Theta_{0,c} =0.93807(0.67573)=0.63388$$𝑇𝑂=𝑇∞+Θ𝑂(𝑇𝑖−𝑇∞)=310+0.63388(645−310)=522.4 K$$T_O=T_\infty+\Theta_O(T_i-T_\infty) =310+0.63388(645-310)=522.4\ \mathrm{K}$$

At the centre of the top face, 𝑥 =𝐿𝑤$x=L_w$ and 𝑟 =0$r=0$. The wall factor includes cos⁡𝜆𝑤$\cos\lambda_w$, while the radial position factor remains one:

Θ𝑇=Θ0,𝑤cos⁡𝜆𝑤 Θ0,𝑐=0.63388cos⁡(0.472701)=0.56437$$\Theta_T=\Theta_{0,w}\cos\lambda_w\,\Theta_{0,c} =0.63388\cos(0.472701)=0.56437$$

Convert the product to temperature:

𝑇𝑇=𝑇∞+Θ𝑇(𝑇𝑖−𝑇∞)=310+0.56437(645−310)=499.1 K$$T_T=T_\infty+\Theta_T(T_i-T_\infty) =310+0.56437(645-310)=499.1\ \mathrm{K}$$

The face centre is cooler than the body centre and both temperatures remain between the initial and bath temperatures. A position chart supplies a ratio relative to its own centre value. Multiplying a position ratio alone by the other centre factor would omit part of the cooling history.

Answer

After one hour, the body centre is at approximately 522 K$522\ \mathrm{K}$ and the centre of the top face is at approximately 499 K$499\ \mathrm{K}$.

Transient heat conduction

Heating an egg

An egg, initially at a uniform temperature of 5 °C$5\ \degree\mathrm{C}$, is dropped into boiling water at 95 °C$95\ \degree\mathrm{C}$. Taking the convection heat transfer coefficient to be 1200 W/(m2⋅°C)$1200\ \mathrm{W/(m^2\cdot\degree\mathrm{C})}$, and approximating the egg to be a 5-cm diameter sphere, determine how long it will take for the centre of the egg to reach 70 °C$70\ \degree\mathrm{C}$.

Worked solution

Figure 6: The egg is modelled as a sphere heated by surrounding water. The requested temperature is at the centre, which can lag behind the surface.

Data

Quantity	Value
Model sphere diameter	𝐷 =0.050 m$D=0.050\ \mathrm m$
Initial temperature	𝑇𝑖 =5 °C$T_i=5\ \degree\mathrm C$
Target centre temperature	𝑇0 =70 °C$T_0=70\ \degree\mathrm C$
Water temperature	𝑇∞ =95 °C$T_\infty=95\ \degree\mathrm C$
Heat transfer coefficient	ℎ =1200 W m−2 K−1$h=1200\ \mathrm{W\,m^{-2}\,K^{-1}}$
Conductivity	𝑘 =0.627 W m−1 K−1$k=0.627\ \mathrm{W\,m^{-1}\,K^{-1}}$
Thermal diffusivity	𝛼 =0.151 ×10−6 m2 s−1$\alpha=0.151\times10^{-6}\ \mathrm{m^2\,s^{-1}}$

The conductivity and diffusivity come from the lecturer’s notes.

Assumptions

• Homogeneous spherical body with a uniform initial temperature.
• Radial conduction.
• Constant properties and heat transfer coefficient.
• Constant water temperature.
• No internal heat generation.
• Shell resistance and energy changes due to cooking are neglected.

Calculations

Test lumped capacitance using 𝐿𝑐 =𝐷/6$L_c=D/6$:

Bi𝑐=ℎ𝐷6𝑘=1200(0.050)6(0.627)=15.95$$\mathrm{Bi}_c=\frac{hD}{6k} =\frac{1200(0.050)}{6(0.627)}=15.95$$

The large Biot number rules out a uniform internal temperature. The sphere solution uses the radius 𝑟𝑜 =𝐷/2 =0.025 m$r_o=D/2=0.025\ \mathrm{m}$ instead:

Bi=ℎ𝑟𝑜𝑘=1200(0.025)0.627=47.85$$\mathrm{Bi}=\frac{hr_o}{k} =\frac{1200(0.025)}{0.627}=47.85$$

At the centre, the one-term solution is

Θ0=𝐴1𝑒−𝜆21Fo.$$\Theta_0=A_1e^{-\lambda_1^2\mathrm{Fo}}.$$

Its target value is

Θ0=𝑇0−𝑇∞𝑇𝑖−𝑇∞=70−955−95=0.27778$$\Theta_0=\frac{T_0-T_\infty}{T_i-T_\infty} =\frac{70-95}{5-95}=0.27778$$

The sphere eigenvalue relation Eq. 20 is 1 −𝜆1cot⁡𝜆1 =Bi$1-\lambda_1\cot\lambda_1=\mathrm{Bi}$. With Bi =47.85$\mathrm{Bi}=47.85$, its first root is 𝜆1 =3.07603$\lambda_1=3.07603$. The corresponding coefficient is

𝐴1=4(sin⁡𝜆1−𝜆1cos⁡𝜆1)2𝜆1−sin⁡(2𝜆1)=1.99588$$A_1=\frac{4(\sin\lambda_1-\lambda_1\cos\lambda_1)} {2\lambda_1-\sin(2\lambda_1)}=1.99588$$

Rearrange the centre-temperature relation to find the Fourier number:

Fo=ln⁡(𝐴1Θ0)𝜆21=ln⁡(1.995880.27778)3.076032=0.20842$$\mathrm{Fo}=\frac{\ln\left(\frac{A_1}{\Theta_0}\right)}{\lambda_1^2} =\frac{\ln\left(\frac{1.99588}{0.27778}\right)}{3.07603^2}=0.20842$$

This just exceeds 0.2, so the usual one-term criterion is satisfied. The time is

𝑡=Fo𝑟2𝑜𝛼=0.20842(0.025)20.151×10−6=863 s=14.4 min$$t=\frac{\mathrm{Fo}r_o^2}{\alpha} =\frac{0.20842(0.025)^2}{0.151\times10^{-6}} =863\ \mathrm{s}=14.4\ \mathrm{min}$$

A chart reading of Fo ≃0.20$\mathrm{Fo}\simeq0.20$ gives about 13.8 minutes, consistent with this estimate. The large Biot number means that internal conduction limits the response. Increasing the external coefficient has less effect than it would in a lumped body. Real eggs have a nonspherical shape and different internal regions, so the calculation is an idealised heating estimate.

Answer

The centre reaches 70 °C$70\,\degree\mathrm{C}$ after approximately 14 min$14\ \mathrm{min}$ in the spherical, constant-property model.

Transient heat conduction

Soil frost depth

What minimum burial depth would you recommend to avoid freezing under conditions for which the soil, initially at a uniform temperature of 15 °C$15\ \degree\mathrm{C}$, is subjected to a constant surface temperature of −10 °C$-10\ \degree\mathrm{C}$ for 3 months?

Properties of soil:

• 𝑘 =0.52 W/(m⋅K)$k=0.52\ \mathrm{W/(m\cdot K)}$
• 𝐶𝑝 =1840 J/(kg⋅K)$C_p=1840\ \mathrm{J/(kg\cdot K)}$
• 𝜌 =2050 kg/m3$\rho=2050\ \mathrm{kg/m^3}$

Worked solution

Measure depth downwards from the cold ground surface. The pipe must lie below the calculated zero-degree isotherm.

Data

Quantity	Value
Initial soil temperature	𝑇𝑖 =15 °C$T_i=15\ \degree\mathrm C$
Prescribed surface temperature	𝑇𝑠 = −10 °C$T_s=-10\ \degree\mathrm C$
Temperature at the required depth	𝑇 =0 °C$T=0\ \degree\mathrm C$
Elapsed time	𝑡 =90 days$t=90\ \mathrm{days}$
Soil conductivity	𝑘 =0.52 W m−1 K−1$k=0.52\ \mathrm{W\,m^{-1}\,K^{-1}}$
Soil density	𝜌 =2050 kg m−3$\rho=2050\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1840 J kg−1 K−1$C_p=1840\ \mathrm{J\,kg^{-1}\,K^{-1}}$

For this calculation, take three months as 90 days.

Assumptions

• Uniform initial soil temperature.
• Unidirectional conduction normal to a plane ground surface.
• Semi-infinite soil depth over the time considered.
• Constant thermal properties.
• Constant prescribed surface temperature.
• No internal heat generation.
• Soil moisture phase change is neglected.

Calculations

The surface temperature is supplied directly, so use the prescribed-temperature semi-infinite solution rather than a convection boundary condition. The thermal diffusivity is

𝛼=𝑘𝜌𝐶𝑝=0.522050(1840)=1.3786×10−7 m2 s−1$$\alpha=\frac{k}{\rho C_p} =\frac{0.52}{2050(1840)}=1.3786\times10^{-7}\ \mathrm{m^2\,s^{-1}}$$

Convert the duration to seconds:

𝑡=90(24)(3600)=7.776×106 s$$t=90(24)(3600)=7.776\times10^6\ \mathrm{s}$$

The diffusion distance provides a useful estimate before solving for depth:

√𝛼𝑡=√1.3786×10−7(7.776×106)=1.0354 m$$\sqrt{\alpha t}=\sqrt{1.3786\times10^{-7}(7.776\times10^6)} =1.0354\ \mathrm{m}$$

The temperature relation from Eq. 31 is

𝑇(𝑥,𝑡)−𝑇𝑖𝑇𝑠−𝑇𝑖=erfc⁡(𝑥2√𝛼𝑡)$$\frac{T(x,t)-T_i}{T_s-T_i} =\operatorname{erfc}\left(\frac{x}{2\sqrt{\alpha t}}\right)$$

At the limiting temperature of 0 °C$0\,\degree\mathrm{C}$,

erfc⁡(𝜉)=0−15−10−15=0.600$$\operatorname{erfc}(\xi)=\frac{0-15}{-10-15}=0.600$$

Inverting the complementary error function gives 𝜉 =0.370807$\xi=0.370807$. The corresponding depth is

𝑥=2𝜉√𝛼𝑡=2(0.370807)(1.0354)=0.7679 m$$x=2\xi\sqrt{\alpha t}=2(0.370807)(1.0354)=0.7679\ \mathrm{m}$$

Temperature increases with depth during cooling, so a pipe deeper than this isotherm remains above 0 °C$0\,\degree\mathrm{C}$ within the model. The calculated depth is consistent with the metre-scale diffusion distance. The far boundary must be well below the disturbed region for the semi-infinite approximation to hold.

Soil moisture releases latent heat on freezing, which this calculation omits. Snow cover and a changing surface temperature would also alter the temperature profile. The result is a classroom estimate based on the specified temperature history.

Answer

The 0 °C$0\,\degree\mathrm{C}$ isotherm reaches approximately 0.77 m$0.77\ \mathrm{m}$. A burial depth of at least 0.8 m$0.8\ \mathrm{m}$ lies below it in this model; 1 m$1\ \mathrm{m}$ provides a larger margin.

Transient heat conduction

Centre temperature and energy released by a cooling slab

A large slab of full thickness 0.100 m starts uniformly at 200 degrees Celsius. Both broad faces cool in fluid at 20 degrees Celsius with ℎ =20$h=20$ W m⁻² K⁻¹. Use 𝑘 =1.0$k=1.0$ W m⁻¹ K⁻¹, 𝜌 =2000$\rho=2000$ kg m⁻³ and 𝐶𝑝 =1000$C_p=1000$ J kg⁻¹ K⁻¹. Neglect edges, radiation and property variation.

After 30 minutes, find the centre, surface and volume-average temperatures, and the energy released per square metre of slab plan area. For the plane-wall one-term solution at Bi =1$\mathrm{Bi}=1$, use 𝜆1 =0.860334$\lambda_1=0.860334$ and 𝐴1 =1.11913$A_1=1.11913$. Check that this approximation is appropriate.

Worked solution

Figure 8: Both slab faces cool under the same conditions. The solution uses half-thickness L, measured from the symmetry plane to either exposed surface.

Data

Quantity	Value
Full slab thickness	2𝐿 =0.100 m$2L=0.100\ \mathrm m$
Half-thickness	𝐿 =0.050 m$L=0.050\ \mathrm m$
Initial temperature	𝑇𝑖 =200 °C$T_i=200\ \degree\mathrm C$
Fluid temperature	𝑇∞ =20 °C$T_\infty=20\ \degree\mathrm C$
Cooling time	𝑡 =1800 s$t=1800\ \mathrm s$
Heat transfer coefficient	ℎ =20 W m−2 K−1$h=20\ \mathrm{W\,m^{-2}\,K^{-1}}$
Conductivity	𝑘 =1.0 W m−1 K−1$k=1.0\ \mathrm{W\,m^{-1}\,K^{-1}}$
Density	𝜌 =2000 kg m−3$\rho=2000\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1000 J kg−1 K−1$C_p=1000\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Supplied first eigenvalue	𝜆1 =0.860334$\lambda_1=0.860334$
Supplied first coefficient	𝐴1 =1.11913$A_1=1.11913$

Assumptions

• Uniform initial temperature.
• Unidirectional conduction through the slab thickness.
• Identical convection conditions on both broad faces.
• Constant properties, fluid temperature and heat transfer coefficient.
• Edge heat transfer and radiation are neglected.
• No internal heat generation.

Calculations

Both broad faces transfer heat. If 𝐴plan$A_{\mathrm{plan}}$ is the area of one face, 𝑉 =2𝐿𝐴plan$V=2LA_{\mathrm{plan}}$ and 𝐴𝑠 =2𝐴plan$A_s=2A_{\mathrm{plan}}$. Therefore

𝐿𝑐=𝑉𝐴𝑠=𝐿=0.050 m.$$L_c=\frac{V}{A_s}=L=0.050\ \mathrm{m}.$$

The Biot number is

Bi𝑐=ℎ𝐿𝑐𝑘=20(0.050)1.0=1.0$$\mathrm{Bi}_c=\frac{hL_c}{k}=\frac{20(0.050)}{1.0}=1.0$$

Lumped capacitance is unsuitable. Use the spatial wall solution, with 𝑥 =0$x=0$ at the midplane. First calculate the diffusivity:

𝛼=𝑘𝜌𝐶𝑝=1.02000(1000)=5.00×10−7 m2 s−1$$\alpha=\frac{k}{\rho C_p} =\frac{1.0}{2000(1000)}=5.00\times10^{-7}\ \mathrm{m^2\,s^{-1}}$$

The Fourier number uses the half-thickness:

Fo=𝛼𝑡𝐿2=5.00×10−7(1800)0.0502=0.360$$\mathrm{Fo}=\frac{\alpha t}{L^2} =\frac{5.00\times10^{-7}(1800)}{0.050^2}=0.360$$

Since Fo >0.2$\mathrm{Fo}>0.2$, the one-term approximation is appropriate. The wall profile is

Θ=Θ0cos⁡(𝜆1𝑥𝐿),$$\Theta=\Theta_0\cos\left(\frac{\lambda_1x}{L}\right),$$

where

Θ=𝑇−𝑇∞𝑇𝑖−𝑇∞.$$\Theta=\frac{T-T_\infty}{T_i-T_\infty}.$$

Its centre value is

Θ0=𝐴1𝑒−𝜆21Fo=1.11913𝑒−0.8603342(0.360)=0.8574$$\Theta_0=A_1e^{-\lambda_1^2\mathrm{Fo}} =1.11913e^{-0.860334^2(0.360)}=0.8574$$

Convert the centre ratio to temperature:

𝑇𝑐=𝑇∞+Θ0(𝑇𝑖−𝑇∞)=20+0.8574(200−20)=174.3 °C$$T_c=T_\infty+\Theta_0(T_i-T_\infty) =20+0.8574(200-20)=174.3\,\degree\mathrm{C}$$

At either face, 𝑥/𝐿 =1$x/L=1$. The surface ratio and temperature are

Θ𝑠=Θ0cos⁡𝜆1=0.8574cos⁡(0.860334)=0.5592$$\Theta_s=\Theta_0\cos\lambda_1 =0.8574\cos(0.860334)=0.5592$$𝑇𝑠=𝑇∞+Θ𝑠(𝑇𝑖−𝑇∞)=20+0.5592(180)=120.6 °C$$T_s=T_\infty+\Theta_s(T_i-T_\infty) =20+0.5592(180)=120.6\,\degree\mathrm{C}$$

Stored energy depends on the volume-average temperature. Integrating the symmetric profile across the thickness gives

¯¯¯¯¯Θ=Θ0∫10cos⁡(𝜆1𝑋) 𝑑𝑋=Θ0sin⁡𝜆1𝜆1=0.8574sin⁡(0.860334)0.860334=0.7554$$\bar\Theta=\Theta_0\int_0^1\cos(\lambda_1X)\,dX =\Theta_0\frac{\sin\lambda_1}{\lambda_1} =0.8574\frac{\sin(0.860334)}{0.860334}=0.7554$$

The average temperature is therefore

¯¯¯¯𝑇=𝑇∞+¯¯¯¯¯Θ(𝑇𝑖−𝑇∞)=20+0.7554(180)=156.0 °C$$\bar T=T_\infty+\bar\Theta(T_i-T_\infty) =20+0.7554(180)=156.0\,\degree\mathrm{C}$$

The released energy equals the decrease in stored thermal energy. Per unit plan area,

𝐸out𝐴plan=𝜌𝐶𝑝(2𝐿)(𝑇𝑖−¯¯¯¯𝑇)=2000(1000)(0.100)(200−156.0)=8.80×106 J m−2$$\frac{E_{\mathrm{out}}}{A_{\mathrm{plan}}} =\rho C_p(2L)(T_i-\bar T) =2000(1000)(0.100)(200-156.0) =8.80\times10^6\ \mathrm{J\,m^{-2}}$$

This includes heat leaving both faces. The ordering 20 <𝑇𝑠 <¯¯¯¯𝑇 <𝑇𝑐 <200 °C$20<T_s<\bar T<T_c<200\,\degree\mathrm{C}$ is consistent with cooling from the surfaces. Using the centre temperature for the energy balance would underestimate the heat released, because most of the slab is cooler than its centre.

Answer

After 30 minutes, 𝑇𝑐 =174.3 °C$T_c=174.3\,\degree\mathrm{C}$, 𝑇𝑠 =120.6 °C$T_s=120.6\,\degree\mathrm{C}$ and ¯¯¯¯𝑇 =156.0 °C$\bar T=156.0\,\degree\mathrm{C}$. The energy released is 8.80 MJ m−2$8.80\ \mathrm{MJ\,m^{-2}}$ of plan area.

Forced convection in external flow

Low-pressure plate

Air at a pressure of 6 kN/m2$6\ \mathrm{kN/m^2}$ and temperature of 300 °C$300\ \degree\mathrm{C}$ flows with a velocity of 10 m/s$10\ \mathrm{m/s}$ over a flat plate 0.5 m long. Estimate the cooling rate per unit width of the plate needed to maintain the surface at a temperature of 27 °C$27\ \degree\mathrm{C}$.

• 𝜈@1 atm =30.84 ×10−6 m2/s$\nu_{\text{@1 atm}}=30.84\times10^{-6}\ \mathrm{m^2/s}$
• 𝑘air at 165 °C =3.6427 ×10−2 W/(m⋅K)$k_{\text{air at }165\,\degree\mathrm{C}}=3.6427\times10^{-2}\ \mathrm{W/(m\cdot K)}$
• Prair at 165 °C ≃0.684$\mathrm{Pr}_{\text{air at }165\,\degree\mathrm{C}}\simeq0.684$

Solution

Figure 2: Air flows along the plate, and the velocity and thermal boundary layers grow from its leading edge. Use the full plate length for the average coefficient.

The plate geometry is the boundary-layer geometry in Figure 1. The stream runs along the 0.50 m length, and the cooling rate is requested per metre of width.

Data

• Air temperature 𝑇∞ =300 °C$T_\infty=300\ \degree\mathrm C$
• Plate temperature 𝑇𝑠 =27 °C$T_s=27\ \degree\mathrm C$
• Plate length 𝐿 =0.50 m$L=0.50\ \mathrm m$
• Air speed 𝑢∞ =10 m s−1$u_\infty=10\ \mathrm{m\,s^{-1}}$
• Absolute pressure 𝑝 =6 kPa$p=6\ \mathrm{kPa}$

The supplied conductivity, Prandtl number and one-atmosphere viscosity are evaluated near the film temperature.

Assumptions

• Steady state.
• Smooth plate aligned with the free stream.
• Constant fluid properties at the film temperature.
• Ideal-gas density and pressure-independent dynamic viscosity.
• Radiation and buoyancy effects are neglected.

Calculations

The hot air heats the colder plate. The required cooling rate is the heat received by one exposed plate face. First evaluate the film temperature:

𝑇𝑓=𝑇𝑠+𝑇∞2=27+3002=163.5 °C$$T_f=\frac{T_s+T_\infty}{2} =\frac{27+300}{2}=163.5\ \degree\mathrm{C}$$

The supplied properties at 165 °C$165\ \degree\mathrm{C}$ approximate this temperature. At fixed temperature, ideal-gas density is proportional to absolute pressure. Since 𝜈 =𝜇/𝜌$\nu=\mu/\rho$ and the dynamic viscosity is nearly pressure independent, the kinematic viscosity increases as pressure falls:

𝜈=𝜈1 atm𝑝1 atm𝑝=30.84×10−6101.3256=5.2081×10−4 m2 s−1$$\nu=\nu_{\mathrm{1\ atm}}\frac{p_{\mathrm{1\ atm}}}{p} =30.84\times10^{-6}\frac{101.325}{6} =5.2081\times10^{-4}\ \mathrm{m^2\,s^{-1}}$$(14)

The Reynolds number based on the full plate length determines the flow regime:

Re𝐿=𝑢∞𝐿𝜈=10(0.50)5.2081×10−4=9600.4$$\mathrm{Re}_L=\frac{u_\infty L}{\nu} =\frac{10(0.50)}{5.2081\times10^{-4}}=9600.4$$

This is well below the nominal transition value 5 ×105$5\times10^5$. Use the laminar, isothermal-plate average in Eq. 7:

¯¯¯¯¯¯¯¯Nu𝐿=0.664Re1/2𝐿Pr1/3=0.664(9600.4)1/2(0.684)1/3=57.323$$\overline{\mathrm{Nu}}_L =0.664\mathrm{Re}_L^{1/2}\mathrm{Pr}^{1/3} =0.664(9600.4)^{1/2}(0.684)^{1/3}=57.323$$

Recover the dimensional coefficient using the same length:

¯¯¯¯ℎ=𝑘¯¯¯¯¯¯¯¯Nu𝐿𝐿=(3.6427×10−2)(57.323)0.50=4.1762 W m−2 K−1$$\bar h=\frac{k\overline{\mathrm{Nu}}_L}{L} =\frac{(3.6427\times10^{-2})(57.323)}{0.50} =4.1762\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For plate width 𝑤$w$, the exposed area is 𝐿𝑤$Lw$. Newton’s relation therefore gives

˙𝑄removed𝑤=¯¯¯¯ℎ𝐿(𝑇∞−𝑇𝑠)=4.1762(0.50)(300−27)=570.1 W m−1$$\frac{\dot Q_{\mathrm{removed}}}{w} =\bar hL(T_\infty-T_s) =4.1762(0.50)(300-27) =570.1\ \mathrm{W\,m^{-1}}$$(15)

As a geometry check, the laminar thickness estimate gives

𝛿(𝐿)≃5𝐿√Re𝐿=0.0255 m,$$\delta(L)\simeq\frac{5L}{\sqrt{\mathrm{Re}_L}}=0.0255\ \mathrm m,$$

about five percent of the plate length. Ignoring the low-pressure viscosity correction would overestimate ℎ$h$ by a factor of about 4.11. The estimated molecular mean free path is much smaller than this boundary-layer thickness, so the low pressure does not by itself invalidate the continuum model.

Answer

The cooling system must remove approximately 570 W$570\ \mathrm W$ per metre of plate width from the exposed face.

Forced convection in external flow

Heated cylinder measurement

A metallic cylinder (12.7 mm in diameter and 94 mm long) is heated internally by an electrical heater and is subjected to a cross flow of air. The air velocity and temperature are maintained at 10 m/s$10\ \mathrm{m/s}$ and 26.2 °C$26.2\ \degree\mathrm{C}$. The heater power dissipation is measured to be 39.1 W, while the average cylinder surface temperature is 128.4 °C$128.4\ \degree\mathrm{C}$. Compare the heat transfer coefficient from experimental observations and correlation predictions.

Solution

Figure 4: Air flows across the cylinder, perpendicular to its axis. The convecting lateral area is pi D L when the end faces are neglected.

Data

Quantity	Value
Cylinder diameter	𝐷 =0.0127 m$D=0.0127\ \mathrm m$
Cylinder length	𝐿 =0.094 m$L=0.094\ \mathrm m$
Air speed	𝑢∞ =10 m s−1$u_\infty=10\ \mathrm{m\,s^{-1}}$
Air temperature	𝑇∞ =26.2 °C$T_\infty=26.2\ \degree\mathrm C$
Surface temperature	𝑇𝑠 =128.4 °C$T_s=128.4\ \degree\mathrm C$
Measured heat dissipation	˙𝑄 =39.1 W$\dot Q=39.1\ \mathrm W$
Air density	𝜌 =0.995 kg m−3$\rho=0.995\ \mathrm{kg\,m^{-3}}$
Dynamic viscosity	𝜇 =2.082 ×10−5 Pa s$\mu=2.082\times10^{-5}\ \mathrm{Pa\,s}$
Air conductivity	𝑘 =0.030 W m−1 K−1$k=0.030\ \mathrm{W\,m^{-1}\,K^{-1}}$
Prandtl number	Pr =0.700$\mathrm{Pr}=0.700$

The film temperature is the mean of the surface and air temperatures, 77.3 degrees Celsius or 350.45 K. Use the lecturer’s property set near 350 K.

Assumptions

• Steady state.
• Uniform mean surface temperature.
• Constant fluid properties at the film temperature.
• The stated 39.1 W$39.1\ \mathrm W$ leaves through the curved cylinder surface.
• Radiation and heat transfer through the ends are neglected in the comparison.
• The long-cylinder cross-flow correlations approximate the finite cylinder.

Calculations

The measured coefficient follows from Newton’s relation. The curved area is

𝐴=𝜋𝐷𝐿=𝜋(0.0127)(0.094)=0.00375043 m2,$$A=\pi DL=\pi(0.0127)(0.094)=0.00375043\ \mathrm{m^2},$$

giving

ℎexp=˙𝑄𝐴(𝑇𝑠−𝑇∞)=39.10.00375043(128.4−26.2)=102.01 W m−2 K−1$$h_{\mathrm{exp}}=\frac{\dot Q}{A(T_s-T_\infty)} =\frac{39.1}{0.00375043(128.4-26.2)} =102.01\ \mathrm{W\,m^{-2}\,K^{-1}}$$(23)

The flow correlation uses diameter, rather than cylinder length:

Re𝐷=𝜌𝑢∞𝐷𝜇=0.995(10)(0.0127)2.082×10−5=6069.4$$\mathrm{Re}_D=\frac{\rho u_\infty D}{\mu} =\frac{0.995(10)(0.0127)}{2.082\times10^{-5}}=6069.4$$

This lies in the interval 4000 <Re𝐷 <40000$4000<\mathrm{Re}_D<40000$, so Table 1 specifies 𝐶 =0.193$C=0.193$ and 𝑚 =0.618$m=0.618$. The power law in Eq. 21 gives

¯¯¯¯¯¯¯¯Nu𝐷=𝐶Re𝑚𝐷Pr1/3=0.193(6069.4)0.618(0.700)1/3=37.317$$\overline{\mathrm{Nu}}_D=C\mathrm{Re}_D^m\mathrm{Pr}^{1/3} =0.193(6069.4)^{0.618}(0.700)^{1/3}=37.317$$

The corresponding coefficient is

¯¯¯¯ℎ=𝑘¯¯¯¯¯¯¯¯Nu𝐷𝐷=0.030(37.317)0.0127=88.15 W m−2 K−1$$\bar h=\frac{k\overline{\mathrm{Nu}}_D}{D} =\frac{0.030(37.317)}{0.0127} =88.15\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For an independent correlation comparison, substitute the same Re and Pr into the Churchill-Bernstein relation in Eq. 22:

¯¯¯¯¯¯¯¯Nu𝐷=0.3+0.62(6069.4)1/2(0.700)1/3[1+(0.40.700)2/3]1/4[1+(6069.4282000)5/8]4/5=40.631$$\overline{\mathrm{Nu}}_D =0.3+\frac{0.62(6069.4)^{1/2}(0.700)^{1/3}} {\left[1+\left(\frac{0.4}{0.700}\right)^{2/3}\right]^{1/4}} \left[1+\left(\frac{6069.4}{282000}\right)^{5/8}\right]^{4/5} =40.631$$

Converting this prediction to a coefficient gives

¯¯¯¯ℎCB=0.030(40.631)0.0127=95.98 W m−2 K−1$$\bar h_{\mathrm{CB}}=\frac{0.030(40.631)}{0.0127} =95.98\ \mathrm{W\,m^{-2}\,K^{-1}}$$

The power law is 13.6 percent below the measured coefficient; Churchill and Bernstein is 5.9 percent below it. The experimental value is an effective coefficient under the assumed heat balance. The original slide diagram also shows 46 W$46\ \mathrm W$ heater input and 6.9 W$6.9\ \mathrm W$ losses, whose difference is 39.1 W$39.1\ \mathrm W$. Using the full 46 W$46\ \mathrm W$ with the curved area would include heat leaving by another route.

Radiation would reduce the heat assigned to convection. For illustration, 𝜖 =0.5$\epsilon=0.5$ and surroundings at the air temperature give

˙𝑄rad=𝜖𝜎𝐴(𝑇4𝑠−𝑇4surr)=0.5(5.6704×10−8)(0.00375043)[(401.55)4−(299.35)4]=1.91 W$$\begin{aligned} \dot Q_{\mathrm{rad}}&=\epsilon\sigma A(T_s^4-T_{\mathrm{surr}}^4)\\ &=0.5(5.6704\times10^{-8})(0.00375043) [(401.55)^4-(299.35)^4]=1.91\ \mathrm W \end{aligned}$$

Subtracting this illustrative loss gives a convection-only measured coefficient of about 97.0 W m−2 K−1$97.0\ \mathrm{W\,m^{-2}\,K^{-1}}$. The emissivity is not supplied, so this is a sensitivity check. The finite aspect ratio 𝐿/𝐷 =7.40$L/D=7.40$ and uncertain support losses also limit the comparison.

Answer

The measured effective coefficient is approximately 102 W m−2 K−1$102\ \mathrm{W\,m^{-2}\,K^{-1}}$. The cylinder power law predicts 88.2 W m−2 K−1$88.2\ \mathrm{W\,m^{-2}\,K^{-1}}$ and Churchill-Bernstein predicts 96.0 W m−2 K−1$96.0\ \mathrm{W\,m^{-2}\,K^{-1}}$ under the stated assumptions.

Forced convection in external flow

Heat transfer from a skin-friction measurement

Air flows along a smooth flat plate at 15 m s⁻¹. The measured average skin-friction coefficient is ¯¯¯¯𝐶𝑓 =0.0040$\bar C_f=0.0040$. At the film temperature, use 𝜌 =1.18$\rho=1.18$ kg m⁻³, 𝐶𝑝 =1005$C_p=1005$ J kg⁻¹ K⁻¹ and Pr =0.72$\mathrm{Pr}=0.72$. Estimate the average heat transfer coefficient with the Colburn analogy. Explain why the uncorrected Reynolds analogy gives a different answer. Assume matching flow and thermal conditions for the analogy and negligible pressure drag.

Solution

Data

• Air speed 𝑢∞ =15 m s−1$u_\infty=15\ \mathrm{m\,s^{-1}}$
• Average skin-friction coefficient ¯¯¯¯𝐶𝑓 =0.0040$\bar C_f=0.0040$
• Film density 𝜌 =1.18 kg m−3$\rho=1.18\ \mathrm{kg\,m^{-3}}$
• Specific heat capacity 𝐶𝑝 =1005 J kg−1 K−1$C_p=1005\ \mathrm{J\,kg^{-1}\,K^{-1}}$
• Prandtl number Pr =0.72$\mathrm{Pr}=0.72$

Assumptions

• Matching flow and thermal conditions for the Colburn analogy.
• Negligible pressure drag.
• Constant fluid properties.

Calculations

Use average quantities throughout. Equation Eq. 13 gives

¯¯¯¯¯¯St=¯¯¯¯𝐶𝑓2Pr−2/3=0.0020(0.72)−2/3=0.002490$$\overline{\mathrm{St}}=\frac{\bar C_f}{2}\mathrm{Pr}^{-2/3} =0.0020(0.72)^{-2/3}=0.002490$$

Recover the dimensional coefficient from Stanton number:

¯¯¯¯ℎ=¯¯¯¯¯¯St𝜌𝑢∞𝐶𝑝=0.002490(1.18)(15)(1005)=44.3 W m−2 K−1$$\bar h=\overline{\mathrm{St}}\rho u_\infty C_p =0.002490(1.18)(15)(1005)=44.3\ \mathrm{W\,m^{-2}\,K^{-1}}$$

The plain Reynolds analogy would set ¯¯¯¯¯¯St =0.0020$\overline{\mathrm{St}}=0.0020$ and give 35.6 W m⁻² K⁻¹. It assumes Pr equal to one. At the supplied Pr, the Colburn correction increases the estimate by a factor of 1.245.

No viscosity or plate length is required because skin friction is measured. Using a local friction coefficient with an average heat transfer coefficient would mix two different quantities. Total drag on a sphere or cylinder also cannot replace the plate skin-friction coefficient in this calculation.

Answer

The Colburn estimate is ¯¯¯¯ℎ =44.3 W m−2 K−1$\bar h=44.3\ \mathrm{W\,m^{-2}\,K^{-1}}$. The uncorrected Reynolds analogy gives 35.6 W m−2 K−1$35.6\ \mathrm{W\,m^{-2}\,K^{-1}}$, because it omits the Prandtl-number correction.

Forced convection in external flow

Where does a local plate coefficient fall to half its value?

Air flows at 0.80 m s⁻¹ over an isothermal flat plate. In the laminar region, the local coefficient at 𝑥1 =0.25$x_1=0.25$ m is ℎ𝑥1 =12.0$h_{x_1}=12.0$ W m⁻² K⁻¹. Use 𝜈 =1.60 ×10−5$\nu=1.60\times10^{-5}$ m² s⁻¹ and constant properties. At what distance will the local coefficient be 6.0 W m⁻² K⁻¹? Find the average coefficient between the leading edge and that distance, and check the laminar assumption.

Solution

Data

• Air speed 𝑢∞ =0.80 m s−1$u_\infty=0.80\ \mathrm{m\,s^{-1}}$
• First position 𝑥1 =0.25 m$x_1=0.25\ \mathrm m$
• Coefficient at the first position ℎ𝑥1 =12.0 W m−2 K−1$h_{x_1}=12.0\ \mathrm{W\,m^{-2}\,K^{-1}}$
• Target coefficient ℎ𝑥2 =6.0 W m−2 K−1$h_{x_2}=6.0\ \mathrm{W\,m^{-2}\,K^{-1}}$
• Kinematic viscosity 𝜈 =1.60 ×10−5 m2 s−1$\nu=1.60\times10^{-5}\ \mathrm{m^2\,s^{-1}}$

Assumptions

• Steady laminar flow over an isothermal plate.
• Constant fluid properties.

Calculations

For fixed speed and properties, the local relation has Nu𝑥 ∝Re1/2𝑥 ∝𝑥1/2$\mathrm{Nu}_x\propto\mathrm{Re}_x^{1/2}\propto x^{1/2}$. Because ℎ𝑥 =Nu𝑥𝑘/𝑥$h_x=\mathrm{Nu}_xk/x$, it follows that ℎ𝑥 ∝𝑥−1/2$h_x\propto x^{-1/2}$:

ℎ𝑥2ℎ𝑥1=(𝑥1𝑥2)1/2𝑥2=𝑥1(12.06.0)2=1.00 m$$\begin{gathered} \frac{h_{x_2}}{h_{x_1}}=\left(\frac{x_1}{x_2}\right)^{1/2}\\ x_2=x_1\left(\frac{12.0}{6.0}\right)^2=1.00\ \mathrm m \end{gathered}$$

Halving a local coefficient requires four times the distance, not twice. The layer thickens downstream and the wall temperature gradient decreases.

The length average is a different quantity. Integrating the local law gives

¯¯¯¯ℎ𝐿=1𝐿∫𝐿0𝐶𝑥−1/2 𝑑𝑥=2𝐶𝐿−1/2=2ℎ𝐿$$\bar h_L=\frac1L\int_0^L Cx^{-1/2}\,dx =2CL^{-1/2}=2h_L$$

Thus the average over the first 1.00 m is 12.0 W m⁻² K⁻¹ even though the local coefficient at its trailing edge is 6.0 W m⁻² K⁻¹. The ideal leading-edge singularity is integrable.

At 𝑥2$x_2$,

Re𝑥2=0.80(1.00)𝜈=50000,$$\mathrm{Re}_{x_2}=\frac{0.80(1.00)}{\nu}=50000,$$

below the transition range discussed above. The whole calculation stays in the laminar region. A measured halving distance beyond transition would need the appropriate changing flow model.

Answer

The local coefficient falls to half at 𝑥2 =1.00 m$x_2=1.00\ \mathrm m$. The average coefficient over that length is 12.0 W m−2 K−1$12.0\ \mathrm{W\,m^{-2}\,K^{-1}}$. The Reynolds number is 50000$50000$, consistent with laminar flow.

Forced convection in internal flow

Electrically heated air tube

Air (1 atm, 77 °C$77\ \degree\mathrm{C}$) enters a 5 mm I.D. tube with a bulk average velocity of 2.5 m/s$2.5\ \mathrm{m/s}$. The velocity profile is developed and the thermal profile is “developing”. The tube length is 1.0 m, and a constant (uniform) heat flux is imposed at the tube surface on the air over the entire length. The bulk average exit temperature of the air stream is 127 °C$127\ \degree\mathrm{C}$. Determine:

• The heat transfer coefficient at the exit (ℎ𝐿$h_L$)
• The heat flux imposed at the surface of the tube (𝑞𝑠$q_s$)
• The surface temperature (𝑇𝑠$T_s$) of the tube at the exit

Use ideal-gas air with 𝑅air =287.05 J/(kg K)$R_{\mathrm{air}}=287.05\ \mathrm{J/(kg\,K)}$ and 𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$. At the mean bulk temperature of 375.15 K, use 𝜇 =2.1805 ×10−5 Pa s$\mu=2.1805\times10^{-5}\ \mathrm{Pa\,s}$, 𝑘 =0.03184 W/(m K)$k=0.03184\ \mathrm{W/(m\,K)}$ and 𝐶𝑝 =1011.5 J/(kg K)$C_p=1011.5\ \mathrm{J/(kg\,K)}$.

Solution

Figure 3: Heat enters through the tube wall while air flows from inlet Ti to outlet Te. In this example the inward wall flux is uniform.

Data

Quantity	Value
Inside diameter	𝐷 =0.0050 m$D=0.0050\ \mathrm m$
Heated length	𝐿 =1.0 m$L=1.0\ \mathrm m$
Inlet mean speed	𝑢𝑚 =2.5 m s−1$u_m=2.5\ \mathrm{m\,s^{-1}}$
Inlet bulk temperature	𝑇𝑖 =77 °C$T_i=77\ \degree\mathrm C$
Outlet bulk temperature	𝑇𝑒 =127 °C$T_e=127\ \degree\mathrm C$
Inlet air density	𝜌𝑖 =1.00810 kg m−3$\rho_i=1.00810\ \mathrm{kg\,m^{-3}}$
Dynamic viscosity	𝜇 =2.1805 ×10−5 Pa s$\mu=2.1805\times10^{-5}\ \mathrm{Pa\,s}$
Air conductivity	𝑘 =0.03184 W m−1 K−1$k=0.03184\ \mathrm{W\,m^{-1}\,K^{-1}}$
Specific heat capacity	𝐶𝑝 =1011.5 J kg−1 K−1$C_p=1011.5\ \mathrm{J\,kg^{-1}\,K^{-1}}$

The wall heat flux is uniform, and the inlet velocity profile is developed. The mean bulk temperature is 102 degrees Celsius or 375.15 K. Use transport properties near 375 K, but pair inlet velocity with inlet density:

𝜌𝑖=𝑃𝑅air𝑇𝑖=101325287.05×350.15=1.00810 kg m−3.$$\begin{aligned} \rho_i&=\frac{P}{R_{\mathrm{air}}T_i}\\ &=\frac{101325}{287.05\times350.15}\\ &=1.00810\ \mathrm{kg\,m^{-3}}. \end{aligned}$$

Assumptions

• Steady state.
• Constant viscosity, conductivity and heat capacity; ideal-gas density varies with temperature at approximately constant pressure.
• Heat supplied at the wall enters the air.
• Axial conduction, viscous heating and kinetic and potential energy changes are neglected.
• The tube is circular and the flow remains laminar.

Calculations

First check the flow regime and whether the temperature profile has developed by the exit. Using the diameter in the Reynolds number, as defined in Eq. 18, gives

Re𝐷=𝜌𝑖𝑢𝑚,𝑖𝐷𝜇=1.00810(2.5)(0.0050)2.1805×10−5=577.91$$\mathrm{Re}_D=\frac{\rho_i u_{m,i}D}{\mu} =\frac{1.00810(2.5)(0.0050)}{2.1805\times10^{-5}}=577.91$$

The Prandtl number is

Pr=𝐶𝑝𝜇𝑘=1011.5(2.1805×10−5)0.03184=0.69271$$\mathrm{Pr}=\frac{C_p\mu}{k} =\frac{1011.5(2.1805\times10^{-5})}{0.03184}=0.69271$$

The flow is laminar. The thermal-entry estimate from Eq. 19 is

𝐿𝑡≃0.0575Re𝐷Pr𝐷=0.0575(577.91)(0.69271)(0.0050)=0.1151 m$$L_t\simeq0.0575\mathrm{Re}_D\mathrm{Pr}D =0.0575(577.91)(0.69271)(0.0050)=0.1151\ \mathrm m$$

The temperature profile develops after heating starts, but the exit at 1.0 m$1.0\ \mathrm m$ lies well beyond this estimated entrance length. Use the developed local constant-flux result from Eq. 12, Nu𝐷 =48/11$\mathrm{Nu}_D=48/11$:

ℎ𝐿=Nu𝐷𝑘𝐷=48(0.03184)11(0.0050)=27.79 W m−2 K−1$$h_L=\frac{\mathrm{Nu}_Dk}{D} =\frac{48(0.03184)}{11(0.0050)} =27.79\ \mathrm{W\,m^{-2}\,K^{-1}}$$(25)

The total energy balance determines the wall heat flux independently of the local entrance coefficient. First calculate the air mass flow:

˙𝑚=𝜌𝑖𝑢𝑚,𝑖𝜋𝐷24=1.00810(2.5)𝜋(0.0050)24=4.94851×10−5 kg s−1$$\dot m=\rho_i u_{m,i}\frac{\pi D^2}{4} =1.00810(2.5)\frac{\pi(0.0050)^2}{4} =4.94851\times10^{-5}\ \mathrm{kg\,s^{-1}}$$

The enthalpy rise is the total heat input:

˙𝑄=˙𝑚𝐶𝑝(𝑇𝑒−𝑇𝑖)=(4.94851×10−5)(1011.5)(127−77)=2.5027 W$$\dot Q=\dot mC_p(T_e-T_i) =(4.94851\times10^{-5})(1011.5)(127-77) =2.5027\ \mathrm W$$

Since the flux is uniform over area 𝜋𝐷𝐿$\pi DL$,

𝑞𝑠=˙𝑄𝜋𝐷𝐿=2.5027𝜋(0.0050)(1.0)=159.33 W m−2$$q_s=\frac{\dot Q}{\pi DL} =\frac{2.5027}{\pi(0.0050)(1.0)} =159.33\ \mathrm{W\,m^{-2}}$$(26)

The local wall relation in Eq. 5,

𝑞𝑠=ℎ𝐿[𝑇𝑠(𝐿)−𝑇𝑒],$$q_s=h_L[T_s(L)-T_e],$$

then gives

𝑇𝑠(𝐿)=𝑇𝑒+𝑞𝑠ℎ𝐿=127+159.3327.79=132.73 °C$$T_s(L)=T_e+\frac{q_s}{h_L} =127+\frac{159.33}{27.79} =132.73\ \degree\mathrm{C}$$(27)

The wall is 5.73 K$5.73\ \mathrm K$ hotter than the exit air, as required for inward heat flow. Near the inlet the coefficient is larger, so the same heat flux needs a smaller wall-to-bulk difference. The developed coefficient must not be applied over that entrance region. The large axial Peclet product

Re Pr𝐿𝐷≃8.01×104$$\mathrm{Re}\,\mathrm{Pr}\frac{L}{D}\simeq8.01\times10^4$$

also supports neglecting axial conduction. At the same volume flow and temperature rise, replacing air by water would require a much larger heat input because water has a much larger volumetric heat capacity.

Answer

At the exit, ℎ𝐿 ≃27.8 W m−2 K−1$h_L\simeq27.8\ \mathrm{W\,m^{-2}\,K^{-1}}$. The imposed heat flux is 𝑞𝑠 ≃159 W m−2$q_s\simeq159\ \mathrm{W\,m^{-2}}$, and the exit wall temperature is 𝑇𝑠(𝐿) ≃132.7 °C$T_s(L)\simeq132.7\ \degree\mathrm{C}$.

Forced convection in internal flow

Steam condenser tube sizing

A horizontal tubular condenser is used to condense saturated steam at 80 °C$80\ \degree\mathrm{C}$. The condenser is a shell and tube one with brass tubes (𝑘 =110 W/(m⋅K)$k=110\ \mathrm{W/(m\cdot K)}$, 1.59 cm OD, 1.34 cm ID). Steam is outside of the tubes (ℎ =10760 W/(m2⋅K)$h=10760\ \mathrm{W/(m^2\cdot K)}$) and cooling water enters the tube at 20 °C$20\ \degree\mathrm{C}$ with a velocity of 1.4 m/s$1.4\ \mathrm{m/s}$. Water leaves the tubes at 40 °C$40\ \degree\mathrm{C}$. If the rate of cooling water supply is 55000 kg/h and the latent heat of condensation of steam at 80 °C$80\ \degree\mathrm{C}$ is 2304 kJ/kg, calculate:

• The number of tubes required
• The length of each tube

Use inlet water density 𝜌𝑖 =998.2 kg/m3$\rho_i=998.2\ \mathrm{kg/m^3}$ at 20 degrees Celsius. At the bulk mean of 30 degrees Celsius, use 𝜇 =7.98 ×10−4 Pa s$\mu=7.98\times10^{-4}\ \mathrm{Pa\,s}$, 𝑘 =0.615 W/(m K)$k=0.615\ \mathrm{W/(m\,K)}$ and 𝐶𝑝 =4180 J/(kg K)$C_p=4180\ \mathrm{J/(kg\,K)}$. Treat 1.4 m/s as the nominal inlet speed; after rounding the number of tubes up, use the exact total mass flow to calculate the duty and Reynolds number of each tube.

Solution

Figure 4: Water is heated as it passes through a condenser tube. The heat flux is inward; the wall temperature and local flux need not be uniform. The steam film and brass wall are included in the overall coefficient.

Data

Quantity	Value
Steam temperature	𝑇steam =80 °C$T_{\mathrm{steam}}=80\ \degree\mathrm C$
Outer diameter	𝐷𝑜 =0.0159 m$D_o=0.0159\ \mathrm m$
Inner diameter	𝐷𝑖 =0.0134 m$D_i=0.0134\ \mathrm m$
Brass conductivity	𝑘𝑤 =110 W m−1 K−1$k_w=110\ \mathrm{W\,m^{-1}\,K^{-1}}$
Outside coefficient	ℎ𝑜 =10760 W m−2 K−1$h_o=10760\ \mathrm{W\,m^{-2}\,K^{-1}}$
Water inlet temperature	𝑇𝑖 =20 °C$T_i=20\ \degree\mathrm C$
Water outlet temperature	𝑇𝑒 =40 °C$T_e=40\ \degree\mathrm C$
Nominal inlet water speed	𝑢𝑚 =1.4 m s−1$u_m=1.4\ \mathrm{m\,s^{-1}}$
Total water mass flow	˙𝑚 =55000 kg h−1$\dot m=55000\ \mathrm{kg\,h^{-1}}$
Inlet water density	𝜌𝑖 =998.2 kg m−3$\rho_i=998.2\ \mathrm{kg\,m^{-3}}$
Dynamic viscosity	𝜇 =7.98 ×10−4 Pa s$\mu=7.98\times10^{-4}\ \mathrm{Pa\,s}$
Water conductivity	𝑘 =0.615 W m−1 K−1$k=0.615\ \mathrm{W\,m^{-1}\,K^{-1}}$
Specific heat capacity	𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$

The rounded water properties are consistent with the inlet and bulk mean states. See the IAPWS liquid-water property release. The inlet density determines flow area; mean-temperature viscosity, conductivity and heat capacity determine heat transfer.

Assumptions

• Steady state.
• Equal flow through parallel tubes.
• Constant properties and heat transfer coefficients.
• Developed turbulent flow in smooth tubes.
• Clean walls, with fouling resistance neglected.
• Negligible heat loss to the surroundings.

Calculations

The nominal inlet speed first determines a trial mass flow per tube:

˙𝑚trial=𝜌𝑖𝑢𝑚,𝑖𝜋𝐷2𝑖4=998.2(1.4)𝜋(0.0134)24=0.19708 kg s−1$$\dot m_{\mathrm{trial}}=\rho_i u_{m,i}\frac{\pi D_i^2}{4} =998.2(1.4)\frac{\pi(0.0134)^2}{4} =0.19708\ \mathrm{kg\,s^{-1}}$$

Divide total flow by this value and round up:

𝑁∗=˙𝑚total˙𝑚trial=55000/36000.19708=77.52$$N_* =\frac{\dot m_{\mathrm{total}}}{\dot m_{\mathrm{trial}}} =\frac{55000/3600}{0.19708}=77.52$$(28)

Use 78 tubes. The exact mass flow per tube is then

˙𝑚tube=550003600(78)=0.195869 kg s−1.$$\dot m_{\mathrm{tube}}=\frac{55000}{3600(78)} =0.195869\ \mathrm{kg\,s^{-1}}.$$

This gives an inlet speed of 1.3914 m s−1$1.3914\ \mathrm{m\,s^{-1}}$, close to the nominal value. Use this mass flow throughout the remaining calculation.

The inside coefficient follows from the mass flow and mean-temperature properties:

Re𝑖=4˙𝑚tube𝜋𝐷𝑖𝜇=4(0.195869)𝜋(0.0134)(7.98×10−4)=23322$$\mathrm{Re}_i=\frac{4\dot m_{\mathrm{tube}}}{\pi D_i\mu} =\frac{4(0.195869)}{\pi(0.0134)(7.98\times10^{-4})}=23322$$Pr=𝐶𝑝𝜇𝑘=4180(7.98×10−4)0.615=5.4238$$\mathrm{Pr}=\frac{C_p\mu}{k} =\frac{4180(7.98\times10^{-4})}{0.615}=5.4238$$

The flow is turbulent. Since the water is heated, use 𝑛 =0.4$n=0.4$ in the Dittus-Boelter relation from Eq. 24:

Nu𝑖=0.023Re0.8𝑖Pr0.4=0.023(23322)0.8(5.4238)0.4=141.14$$\mathrm{Nu}_i=0.023\mathrm{Re}_i^{0.8}\mathrm{Pr}^{0.4} =0.023(23322)^{0.8}(5.4238)^{0.4}=141.14$$

Convert Nusselt number using the inside diameter:

ℎ𝑖=Nu𝑖𝑘𝐷𝑖=141.14(0.615)0.0134=6478 W m−2 K−1$$h_i=\frac{\mathrm{Nu}_ik}{D_i} =\frac{141.14(0.615)}{0.0134} =6478\ \mathrm{W\,m^{-2}\,K^{-1}}$$(29)

Heat passes through the steam film, the brass wall and the water film in series. The steam temperature is fixed, but the inner wall is cooler. Using the cylindrical conduction resistance, the overall coefficient on inside area satisfies

1𝑈𝑖=𝐷𝑖𝐷𝑜ℎ𝑜+𝐷𝑖ln⁡(𝐷𝑜𝐷𝑖)2𝑘𝑤+1ℎ𝑖=0.01340.0159(10760)+0.0134ln⁡(0.01590.0134)2(110)+16478=2.4311×10−4 m2 K W−1$$\begin{aligned} \frac{1}{U_i} &=\frac{D_i}{D_oh_o} +\frac{D_i\ln\left(\frac{D_o}{D_i}\right)}{2k_w}+\frac1{h_i}\\ &=\frac{0.0134}{0.0159(10760)} +\frac{0.0134\ln\left(\frac{0.0159}{0.0134}\right)}{2(110)}+\frac1{6478}\\ &=2.4311\times10^{-4}\ \mathrm{m^2\,K\,W^{-1}} \end{aligned}$$(30)

Thus 𝑈𝑖 =4113 W m−2 K−1$U_i=4113\ \mathrm{W\,m^{-2}\,K^{-1}}$. The water film supplies about 63.5 percent of the total resistance, the steam film 32.2 percent and the brass wall 4.3 percent. Changing the wall conductivity alone would therefore have little effect. Area bases must remain consistent, with 𝑈𝑖𝐴𝑖 =𝑈𝑜𝐴𝑜$U_iA_i=U_oA_o$.

The driving differences are 80 −20 =60 K$80-20=60\ \mathrm K$ at the water inlet and 80 −40 =40 K$80-40=40\ \mathrm K$ at the outlet. Their logarithmic mean, defined in Eq. 15, is

Δ𝑇ln⁡=Δ𝑇𝑖−Δ𝑇𝑒ln⁡(Δ𝑇𝑖Δ𝑇𝑒)=60−40ln⁡(6040)=49.326 K$$\Delta T_{\ln}=\frac{\Delta T_i-\Delta T_e} {\ln\left(\frac{\Delta T_i}{\Delta T_e}\right)} =\frac{60-40}{\ln\left(\frac{60}{40}\right)}=49.326\ \mathrm K$$

The per-tube heat duty follows from the water enthalpy rise:

˙𝑄tube=˙𝑚tube𝐶𝑝(𝑇𝑒−𝑇𝑖)=0.195869(4180)(40−20)=16375 W$$\dot Q_{\mathrm{tube}}=\dot m_{\mathrm{tube}}C_p(T_e-T_i) =0.195869(4180)(40-20)=16375\ \mathrm W$$

Set this equal to 𝑈𝑖𝜋𝐷𝑖𝐿Δ𝑇ln⁡$U_i\pi D_iL\Delta T_{\ln}$ and solve for length:

𝐿=˙𝑄tube𝑈𝑖𝜋𝐷𝑖Δ𝑇ln⁡=163754113𝜋(0.0134)(49.326)=1.92 m$$L=\frac{\dot Q_{\mathrm{tube}}}{U_i\pi D_i\Delta T_{\ln}} =\frac{16375}{4113\pi(0.0134)(49.326)}=1.92\ \mathrm m$$(31)

The ratio 𝐿/𝐷𝑖 ≃143$L/D_i\simeq143$ supports the developed turbulent approximation. The ideal isothermal steam source needs no flow-arrangement correction to the logarithmic mean. The arithmetic mean 50 K$50\ \mathrm K$ would overestimate the driving force by about 1.37 percent.

As a steam-side check, the total water duty and the stated latent heat give

˙𝑄total=550003600(4180)(40−20)=1.2772×106 W$$\dot Q_{\mathrm{total}}=\frac{55000}{3600}(4180)(40-20) =1.2772\times10^6\ \mathrm W$$˙𝑚steam=˙𝑄totalℎ𝑓𝑔=1.2772×1062.304×106=0.5544 kg s−1≃1996 kg h−1$$\dot m_{\mathrm{steam}}=\frac{\dot Q_{\mathrm{total}}}{h_{fg}} =\frac{1.2772\times10^6}{2.304\times10^6} =0.5544\ \mathrm{kg\,s^{-1}}\simeq1996\ \mathrm{kg\,h^{-1}}$$

The latent heat checks the steam requirement; it does not determine tube count. These are clean-tube estimates using the stated correlation. A practical design also needs allowances for fouling, pressure drop and available tube sizes.

Answer

Using properties at the stated water temperatures, provide 78 parallel tubes of approximately 1.92 m$1.92\ \mathrm m$ heated length each.

Forced convection in internal flow

Which pipe profile develops first?

Oil enters a 16 mm diameter pipe with Re =900$\mathrm{Re}=900$ and Pr =60$\mathrm{Pr}=60$. Estimate the hydrodynamic and thermal entrance lengths. Is a 5 m pipe long enough for both profiles to develop? Assume a uniform velocity and temperature at the inlet and heating beginning there.

Solution

Data

• Diameter 𝐷 =0.016 m$D=0.016\ \mathrm m$
• Pipe length 𝐿 =5.0 m$L=5.0\ \mathrm m$
• Reynolds number Re =900$\mathrm{Re}=900$
• Prandtl number Pr =60$\mathrm{Pr}=60$

Assumptions

• Uniform inlet velocity and temperature.
• Heating begins at the pipe inlet.
• Constant properties.

Calculations

The Reynolds number is below 2300, so use the laminar estimates in Eq. 19, converting the diameter to 0.016 m:

𝐿ℎ=0.0575(900)(0.016)=0.828 m𝐿𝑡=0.0575(900)(60)(0.016)=49.7 m$$L_h=0.0575(900)(0.016)=0.828\ \mathrm m \qquad L_t=0.0575(900)(60)(0.016)=49.7\ \mathrm m$$

The velocity profile develops within the pipe, but the temperature profile does not. The ratio 𝐿𝑡/𝐿ℎ =Pr =60$L_t/L_h=\mathrm{Pr}=60$ expresses the slower thermal diffusion relative to momentum diffusion in this oil. A heat transfer calculation for the 5 m pipe needs a thermal-entry correlation. A developed velocity profile alone does not justify a developed Nusselt number.

If heating starts after a long unheated approach, measure thermal entry from the start of heating. The two entrance lengths then have different origins.

Answer

The hydrodynamic entrance length is 𝐿ℎ =0.828 m$L_h=0.828\ \mathrm m$ and the thermal entrance length is 𝐿𝑡 =49.7 m$L_t=49.7\ \mathrm m$. The 5.0 m$5.0\ \mathrm m$ pipe permits velocity development but remains thermally developing.

Forced convection in internal flow

Compare constant wall temperature and constant wall heat flux

Water enters a 2.0 m$2.0\ \mathrm m$ heated tube at 20 °C$20\ \degree\mathrm{C}$. The inside diameter is 0.020 m$0.020\ \mathrm m$, mass flow is 0.020 kg s−1$0.020\ \mathrm{kg\,s^{-1}}$ and 𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$. Adopt a constant effective coefficient ℎ =400 W m−2 K−1$h=400\ \mathrm{W\,m^{-2}\,K^{-1}}$ to isolate the effect of the wall condition. Neglect axial conduction and heat loss.

Find the outlet temperature and total heat input for two separate cases: a wall maintained at 80 °C$80\ \degree\mathrm{C}$, and a uniform inward wall heat flux of 10000 W m−2$10000\ \mathrm{W\,m^{-2}}$. Explain how the wall-to-bulk temperature difference changes with distance in each case.

Solution

Figure 5: The same flowing stream is heated in two separate cases. One fixes the wall temperature; the other fixes inward heat flux. The energy balance relates inlet and outlet bulk temperatures in both cases.

Data

Quantity	Value
Tube length	𝐿 =2.0 m$L=2.0\ \mathrm m$
Diameter	𝐷 =0.020 m$D=0.020\ \mathrm m$
Mass flow	˙𝑚 =0.020 kg s−1$\dot m=0.020\ \mathrm{kg\,s^{-1}}$
Inlet temperature	𝑇𝑖 =20 °C$T_i=20\ \degree\mathrm C$
Specific heat capacity	𝐶𝑝 =4180 J kg−1 K−1$C_p=4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Supplied effective coefficient	ℎ =400 W m−2 K−1$h=400\ \mathrm{W\,m^{-2}\,K^{-1}}$
Wall temperature in the first case	𝑇𝑠 =80 °C$T_s=80\ \degree\mathrm C$
Wall heat flux in the second case	𝑞𝑠 =10000 W m−2$q_s=10000\ \mathrm{W\,m^{-2}}$

Assumptions

• Steady state.
• Constant fluid properties.
• Axial conduction and heat loss are neglected.

Calculations

The heated area is 𝐴𝑠 =𝜋𝐷𝐿 =0.12566$A_s=\pi DL=0.12566$ m² and the stream heat-capacity rate is ˙𝑚𝐶𝑝 =83.6$\dot mC_p=83.6$ W K⁻¹.

For constant wall temperature, Eq. 14 gives

ℎ𝐴𝑠˙𝑚𝐶𝑝=400[𝜋(0.020)(2.0)]0.020(4180)=0.6013$$\frac{hA_s}{\dot mC_p}=\frac{400[\pi(0.020)(2.0)]}{0.020(4180)}=0.6013$$𝑇𝑒=𝑇𝑠−(𝑇𝑠−𝑇𝑖)𝑒−ℎ𝐴𝑠˙𝑚𝐶𝑝=80−(80−20)𝑒−0.6013=47.11 °C$$T_e=T_s-(T_s-T_i)e^{-\frac{hA_s}{\dot m C_p}} =80-(80-20)e^{-0.6013}=47.11\ \degree\mathrm{C}$$˙𝑄=83.6(47.11−20)=2267 W$$\dot Q=83.6(47.11-20)=2267\ \mathrm W$$

The driving difference decreases from 60 K to 32.89 K. Even with a constant coefficient, the heat flux falls from 24000 to 13155 W m⁻². Fully developed flow does not require a constant temperature difference.

For constant wall heat flux, the overall balance instead gives

˙𝑄=𝑞𝑠𝐴𝑠=10000[𝜋(0.020)(2.0)]=1257 W$$\dot Q=q_s A_s=10000[\pi(0.020)(2.0)]=1257\ \mathrm W$$𝑇𝑒=𝑇𝑖+˙𝑄˙𝑚𝐶𝑝=20+125783.6=35.03 °C$$T_e=T_i+\frac{\dot Q}{\dot m C_p}=20+\frac{1257}{83.6}=35.03\ \degree\mathrm{C}$$𝑇𝑠−𝑇𝑚=𝑞𝑠ℎ=25.0 K$$T_s-T_m=\frac{q_s}{h}=25.0\ \mathrm K$$

The bulk temperature rises linearly, and the wall temperature rises with it, from 45.0 °C$45.0\ \degree\mathrm{C}$ to 60.03 °C$60.03\ \degree\mathrm{C}$. Their difference stays constant under the stipulated constant coefficient. In a real entrance region, ℎ(𝑥)$h(x)$ varies and a constant imposed flux need not give a constant difference. The two cases prescribe different heat inputs, so their duties need not agree.

Answer

For fixed wall temperature, 𝑇𝑒 =47.1 °C$T_e=47.1\ \degree\mathrm{C}$ and ˙𝑄 =2.27 ×103 W$\dot Q=2.27\times10^3\ \mathrm W$, with a decreasing wall-to-bulk difference. For fixed heat flux, 𝑇𝑒 =35.0 °C$T_e=35.0\ \degree\mathrm{C}$ and ˙𝑄 =1.26 ×103 W$\dot Q=1.26\times10^3\ \mathrm W$, with a constant 25.0 K$25.0\ \mathrm K$ wall-to-bulk difference.

Natural convection

Heating a one-sided brick wall

A flat wall of fire-clay brick (0.5 m thick) initially has a uniform temperature of 200K. One of its faces is suddenly exposed to hot gas at 1200K. If the heat transfer coefficient on the hot side is 7.38 W/m2.K$7.38\ \mathrm{W/m^2.K}$ and the other face of the wall is insulated, determine:

• The time it takes for the centre of the wall to reach 600K

Properties of fire-clay brick: 𝑘 =1.125 W/m.K$k=1.125\ \mathrm{W/m.K}$; 𝐶𝑝 =919 J/kg.K$C_p=919\ \mathrm{J/kg.K}$; 𝜌 =2310 kg/m3$\rho=2310\ \mathrm{kg/m^3}$

Estimate the heat transfer coefficient, assuming that it stays constant throughout the transient stage! Does the number quoted above make sense?

Solution

Reflecting the wall across its insulated face allows us to use the symmetric plane-wall solution. The half-thickness labelled 𝐿$L$ in the schematic is denoted 𝑏$b$ below, to distinguish it from the vertical convection length.

Data

Quantity	Value
Actual wall thickness	𝑏 =0.50 m$b=0.50\ \mathrm m$
Initial brick temperature	𝑇𝑖 =200 K$T_i=200\ \mathrm K$
Hot-gas temperature	𝑇∞ =1200 K$T_\infty=1200\ \mathrm K$
Target temperature	𝑇 =600 K$T=600\ \mathrm K$
Brick conductivity	𝑘𝑤 =1.125 W m−1 K−1$k_w=1.125\ \mathrm{W\,m^{-1}\,K^{-1}}$
Brick density	𝜌𝑤 =2310 kg m−3$\rho_w=2310\ \mathrm{kg\,m^{-3}}$
Brick specific heat capacity	𝐶𝑝,𝑤 =919 J kg−1 K−1$C_{p,w}=919\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Specified transient coefficient	ℎ =7.38 W m−2 K−1$h=7.38\ \mathrm{W\,m^{-2}\,K^{-1}}$
Air conductivity at 700 K	𝑘 =0.0530 W m−1 K−1$k=0.0530\ \mathrm{W\,m^{-1}\,K^{-1}}$
Air kinematic viscosity at 700 K	𝜈 =6.58 ×10−5 m2 s−1$\nu=6.58\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Air Prandtl number at 700 K	Pr =0.697$\mathrm{Pr}=0.697$

The lecturer’s initial-film air properties allow an independent check of the specified coefficient. The slide supplies no wall height for this convection estimate.

Assumptions

• One-dimensional transient conduction through the wall.
• Constant solid properties and heat transfer coefficient.
• No internal heat generation.
• Heat transfer through the edges is neglected.
• Radiation is neglected in the classroom model.
• A vertical panel of height 𝐿 =1.0 m$L=1.0\ \mathrm m$ is used only to estimate the convection coefficient.

Calculations

First check the supplied coefficient. Initially the exposed wall is at 200 K$200\ \mathrm K$, giving

𝑇𝑓=1200+2002=700 K$$T_f=\frac{1200+200}{2}=700\ \mathrm K$$

and 𝛽 =1/𝑇𝑓$\beta=1/T_f$ for ideal-gas air. The vertical-plate Rayleigh number is

Ra𝐿=𝑔𝛽(𝑇∞−𝑇𝑠)𝐿3𝜈2Pr=9.81(1700)(1200−200)(1.0)3(6.58×10−5)2(0.697)=2.256×109$$\mathrm{Ra}_L=\frac{g\beta(T_\infty-T_s)L^3}{\nu^2}\mathrm{Pr} =\frac{9.81\left(\frac{1}{700}\right)(1200-200)(1.0)^3} {(6.58\times10^{-5})^2}(0.697) =2.256\times10^9$$

This exceeds the upper limit of the laminar quarter-power correlation in Eq. 18. Use the vertical-plate turbulent estimate for 109 ≲Ra𝐿 ≲1013$10^9\lesssim\mathrm{Ra}_L\lesssim10^{13}$:

¯¯¯¯¯¯¯¯Nu𝐿=0.10Ra1/3𝐿=0.10(2.256×109)1/3=131.15$$\overline{\mathrm{Nu}}_L=0.10\mathrm{Ra}_L^{1/3} =0.10(2.256\times10^9)^{1/3}=131.15$$

The resulting coefficient is

¯¯¯¯ℎ=𝑘¯¯¯¯¯¯¯¯Nu𝐿𝐿=0.0530(131.15)1.0=6.95 W m−2 K−1$$\bar h=\frac{k\overline{\mathrm{Nu}}_L}{L} =\frac{0.0530(131.15)}{1.0} =6.95\ \mathrm{W\,m^{-2}\,K^{-1}}$$(23)

Churchill-Chu gives about 8.36 W m−2 K−1$8.36\ \mathrm{W\,m^{-2}\,K^{-1}}$ for the same conditions. The supplied value 7.38 W m−2 K−1$7.38\ \mathrm{W\,m^{-2}\,K^{-1}}$ is therefore a plausible convection estimate. The very large temperature range makes constant-property natural-convection calculations approximate. Use the supplied ℎ =7.38 W m−2 K−1$h=7.38\ \mathrm{W\,m^{-2}\,K^{-1}}$ for the requested transient calculation.

Measure 𝑥$x$ from the insulated face, with the exposed face at 𝑥 =𝑏$x=b$. The physical centre of the brick is at 𝑥 =𝑏/2$x=b/2$. Reflection across the insulated face produces an equivalent symmetric slab of thickness 2𝑏 =1.0 m$2b=1.0\ \mathrm m$. Its half-thickness is 𝑏$b$, so it is 0.50 m$0.50\ \mathrm m$, rather than 0.25 m$0.25\ \mathrm m$, that belongs in the slab Biot and Fourier numbers.

The solid diffusivity is

𝛼𝑤=𝑘𝑤𝜌𝑤𝐶𝑝,𝑤=1.1252310(919)=5.29938×10−7 m2 s−1$$\alpha_w=\frac{k_w}{\rho_wC_{p,w}} =\frac{1.125}{2310(919)} =5.29938\times10^{-7}\ \mathrm{m^2\,s^{-1}}$$

The target temperature ratio is

Θ=𝑇−𝑇∞𝑇𝑖−𝑇∞=600−1200200−1200=0.600$$\Theta=\frac{T-T_\infty}{T_i-T_\infty} =\frac{600-1200}{200-1200}=0.600$$

The slab Biot number is

Bi𝑏=ℎ𝑏𝑘𝑤=7.38(0.50)1.125=3.28$$\mathrm{Bi}_b=\frac{hb}{k_w} =\frac{7.38(0.50)}{1.125}=3.28$$(24)

Internal resistance is significant, so lumped capacitance is unsuitable. For the reflected slab, the first-term plane-wall solution is

Θ(𝑥,𝑡)=𝐴1𝑒−𝜆21Focos⁡(𝜆1𝑥𝑏)$$\Theta(x,t)=A_1e^{-\lambda_1^2\mathrm{Fo}} \cos\left(\frac{\lambda_1x}{b}\right)$$(25)

The eigenvalue and amplitude satisfy

𝜆1tan⁡𝜆1=Bi𝑏𝐴1=4sin⁡𝜆12𝜆1+sin⁡(2𝜆1)$$\lambda_1\tan\lambda_1=\mathrm{Bi}_b \qquad A_1=\frac{4\sin\lambda_1}{2\lambda_1+\sin(2\lambda_1)}$$

Solving with Bi𝑏 =3.28$\mathrm{Bi}_b=3.28$ gives 𝜆1 ≃1.21582$\lambda_1\simeq1.21582$ and 𝐴1 ≃1.21638$A_1\simeq1.21638$. At the physical centre 𝑥/𝑏 =0.50$x/b=0.50$, rearrange Eq. 25:

Fo=ln⁡[𝐴1cos⁡(𝜆12)Θ]𝜆21=ln⁡[1.21638cos⁡(1.215822)0.600](1.21582)2≃0.3445$$\mathrm{Fo}=\frac{\ln\left[\frac{A_1\cos\left(\frac{\lambda_1}{2}\right)}{\Theta}\right]}{\lambda_1^2} =\frac{\ln\left[\frac{1.21638\cos\left(\frac{1.21582}{2}\right)}{0.600}\right]}{(1.21582)^2} \simeq0.3445$$(26)

The dimensional time follows from Fo =𝛼𝑤𝑡/𝑏2$\mathrm{Fo}=\alpha_wt/b^2$:

𝑡=Fo𝑏2𝛼𝑤≃0.3445(0.50)25.29938×10−7≃1.625×105 s≃45.15 h$$t=\frac{\mathrm{Fo}b^2}{\alpha_w} \simeq\frac{0.3445(0.50)^2}{5.29938\times10^{-7}} \simeq1.625\times10^5\ \mathrm s \simeq45.15\ \mathrm h$$

The calculated Fourier number exceeds 0.2, supporting the first-term approximation. The insulated face is the centre of the reflected slab, but it is not the physical centre of the original brick. At 𝑥 =0$x=0$ the cosine is one, giving about 62.65 h$62.65\ \mathrm h$ to reach 600 K$600\ \mathrm K$. These times describe different positions. If the estimated ℎ =6.95 W m−2 K−1$h=6.95\ \mathrm{W\,m^{-2}\,K^{-1}}$ replaces the stipulated value, the physical-centre result is about 46.51 h$46.51\ \mathrm h$. The diffusion scale 𝑏2/𝛼𝑤 ≃131 h$b^2/\alpha_w\simeq131\ \mathrm h$ is consistent with these heating times.

Freezing ℎ$h$ is approximate because both the driving temperature difference and fluid properties change during heating. Radiation can matter much more. For example, with emissivity 0.8, a 600 K$600\ \mathrm K$ surface facing 1200 K$1200\ \mathrm K$ surroundings has an effective radiation coefficient about 147 W m−2 K−1$147\ \mathrm{W\,m^{-2}\,K^{-1}}$. The convection-only model would therefore overestimate heating time in a furnace with those radiative conditions.

Answer

Using the supplied constant coefficient, the physical centre of the wall reaches 600 K$600\ \mathrm K$ after approximately 45.2 h$45.2\ \mathrm h$. The separate natural-convection estimate gives ℎ ≃6.95 W m−2 K−1$h\simeq6.95\ \mathrm{W\,m^{-2}\,K^{-1}}$, supporting the order of magnitude of the stated 7.38 W m−2 K−1$7.38\ \mathrm{W\,m^{-2}\,K^{-1}}$.

Natural convection

Surface temperature of a transformer tank

Determine the surface temperature of a cylindrical tank (0.75 m Diameter, 1.2m high), which contains a transformer immersed in an oil bath, producing a uniform surface temperature condition. All heat loss from the surface may be assumed to be due to natural convection to surrounding air at 295 K. The heat dissipation rate from the transformer is constant at 1.5 kW.

(Welty et al. 2001)

Solution

The tank side and top have different natural-convection characteristic lengths. The base is taken to rest on the floor, as in the lecturer’s calculation.

Figure 3: Cylindrical transformer tank with diameter, height, ambient temperature and separate characteristic lengths for its side and top.

Data

Quantity	Value
Tank diameter	𝐷 =0.75 m$D=0.75\ \mathrm m$
Tank height	𝐻 =1.2 m$H=1.2\ \mathrm m$
Heat dissipation	˙𝑄 =1500 W$\dot Q=1500\ \mathrm W$
Ambient air temperature	𝑇∞ =295 K$T_\infty=295\ \mathrm K$
Air conductivity at 347.5 K	𝑘 =0.0294 W m−1 K−1$k=0.0294\ \mathrm{W\,m^{-1}\,K^{-1}}$
Air kinematic viscosity at 347.5 K	𝜈 =2.05 ×10−5 m2 s−1$\nu=2.05\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Air Prandtl number at 347.5 K	Pr =0.718$\mathrm{Pr}=0.718$
Air conductivity at 340 K	𝑘 =0.0289 W m−1 K−1$k=0.0289\ \mathrm{W\,m^{-1}\,K^{-1}}$
Air kinematic viscosity at 340 K	𝜈 =1.98 ×10−5 m2 s−1$\nu=1.98\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Air Prandtl number at 340 K	Pr =0.720$\mathrm{Pr}=0.720$
Ideal-gas expansion coefficient	𝛽 =1/𝑇𝑓$\beta=1/T_f$

The oil bath gives an approximately uniform surface temperature. Use the lecturer’s air properties at the two trial film temperatures.

Assumptions

• Steady state.
• The floor-supported base is adiabatic.
• The side behaves as a vertical plate.
• Fluid properties are constant at each trial film temperature.

Calculations

The side area and vertical characteristic length are

𝐴side=𝜋𝐷𝐻=𝜋(0.75)(1.2)=2.82743 m2$$A_{\mathrm{side}}=\pi DH=\pi(0.75)(1.2)=2.82743\ \mathrm{m^2}$$

The side characteristic length is

𝐿side=𝐻=1.2 m$$L_{\mathrm{side}}=H=1.2\ \mathrm m$$

For the upward-facing top, use area divided by perimeter:

𝐴top=𝜋𝐷24=𝜋(0.75)24=0.441786 m2$$A_{\mathrm{top}}=\frac{\pi D^2}{4} =\frac{\pi(0.75)^2}{4}=0.441786\ \mathrm{m^2}$$

The top characteristic length is

𝐿top=𝐴top𝜋𝐷=𝐷4=0.1875 m$$L_{\mathrm{top}}=\frac{A_{\mathrm{top}}}{\pi D} =\frac D4=0.1875\ \mathrm m$$

The two surface heat rates add. Define their combined conductance 𝐺$G$:

𝐺=¯¯¯¯ℎside𝐴side+¯¯¯¯ℎtop𝐴top$$G=\bar h_{\mathrm{side}}A_{\mathrm{side}} +\bar h_{\mathrm{top}}A_{\mathrm{top}}$$

˙𝑄=𝐺(𝑇𝑠−𝑇∞)$$\dot Q=G(T_s-T_\infty)$$(27)

The coefficients depend on the unknown surface temperature through Rayleigh number and film properties. Guess 𝑇𝑠$T_s$, calculate the heat loss and adjust the guess until the heat loss equals 1500 W$1500\ \mathrm W$.

For the first trial, let 𝑇𝑠 =400 K$T_s=400\ \mathrm K$. Then Δ𝑇 =105 K$\Delta T=105\ \mathrm K$ and

𝑇𝑓=400+2952=347.5 K.$$T_f=\frac{400+295}{2}=347.5\ \mathrm K.$$

Using 𝐻$H$ for the side gives

Ra𝐻=𝑔𝛽Δ𝑇𝐻3𝜈2Pr=9.81(1347.5)(105)(1.2)3(2.05×10−5)2(0.718)=8.751×109$$\mathrm{Ra}_H=\frac{g\beta\Delta T H^3}{\nu^2}\mathrm{Pr} =\frac{9.81\left(\frac{1}{347.5}\right)(105)(1.2)^3}{(2.05\times10^{-5})^2}(0.718) =8.751\times10^9$$

Since both surfaces see the same temperature difference and fluid, the top Rayleigh number follows by scaling length cubed:

Ra𝐷/4=Ra𝐻(𝐷4𝐻)3=(8.751×109)(0.18751.2)3=3.338×107$$\mathrm{Ra}_{D/4}=\mathrm{Ra}_H\left(\frac{D}{4H}\right)^3 =(8.751\times10^9)\left(\frac{0.1875}{1.2}\right)^3 =3.338\times10^7$$

For the side, use the turbulent vertical-plate correlation ¯¯¯¯¯¯¯¯Nu𝐻 =0.10Ra1/3𝐻$\overline{\mathrm{Nu}}_H=0.10\mathrm{Ra}_H^{1/3}$, applicable here above Ra𝐻 ≃109$\mathrm{Ra}_H\simeq10^9$. For the hot upward-facing top, use ¯¯¯¯¯¯¯¯Nu𝐷/4 =0.15Ra1/3𝐷/4$\overline{\mathrm{Nu}}_{D/4}=0.15\mathrm{Ra}_{D/4}^{1/3}$ in the range 107 ≲Ra𝐷/4 ≲1011$10^7\lesssim\mathrm{Ra}_{D/4}\lesssim10^{11}$.

ℎside=𝑘𝐻(0.10Ra1/3𝐻)=0.02941.2[0.10(8.751×109)1/3]=5.0488 W m−2 K−1ℎtop=4𝑘𝐷(0.15Ra1/3𝐷/4)=0.02940.1875[0.15(3.338×107)1/3]=7.5732 W m−2 K−1$$\begin{aligned} h_{\mathrm{side}}&=\frac{k}{H}(0.10\mathrm{Ra}_H^{1/3})\\ &=\frac{0.0294}{1.2}[0.10(8.751\times10^9)^{1/3}] =5.0488\ \mathrm{W\,m^{-2}\,K^{-1}}\\ h_{\mathrm{top}}&=\frac{4k}{D}(0.15\mathrm{Ra}_{D/4}^{1/3})\\ &=\frac{0.0294}{0.1875}[0.15(3.338\times10^7)^{1/3}] =7.5732\ \mathrm{W\,m^{-2}\,K^{-1}} \end{aligned}$$(28)

Substitute the coefficients and areas into Eq. 27:

˙𝑄400=[5.0488(2.82743)+7.5732(0.441786)](400−295)=1850.2 W$$\dot Q_{400}=[5.0488(2.82743)+7.5732(0.441786)](400-295) =1850.2\ \mathrm W$$

The first guess loses more than 1500 W$1500\ \mathrm W$, so reduce the surface temperature. At 𝑇𝑠 =385 K$T_s=385\ \mathrm K$, the film is at 340 K$340\ \mathrm K$ and Δ𝑇 =90 K$\Delta T=90\ \mathrm K$. Repeating the same relations gives

Ra𝐻=9.81(1340)(90)(1.2)3(1.98×10−5)2(0.720)=8.241×109Ra𝐷/4=(8.241×109)(0.18751.2)3=3.144×107ℎside=0.02891.2[0.10(8.241×109)1/3]=4.8646 W m−2 K−1ℎtop=0.02890.1875[0.15(3.144×107)1/3]=7.2968 W m−2 K−1˙𝑄385=[4.8646(2.82743)+7.2968(0.441786)](90)=1528.0 W$$\begin{aligned} \mathrm{Ra}_H&=\frac{9.81\left(\frac{1}{340}\right)(90)(1.2)^3}{(1.98\times10^{-5})^2}(0.720) =8.241\times10^9\\ \mathrm{Ra}_{D/4}&=(8.241\times10^9)\left(\frac{0.1875}{1.2}\right)^3 =3.144\times10^7\\ h_{\mathrm{side}}&=\frac{0.0289}{1.2}[0.10(8.241\times10^9)^{1/3}] =4.8646\ \mathrm{W\,m^{-2}\,K^{-1}}\\ h_{\mathrm{top}}&=\frac{0.0289}{0.1875}[0.15(3.144\times10^7)^{1/3}] =7.2968\ \mathrm{W\,m^{-2}\,K^{-1}}\\ \dot Q_{385}&=[4.8646(2.82743)+7.2968(0.441786)](90) =1528.0\ \mathrm W \end{aligned}$$(29)

For the small final correction, hold the 340 K$340\ \mathrm K$ property set fixed. Both correlations have exponent 1/3$1/3$, so ℎ ∝Δ𝑇1/3$h\propto\Delta T^{1/3}$ and the heat balance gives ˙𝑄 ∝Δ𝑇4/3$\dot Q\propto\Delta T^{4/3}$. Therefore,

Δ𝑇=90(15001528.0)3/4=88.760 K$$\Delta T=90\left(\frac{1500}{1528.0}\right)^{3/4} =88.760\ \mathrm K$$(30)

The surface temperature is

𝑇𝑠=295+88.760=383.760 K.$$T_s=295+88.760=383.760\ \mathrm K.$$

Its film temperature is 339.38 K$339.38\ \mathrm K$, close to the property-table value used for the final correction. Further precision would require updating those properties. The side supplies about 81 percent of the heat loss.

Check the vertical-cylinder approximation using Eq. 20. At the corrected temperature, Gr𝐻 ≃1.13 ×1010$\mathrm{Gr}_H\simeq1.13\times10^{10}$, so the required diameter is

𝐷min⁡=35𝐻Gr1/4𝐻≃35(1.2)(1.13×1010)1/4≃0.129 m$$D_{\min}=\frac{35H}{\mathrm{Gr}_H^{1/4}} \simeq\frac{35(1.2)}{(1.13\times10^{10})^{1/4}} \simeq0.129\ \mathrm m$$

The actual diameter 0.75 m$0.75\ \mathrm m$ exceeds this value. Radiation, excluded by the question, could be significant in an actual tank. For 𝜖 =0.8$\epsilon=0.8$ and surroundings at 295 K$295\ \mathrm K$, the radiation coefficient near the calculated temperature is about 7.2 W m−2 K−1$7.2\ \mathrm{W\,m^{-2}\,K^{-1}}$, comparable to convection. Including radiation would reduce the surface temperature. A raised tank would also lose heat from its downward-facing base, requiring that surface’s own natural-convection correlation.

Answer

The convection-only estimate is 𝑇𝑠 ≃384 K$T_s\simeq384\ \mathrm K$, or 111 °C$111\ \degree\mathrm{C}$, for an adiabatic floor-supported base.

Natural convection

Cooling a horizontal metal cylinder in still air

A long horizontal metal cylinder of diameter 0.180 m$0.180\ \mathrm m$ starts uniformly at 180 °C$180\ \degree\mathrm{C}$ in still air at 30 °C$30\ \degree\mathrm{C}$. Use solid properties 𝑘𝑤 =45 W m−1 K−1$k_w=45\ \mathrm{W\,m^{-1}\,K^{-1}}$, 𝜌 =7800 kg m−3$\rho=7800\ \mathrm{kg\,m^{-3}}$ and 𝐶𝑝 =500 J kg−1 K−1$C_p=500\ \mathrm{J\,kg^{-1}\,K^{-1}}$. At the initial film temperature, use air properties 𝑘 =0.032 W m−1 K−1$k=0.032\ \mathrm{W\,m^{-1}\,K^{-1}}$, 𝜈 =2.40 ×10−5 m2 s−1$\nu=2.40\times10^{-5}\ \mathrm{m^2\,s^{-1}}$ and Pr =0.70$\mathrm{Pr}=0.70$.

Estimate the natural-convection coefficient and the centre temperature after 20 minutes. Neglect end effects and radiation, and hold the estimated coefficient and properties constant over the cooling interval. If a fan instead produces a coefficient of 30 W m−2 K−1$30\ \mathrm{W\,m^{-2}\,K^{-1}}$, can the same transient model still be used?

Solution

Figure 4: The horizontal cylinder loses heat from its curved surface. In the still-air case there is no imposed air speed; the fan case supplies an effective coefficient. End losses are neglected.

Data

Quantity	Value
Cylinder diameter	𝐷 =0.180 m$D=0.180\ \mathrm m$
Initial temperature	𝑇𝑖 =180 °C$T_i=180\ \degree\mathrm C$
Air temperature	𝑇∞ =30 °C$T_\infty=30\ \degree\mathrm C$
Cooling time	𝑡 =1200 s$t=1200\ \mathrm s$
Solid conductivity	𝑘𝑤 =45 W m−1 K−1$k_w=45\ \mathrm{W\,m^{-1}\,K^{-1}}$
Solid density	𝜌𝑤 =7800 kg m−3$\rho_w=7800\ \mathrm{kg\,m^{-3}}$
Solid specific heat capacity	𝐶𝑝,𝑤 =500 J kg−1 K−1$C_{p,w}=500\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Air conductivity	𝑘 =0.032 W m−1 K−1$k=0.032\ \mathrm{W\,m^{-1}\,K^{-1}}$
Air kinematic viscosity	𝜈 =2.40 ×10−5 m2 s−1$\nu=2.40\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Air Prandtl number	Pr =0.70$\mathrm{Pr}=0.70$
Coefficient with the fan	ℎ =30 W m−2 K−1$h=30\ \mathrm{W\,m^{-2}\,K^{-1}}$

Assumptions

• Uniform initial solid temperature.
• Constant properties and estimated heat transfer coefficient.
• End effects and radiation are neglected.
• Ideal-gas air for the expansion coefficient.

Calculations

Still air permits buoyancy-driven motion, so use natural convection around a horizontal cylinder. The initial film temperature is

𝑇𝑓=180+302+273.15=378.15 K.$$T_f=\frac{180+30}{2}+273.15=378.15\,\mathrm{K}.$$

For ideal-gas air, 𝛽 =1/𝑇𝑓 =0.002644$\beta=1/T_f=0.002644$ K⁻¹.

The convection correlation uses the cylinder diameter:

Ra𝐷=𝑔𝛽(𝑇𝑖−𝑇∞)𝐷3𝜈2Pr=9.81(1378.15)(180−30)(0.180)3(2.40×10−5)2(0.70)=2.758×107$$\mathrm{Ra}_D=\frac{g\beta(T_i-T_\infty)D^3}{\nu^2}\mathrm{Pr} =\frac{9.81\left(\frac{1}{378.15}\right)(180-30)(0.180)^3}{(2.40\times10^{-5})^2}(0.70) =2.758\times10^7$$

This is within the range of Eq. 21. Substitution gives

¯¯¯¯¯¯¯¯Nu𝐷=[0.60+0.387(2.758×107)1/6[1+(0.5590.70)9/16]8/27]2=38.17$$\overline{\mathrm{Nu}}_D= \left[0.60+\frac{0.387(2.758\times10^7)^{1/6}}{\left[1+\left(\frac{0.559}{0.70}\right)^{9/16}\right]^{8/27}}\right]^2=38.17$$

The corresponding coefficient is

¯¯¯¯ℎ=𝑘¯¯¯¯¯¯¯¯Nu𝐷𝐷=38.17(0.032)0.180=6.786 W m−2 K−1$$\bar h=\frac{k\overline{\mathrm{Nu}}_D}{D}=\frac{38.17(0.032)}{0.180} =6.786\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For the lumped-model check, use exposed volume divided by area. Ignoring ends,

𝑉𝐴𝑠=𝐷4=0.0450 m,$$\frac{V}{A_s}=\frac{D}{4}=0.0450\ \mathrm m,$$

so

Bi𝑐=¯¯¯¯ℎ(𝑉𝐴𝑠)𝑘𝑤=6.786(0.0450)45=0.00679<0.1$$\mathrm{Bi}_c=\frac{\bar h\left(\frac{V}{A_s}\right)}{k_w}=\frac{6.786(0.0450)}{45}=0.00679<0.1$$

The centre is approximately at the uniform body temperature. The lumped energy balance is

𝜌𝑤𝐶𝑝,𝑤𝑉𝑑𝑇𝑑𝑡=−¯¯¯¯ℎ𝐴𝑠(𝑇−𝑇∞)$$\rho_w C_{p,w}V\frac{dT}{dt}=-\bar h A_s(T-T_\infty)$$

With constant ¯¯¯¯ℎ$\bar h$, integration gives

𝑇−𝑇∞=(𝑇𝑖−𝑇∞)𝑒−𝑡𝜏,$$T-T_\infty=(T_i-T_\infty)e^{-\frac{t}{\tau}},$$

where the time constant is

𝜏=𝜌𝑤𝐶𝑝,𝑤(𝑉𝐴𝑠)¯¯¯¯ℎ=7800(500)(0.0450)6.786=25862 s$$\tau=\frac{\rho_w C_{p,w}\left(\frac{V}{A_s}\right)}{\bar h} =\frac{7800(500)(0.0450)}{6.786}=25862\ \mathrm s$$

At 𝑡 =1200 s$t=1200\ \mathrm s$, the temperature is

𝑇(1200)=30+(180−30)𝑒−120025862=173.2 °C$$T(1200)=30+(180-30)e^{-\frac{1200}{25862}} =173.2\ \degree\mathrm{C}$$

Diameter in Rayleigh number and 𝐷/4$D/4$ in the lumped Biot number answer different questions. Rayleigh number describes buoyant flow around the body; the Biot number compares internal and external thermal resistances.

With the fan, recalculate

Bi𝑐=30(0.045)45=0.0300.$$\mathrm{Bi}_c=\frac{30(0.045)}{45}=0.0300.$$

Lumped capacitance remains valid, and the same exponential gives 152.2 °C$152.2\ \degree\mathrm{C}$ after 20 minutes. If the forced coefficient had not been supplied, it would require a suitable cross-flow correlation and a new Biot check. One cannot assume that every increase in ℎ$h$ preserves lumped behaviour.

Natural-convection ℎ$h$ changes as the cylinder cools. The small 6.8 K temperature change supports this initial-coefficient estimate over 20 minutes. For a larger change, update properties and ℎ(𝑇)$h(T)$ in the transient balance. Radiation is excluded by the problem; this is a convection-only estimate.

Answer

The initial natural-convection coefficient is ¯¯¯¯ℎ =6.79 W m−2 K−1$\bar h=6.79\ \mathrm{W\,m^{-2}\,K^{-1}}$, giving a centre temperature of approximately 173.2 °C$173.2\ \degree\mathrm{C}$ after 20 minutes. With the supplied fan coefficient, lumped capacitance remains valid and gives 152.2 °C$152.2\ \degree\mathrm{C}$.

Fundamentals of mass diffusion

Water evaporation

Water in the bottom of a narrow metal tube is held at a constant temperature of 293 K. The total pressure of air (assumed dry) is 1.01325 ×105 Pa$1.01325\times10^5\ \mathrm{Pa}$ and the temperature is 293 K. Water evaporates and diffuses through the air in the tube, and the diffusion path 𝑧2 −𝑧1$z_2-z_1$ is 0.1524 m. Calculate the rate of evaporation at steady state in kgmol/(s ⋅m2)$\mathrm{kgmol/(s\cdot m^2)}$. The diffusivity of water vapour at 293 K and 1 atm pressure is 0.250 ×10−4 m2/s$0.250\times10^{-4}\ \mathrm{m^2/s}$. Assume that the system is isothermal. The vapour pressure of water at 1 atm and 293 K is 17.535 mmHg.

Additional note: Moving boundaries or pseudo-steady state diffusion. How long would it take for the water level to decrease by 0.1 m?

Solution

Figure 3: Water vapour diffuses through the gas above the liquid to the tube mouth. As the liquid surface falls, the gas diffusion distance increases.

The evaporation path is the gas space between the water surface and the tube mouth, as shown in Figure 2. Water vapour moves upwards. Air has zero net flux across the liquid interface.

Data

Quantity	Value
Initial gas diffusion length	𝐿0 =0.1524 m$L_0=0.1524\ \mathrm m$
Temperature	𝑇 =293 K$T=293\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =2.50 ×10−5 m2 s−1$D_{AB}=2.50\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface vapour pressure	𝑝𝐴,𝑠 =17.535 mmHg$p_{A,s}=17.535\ \mathrm{mmHg}$
Liquid density for the moving-boundary calculation	𝜌𝐿 =1000 kg m−3$\rho_L=1000\ \mathrm{kg\,m^{-3}}$
Water molar mass	𝑀𝐴 =18 kg kmol−1$M_A=18\ \mathrm{kg\,kmol^{-1}}$
Final gas diffusion length	𝐿𝑓 =0.1524 +0.1000 =0.2524 m$L_f=0.1524+0.1000=0.2524\ \mathrm m$

The source unit kgmol is equivalent to kmol. Liquid density and molar mass are supplementary inputs needed to convert liquid-depth loss to an amount evaporated.

Assumptions

• Unidirectional diffusion through the gas space.
• Steady gas profile for the initial flux calculation.
• Pseudo-steady gas profile as the liquid surface recedes.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• Vapour-liquid equilibrium at the water surface.
• Zero net air flux through the liquid surface.
• Negligible natural convection.

Calculations

At the liquid surface, equilibrium fixes the vapour partial pressure. At the mouth, the dry air gives 𝑝𝐴2 =0$p_{A2}=0$. Converting the surface vapour pressure to Pa gives

𝑝𝐴1=17.535760(101325)=2337.8 Pa$$p_{A1}=\frac{17.535}{760}(101325)=2337.8\ \mathrm{Pa}$$

The surface mole fraction and total gas concentration are

𝑦𝐴1=𝑝𝐴1𝑃=2337.8101325=0.0230724$$y_{A1}=\frac{p_{A1}}P=\frac{2337.8}{101325}=0.0230724$$𝑐=𝑃𝑅𝑢𝑇=1013258314(293)=0.0415948 kmol m−3$$c=\frac{P}{R_uT}=\frac{101325}{8314(293)}=0.0415948\ \mathrm{kmol\,m^{-3}}$$

Because air has zero net flux, the appropriate governing relation is the stagnant-carrier result, Eq. 9,

𝑁𝐴=𝑐𝐷𝐴𝐵𝐿0ln⁡1−𝑦𝐴21−𝑦𝐴1$$N_A=\frac{cD_{AB}}{L_0}\ln\frac{1-y_{A2}}{1-y_{A1}}$$

Substituting the boundary fractions and the initial gas length gives

𝑁𝐴=0.0415948(2.50×10−5)0.1524ln⁡11−0.0230724=1.593×10−7 kmol m−2 s−1$$N_A=\frac{0.0415948(2.50\times10^{-5})}{0.1524} \ln\frac{1}{1-0.0230724} =1.593\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}}$$(12)

Multiplying by molar mass converts the molar flux to mass flux,

𝑛𝐴=𝑀𝐴𝑁𝐴=18(1.593×10−7)=2.867×10−6 kg m−2 s−1$$n_A=M_AN_A=18(1.593\times10^{-7})=2.867\times10^{-6}\ \mathrm{kg\,m^{-2}\,s^{-1}}$$

The initial rate is about 0.248 kg m−2$0.248\ \mathrm{kg\,m^{-2}}$ per day. The stagnant-carrier flux is only 1.17 percent above the equimolar estimate because the water vapour is dilute.

For the falling liquid level, the gas length increases and the evaporation flux decreases. Define the constant logarithmic driving force,

Λ=ln⁡1−𝑦𝐴21−𝑦𝐴1=ln⁡11−0.0230724=0.0233427$$\Lambda=\ln\frac{1-y_{A2}}{1-y_{A1}}=\ln\frac{1}{1-0.0230724}=0.0233427$$

The liquid molar density is

𝑐𝐿=𝜌𝐿𝑀𝐴=100018=55.5556 kmol m−3$$c_L=\frac{\rho_L}{M_A}=\frac{1000}{18}=55.5556\ \mathrm{kmol\,m^{-3}}$$

A decrease in liquid depth 𝑑𝐿$dL$ removes 𝑐𝐿𝑑𝐿$c_LdL$ kmol per unit tube area. Equating that loss to the instantaneous evaporation flux gives

𝑐𝐿𝑑𝐿𝑑𝑡=𝑐𝐷𝐴𝐵Λ𝐿$$c_L\frac{dL}{dt}=\frac{cD_{AB}\Lambda}{L}$$

Separating variables and integrating between the initial and final levels gives the moving-boundary relation, Eq. 11,

𝑡=𝑐𝐿(𝐿2𝑓−𝐿20)2𝑐𝐷𝐴𝐵Λ$$t=\frac{c_L(L_f^2-L_0^2)}{2cD_{AB}\Lambda}$$

Substitute the consistent kmol-based concentrations,

𝑡=55.5556(0.25242−0.15242)2(0.0415948)(2.50×10−5)(0.0233427)=4.632×107 s=536 days$$t=\frac{55.5556(0.2524^2-0.1524^2)} {2(0.0415948)(2.50\times10^{-5})(0.0233427)} =4.632\times10^7\ \mathrm s=536\ \mathrm{days}$$(13)

As a check, removing 0.1 m$0.1\ \mathrm m$ of water requires 5.556 kmol m−2$5.556\ \mathrm{kmol\,m^{-2}}$. Even if the initial flux stayed constant, evaporation would take

5.5561.593×10−7=3.49×107 s.$$\frac{5.556}{1.593\times10^{-7}}=3.49\times10^7\ \mathrm s.$$

The falling level lengthens the diffusion path, so the calculated time must be greater than that lower bound.

The gas diffusion time is 𝐿20/𝐷𝐴𝐵 ≃929 s$L_0^2/D_{AB}\simeq929\ \mathrm s$, much shorter than the liquid retreat time. This supports the pseudo-steady approximation. Maintaining the stipulated isothermal, convection-free conditions for 536 days would be difficult in an experiment.

Answer

The initial evaporation flux is 𝑁𝐴 =1.59 ×10−7 kgmol m−2 s−1$N_A=1.59\times10^{-7}\ \mathrm{kgmol\,m^{-2}\,s^{-1}}$, directed towards the tube mouth. The corresponding mass flux is 2.87 ×10−6 kg m−2 s−1$2.87\times10^{-6}\ \mathrm{kg\,m^{-2}\,s^{-1}}$.

Under the pseudo-steady moving-boundary model, the water level takes 4.63 ×107 s$4.63\times10^7\ \mathrm s$, or about 536 days, to fall by 0.1 m$0.1\ \mathrm m$.

Fundamentals of mass diffusion

Butanol in water

Calculate the rate of diffusion of butanol at 20 °C$20\ \degree\mathrm{C}$ under unidirectional steady state conditions through a 0.1 cm thick film of water when the concentrations of butanol at the opposite sides of the film are, 10% and 4% butanol by weight, respectively.

𝐷𝐴𝐵$D_{AB}$ (butanol in water solution)=5.9 ×10−6 cm2/s$5.9\times10^{-6}\ \mathrm{cm^2/s}$. The densities of 10% and 4% butanol solutions at 20 °C$20\ \degree\mathrm{C}$ may be taken as 0.971 and 0.992 g/cc, respectively. Molecular weight of Butanol is 74, and that of water is 18.

Solution

Data

Quantity	Value
Film thickness	𝐿 =0.1 cm$L=0.1\ \mathrm{cm}$
Diffusivity	𝐷𝐴𝐵 =5.9 ×10−6 cm2 s−1$D_{AB}=5.9\times10^{-6}\ \mathrm{cm^2\,s^{-1}}$
Butanol mass fraction at side 1	𝑤𝐴1 =0.10$w_{A1}=0.10$
Density at side 1	𝜌1 =0.971 g cm−3$\rho_1=0.971\ \mathrm{g\,cm^{-3}}$
Butanol mass fraction at side 2	𝑤𝐴2 =0.04$w_{A2}=0.04$
Density at side 2	𝜌2 =0.992 g cm−3$\rho_2=0.992\ \mathrm{g\,cm^{-3}}$
Butanol molar mass	𝑀𝐴 =74 g mol−1$M_A=74\ \mathrm{g\,mol^{-1}}$
Water molar mass	𝑀𝐵 =18 g mol−1$M_B=18\ \mathrm{g\,mol^{-1}}$

Assumptions

• Steady state.
• Unidirectional diffusion across the film.
• Isothermal conditions at 20 °C$20\ \degree\mathrm{C}$.
• No chemical reaction.
• Zero net water flux.
• Constant diffusivity.
• Total molar concentration approximated by its arithmetic mean across the film.

Calculations

Let A be butanol and B be water, with positive flux from the 10 percent side to the 4 percent side. The stagnant-water model uses mole fractions, so first convert the supplied mass fractions. On a 100 g$100\ \mathrm g$ basis, side 1 contains 10/74$10/74$ mol butanol and 90/18$90/18$ mol water. Hence

𝑥𝐴1=10741074+9018=0.0263158$$x_{A1}=\frac{\frac{10}{74}}{\frac{10}{74}+\frac{90}{18}}=0.0263158$$

For side 2, the same conversion gives

𝑥𝐴2=474474+9618=0.0100334$$x_{A2}=\frac{\frac{4}{74}}{\frac{4}{74}+\frac{96}{18}}=0.0100334$$

The total molar concentration equals the total moles divided by the solution volume. Since a 100 g$100\ \mathrm g$ sample occupies 100/𝜌𝑖 cm3$100/\rho_i\ \mathrm{cm^3}$,

𝑐1=1074+90181000.971=0.0498622 mol cm−3$$c_1=\frac{\frac{10}{74}+\frac{90}{18}}{\frac{100}{0.971}}=0.0498622\ \mathrm{mol\,cm^{-3}}$$𝑐2=474+96181000.992=0.0534429 mol cm−3$$c_2=\frac{\frac{4}{74}+\frac{96}{18}}{\frac{100}{0.992}}=0.0534429\ \mathrm{mol\,cm^{-3}}$$

Their arithmetic mean is

¯¯¯¯𝑐=𝑐1+𝑐22=0.0498622+0.05344292=0.0516525 mol cm−3$$\overline c=\frac{c_1+c_2}{2}=\frac{0.0498622+0.0534429}{2}=0.0516525\ \mathrm{mol\,cm^{-3}}$$(14)

The total molar concentration changes by about 7 percent across the film. Treating it as uniform is an approximation for this liquid mixture. Constant temperature and pressure alone do not make a liquid’s molar concentration constant.

With zero net water flux, the governing total-flux relation is

𝑁𝐴=−¯¯¯¯𝑐𝐷𝐴𝐵1−𝑥𝐴𝑑𝑥𝐴𝑑𝑧$$N_A=-\frac{\overline cD_{AB}}{1-x_A}\frac{dx_A}{dz}$$

Integrating over the film gives the liquid version of Eq. 9,

𝑁𝐴=¯¯¯¯𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑥𝐴21−𝑥𝐴1$$N_A=\frac{\overline cD_{AB}}{L}\ln\frac{1-x_{A2}}{1-x_{A1}}$$

Substituting in centimetre and mol units gives

𝑁𝐴=0.0516525(5.9×10−6)0.1ln⁡1−0.01003341−0.0263158=5.054×10−8 mol cm−2 s−1=5.054×10−7 kmol m−2 s−1$$\begin{aligned} N_A&=\frac{0.0516525(5.9\times10^{-6})}{0.1} \ln\frac{1-0.0100334}{1-0.0263158}\\ &=5.054\times10^{-8}\ \mathrm{mol\,cm^{-2}\,s^{-1}}\\ &=5.054\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}} \end{aligned}$$(15)

The mass flux follows by multiplying by the molar mass of butanol,

𝑛𝐴=𝑀𝐴𝑁𝐴=74(5.054×10−8)=3.740×10−6 g cm−2 s−1$$n_A=M_AN_A=74(5.054\times10^{-8})=3.740\times10^{-6}\ \mathrm{g\,cm^{-2}\,s^{-1}}$$

As a check, neglecting the stagnant-water correction gives

𝑁𝐴,eq=¯¯¯¯𝑐𝐷𝐴𝐵(𝑥𝐴1−𝑥𝐴2)𝐿=4.962×10−8 mol cm−2 s−1.$$N_{A,\mathrm{eq}}=\frac{\overline cD_{AB}(x_{A1}-x_{A2})}{L}=4.962\times10^{-8}\ \mathrm{mol\,cm^{-2}\,s^{-1}}.$$

The full result is 1.85 percent larger, consistent with dilute butanol.

The flux is about 3.23 kg m−2$3.23\ \mathrm{kg\,m^{-2}}$ per day. Comparing it with the water-evaporation example requires the concentration difference and film thickness as well as the diffusivity.

Answer

Butanol diffuses from the 10 percent side to the 4 percent side at 𝑁𝐴 =5.05 ×10−7 kmol m−2 s−1$N_A=5.05\times10^{-7}\ \mathrm{kmol\,m^{-2}\,s^{-1}}$, equivalent to a mass flux of 3.74 ×10−5 kg m−2 s−1$3.74\times10^{-5}\ \mathrm{kg\,m^{-2}\,s^{-1}}$.

Fundamentals of mass diffusion

Compare equimolar diffusion and a stagnant carrier

A binary ideal gas is at 300 K and 100 kPa. Across a 5.0 mm plane film, the mole fraction of A falls from 0.30 to zero. Use 𝐷𝐴𝐵 =2.0 ×10−5$D_{AB}=2.0\times10^{-5}$ m² s⁻¹ and 𝑅 =8.314$R=8.314$ J mol⁻¹ K⁻¹. Compare the steady molar flux of A for equimolar counter-diffusion and for zero net flux of B. Take positive flux from the high-A face to the low-A face. Explain the difference and determine the flux ratio if the high-face mole fraction is reduced to 0.010, with all other data unchanged.

Solution

Data

Quantity	Value
Temperature	𝑇 =300 K$T=300\ \mathrm K$
Pressure	𝑃 =100000 Pa$P=100000\ \mathrm{Pa}$
Film thickness	𝐿 =0.0050 m$L=0.0050\ \mathrm m$
Diffusivity	𝐷𝐴𝐵 =2.0 ×10−5 m2 s−1$D_{AB}=2.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Gas constant	𝑅 =8.314 J mol−1 K−1$R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$
Side 1 mole fraction in the first comparison	𝑦𝐴1 =0.30$y_{A1}=0.30$
Side 2 mole fraction	𝑦𝐴2 =0$y_{A2}=0$
Side 1 mole fraction in the second comparison	𝑦𝐴1 =0.010$y_{A1}=0.010$

Assumptions

• Steady state.
• One-dimensional diffusion across a plane film.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• No reaction in the film.

Calculations

Positive flux points from the high-A face towards the low-A face. The total concentration is

𝑐=𝑃𝑅𝑇=1000008.314(300)=40.09 mol m−3$$c=\frac{P}{RT}=\frac{100000}{8.314(300)}=40.09\ \mathrm{mol\,m^{-3}}$$

For equimolar counter-diffusion, 𝑁𝐴 +𝑁𝐵 =0$N_A+N_B=0$. As in Eq. 8, the bulk contribution vanishes and the governing relation is

𝑁𝐴=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧.$$N_A=-cD_{AB}\frac{dy_A}{dz}.$$

Integrating between the film faces gives

𝑁𝐴,eq=𝑐𝐷𝐴𝐵𝐿(𝑦𝐴1−𝑦𝐴2)=40.09(2.0×10−5)0.0050(0.30−0)=0.04811 mol m−2 s−1$$\begin{aligned} N_{A,\mathrm{eq}}&=\frac{cD_{AB}}L(y_{A1}-y_{A2})\\ &=\frac{40.09(2.0\times10^{-5})}{0.0050}(0.30-0)\\ &=0.04811\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

For zero net flux of B, the total flux satisfies

𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑧.$$N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dz}.$$

Integrating this relation gives

𝑁𝐴,stag=𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑦𝐴21−𝑦𝐴1=40.09(2.0×10−5)0.0050ln⁡10.70=0.05720 mol m−2 s−1$$\begin{aligned} N_{A,\mathrm{stag}}&=\frac{cD_{AB}}L\ln\frac{1-y_{A2}}{1-y_{A1}}\\ &=\frac{40.09(2.0\times10^{-5})}{0.0050}\ln\frac{1}{0.70}\\ &=0.05720\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

A enters the film without a compensating net flux of B, producing molar flow in A’s direction. That bulk flow transports additional A. B still diffuses towards the high-A side, but its bulk contribution cancels its diffusion.

With 𝑦𝐴2 =0$y_{A2}=0$, dividing the two fluxes gives

𝑁𝐴,stag𝑁𝐴,eq=−ln⁡(1−𝑦𝐴1)𝑦𝐴1$$\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}} =\frac{-\ln(1-y_{A1})}{y_{A1}}$$

The common concentration, diffusivity and film thickness cancel. The ratios at the two supplied compositions are

[𝑁𝐴,stag𝑁𝐴,eq]𝑦𝐴1=0.30=−ln⁡(0.70)0.30=1.189$$\left[\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}}\right]_{y_{A1}=0.30} =\frac{-\ln(0.70)}{0.30}=1.189$$[𝑁𝐴,stag𝑁𝐴,eq]𝑦𝐴1=0.010=−ln⁡(0.990)0.010=1.0050$$\left[\frac{N_{A,\mathrm{stag}}}{N_{A,\mathrm{eq}}}\right]_{y_{A1}=0.010} =\frac{-\ln(0.990)}{0.010}=1.0050$$

As a dilute-limit check,

−ln⁡(1−𝑦)=𝑦+𝑦22+⋯ ,$$-\ln(1-y)=y+\frac{y^2}{2}+\cdots,$$

so the ratio approaches one as 𝑦 →0$y\to0$. The two models then give nearly the same flux.

Answer

At 𝑦𝐴1 =0.30$y_{A1}=0.30$, the equimolar flux is 0.0481 mol m−2 s−1$0.0481\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the stagnant-carrier flux is 0.0572 mol m−2 s−1$0.0572\ \mathrm{mol\,m^{-2}\,s^{-1}}$. The latter is 18.9 percent larger. At 𝑦𝐴1 =0.010$y_{A1}=0.010$, their ratio falls to 1.005.

Species balances and diffusion with reaction

Cylindrical wall

A given gas A can diffuse through the solid wall of a tube in which it flows. Find the rate of gas diffusion in terms of the tube dimensions (inside wall radius 𝑅𝑖$R_i$, outside wall radius 𝑅𝑜$R_o$), the gas-solid diffusivity, and the gas concentrations at the walls.

Assumptions:

• steady state
• flux in 𝑟$r$ direction only
• no chemical reaction
• A concentration in gas is quite dilute →𝑥𝐴 ∼0$\rightarrow x_A\sim0$
• Constant 𝐷𝐴𝐵$D_{AB}$

Solution

The radial shell in Figure 3 shows why a cylindrical balance must account for the changing area.

Figure 3: A stationary cylindrical wall and an enlarged radial shell. The molar rate equals the local flux multiplied by the cylindrical area 2πrL.

Data

Quantity	Value
Inner wall radius	𝑅𝑖$R_i$
Outer wall radius	𝑅𝑜$R_o$
Axial length	𝐿$L$
Constant gas-solid diffusivity	𝐷𝐴𝐵$D_{AB}$
Dissolved concentration at the inner face	𝑐𝐴𝑖$c_{Ai}$ in kmol m−3$\mathrm{kmol\,m^{-3}}$
Dissolved concentration at the outer face	𝑐𝐴𝑜$c_{Ao}$ in kmol m−3$\mathrm{kmol\,m^{-3}}$

Assumptions

• Steady state.
• Radial diffusion only.
• No chemical reaction.
• Dilute dissolved A in a stationary solid matrix.
• Constant diffusivity.
• Negligible end effects.

Calculations

Choose positive flux radially outwards. A steady shell has no accumulation or reaction, so its inward and outward molar rates are equal. From Eq. 10, the governing radial balance is

𝑑𝑑𝑟(𝑟𝑁𝐴𝑟)=0$$\frac{d}{dr}(rN_{Ar})=0$$

Consequently, the total outward molar rate is independent of radius,

˙𝑛𝐴=2𝜋𝑟𝐿𝑁𝐴𝑟$$\dot n_A=2\pi rL N_{Ar}$$

The source denotes this rate by 𝑊𝐴$W_A$. In a stationary solid, transport of dilute dissolved A follows Fick’s law,

𝑁𝐴𝑟=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑟$$N_{Ar}=-D_{AB}\frac{dc_A}{dr}$$

Substitute the flux into the rate expression, separate the variables and integrate between the two wall faces,

˙𝑛𝐴∫𝑅𝑜𝑅𝑖𝑑𝑟𝑟=−2𝜋𝐿𝐷𝐴𝐵∫𝑐𝐴𝑜𝑐𝐴𝑖𝑑𝑐𝐴˙𝑛𝐴ln⁡𝑅𝑜𝑅𝑖=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)$$\begin{aligned} \dot n_A\int_{R_i}^{R_o}\frac{dr}{r} &=-2\pi LD_{AB}\int_{c_{Ai}}^{c_{Ao}}dc_A\\ \dot n_A\ln\frac{R_o}{R_i} &=2\pi LD_{AB}(c_{Ai}-c_{Ao}) \end{aligned}$$

Solving for the rate gives

˙𝑛𝐴=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)ln⁡(𝑅𝑜𝑅𝑖)$$\dot n_A=\frac{2\pi LD_{AB}(c_{Ai}-c_{Ao})}{\ln\left(\frac{R_o}{R_i}\right)}$$(14)

The concentration profile follows from the same differential equation. Since 𝑟 𝑑𝑐𝐴/𝑑𝑟$r\,dc_A/dr$ is constant, integration from 𝑅𝑖$R_i$ to an arbitrary radius 𝑟$r$ gives

𝑐𝐴(𝑟)−𝑐𝐴𝑖=𝐶1ln⁡(𝑟𝑅𝑖)$$c_A(r)-c_{Ai}=C_1\ln\left(\frac{r}{R_i}\right)$$

At 𝑟 =𝑅𝑜$r=R_o$, impose 𝑐𝐴 =𝑐𝐴𝑜$c_A=c_{Ao}$ to obtain

𝐶1=𝑐𝐴𝑜−𝑐𝐴𝑖ln⁡(𝑅𝑜𝑅𝑖).$$C_1=\frac{c_{Ao}-c_{Ai}}{\ln\left(\frac{R_o}{R_i}\right)}.$$

Therefore

𝑐𝐴−𝑐𝐴𝑖𝑐𝐴𝑜−𝑐𝐴𝑖=ln⁡(𝑟𝑅𝑖)ln⁡(𝑅𝑜𝑅𝑖)$$\frac{c_A-c_{Ai}}{c_{Ao}-c_{Ai}} =\frac{\ln\left(\frac{r}{R_i}\right)}{\ln\left(\frac{R_o}{R_i}\right)}$$(15)

The profile recovers both specified wall concentrations. When 𝑐𝐴𝑖 >𝑐𝐴𝑜$c_{Ai}>c_{Ao}$, its gradient is negative and the rate is positive outwards. The units of Eq. 14 are

m (m2 s−1)(kmol m−3)=kmol s−1.$$\mathrm m\,(\mathrm{m^2\,s^{-1}})(\mathrm{kmol\,m^{-3}})=\mathrm{kmol\,s^{-1}}.$$

The concentration-based resistance is

ln⁡(𝑅𝑜𝑅𝑖)2𝜋𝐿𝐷𝐴𝐵,$$\frac{\ln\left(\frac{R_o}{R_i}\right)}{2\pi LD_{AB}},$$

in s m−3$\mathrm{s\,m^{-3}}$. Define the log-mean area,

𝐴lm=2𝜋𝐿(𝑅𝑜−𝑅𝑖)ln⁡(𝑅𝑜𝑅𝑖)$$A_{\mathrm{lm}}=\frac{2\pi L(R_o-R_i)}{\ln\left(\frac{R_o}{R_i}\right)}$$

The rate can then be written as

˙𝑛𝐴=𝐷𝐴𝐵𝐴lm(𝑐𝐴𝑖−𝑐𝐴𝑜)𝑅𝑜−𝑅𝑖,$$\dot n_A=\frac{D_{AB}A_{\mathrm{lm}}(c_{Ai}-c_{Ao})}{R_o-R_i},$$

just as cylindrical conduction can be written using a log-mean area.

The specified concentrations must be concentrations dissolved in the solid. A bulk gas concentration requires a solubility relation before it can be used as a wall boundary condition. For example, 𝑐𝐴 =𝑆𝐴𝑝𝐴$c_A=S_Ap_A$, with 𝑆𝐴$S_A$ in kmol m−3 Pa−1$\mathrm{kmol\,m^{-3}\,Pa^{-1}}$, gives the permeability 𝐷𝐴𝐵𝑆𝐴$D_{AB}S_A$.

Answer

The outward molar rate is

˙𝑛𝐴=2𝜋𝐿𝐷𝐴𝐵(𝑐𝐴𝑖−𝑐𝐴𝑜)ln⁡(𝑅𝑜𝑅𝑖).$$\dot n_A=\frac{2\pi LD_{AB}(c_{Ai}-c_{Ao})}{\ln\left(\frac{R_o}{R_i}\right)}.$$

It is positive when the dissolved concentration is greater at the inner wall. The logarithmic concentration profile is given by Eq. 15.

Species balances and diffusion with reaction

Naphthalene sphere

A spherical ball of solid, nonporous naphthalene is suspended in still air. The naphthalene ball slowly sublimes, releasing naphthalene into the surrounding air. It may be assumed that the sublimation occurs as a molecular diffusion-limited process. Through consideration of the diffusion equation, estimate the time required to reduce the diameter of the ball from 2 to 0.5 cm when the surrounding air is at 347 K and 1 atm.

Data for naphthalene:

• MW = 128 g/mol
• Density = 1.145 g/cm3$^3$
• Diffusivity in air = 8.19 ×10−6 m2/s$8.19\times10^{-6}\ \mathrm{m^2/s}$
• Vapour pressure = 666 Pa at 347 K

Solution

The spherical geometry in Figure 2 applies at each instantaneous ball radius. Vapour diffuses outwards through air while the solid surface slowly recedes.

Data

Quantity	Value
Temperature	𝑇 =347 K$T=347\ \mathrm K$
Pressure	𝑃 =1 atm =101325 Pa$P=1\ \mathrm{atm}=101325\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =8.19 ×10−6 m2 s−1$D_{AB}=8.19\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Saturation pressure	𝑝sat𝐴 =666 Pa$p_A^{\mathrm{sat}}=666\ \mathrm{Pa}$
Solid density	𝜌𝑠 =1145 kg m−3$\rho_s=1145\ \mathrm{kg\,m^{-3}}$
Molar mass	𝑀𝐴 =128 kg kmol−1$M_A=128\ \mathrm{kg\,kmol^{-1}}$
Initial radius	𝑅𝑖 =0.010 m$R_i=0.010\ \mathrm m$
Final radius	𝑅𝑓 =0.0025 m$R_f=0.0025\ \mathrm m$

The supplied dimensions are diameters. The table gives the corresponding radii.

Assumptions

• Spherically symmetric diffusion.
• Pseudo-steady gas profile as the surface recedes.
• Isothermal sublimation at 347 K$347\ \mathrm K$.
• Ideal gas behaviour.
• Constant pressure, diffusivity and solid density.
• Vapour-solid equilibrium at the surface.
• Negligible naphthalene concentration far from the ball.
• Zero net air flux through the solid surface.
• Negligible natural convection and gas-phase reaction.

Calculations

From Eq. 11 at a fixed instantaneous radius, the steady radial balance gives

𝑑(𝑟2𝑁𝐴𝑟)𝑑𝑟=0.$$\frac{d(r^2N_{Ar})}{dr}=0.$$

The same total rate crosses each concentric sphere,

˙𝑛𝐴=4𝜋𝑟2𝑁𝐴𝑟$$\dot n_A=4\pi r^2N_{Ar}$$

Air cannot cross the solid surface, so 𝑁𝐵𝑟 =0$N_{Br}=0$. Writing Eq. 12 in radial form retains the resulting molar flow,

𝑁𝐴𝑟=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑟$$N_{Ar}=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{dr}$$

Substituting the rate and integrating between radii 𝑅1$R_1$ and 𝑅2$R_2$ gives

˙𝑛𝐴4𝜋𝑐𝐷𝐴𝐵∫𝑅2𝑅1𝑑𝑟𝑟2=−∫𝑦𝐴2𝑦𝐴1𝑑𝑦𝐴1−𝑦𝐴$$\frac{\dot n_A}{4\pi cD_{AB}}\int_{R_1}^{R_2}\frac{dr}{r^2} =-\int_{y_{A1}}^{y_{A2}}\frac{dy_A}{1-y_A}$$

Evaluating both integrals and solving for the rate gives

˙𝑛𝐴=4𝜋𝑐𝐷𝐴𝐵1𝑅1−1𝑅2ln⁡1−𝑦𝐴21−𝑦𝐴1$$\dot n_A=\frac{4\pi cD_{AB}}{\frac{1}{R_1}-\frac{1}{R_2}}\ln\frac{1-y_{A2}}{1-y_{A1}}$$(16)

For an effectively infinite surrounding gas, take 𝑅1 =𝑅$R_1=R$, 𝑅2 →∞$R_2\to\infty$ and 𝑦𝐴2 =0$y_{A2}=0$. Surface equilibrium and the ideal gas relation give

𝑦𝐴𝑠=𝑝sat𝐴𝑃=666101325=0.00657291$$y_{As}=\frac{p_A^{\mathrm{sat}}}{P}=\frac{666}{101325}=0.00657291$$𝑐=𝑃𝑅𝑢𝑇=1013258314(347)=0.0351218 kmol m−3$$c=\frac{P}{R_uT}=\frac{101325}{8314(347)}=0.0351218\ \mathrm{kmol\,m^{-3}}$$

Define the constant composition factor,

𝐹=ln⁡11−𝑦𝐴𝑠=0.00659461$$F=\ln\frac{1}{1-y_{As}}=0.00659461$$

The instantaneous sublimation rate is then ˙𝑛𝐴 =4𝜋𝑅𝑐𝐷𝐴𝐵𝐹$\dot n_A=4\pi RcD_{AB}F$. The sphere loses moles as its radius decreases. Differentiating its solid inventory gives

˙𝑛𝐴=−𝑑𝑑𝑡(𝜌𝑠𝑀𝐴4𝜋𝑅33)=−𝜌𝑠𝑀𝐴4𝜋𝑅2𝑑𝑅𝑑𝑡$$\dot n_A=-\frac{d}{dt}\left(\frac{\rho_s}{M_A}\frac{4\pi R^3}{3}\right) =-\frac{\rho_s}{M_A}4\pi R^2\frac{dR}{dt}$$

Equate surface loss to diffusion into the gas. After cancelling 4𝜋𝑅$4\pi R$ and separating variables,

𝑅 𝑑𝑅=−𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠 𝑑𝑡∫𝑅𝑓𝑅𝑖𝑅 𝑑𝑅=−𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠∫𝑡0𝑑𝑡𝑡=𝜌𝑠(𝑅2𝑖−𝑅2𝑓)2𝑀𝐴𝑐𝐷𝐴𝐵𝐹$$\begin{aligned} R\,dR&=-\frac{M_AcD_{AB}F}{\rho_s}\,dt\\ \int_{R_i}^{R_f}R\,dR&=-\frac{M_AcD_{AB}F}{\rho_s}\int_0^t dt\\ t&=\frac{\rho_s(R_i^2-R_f^2)}{2M_AcD_{AB}F} \end{aligned}$$(17)

Substitute the radii and properties on a consistent kmol basis,

𝑡=1145(0.0102−0.00252)2(128)(0.0351218)(8.19×10−6)(0.00659461)=2.210×105 s=61.40 h$$t=\frac{1145(0.010^2-0.0025^2)} {2(128)(0.0351218)(8.19\times10^{-6})(0.00659461)} =2.210\times10^5\ \mathrm s=61.40\ \mathrm h$$(18)

The gas adjustment time is

𝑅2𝑖𝐷𝐴𝐵=0.01028.19×10−6=12.2 s,$$\frac{R_i^2}{D_{AB}}=\frac{0.010^2}{8.19\times10^{-6}}=12.2\ \mathrm s,$$

much shorter than the sublimation time. This supports a pseudo-steady gas profile. Replacing 𝐹$F$ by the dilute approximation 𝑦𝐴𝑠$y_{As}$ gives 61.60 h, only 0.33 percent higher.

The integrated balance can also be expressed as a diameter-square law,

𝑑2=𝑑2𝑖−𝐾𝑡$$d^2=d_i^2-Kt$$𝐾=8𝑀𝐴𝑐𝐷𝐴𝐵𝐹𝜌𝑠$$K=\frac{8M_AcD_{AB}F}{\rho_s}$$

Here 𝐾$K$ has units m2 s−1$\mathrm{m^2\,s^{-1}}$. The composition factor must remain in 𝐾$K$ because the vapour pressure determines the driving force. A different thermal model would be needed if sublimation cooling changes the surface temperature appreciably.

Answer

The diffusion-limited model predicts 𝑡 =2.21 ×105 s$t=2.21\times10^5\ \mathrm s$, or about 61.4 h, for the diameter to decrease from 2 cm to 0.5 cm.

Species balances and diffusion with reaction

Catalytic tube

Consider a catalytic tubular reactor, where a dilute solution of solute A in solvent S is in fully developed, laminar flow in the region 𝑧 <0$z<0$. It encounters the catalytic wall in the region 0 ≤𝑧 ≤𝐿$0\le z\le L$, where solute A is instantaneously and irreversibly rearranged to an isomer B. Write the appropriate diffusion equation (in differential form) for this problem, along with the boundary conditions. Assume that the flow is isothermal and neglect the presence of B.

Solution

Figure 7: Axial liquid flow in a tube with a catalytic wall. For the instantaneous wall reaction, the labelled wall concentration is zero. Species A is consumed at the wall as the liquid flows downstream.

In this diagram, 𝑑 =2𝑅$d=2R$ and 𝑄 =¯¯¯¯𝑣𝜋𝑅2$Q=\bar v\pi R^2$. Its concentration labels denote the molar concentration 𝑐𝐴$c_A$ used below.

Data

• Tube radius 𝑅$R$
• Catalyst-coated length 0 ≤𝑧 ≤𝐿$0\le z\le L$
• Mean axial velocity ¯¯¯¯𝑣$\bar v$
• Uniform inlet concentration 𝑐𝐴0$c_{A0}$
• Wall reaction A converts instantaneously to its isomer B

The dilute liquid enters in fully developed laminar flow. The question asks for a model and boundary conditions, so no numerical solution is required.

Assumptions

• Steady state.
• Axisymmetric flow and concentration field.
• Constant properties.
• No radial velocity.
• No homogeneous reaction in the liquid.
• Uniform inlet concentration.
• Axial diffusion neglected only when axial convection dominates over the length of interest.

The specified isothermal flow and instantaneous wall reaction are supplied features of the model.

Calculations

Reaction occurs on the wall, so it enters through a boundary condition. It does not supply a volumetric 𝑅𝐴$R_A$ term in the liquid. The steady, axisymmetric form of Eq. 10 is

0=−1𝑟∂(𝑟𝑁𝐴𝑟)∂𝑟−∂𝑁𝐴𝑧∂𝑧$$0=-\frac1r\frac{\partial(rN_{Ar})}{\partial r}-\frac{\partial N_{Az}}{\partial z}$$

There is no radial convection. Axial convection remains present even though the solute is dilute. The two total-flux components are

𝑁𝐴𝑟=−𝐷𝐴𝐵∂𝑐𝐴∂𝑟$$N_{Ar}=-D_{AB}\frac{\partial c_A}{\partial r}$$𝑁𝐴𝑧=𝑣𝑧(𝑟)𝑐𝐴−𝐷𝐴𝐵∂𝑐𝐴∂𝑧$$N_{Az}=v_z(r)c_A-D_{AB}\frac{\partial c_A}{\partial z}$$

Substituting the fluxes into the balance gives the full convection-diffusion equation,

𝑣𝑧(𝑟)∂𝑐𝐴∂𝑧=𝐷𝐴𝐵𝑟∂∂𝑟(𝑟∂𝑐𝐴∂𝑟)+𝐷𝐴𝐵∂2𝑐𝐴∂𝑧2$$v_z(r)\frac{\partial c_A}{\partial z} =\frac{D_{AB}}r\frac{\partial}{\partial r}\left(r\frac{\partial c_A}{\partial r}\right) +D_{AB}\frac{\partial^2c_A}{\partial z^2}$$(30)

For fully developed laminar flow in a circular tube, the velocity profile is

𝑣𝑧(𝑟)=2¯¯¯¯𝑣[1−(𝑟𝑅)2]$$v_z(r)=2\bar v\left[1-\left(\frac{r}{R}\right)^2\right]$$

To judge axial diffusion, compare the convection scale ¯¯¯¯𝑣𝑐𝐴/𝐿$\bar v c_A/L$ with the diffusion scale 𝐷𝐴𝐵𝑐𝐴/𝐿2$D_{AB}c_A/L^2$. Their ratio is the axial Péclet number,

Pe𝐿=¯¯¯¯𝑣𝐿𝐷𝐴𝐵$$\mathrm{Pe}_L=\frac{\bar vL}{D_{AB}}$$

For illustration, use ¯¯¯¯𝑣 =0.1 m s−1$\bar v=0.1\ \mathrm{m\,s^{-1}}$, 𝐿 =0.1 m$L=0.1\ \mathrm m$ and 𝐷𝐴𝐵 =10−9 m2 s−1$D_{AB}=10^{-9}\ \mathrm{m^2\,s^{-1}}$. These are supplementary scale values rather than data supplied in the question. They give

Pe𝐿=0.1(0.1)10−9=107$$\mathrm{Pe}_L=\frac{0.1(0.1)}{10^{-9}}=10^7$$

Axial convection then dominates axial diffusion over this length, away from axial end regions. The resulting working equation is

𝑣𝑧(𝑟)∂𝑐𝐴∂𝑧=𝐷𝐴𝐵𝑟∂∂𝑟(𝑟∂𝑐𝐴∂𝑟)$$v_z(r)\frac{\partial c_A}{\partial z} =\frac{D_{AB}}r\frac{\partial}{\partial r}\left(r\frac{\partial c_A}{\partial r}\right)$$(31)

The reduced equation needs inlet data and two radial conditions. The inlet supplies uniform A. Instantaneous reaction holds the catalytic-wall concentration at zero. Centreline symmetry gives zero radial gradient. These conditions form the system

𝑐𝐴(𝑟,0)=𝑐𝐴0,0≤𝑟<𝑅𝑐𝐴(𝑅,𝑧)=0,0<𝑧≤𝐿[∂𝑐𝐴∂𝑟]𝑟=0=0,0<𝑧≤𝐿$$\begin{aligned} c_A(r,0)&=c_{A0}, &&0\le r<R\\ c_A(R,z)&=0, &&0<z\le L\\ \left[\frac{\partial c_A}{\partial r}\right]_{r=0}&=0, &&0<z\le L \end{aligned}$$

The inlet and wall data are interpreted as one-sided conditions at the inlet-wall corner, where the idealised concentration jumps. Every term in the governing equation has units kmol m−3 s−1$\mathrm{kmol\,m^{-3}\,s^{-1}}$.

If axial diffusion is retained in Eq. 30, the axial equation is second order. A finite-domain model then also needs an appropriate outlet condition, such as zero axial concentration gradient at a sufficiently downstream boundary. The Péclet estimate supports the reduced model over a specified scale; it is not a pointwise argument at the wall, where the velocity is zero.

For a finite first-order surface reaction, the wall condition becomes

−𝐷𝐴𝐵∂𝑐𝐴∂𝑟=𝑘′′1𝑐𝐴$$-D_{AB}\frac{\partial c_A}{\partial r}=k_1''c_A$$

at 𝑟 =𝑅$r=R$, because diffusive arrival must equal surface consumption.

This formulation is the mass-transfer Graetz problem. In the fully developed concentration region, a constant wall concentration gives Sh =𝑘𝑐𝑑𝑡/𝐷𝐴𝐵 =3.66$\mathrm{Sh}=k_cd_t/D_{AB}=3.66$, where 𝑑𝑡 =2𝑅$d_t=2R$ and 𝑘𝑐$k_c$ uses the difference between mixing-cup and wall concentrations. The entrance-region coefficient is larger. The thermal problem with the same geometry and boundary conditions gives Nu =3.66$\mathrm{Nu}=3.66$.

Answer

Use Eq. 31 with

𝑣𝑧=2¯¯¯¯𝑣[1−(𝑟𝑅)2]$$v_z=2\bar v\left[1-\left(\frac{r}{R}\right)^2\right]$$

when axial diffusion can be neglected. Apply uniform inlet concentration, zero A concentration at the catalytic wall, and zero radial concentration gradient at the centreline, as stated above.

If axial diffusion is significant, use Eq. 30 and provide an additional axial boundary condition.

Species balances and diffusion with reaction

Derive a film balance with finite surface reaction

A dilute gas species A crosses a plane film to a catalyst, where it undergoes an isomerisation 𝐴 ⟶𝐵$A\longrightarrow B$. The gas is at 500 K and 100 kPa. The bulk mole fraction is 0.020, film thickness is 1.5 mm, diffusivity is 6.0 ×10−5$6.0\times10^{-5}$ m² s⁻¹ and surface rate constant is 𝑘𝑠 =0.015$k_s=0.015$ m s⁻¹. The catalyst area is 0.040 m². Assume steady one-dimensional isothermal transport with constant pressure and diffusivity, no gas-phase reaction, and negligible normal bulk velocity. Let 𝑧 =0$z=0$ be the bulk edge and 𝑧 =𝐿$z=L$ the catalyst.

Derive the concentration profile from a shell balance. Find the surface mole fraction and molar removal rate. Which resistance is greater, and what happens in the limits of very slow and very fast surface reaction?

Solution

Figure 8: A diffuses from the bulk through a nonreacting film to the catalyst. A finite reaction rate leaves a nonzero surface concentration, determined by matching film transport and surface consumption.

Data

Quantity	Value
Temperature	𝑇 =500 K$T=500\ \mathrm K$
Pressure	𝑃 =100000 Pa$P=100000\ \mathrm{Pa}$
Bulk mole fraction	𝑦𝐴𝑏 =0.020$y_{Ab}=0.020$
Film thickness	𝐿 =0.0015 m$L=0.0015\ \mathrm m$
Diffusivity	𝐷𝐴𝐵 =6.0 ×10−5 m2 s−1$D_{AB}=6.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface reaction coefficient	𝑘𝑠 =0.015 m s−1$k_s=0.015\ \mathrm{m\,s^{-1}}$
Surface area	𝐴𝑠 =0.040 m2$A_s=0.040\ \mathrm{m^2}$

Assumptions

• Steady state.
• One-dimensional transport across a plane film.
• Ideal gas behaviour.
• Constant temperature, pressure and diffusivity.
• No gas-phase reaction.
• Negligible normal bulk velocity.

Calculations

For a film slice of area 𝑆$S$ and thickness Δ𝑧$\Delta z$, no accumulation or volumetric reaction occurs. The shell balance is

𝑆𝑁𝐴(𝑧)−𝑆𝑁𝐴(𝑧+Δ𝑧)=0⟹𝑑𝑁𝐴𝑑𝑧=0$$SN_A(z)-SN_A(z+\Delta z)=0 \quad\Longrightarrow\quad \frac{dN_A}{dz}=0$$

With

𝑁𝐴=−𝐷𝐴𝐵𝑑𝑐𝐴𝑑𝑧$$N_A=-D_{AB}\frac{dc_A}{dz}$$

and constant diffusivity, 𝑑2𝑐𝐴/𝑑𝑧2 =0$d^2c_A/dz^2=0$. Integrating twice and imposing 𝑐𝐴(0) =𝑐𝐴𝑏$c_A(0)=c_{Ab}$ and 𝑐𝐴(𝐿) =𝑐𝐴𝑠$c_A(L)=c_{As}$ gives

𝑐𝐴(𝑧)=𝑐𝐴𝑏+(𝑐𝐴𝑠−𝑐𝐴𝑏)𝑧𝐿,𝑁𝐴=𝐷𝐴𝐵𝐿(𝑐𝐴𝑏−𝑐𝐴𝑠)$$c_A(z)=c_{Ab}+(c_{As}-c_{Ab})\frac zL, \qquad N_A=\frac{D_{AB}}L(c_{Ab}-c_{As})$$

Reaction occurs at the boundary. It is absent from the film differential equation. At steady state delivery equals surface consumption:

𝐷𝐴𝐵𝐿(𝑐𝐴𝑏−𝑐𝐴𝑠)=𝑘𝑠𝑐𝐴𝑠,𝑐𝐴𝑠=𝑐𝐴𝑏1+𝑘𝑠𝐿𝐷𝐴𝐵$$\frac{D_{AB}}L(c_{Ab}-c_{As})=k_sc_{As}, \qquad c_{As}=\frac{c_{Ab}}{1+\frac{k_sL}{D_{AB}}}$$

Total gas concentration is uniform, so mole fractions have the same ratio:

𝑘𝑠𝐿𝐷𝐴𝐵=0.375$$\frac{k_sL}{D_{AB}}=0.375$$𝑦𝐴𝑠=0.0201.375=0.01455$$y_{As}=\frac{0.020}{1.375}=0.01455$$𝑐=1000008.314(500)=24.06 mol m−3$$c=\frac{100000}{8.314(500)}=24.06\ \mathrm{mol\,m^{-3}}$$𝑁𝐴=𝑘𝑠𝑐𝑦𝐴𝑠=0.015(24.06)(0.01455)=0.005249 mol m−2 s−1$$N_A=k_scy_{As}=0.015(24.06)(0.01455) =0.005249\ \mathrm{mol\,m^{-2}\,s^{-1}}$$˙𝑛𝐴=𝐴𝑠𝑁𝐴=0.040(0.005249)=2.099×10−4 mol s−1$$\dot n_A=A_sN_A=0.040(0.005249) =2.099\times10^{-4}\ \mathrm{mol\,s^{-1}}$$

Both concentration-based resistances have units s m⁻¹: 𝐿/𝐷𝐴𝐵 =25.0$L/D_{AB}=25.0$ and 1/𝑘𝑠 =66.7$1/k_s=66.7$. Their sum gives

𝑁𝐴=𝑐𝐴𝑏𝐿𝐷𝐴𝐵+1𝑘𝑠.$$N_A=\frac{c_{Ab}}{\frac{L}{D_{AB}}+\frac{1}{k_s}}.$$

Surface reaction supplies the larger resistance, but neither resistance is negligible.

As 𝑘𝑠 →0$k_s\to0$, 𝑐𝐴𝑠 →𝑐𝐴𝑏$c_{As}\to c_{Ab}$ and removal tends to zero. As 𝑘𝑠 →∞$k_s\to\infty$, 𝑐𝐴𝑠 →0$c_{As}\to0$ and removal approaches the diffusion-limited value 𝐷𝐴𝐵𝑐𝐴𝑏/𝐿$D_{AB}c_{Ab}/L$. The isomerisation exchanges one gas mole for one gas mole, consistent with zero net molar flow. Other stoichiometries require checking that assumption.

For a cylindrical reactor with axial flow, the earlier catalytic-tube example retains axial convection. Its finite-rate wall condition uses the same principle: arriving diffusive flux equals surface consumption. Its radial profile must be found with the axial concentration field, rather than assumed linear as in this plane, nonconvecting film.

Answer

The surface mole fraction is 𝑦𝐴𝑠 =0.0146$y_{As}=0.0146$ and the molar removal rate is ˙𝑛𝐴 =2.10 ×10−4 mol s−1$\dot n_A=2.10\times10^{-4}\ \mathrm{mol\,s^{-1}}$. Surface reaction supplies the larger resistance, 66.7 s m−1$66.7\ \mathrm{s\,m^{-1}}$, compared with 25.0 s m−1$25.0\ \mathrm{s\,m^{-1}}$ for the film. The slow-reaction limit gives zero removal; the fast-reaction limit gives the diffusion-limited flux 𝐷𝐴𝐵𝑐𝐴𝑏/𝐿$D_{AB}c_{Ab}/L$.

Species balances and diffusion with reaction

Use reaction stoichiometry to determine flux directions

At a catalyst, gas R reacts by 𝑅 ⟶2𝑃$R\longrightarrow2P$. Take the coordinate direction towards the catalyst and suppose 𝑁𝑅 =0.010$N_R=0.010$ mol m⁻² s⁻¹. Find 𝑁𝑃$N_P$ and the total molar flux. Is equimolar counter-diffusion appropriate? Compare with a reaction 𝑅 ⟶𝑃$R\longrightarrow P$ at the same reactant flux.

Solution

Data

• Surface reaction 𝑅 ⟶2𝑃$R\longrightarrow2P$
• Reactant flux toward the catalyst 𝑁𝑅 =0.010 mol m−2 s−1$N_R=0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$
• Comparison reaction 𝑅 ⟶𝑃$R\longrightarrow P$

Positive flux points toward the catalyst.

Assumptions

• Steady surface reaction.
• No surface accumulation.
• Only the stated reaction produces or consumes R and P.

Calculations

Products move away from the surface. Two moles of P leave for each mole of R arriving, so

𝑁𝑃=−2𝑁𝑅=−2(0.010)=−0.020 mol m−2 s−1$$N_P=-2N_R=-2(0.010)=-0.020\ \mathrm{mol\,m^{-2}\,s^{-1}}$$𝑁𝑅+𝑁𝑃=0.010−0.020=−0.010 mol m−2 s−1$$N_R+N_P=0.010-0.020=-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

The net molar flow is away from the catalyst, so the equimolar model does not apply. In the total-flux expression

𝑁𝑅=−𝑐𝐷𝑅𝑃 𝑑𝑦𝑅𝑑𝑧+𝑦𝑅(𝑁𝑅+𝑁𝑃),$$N_R=-cD_{RP}\,\frac{dy_R}{dz}+y_R(N_R+N_P),$$

the bulk term is negative. It opposes R’s inward diffusive transport.

The flux relation can also be written 𝑁𝑅 = −𝑁𝑃/2$N_R=-N_P/2$. The negative sign expresses opposing directions; the factor comes from stoichiometry.

For 𝑅 ⟶𝑃$R\longrightarrow P$, 𝑁𝑃 = −𝑁𝑅$N_P=-N_R$ and the total molar flux is zero. That reaction permits the equimolar model under the other stated binary-film assumptions. Constant temperature and pressure alone do not determine the flux ratio.

Answer

For 𝑅 ⟶2𝑃$R\longrightarrow2P$, 𝑁𝑃 = −0.020 mol m−2 s−1$N_P=-0.020\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the total molar flux is −0.010 mol m−2 s−1$-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$, directed away from the catalyst. Equimolar counter-diffusion does not apply. For 𝑅 ⟶𝑃$R\longrightarrow P$, 𝑁𝑃 = −0.010 mol m−2 s−1$N_P=-0.010\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and the total molar flux is zero.

Species balances and diffusion with reaction

Distinguish reaction throughout a liquid from reaction at a wall

An idealised stationary water layer of depth 0.010 m absorbs oxygen at its upper face. That face is held at 𝑐𝐴0 =0.25$c_{A0}=0.25$ mol m⁻³. Oxygen consumption throughout the liquid follows 𝑅𝐴 = −𝑘′′′1𝑐𝐴$R_A=-k_1'''c_A$, with 𝑘′′′1 =0.0020$k_1'''=0.0020$ s⁻¹ and 𝐷𝐴𝐵 =2.0 ×10−9$D_{AB}=2.0\times10^{-9}$ m² s⁻¹. The bottom is impermeable. Assume a steady one-dimensional profile and constant properties.

Formulate the problem and estimate the oxygen influx. Explain how the model would change if consumption occurred only on a catalytic bottom instead.

Solution

Data

• Liquid depth 𝐿 =0.010 m$L=0.010\ \mathrm m$
• Fixed top concentration 𝑐𝐴0 =0.25 mol m−3$c_{A0}=0.25\ \mathrm{mol\,m^{-3}}$
• First-order volumetric rate constant 𝑘′′′1 =0.0020 s−1$k_1^{\prime\prime\prime}=0.0020\ \mathrm{s^{-1}}$
• Diffusivity 𝐷𝐴𝐵 =2.0 ×10−9 m2 s−1$D_{AB}=2.0\times10^{-9}\ \mathrm{m^2\,s^{-1}}$

The bottom is impermeable.

Assumptions

• Steady state.
• One-dimensional diffusion.
• Stationary liquid.
• Dilute oxygen.
• Constant properties.
• First-order homogeneous oxygen consumption.

Calculations

Measure 𝑥$x$ downwards from the upper surface. Consumption occurs throughout the liquid, so it is a homogeneous reaction and belongs in the local balance:

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2−𝑘′′′1𝑐𝐴=0,𝑐𝐴(0)=𝑐𝐴0,[𝑑𝑐𝐴𝑑𝑥]𝑥=𝐿=0$$D_{AB}\frac{d^2c_A}{dx^2}-k_1'''c_A=0, \qquad c_A(0)=c_{A0}, \qquad \left[\frac{dc_A}{dx}\right]_{x=L}=0$$

The reaction penetration length is

√𝐷𝐴𝐵𝑘′′′1=0.0010 m,$$\sqrt{\frac{D_{AB}}{k_1'''}}=0.0010\,\mathrm{m},$$

and

Ha=𝐿√𝑘′′′1𝐷𝐴𝐵=10.$$\mathrm{Ha}=L\sqrt{\frac{k_1'''}{D_{AB}}}=10.$$

From Eq. 29,

𝑁𝐴(0)=𝑐𝐴0√𝑘′′′1𝐷𝐴𝐵tanh⁡(Ha)=0.25√0.0020(2.0×10−9)tanh⁡(10)≃5.00×10−7 mol m−2 s−1$$\begin{aligned} N_A(0)&=c_{A0}\sqrt{k_1'''D_{AB}}\tanh(\mathrm{Ha})\\ &=0.25\sqrt{0.0020(2.0\times10^{-9})}\tanh(10)\\ &\simeq5.00\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}} \end{aligned}$$

Most consumption takes place near the upper surface. Adding more depth has little effect in this large-Ha limit.

With reaction only at a catalytic bottom, the interior has no volumetric source:

𝐷𝐴𝐵𝑑2𝑐𝐴𝑑𝑥2=0.$$D_{AB}\frac{d^2c_A}{dx^2}=0.$$

Replace the impermeable-bottom condition by

−𝐷𝐴𝐵𝑐′𝐴(𝐿)=𝑘𝑠𝑐𝐴(𝐿)$$-D_{AB}c_A'(L)=k_sc_A(L)$$

for a first-order surface reaction. The two rate constants have different units, s⁻¹ in the liquid and m s⁻¹ at the surface.

A pond can require an unsteady absorption model or a steady reaction model. Its size alone does not decide. The stated consumption, flow, initial state and boundary conditions determine which model is appropriate.

Answer

The oxygen influx is 5.00 ×10−7 mol m−2 s−1$5.00\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}}$ into the liquid. If reaction occurs only at the bottom, use 𝐷𝐴𝐵𝑐′′𝐴 =0$D_{AB}c_A''=0$ in the interior and the bottom condition

−𝐷𝐴𝐵𝑐′𝐴(𝐿)=𝑘𝑠𝑐𝐴(𝐿).$$-D_{AB}c_A'(L)=k_sc_A(L).$$

Species balances and diffusion with reaction

An expanding reaction mixture changes the film profile

A catalyst converts gas A instantaneously by 𝐴 ⟶3𝐵$A\longrightarrow3B$. A plane gas film of thickness 𝛿 =2.0$\delta=2.0$ mm separates the surface from a well-mixed bulk gas with 𝑦𝐴0 =0.20$y_{A0}=0.20$. The binary gas is at 500 K and 1.5 bar, with 𝐷𝐴𝐵 =2.0 ×10−5$D_{AB}=2.0\times10^{-5}$ m² s⁻¹. Assume steady, one-dimensional transport, constant temperature, pressure and diffusivity, and no reaction inside the film. The catalyst is impermeable except for the reactant and product exchange specified by the reaction.

Taking 𝑧$z$ from the bulk edge towards the catalyst, derive 𝑦𝐴(𝑧)$y_A(z)$ and the inward flux of A. Compare with equimolar counter-diffusion across the same film. Does a constant species flux imply a linear mole-fraction profile?

Solution

Figure 9: A crosses the gas film toward the catalyst. The surface reaction produces B, whose outward flux changes the mixture molar flow.

Data

Quantity	Value
Film thickness	𝛿 =0.0020 m$\delta=0.0020\ \mathrm m$
Bulk mole fraction	𝑦𝐴0 =0.20$y_{A0}=0.20$
Temperature	𝑇 =500 K$T=500\ \mathrm K$
Pressure	𝑃 =150000 Pa$P=150000\ \mathrm{Pa}$
Diffusivity	𝐷𝐴𝐵 =2.0 ×10−5 m2 s−1$D_{AB}=2.0\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Instantaneous catalyst reaction	𝐴 ⟶3𝐵$A\longrightarrow3B$

Assumptions

• Steady state.
• One-dimensional transport across a plane film.
• Binary ideal gas mixture.
• Constant temperature, pressure and diffusivity.
• No gas-phase reaction.
• Surface exchange occurs only through the stated reaction.

Calculations

A film shell balance gives 𝑑𝑁𝐴/𝑑𝑧 =0$dN_A/dz=0$ because there is no accumulation or gas-phase reaction. The boundary conditions are 𝑦𝐴(0) =𝑦𝐴0$y_A(0)=y_{A0}$ and 𝑦𝐴(𝛿) =0$y_A(\delta)=0$. Surface stoichiometry fixes 𝑁𝐵 = −3𝑁𝐴$N_B=-3N_A$, so the total flux is 𝑁𝐴 +𝑁𝐵 = −2𝑁𝐴$N_A+N_B=-2N_A$. Product expansion creates outward molar flow even though there is no externally imposed flow through the film.

Keep that flow in the total-flux equation:

𝑁𝐴=−𝑐𝐷𝐴𝐵𝑑𝑦𝐴𝑑𝑧−2𝑦𝐴𝑁𝐴𝑑𝑦𝐴1+2𝑦𝐴=−𝑁𝐴𝑐𝐷𝐴𝐵 𝑑𝑧$$\begin{gathered} N_A=-cD_{AB}\frac{dy_A}{dz}-2y_AN_A\\ \frac{dy_A}{1+2y_A}=-\frac{N_A}{cD_{AB}}\,dz \end{gathered}$$

Integrating from the bulk edge to position 𝑧$z$ gives

12ln⁡1+2𝑦𝐴(𝑧)1+2𝑦𝐴0=−𝑁𝐴𝑧𝑐𝐷𝐴𝐵$$\frac12\ln\frac{1+2y_A(z)}{1+2y_{A0}} =-\frac{N_Az}{cD_{AB}}$$

Using the zero surface concentration at 𝑧 =𝛿$z=\delta$ determines the flux. Substitution back into the integrated equation gives the profile:

𝑁𝐴=𝑐𝐷𝐴𝐵2𝛿ln⁡(1+2𝑦𝐴0)𝑦𝐴(𝑧)=12[(1+2𝑦𝐴0)1−𝑧𝛿−1]$$\begin{aligned} N_A&=\frac{cD_{AB}}{2\delta}\ln(1+2y_{A0})\\ y_A(z)&=\frac12\left[(1+2y_{A0})^{1-\frac{z}{\delta}}-1\right] \end{aligned}$$

The numerical concentration is 𝑐 =𝑃/(𝑅𝑇) =36.08$c=P/(RT)=36.08$ mol m⁻³, so

𝑁𝐴=36.08(2.0×10−5)2(0.0020)ln⁡[1+2(0.20)]=0.06071 mol m−2 s−1$$N_A=\frac{36.08(2.0\times10^{-5})}{2(0.0020)}\ln[1+2(0.20)] =0.06071\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

An equimolar model would give

𝑁𝐴,𝑒𝑞=𝑐𝐷𝐴𝐵𝑦𝐴0𝛿=0.07217 mol m−2 s−1.$$N_{A,eq}=\frac{cD_{AB}y_{A0}}{\delta}=0.07217\,\mathrm{mol\,m^{-2}\,s^{-1}}.$$

The actual flux in this film model is 15.9 percent below that value. Outward bulk flow opposes the inward transport of A. This reverses the effect of the inward molar flow in the contracting 2𝐴 ⟶𝐵$2A\longrightarrow B$ case above.

At 𝑧 =𝛿/2$z=\delta/2$,

𝑦𝐴=√1.4−12=0.09161,$$y_A=\frac{\sqrt{1.4}-1}{2}=0.09161,$$

compared with 0.100 for a straight line. The species flux is constant, but its diffusive and bulk contributions vary with composition. Constant flux therefore does not require a linear profile. Both imposed endpoints are recovered exactly. As 𝑦𝐴0 →0$y_{A0}\to0$, the logarithm approaches 2𝑦𝐴0$2y_{A0}$ and the dilute equimolar estimate is recovered.

Answer

The inward flux is 𝑁𝐴 =0.0607 mol m−2 s−1$N_A=0.0607\ \mathrm{mol\,m^{-2}\,s^{-1}}$, 15.9 percent below the equimolar estimate. The profile is

𝑦𝐴(𝑧)=12[(1+2𝑦𝐴0)1−𝑧𝛿−1].$$y_A(z)=\frac{1}{2}\left[(1+2y_{A0})^{1-\frac{z}{\delta}}-1\right].$$

The profile is curved despite the constant species flux.

Transient mass diffusion

Worked coal burnout model

For the reaction and air composition above, find the particle radius and burnout time in terms of 𝑅0$R_0$, 𝑐$c$, 𝐷$D$, 𝜌𝐶$\rho_C$ and 𝑀𝐶$M_C$. Quantify the flux correction relative to pure diffusion.

Calculation and answer

Data

Quantity	Value
Initial particle radius	𝑅0$R_0$
Gas total molar concentration	𝑐$c$
Effective oxygen diffusivity	𝐷$D$
Carbon density	𝜌𝐶$\rho_C$
Carbon molar mass	𝑀𝐶$M_C$
Ambient oxygen mole fraction	𝑦𝑂2,∞ =0.21$y_{O_2,\infty}=0.21$
Reaction stoichiometry	2.5 mol oxygen consumed per 3 mol carbon

Assumptions

• Spherical particle with uniform solid density.
• Pseudo steady diffusion in the surrounding gas.
• Instantaneous surface reaction and no gas-phase reaction.
• Constant temperature, total concentration and effective diffusivity.

Calculations

The steady gas calculation supplies the oxygen flux at each current radius. Writing Eq. 5 for the inward magnitude gives

∣𝑁𝑂2,𝑟(𝑅)∣=5𝑐𝐷𝑅ln⁡(1+0.2𝑦∞)$$|N_{O_2,r}(R)|=\frac{5cD}{R}\ln(1+0.2y_\infty)$$

Substituting the ambient mole fraction gives

∣𝑁𝑂2,𝑟(𝑅)∣=5𝑐𝐷𝑅ln⁡(1.042)=0.20571𝑐𝐷𝑅$$|N_{O_2,r}(R)|=\frac{5cD}{R}\ln(1.042) =0.20571\frac{cD}{R}$$

Multiply the oxygen rate by the stoichiometric carbon-to-oxygen ratio:

˙𝑛𝐶=32.5 4𝜋𝑅2∣𝑁𝑂2,𝑟(𝑅)∣=24𝜋𝑐𝐷ln⁡(1.042)𝑅$$\dot n_C=\frac{3}{2.5}\,4\pi R^2|N_{O_2,r}(R)| =24\pi cD\ln(1.042)R$$

The carbon inventory is

𝑛𝐶=(𝜌𝐶𝑀𝐶)4𝜋𝑅33.$$n_C=\left(\frac{\rho_C}{M_C}\right)\frac{4\pi R^3}{3}.$$

Its decrease equals the positive consumption rate:

𝜌𝐶𝑀𝐶4𝜋𝑅2𝑑𝑅𝑑𝑡=−24𝜋𝑐𝐷ln⁡(1.042)𝑅$$\frac{\rho_C}{M_C}4\pi R^2\frac{dR}{dt} =-24\pi cD\ln(1.042)R$$

Cancel a factor of radius and integrate from 𝑅0$R_0$ at zero time to 𝑅$R$ at time 𝑡$t$:

∫𝑅𝑅0𝑅 𝑑𝑅=−6𝑐𝐷𝑀𝐶𝜌𝐶ln⁡(1.042)∫𝑡0𝑑𝑡$$\int_{R_0}^{R}R\,dR =-\frac{6cDM_C}{\rho_C}\ln(1.042)\int_0^t dt$$

Thus 𝑅2 =𝑅20 −𝐾𝑡$R^2=R_0^2-Kt$, with

𝐾=12𝑐𝐷𝑀𝐶ln⁡(1.042)𝜌𝐶.$$K=\frac{12cDM_C\ln(1.042)}{\rho_C}.$$

The dimensions of 𝐾$K$ are m2 s−1$\mathrm{m^{2}\,s^{-1}}$, as required for a radius-squared decay law.

Without outward mixture flow, the oxygen flux magnitude would be 0.21𝑐𝐷/𝑅$0.21cD/R$. The relative reduction is

1−0.205710.21=0.02043$$1-\frac{0.20571}{0.21}=0.02043$$

The outward product flow therefore reduces the oxygen supply by 2.04 percent. Doubling 𝑅0$R_0$ quadruples burnout time within this model.

Answer

The radius and burnout time are

𝑅(𝑡)=√𝑅20−𝐾𝑡,𝐾=0.49370𝑐𝐷𝑀𝐶𝜌𝐶$$R(t)=\sqrt{R_0^2-Kt},\qquad K=0.49370\frac{cDM_C}{\rho_C}$$𝑡𝑏=𝜌𝐶𝑅200.49370𝑐𝐷𝑀𝐶$$t_b=\frac{\rho_CR_0^2}{0.49370cDM_C}$$

The radius expression applies for 0 ≤𝑡 ≤𝑡𝑏$0\leq t\leq t_b$. The oxygen flux is 2.04 percent smaller than the pure-diffusion estimate. No numerical burnout time is determined without the carbon density and operating gas properties.

Transient mass diffusion

Drug capsule

A spherical drug capsule (0.652 cm diameter) has an initial concentration of 68.9 mg/cm3$^3$ Dramamine.

• What is the residual concentration of Dramamine at the centre of the spherical bead after 48 hours?

The capsule is intended to release the drug in the gastrointestinal system. Its nontoxic gelatinous material allows the drug to diffuse out gradually. The drug within the bead is the source of A; the surrounding fluid is the sink.

Data: 𝐷𝐴𝐵$D_{AB}$ (Dramamine in the gel matrix of the capsule) = 3 ×10−7 cm2/s$3\times10^{-7}\ \mathrm{cm^2/s}$ at body temperature of 37 °C$37\ \degree\mathrm{C}$.

Calculation and answer

Figure 5: Drug initially fills the spherical bead uniformly. It diffuses radially to a sink at the surface. Total release requires the concentration averaged over the whole bead.

The spherical diffusion model is the one shown in Figure 4. The maintained sink fixes the surface concentration, while the centre concentration is the requested unknown.

Data

Quantity	Value
Bead diameter	𝑑 =0.652 cm$d=0.652\ \mathrm{cm}$
Bead radius	𝑅𝑏 =0.326 cm$R_b=0.326\ \mathrm{cm}$
Initial drug concentration	𝐶𝑚𝑖 =68.9 mg cm−3$C_{mi}=68.9\ \mathrm{mg\,cm^{-3}}$
Drug diffusivity	𝐷 =3 ×10−7 cm2 s−1$D=3\times10^{-7}\ \mathrm{cm^2\,s^{-1}}$
Duration	𝑡 =48 h =172800 s$t=48\ \mathrm h=172800\ \mathrm s$
Body temperature	𝑇 =37 °C$T=37\ \degree\mathrm C$

Assumptions

• Spherical, stationary gel with constant radius.
• Uniform initial drug concentration.
• Well-stirred sink maintains zero surface concentration.
• No drug reaction within the gel.
• Constant diffusivity.

Calculations

The sink removes drug from the bead surface while internal diffusion supplies it from the centre. A prescribed surface concentration corresponds to infinite mass Biot number. First calculate the Fourier number using the radius:

Fo𝑚=𝐷𝑡𝑅2𝑏=3×10−7(172800)0.3262=0.48779$$\mathrm{Fo}_m=\frac{Dt}{R_b^2} =\frac{3\times10^{-7}(172800)}{0.326^2} =0.48779$$

The Fourier number exceeds 0.2, so the first spherical eigenmode is sufficient. For zero surface concentration the centre ratio is

Θ0=𝐶𝑚𝑐𝐶𝑚𝑖≃2𝑒−𝜋2Fo𝑚$$\Theta_0=\frac{C_{mc}}{C_{mi}}\simeq2e^{-\pi^2\mathrm{Fo}_m}$$

Substitution gives

Θ0≃2𝑒−𝜋2(0.48779)=0.016226$$\Theta_0\simeq2e^{-\pi^2(0.48779)}=0.016226$$

Multiply this ratio by the initial concentration to recover the centre concentration:

𝐶𝑚𝑐=𝐶𝑚𝑖Θ0=68.9(0.016226)=1.1180 mg cm−3$$C_{mc}=C_{mi}\Theta_0 =68.9(0.016226)=1.1180\ \mathrm{mg\,cm^{-3}}$$(19)

As an additional check on total release, the spherical volume-average ratio is ¯¯¯¯¯Θ ≃3Θ0/𝜋2$\bar\Theta\simeq3\Theta_0/\pi^2$. Hence

𝐹=1−¯¯¯¯¯Θ≃1−3(0.016226)𝜋2=0.99507$$F=1-\bar\Theta\simeq1-\frac{3(0.016226)}{\pi^2} =0.99507$$

The centre retains 1.62 percent of its initial concentration while only 0.493 percent of the initial drug remains in the entire bead. The centre has the largest residual concentration, so these two results have the expected order. Finite external resistance or swelling-dependent diffusivity requires a different model.

Answer

After 48 hours, the concentration at the centre is approximately

𝐶𝑚𝑐=1.12 mg cm−3$$C_{mc}=1.12\ \mathrm{mg\,cm^{-3}}$$

The corresponding additional estimate is 99.5 percent of the drug released.

Transient mass diffusion

Case study 2: phosphorus in silicon

Chemical vapour deposition takes place at 1100 °C$1100\ \degree\mathrm C$, where phosphorus diffuses into silicon. The surface concentration is maintained at 𝐶𝐴𝑆 =2.5 ×1020$C_{AS}=2.5\times10^{20}$ phosphorus atoms per cm3$^3$ of solid silicon. For comparison, pure silicon contains about 5 ×1022$5\times10^{22}$ silicon atoms per cm3$^3$, so the phosphorus concentration is dilute.

Treat the phosphorus-rich coating as an effectively infinite source relative to the amount transferred. Predict the depth at which the phosphorus concentration reaches one percent of the surface value after one hour, 2.5 ×1018$2.5\times10^{18}$ phosphorus atoms per cm3$^3$. Also predict the concentration profile after one hour.

Calculation and answer

The coating and diffusion direction are shown in Figure 7. The calculated profile in Figure 8 can be used to check the recovered one-percent depth.

Data

• Surface phosphorus concentration 𝐶𝑛𝑠 =2.5 ×1020 atoms cm−3$C_{ns}=2.5\times10^{20}\ \mathrm{atoms\,cm^{-3}}$

• Initial phosphorus concentration 𝐶𝑛𝑖 =0$C_{ni}=0$

• Diffusivity at 1100 degrees Celsius 𝐷 =6.5 ×10−13 cm2 s−1$D=6.5\times10^{-13}\ \mathrm{cm^2\,s^{-1}}$

• Duration 𝑡 =3600 s$t=3600\ \mathrm s$

• Target concentration

  0.01𝐶𝑛𝑠=2.5×1018 atoms cm−3$$0.01C_{ns}=2.5\times10^{18}\ \mathrm{atoms\,cm^{-3}}$$

Assumptions

• One-dimensional diffusion into stationary silicon.
• Uniform temperature and constant diffusivity.
• No reaction in the silicon.
• An effectively infinite coating maintains the surface concentration.
• The wafer is sufficiently thick for the semi-infinite approximation.

Calculations

Phosphorus enters from the maintained coating and diffuses into the initially phosphorus-free silicon. The semi-infinite diffusion solution on an atomic-concentration basis is

𝐶𝑛(𝑥,𝑡)=𝐶𝑛𝑠erfc⁡(𝑥2√𝐷𝑡)$$C_n(x,t)=C_{ns}\operatorname{erfc}\left(\frac{x}{2\sqrt{Dt}}\right)$$

At the target concentration,

erfc⁡(𝜂)=2.5×10182.5×1020=0.01$$\operatorname{erfc}(\eta)=\frac{2.5\times10^{18}}{2.5\times10^{20}}=0.01$$

Therefore erf⁡(𝜂) =0.99$\operatorname{erf}(\eta)=0.99$, and the inverse error function gives 𝜂 =1.821386$\eta=1.821386$. The source table reading of 1.82 gives the same depth to three significant figures.

The one-hour diffusion length is

2√𝐷𝑡=2√6.5×10−13(3600)=9.67471×10−5 cm$$2\sqrt{Dt}=2\sqrt{6.5\times10^{-13}(3600)} =9.67471\times10^{-5}\ \mathrm{cm}$$

The requested depth follows from the definition of 𝜂$\eta$:

𝑥=2𝜂√𝐷𝑡=1.821386(9.67471×10−5)=1.76214×10−4 cm=1.76214 𝜇m$$\begin{aligned} x&=2\eta\sqrt{Dt} =1.821386(9.67471\times10^{-5}) \\&=1.76214\times10^{-4}\ \mathrm{cm} =1.76214\ \mathrm{\mu m} \end{aligned}$$(26)

Substituting the same diffusion length into the concentration solution gives the complete profile:

𝐶𝑛(𝑥)=2.5×1020erfc⁡(𝑥0.967471 𝜇m) atoms cm−3$$C_n(x)=2.5\times10^{20}\operatorname{erfc} \left(\frac{x}{0.967471\ \mathrm{\mu m}}\right) \ \mathrm{atoms\,cm^{-3}}$$(27)

At 𝑥 =0.967471 𝜇m$x=0.967471\ \mathrm{\mu m}$, the profile gives 3.93248 ×1019 atoms cm−3$3.93248\times10^{19}\ \mathrm{atoms\,cm^{-3}}$. At 𝑥 =2 𝜇m$x=2\ \mathrm{\mu m}$ it gives 8.6523 ×1017 atoms cm−3$8.6523\times10^{17}\ \mathrm{atoms\,cm^{-3}}$. Substitution of the calculated target depth recovers one percent of the surface concentration.

For an illustrative 300 𝜇m$300\ \mathrm{\mu m}$ wafer, the disturbance-depth estimate is 4√𝐷𝑡 =1.93494 𝜇m$4\sqrt{Dt}=1.93494\ \mathrm{\mu m}$ and the thickness-based Fourier number is 2.6 ×10−6$2.6\times10^{-6}$. The far boundary is therefore well beyond the diffusion region. The one-percent criterion defines a concentration depth; an electrical junction depth also requires the wafer’s background dopant concentration.

Answer

After one hour the one-percent concentration depth is 1.76 𝜇m$1.76\ \mathrm{\mu m}$. The one-hour phosphorus concentration profile is given by Eq. 27.

Transient mass diffusion

Concentration, instantaneous flux and total uptake

A stationary polymer initially contains a uniform solute concentration of 0.20 mol m⁻³. At 𝑡 =0$t=0$ one face is held at 2.0 mol m⁻³ by a maintained reservoir. Use constant diffusivity 𝐷𝐴𝐵 =1.0 ×10−10$D_{AB}=1.0\times10^{-10}$ m² s⁻¹, negligible external resistance and no reaction. The polymer is 0.020 m thick and transport is normal to the exposed face; neglect edges.

After one hour, find the concentration 0.60 mm below the surface, the instantaneous surface flux and the cumulative uptake per unit exposed area. Check the semi-infinite approximation, and describe the effect of increasing the duration to four hours.

Solution

Figure 9: Solute enters an initially uniform body from a maintained surface concentration. Position is measured inward from the exposed face.

Data

Quantity	Value
Initial concentration	𝑐𝐴𝑖 =0.20 mol m−3$c_{Ai}=0.20\ \mathrm{mol\,m^{-3}}$
Surface concentration	𝑐𝐴𝑠 =2.0 mol m−3$c_{As}=2.0\ \mathrm{mol\,m^{-3}}$
Diffusivity	𝐷𝐴𝐵 =10−10 m2 s−1$D_{AB}=10^{-10}\ \mathrm{m^2\,s^{-1}}$
Evaluation depth	𝑥 =0.000600 m$x=0.000600\ \mathrm m$
Evaluation time	𝑡 =3600 s$t=3600\ \mathrm s$
Body thickness	𝐻 =0.020 m$H=0.020\ \mathrm m$

Assumptions

• One-dimensional diffusion into a stationary polymer.
• Uniform initial concentration and maintained surface concentration.
• Negligible external resistance and edge effects.
• No reaction and constant diffusivity.

Calculations

At 3600 s, √𝐷𝐴𝐵𝑡 =0.000600$\sqrt{D_{AB}t}=0.000600$ m. The estimated disturbance depth is 4√𝐷𝐴𝐵𝑡 =0.00240$4\sqrt{D_{AB}t}=0.00240$ m, well short of the 0.020 m far boundary. The semi-infinite solution is appropriate on this scale.

At 𝑥 =0.000600$x=0.000600$ m,

𝜂=𝑥2√𝐷𝐴𝐵𝑡=0.5.$$\eta=\frac{x}{2\sqrt{D_{AB}t}}=0.5.$$

The concentration relation is

𝑐𝐴=𝑐𝐴𝑖+(𝑐𝐴𝑠−𝑐𝐴𝑖)erfc⁡(𝜂)$$c_A=c_{Ai}+(c_{As}-c_{Ai})\operatorname{erfc}(\eta)$$

Since erfc⁡(0.5) =0.479500$\operatorname{erfc}(0.5)=0.479500$, substitution gives

𝑐𝐴=0.20+(2.0−0.20)(0.479500)=1.063 mol m−3$$c_A=0.20+(2.0-0.20)(0.479500)=1.063\ \mathrm{mol\,m^{-3}}$$

The instantaneous surface flux follows from the surface gradient:

𝑁𝐴(0,𝑡)=(𝑐𝐴𝑠−𝑐𝐴𝑖)√𝐷𝐴𝐵𝜋𝑡$$N_A(0,t)=(c_{As}-c_{Ai})\sqrt{\frac{D_{AB}}{\pi t}}$$

Substituting the concentration difference and elapsed time into Eq. 23 gives

𝑁𝐴(0,𝑡)=(2.0−0.20)√10−10𝜋(3600)=1.693×10−7 mol m−2 s−1$$N_A(0,t)=(2.0-0.20)\sqrt{\frac{10^{-10}}{\pi(3600)}} =1.693\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

The cumulative uptake is the time integral of surface flux:

𝑞𝐴(𝑡)=2(𝑐𝐴𝑠−𝑐𝐴𝑖)√𝐷𝐴𝐵𝑡𝜋$$q_A(t)=2(c_{As}-c_{Ai})\sqrt{\frac{D_{AB}t}{\pi}}$$

Substituting the same data gives

𝑞𝐴(𝑡)=2(2.0−0.20)√10−10(3600)𝜋=1.219×10−3 mol m−2$$q_A(t)=2(2.0-0.20)\sqrt{\frac{10^{-10}(3600)}\pi} =1.219\times10^{-3}\ \mathrm{mol\,m^{-2}}$$

Multiplying the final flux by the duration would give half this uptake. The surface flux was greater earlier and falls as 𝑡−1/2$t^{-1/2}$.

After four hours, the disturbance depth and cumulative uptake both double, while instantaneous surface flux halves. The disturbance depth is then 0.00480 m, still below the far boundary. At a fixed depth the concentration also changes; only points at equal 𝑥/√𝑡$x/\sqrt t$ have the same concentration ratio. These results require the maintained surface concentration and constant diffusivity specified in the problem.

Answer

After one hour, 𝑐𝐴 =1.06 mol m−3$c_A=1.06\ \mathrm{mol\,m^{-3}}$ at 0.60 mm, 𝑁𝐴(0,𝑡) =1.69 ×10−7 mol m−2 s−1$N_A(0,t)=1.69\times10^{-7}\ \mathrm{mol\,m^{-2}\,s^{-1}}$ and 𝑞𝐴 =1.22 ×10−3 mol m−2$q_A=1.22\times10^{-3}\ \mathrm{mol\,m^{-2}}$. At four hours the uptake and disturbance depth double and the surface flux halves.

Transient mass diffusion

Use Henry’s law to set a transient diffusion boundary

A 0.10 m deep layer of still water at 298 K initially contains dissolved molecular oxygen uniformly at 10 g m⁻³. At 𝑡 =0$t=0$ a regulated pure-oxygen headspace maintains an oxygen partial pressure of 2.0 atm above the water. Use Henry’s law in the form 𝑝𝐴 =𝐻𝑐𝐴𝑠$p_A=Hc_{As}$ with 𝐻 =0.80$H=0.80$ atm m³ mol⁻¹, and 𝐷𝐴𝐵 =2.1 ×10−9$D_{AB}=2.1\times10^{-9}$ m² s⁻¹. The molar mass of molecular oxygen is 32.0 g mol⁻¹.

How long does the concentration 5.0 mm below the interface take to reach 20 g m⁻³? Assume negligible gas-side resistance, rapid interfacial equilibration, constant properties, no reaction and no liquid motion. Check whether a semi-infinite approximation is consistent with the result. For the calculation,

erf⁡(1.03608)=0.857143.$$\operatorname{erf}(1.03608)=0.857143.$$

Solution

Figure 10: Oxygen enters water through the top surface. Henry's law sets the dissolved surface concentration; the target is at a specified depth below that surface.

Data

Quantity	Value
Water depth	𝐿 =0.10 m$L=0.10\ \mathrm m$
Oxygen diffusivity	𝐷𝐴𝐵 =2.1 ×10−9 m2 s−1$D_{AB}=2.1\times10^{-9}\ \mathrm{m^2\,s^{-1}}$
Oxygen partial pressure	𝑝𝐴 =2.0 atm$p_A=2.0\ \mathrm{atm}$
Henry’s constant	𝐻 =0.80 atm m3 mol−1$H=0.80\ \mathrm{atm\,m^3\,mol^{-1}}$
Initial mass concentration	10 g m−3$10\ \mathrm{g\,m^{-3}}$
Target mass concentration	20 g m−3$20\ \mathrm{g\,m^{-3}}$
Evaluation depth	𝑥 =0.0050 m$x=0.0050\ \mathrm m$
Oxygen molar mass	𝑀𝐴 =32.0 g mol−1$M_A=32.0\ \mathrm{g\,mol^{-1}}$

Assumptions

• One-dimensional diffusion in stationary water.
• Uniform initial concentration.
• Maintained gas partial pressure and rapid interfacial equilibrium.
• Negligible gas-side resistance.
• Constant properties and no reaction.

Calculations

The initial concentration and the interfacial equilibrium concentration are different. Henry’s law fixes the latter on a molar basis:

𝑐𝐴𝑠=𝑝𝐴𝐻=2.00.80=2.50 mol m−3=80.0 g m−3$$c_{As}=\frac{p_A}{H}=\frac{2.0}{0.80} =2.50\ \mathrm{mol\,m^{-3}}=80.0\ \mathrm{g\,m^{-3}}$$

The regulated headspace makes this a maintained surface boundary, as required by Eq. 22. The surface value would change if gas were depleted from an unregulated finite headspace.

Using mass concentration 𝐶𝑚$C_m$, which is proportional to molar concentration, the semi-infinite solution is

𝐶𝑚−𝐶𝑚𝑖𝐶𝑚𝑠−𝐶𝑚𝑖=erfc⁡(𝑥2√𝐷𝐴𝐵𝑡)$$\frac{C_m-C_{mi}}{C_{ms}-C_{mi}} =\operatorname{erfc}\left(\frac{x}{2\sqrt{D_{AB}t}}\right)$$

At the target, this ratio is

20−1080−10=17.$$\frac{20-10}{80-10}=\frac{1}{7}.$$

Thus erf⁡(𝜂) =6/7$\operatorname{erf}(\eta)=6/7$ and 𝜂 =1.03608$\eta=1.03608$. Solving for time,

𝑡=𝑥24𝐷𝐴𝐵𝜂2=(0.0050)24(2.1×10−9)(1.03608)2=2773 s=46.2 min$$\begin{aligned} t&=\frac{x^2}{4D_{AB}\eta^2}\\ &=\frac{(0.0050)^2}{4(2.1\times10^{-9})(1.03608)^2}\\ &=2773\ \mathrm s=46.2\ \mathrm{min} \end{aligned}$$

The disturbance-depth estimate is 4√𝐷𝐴𝐵𝑡 =0.00965$4\sqrt{D_{AB}t}=0.00965$ m, much less than the 0.10 m water depth. The far boundary is therefore unimportant on this time scale within the stated diffusion-only model.

At twice the target depth, reaching the same concentration requires four times the duration, about 3.08 hours. Using 16 g mol⁻¹ would describe atomic oxygen and impose the wrong surface concentration for this problem. Natural convection, mixing or biological consumption would require revisiting the stated model rather than changing only the diffusion time.

Answer

The target concentration is reached after approximately 2.77 ×103 s =46.2 min$2.77\times10^3\ \mathrm{s}=46.2\ \mathrm{min}$. The disturbance depth is about 9.65 mm, so the semi-infinite approximation is consistent with the 100 mm water depth.

Convective mass transfer

Volatile solid

Air flows at 2 m s⁻¹ over a hypothetical volatile solid A with an exposed area of 10−3 m2$10^{-3}\ \mathrm{m^2}$. The measured sublimation rate is 2.29 ×10−4 mol s−1$2.29\times10^{-4}\ \mathrm{mol\,s^{-1}}$. Both the air and the surface are maintained at 293 K; external heating supplies the latent heat. The pressure is 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$ and incoming air contains negligible A.

For this model solid, use the stipulated equilibrium vapour pressure 𝑝𝐴,𝑠 =4740 Pa$p_{A,s}=4740\ \mathrm{Pa}$ and diffusivity in air 𝐷𝐴𝐵 =1.5 ×10−5 m2 s−1$D_{AB}=1.5\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$. Use air properties 𝜈 =1.55 ×10−5 m2 s−1$\nu=1.55\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$, Pr =0.71$\mathrm{Pr}=0.71$, 𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$ and molar mass 𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$.

Determine the concentration-basis mass transfer coefficient and estimate the heat transfer coefficient using the Chilton-Colburn analogy. This adaptation replaces the inconsistent CO2$_2$ interface state in slide 17.

Solution

Data

Quantity	Value
Exposed area	𝐴𝑠 =10−3 m2$A_s=10^{-3}\ \mathrm{m^2}$
Measured molar rate	˙𝑛𝐴 =2.29 ×10−4 mol s−1$\dot n_A=2.29\times10^{-4}\ \mathrm{mol\,s^{-1}}$
Gas temperature	𝑇 =293 K$T=293\ \mathrm K$
Gas pressure	𝑃 =101300 Pa$P=101300\ \mathrm{Pa}$
Species diffusivity	𝐷𝐴𝐵 =1.5 ×10−5 m2 s−1$D_{AB}=1.5\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Kinematic viscosity	𝜈 =1.55 ×10−5 m2 s−1$\nu=1.55\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface equilibrium pressure	𝑝𝐴,𝑠 =4740 Pa$p_{A,s}=4740\ \mathrm{Pa}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Specific heat capacity	𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Air molar mass	𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$

These values are stipulated for the hypothetical solid and air.

Assumptions

• Steady sublimation over the exposed area.
• Negligible A in the bulk air.
• Ideal gas behaviour and constant supplied properties.
• Low normal mass flux and matching heat and mass transfer conditions.

Calculations

The gas-side surface concentration follows from ideal-gas equilibrium:

𝑐𝐴,𝑠=𝑝𝐴,𝑠𝑅𝑇=47408.314(293)=1.9458 mol m−3$$c_{A,s}=\frac{p_{A,s}}{RT} =\frac{4740}{8.314(293)}=1.9458\ \mathrm{mol\,m^{-3}}$$

The measured rate equals coefficient times area times concentration driving force:

˙𝑛𝐴=𝑘𝑐𝐴𝑠(𝑐𝐴,𝑠−𝑐𝐴,∞)$$\dot n_A=k_cA_s(c_{A,s}-c_{A,\infty})$$

Since the bulk concentration is negligible, solve for 𝑘𝑐$k_c$.

From Eq. 3,

𝑘𝑐=2.29×10−410−3×1.9458=0.1177 m s−1$$k_c=\frac{2.29\times10^{-4}}{10^{-3}\times1.9458} =0.1177\ \mathrm{m\,s^{-1}}$$

The Schmidt number uses momentum and molecular diffusivities:

Sc=𝜈𝐷𝐴𝐵=1.55×10−51.5×10−5=1.0333$$\mathrm{Sc}=\frac{\nu}{D_{AB}} =\frac{1.55\times10^{-5}}{1.5\times10^{-5}}=1.0333$$𝜌=101300×0.0298.314×293$$\rho=\frac{101300\times0.029}{8.314\times293}$$

𝜌=1.2059 kg m−3$$\rho=1.2059\ \mathrm{kg\,m^{-3}}$$(12)

The Chilton-Colburn relation converts the measured mass coefficient to a heat coefficient:

ℎ=𝜌𝐶𝑝𝑘𝑐(ScPr)2/3$$h=\rho C_pk_c\left(\frac{\mathrm{Sc}}{\mathrm{Pr}}\right)^{2/3}$$

Substituting the air properties gives

ℎ=1.2059(1007)(0.1177)(1.03330.71)2/3=183.55 W m−2 K−1$$h=1.2059(1007)(0.1177)\left(\frac{1.0333}{0.71}\right)^{2/3} =183.55\ \mathrm{W\,m^{-2}\,K^{-1}}$$

The surface fraction is about 0.0468, so the low-flux approximation is plausible but not exact.

External heating maintains the specified surface temperature. The estimated heat coefficient describes how this flow would respond to a small temperature difference; it does not supply latent heat when the air and surface are isothermal.

Answer

The concentration-basis mass transfer coefficient is 𝑘𝑐 ≃0.118 m s−1$k_c\simeq0.118\ \mathrm{m\,s^{-1}}$. The corresponding heat transfer coefficient is ℎ ≃184 W m−2 K−1$h\simeq184\ \mathrm{W\,m^{-2}\,K^{-1}}$.

Convective mass transfer

Naphthalene sheet

A stream of air at 100 kPa pressure and 290 K is flowing on the top surface of a thin flat sheet of solid naphthalene of length 1 m with a velocity of 20 m/s and surface temperature of 310K. Calculate:

• The average mass transfer coefficient over the flat plate
• The rate of loss of naphthalene from the surface per unit width

Data @ 𝑇𝑓 =300 K$T_f=300\ \mathrm{K}$:

• 𝐷𝐴𝐵 =6 ×10−6 m2/s$D_{AB}=6\times10^{-6}\ \mathrm{m^2/s}$ (naphthalene vapour in air)
• 𝜈 =1.5 ×10−5 m2/s$\nu=1.5\times10^{-5}\ \mathrm{m^2/s}$ (kinematic viscosity of air)
• 𝐶𝐴𝑠 =1 ×10−5 kmol/m3$C_{As}=1\times10^{-5}\ \mathrm{kmol/m^3}$ (conc. of naphthalene at the air-solid interface)

Solution

Figure 3: Air flows along the naphthalene sheet. The velocity boundary layer is the same type as for a heated plate; the concentration boundary layer replaces the thermal boundary layer in the mass transfer analogy.

Use the concentration boundary layer in Figure 2, with the leading edge at the start of the sheet. The one-metre dimension is along the flow; width remains unspecified.

Data

• Sheet length 𝐿 =1 m$L=1\ \mathrm m$

• Air speed 𝑈 =20 m s−1$U=20\ \mathrm{m\,s^{-1}}$

• Film diffusivity 𝐷𝐴𝐵 =6 ×10−6 m2 s−1$D_{AB}=6\times10^{-6}\ \mathrm{m^2\,s^{-1}}$

• Film kinematic viscosity 𝜈 =1.5 ×10−5 m2 s−1$\nu=1.5\times10^{-5}\ \mathrm{m^2\,s^{-1}}$

• Surface concentration

  𝑐𝐴,𝑠=10−5 kmol m−3=0.0100 mol m−3$$c_{A,s}=10^{-5}\ \mathrm{kmol\,m^{-3}}=0.0100\ \mathrm{mol\,m^{-3}}$$

Assumptions

• Steady transfer from a uniform saturated surface.
• Negligible naphthalene in the incoming air.
• Constant properties and negligible normal injection.
• The source estimate treats the boundary layer as turbulent from the leading edge. A separate estimate allows natural transition.

Calculations

Use length-based flow and transfer groups:

Re𝐿=𝑈𝐿𝜈,Sc=𝜈𝐷𝐴𝐵$$\mathrm{Re}_L=\frac{UL}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}$$

The groups are

Re𝐿=201.5×10−5=1.3333×106$$\mathrm{Re}_L=\frac{20}{1.5\times10^{-5}}=1.3333\times10^6$$

and Sc =2.5$\mathrm{Sc}=2.5$.

For the source’s fully turbulent model, first determine the average Sherwood number, then recover the coefficient from

¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿.$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L}.$$

The transfer area per unit width is 𝐿$L$, so the loss per unit width is

˙𝑛𝐴𝑏=¯¯¯¯𝑘𝑐𝐿(𝑐𝐴,𝑠−𝑐𝐴,∞).$$\frac{\dot n_A}{b}=\bar k_cL(c_{A,s}-c_{A,\infty}).$$

Substitution gives

¯¯¯¯¯¯¯Sh𝐿=0.0366Re0.8𝐿Sc1/3$$\overline{\mathrm{Sh}}_L=0.0366\mathrm{Re}_L^{0.8}\mathrm{Sc}^{1/3}$$¯¯¯¯¯¯¯Sh𝐿=3945.3$$\overline{\mathrm{Sh}}_L=3945.3$$

Use the same one-metre length in the dimensional conversion:

¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿=3945.3(6×10−6)1.0=0.02367 m s−1$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L} =\frac{3945.3(6\times10^{-6})}{1.0}=0.02367\ \mathrm{m\,s^{-1}}$$˙𝑛𝐴width=¯¯¯¯𝑘𝑐𝐿𝑐𝐴,𝑠$$\frac{\dot n_A}{\text{width}}=\bar k_c Lc_{A,s}$$

˙𝑛𝐴𝑏=2.367×10−4 mol s−1 m−1$$\frac{\dot n_A}{b}=2.367\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$$(13)

Here 𝑏$b$ denotes sheet width in metres.

The table uses 0.037, which changes this estimate by about one percent.

For an untripped plate, the laminar starting length is

𝑥𝑡𝑟=5×105𝜈𝑈=0.375 m.$$x_{tr}=\frac{5\times10^5\nu}{U}=0.375\,\mathrm{m}.$$

The mixed average relation

¯¯¯¯¯¯¯Sh𝐿=(0.037Re0.8𝐿−871)Sc1/3$$\overline{\mathrm{Sh}}_L=(0.037\mathrm{Re}_L^{0.8}-871)\mathrm{Sc}^{1/3}$$

gives 2806.4, ¯¯¯¯𝑘𝑐 =0.01684$\bar k_c=0.01684$ m s⁻¹ and ˙𝑛𝐴/width =1.684 ×10−4$\dot n_A/\text{width}=1.684\times10^{-4}$ mol s⁻¹ m⁻¹.

This is the preferred estimate when transition occurs naturally at the stated Reynolds number.

Both answers are conditional on the specified flow model; they are not interchangeable rounding choices.

With 𝑀𝐴 =0.128$M_A=0.128$ kg mol⁻¹ the source fully turbulent estimate corresponds to 2.62 kg day⁻¹ per metre of width.

Answer

With the source’s fully turbulent model, ¯¯¯¯𝑘𝑐 =0.0237 m s−1$\bar k_c=0.0237\ \mathrm{m\,s^{-1}}$ and the loss per unit width is 2.37 ×10−4 mol s−1 m−1$2.37\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$. For an untripped plate with natural transition, the preferred estimates are 0.0168 m s−1$0.0168\ \mathrm{m\,s^{-1}}$ and 1.68 ×10−4 mol s−1 m−1$1.68\times10^{-4}\ \mathrm{mol\,s^{-1}\,m^{-1}}$.

Convective mass transfer

Wet-bulb thermometer

Dry air at 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$ blows across a thermometer whose bulb has been covered with a dampened wick. The classical “wet bulb” thermometer indicates a steady-state temperature is reached by a small amount of liquid water evaporating into a large reservoir of unsaturated vapour-gas mixture. The thermometer reads 290 K. At this temperature, the following properties were evaluated:

Property	Value
Vapour pressure of water	1.94 ×103 Pa$1.94\times10^3\ \mathrm{Pa}$
Density of air	1.219 kg/m3$^3$
Latent heat of vaporisation of water	2461 kJ/kg
Prandtl number	0.71
Schmidt number	0.61
Specific heat of air	1.006 kJ/kg.K

What is the temperature of the dry air?

Solution

Data

Quantity	Value
Wet-bulb temperature	𝑇𝑠 =290 K$T_s=290\ \mathrm K$
Surface vapour pressure	𝑝𝐴,𝑠 =1940 Pa$p_{A,s}=1940\ \mathrm{Pa}$
Air density	𝜌 =1.219 kg m−3$\rho=1.219\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1006 J kg−1 K−1$C_p=1006\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Schmidt number	Sc =0.61$\mathrm{Sc}=0.61$
Latent heat	𝜆 =2.461 ×106 J kg−1$\lambda=2.461\times10^6\ \mathrm{J\,kg^{-1}}$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

Assumptions

• Steady wick temperature and evaporation rate.
• Dry incoming air and saturated gas at the wet surface.
• Radiation and conduction through the thermometer stem are neglected.
• Sensible energy transport by injected vapour and blowing are neglected.
• Constant properties and matching heat and mass transfer conditions.

Calculations

At steady state convection supplies evaporation’s latent heat.

With zero incoming vapour,

ℎ(𝑇∞−𝑇𝑠)=𝜆𝑀𝐴𝑘𝑐𝑐𝐴,𝑠$$h(T_\infty-T_s)=\lambda M_A k_c c_{A,s}$$

𝑘𝑐ℎ=1𝜌𝐶𝑝(PrSc)2/3$$\frac{k_c}{h}=\frac1{\rho C_p} \left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3}$$(14)

The ratio is 9.023 ×10−4$9.023\times10^{-4}$ m3$^3$ K/J, and

𝑐𝐴,𝑠=19408.314×290=0.80465 mol m−3.$$c_{A,s}=\frac{1940}{8.314\times290}=0.80465\,\mathrm{mol\,m^{-3}}.$$

The energy balance gives the temperature rise above the wet bulb:

𝑇∞−𝑇𝑠=𝜆𝑀𝐴𝑐𝐴,𝑠𝑘𝑐ℎ$$T_\infty-T_s=\lambda M_Ac_{A,s}\frac{k_c}{h}$$

Substituting the latent heat, concentration and coefficient ratio gives

𝑇∞−𝑇𝑠=2.461×106(0.018)(0.80465)(9.023×10−4)=32.161 K$$T_\infty-T_s=2.461\times10^6(0.018)(0.80465)(9.023\times10^{-4}) =32.161\ \mathrm{K}$$

Adding this temperature difference to the wet-bulb reading gives

𝑇∞=290+32.161=322.16 K$$T_\infty=290+32.161=322.16\ \mathrm{K}$$

Velocity cancels because the same flow affects both coefficients.

This cancellation holds under the stated analogy and energy assumptions; it does not remove ventilation requirements or radiative errors of a real thermometer.

Answer

The estimated dry-air temperature is 𝑇∞ =322.2 K$T_\infty=322.2\ \mathrm{K}$, approximately 49.0 °C$49.0\,\degree\mathrm{C}$.

Convective mass transfer

Heat and mass analogy

Water evaporates from a wetted surface of rectangular shape, with an air stream (1 atm, 35 °C$35\ \degree\mathrm{C}$) blown over the surface at a velocity of 15 m/s. Heat transfer measurements indicate that for air at 1 atm and 35 °C$35\ \degree\mathrm{C}$, the average heat transfer coefficient is given by the following empirical relation:

⟨ℎ⟩=21𝑉0.6∞$$\langle h\rangle=21V_\infty^{0.6}$$

where ⟨ℎ⟩$\langle h\rangle$ is in W/(m2$^2$ K) and 𝑉∞$V_\infty$ is in m/s. Estimate the mass transfer coefficient and evaporation rate for an area of 1.5 m2$^2$. Assume dry incoming air and a surface maintained at 35 °C$35\ \degree\mathrm C$ by external heating. Use 𝑝sat𝐴 =5624 Pa$p_A^{\mathrm{sat}}=5624\ \mathrm{Pa}$ at 𝑇𝑠 =308.15 K$T_s=308.15\ \mathrm K$ and the following rounded, stipulated transport properties: 𝜌 =1.128 kg/m3$\rho=1.128\ \mathrm{kg/m^3}$, 𝐶𝑝 =1007 J/(kg K)$C_p=1007\ \mathrm{J/(kg\,K)}$, 𝜈 =15.89 ×10−6 m2/s$\nu=15.89\times10^{-6}\ \mathrm{m^2/s}$, 𝐷𝐴𝐵 =2.6 ×10−5 m2/s$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2/s}$ and Pr =0.71$\mathrm{Pr}=0.71$.

Solution

Data

Quantity	Value
Air speed	𝑈 =15 m s−1$U=15\ \mathrm{m\,s^{-1}}$
Wet area	𝐴𝑠 =1.5 m2$A_s=1.5\ \mathrm{m^2}$
Supplied heat correlation	¯¯¯¯ℎ =21𝑈0.6$\bar h=21U^{0.6}$
Air density	𝜌 =1.128 kg m−3$\rho=1.128\ \mathrm{kg\,m^{-3}}$
Specific heat capacity	𝐶𝑝 =1007 J kg−1 K−1$C_p=1007\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Kinematic viscosity	𝜈 =15.89 ×10−6 m2 s−1$\nu=15.89\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.6 ×10−5 m2 s−1$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Prandtl number	Pr =0.71$\mathrm{Pr}=0.71$
Surface saturation pressure	𝑝sat𝐴 =5624 Pa$p_A^{\mathrm{sat}}=5624\ \mathrm{Pa}$
Surface temperature	𝑇𝑠 =308.15 K$T_s=308.15\ \mathrm K$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

The transport properties are rounded model inputs. The surface saturation pressure is consistent with 35 degrees Celsius using the NIST Antoine correlation.

Assumptions

• Steady transfer from a uniformly wet surface.
• Dry incoming air.
• The surface is maintained at the stated air temperature.
• Constant supplied properties.
• Heat and mass transfer share the same flow and surface conditions.
• Normal evaporation flow is small.

Calculations

The measurement gives

¯¯¯¯ℎ=21×150.6=106.63 W m−2 K−1.$$\bar h=21\times15^{0.6}=106.63\,\mathrm{W\,m^{-2}\,K^{-1}}.$$

Compute Sc =0.61115$\mathrm{Sc}=0.61115$.

The matching-flow Chilton-Colburn relation is

¯¯¯¯𝑘𝑐=¯¯¯¯ℎ𝜌𝐶𝑝(PrSc)2/3$$\bar k_c=\frac{\bar h}{\rho C_p}\left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3}$$

Substituting the coefficient and property values gives

¯¯¯¯𝑘𝑐=106.631.128×1007(0.710.61115)2/3¯¯¯¯𝑘𝑐=0.10374 m s−1$$\begin{gathered} \bar k_c=\frac{106.63}{1.128\times1007} \left(\frac{0.71}{0.61115}\right)^{2/3} \\ \bar k_c=0.10374\ \mathrm{m\,s^{-1}} \end{gathered}$$(15)

At 308.15 K,

𝑐𝐴,𝑠=56248.314×308.15=2.1952 mol m−3.$$c_{A,s}=\frac{5624}{8.314\times308.15}=2.1952\,\mathrm{mol\,m^{-3}}.$$

The evaporation rate is coefficient times area times concentration difference:

˙𝑛𝐴=¯¯¯¯𝑘𝑐𝐴𝑠(𝑐𝐴,𝑠−𝑐𝐴,∞)$$\dot n_A=\bar k_cA_s(c_{A,s}-c_{A,\infty})$$

With dry bulk air, substitution gives

˙𝑛𝐴=0.10374(1.5)(2.1952)=0.3416 mol s−1$$\dot n_A=0.10374(1.5)(2.1952)=0.3416\ \mathrm{mol\,s^{-1}}$$

Convert the molar rate to mass rate:

˙𝑚𝐴=𝑀𝐴˙𝑛𝐴=0.018(0.3416)=0.00615 kg s−1=22.1 kg h−1$$\dot m_A=M_A\dot n_A=0.018(0.3416) =0.00615\ \mathrm{kg\,s^{-1}}=22.1\ \mathrm{kg\,h^{-1}}$$

Dryness and maintained surface temperature are explicit additions to the source question.

At 50 percent relative humidity with equal surface and ambient temperature the driving force and rate halve.

The saturated surface fraction is about 0.0555, exceeding the source’s 0.05 rule of thumb; a blowing correction may matter.

Maintaining the warm surface requires heat input.

This is not an adiabatic wet-bulb calculation.

Answer

Using the supplied additional properties, ¯¯¯¯𝑘𝑐 ≃0.104 m s−1$\bar k_c\simeq0.104\ \mathrm{m\,s^{-1}}$ and

˙𝑚𝐴≃0.00615 kg s−1=22.1 kg h−1.$$\dot m_A\simeq0.00615\ \mathrm{kg\,s^{-1}}=22.1\ \mathrm{kg\,h^{-1}}.$$

These estimates require the stated dry-air and maintained-surface-temperature assumptions.

Convective mass transfer

Annular transfer

The following correlation has been proposed to characterise the heat transfer of fluids flowing in the annular section between concentric tubes:

Nu𝑑1=ℎ𝑑1𝑘=𝐶Re0.8𝑑1Pr1/3(𝜇𝜇𝑠)0.14(𝑑2𝑑1)0.15$$\mathrm{Nu}_{d_1}=\frac{hd_1}{k} =C\mathrm{Re}_{d_1}^{0.8}\mathrm{Pr}^{1/3} \left(\frac{\mu}{\mu_s}\right)^{0.14} \left(\frac{d_2}{d_1}\right)^{0.15}$$(16)

Show how this equation can be used to predict the mass-transfer coefficient from a naphthalene rod that is used to form the inner tube of an annular duct. 𝑑1$d_1$ and 𝑑2$d_2$ are the diameters of the inner and outer tubes, respectively, whereas 𝜇$\mu$ and 𝜇𝑠$\mu_s$ are the viscosity of the fluid at the bulk and surface temperatures.

Solution

Data

• Inner diameter 𝑑1$d_1$
• Outer diameter 𝑑2$d_2$
• Reynolds number based on inner diameter Re𝑑1$\mathrm{Re}_{d_1}$
• Empirical correlation constant 𝐶$C$

No numerical geometry, velocity or value of the correlation constant is supplied. Here the symbol for the correlation constant does not denote concentration.

Assumptions

• The heat and mass problems have matching geometry and flow regime.
• Uniform surface concentration corresponds to the heat correlation’s wall condition.
• Low mass flux leaves the velocity field essentially unchanged.
• The gas is dilute and isothermal, so wall and bulk viscosities are approximately equal.
• The translated correlation is used within the applicable Reynolds and Schmidt ranges.

Calculations

Keep the same geometry, Reynolds definition and experimental range.

Replace Nusselt by Sherwood and Prandtl by Schmidt.

At isothermal dilute conditions 𝜇𝑠 ≈𝜇$\mu_s\approx\mu$, giving

Sh𝑑1=𝐶Re0.8𝑑1Sc1/3(𝑑2𝑑1)0.15$$\mathrm{Sh}_{d_1}=C\mathrm{Re}_{d_1}^{0.8}\mathrm{Sc}^{1/3}\left(\frac{d_2}{d_1}\right)^{0.15}$$

𝑘𝑐=𝐷𝐴𝐵𝑑1Sh𝑑1$$k_c=\frac{D_{AB}}{d_1}\mathrm{Sh}_{d_1}$$(17)

The viscosity factor can be retained if wall temperature or composition changes viscosity appreciably, but the heat correlation alone does not validate a mass-composition correction.

The Reynolds and diameter-ratio factors describe flow and geometry, so they remain unchanged. The viscosity correction becomes unity under the stated isothermal dilute approximation.

Answer

For the stated isothermal, dilute model,

𝑘𝑐=𝐷𝐴𝐵𝑑1𝐶Re0.8𝑑1Sc1/3(𝑑2𝑑1)0.15$$k_c=\frac{D_{AB}}{d_1}C\mathrm{Re}_{d_1}^{0.8}\mathrm{Sc}^{1/3} \left(\frac{d_2}{d_1}\right)^{0.15}$$

The answer remains symbolic because the required numerical data are not supplied.

Convective mass transfer

Wet felt pipe

Dry air (310 K, 1.013 ×105 Pa$1.013\times10^5\ \mathrm{Pa}$, 1.5 m/s) is flowing through a pipe lined with a felt material that is continuously saturated with water at 290 K. The pipe is 0.15 m in diameter and 6 m long. The saturated felt material has a roughness index, 𝑒/𝐷$e/D$, of 0.00001. Assuming that the temperature of the air and pipe wall is constant, determine the rate at which water must be added to keep the felt continuously saturated.

Solution

The inlet, outlet and wetted wall are shown in Figure 4. Concentration rises along the pipe, so the downstream driving force is smaller than the inlet driving force.

Data

Quantity	Value
Pipe diameter	𝑑 =0.15 m$d=0.15\ \mathrm m$
Pipe length	𝐿 =6 m$L=6\ \mathrm m$
Bulk air speed	𝑈 =1.5 m s−1$U=1.5\ \mathrm{m\,s^{-1}}$
Dry inlet temperature	𝑇𝑖 =310 K$T_i=310\ \mathrm K$
Wet wall temperature	𝑇𝑠 =290 K$T_s=290\ \mathrm K$
Surface vapour pressure	𝑝𝐴,𝑠 =1917 Pa$p_{A,s}=1917\ \mathrm{Pa}$
Kinematic viscosity	𝜈 =1.569 ×10−5 m2 s−1$\nu=1.569\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.626 ×10−5 m2 s−1$D_{AB}=2.626\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Fanning friction factor	𝐶𝑓 =0.007$C_f=0.007$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$

The vapour pressure, transport properties and friction factor come from the lecturer’s notes.

Assumptions

• Steady flow and evaporation.
• Constant bulk velocity and supplied properties.
• Uniform saturated wall concentration.
• Each cross-section has a representative mixed bulk concentration.
• Axial diffusion and normal-flow effects are neglected.
• A developed-flow average coefficient approximates the pipe transfer.

Calculations

The surface concentration is

𝑐𝐴,𝑠=19178.314×290=0.79508 mol m−3,$$c_{A,s}=\frac{1917}{8.314\times290}=0.79508\,\mathrm{mol\,m^{-3}},$$

and 𝑐𝐴,𝑖 =0$c_{A,i}=0$.

The groups are

Re𝑑=1.5×0.15𝜈=14340$$\mathrm{Re}_d=\frac{1.5\times0.15}{\nu}=14340$$

and Sc =0.59749$\mathrm{Sc}=0.59749$.

The pipe spans 40 diameters; the roughly ten-diameter turbulent development estimate motivates a developed average approximation, not exact development at the inlet.

The concentration model treats the maintained wall and gas temperatures through the supplied surface concentration and constant properties. It does not solve the axial thermal field. The friction analogy gives the coefficient as

𝑘𝑐=𝐶𝑓𝑈2Sc2/3=0.0074007 m s−1$$k_c=\frac{C_f U}{2\mathrm{Sc}^{2/3}} =0.0074007\ \mathrm{m\,s^{-1}}$$(19)

The source table’s Dittus-Boelter-type mass row gives about 0.00688 m s⁻¹, but Sc =0.5975$\mathrm{Sc}=0.5975$ falls below its stated 0.7 limit.

That comparison is an extrapolation and is not the governing answer.

The primary friction estimate also sits at the approximate lower edge of the source’s Chilton-Colburn range; treat the result as an engineering estimate.

For a slice 𝑑𝑥$dx$, steady balance is inflow minus outflow plus wall input equals zero.

Inflow is 𝑈𝐴𝑐𝑐𝐴(𝑥)$U A_c c_A(x)$, outflow is 𝑈𝐴𝑐𝑐𝐴(𝑥 +𝑑𝑥)$U A_c c_A(x+dx)$, and wall input is

𝑘𝑐𝜋𝑑 𝑑𝑥(𝑐𝐴,𝑠−𝑐𝐴),$$k_c\pi d\,dx(c_{A,s}-c_A),$$

where 𝐴𝑐 =𝜋𝑑2/4$A_c=\pi d^2/4$.

Dividing and taking the limit yields

𝑈𝑑4𝑑𝑐𝐴𝑑𝑥=𝑘𝑐(𝑐𝐴,𝑠−𝑐𝐴)$$\frac{Ud}{4}\frac{dc_A}{dx}=k_c(c_{A,s}-c_A)$$(20)

Integrate from inlet to exit.

For this circular pipe the Number of transfer units is

𝒩=4𝑘𝑐𝐿𝑈𝑑=𝑘𝑐𝐴𝑠𝑄,$$\mathcal N=\frac{4k_cL}{Ud}=\frac{k_cA_s}{Q},$$

using 𝐴𝑠 =𝜋𝑑𝐿$A_s=\pi dL$ and 𝑄 =𝑈𝜋𝑑2/4$Q=U\pi d^2/4$. Small 𝒩$\mathcal N$ leaves the gas close to its inlet composition; large 𝒩$\mathcal N$ brings it towards the maintained wall value. For 𝑘𝑐 =0.01 m s−1$k_c=0.01\ \mathrm{m\,s^{-1}}$, 𝐿 =1 m$L=1\ \mathrm m$, 𝑈 =2 m s−1$U=2\ \mathrm{m\,s^{-1}}$ and 𝑑 =0.02 m$d=0.02\ \mathrm m$, 𝒩 =1$\mathcal N=1$ and the remaining concentration difference is 𝑒−1 =0.368$e^{-1}=0.368$ of its inlet value. Omitting the factor four would underpredict the contact effect.

Then

𝑐𝐴,𝑠−𝑐𝐴,𝑒𝑐𝐴,𝑠−𝑐𝐴,𝑖=𝑒−𝒩𝒩=0.78941𝑐𝐴,𝑒=0.4340 mol m−3$$\begin{gathered} \frac{c_{A,s}-c_{A,e}}{c_{A,s}-c_{A,i}}=e^{-\mathcal N}\\ \mathcal N=0.78941\\ c_{A,e}=0.4340\ \mathrm{mol\,m^{-3}} \end{gathered}$$(21)

For the dry inlet, the integrated balance can be evaluated directly:

𝒩=4(0.0074007)(6)1.5(0.15)=0.78941$$\mathcal N=\frac{4(0.0074007)(6)}{1.5(0.15)}=0.78941$$𝑐𝐴,𝑒=𝑐𝐴,𝑠−(𝑐𝐴,𝑠−𝑐𝐴,𝑖)𝑒−𝒩=0.79508(1−𝑒−0.78941)=0.4340 mol m−3$$c_{A,e}=c_{A,s}-(c_{A,s}-c_{A,i})e^{-\mathcal N} =0.79508(1-e^{-0.78941})=0.4340\ \mathrm{mol\,m^{-3}}$$

The exit is about 54.6 percent of surface saturation on this concentration basis.

Let Δ𝑐𝑖 =𝑐𝐴,𝑠 −𝑐𝐴,𝑖$\Delta c_i=c_{A,s}-c_{A,i}$ and Δ𝑐𝑒 =𝑐𝐴,𝑠 −𝑐𝐴,𝑒$\Delta c_e=c_{A,s}-c_{A,e}$.

The logarithmic mean is

Δ𝑐𝑙𝑚=Δ𝑐𝑖−Δ𝑐𝑒ln⁡(Δ𝑐𝑖Δ𝑐𝑒)Δ𝑐𝑙𝑚=0.54981 mol m−3$$\begin{gathered} \Delta c_{lm}=\frac{\Delta c_i-\Delta c_e}{\ln\left(\frac{\Delta c_i}{\Delta c_e}\right)}\\ \Delta c_{lm}=0.54981\ \mathrm{mol\,m^{-3}} \end{gathered}$$(22)

It is the exact mean driving force for this exponential model.

At equal end differences its limiting value is that common difference.

Apply the mean driving force over wall area 𝜋𝑑𝐿$\pi dL$:

˙𝑛𝐴=𝑘𝑐𝜋𝑑𝐿Δ𝑐𝑙𝑚=0.0074007𝜋(0.15)(6)(0.54981)=0.01150 mol s−1$$\dot n_A=k_c\pi dL\Delta c_{lm} =0.0074007\pi(0.15)(6)(0.54981) =0.01150\ \mathrm{mol\,s^{-1}}$$

An independent overall balance gives

˙𝑛𝐴=𝑈𝐴𝑐(𝑐𝐴,𝑒−𝑐𝐴,𝑖)=1.5𝜋(0.15)24(0.4340−0)=0.01150 mol s−1$$\dot n_A=UA_c(c_{A,e}-c_{A,i}) =1.5\frac{\pi(0.15)^2}{4}(0.4340-0) =0.01150\ \mathrm{mol\,s^{-1}}$$

Agreement confirms that the wall-transfer calculation and outlet concentration satisfy the same material balance.

Thus the final replenishment rate is

˙𝑚𝐴=0.018˙𝑛𝐴=2.07×10−4 kg s−1,$$\dot m_A=0.018\dot n_A=2.07\times10^{-4}\,\mathrm{kg\,s^{-1}},$$

or 0.745 kg h⁻¹.

Answer

The required water replenishment rate is approximately 2.07 ×10−4 kg s−1$2.07\times10^{-4}\ \mathrm{kg\,s^{-1}}$, or 0.745 kg h−1$0.745\ \mathrm{kg\,h^{-1}}$. The exit bulk concentration is 0.434 mol m−3$0.434\ \mathrm{mol\,m^{-3}}$.

Convective mass transfer

Warm water pan

A 225 mm diameter pan of water is maintained at 37 °C$37\ \degree\mathrm{C}$ is exposed to a dry, still atmosphere at 17 °C$17\ \degree\mathrm{C}$. Estimate the heat and mass transfer coefficients, as well as the evaporation rate from the pan,

a. Assuming the convection is driven by species transfer; b. Assuming that the convection is driven by temperature gradients.

Solution

Data

Quantity	Value
Pan diameter	𝑑 =0.225 m$d=0.225\ \mathrm m$
Rounded surface temperature	𝑇𝑠 =310 K$T_s=310\ \mathrm K$
Rounded air temperature	𝑇∞ =290 K$T_\infty=290\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Gravitational acceleration	𝑔 =9.81 m s−2$g=9.81\ \mathrm{m\,s^{-2}}$
Kinematic viscosity	𝜈 =15.89 ×10−6 m2 s−1$\nu=15.89\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.6 ×10−5 m2 s−1$D_{AB}=2.6\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Prandtl number	Pr =0.707$\mathrm{Pr}=0.707$
Specific heat capacity	𝐶𝑝 =1006 J kg−1 K−1$C_p=1006\ \mathrm{J\,kg^{-1}\,K^{-1}}$
Air conductivity	𝑘 =0.0263 W m−1 K−1$k=0.0263\ \mathrm{W\,m^{-1}\,K^{-1}}$
Film density	𝜌 =1.1769 kg m−3$\rho=1.1769\ \mathrm{kg\,m^{-3}}$
Expansion coefficient	𝛽 =1/300 K−1$\beta=1/300\ \mathrm{K^{-1}}$
Surface saturation pressure	𝑝sat𝐴 =6220 Pa$p_A^{\mathrm{sat}}=6220\ \mathrm{Pa}$
Water molar mass	𝑀𝐴 =0.018 kg mol−1$M_A=0.018\ \mathrm{kg\,mol^{-1}}$
Air molar mass	𝑀𝐵 =0.029 kg mol−1$M_B=0.029\ \mathrm{kg\,mol^{-1}}$

Use the source’s rounded temperatures and the supplementary properties from the lecturer’s notes.

Assumptions

• Steady natural convection above an upward-facing wet surface.
• Dry bulk air and saturated gas at the water surface.
• Uniform maintained surface temperature.
• Constant supplied film properties and ideal gas behaviour.
• Low normal evaporation flux.
• Composition buoyancy and thermal buoyancy are considered separately, as requested.

Calculations

Area is

𝐴𝑠=𝜋(0.225)24=0.039761 m2$$A_s=\frac{\pi(0.225)^2}{4}=0.039761\,\mathrm{m^2}$$

and horizontal characteristic length is

𝐿=𝐴𝑠𝑃𝑠=𝑑4=0.05625 m.$$L=\frac{A_s}{P_s}=\frac{d}{4}=0.05625\,\mathrm{m}.$$

Using diameter instead gives a Grashof number 64 times too large.

The surface water mass concentration is

𝜌𝐴,𝑠=𝑝𝐴,𝑠𝑀𝐴𝑅𝑇𝑠=0.043440 kg m−3$$\rho_{A,s}=\frac{p_{A,s}M_A}{RT_s}=0.043440\,\mathrm{kg\,m^{-3}}$$

and bulk water concentration is zero.

The Schmidt number is 0.61115.

For composition alone compare dry and saturated air at the same 310 K.

The densities are

𝜌𝐵,310=𝑃𝑀𝐵𝑅𝑇𝑠=1.1401$$\rho_{B,310}=\frac{PM_B}{RT_s}=1.1401$$

and

𝜌𝑠,310=𝜌𝐴,𝑠+𝜌𝐵,𝑠$$\rho_{s,310}=\rho_{A,s}+\rho_{B,s}$$𝜌𝐴,𝑠=6220(0.018)8.314(310)$$\rho_{A,s}=\frac{6220(0.018)}{8.314(310)}$$𝜌𝐵,𝑠=(101325−6220)(0.029)8.314(310)$$\rho_{B,s}=\frac{(101325-6220)(0.029)}{8.314(310)}$$

𝜌𝑠,310=1.1136 kg m−3$$\rho_{s,310}=1.1136\ \mathrm{kg\,m^{-3}}$$(23)

Thus Δ𝜌𝐷 =0.02655$\Delta\rho_D=0.02655$ kg m⁻³.

Use their mean density 1.12685 kg m⁻³ in Eq. 18.

The composition Grashof number is

Gr𝐷=𝑔Δ𝜌𝐷𝐿3𝜌𝜈2$$\mathrm{Gr}_D=\frac{g\Delta\rho_DL^3}{\rho\nu^2}$$

Substituting the same-temperature density difference gives

Gr𝐷=9.81(0.02655)(0.05625)31.12685(15.89×10−6)2=1.6291×105$$\mathrm{Gr}_D=\frac{9.81(0.02655)(0.05625)^3} {1.12685(15.89\times10^{-6})^2} =1.6291\times10^5$$

The composition Rayleigh number is Ra𝐷 =Gr𝐷Sc =9.9562 ×104$\mathrm{Ra}_D=\mathrm{Gr}_D\mathrm{Sc}=9.9562\times10^4$, within the specified horizontal-plate range.

The correlation gives

Sh=0.54Ra1/4𝐷=0.54(9.9562×104)1/4=9.5922$$\mathrm{Sh}=0.54\mathrm{Ra}_D^{1/4}=0.54(9.9562\times10^4)^{1/4}=9.5922$$𝑘𝑐=Sh𝐷𝐴𝐵𝐿=9.5922(2.6×10−5)0.05625=0.0044337 m s−1$$k_c=\frac{\mathrm{Sh}D_{AB}}L=\frac{9.5922(2.6\times10^{-5})}{0.05625}=0.0044337\ \mathrm{m\,s^{-1}}$$˙𝑚𝐴=𝑘𝑐𝐴𝑠𝜌𝐴,𝑠=0.0044337(0.039761)(0.043440)$$\dot m_A=k_cA_s\rho_{A,s}=0.0044337(0.039761)(0.043440)$$

˙𝑚𝐴=7.6580×10−6 kg s−1$$\dot m_A=7.6580\times10^{-6}\ \mathrm{kg\,s^{-1}}$$(24)

The composition-only estimate is 27.6 g h⁻¹.

Under the matching-flow transfer analogy, the corresponding heat coefficient is

ℎ=𝜌𝐶𝑝𝑘𝑐(ScPr)2/3=1.1769(1006)(0.0044337)(0.611150.707)2/3≃4.76 W m−2 K−1$$h=\rho C_pk_c\left(\frac{\mathrm{Sc}}{\mathrm{Pr}}\right)^{2/3} =1.1769(1006)(0.0044337) \left(\frac{0.61115}{0.707}\right)^{2/3} \simeq4.76\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For thermal buoyancy alone, use the thermal form of Eq. 18:

Gr𝑇=𝑔𝛽(𝑇𝑠−𝑇∞)𝐿3𝜈2$$\mathrm{Gr}_T=\frac{g\beta(T_s-T_\infty)L^3}{\nu^2}$$

Substitution gives

Gr𝑇=9.81(1300)(310−290)(0.05625)3(15.89×10−6)2=4.6100×105$$\mathrm{Gr}_T=\frac{9.81\left(\frac{1}{300}\right)(310-290)(0.05625)^3} {(15.89\times10^{-6})^2}=4.6100\times10^5$$

The thermal Rayleigh number is Ra𝑇 =Gr𝑇Pr =3.2593 ×105$\mathrm{Ra}_T=\mathrm{Gr}_T\mathrm{Pr}=3.2593\times10^5$. The matching heat correlation gives

Nu=0.54Ra1/4𝑇=0.54(3.2593×105)1/4=12.903$$\mathrm{Nu}=0.54\mathrm{Ra}_T^{1/4} =0.54(3.2593\times10^5)^{1/4}=12.903$$

Recover the heat transfer coefficient using the thermal conductivity:

ℎ=Nu𝑘𝐿=12.903(0.0263)0.05625=6.033 W m−2 K−1$$h=\frac{\mathrm{Nu}k}{L} =\frac{12.903(0.0263)}{0.05625} =6.033\ \mathrm{W\,m^{-2}\,K^{-1}}$$

For the thermally driven flow, convert ℎ$h$ to a mass coefficient:

𝑘𝑐=ℎ𝜌𝐶𝑝(PrSc)2/3=6.0331.1769(1006)(0.7070.61115)2/3=0.005615 m s−1$$k_c=\frac{h}{\rho C_p}\left(\frac{\mathrm{Pr}}{\mathrm{Sc}}\right)^{2/3} =\frac{6.033}{1.1769(1006)} \left(\frac{0.707}{0.61115}\right)^{2/3} =0.005615\ \mathrm{m\,s^{-1}}$$

The evaporation rate then follows from the surface-to-bulk mass-concentration difference:

˙𝑚𝐴=𝑘𝑐𝐴𝑠(𝜌𝐴,𝑠−𝜌𝐴,∞)=0.005615(0.039761)(0.043440)=9.698×10−6 kg s−1=34.91 g h−1$$\begin{aligned} \dot m_A&=k_cA_s(\rho_{A,s}-\rho_{A,\infty})\\ &=0.005615(0.039761)(0.043440)\\ &=9.698\times10^{-6}\ \mathrm{kg\,s^{-1}}=34.91\ \mathrm{g\,h^{-1}} \end{aligned}$$

These are the two requested isolated-mechanism answers.

For comparison, using dry-air density at 290 K and saturated-air density at 310 K gives a total density difference that includes both mechanisms.

The mean density 1.16616 and total difference 0.10515 kg m⁻³ give Gr =6.2366 ×105$\mathrm{Gr}=6.2366\times10^5$, Sh =13.419$\mathrm{Sh}=13.419$ and 𝑘𝑐 =0.006202$k_c=0.006202$ m s⁻¹.

The corresponding total-density estimate is ˙𝑚𝐴 =1.0712 ×10−5$\dot m_A=1.0712\times10^{-5}$ kg s⁻¹, about 38.6 g h⁻¹.

This approximately 39 g h⁻¹ estimate already includes thermal expansion. Adding a thermal Grashof contribution would count that expansion twice.

The real pan has reinforcing thermal and composition buoyancy.

The total-density route is a possible single-group engineering approximation; an exact coupled answer requires a validated heat-and-species natural-convection model.

The two isolated evaporation rates cannot be added to predict the combined rate, because the same velocity field transports both heat and species.

The saturation fraction is 0.0614, so low-flux and dry-air-property assumptions remain approximate.

The maintained water temperature implies external heat supply; no latent-heat constraint has been omitted from an adiabatic balance because this pan is not adiabatic.

Answer

For composition-driven convection alone, 𝑘𝑐 =0.00443 m s−1$k_c=0.00443\ \mathrm{m\,s^{-1}}$, ℎ ≃4.76 W m−2 K−1$h\simeq4.76\ \mathrm{W\,m^{-2}\,K^{-1}}$ and ˙𝑚𝐴 ≃27.6 g h−1$\dot m_A\simeq27.6\ \mathrm{g\,h^{-1}}$.

For temperature-driven convection alone, ℎ =6.03 W m−2 K−1$h=6.03\ \mathrm{W\,m^{-2}\,K^{-1}}$, 𝑘𝑐 =0.00562 m s−1$k_c=0.00562\ \mathrm{m\,s^{-1}}$ and ˙𝑚𝐴 ≃34.9 g h−1$\dot m_A\simeq34.9\ \mathrm{g\,h^{-1}}$.

Convective mass transfer

Evaporation from a tray and an outlet-humidity check

A water tray is flush with the floor of a rectangular duct. The tray is 0.60 m long along the air flow and 0.25 m wide. The duct is 0.25 m wide and 0.12 m high. Initially dry air enters at 2.0 m s⁻¹, 30 °C$30\,\degree\mathrm{C}$ and 101325 Pa. A heater maintains the water at 30 °C$30\,\degree\mathrm{C}$. Assume a uniform air stream over the tray and a concentration boundary layer beginning at its leading edge. Neglect upstream velocity-profile effects.

Use 𝜈 =1.60 ×10−5$\nu=1.60\times10^{-5}$ m² s⁻¹, 𝐷𝐴𝐵 =2.60 ×10−5$D_{AB}=2.60\times10^{-5}$ m² s⁻¹, water saturation pressure 4240 Pa, 𝑀𝐴 =0.0180$M_A=0.0180$ kg mol⁻¹ and 𝑅 =8.314$R=8.314$ J mol⁻¹ K⁻¹. Calculate the average mass transfer coefficient and evaporation rate. Estimate the outlet relative humidity and use it to check the assumption of dry bulk air over the entire tray. Use the dilute-transfer flat-plate model without a Stefan-flow correction.

Solution

Data

Quantity	Value
Tray length	𝐿 =0.60 m$L=0.60\ \mathrm m$
Tray width	𝑏 =0.25 m$b=0.25\ \mathrm m$
Duct height	𝐻 =0.12 m$H=0.12\ \mathrm m$
Air speed	𝑈 =2.0 m s−1$U=2.0\ \mathrm{m\,s^{-1}}$
Temperature	𝑇 =303.15 K$T=303.15\ \mathrm K$
Pressure	𝑃 =101325 Pa$P=101325\ \mathrm{Pa}$
Kinematic viscosity	𝜈 =1.60 ×10−5 m2 s−1$\nu=1.60\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵 =2.60 ×10−5 m2 s−1$D_{AB}=2.60\times10^{-5}\ \mathrm{m^2\,s^{-1}}$
Surface saturation pressure	𝑝sat𝐴 =4240 Pa$p_A^{\mathrm{sat}}=4240\ \mathrm{Pa}$
Water molar mass	𝑀𝐴 =0.0180 kg mol−1$M_A=0.0180\ \mathrm{kg\,mol^{-1}}$

Assumptions

• Steady external transfer with a boundary layer starting at the tray leading edge.
• Uniform incoming velocity and negligible upstream profile effects.
• Maintained saturated surface at the air temperature.
• Constant properties and dilute transfer without a Stefan-flow correction.
• Dry bulk air is used for the initial rate estimate and then checked.

Calculations

The concentration boundary layer grows along the tray, so use its length 𝐿 =0.60$L=0.60$ m for external transfer. The duct hydraulic diameter is not the length in this flat-plate correlation:

Re𝐿=2.0(0.60)1.60×10−5=75000Sc=1.60×10−52.60×10−5=0.6154$$\mathrm{Re}_L=\frac{2.0(0.60)}{1.60\times10^{-5}}=75000 \qquad \mathrm{Sc}=\frac{1.60\times10^{-5}}{2.60\times10^{-5}}=0.6154$$

The layer is laminar, and Sc exceeds 0.5. The average flat-plate correlation and coefficient definition are

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3$$\overline{\mathrm{Sh}}_L=0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3}$$¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}{L}$$

Substitution gives

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3=154.7$$\overline{\mathrm{Sh}}_L=0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3} =154.7$$¯¯¯¯𝑘𝑐=¯¯¯¯¯¯¯Sh𝐿𝐷𝐴𝐵𝐿=0.006702 m s−1$$\bar k_c=\frac{\overline{\mathrm{Sh}}_LD_{AB}}L =0.006702\ \mathrm{m\,s^{-1}}$$

At the interface the vapour is saturated. With 𝑇 =303.15$T=303.15$ K,

𝑐𝐴𝑠=42408.314𝑇=1.682 mol m−3.$$c_{As}=\frac{4240}{8.314T}=1.682\,\mathrm{mol\,m^{-3}}.$$

Approximate the bulk concentration as zero and use transfer area

𝐴𝑠=0.60(0.25)=0.150 m2:$$A_s=0.60(0.25)=0.150\,\mathrm{m^2}:$$˙𝑛𝐴=¯¯¯¯𝑘𝑐𝐴𝑠(𝑐𝐴𝑠−𝑐𝐴,∞)=0.006702(0.150)(1.682−0)=1.691×10−3 mol s−1˙𝑚𝐴=𝑀𝐴˙𝑛𝐴=0.1096 kg h−1$$\begin{aligned} \dot n_A&=\bar k_cA_s(c_{As}-c_{A,\infty})=0.006702(0.150)(1.682-0)=1.691\times10^{-3}\ \mathrm{mol\,s^{-1}}\\ \dot m_A&=M_A\dot n_A=0.1096\ \mathrm{kg\,h^{-1}} \end{aligned}$$

For the outlet balance use the duct flow area,

𝐴𝑑𝑢𝑐𝑡=0.25(0.12)=0.0300 m2.$$A_{duct}=0.25(0.12)=0.0300\,\mathrm{m^2}.$$

The dry-air molar flow is

˙𝑛𝑎𝑖𝑟=𝑢𝐴𝑑𝑢𝑐𝑡𝑃𝑅𝑇=2.412 mol s−1$$\dot n_{air}=uA_{duct}\frac P{RT}=2.412\ \mathrm{mol\,s^{-1}}$$

The outlet contains both dry air and the added vapour:

𝑦𝐴,𝑜𝑢𝑡=˙𝑛𝐴˙𝑛𝑎𝑖𝑟+˙𝑛𝐴=7.007×10−4$$y_{A,out}=\frac{\dot n_A}{\dot n_{air}+\dot n_A} =7.007\times10^{-4}$$RH𝑜𝑢𝑡=𝑦𝐴,𝑜𝑢𝑡𝑃𝑝𝑠𝑎𝑡𝐴=0.01674=1.67 percent$$\mathrm{RH}_{out}=\frac{y_{A,out}P}{p_A^{sat}} =0.01674=1.67\text{ percent}$$

Under the assumed mixed-bulk description, the bulk concentration reaches only 1.67 percent of the surface saturation concentration. The driving force therefore falls by less than 2 percent from inlet to outlet. This supports using dry bulk air in the first evaporation estimate. A much wetter outlet would require coupling local transfer to the changing bulk humidity.

This humidity check does not remove all model approximations. The surface mole fraction is

4240101325=0.04185,$$\frac{4240}{101325}=0.04185,$$

so a Stefan-flow correction can also change the dilute estimate by a few percent. The imposed heater avoids an evaporative-cooling calculation; without it, the surface temperature and vapour pressure would need a coupled energy balance.

Answer

The average mass transfer coefficient is ¯¯¯¯𝑘𝑐 =0.00670 m s−1$\bar k_c=0.00670\ \mathrm{m\,s^{-1}}$ and the evaporation rate is 0.110 kg h−1$0.110\ \mathrm{kg\,h^{-1}}$. The estimated outlet relative humidity is 1.67 percent, supporting the dry-bulk approximation.

Convective mass transfer

Drying a thin coating on a spherical pellet

A rigid spherical pellet of diameter 0.010 m has a thin wet coating containing 0.0020 g of volatile solvent per cm² of pellet surface. Solvent-free air flows past it at 0.10 m s⁻¹. Both the air and pellet surface are maintained at 300 K and 101325 Pa. External temperature control supplies the evaporation heat.

Use air properties 𝜌 =1.18$\rho=1.18$ kg m⁻³, 𝜇 =1.85 ×10−5$\mu=1.85\times10^{-5}$ Pa s, 𝑘 =0.0263$k=0.0263$ W m⁻¹ K⁻¹ and Pr =0.707$\mathrm{Pr}=0.707$. The solvent has 𝐷𝐴𝐵 =9.62 ×10−6$D_{AB}=9.62\times10^{-6}$ m² s⁻¹, saturation pressure 1270 Pa at 300 K, molar mass 0.0780 kg mol⁻¹ and liquid density 876.5 kg m⁻³. For this low-Re sphere, adopt the matching correlations

Nu𝐷=2+0.6Re1/2𝐷Pr1/3Sh𝐷=2+0.6Re1/2𝐷Sc1/3$$\begin{aligned} \mathrm{Nu}_D&=2+0.6\mathrm{Re}_D^{1/2}\mathrm{Pr}^{1/3}\\ \mathrm{Sh}_D&=2+0.6\mathrm{Re}_D^{1/2}\mathrm{Sc}^{1/3} \end{aligned}$$

Estimate the heat and mass transfer coefficients and the drying time. Assume the exposed area stays saturated at the stated vapour pressure until the solvent is exhausted, coating resistance is negligible, and normal evaporation flow is small. Check the thin-coating assumption.

Solution

Figure 5: A thin solvent coating surrounds the pellet. Its surface concentration is fixed by vapour equilibrium, and solvent transfers into the surrounding air. The diagram shows the external gas film; the model tracks coating loss rather than diffusion inside the pellet.

Data

Quantity	Value
Pellet diameter	𝐷 =0.010 m$D=0.010\ \mathrm m$
Air speed	𝑢 =0.10 m s−1$u=0.10\ \mathrm{m\,s^{-1}}$
Initial solvent loading	𝑚𝐴/𝐴𝑠 =0.0200 kg m−2$m_A/A_s=0.0200\ \mathrm{kg\,m^{-2}}$
Solvent diffusivity	𝐷𝐴𝐵 =9.62 ×10−6 m2 s−1$D_{AB}=9.62\times10^{-6}\ \mathrm{m^2\,s^{-1}}$
Solvent saturation pressure	𝑝sat𝐴 =1270 Pa$p_A^{\mathrm{sat}}=1270\ \mathrm{Pa}$
Solvent molar mass	𝑀𝐴 =0.0780 kg mol−1$M_A=0.0780\ \mathrm{kg\,mol^{-1}}$
Liquid density	𝜌𝐿 =876.5 kg m−3$\rho_L=876.5\ \mathrm{kg\,m^{-3}}$
Air and surface temperature	𝑇 =𝑇𝑠 =300 K$T=T_s=300\ \mathrm K$

Use the supplied air properties.

Assumptions

• Quasi steady external transfer and constant supplied properties.
• Negligible solvent in the bulk air.
• The exposed coating remains saturated until the solvent is exhausted.
• Negligible coating resistance and normal evaporation flow.
• Surface temperature is maintained by external heat supply.
• Fixed pellet diameter is used and the coating thickness is checked.

Calculations

Use pellet diameter for both the flow Reynolds number and the transfer groups:

Re𝐷=𝜌𝑢𝐷𝜇=1.18(0.10)(0.010)1.85×10−5=63.78$$\mathrm{Re}_D=\frac{\rho uD}{\mu}=\frac{1.18(0.10)(0.010)}{1.85\times10^{-5}}=63.78$$Sc=𝜇𝜌𝐷𝐴𝐵=1.85×10−51.18(9.62×10−6)=1.630$$\mathrm{Sc}=\frac{\mu}{\rho D_{AB}}=\frac{1.85\times10^{-5}}{1.18(9.62\times10^{-6})}=1.630$$Nu𝐷=2+0.6(63.78)1/2(0.707)1/3=6.269$$\mathrm{Nu}_D=2+0.6(63.78)^{1/2}(0.707)^{1/3}=6.269$$Sh𝐷=2+0.6(63.78)1/2(1.630)1/3=7.639$$\mathrm{Sh}_D=2+0.6(63.78)^{1/2}(1.630)^{1/3}=7.639$$

This Reynolds number is within the low-Re use described for the sphere correlation in the external forced-convection notes. The two transfer groups differ because Schmidt and Prandtl numbers differ. Their additive two is the stagnant-sphere limit and must be retained.

Recover the dimensional coefficients with the corresponding transport property:

ℎ=Nu𝐷𝑘𝐷=16.49 W m−2 K−1𝑘𝑐=Sh𝐷𝐷𝐴𝐵𝐷=0.007349 m s−1$$\begin{aligned} h&=\frac{\mathrm{Nu}_Dk}{D}=16.49\ \mathrm{W\,m^{-2}\,K^{-1}}\\ k_c&=\frac{\mathrm{Sh}_DD_{AB}}D=0.007349\ \mathrm{m\,s^{-1}} \end{aligned}$$

Since temperature is controlled, the heat coefficient need not determine the evaporation rate. The supplied vapour pressure fixes

𝑐𝐴𝑠=12708.314(300)=0.5092 mol m−3.$$c_{As}=\frac{1270}{8.314(300)}=0.5092\,\mathrm{mol\,m^{-3}}.$$

With dry bulk air,

𝑁𝐴=𝑘𝑐(𝑐𝐴𝑠−𝑐𝐴,∞)=0.007349(0.5092−0)=0.003742 mol m−2 s−1$$N_A=k_c(c_{As}-c_{A,\infty})=0.007349(0.5092-0)=0.003742\ \mathrm{mol\,m^{-2}\,s^{-1}}$$

The spherical area is

𝐴𝑠=𝜋𝐷2=3.1416×10−4 m2.$$A_s=\pi D^2=3.1416\times10^{-4}\,\mathrm{m^2}.$$

The loading converts as 0.0020$0.0020$ g cm⁻² =0.0200$=0.0200$ kg m⁻², so the initial solvent mass is 𝑚𝐴0 =6.283 ×10−6$m_{A0}=6.283\times10^{-6}$ kg. The constant-rate estimate is

𝑡𝑑𝑟𝑦=𝑚𝐴0𝑀𝐴𝑁𝐴𝐴𝑠=0.02000.0780(0.003742)=68.5 s$$t_{dry}=\frac{m_{A0}}{M_AN_AA_s} =\frac{0.0200}{0.0780(0.003742)}=68.5\ \mathrm s$$

The area cancels because both inventory and evaporation rate scale with it. Diameter still affects drying time through the transfer coefficient.

The solvent-equivalent coating thickness is

0.0200876.5=2.28×10−5 m,$$\frac{0.0200}{876.5}=2.28\times10^{-5}\ \mathrm m,$$

only 0.46 percent of the pellet radius. Treating the diameter as fixed is consistent with a thin coating provided the nonvolatile coating is also thin. The surface mole fraction is 0.01253, supporting the dilute approximation. Actual paint can dry more slowly as solvent activity falls or diffusion through the coating becomes limiting; the calculated time uses the stated constant-activity model.

Answer

The heat transfer coefficient is ℎ =16.5 W m−2 K−1$h=16.5\ \mathrm{W\,m^{-2}\,K^{-1}}$, the mass transfer coefficient is 𝑘𝑐 =0.00735 m s−1$k_c=0.00735\ \mathrm{m\,s^{-1}}$ and the constant-rate drying time is approximately 68.5 s. The solvent-equivalent thickness is 22.8 𝜇m$22.8\ \mathrm{\mu m}$, only 0.46 percent of the pellet radius.

Workshop 5: Steady heat conduction

Heat loss through an insulated pipe

Saturated steam at 130 ∘C$130\,{}^\circ\mathrm{C}$ is flowing inside a steel pipe having an ID of 2.09 cm and an OD of 2.67 cm. The pipe is insulated with 3.8 cm of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated to be ℎ𝑖$h_i$ = 5680 W/m2.K$\mathrm{W/m^2.K}$, while on the outside of the insulation it is estimated to be ℎ𝑜$h_o$ = 11.4 W/m2.K$\mathrm{W/m^2.K}$. The mean thermal conductivities of the metal and the insulation material are 45 W/m.K$\mathrm{W/m.K}$ and 0.064 W/m.K$\mathrm{W/m.K}$, respectively. Calculate the heat loss for 1 m of pipe using the method of resistances if the surrounding air is at 26 ∘C$26\,{}^\circ\mathrm{C}$.

Worked solution

Heat flows radially from the steam, through the steel and insulation, and then into the air. The same heat rate passes through four resistances.

Figure 1: Cross-section of an insulated pipe. The inner and outer steel radii are r1 and r2; insulation extends to r3. Heat crosses the inner film, steel, insulation and outer film in series.

Data

Quantity	Symbol	Value
Pipe length	𝐿$L$	1.0 m$1.0\ \mathrm m$
Inside diameter	𝐷1$D_1$	0.0209 m$0.0209\ \mathrm m$
Outside steel diameter	𝐷2$D_2$	0.0267 m$0.0267\ \mathrm m$
Radial insulation thickness	𝑡$t$	0.0380 m$0.0380\ \mathrm m$
Steel conductivity	𝑘𝑠$k_s$	45 W m−1 K−1$45\ \mathrm{W\,m^{-1}\,K^{-1}}$
Insulation conductivity	𝑘𝑖$k_i$	0.064 W m−1 K−1$0.064\ \mathrm{W\,m^{-1}\,K^{-1}}$
Steam and air temperatures	𝑇𝑠𝑡𝑒𝑎𝑚,𝑇𝑎𝑖𝑟$T_{steam},T_{air}$	130,26 ∘C$130,26\,{}^\circ\mathrm C$
Inside and outside coefficients	ℎ𝑖,ℎ𝑜$h_i,h_o$	5680,11.4 W m−2 K−1$5680,11.4\ \mathrm{W\,m^{-2}\,K^{-1}}$

We need the heat loss from the specified one-metre length.

Assumptions

• Steady radial heat flow with negligible end losses.
• Constant conductivities and heat transfer coefficients.
• Perfect thermal contact between steel and insulation.
• No separate radiation or fouling resistance.

Calculations

First convert diameters to radii. Insulation thickness is already a radial length, so it is added directly to the outside steel radius:

𝑟1=𝐷12=0.01045 m,𝑟2=𝐷22=0.01335 m$$r_1=\frac{D_1}{2}=0.01045\ \mathrm m,\qquad r_2=\frac{D_2}{2}=0.01335\ \mathrm m$$𝑟3=𝑟2+𝑡=0.01335+0.0380=0.05135 m$$r_3=r_2+t=0.01335+0.0380=0.05135\ \mathrm m$$

Each film uses its own cylindrical area, 2𝜋𝑟𝐿$2\pi rL$. Conduction through a cylindrical layer uses both bounding radii. The four resistances are

Element	Resistance relation	Value, K W−1$\mathrm{K\,W^{-1}}$
Inside steam film	𝑅𝑐𝑖 =1/(ℎ𝑖2𝜋𝑟1𝐿)$R_{ci}=1/(h_i2\pi r_1L)$	0.00268$0.00268$
Steel wall	𝑅𝑠 =ln⁡(𝑟2/𝑟1)/(2𝜋𝑘𝑠𝐿)$R_s=\ln(r_2/r_1)/(2\pi k_sL)$	0.000866$0.000866$
Insulation	𝑅𝑖 =ln⁡(𝑟3/𝑟2)/(2𝜋𝑘𝑖𝐿)$R_i=\ln(r_3/r_2)/(2\pi k_iL)$	3.35$3.35$
Outside air film	𝑅𝑐𝑜 =1/(ℎ𝑜2𝜋𝑟3𝐿)$R_{co}=1/(h_o2\pi r_3L)$	0.272$0.272$

For example, the largest resistance is obtained from

𝑅𝑖=ln⁡(0.051350.01335)2𝜋(0.064)(1.0)≃3.35 K W−1$$R_i=\frac{\ln\left(\frac{0.05135}{0.01335}\right)}{2\pi(0.064)(1.0)} \simeq3.35\ \mathrm{K\,W^{-1}}$$

The elements are in series, so

𝑅𝑡𝑜𝑡𝑎𝑙=𝑅𝑐𝑖+𝑅𝑠+𝑅𝑖+𝑅𝑐𝑜≃3.626 K W−1$$R_{total}=R_{ci}+R_s+R_i+R_{co}\simeq3.626\ \mathrm{K\,W^{-1}}$$

Use the steam-to-air temperature difference across the complete network:

˙𝑄=𝑇𝑠𝑡𝑒𝑎𝑚−𝑇𝑎𝑖𝑟𝑅𝑡𝑜𝑡𝑎𝑙=130−263.626≃28.7 W$$\dot Q=\frac{T_{steam}-T_{air}}{R_{total}} =\frac{130-26}{3.626}\simeq28.7\ \mathrm W$$

The outer insulation temperature provides a useful direction check:

𝑇𝑠,𝑜=𝑇𝑎𝑖𝑟+˙𝑄𝑅𝑐𝑜≃26+28.7(0.272)=33.8 ∘C$$T_{s,o}=T_{air}+\dot QR_{co} \simeq26+28.7(0.272)=33.8\,{}^\circ\mathrm C$$

It lies above the air temperature and below the steam temperature. The insulation supplies about 92 percent of the total resistance. Its large temperature drop is consistent with the small heat loss.

Answer

The heat loss is approximately 28.7 W$28.7\ \mathrm W$ from one metre of pipe, or 28.7 W m−1$28.7\ \mathrm{W\,m^{-1}}$. This is a heat rate per pipe length, not a heat flux per surface area.

Workshop 5: Steady heat conduction

Insulation surface temperature

In a laboratory experiment, an electric current is passed through a slab of thickness 2𝛿$2\delta$. As a result, heat generation 𝑆𝑒$S_e$ takes place in the slab. The slab is insulated on both sides (thickness = t) to satisfy the safety considerations related to the danger stemming from the fact that an electric current is passing. In addition, the insulated slab is cooled on both sides by a fluid of temperature 𝑇𝑐$T_c$, with convective coefficient of h. Determine the temperature at the surface of the insulation, assuming that the thermal conductivities for the slab and the insulation material are constants, denoted by 𝑘𝑠$k_s$ and 𝑘𝑖$k_i$, respectively.

Worked solution

Choose 𝑥 =0$x=0$ at the slab centre and 𝑥 =𝛿$x=\delta$ at its right face.

Symmetry gives 𝑑𝑇/𝑑𝑥 =0$dT/dx=0$ at the centre.

The slab energy balance is

𝑘𝑠𝑑2𝑇𝑑𝑥2+𝑆𝑒=0.$$k_s\frac{d^2T}{dx^2}+S_e=0.$$

Integrating with the symmetry condition gives

𝑑𝑇𝑑𝑥=−𝑆𝑒𝑥𝑘𝑠.$$\frac{dT}{dx}=-\frac{S_ex}{k_s}.$$

The heat flux leaving either slab face, in W m−2$\mathrm{W\,m^{-2}}$, is therefore

𝑞′′=−𝑘𝑠[𝑑𝑇𝑑𝑥]𝛿=𝑆𝑒𝛿.$$q''=-k_s\left[\frac{dT}{dx}\right]_{\delta}=S_e\delta.$$

Each face removes half the generation per unit area of the whole slab.

The insulation carries this constant flux, so

𝑆𝑒𝛿=ℎ(𝑇𝑜−𝑇𝑐)=𝑘𝑖𝑡(𝑇𝑖−𝑇𝑜).$$S_e\delta=h(T_o-T_c) =\frac{k_i}{t}(T_i-T_o).$$

Hence the outer surface and interface temperatures are

𝑇𝑜=𝑇𝑐+𝑆𝑒𝛿ℎ,$$T_o=T_c+\frac{S_e\delta}{h},$$𝑇𝑖=𝑇𝑐+𝑆𝑒𝛿ℎ+𝑆𝑒𝛿𝑡𝑘𝑖.$$T_i=T_c+\frac{S_e\delta}{h} +\frac{S_e\delta t}{k_i}.$$

Both expressions return kelvin.

The overall generation fixes the surface heat flux, so 𝑘𝑠$k_s$ affects the temperature gradient inside the slab rather than 𝑇𝑜$T_o$.

For example the centre temperature is

𝑇𝑖+𝑆𝑒𝛿22𝑘𝑠,$$T_i+\frac{S_e\delta^2}{2k_s},$$

also in kelvin.

Workshop 5: Steady heat conduction

Temperature-dependent conductivity

A pipe designed to transport liquid metal is to be embedded in a 1.2 m thick wall. The inside surface of the wall has a constant surface temperature of 925 K, whereas its external surface is exposed to air with a convective heat transfer coefficient of 23 W/m2.K$\mathrm{W/m^2.K}$ and 300 K. The thermal conductivity of the wall varies with temperature according to 𝑘 =0.073 +3.942 ×10−4𝑇$k = 0.073 + 3.942\times10^{-4}T$, where T is temperature in Kelvin, and k is in W/m.K$\mathrm{W/m.K}$. It is desired that the pipe is located at a point in the wall where the temperature is < 650 K. Calculate how far from the hot surface should the pipe be located and the heat flux for the wall.

Worked solution

Data

Quantity	Symbol	Value
Wall thickness	𝐿$L$	1.2 m$1.2\ \mathrm m$
Hot-face temperature	𝑇ℎ$T_h$	925 K$925\ \mathrm K$
Outside air	𝑇∞$T_\infty$	300 K$300\ \mathrm K$
Outside coefficient	ℎ$h$	23 W m−2 K−1$23\ \mathrm{W\,m^{-2}\,K^{-1}}$
Conductivity	𝑘(𝑇)$k(T)$	0.073 +3.942 ×10−4𝑇 W m−1 K−1$0.073+3.942\times10^{-4}T\ \mathrm{W\,m^{-1}\,K^{-1}}$
Pipe temperature limit	𝑇𝑙𝑖𝑚𝑖𝑡$T_{limit}$	650 K$650\ \mathrm K$

We need the outward heat flux and the depth at which the wall reaches the pipe-temperature limit.

Assumptions

• Steady one-dimensional conduction with no heat generation.
• Negligible disturbance of the wall temperature field by the embedded pipe.
• The given conductivity relation applies throughout the temperature range.

Calculations

Write 𝑘(𝑇) =𝑎 +𝑏𝑇$k(T)=a+bT$, where 𝑎 =0.073 W m−1 K−1$a=0.073\,\mathrm{W\,m^{-1}\,K^{-1}}$ and 𝑏 =3.942 ×10−4 W m−1 K−2$b=3.942\times10^{-4}\,\mathrm{W\,m^{-1}\,K^{-2}}$.

Let 𝑥$x$ increase from the hot face towards the air, 𝐿 =1.2 m$L=1.2\,\mathrm{m}$, and 𝑇𝑜 =𝑇(𝐿)$T_o=T(L)$.

With no generation the positive heat flux is constant:

𝑞′′=−𝑘(𝑇)𝑑𝑇𝑑𝑥.$$q''=-k(T)\frac{dT}{dx}.$$

Integrating across the wall and applying convection gives

𝑞′′𝐿=𝑎(925−𝑇𝑜)+𝑏2(9252−𝑇2𝑜),𝑞′′=23(𝑇𝑜−300).$$\begin{aligned} q''L&=a(925-T_o) +\frac b2(925^2-T_o^2),\\ q''&=23(T_o-300). \end{aligned}$$(1)

Substituting the second relation into the first produces

1.971×10−4𝑇2𝑜+27.673𝑇𝑜−8516.17=0.$$1.971\times10^{-4}T_o^2+27.673T_o -8516.17=0.$$

The physical root is 𝑇𝑜 ≈307.07 K$T_o\approx307.07\,\mathrm{K}$.

The other root is negative and lies outside the specified temperature interval.

From Eq. 1, the outward heat flux is

𝑞′′=23(𝑇𝑜−300)≈162.6 W m−2.$$q''=23(T_o-300) \approx162.6\,\mathrm{W\,m^{-2}}.$$

Integrating from 925 K$925\,\mathrm{K}$ to 650 K$650\,\mathrm{K}$ gives

𝑞′′𝑥650=𝑎(925−650)+𝑏2(9252−6502),$$q''x_{650}=a(925-650) +\frac b2(925^2-650^2),$$𝑥650≈0.648 m.$$x_{650}\approx0.648\,\mathrm{m}.$$

Since 𝑘(𝑇) >0$k(T)>0$ throughout the wall, the temperature decreases with 𝑥$x$.

The pipe must lie farther than 0.648 m$0.648\,\mathrm{m}$ from the hot face towards the cold face.

Equality gives 650 K$650\,\mathrm{K}$ and does not meet the strict limit.

A finite pipe must have its entire occupied region on the cold side if every part must remain below the limit.

Answer

The outward heat flux is 163 W m−2$163\ \mathrm{W\,m^{-2}}$. The pipe must be placed more than 0.648 m$0.648\ \mathrm m$ from the hot face towards the cold face to remain below 650 K$650\ \mathrm K$.

Workshop 6: Transient heat conduction

Plate surface temperature

In a production facility, 4-cm-thick large plates (k = 107 W/m ⋅∘C$\mathrm{W/m\cdot{}^\circ C}$, 𝜌$\rho$ = 8520 kg/m3$\mathrm{kg/m^3}$, 𝐶𝑝$C_p$ = 381.2 J/kg ⋅∘C$\mathrm{J/kg\cdot{}^\circ C}$) that are initially at a uniform temperature of 20 ∘C$20\,{}^\circ\mathrm{C}$ are heated by passing them through an oven maintained at 500 ∘C$500\,{}^\circ\mathrm{C}$. The plates remain in the oven for a period of 7 min. Taking the convection heat transfer coefficient to be h = 120 W/m2 ⋅∘C$\mathrm{W/m^2\cdot{}^\circ C}$, determine the surface temperature of the plates when they come out of the oven. Assume L = W = 1 m.

Worked solution

Figure 1: Both faces of the plate are heated by the oven air. Use half the plate thickness for the plane-wall solution, not its one-metre plan dimension.

Data

Quantity	Symbol	Value
Plate dimensions	𝐻,𝑊,2𝐿$H,W,2L$	1.0,1.0,0.040 m$1.0,1.0,0.040\ \mathrm m$
Initial and oven temperatures	𝑇𝑖,𝑇∞$T_i,T_\infty$	20,500 ∘C$20,500\,{}^\circ\mathrm C$
Heating duration	𝑡$t$	420 s$420\ \mathrm s$
Convection coefficient	ℎ$h$	120 W m−2 K−1$120\ \mathrm{W\,m^{-2}\,K^{-1}}$
Conductivity	𝑘$k$	107 W m−1 K−1$107\ \mathrm{W\,m^{-1}\,K^{-1}}$
Density	𝜌$\rho$	8520 kg m−3$8520\ \mathrm{kg\,m^{-3}}$
Specific heat	𝑐𝑝$c_p$	381.2 J kg−1 K−1$381.2\ \mathrm{J\,kg^{-1}\,K^{-1}}$

Here 𝐿 =0.020 m$L=0.020\ \mathrm m$ denotes half the plate thickness in the plane-wall model. The one-metre dimensions in the question set the face area.

Assumptions

• Both broad faces see the same oven temperature and coefficient.
• Edge heat transfer and radiation are neglected in the plane-wall comparison.
• Properties remain constant and the initial temperature is uniform.

Calculations

The worked solution compares lumped capacitance with the one-term plane-wall solution.

For the comparison, solve 𝜆1tan⁡𝜆1 =𝐵𝑖$\lambda_1\tan\lambda_1=Bi$ with 𝐵𝑖 =ℎ𝐿/𝑘$Bi=hL/k$ and use

Θ(𝑥,𝑡)=𝐴1𝑒−𝜆21𝐹𝑜cos⁡(𝜆1𝑥𝐿),𝐴1=4sin⁡𝜆12𝜆1+sin⁡(2𝜆1),𝐹𝑜=𝛼𝑡𝐿2,𝛼=𝑘𝜌𝑐𝑝.$$\begin{aligned} \Theta(x,t)&=A_1 e^{-\lambda_1^2Fo} \cos\left(\frac{\lambda_1 x}{L}\right),\\ A_1&=\frac{4\sin\lambda_1}{2\lambda_1+\sin(2\lambda_1)},\\ Fo&=\frac{\alpha t}{L^2},\qquad \alpha=\frac{k}{\rho c_p}. \end{aligned}$$

Here 𝑥 =0$x=0$ is the centre and

Θ=𝑇−𝑇∞𝑇𝑖−𝑇∞.$$\Theta=\frac{T-T_\infty}{T_i-T_\infty}.$$

The plate characteristic length is its volume divided by its heat-transfer area.

With only the two broad faces transferring heat,

𝑉=(1)(1)(0.040)=0.040 m3$$V=(1)(1)(0.040)=0.040\ \mathrm{m^3}$$

and 𝐴𝑠 =2(1)(1) =2.0 m2$A_s=2(1)(1)=2.0\ \mathrm{m^2}$. The half-thickness and volume-to-area length are equal for this model.

Thus 𝐿𝑐 =𝑉/𝐴𝑠 =0.02 m$L_c=V/A_s=0.02\,\mathrm{m}$ and

𝐵𝑖𝑐=ℎ𝐿𝑐𝑘=0.02243<0.1.$$Bi_c=\frac{hL_c}{k}=0.02243<0.1.$$

Lumped capacitance is a reasonable approximation.

The uniform-temperature energy balance gives

Θ=𝑒−ℎ𝐴𝑠𝑡𝜌𝑐𝑝𝑉.$$\Theta=e^{-\frac{hA_st}{\rho c_pV}}.$$

Using 𝑡 =420 s$t=420\,\mathrm{s}$ gives Θ =0.46029$\Theta=0.46029$ and

𝑇=500−480Θ=279.06 ∘C.$$T=500-480\Theta=279.06\,^{\circ}\mathrm{C}.$$

Within this approximation the surface and centre have the same temperature.

For the plane-wall comparison, 𝛼 =3.2945 ×10−5 m2 s−1$\alpha=3.2945\times10^{-5}\,\mathrm{m^2\,s^{-1}}$ and 𝐹𝑜 =34.592$Fo=34.592$.

The eigenvalue is 𝜆1 =0.149209$\lambda_1=0.149209$ and 𝐴1 =1.003709$A_1=1.003709$.

Evaluating the supplied relation at 𝑥 =𝐿$x=L$ gives

Θ𝑠=1.003709 𝑒−(0.149209)2(34.592)×cos⁡(0.149209)=0.459504,$$\Theta_s=1.003709\,e^{-(0.149209)^2(34.592)} \quad\times\cos(0.149209)=0.459504,$$𝑇𝑠=279.44 ∘C.$$T_s=279.44\,^{\circ}\mathrm{C}.$$

The finite-wall result is about 0.38 degrees above the lumped estimate.

The large Fourier number makes higher eigenmodes negligible.

The temperature is not exactly uniform, but the small Biot number explains the close agreement.

Answer

The one-term plane-wall estimate gives a surface temperature of 279.4 ∘C$279.4\,{}^\circ\mathrm C$. The simpler lumped estimate is 279.1 ∘C$279.1\,{}^\circ\mathrm C$, consistent with the small Biot number.

Workshop 6: Transient heat conduction

Maximum time in air

During the quenching process of stainless steel ball bearings, it is known that the temperature of the bearings should not reach below 850 ∘C$850\,{}^\circ\mathrm{C}$ prior to entering the water to minimise the developments of surface defects. Ball bearings with diameter of 1.2 cm (𝜌$\rho$ = 8085 kg/m3$\mathrm{kg/m^3}$, 𝐶𝑝$C_p$ = 0.480 kJ/kg ⋅∘C$\mathrm{kJ/kg\cdot{}^\circ C}$, k = 15.1 W/m ⋅∘C$\mathrm{W/m\cdot{}^\circ C}$ and 𝛼$\alpha$ = 3.91 × 10−6$10^{-6}$ m2/s$\mathrm{m^2/s}$) are taken out of the oven at a uniform temperature of 900 ∘C$900\,{}^\circ\mathrm{C}$, and are exposed to 30 ∘C$30\,{}^\circ\mathrm{C}$ air for a while before they are dropped into the water. Assuming that the heat transfer coefficient of the air is 125 W/m2.K$\mathrm{W/m^2.K}$, how long can the spheres be left in air before quenching?

Worked solution

Figure 2: A small spherical bearing cools in air. Its radius is half the supplied diameter, and the exposed area is the full spherical surface.

For a sphere the volume-to-area length is one third of its radius.

With 𝑅 =0.006 m$R=0.006\,\mathrm{m}$, 𝐿𝑐 =𝑅/3 =0.002 m$L_c=R/3=0.002\,\mathrm{m}$ and

𝐵𝑖𝑐=ℎ𝐿𝑐𝑘=0.01656<0.1.$$Bi_c=\frac{hL_c}{k}=0.01656<0.1.$$

Use the lumped cooling relation to solve for time:

𝑡=𝜌𝑐𝑝𝑅3ℎln⁡(𝑇𝑖−𝑇∞𝑇−𝑇∞).$$t=\frac{\rho c_p R}{3h} \ln\left(\frac{T_i-T_\infty}{T-T_\infty}\right).$$

At the limiting temperature,

𝑡max⁡=8085(480)(0.006)3(125)ln⁡(870820)=3.675 s.$$t_{\max}=\frac{8085(480)(0.006)}{3(125)} \ln\left(\frac{870}{820}\right) =3.675\,\mathrm{s}.$$

The lumped model permits at most about 3.68 s$3.68\,\mathrm{s}$ in air.

This estimate treats the whole sphere as isothermal.

It does not guarantee a strict local surface limit: a real cooling sphere has a colder surface than centre.

Such a guarantee needs a spatial transient model and allowances for radiation and uncertain properties at these high temperatures.

Workshop 6: Transient heat conduction

Centre and surface temperatures

In the Betty Crocker Cookbook, it is stated that it takes 2 h 45 minutes to roast a 3.2 kg ribs that is initially at 4.5 ∘C$4.5\,{}^\circ\mathrm{C}$ in an oven that is maintained at 163 ∘C$163\,{}^\circ\mathrm{C}$. It is recommended that a meat thermometer be used to monitor the cooking, and the rib is considered medium-rare when the thermometer, inserted into the centre of the thickest part of the meat, registers 60 ∘C$60\,{}^\circ\mathrm{C}$. If the ribs is approximated as a homogeneous spherical object with 𝜌$\rho$ = 1200 kg/m3$\mathrm{kg/m^3}$, 𝐶𝑝$C_p$ = 4.1 kJ/kg ⋅∘C$\mathrm{kJ/kg\cdot{}^\circ C}$, k = 0.45 W/m ⋅∘C$\mathrm{W/m\cdot{}^\circ C}$ and 𝛼$\alpha$ = 0.91 × 10−7$10^{-7}$ m2/s$\mathrm{m^2/s}$, determine:

a. The heat transfer coefficient at the surface of the ribs

b. The temperature at the outer surface of the ribs when it is done (medium rare)

Worked solution

Figure 3: A spherical body has distinct centre and surface temperatures during heating. The measured centre temperature is used to infer the coefficient at the outer boundary.

Data

Quantity	Symbol	Value
Mass	𝑚$m$	3.2 kg$3.2\ \mathrm{kg}$
Density	𝜌$\rho$	1200 kg m−3$1200\ \mathrm{kg\,m^{-3}}$
Initial and oven temperatures	𝑇𝑖,𝑇∞$T_i,T_\infty$	4.5,163 ∘C$4.5,163\,{}^\circ\mathrm C$
Observed centre temperature	𝑇𝑐$T_c$	60 ∘C$60\,{}^\circ\mathrm C$
Elapsed time	𝑡$t$	9900 s$9900\ \mathrm s$
Conductivity	𝑘$k$	0.45 W m−1 K−1$0.45\ \mathrm{W\,m^{-1}\,K^{-1}}$
Supplied diffusivity	𝛼$\alpha$	0.91 ×10−7 m2 s−1$0.91\times10^{-7}\ \mathrm{m^2\,s^{-1}}$

Assumptions

Use the given rounded diffusivity for the transient calculation. Assume radial conduction, constant properties, no evaporation, no generation, and no separate radiative heat flux. The effective coefficient describes the stated idealised boundary, not a measured material property.

Calculations

The first step is to infer the radius from

𝑚𝜌=4𝜋𝑅33.$$\frac{m}{\rho}=\frac{4\pi R^3}{3}.$$

For 𝐵𝑖𝑅 =ℎ𝑅/𝑘$Bi_R=hR/k$, the positive eigenvalues satisfy 1 −𝜆𝑛cot⁡𝜆𝑛 =𝐵𝑖𝑅$1-\lambda_n\cot\lambda_n=Bi_R$. The full spherical solution is

Θ(𝑟,𝑡)=∞∑𝑛=1𝐴𝑛𝑒−𝜆2𝑛𝐹𝑜𝑓𝑛(𝑟),𝐴𝑛=4(sin⁡𝜆𝑛−𝜆𝑛cos⁡𝜆𝑛)2𝜆𝑛−sin⁡(2𝜆𝑛),𝑓𝑛(𝑟)=sin⁡(𝜆𝑛𝑟𝑅)𝜆𝑛𝑟𝑅,𝐹𝑜=𝛼𝑡𝑅2.$$\begin{aligned} \Theta(r,t)&=\sum_{n=1}^{\infty} A_n e^{-\lambda_n^2Fo}f_n(r),\\ A_n&=\frac{4(\sin\lambda_n-\lambda_n\cos\lambda_n)} {2\lambda_n-\sin(2\lambda_n)},\\ f_n(r)&=\frac{\sin\left(\frac{\lambda_n r}{R}\right)}{\frac{\lambda_n r}{R}},\\ Fo&=\frac{\alpha t}{R^2}. \end{aligned}$$(1)

Use 𝑓𝑛(0) =1$f_n(0)=1$. Retain enough terms for convergence; the usual one-term criterion 𝐹𝑜 >0.2$Fo>0.2$ may not hold here. The first eigenvalue lies between zero and 𝜋$\pi$, with each subsequent one between adjacent multiples of 𝜋$\pi$.

The mass and density fix the radius without a geometric guess.

𝑉=3.21200=0.0026667 m3$$V=\frac{3.2}{1200}=0.0026667\,\mathrm{m^3}$$

and 𝑅 =0.0860254 m$R=0.0860254\,\mathrm{m}$.

With 𝑡 =9900 s$t=9900\,\mathrm{s}$, 𝐹𝑜 =0.121737$Fo=0.121737$.

The observed centre temperature requires

Θ0=60−1634.5−163=0.649842.$$\Theta_0=\frac{60-163}{4.5-163}=0.649842.$$

For a trial 𝐵𝑖𝑅$Bi_R$, solve the eigenvalue equation in each interval and evaluate Eq. 1 at the centre.

Adjust 𝐵𝑖𝑅$Bi_R$ until the sum equals 0.649842$0.649842$.

The following trials show how the inverse calculation brackets the measured centre temperature. Each row uses its own eigenvalues in the same series.

Trial 𝐵𝑖𝑅$Bi_R$	First eigenvalue 𝜆1$\lambda_1$	Centre ratio Θ0$\Theta_0$	Predicted centre temperature
10.0$10.0$	2.83630$2.83630$	0.690587$0.690587$	53.54 ∘C$53.54\,{}^\circ\mathrm C$
20.0$20.0$	2.98572$2.98572$	0.643649$0.643649$	60.98 ∘C$60.98\,{}^\circ\mathrm C$
17.8639$17.8639$	2.96741$2.96741$	0.649842$0.649842$	60.00 ∘C$60.00\,{}^\circ\mathrm C$

The first trial heats the centre too slowly; the second heats it too quickly. Solving between them gives the fitted Biot number. Once it is fitted, use that same boundary coefficient to predict the surface temperature.

Five terms already agree with ten terms to better than 10−10$10^{-10}$ in the temperature ratio at the fitted coefficient.

This gives

𝐵𝑖𝑅=17.8639ℎ=𝐵𝑖𝑅𝑘𝑅=93.45 W m−2 K−1$$Bi_R=17.8639 \qquad h=\frac{Bi_R k}{R} =93.45\,\mathrm{W\,m^{-2}\,K^{-1}}$$

The fitted spherical series gives a surface ratio of 0.0408906.

At 𝑟 =𝑅$r=R$, retain the same eigenvalues and coefficients but multiply each term by sin⁡𝜆𝑛/𝜆𝑛$\sin\lambda_n/\lambda_n$.

Therefore

𝑇𝑠=163+(4.5−163)(0.0408906)=156.52 ∘C.$$T_s=163+(4.5-163)(0.0408906) =156.52\,^{\circ}\mathrm{C}.$$

The surface is much hotter than the centre during heating.

Lumped capacitance is unsuitable because 𝐵𝑖𝑐 =𝐵𝑖𝑅/3 =5.95$Bi_c=Bi_R/3=5.95$.

A one-term inverse calculation also introduces avoidable error at this Fourier number.

The handwritten source’s trial value 𝐵𝑖𝑅 ≈30$Bi_R\approx30$ does not reproduce the stated centre temperature with a converged series; its coefficient is not used as the answer.

Answer

The fitted effective coefficient is ℎ =93.5 W m−2 K−1$h=93.5\ \mathrm{W\,m^{-2}\,K^{-1}}$ and the surface temperature is 156.5 ∘C$156.5\,{}^\circ\mathrm C$. These values use the converged spherical series and the stipulated constant-property model.

Workshop 7: Forced and natural convection

Uniform heat flux

Water at 16 ∘C$16\,{}^\circ\mathrm{C}$ enters a 2.54 cm ID tube that is used to cool a nuclear reactor. The water flow rate is 0.114 m3/min$\mathrm{m^3/min}$. Using the Colburn analogy, determine the total rate of heat transfer, the exiting water temperature and the wall temperature at the exit, if the length of the tube is 4.5 m and the wall condition is one of uniform heat flux of 5.7 × 103$10^3$ kJ/hr.m2$\mathrm{kJ/hr.m^2}$.

Worked solution

Figure 1: Water flows through the heated tube. The specified inward heat flux supplies the bulk enthalpy rise; the local outlet coefficient determines the exit wall temperature.

Data

Quantity	Symbol	Value
Tube diameter and length	𝐷,𝐿$D,L$	0.0254,4.5 m$0.0254,4.5\ \mathrm m$
Volume flow	˙𝑉$\dot V$	0.114 m3 min−1$0.114\ \mathrm{m^{3}\,min^{-1}}$
Inlet temperature	𝑇𝑖𝑛$T_{in}$	16 ∘C$16\,{}^\circ\mathrm C$
Uniform wall flux	𝑞′′$q^{\prime\prime}$	5.7 ×103 kJ h−1 m−2$5.7\times10^3\ \mathrm{kJ\,h^{-1}\,m^{-2}}$

Water property	Constant value
Density 𝜌$\rho$	1000 kg m−3$1000\,\mathrm{kg\,m^{-3}}$
Specific heat 𝑐𝑝$c_p$	4180 J kg−1 K−1$4180\,\mathrm{J\,kg^{-1}\,K^{-1}}$
Dynamic viscosity 𝜇$\mu$	0.001 Pa s$0.001\,\mathrm{Pa\,s}$
Prandtl number 𝑃𝑟$Pr$	7.0$7.0$

Assumptions

Use the source solution’s rounded water properties throughout. Assume steady single-phase flow, fully developed turbulent heat transfer, negligible axial conduction and no heat loss elsewhere. Ignore entrance effects. The prescribed heat flux acts on the inner wetted surface.

Calculations

For this smooth tube use

𝑅𝑒=𝜌𝑢𝐷𝜇,𝑓𝐹=0.0791𝑅𝑒−1/4,𝑆𝑡 𝑃𝑟2/3=𝑓𝐹2,𝑆𝑡=ℎ𝜌𝑢𝑐𝑝.$$\begin{aligned} Re&=\frac{\rho uD}{\mu},\qquad f_F=0.0791Re^{-1/4},\\ St\,Pr^{2/3}&=\frac{f_F}{2},\qquad St=\frac{h}{\rho u c_p}. \end{aligned}$$(1)

Here 𝑓𝐹$f_F$ is the Fanning friction factor, one quarter of the Darcy factor. The correlation applies to smooth turbulent flow in the approximate range 4000 <𝑅𝑒 <105$4000<Re<10^5$. Use

˙𝑄=˙𝑚𝑐𝑝(𝑇𝑜𝑢𝑡−𝑇𝑖𝑛)$$\dot Q=\dot m c_p(T_{out}-T_{in})$$

and the local wall relation 𝑞′′ =ℎ(𝑇𝑤 −𝑇𝑏)$q''=h(T_w-T_b)$.

The imposed heat flux fixes the duty before a convection coefficient is needed.

The supplied flux is in kJ per hour per square metre. Convert energy and time together:

𝑞′′=5.7×10310003600=1583.33 W m−2$$q''=5.7\times10^3\frac{1000}{3600}=1583.33\ \mathrm{W\,m^{-2}}$$

The wetted area is 𝜋𝐷𝐿 =0.359084 m2$\pi DL=0.359084\,\mathrm{m^2}$, hence

˙𝑄=𝑞′′𝜋𝐷𝐿=568.55 W.$$\dot Q=q''\pi DL=568.55\,\mathrm W.$$

The flow rate is 0.0019 m3 s−1$0.0019\,\mathrm{m^3\,s^{-1}}$ and ˙𝑚 =1.900 kg s−1$\dot m=1.900\,\mathrm{kg\,s^{-1}}$.

The bulk energy balance gives

𝑇𝑜𝑢𝑡=16+568.551.900(4180)=16.0716 ∘C.$$T_{out}=16+\frac{568.55}{1.900(4180)} =16.0716\,^{\circ}\mathrm C.$$

The cross-sectional area gives 𝑢 =3.7497 m s−1$u=3.7497\,\mathrm{m\,s^{-1}}$, 𝑅𝑒 =95242$Re=95242$ and 𝑓𝐹 =0.0045027$f_F=0.0045027$.

This is turbulent and within the stated smooth-tube correlation range.

Equation Eq. 1 gives

ℎ=𝑓𝐹2𝜌𝑢𝑐𝑝𝑃𝑟−2/3=9643 W m−2 K−1,$$h=\frac{f_F}{2}\rho u c_pPr^{-2/3} =9643\,\mathrm{W\,m^{-2}\,K^{-1}},$$𝑇𝑤,𝑜𝑢𝑡=𝑇𝑜𝑢𝑡+𝑞′′ℎ=16.2358 ∘C.$$T_{w,out}=T_{out}+\frac{q''}{h} =16.2358\,^{\circ}\mathrm C.$$

The wall is hotter than the local bulk water, as required for inward heat transfer.

The source’s chart estimate 𝑓𝐹 =0.0046$f_F=0.0046$ gives a slightly different coefficient.

The supplied smooth-tube formula removes chart-reading ambiguity.

Answer

The water receives 569 W$569\ \mathrm W$, exits at 16.07 ∘C$16.07\,{}^\circ\mathrm C$, and has an exit wall temperature of 16.24 ∘C$16.24\,{}^\circ\mathrm C$.

The water temperatures retain two decimal places to show the small temperature rise.

Workshop 7: Forced and natural convection

Thermally developing laminar flow

Consider the flow of oil in a 30 cm diameter pipeline at an average velocity of 2 m/s and temperature of 20 ∘C$20\,{}^\circ\mathrm{C}$. A section of the pipeline (200 m long) passes through icy waters of a lake at 0 ∘C$0\,{}^\circ\mathrm{C}$. Measurements indicate that the surface temperature of the pipe is very nearly 0 ∘C$0\,{}^\circ\mathrm{C}$. Disregarding the thermal resistance of the pipe material, determine:

a. The temperature of the oil when the pipe leaves the lake;

b. The rate of heat transfer from the oil.

Properties of oil (@ 20 ∘C$20\,{}^\circ\mathrm{C}$): 𝜌$\rho$ = 888 kg/m3$\mathrm{kg/m^3}$; k = 0.145 W/m.K$\mathrm{W/m.K}$; 𝜈$\nu$ = 901 × 10−6$10^{-6}$ m2/s$\mathrm{m^2/s}$; 𝑐𝑝$c_p$ = 1880 J/kg.K$\mathrm{J/kg.K}$.

Worked solution

Data

Quantity	Symbol	Value
Cold-section diameter and length	𝐷,𝐿$D,L$	0.30,200 m$0.30,200\ \mathrm m$
Mean speed	𝑢$u$	2.0 m s−1$2.0\ \mathrm{m\,s^{-1}}$
Inlet and wall temperatures	𝑇𝑖𝑛,𝑇𝑠$T_{in},T_s$	20,0 ∘C$20,0\,{}^\circ\mathrm C$
Density	𝜌$\rho$	888 kg m−3$888\ \mathrm{kg\,m^{-3}}$
Conductivity	𝑘$k$	0.145 W m−1 K−1$0.145\ \mathrm{W\,m^{-1}\,K^{-1}}$
Kinematic viscosity	𝜈$\nu$	901 ×10−6 m2 s−1$901\times10^{-6}\ \mathrm{m^{2}\,s^{-1}}$
Specific heat	𝑐𝑝$c_p$	1880 J kg−1 K−1$1880\ \mathrm{J\,kg^{-1}\,K^{-1}}$

Assumptions

Assume steady hydrodynamically developed flow at the start of the cold section, a uniform inlet bulk temperature, constant properties and negligible viscous heating and axial conduction. Neglect pipe-wall resistance, so the inner wall is also isothermal at 𝑇𝑠 =0 ∘C$T_s=0\,^{\circ}\mathrm C$. The thermal boundary layer starts at the lake entrance. Check 𝑅𝑒 =𝑢𝐷/𝜈$Re=uD/\nu$ and 𝑃𝑟 =𝜌𝜈𝑐𝑝/𝑘$Pr=\rho\nu c_p/k$.

Calculations

For laminar flow, estimate the thermal entry length with

𝐿𝑡=0.0575𝑅𝑒 𝑃𝑟 𝐷.$$L_t=0.0575Re\,Pr\,D.$$

Use the length-average laminar entrance correlation and integrated bulk balance for constant wall temperature:

𝐺𝑧=𝑅𝑒 𝑃𝑟 𝐷𝐿,¯¯¯¯¯¯¯¯𝑁𝑢𝐷=3.66+0.065𝐺𝑧1+0.04𝐺𝑧2/3,¯¯¯¯ℎ=¯¯¯¯¯¯¯¯𝑁𝑢𝐷𝑘𝐷,𝑇𝑜𝑢𝑡−𝑇𝑠𝑇𝑖𝑛−𝑇𝑠=exp⁡(−¯¯¯¯ℎ𝜋𝐷𝐿˙𝑚𝑐𝑝).$$\begin{aligned} Gz&=Re\,Pr\,\frac{D}{L},\\ \overline{Nu}_D&=3.66+\frac{0.065Gz}{1+0.04Gz^{2/3}},\\ \overline h&=\frac{\overline{Nu}_D k}{D},\\ \frac{T_{out}-T_s}{T_{in}-T_s} &=\exp\left(-\frac{\overline h\pi DL}{\dot m c_p}\right). \end{aligned}$$(2)

The correlation models thermal development with a developed velocity profile; ignore wall-to-bulk viscosity corrections under the stated constant-property assumption. Do not substitute the fully developed value 𝑁𝑢 =3.66$Nu=3.66$ prematurely.

The oil velocity profile is laminar but its thermal boundary layer is still developing.

The supplied properties give 𝑅𝑒 =665.93$Re=665.93$, 𝑃𝑟 =10373.55$Pr=10373.55$ and 𝐿𝑡 =1.1916 ×105 m$L_t=1.1916\times10^5\,\mathrm m$, far longer than the cold section.

The large Prandtl number explains the long thermal entry region.

Equation Eq. 2 gives 𝐺𝑧 =10362.04$Gz=10362.04$, ¯¯¯¯¯¯¯¯𝑁𝑢𝐷 =37.3168$\overline{Nu}_D=37.3168$ and ¯¯¯¯ℎ =18.0364 W m−2 K−1$\overline h=18.0364\,\mathrm{W\,m^{-2}\,K^{-1}}$.

The mass flow rate is

˙𝑚=𝜌𝑢𝜋𝐷24=125.538 kg s−1.$$\dot m=\rho u\frac{\pi D^2}{4} =125.538\,\mathrm{kg\,s^{-1}}.$$

The dimensionless exponent compares the wall conductance with the heat-capacity flow rate:

¯¯¯¯ℎ𝜋𝐷𝐿˙𝑚𝑐𝑝=18.0364𝜋(0.30)(200)125.538(1880)=0.0144052$$\frac{\overline h\pi DL}{\dot m c_p}=\frac{18.0364\pi(0.30)(200)}{125.538(1880)}=0.0144052$$

The exponential bulk balance then gives

𝑇𝑜𝑢𝑡=20exp⁡(−0.0144052)=19.7140 ∘C,$$T_{out}=20\exp(-0.0144052) =19.7140\,^{\circ}\mathrm C,$$˙𝑄𝑙𝑜𝑠𝑠=˙𝑚𝑐𝑝(20−𝑇𝑜𝑢𝑡)=67508 W=67.51 kW.$$\dot Q_{loss}=\dot m c_p(20-T_{out}) =67508\,\mathrm W=67.51\,\mathrm{kW}.$$

The oil cools only 0.286 degrees because the heat-capacity flow rate is large.

This positive heat-loss convention is the magnitude of the negative heat input to the oil.

Retaining intermediate precision avoids the handwritten source’s rounded outlet temperature changing the reported duty to about 68.4 kW.

Answer

The oil exits at 19.7 ∘C$19.7\,{}^\circ\mathrm C$ and loses 67.5 kW$67.5\ \mathrm{kW}$ over the cold section. The unrounded energy balance uses a bulk temperature drop of 0.286 K. The flow is laminar but thermally developing.

Workshop 7: Forced and natural convection

Convection resistances per unit length

A thick-walled steel pipe (k = 60 W/m.K$\mathrm{W/m.K}$) carrying hot water is cooled externally by a cross-flow airstream at a velocity of 20 m/s and a temperature of 25 ∘C$25\,{}^\circ\mathrm{C}$. The inner and outer diameters of the pipe are 20 mm and 25 mm, respectively. At a certain location along the pipe, the mean temperature of the water is 80 ∘C$80\,{}^\circ\mathrm{C}$. Assuming the flow inside the tube is fully developed with a Reynolds number of 20,000, find the heat transfer rate to the air stream per unit pipe length. What will be the heat transfer rate if the surrounding air is stagnant?

Worked solution

Figure 2: Heat passes from the water through the inside film, steel wall and outside air film. The inside and outside coefficients act on different cylindrical areas.

Data

Quantity	Symbol	Value
Inside and outside diameters	𝐷𝑖,𝐷𝑜$D_i,D_o$	0.020,0.025 m$0.020,0.025\ \mathrm m$
Steel conductivity	𝑘𝑠𝑡𝑒𝑒𝑙$k_{steel}$	60 W m−1 K−1$60\ \mathrm{W\,m^{-1}\,K^{-1}}$
Water and ambient air temperatures	𝑇𝑏,𝑇∞$T_b,T_\infty$	80,25 ∘C$80,25\,{}^\circ\mathrm C$
Internal Reynolds number	Re𝑖$\mathrm{Re}_i$	20000$20000$
Cross-flow speed	𝑢∞$u_\infty$	20 m s−1$20\ \mathrm{m\,s^{-1}}$

The property approximations are

Property	Water	Moving air	Stagnant air film
Conductivity 𝑘$k$, W m−1 K−1$\mathrm{W\,m^{-1}\,K^{-1}}$	0.67$0.67$	0.0263$0.0263$	0.0282$0.0282$
Prandtl number 𝑃𝑟$Pr$	2.2$2.2$	0.707$0.707$	0.707$0.707$
Kinematic viscosity 𝜈$\nu$, m2 s−1$\mathrm{m^2\,s^{-1}}$	Not needed	15.89 ×10−6$15.89\times10^{-6}$	18.2 ×10−6$18.2\times10^{-6}$

Assumptions

• Steady radial wall conduction with negligible fouling and axial wall conduction.
• An isolated long horizontal pipe with radiation excluded.
• The local bulk water temperature is fixed in both air cases.

Water and moving-air values follow the source solution. The stagnant-air values approximate air at 325 K and stay fixed while the film temperature in 𝛽 =1/𝑇𝑓$\beta=1/T_f$ is iterated. Use 𝑔 =9.81 m s−2$g=9.81\,\mathrm{m\,s^{-2}}$.

Calculations

For internal water use the source’s Colburn form; for cylinder cross-flow use the stated Zukauskas approximation with the surface-Prandtl correction set to one:

𝑁𝑢𝑖=0.023𝑅𝑒0.8𝑖𝑃𝑟1/3𝑖,ℎ𝑖=𝑁𝑢𝑖𝑘𝑖𝐷𝑖,𝑅𝑒𝑜=𝑢∞𝐷𝑜𝜈𝑜,𝑁𝑢𝑜=0.26𝑅𝑒0.6𝑜𝑃𝑟0.37𝑜,ℎ𝑜=𝑁𝑢𝑜𝑘𝑜𝐷𝑜.$$\begin{aligned} Nu_i&=0.023Re_i^{0.8}Pr_i^{1/3},\\ h_i&=\frac{Nu_i k_i}{D_i},\\ Re_o&=\frac{u_\infty D_o}{\nu_o},\\ Nu_o&=0.26Re_o^{0.6}Pr_o^{0.37},\\ h_o&=\frac{Nu_o k_o}{D_o}. \end{aligned}$$

The outside correlation applies to 1000 <𝑅𝑒𝑜 <2 ×105$1000<Re_o<2\times10^5$ and 𝑃𝑟𝑜 <10$Pr_o<10$. This explicitly uses the source’s 1/3$1/3$ internal exponent rather than the Dittus-Boelter cooling exponent. Resistances per unit length have units m K W−1$\mathrm{m\,K\,W^{-1}}$:

𝑅′𝑖=1ℎ𝑖𝜋𝐷𝑖,𝑅′𝑤=ln⁡(𝐷𝑜𝐷𝑖)2𝜋𝑘𝑠𝑡𝑒𝑒𝑙,𝑅′𝑜=1ℎ𝑜𝜋𝐷𝑜,𝑞′=80−25𝑅′𝑖+𝑅′𝑤+𝑅′𝑜.$$\begin{aligned} R_i'&=\frac{1}{h_i\pi D_i},\\ R_w'&=\frac{\ln\left(\frac{D_o}{D_i}\right)}{2\pi k_{steel}},\\ R_o'&=\frac{1}{h_o\pi D_o},\\ q'&=\frac{80-25}{R_i'+R_w'+R_o'}. \end{aligned}$$(3)

For stagnant air, the horizontal-cylinder Churchill-Chu relation is

𝑁𝑢𝐷=[0.60+0.387𝑅𝑎1/6𝐷[1+(0.559𝑃𝑟)9/16]8/27]2.$$Nu_D=\left[0.60+\frac{0.387Ra_D^{1/6}} {\left[1+\left(\frac{0.559}{Pr}\right)^{9/16}\right]^{8/27}}\right]^2.$$

Use

𝑅𝑎𝐷=𝑔𝛽(𝑇𝑠,𝑜−25)𝐷3𝑜𝑃𝑟𝜈2$$Ra_D=\frac{g\beta(T_{s,o}-25)D_o^3Pr}{\nu^2}$$

within the range 𝑅𝑎𝐷 <1012$Ra_D<10^{12}$, and recover ℎ𝑜 =𝑁𝑢𝐷𝑘/𝐷𝑜$h_o=Nu_D k/D_o$. Evaluate the properties at

𝑇𝑓=𝑇𝑠,𝑜+252+273.15 K.$$T_f=\frac{T_{s,o}+25}{2}+273.15\ \mathrm K.$$

The outer surface temperature must satisfy both

𝑞′=80−𝑇𝑠,𝑜𝑅′𝑖+𝑅′𝑤$$q'=\frac{80-T_{s,o}}{R_i'+R_w'}$$

and

𝑞′=ℎ𝑜𝜋𝐷𝑜(𝑇𝑠,𝑜−25).$$q'=h_o\pi D_o(T_{s,o}-25).$$

Solve iteratively.

The external air film controls most of the forced-convection resistance.

The internal calculation gives 𝑁𝑢𝑖 =82.5454$Nu_i=82.5454$ and ℎ𝑖 =2765.27 W m−2 K−1$h_i=2765.27\,\mathrm{W\,m^{-2}\,K^{-1}}$.

Outside, 𝑅𝑒𝑜 =31466.3$Re_o=31466.3$ is within the stated range, 𝑁𝑢𝑜 =114.279$Nu_o=114.279$ and ℎ𝑜 =120.221 W m−2 K−1$h_o=120.221\,\mathrm{W\,m^{-2}\,K^{-1}}$.

Using Eq. 3 gives

𝑅′𝑖=0.0057555 m K W−1,$$R_i'=0.0057555\,\mathrm{m\,K\,W^{-1}},$$𝑅′𝑤=0.00059191 m K W−1,$$R_w'=0.00059191\,\mathrm{m\,K\,W^{-1}},$$𝑅′𝑜=0.105908 m K W−1,$$R_o'=0.105908\,\mathrm{m\,K\,W^{-1}},$$𝑞′𝑓𝑜𝑟𝑐𝑒𝑑=489.95 W m−1.$$q'_{forced}=489.95\,\mathrm{W\,m^{-1}}.$$

The outer surface is

25+𝑞′𝑅′𝑜=76.8901 ∘C.$$25+q'R_o'=76.8901\,^{\circ}\mathrm C.$$

The handwritten resistance labels transpose the inner and outer coefficients; their sum and approximate 489.5 W per metre result are nevertheless consistent.

Here each resistance uses its correct coefficient and diameter.

For stagnant air, trial an outer surface temperature between 25 and 80 degrees and equate the conductive/internal flux to the natural-convection flux.

The converged values are

𝑇𝑠,𝑜=79.7947 ∘C𝑇𝑓=325.547 K$$T_{s,o}=79.7947\,^{\circ}\mathrm C \qquad T_f=325.547\,\mathrm K$$𝑅𝑎𝐷=55066.9𝑁𝑢𝐷=6.66158$$Ra_D=55066.9 \qquad Nu_D=6.66158$$ℎ𝑜=7.5143 W m−2 K−1𝑞′𝑛𝑎𝑡𝑢𝑟𝑎𝑙=32.338 W m−1$$h_o=7.5143\,\mathrm{W\,m^{-2}\,K^{-1}} \qquad q'_{natural}=32.338\,\mathrm{W\,m^{-1}}$$

Both heat-rate calculations agree:

ℎ𝑜𝜋𝐷𝑜(𝑇𝑠,𝑜−25)=80−𝑇𝑠,𝑜𝑅′𝑖+𝑅′𝑤=32.338 W m−1.$$h_o\pi D_o(T_{s,o}-25)=\frac{80-T_{s,o}}{R_i'+R_w'}=32.338\ \mathrm{W\,m^{-1}}.$$

The natural-convection Rayleigh number lies within the specified range.

The final film is close to the supplied 325 K property approximation.

Stagnant air removes much less heat and leaves the wall closer to the water temperature.

Radiation was excluded, so this is a convection-only comparison.

Answer

The heat loss is approximately 490 W m−1$490\ \mathrm{W\,m^{-1}}$ in moving air and 32.3 W m−1$32.3\ \mathrm{W\,m^{-1}}$ in stagnant air. The latter is the convection-only result for the specified horizontal pipe.

Workshop 8: Evaporation and stagnant-gas diffusion

Planar diffusion through stagnant air

A tank (diameter = 1m, height = 3 m) containing liquid methanol (MW = 32 g/mol) has its top open to the atmosphere. The level of methanol in the tank is maintained at 0.5 m. The gas space inside the tank is stagnant and the methanol vapours are immediately dispersed once they exit the tank. Engineers on‐site are concerned that this open tank may be emitting a considerable amount of methanol vapour.

a. What is the emission rate of methanol vapour from the tank (in kg methanol/day$\mathrm{kg\ methanol/day}$) when the temperature of the tank is 30 ∘C$30\,{}^\circ\mathrm{C}$? State all assumptions and boundary conditions.

b. If the temperature of the tank is raised to 40 ∘C$40\,{}^\circ\mathrm{C}$, what is the new emission rate?

Data: Vapour pressure of MeOH = 163 mmHg (30 ∘C$30\,{}^\circ\mathrm{C}$); 265 mmHg (40 ∘C$40\,{}^\circ\mathrm{C}$)

𝐷𝐴𝐵$D_{AB}$ (methanol/air) @ 298 K = 0.162 × 10−4$10^{-4}$ m2/s$\mathrm{m^2/s}$.

Worked solution

Figure 1: The open tank has the same one-dimensional gas-space model as this evaporation tube. The diffusion distance is the tank height minus the maintained liquid depth.

Data

Quantity	Symbol	Value
Tank diameter	𝑑$d$	1.0 m$1.0\ \mathrm m$
Gas depth above maintained liquid level	𝐿$L$	3.0 −0.5 =2.50 m$3.0-0.5=2.50\ \mathrm m$
Atmospheric pressure	𝑃$P$	101325 Pa =760 mmHg$101325\ \mathrm{Pa}=760\ \mathrm{mmHg}$
Methanol molar mass	𝑀𝐴$M_A$	0.032 kg mol−1$0.032\ \mathrm{kg\,mol^{-1}}$
Reference diffusivity	𝐷𝐴𝐵(298)$D_{AB}(298)$	1.62 ×10−5 m2 s−1$1.62\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$
Saturation pressure at 30 and 40 degrees Celsius	𝑝𝑠𝑎𝑡$p_{sat}$	163,265 mmHg$163,265\ \mathrm{mmHg}$

We need the total daily emission from the open tank at each temperature.

Assumptions

• Steady one-dimensional diffusion through the gas space.
• Uniform temperature in each calculation and ideal-gas behaviour.
• Constant pressure, maintained liquid level and saturated vapour at the surface.
• Zero net air flux through the liquid; dry gas at the tank mouth.
• Natural convection, wind penetration and transport through the walls are neglected.

Calculations

Let A denote methanol and B denote air. Coordinate 𝑠$s$ points upwards from the liquid. Zero net air flux permits Stefan flow of the mixture, so use

𝑁𝐴=−𝑐𝐷𝐴𝐵1−𝑦𝐴𝑑𝑦𝐴𝑑𝑠,𝑐=𝑃𝑅𝑇$$N_A=-\frac{cD_{AB}}{1-y_A}\frac{dy_A}{ds},\qquad c=\frac{P}{RT}$$(1)

Here 𝑅 =8.314 Pa m3 mol−1 K−1$R=8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}$. The boundary values are 𝑦𝐴(0) =𝑝𝑠𝑎𝑡/𝑃$y_A(0)=p_{sat}/P$ and 𝑦𝐴(𝐿) =0$y_A(L)=0$. At the same pressure, estimate the change in diffusivity from

𝐷𝐴𝐵(𝑇)=𝐷𝐴𝐵(298)(𝑇298)3/2$$D_{AB}(T)=D_{AB}(298)\left(\frac{T}{298}\right)^{3/2}$$

Integrate Eq. 1 across the gas space and multiply the flux by the open cross-sectional area, molar mass and seconds per day.

The maintained liquid level fixes the gas diffusion distance throughout each calculation.

Since 𝑁𝐴$N_A$ is constant over the planar gas space, integration gives

𝑁𝐴=𝑐𝐷𝐴𝐵𝐿ln⁡1−𝑦𝐴(𝐿)1−𝑦𝐴(0).$$N_A=\frac{cD_{AB}}{L}\ln\frac{1-y_A(L)}{1-y_A(0)}.$$(2)

At 30 degrees, 𝑇 =303.15 K$T=303.15\,\mathrm K$,

𝑦𝐴(0)=163760=0.214474,$$y_A(0)=\frac{163}{760}=0.214474,$$

𝑐 =40.2021 mol m−3$c=40.2021\,\mathrm{mol\,m^{-3}}$ and 𝐷𝐴𝐵 =1.66218 ×10−5 m2 s−1$D_{AB}=1.66218\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

Using Eq. 2 gives 𝑁𝐴 =6.45246 ×10−5 mol m−2 s−1$N_A=6.45246\times10^{-5}\,\mathrm{mol\,m^{-2}\,s^{-1}}$.

The emitting area is

𝐴=𝜋(1.0)24=0.785398 m2,$$A=\frac{\pi(1.0)^2}{4}=0.785398\,\mathrm{m^2},$$

hence

˙𝑚𝐴=𝑀𝐴𝐴𝑁𝐴=1.62168×10−6 kg s−1,$$\dot m_A=M_A A N_A =1.62168\times10^{-6}\,\mathrm{kg\,s^{-1}},$$𝑚𝐴,𝑑𝑎𝑦=86400˙𝑚𝐴=0.1401 kg day−1.$$m_{A,day}=86400\dot m_A =0.1401\,\mathrm{kg\,day^{-1}}.$$

At 40 degrees, 𝑇 =313.15 K$T=313.15\,\mathrm K$, 𝑦𝐴(0) =0.348684$y_A(0)=0.348684$, 𝑐 =38.9183 mol m−3$c=38.9183\,\mathrm{mol\,m^{-3}}$ and 𝐷𝐴𝐵 =1.74510 ×10−5 m2 s−1$D_{AB}=1.74510\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

The same integration yields

𝑁𝐴=1.16479×10−4 mol m−2 s−1𝑚𝐴,𝑑𝑎𝑦=0.2529 kg day−1$$N_A=1.16479\times10^{-4}\,\mathrm{mol\,m^{-2}\,s^{-1}} \qquad m_{A,day}=0.2529\,\mathrm{kg\,day^{-1}}$$

The predicted emission increases by a factor of 1.805.

Increased saturation vapour pressure is the main cause, with a smaller diffusivity increase and a slight gas-concentration decrease.

These rates require the assumed stagnant gas space; ventilation or buoyant circulation would change the model.

Answer

The methanol emission is 0.140 kg day−1$0.140\ \mathrm{kg\,day^{-1}}$ at 30 ∘C$30\,{}^\circ\mathrm C$ and 0.253 kg day−1$0.253\ \mathrm{kg\,day^{-1}}$ at 40 ∘C$40\,{}^\circ\mathrm C$, under the stagnant-gas model.

Workshop 8: Evaporation and stagnant-gas diffusion

Spherical diffusion and a shrinking liquid balance

Consider the evaporation of water droplets (diameter ~ 0.1 mm) in still air at 50 ∘C$50\,{}^\circ\mathrm{C}$, 1 atm pressure, with 20% relative humidity. As the droplets are very small, it can be considered that their terminal velocity is negligible, i.e. they are suspended in air. The droplet temperature can be assumed to be at 50 ∘C$50\,{}^\circ\mathrm{C}$ and its vapour pressure is 12 kPa. If the diffusion coefficient of water vapour in air is 0.288 cm2/s$\mathrm{cm^2/s}$,

a. Calculate the initial rate of evaporation of a single droplet, assuming steady state condition.

b. Determine the time required for the water droplet to evaporate completely.

Worked solution

Figure 2: The droplet evaporates into air with a fixed far-field humidity. Its surface concentration is saturated, while its radius decreases as liquid is lost. The dashed outer boundary is taken to infinity in this problem.

Data

Property	Value
Total pressure 𝑃$P$	101325 Pa$101325\,\mathrm{Pa}$
Temperature 𝑇$T$	323.15 K$323.15\,\mathrm K$
Water saturation vapour pressure 𝑝𝑠𝑎𝑡$p_{sat}$	12000 Pa$12000\,\mathrm{Pa}$
Water-air diffusivity 𝐷𝐴𝐵$D_{AB}$	0.288 cm2 s−1$0.288\,\mathrm{cm^2\,s^{-1}}$
Liquid water density 𝜌𝐿$\rho_L$	1000 kg m−3$1000\,\mathrm{kg\,m^{-3}}$
Water molar mass 𝑀𝐴$M_A$	0.018 kg mol−1$0.018\,\mathrm{kg\,mol^{-1}}$

Assumptions

Assume spherical symmetry, negligible droplet velocity, constant properties and no air dissolving in the droplet. The gas profile adjusts much faster than the droplet shrinks, so treat it as steady at each instantaneous radius 𝑎(𝑡)$a(t)$. An external heat supply maintains the liquid and gas at 50 degrees; ignore evaporative cooling. Far-field humidity stays fixed in an effectively infinite gas domain. At 𝑟 =𝑎$r=a$, 𝑦𝑠 =𝑝𝑠𝑎𝑡/𝑃$y_s=p_{sat}/P$. At 𝑟 →∞$r\to\infty$, 𝑦∞ =0.20𝑝𝑠𝑎𝑡/𝑃$y_\infty=0.20p_{sat}/P$. Relative humidity is a ratio of water partial pressure to saturation pressure, not a gas mole fraction.

Calculations

Derive 𝑟2𝑁𝐴 =constant$r^2N_A=\text{constant}$ from a spherical gas shell balance. Integrate Eq. 1 radially using 𝑁𝐵 =0$N_B=0$. Then use the shrinking liquid balance

−𝜌𝐿4𝜋𝑎2𝑑𝑎𝑑𝑡=𝑀𝐴˙𝑛𝐴,$$-\rho_L4\pi a^2\frac{da}{dt}=M_A\dot n_A,$$

where ˙𝑛𝐴 =4𝜋𝑟2𝑁𝐴$\dot n_A=4\pi r^2N_A$ is the molar evaporation rate. Use the instantaneous rate as the radius decreases; do not divide the initial mass by the initial rate to obtain the lifetime.

The spherical area increases outward while the total molar evaporation rate stays constant.

A steady shell balance gives

𝑑(𝑟2𝑁𝐴)𝑑𝑟=0.$$\frac{d(r^2N_A)}{dr}=0.$$

Write 𝑟2𝑁𝐴 =𝐶$r^2N_A=C$ and integrate the flux law between 𝑎$a$ and infinity:

𝐶𝑎=𝑐𝐷𝐴𝐵Λ,Λ=ln⁡1−𝑦∞1−𝑦𝑠,˙𝑛𝐴=4𝜋𝑎𝑐𝐷𝐴𝐵Λ.$$\begin{aligned} \frac{C}{a}&=cD_{AB}\Lambda,\\ \Lambda&=\ln\frac{1-y_\infty}{1-y_s},\\ \dot n_A&=4\pi a cD_{AB}\Lambda. \end{aligned}$$(3)

Convert 𝐷𝐴𝐵 =2.88 ×10−5 m2 s−1$D_{AB}=2.88\times10^{-5}\,\mathrm{m^2\,s^{-1}}$ and

𝑎0=𝑑02=5.00×10−5 m.$$a_0=\frac{d_0}{2}=5.00\times10^{-5}\,\mathrm m.$$

The boundary fractions are 𝑦𝑠 =0.118431$y_s=0.118431$ and 𝑦∞ =0.0236862$y_\infty=0.0236862$, so Λ =0.102081$\Lambda=0.102081$ and 𝑐 =37.7140 mol m−3$c=37.7140\,\mathrm{mol\,m^{-3}}$.

Using the initial radius in Eq. 3 gives

˙𝑛𝐴(0)=4𝜋(5.00×10−5)(37.7140)(2.88×10−5)(0.102081)$$\dot n_A(0)=4\pi(5.00\times10^{-5})(37.7140)(2.88\times10^{-5})(0.102081)$$

Evaluating the initial rate gives

˙𝑛𝐴(0)=6.96655×10−8 mol s−1,$$\dot n_A(0)=6.96655\times10^{-8}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝐴(0)=𝑀𝐴˙𝑛𝐴(0)=1.2540×10−9 kg s−1.$$\dot m_A(0)=M_A\dot n_A(0) =1.2540\times10^{-9}\,\mathrm{kg\,s^{-1}}.$$

Substitute the instantaneous rate into the liquid balance, cancel 4𝜋𝑎$4\pi a$ and integrate:

−𝜌𝐿𝑎𝑑𝑎𝑑𝑡=𝑀𝐴𝑐𝐷𝐴𝐵Λ,𝑎2(𝑡)=𝑎20−2𝑀𝐴𝑐𝐷𝐴𝐵Λ𝜌𝐿𝑡,𝑡𝑓=𝜌𝐿𝑎202𝑀𝐴𝑐𝐷𝐴𝐵Λ=0.6263 s.$$\begin{aligned} -\rho_L a\frac{da}{dt}&=M_AcD_{AB}\Lambda,\\ a^2(t)&=a_0^2-\frac{2M_AcD_{AB}\Lambda}{\rho_L}t,\\ t_f&=\frac{\rho_La_0^2}{2M_AcD_{AB}\Lambda}\\ &=0.6263\,\mathrm s. \end{aligned}$$

The initial liquid mass is 5.2360 ×10−10 kg$5.2360\times10^{-10}\,\mathrm{kg}$.

Dividing it by the initial mass rate gives 0.4175 s and underestimates the lifetime because the rate decreases with radius.

The gas diffusion time

𝑎20𝐷𝐴𝐵=8.68×10−5 s$$\frac{a_0^2}{D_{AB}}=8.68\times10^{-5}\,\mathrm s$$

is much shorter than 𝑡𝑓$t_f$, which supports the quasi-steady approximation.

Complete evaporation is the 𝑎 →0$a\to0$ limit of this continuum model.

Cooling and molecular-scale effects are outside the assumptions.

Answer

The initial mass evaporation rate is 1.25 ×10−9 kg s−1$1.25\times10^{-9}\ \mathrm{kg\,s^{-1}}$ per droplet. The shrinking-sphere model gives a lifetime of 0.626 s$0.626\ \mathrm s$, with the droplet temperature maintained at 50 ∘C$50\,{}^\circ\mathrm C$.

Workshop 9: Diffusion with surface reaction

Cylindrical diffusion to a shrinking rod

A cylindrical graphite (pure carbon) rod (length = 25 cm, initial diameter = 2 cm) is placed in a combustion chamber with flowing air stream at 1100 K and 2 atm total pressure. A stagnant gas boundary layer 5 mm thick is formed around the surface of the rod. At 1100 K, solid carbon oxidises to carbon dioxide gas,

C (s)+O2 (g)→CO2 (g)$$\mathrm{C\ (s) + O_2\ (g) \rightarrow CO_2\ (g)}$$

Assuming that the oxidation is limited by molecular diffusion of O2$\mathrm{O_2}$ through the stagnant gas film, and outside of the gas film, the bulk composition of air stream prevails,

a) Estimate the initial rate of CO2$\mathrm{CO_2}$ production from the rod;

b) Write an equation for the rate of change in the radius of the rod as a function of time.

Data: Density of solid graphite = 2.25 g/cm3$\mathrm{g/cm^3}$; 𝐷𝐴𝐵$D_{AB}$ (CO2$\mathrm{CO_2}$-air) @ 273 K and 1 atm = 0.136 cm2/s$\mathrm{cm^2/s}$

Worked solution

Figure 1: Radial diffusion through a gas film around the graphite rod. The radius a decreases during oxidation, while the film thickness remains fixed. The drawing is a cylindrical cross-section; oxygen diffuses inward and carbon dioxide outward.

Data

Quantity	Symbol	Value
Initial radius and rod length	𝑎0,ℓ$a_0,\ell$	0.010,0.25 m$0.010,0.25\ \mathrm m$
Gas-film thickness	𝛿$\delta$	0.0050 m$0.0050\ \mathrm m$
Gas temperature and pressure	𝑇,𝑃$T,P$	1100 K,2 atm$1100\ \mathrm K,2\ \mathrm{atm}$

Property	Value
Graphite density 𝜌𝐶$\rho_C$	2250 kg m−3$2250\,\mathrm{kg\,m^{-3}}$
Carbon molar mass 𝑀𝐶$M_C$	0.012 kg mol−1$0.012\,\mathrm{kg\,mol^{-1}}$
Carbon-dioxide molar mass 𝑀𝐶𝑂2$M_{CO_2}$	0.044 kg mol−1$0.044\,\mathrm{kg\,mol^{-1}}$
Bulk oxygen mole fraction 𝑦𝑏$y_b$	0.21
Reference effective gas diffusivity 𝐷0$D_0$	0.136 cm2 s−1$0.136\,\mathrm{cm^2\,s^{-1}}$ at 273 K and 1 atm

Assumptions

Treat the source’s CO2-air diffusivity as a common effective diffusivity for the oxygen and carbon-dioxide counterdiffusion. This is an approximation to multicomponent air. Nitrogen has zero net flux, and the reaction exchanges one gas mole for one gas mole, so total gas molar flux is zero. Estimate the diffusivity by

𝐷=𝐷01 atm𝑃(𝑇273 K)3/2.$$D=D_0\frac{1\,\mathrm{atm}}{P}\left(\frac{T}{273\,\mathrm K}\right)^{3/2}.$$

Assume isothermal quasi-steady radial diffusion, no gas-phase reaction, constant density and length, and no transport through the ends. The maintained bulk gas has 𝑦𝑂2(𝑎 +𝛿) =𝑦𝑏$y_{O_2}(a+\delta)=y_b$, while the fast reaction gives 𝑦𝑂2(𝑎) =0$y_{O_2}(a)=0$. Keep 𝛿$\delta$ fixed as 𝑎$a$ decreases. Derive the constant 𝑟𝑁𝑂2$rN_{O_2}$ from a cylindrical shell balance, integrate Eq. 1, and relate oxygen consumption to the changing carbon mass. Do not use the initial area as a constant area for the shrinking rod.

Calculations

Use molar flux 𝑁𝐴$N_A$ in mol m−2 s−1$\mathrm{mol\,m^{-2}\,s^{-1}}$ and ideal-gas concentration 𝑐 =𝑃/(𝑅𝑇)$c=P/(RT)$, with 𝑅 =8.314 Pa m3 mol−1 K−1$R=8.314\,\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}}$. With negligible total molar flux, diffusion obeys

𝑁𝐴=−𝑐𝐷𝑑𝑦𝐴𝑑𝑠.$$N_A=-cD\frac{dy_A}{ds}.$$(1)

The coordinate 𝑠$s$ increases away from the reacting surface. Reactant flux is therefore negative. Report removal and production rates as positive magnitudes.

The oxygen consumption balances carbon-dioxide production mole for mole at the rod surface.

A steady cylindrical shell balance gives

𝑑(𝑟𝑁𝑂2)𝑑𝑟=0.$$\frac{d(rN_{O_2})}{dr}=0.$$

The positive oxygen consumption rate 𝑞$q$ satisfies 𝑞 = −2𝜋𝑟ℓ𝑁𝑂2$q=-2\pi r\ell N_{O_2}$.

Integrating Eq. 1 between 𝑎$a$ and 𝑎 +𝛿$a+\delta$ gives

𝑞(𝑎)=2𝜋ℓ𝑐𝐷𝑦𝑏ln⁡(𝑎+𝛿𝑎).$$q(a)=\frac{2\pi\ell cD y_b}{\ln\left(\frac{a+\delta}{a}\right)}.$$(2)

Convert 𝐷0 =1.36 ×10−5 m2 s−1$D_0=1.36\times10^{-5}\,\mathrm{m^2\,s^{-1}}$.

The scaled 𝐷 =5.49989 ×10−5 m2 s−1$D=5.49989\times10^{-5}\,\mathrm{m^2\,s^{-1}}$ and 𝑐 =22.1587 mol m−3$c=22.1587\,\mathrm{mol\,m^{-3}}$.

At 𝑎0$a_0$,

ln⁡(𝑎0+𝛿𝑎0)=ln⁡(1.5)=0.405465.$$\ln\left(\frac{a_0+\delta}{a_0}\right)=\ln(1.5)=0.405465.$$

Using Eq. 2, substitute the length and cylindrical logarithmic factor:

𝑞(𝑎0)=2𝜋(0.25)(22.1587)(5.49989×10−5)(0.21)0.405465$$q(a_0)=\frac{2\pi(0.25)(22.1587)(5.49989\times10^{-5})(0.21)}{0.405465}$$

Evaluating this expression gives

𝑞(𝑎0)=9.91479×10−4 mol s−1,$$q(a_0)=9.91479\times10^{-4}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝐶𝑂2(0)=𝑀𝐶𝑂2𝑞(𝑎0)=4.36251×10−5 kg s−1.$$\dot m_{CO_2}(0)=M_{CO_2}q(a_0) =4.36251\times10^{-5}\,\mathrm{kg\,s^{-1}}.$$

The carbon mass is 𝑚𝐶 =𝜌𝐶𝜋𝑎2ℓ$m_C=\rho_C\pi a^2\ell$.

Its balance is

2𝜋𝜌𝐶𝑎ℓ𝑑𝑎𝑑𝑡=−𝑀𝐶𝑞(𝑎).$$2\pi\rho_C a\ell\frac{da}{dt}=-M_Cq(a).$$

Combining this balance with Eq. 2 yields

𝑑𝑎𝑑𝑡=−𝑀𝐶𝑐𝐷𝑦𝑏𝜌𝐶𝑎ln⁡(𝑎+𝛿𝑎),𝑎(0)=𝑎0.$$\frac{da}{dt}=-\frac{M_C cD y_b}{\rho_C a\ln\left(\frac{a+\delta}{a}\right)}, \qquad a(0)=a_0.$$(3)

Initially

𝑑𝑎𝑑𝑡=−3.36637×10−7 m s−1.$$\frac{da}{dt}=-3.36637\times10^{-7}\,\mathrm{m\,s^{-1}}.$$

The negative sign represents loss of radius.

Carbon consumption uses 𝑀𝐶$M_C$; carbon-dioxide mass production uses 𝑀𝐶𝑂2$M_{CO_2}$.

The gas-film diffusion time 𝛿2/𝐷 =0.455 s$\delta^2/D=0.455\,\mathrm s$ is much shorter than the initial shrinkage scale

𝑎0∣𝑑𝑎𝑑𝑡∣=2.97×104 s,$$\frac{a_0}{\left|\frac{da}{dt}\right|}=2.97\times10^4\,\mathrm s,$$

supporting quasi-steady transport initially.

The model excludes combustion heating, carbon-monoxide formation, end effects and changes in the flow as the rod shrinks.

Answer

The initial carbon-dioxide production rate is 9.91 ×10−4 mol s−1$9.91\times10^{-4}\ \mathrm{mol\,s^{-1}}$, or 4.36 ×10−5 kg s−1$4.36\times10^{-5}\ \mathrm{kg\,s^{-1}}$. Radius evolves according to Eq. 3, starting from 𝑎0 =0.010 m$a_0=0.010\ \mathrm m$; initially

𝑑𝑎𝑑𝑡=−3.37×10−7 m s−1.$$\frac{da}{dt}=-3.37\times10^{-7}\ \mathrm{m\,s^{-1}}.$$

Workshop 9: Diffusion with surface reaction

Film resistance and finite surface reaction

Nitric oxide (NO) emission from automobile exhaust can be reduced by using a catalytic converter. The following reaction occurs at the catalytic surface:

NO+CO→0.5 N2+CO2$$\mathrm{NO + CO \rightarrow 0.5\ N_2 + CO_2}$$

The concentration of NO is reduced by passing the exhaust gases over the surface of the catalyst, and the rate of reduction at the catalyst surface is as follows:

𝑁𝑁𝑂=−𝑘′′𝐶𝑁𝑂(0),$$N_{NO}=-k'' C_{NO(0)},$$

where 𝐶𝑁𝑂(0)$C_{NO(0)}$ is the concentration of NO at the catalyst surface.

As a first approximation, it may be assumed that NO reaches the surface by one-dimensional diffusion through a thin gas film of thickness L that adjoins the surface, and that the effects of bulk motion is negligible. Determine the mole fraction of NO at the catalytic surface and the NO removal rate for a catalytic surface area of 200 cm2$\mathrm{cm^2}$.

Data: The exhaust gas is at 500 ∘C$500\,{}^\circ\mathrm{C}$ and 1.2 bar, the mole fraction of NO at L is 0.15, 𝐷𝐴𝐵$D_{AB}$ = 10−4$10^{-4}$ m2/s$\mathrm{m^2/s}$, 𝑘′′$k''$ = 0.05 m/s and the film thickness is L = 1 mm.

Worked solution

Figure 2: Nitric oxide diffuses through a nonreacting gas film to the catalyst. Its surface concentration is not zero when surface reaction is finite. The solution measures x from the catalyst toward the bulk.

Data

Property	Value
Bulk NO mole fraction 𝑦𝑏$y_b$	0.15
Film thickness 𝐿$L$	0.0010 m$0.0010\,\mathrm m$
Effective NO diffusivity 𝐷$D$	1.0 ×10−4 m2 s−1$1.0\times10^{-4}\,\mathrm{m^2\,s^{-1}}$
Surface rate constant 𝑘𝑠$k_s$	0.050 m s−1$0.050\,\mathrm{m\,s^{-1}}$
Catalytic area 𝐴𝑠$A_s$	200 cm2$200\,\mathrm{cm^2}$
NO molar mass 𝑀𝑁𝑂$M_{NO}$	0.030 kg mol−1$0.030\,\mathrm{kg\,mol^{-1}}$

Assumptions

Assume steady isothermal one-dimensional diffusion with constant pressure and 𝐷$D$, no gas-phase reaction, sufficient CO, and the source’s first approximation of negligible bulk-motion contribution to NO transport. The reaction changes total gas mole number, so this approximation is not exact equimolar diffusion. Do not interpret the 0.15 mole fraction as realistic dilute exhaust data. It is the supplied idealised problem value. At the surface,

𝑁𝑁𝑂(0)=−𝑘𝑠𝑐𝑦𝑠,𝑦𝑠=𝑦𝑁𝑂(0),$$N_{NO}(0)=-k_s c y_s, \qquad y_s=y_{NO}(0),$$

and 𝑦𝑁𝑂(𝐿) =𝑦𝑏$y_{NO}(L)=y_b$. Integrate Eq. 1 in 𝑥$x$ and equate the film flux to the surface reaction flux.

Calculations

The finite catalytic reaction leaves a nonzero nitric-oxide concentration at the surface.

The steady flux is constant and the profile is linear:

𝑦𝑁𝑂(𝑥)=𝑦𝑠+(𝑦𝑏−𝑦𝑠)𝑥𝐿,$$y_{NO}(x)=y_s+(y_b-y_s)\frac{x}{L},$$𝑁𝑁𝑂=−𝑐𝐷𝐿(𝑦𝑏−𝑦𝑠).$$N_{NO}=-\frac{cD}{L}(y_b-y_s).$$

Matching the reaction flux gives

𝐷𝐿(𝑦𝑏−𝑦𝑠)=𝑘𝑠𝑦𝑠,𝑦𝑠=𝑦𝑏1+𝑘𝑠𝐿𝐷=0.151+0.50=0.100.$$\begin{aligned} \frac{D}{L}(y_b-y_s)&=k_s y_s,\\ y_s&=\frac{y_b}{1+\frac{k_sL}{D}}\\ &=\frac{0.15}{1+0.50}=0.100. \end{aligned}$$(4)

Convert 𝑇 =773.15 K$T=773.15\,\mathrm K$, 𝑃 =120000 Pa$P=120000\,\mathrm{Pa}$ and 𝐴𝑠 =0.0200 m2$A_s=0.0200\,\mathrm{m^2}$.

Then 𝑐 =18.6684 mol m−3$c=18.6684\,\mathrm{mol\,m^{-3}}$ and 𝑁𝑁𝑂 = −0.0933421 mol m−2 s−1$N_{NO}=-0.0933421\,\mathrm{mol\,m^{-2}\,s^{-1}}$.

The positive removal rate is

˙𝑛𝑁𝑂=−𝐴𝑠𝑁𝑁𝑂=1.86684×10−3 mol s−1,$$\dot n_{NO}=-A_sN_{NO} =1.86684\times10^{-3}\,\mathrm{mol\,s^{-1}},$$˙𝑚𝑁𝑂=𝑀𝑁𝑂˙𝑛𝑁𝑂=5.60052×10−5 kg s−1.$$\dot m_{NO}=M_{NO}\dot n_{NO} =5.60052\times10^{-5}\,\mathrm{kg\,s^{-1}}.$$

The equivalent concentration-based coefficient is

𝑘𝑜𝑣𝑒𝑟𝑎𝑙𝑙=(𝐿𝐷+1𝑘𝑠)−1.$$k_{overall}=\left(\frac{L}{D}+\frac{1}{k_s}\right)^{-1}.$$

Its value is 0.0333333 m s−1$0.0333333\,\mathrm{m\,s^{-1}}$.

The film resistance 𝐿/𝐷 =10 s m−1$L/D=10\,\mathrm{s\,m^{-1}}$ and reaction resistance 1/𝑘𝑠 =20 s m−1$1/k_s=20\,\mathrm{s\,m^{-1}}$ add.

Reaction supplies twice the film resistance, but neither can be ignored.

The calculated surface fraction in Eq. 4 confirms finite reaction, rather than a perfectly absorbing surface.

These results apply only to the stated film approximation.

Answer

The surface NO mole fraction is 0.100$0.100$. The positive removal rate is 1.87 ×10−3 mol s−1$1.87\times10^{-3}\ \mathrm{mol\,s^{-1}}$, equivalent to 5.60 ×10−5 kg s−1$5.60\times10^{-5}\ \mathrm{kg\,s^{-1}}$, within the stipulated film approximation.

Workshop 10: Transient mass diffusion

Time for case hardening

The surface of a mild steel component is commonly hardened by packing the component in a carbonaceous material in a furnace at a high temperature for a predetermined time. Consider such a component with a uniform initial carbon concentration of 0.15% by mass. The component is packed in a carbonaceous material and is placed in a high‐temperature furnace. The diffusion coefficient of carbon in steel at the furnace temperature is given to be 4.8 × 10−10$10^{-10}$ m2/s$\mathrm{m^2/s}$, and the equilibrium concentration of carbon in the iron at the interface is determined from equilibrium data to be 1.2% by mass. Determine how long the component should be kept in the furnace for the mass concentration of carbon 0.5 mm below the surface to reach 1%.

Worked solution

Figure 1: Carbon enters the steel through a maintained surface concentration. The target is at a depth measured inward from that surface; the far boundary remains unaffected.

Data

Quantity	Symbol	Value
Initial, surface and target mass fractions	𝑤𝑖,𝑤𝑠,𝑤$w_i,w_s,w$	0.0015,0.012,0.010$0.0015,0.012,0.010$
Target depth	𝑥$x$	0.00050 m$0.00050\ \mathrm m$
Diffusivity at furnace temperature	𝐷$D$	4.8 ×10−10 m2 s−1$4.8\times10^{-10}\ \mathrm{m^{2}\,s^{-1}}$

Assumptions

Assume constant diffusivity, no bulk motion or reaction inside each solid, uniform initial concentration and a maintained surface concentration. The surface condition changes at 𝑡 =0$t=0$. For constant density, carbon mass fraction obeys the same diffusion equation as concentration.

Calculations

Define the normalised concentration

Θ=𝐶−𝐶𝑠𝐶𝑖−𝐶𝑠.$$\Theta=\frac{C-C_s}{C_i-C_s}.$$(1)

Use the same ratio with mass fraction 𝑤$w$ replacing 𝐶$C$ in problem 1. The error function is

erf⁡(𝑧)=2√𝜋∫𝑧0𝑒−𝑢2 𝑑𝑢.$$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-u^2}\,du.$$

For interpolation, erf⁡(0.170) =0.189992$\operatorname{erf}(0.170)=0.189992$ and erf⁡(0.171) =0.191088$\operatorname{erf}(0.171)=0.191088$. A calculator with an inverse error function can also be used.

Treat the solid as a semi-infinite medium with a planar surface at 𝑥 =0$x=0$, where 𝑥$x$ increases into the steel. Other surfaces are far beyond the diffusion penetration depth during the treatment. Assume no external transport resistance and constant steel density. Use fractions 0.0015, 0.012 and 0.010, or use percentages consistently. The semi-infinite solution is

𝑤−𝑤𝑖𝑤𝑠−𝑤𝑖=1−erf⁡(𝑥2√𝐷𝑡).$$\frac{w-w_i}{w_s-w_i}=1-\operatorname{erf}\left(\frac{x}{2\sqrt{Dt}}\right).$$(2)

The steel treatment time follows from the inverse error function at the specified depth.

The required fraction of the surface concentration change is

0.010−0.00150.012−0.0015=0.809524.$$\frac{0.010-0.0015}{0.012-0.0015}=0.809524.$$

Hence Eq. 2 gives erf⁡(𝑧) =0.190476$\operatorname{erf}(z)=0.190476$, where

𝑧=𝑥2√𝐷𝑡.$$z=\frac{x}{2\sqrt{Dt}}.$$

Inversion gives 𝑧 =0.170441$z=0.170441$.

Using 𝑥 =5.0 ×10−4 m$x=5.0\times10^{-4}\,\mathrm m$,

𝑡=𝑥24𝐷𝑧2=4482.18 s=74.703 min.$$t=\frac{x^2}{4Dz^2} =4482.18\,\mathrm s=74.703\,\mathrm{min}.$$

Interpolation of the supplied table gives about 4482 s$4482\,\mathrm s$.

The source’s rounded 𝑧 =0.17$z=0.17$ gives about 4505 s$4505\,\mathrm s$.

The characteristic penetration distance 2√𝐷𝑡 =2.93 mm$2\sqrt{Dt}=2.93\,\mathrm{mm}$ must be small relative to the component dimensions for this planar semi-infinite model to apply.

The answer excludes furnace heating time.

Answer

The required diffusion time at furnace temperature is 4.48 ×103 s$4.48\times10^3\ \mathrm s$, approximately 74.7 min$74.7\ \mathrm{min}$. Furnace warm-up time is additional.

Workshop 10: Transient mass diffusion

Centre concentration in a finite slab

A large sheet of material 40 mm thick contains dissolved hydrogen at a uniform concentration of 3 kmol/m3$\mathrm{kmol/m^3}$. The sheet is exposed to a fluid stream, causing the concentration of the dissolved hydrogen to be reduced suddenly to zero at both surfaces. This surface condition is maintained constant thereafter. If the mass diffusivity of hydrogen is 9 × 10−7$10^{-7}$ m2/s$\mathrm{m^2/s}$, how much time is required to bring the density of dissolved hydrogen to 1.2 kg/m3$\mathrm{kg/m^3}$ at the centre of the sheet?

Worked solution

Figure 2: Hydrogen leaves both faces of the sheet under equal surface conditions. Use half-thickness L between the centre plane and either face.

Data

Quantity	Symbol	Value
Sheet thickness and half-thickness	2𝐿,𝐿$2L,L$	0.040,0.020 m$0.040,0.020\ \mathrm m$
Initial molar concentration	𝐶𝑖$C_i$	3.0 kmol m−3$3.0\ \mathrm{kmol\,m^{-3}}$
Target centre mass concentration	𝜌𝐻2,0$\rho_{H_2,0}$	1.2 kg m−3$1.2\ \mathrm{kg\,m^{-3}}$
Surface concentration	𝐶𝑠$C_s$	0$0$
Diffusivity	𝐷$D$	9.0 ×10−7 m2 s−1$9.0\times10^{-7}\ \mathrm{m^{2}\,s^{-1}}$
Molecular hydrogen molar mass	𝑀𝐻2$M_{H_2}$	2.0 kg kmol−1$2.0\ \mathrm{kg\,kmol^{-1}}$

The worked solution uses molecular hydrogen molar mass 𝑀𝐻2 =2.0 kg kmol−1$M_{H_2}=2.0\,\mathrm{kg\,kmol^{-1}}$.

Assumptions

Assume one-dimensional diffusion normal to the faces, negligible edge effects and no external transport resistance. Set 𝑥 =0$x=0$ at the centre, with faces at 𝑥 = ±𝐿$x=\pm L$ and half-thickness 𝐿 =20 mm$L=20\,\mathrm{mm}$. Define 𝐹𝑜 =𝐷𝑡/𝐿2$Fo=Dt/L^2$.

Calculations

The exact centre concentration ratio is

Θ0=4𝜋∞∑𝑛=0(−1)𝑛2𝑛+1exp⁡[−(2𝑛+1)2𝜋2𝐹𝑜4].$$\Theta_0=\frac4\pi\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1} \exp\left[-\frac{(2n+1)^2\pi^2Fo}{4}\right].$$

For 𝐹𝑜 ≳0.2$Fo\gtrsim0.2$, use the first-term approximation

Θ0≃4𝜋exp⁡(−𝜋2𝐹𝑜4).$$\Theta_0\simeq\frac4\pi\exp\left(-\frac{\pi^2Fo}{4}\right).$$(3)

Use Eq. 1 with 𝐶𝑠 =0$C_s=0$. Convert the target mass concentration to molar concentration before forming the ratio. Check that the calculated 𝐹𝑜$Fo$ is consistent with the approximation.

The sheet target is one fifth of its initial hydrogen concentration after unit conversion.

Convert the centre target to

𝐶0=1.22.0=0.60 kmol m−3.$$C_0=\frac{1.2}{2.0}=0.60\,\mathrm{kmol\,m^{-3}}.$$

Then

Θ0=0.603.0=0.20.$$\Theta_0=\frac{0.60}{3.0}=0.20.$$

Solving Eq. 3 gives

𝐹𝑜=4𝜋2ln⁡(4𝜋0.20)=0.750183.$$Fo=\frac4{\pi^2}\ln\left(\frac{\frac{4}{\pi}}{0.20}\right)=0.750183.$$

The time uses the half-thickness, not the total thickness:

𝑡=𝐹𝑜𝐿2𝐷=333.415 s=5.55691 min.$$t=Fo\frac{L^2}{D} =333.415\,\mathrm s=5.55691\,\mathrm{min}.$$

Since 𝐹𝑜 >0.2$Fo>0.2$, the first-term approximation is appropriate.

The next term at this time is about −2.47 ×10−8$-2.47\times10^{-8}$ in Θ0$\Theta_0$, so retaining it does not change the reported time.

The source chart gives 𝐹𝑜 ≈0.75$Fo\approx0.75$ and 𝑡 ≈333 s$t\approx333\,\mathrm s$.

Using the full 40 mm thickness in place of 𝐿$L$ would overestimate time by a factor of four.

Answer

The centre reaches 1.2 kg m−3$1.2\ \mathrm{kg\,m^{-3}}$ of molecular hydrogen after 333 s$333\ \mathrm s$, or 5.56 min$5.56\ \mathrm{min}$. The resulting Fourier number supports the first-term approximation.

Workshop 10: Transient mass diffusion

Drug release from a spherical bead

Consider a small spherical bead (D = 0.1 cm) that initially contains 0.2 mmol/L of the drug griseofulvin (species A). Determine the time it will take for the concentration of griseofulvin to reach 10% of its initial value, if its diffusivity within the bead material is 1.5 × 10−7$10^{-7}$ cm2/s$\mathrm{cm^2/s}$. Upon release from the bead, the drug is immediately consumed so that the surface concentration can be assumed to be zero.

Worked solution

Figure 3: Drug diffuses radially from an initially uniform spherical bead to a sink. The centre concentration and total amount remaining are different measures.

Data

Quantity	Symbol	Value
Diameter and radius	𝑑,𝑎$d,a$	0.10,0.050 cm$0.10,0.050\ \mathrm{cm}$
Initial concentration	𝐶𝑖$C_i$	0.20 mmol L−1$0.20\ \mathrm{mmol\,L^{-1}}$
Target centre concentration	𝐶0$C_0$	0.020 mmol L−1$0.020\ \mathrm{mmol\,L^{-1}}$
Surface concentration	𝐶𝑠$C_s$	0$0$
Diffusivity	𝐷$D$	1.5 ×10−7 cm2 s−1$1.5\times10^{-7}\ \mathrm{cm^{2}\,s^{-1}}$

The source question does not specify the location of the target concentration. This worked solution uses the centre, following the supplied solution.

Assumptions

Assume spherical symmetry, a fixed bead radius 𝑎 =𝑑/2$a=d/2$, constant temperature and no reaction inside the bead. The centre has zero radial concentration gradient; the surface at 𝑟 =𝑎$r=a$ has 𝐶𝑠 =0$C_s=0$. Define 𝐹𝑜 =𝐷𝑡/𝑎2$Fo=Dt/a^2$.

Calculations

The exact centre concentration ratio is

Θ0=2∞∑𝑛=1(−1)𝑛+1exp⁡(−𝑛2𝜋2𝐹𝑜).$$\Theta_0=2\sum_{n=1}^{\infty}(-1)^{n+1}\exp(-n^2\pi^2Fo).$$

For 𝐹𝑜 ≳0.2$Fo\gtrsim0.2$, use

Θ0≃2exp⁡(−𝜋2𝐹𝑜).$$\Theta_0\simeq2\exp(-\pi^2Fo).$$(4)

Check the approximation after computing 𝐹𝑜$Fo$. Centre concentration differs from the volume-averaged concentration. This question uses the centre, matching the coordinator’s worked solution.

The bead centre reaches the specified fraction when its Fourier number is about 0.304.

The target is 𝐶0 =0.020 mmol L−1$C_0=0.020\,\mathrm{mmol\,L^{-1}}$, giving Θ0 =0.10$\Theta_0=0.10$.

Using Eq. 4 yields

𝐹𝑜=ln⁡(20.10)𝜋2=0.303531.$$Fo=\frac{\ln\left(\frac{2}{0.10}\right)}{\pi^2}=0.303531.$$

Use

𝑎=0.050 cm=5.0×10−4 m$$a=0.050\,\mathrm{cm}=5.0\times10^{-4}\,\mathrm m$$

and 𝐷 =1.5 ×10−11 m2 s−1$D=1.5\times10^{-11}\,\mathrm{m^2\,s^{-1}}$.

Then

𝑡=𝐹𝑜𝑎2𝐷=5058.85 s=84.3142 min=1.40524 h.$$t=Fo\frac{a^2}{D} =5058.85\,\mathrm s =84.3142\,\mathrm{min}=1.40524\,\mathrm h.$$

The Fourier number exceeds 0.2.

At this time the second term of the exact centre series is −1.25 ×10−5$-1.25\times10^{-5}$, so the first-term ratio is accurate to about 0.013%.

The exact series gives about 5058.64 s$5058.64\,\mathrm s$.

The source rounds 𝐹𝑜$Fo$ to 0.3 and obtains 5000 s$5000\,\mathrm s$ or 1.39 hours.

The diameter must first be halved.

Initial concentration cancels from the ratio because this constant-property diffusion model is linear.

Answer

The bead centre reaches ten percent of its initial concentration after approximately 5.06 ×103 s$5.06\times10^3\ \mathrm s$, or 84.3 min$84.3\ \mathrm{min}$. This is a centre-concentration result, not a volume-average result.

Workshop 11: Convective mass transfer and dissolution

Solvent flowing over a flat coating

In a manufacturing process, an organic solvent (methyl ethyl ketone, MEK) is used to dissolve a thin coating of polymer film away from a nonporous flat surface (length 20 cm, width 10 cm, see Figure 1). The thickness of the polymer film is initially uniform at 0.2 mm. In the process, MEK is added at a flow rate of 30 cm3/s$\mathrm{cm^3/s}$ to an open flat pan of length 30 cm and width 10 cm. The depth of the liquid MEK in the pan is maintained at 2.0 cm. Assuming that the concentration of the dissolved polymer in the bulk solvent is essentially zero (𝑐𝐴∞ =0$c_{A\infty}=0$), determine the rate of removal of polymer coating.

Data: diffusion coefficient of dissolved polymer in MEK = 3 x 10−6$10^{-6}$ cm2/s$\mathrm{cm^2/s}$; kinematic viscosity of MEK = 6 x 10−3$10^{-3}$ cm2/s$\mathrm{cm^2/s}$; density of polymer material (solid) = 1.05 g/cm3$\mathrm{g/cm^3}$; density of MEK = 0.8 g/cm3$\mathrm{g/cm^3}$; maximum solubility of polymer in MEK = 0.04 g/cm3$\mathrm{g/cm^3}$.

Figure 1: Pan and polymer coating. The pan is 30 cm long and 10 cm wide, the coating is 20 cm long and initially 0.2 mm thick, and the liquid depth is 2 cm. Original redrawing of Figure 1 from the source question sheet.

Worked solution

Data

Quantity	Symbol	Value
Coated length and width	𝐿,𝑏$L,b$	20,10 cm$20,10\ \mathrm{cm}$
Liquid depth and initial coating thickness	𝐻,𝛿$H,\delta$	2.0,0.020 cm$2.0,0.020\ \mathrm{cm}$
Solvent volume flow	𝑄$Q$	30 cm3 s−1$30\ \mathrm{cm^{3}\,s^{-1}}$
Diffusivity	𝐷𝐴𝐵$D_{AB}$	3.0 ×10−6 cm2 s−1$3.0\times10^{-6}\ \mathrm{cm^{2}\,s^{-1}}$
Kinematic viscosity	𝜈$\nu$	0.0060 cm2 s−1$0.0060\ \mathrm{cm^{2}\,s^{-1}}$
Surface and bulk mass concentrations	𝐶𝐴,𝑠,𝐶𝐴,𝑏$C_{A,s},C_{A,b}$	0.040,0 g cm−3$0.040,0\ \mathrm{g\,cm^{-3}}$
Polymer density	𝜌𝑝$\rho_p$	1.05 g cm−3$1.05\ \mathrm{g\,cm^{-3}}$

Assumptions

Use mass concentration 𝐶𝐴$C_A$, so 𝑘𝑐(𝐶𝐴,𝑠 −𝐶𝐴,𝑏)$k_c(C_{A,s}-C_{A,b})$ is a mass flux. In centimetre units, 𝐶𝐴$C_A$ in g cm⁻³ and 𝑘𝑐$k_c$ in cm s⁻¹ give g cm⁻² s⁻¹. Assume isothermal, dilute transport with constant properties, negligible reaction in the liquid and rapid dissolution maintaining saturation at the solid-liquid interface. The coating is thin enough to neglect changes in flow geometry during an instantaneous rate calculation.

Model the exposed upper coating face as a laminar flat plate with a fresh concentration boundary layer at its leading edge. Neglect pan end effects, sidewall transfer and the velocity-profile corrections of a shallow channel. This is the source’s external-flow approximation, not a resolved pan flow.

Calculations

Use superficial velocity

𝑈=𝑄𝑏(𝐻−𝛿),$$U=\frac{Q}{b(H-\delta)},$$

coated area 𝐴 =𝑏𝐿$A=bL$ and

Re𝐿=𝑈𝐿𝜈,Sc=𝜈𝐷𝐴𝐵.$$\mathrm{Re}_L=\frac{UL}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}.$$

The average coefficient follows

¯¯¯¯¯¯¯Sh𝐿=0.664Re1/2𝐿Sc1/3=¯¯¯¯𝑘𝑐𝐿𝐷𝐴𝐵.$$\overline{\mathrm{Sh}}_L =0.664\mathrm{Re}_L^{1/2}\mathrm{Sc}^{1/3} =\frac{\overline{k}_cL}{D_{AB}}.$$(1)

Use Eq. 1 for Re𝐿 <5 ×105$\mathrm{Re}_L<5\times10^5$ and Sc >0.5$\mathrm{Sc}>0.5$. The initial coating mass is 𝜌𝑝𝐴𝛿$\rho_pA\delta$.

The polymer removal rate uses the coated area rather than the total pan area.

Convert thickness to 𝛿 =0.020 cm$\delta=0.020\,\mathrm{cm}$.

The initial liquid cross-section is

10(2.0−0.020)=19.8 cm2,$$10(2.0-0.020)=19.8\,\mathrm{cm^2},$$

giving

𝑈=1.51515 cm s−1Re𝐿=5050.51Sc=2000$$U=1.51515\,\mathrm{cm\,s^{-1}} \qquad \mathrm{Re}_L=5050.51 \qquad \mathrm{Sc}=2000$$

The specified laminar correlation is within its stated group ranges.

Using Eq. 1 gives ¯¯¯¯¯¯¯Sh𝐿 =594.537$\overline{\mathrm{Sh}}_L=594.537$ and ¯¯¯¯𝑘𝑐 =8.91805 ×10−5 cm s−1$\overline{k}_c=8.91805\times10^{-5}\,\mathrm{cm\,s^{-1}}$.

The exposed coating area is

𝐴=𝑏𝐿=10(20)=200 cm2,$$A=bL=10(20)=200\ \mathrm{cm^2},$$

and the dry-bulk concentration difference is 0.040 g cm−3$0.040\ \mathrm{g\,cm^{-3}}$. Substitution gives

˙𝑚𝐴=¯¯¯¯𝑘𝑐(𝐶𝐴,𝑠−𝐶𝐴,𝑏)𝐴=7.13444×10−4 g s−1=7.13444×10−7 kg s−1.$$\dot m_A=\overline{k}_c(C_{A,s}-C_{A,b})A =7.13444\times10^{-4}\,\mathrm{g\,s^{-1}} =7.13444\times10^{-7}\,\mathrm{kg\,s^{-1}}.$$

Initial coating mass is

1.05(200)(0.020)=4.20 g.$$1.05(200)(0.020)=4.20\,\mathrm g.$$

At the assumed constant rate, removal time is

4.20˙𝑚𝐴=5886.93 s=1.63526 h.$$\frac{4.20}{\dot m_A}=5886.93\,\mathrm s=1.63526\,\mathrm h.$$

Report about 1.64 h$1.64\,\mathrm h$.

The 2019 solution reports the same rounded time.

The estimate neglects the small velocity change as the coating shrinks and any change in surface behaviour near complete removal.

The outlet pickup estimated from ˙𝑚𝐴/𝑄$\dot m_A/Q$ is 2.37815 ×10−5 g cm−3$2.37815\times10^{-5}\,\mathrm{g\,cm^{-3}}$, only 0.0595% of saturation.

This supports the approximation of zero bulk concentration.

Solvent density is unnecessary because kinematic viscosity and mass concentration are given.

Answer

The initial polymer removal rate is 7.13 ×10−4 g s−1$7.13\times10^{-4}\ \mathrm{g\,s^{-1}}$. Maintaining that rate gives an estimated coating-removal time of 1.64 h$1.64\ \mathrm h$.

Workshop 11: Convective mass transfer and dissolution

Balance and a conditional diffusivity estimate

An experiment on the dissolution of fatty deposits in blood vessels involves the setup of a tube (0.8 cm inner diameter, 50 cm in length) that is coated on the inner surface with a thin layer (0.01 cm thick) of lipophilic material that represents the fatty deposits, called species A (𝜌$\rho$ = 1.10 g/cm3$\mathrm{g/cm^3}$). A synthetic solution (𝜈$\nu$ = 0.02 cm2/s$\mathrm{cm^2/s}$, 𝜌$\rho$ = 1.04 g/cm3$\mathrm{g/cm^3}$) is pumped through the tube at a rate of 20 cm3/s$\mathrm{cm^3/s}$. The maximum solubility of A in the synthetic solution is 20 mg A/cm3$\mathrm{mg\ A/cm^3}$ solution.

a. Develop a material balance to predict the concentration of A in the solution exiting the tube

b. Experimental measurements indicate that the concentration of A in the synthetic solution exiting the tube is 0.1 mg A/cm3$\mathrm{mg\ A/cm^3}$. Determine the diffusion coefficient of A in the solution.

Worked solution

Figure 2: Liquid flows through a tube with a saturated coated wall. The wall adds dissolved material, so the outlet bulk concentration exceeds the inlet concentration.

Data

Quantity	Symbol	Value
Liquid bore diameter	𝑑$d$	0.80 cm$0.80\ \mathrm{cm}$
Coated length	𝐿$L$	50 cm$50\ \mathrm{cm}$
Volume flow	𝑄$Q$	20 cm3 s−1$20\ \mathrm{cm^{3}\,s^{-1}}$
Kinematic viscosity	𝜈$\nu$	0.020 cm2 s−1$0.020\ \mathrm{cm^{2}\,s^{-1}}$
Saturated wall concentration	𝐶𝐴,𝑠$C_{A,s}$	20 mg cm−3$20\ \mathrm{mg\,cm^{-3}}$
Inlet and measured outlet concentrations	𝐶𝐴,𝑖,𝐶𝐴,𝑜$C_{A,i},C_{A,o}$	0,0.10 mg cm−3$0,0.10\ \mathrm{mg\,cm^{-3}}$

The supplied solution uses Sh𝑑 =𝑘𝑐𝑑/𝐷𝐴𝐵 =3.66$\mathrm{Sh}_d=k_cd/D_{AB}=3.66$ for fully developed laminar mass transfer. The calculation below checks that assumption. It treats the quoted diameter as the initial liquid bore and takes the inlet concentration to be zero.

Assumptions

Assume steady liquid transport during the measurement, negligible axial diffusion, no bulk reaction, constant flow rate and a saturated coated wall. The cross-section balance uses a mixed-mean concentration, not a uniform local radial profile. Assume an upstream uncoated length establishes the laminar velocity profile before the coated section. The source correlation 3.66 additionally requires a developed concentration profile at fixed wall concentration. Reynolds number alone does not establish this condition.

Calculations

𝑈=4𝑄𝜋𝑑2,Re𝑑=𝑈𝑑𝜈,Sc=𝜈𝐷𝐴𝐵.$$U=\frac{4Q}{\pi d^2},\qquad \mathrm{Re}_d=\frac{Ud}{\nu},\qquad \mathrm{Sc}=\frac{\nu}{D_{AB}}.$$

Estimate entrance lengths using

𝐿ℎ≃0.05Re𝑑𝑑,𝐿𝑐≃Sc𝐿ℎ.$$L_h\simeq0.05\mathrm{Re}_d d,\qquad L_c\simeq\mathrm{Sc}L_h.$$(2)

The tube measurement determines an effective coefficient before any diffusivity correlation is chosen.

For a slice of length 𝑑𝑧$dz$, the wall adds mass at rate

𝑘𝑐(𝐶𝐴,𝑠−𝐶𝐴)𝜋𝑑 𝑑𝑧.$$k_c(C_{A,s}-C_A)\pi d\,dz.$$

The steady balance is

𝑄𝑑𝐶𝐴𝑑𝑧=𝑘𝑐𝜋𝑑(𝐶𝐴,𝑠−𝐶𝐴).$$Q\frac{dC_A}{dz}=k_c\pi d(C_{A,s}-C_A).$$(3)

Integrate Eq. 3 with 𝐶𝐴(0) =𝐶𝐴,𝑖 =0$C_A(0)=C_{A,i}=0$.

Define

Λ=𝑘𝑐𝜋𝑑𝐿𝑄,$$\Lambda=\frac{k_c\pi dL}{Q},$$

a dimensionless transfer measure.

Then

𝐶𝐴,𝑠−𝐶𝐴,𝑜𝐶𝐴,𝑠−𝐶𝐴,𝑖=exp⁡(−Λ),𝐶𝐴,𝑜=𝐶𝐴,𝑠[1−exp⁡(−Λ)].$$\begin{aligned} \frac{C_{A,s}-C_{A,o}}{C_{A,s}-C_{A,i}} &=\exp(-\Lambda),\\ C_{A,o}&=C_{A,s}[1-\exp(-\Lambda)]. \end{aligned}$$(4)

If the local coefficient varies, the exponential instead contains

𝜋𝑑𝑄∫𝐿0𝑘𝑐(𝑧) 𝑑𝑧.$$\frac{\pi d}{Q}\int_0^L k_c(z)\,dz.$$

Thus the inferred constant coefficient is the length average under this balance model.

Using Eq. 4 with consistent mg cm⁻³ units gives

𝑘𝑐=−𝑄𝜋𝑑𝐿ln⁡(20−0.1020−0)=7.97771×10−4 cm s−1=7.97771×10−6 m s−1.$$\begin{aligned} k_c&=-\frac{Q}{\pi dL}\ln\left(\frac{20-0.10}{20-0}\right)\\ &=7.97771\times10^{-4}\,\mathrm{cm\,s^{-1}}\\ &=7.97771\times10^{-6}\,\mathrm{m\,s^{-1}}. \end{aligned}$$

The measured mass pickup is

𝑄(𝐶𝐴,𝑜−𝐶𝐴,𝑖)=2.0 mg s−1.$$Q(C_{A,o}-C_{A,i})=2.0\,\mathrm{mg\,s^{-1}}.$$

Velocity is 𝑈 =39.7887 cm s−1$U=39.7887\,\mathrm{cm\,s^{-1}}$ and Re𝑑 =1591.55$\mathrm{Re}_d=1591.55$, consistent with laminar flow.

Conditional on Sh𝑑 =3.66$\mathrm{Sh}_d=3.66$,

𝐷𝐴𝐵=𝑘𝑐𝑑3.66=1.74376×10−4 cm2 s−1=1.74376×10−8 m2 s−1.$$D_{AB}=\frac{k_cd}{3.66} =1.74376\times10^{-4}\,\mathrm{cm^{2}\,s^{-1}} =1.74376\times10^{-8}\,\mathrm{m^{2}\,s^{-1}}.$$

This reproduces the coordinator’s rounded result, but its validity needs a separate check.

The inferred Schmidt number is 114.695$114.695$.

Using Eq. 2 gives 𝐿ℎ =63.662 cm$L_h=63.662\,\mathrm{cm}$ and

𝐿𝑐=7301.68 cm=73.0168 m,$$L_c=7301.68\,\mathrm{cm}=73.0168\,\mathrm m,$$

far greater than the coated length of 0.50 m$0.50\,\mathrm m$.

An upstream uncoated tube can develop the velocity profile, but cannot develop a solute profile before dissolution starts.

The supplied 3.66 model therefore does not provide a justified physical diffusivity for this entrance-region experiment.

The data establish the effective 𝑘𝑐$k_c$ and mass removal rate under the stated balance.

Inferring a reliable diffusivity additionally needs an appropriate developing-flow correlation or a resolved transport model, with the velocity entrance condition specified.

The conditional numerical result must not be reported as a validated measured diffusivity.

Coating thickness and solid density describe the available inventory; liquid density is unnecessary when kinematic viscosity is supplied.

Answer

The measured outlet concentration gives an effective 𝑘𝑐 =7.98 ×10−6 m s−1$k_c=7.98\times10^{-6}\ \mathrm{m\,s^{-1}}$ and a mass pickup of 2.0 mg s−1$2.0\ \mathrm{mg\,s^{-1}}$. Assuming Sh =3.66$\mathrm{Sh}=3.66$ gives 𝐷𝐴𝐵 =1.74 ×10−8 m2 s−1$D_{AB}=1.74\times10^{-8}\ \mathrm{m^{2}\,s^{-1}}$, but the concentration-entry check fails. This diffusivity is conditional and is not validated by the experiment.