Heat transfer modes and steady conduction

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1 Reading list

• Çengel and Ghajar (2015), sections 3.1 to 3.4.
• Bird et al. (2006), sections 10.1, 10.2 and 11.1.

2 Choose a steady-conduction model

Given conditions	Model and checks
Plane layers, fixed end temperatures or surface films	Series resistance, Eq. 1; include each exposed film.
Heat can cross separate side-by-side paths	Combine each series path before putting paths in parallel, Eq. 4.
Hollow cylinder or sphere	Use the radial resistance, Eq. 5 or Eq. 7; area varies with radius.
Internal heat generation	Use the source term in the energy balance and solve for the temperature profile, Eq. 11.

3 Heat transfer mechanisms

3.1 Example: a cup of hot tea

Figure 1: Section through a ceramic cup of hot tea. Arrows show circulation in the tea, rising warm air outside, conduction through the wall and into the bench, radiation to the room, and evaporation from the free surface.

A cup of hot tea cools because it loses heat to its surroundings. Let’s assume the air, bench and room surfaces are cooler than the tea.

There are several modes of heat transfer at work here. Heat passes through the ceramic wall by conduction. The tea near the wall cools, becomes denser and sinks. Warmer tea rises through the middle to replace it, forming the circulation shown in Figure 1. Outside the cup, air in contact with the wall warms and rises. In both cases, temperature differences cause density differences that drive the fluid motion. This phenomenon is called natural convection.

The cup also loses heat by radiation to the cooler room surfaces. So, at the outer wall, convection and radiation occur at the same time. There are other ways for heat to leave the tea as well. Heat passes through the base into the bench by conduction. At the exposed tea surface, water evaporates into the air and takes energy with it. Evaporation therefore involves both heat and mass transfer.

Let’s follow the heat leaving through the side wall. It first passes from the tea to the inner wall by convection, then through the ceramic by conduction. From the outer wall, it leaves by convection to the air and radiation to the room surfaces. These steps help explain why we use thermal resistances in series and in parallel. Heat must pass through the inner convection resistance and the wall resistance in sequence, so these are in series. At the outer wall, convection and radiation provide parallel routes for heat loss. Heat leaving through the tea surface or the base follows separate routes.

How do we calculate the heat transferred? For one-dimensional conduction, we use Fourier’s law, 𝑞 = −𝑘 𝑑𝑇/𝑑𝑥$q=-k\,dT/dx$. Here 𝑞$q$ is the heat transferred per unit area per unit time, or heat flux, measured in W m−2$\mathrm{W\,m^{-2}}$. The thermal conductivity 𝑘$k$ has units of W m−1 K−1$\mathrm{W\,m^{-1}\,K^{-1}}$. The minus sign tells us that heat flows towards lower temperature.

For convection, we use 𝑞 =ℎ(𝑇𝑠 −𝑇∞)$q=h(T_s-T_\infty)$. Here 𝑇𝑠$T_s$ is the surface temperature and 𝑇∞$T_\infty$ is the fluid temperature away from the surface. The heat transfer coefficient ℎ$h$ has units of W m−2 K−1$\mathrm{W\,m^{-2}\,K^{-1}}$. It depends on the fluid, the geometry and how the fluid flows. For example, blowing air over the cup changes ℎ$h$ and can increase the rate of heat loss.

Conduction

Conduction is the transfer of thermal energy within a medium due to a temperature gradient, through microscopic interactions without bulk motion of the medium. It occurs in solids, liquids and gases.

Convection

Convection is the transfer of heat between a surface and a fluid in motion when they are at different temperatures. It involves the combined effects of conduction and bulk fluid motion.

Radiation

Thermal radiation is the emission of energy as electromagnetic waves by matter at a nonzero absolute temperature. Energy transfer by radiation does not require a material medium.

Heat flux

Heat flux is the rate of heat transfer per unit area normal to the direction of transfer. Its SI unit is W m−2$\mathrm{W\,m^{-2}}$. A signed heat flux specifies the direction of transfer relative to the chosen coordinate or surface normal.

Thermal conductivity

Thermal conductivity, 𝑘$k$, is a material property that relates conductive heat flux to the temperature gradient through Fourier’s law. For one-dimensional conduction, 𝑞 = −𝑘 𝑑𝑇/𝑑𝑥$q=-k\,dT/dx$. Its SI unit is W m−1 K−1$\mathrm{W\,m^{-1}\,K^{-1}}$.

Heat transfer coefficient

The heat transfer coefficient, ℎ$h$, is the proportionality coefficient between the convective heat flux at a surface and the temperature difference between the surface and the fluid away from it, 𝑞 =ℎ(𝑇𝑠 −𝑇∞)$q=h(T_s-T_\infty)$. Its SI unit is W m−2 K−1$\mathrm{W\,m^{-2}\,K^{-1}}$. It depends on the fluid properties, surface geometry and flow conditions.

Consider these three cases:

1. Heat passes across a layer of stationary air, with radiation neglected.
2. Water flows over a hot plate and removes heat from it.
3. Heat passes between two surfaces separated by a vacuum.

Which sequence gives the heat transfer mode in each case, in the order listed?

• Convection, conduction, radiation.
• Conduction, convection, radiation. ✓
• Conduction, radiation, convection.
• Radiation, convection, conduction.

Explanation

The air is stationary, so there is no heat transport by bulk fluid motion. With radiation neglected, heat crosses the air layer by conduction. Heat transfer from the plate to the flowing water is convection. It combines conduction near the plate with heat transport by the moving water. Across a vacuum, heat transfers by radiation, which requires no material medium.

A fan blows air over a metal plate. Suppose we turn up the fan so the air flows faster, while keeping the plate and air temperatures the same. What happens to the plate’s thermal conductivity 𝑘$k$ and the heat transfer coefficient ℎ$h$?

• Both 𝑘$k$ and ℎ$h$ stay the same because the temperatures have not changed.
• 𝑘$k$ increases, while ℎ$h$ stays the same.
• 𝑘$k$ stays the same, while ℎ$h$ generally increases. ✓
• Both 𝑘$k$ and ℎ$h$ increase as the air flows faster.

Explanation

Turning up the fan changes the air flow over the plate. It does not change the metal or its temperature, so 𝑘$k$ stays the same. The coefficient ℎ$h$ depends on the flow conditions and generally increases with air speed. More heat can therefore leave the plate for the same temperature difference between the plate and the air.

4 Case 1: 1D steady conduction without heat generation

4.1 A wall with several layers

Thermal resistance

Thermal resistance, 𝑅$R$, is the ratio of the temperature difference across a heat-transfer process to the resulting heat-transfer rate, 𝑅 =Δ𝑇/˙𝑄$R=\Delta T/\dot Q$. Its SI unit is K W−1$\mathrm{K\,W^{-1}}$.

Let’s consider the three-layer wall in Figure 2, with hot fluid on one side and cold fluid on the other. The fluid temperatures are 𝑇ℎ$T_h$ and 𝑇𝑐$T_c$, with 𝑇ℎ >𝑇𝑐$T_h>T_c$. Each layer has the same area 𝐴$A$, but its thickness 𝐿𝑖$L_i$ and thermal conductivity 𝑘𝑖$k_i$ may differ.

Assume steady, one-dimensional conduction through the wall, with no heat generation. The conductivity of each layer is constant, and the layers are in perfect thermal contact.

Heat first passes from the hot fluid to the wall by convection. It then conducts through each layer before leaving the wall by convection to the cold fluid. At steady state, heat enters and leaves each layer at the same rate. If heat entered a layer faster than it left, the layer would store energy and warm up.

Figure 2: Schematic temperature profile from hot fluid to cold fluid through three solid layers, with surface and interface temperatures, conductivities and thicknesses labelled.

The wall surfaces are at 𝑇1$T_1$ and 𝑇4$T_4$, and the two interfaces are at 𝑇2$T_2$ and 𝑇3$T_3$. The heat transfer coefficients on the hot and cold sides are ℎℎ$h_h$ and ℎ𝑐$h_c$, respectively. We can write the heat rate through each part as

Heat-transfer step	Heat rate
Hot fluid to wall	˙𝑄𝑥 =ℎℎ𝐴(𝑇ℎ −𝑇1)$\dot Q_x=h_hA(T_h-T_1)$
Through solid layer 𝑖$i$	˙𝑄𝑥 =𝑘𝑖𝐴𝐿𝑖(𝑇𝑖 −𝑇𝑖+1)$\dot Q_x=\frac{k_iA}{L_i}(T_i-T_{i+1})$
Wall to cold fluid	˙𝑄𝑥 =ℎ𝑐𝐴(𝑇4 −𝑇𝑐)$\dot Q_x=h_cA(T_4-T_c)$

The middle equation applies to all three solid layers, with 𝑖 =1,2,3$i=1,2,3$. Heat passes through each layer and fluid film at the same rate ˙𝑄𝑥$\dot Q_x$. The temperature drop across each layer depends on its thermal resistance.

Rearranging the three heat-rate equations above to express their temperature differences gives

𝑇ℎ−𝑇1=˙𝑄𝑥ℎℎ𝐴,$$T_h-T_1=\frac{\dot Q_x}{h_hA},$$ 𝑇𝑖−𝑇𝑖+1=˙𝑄𝑥𝐿𝑖𝑘𝑖𝐴$$T_i-T_{i+1}=\frac{\dot Q_xL_i}{k_iA}$$

and

𝑇4−𝑇𝑐=˙𝑄𝑥ℎ𝑐𝐴.$$T_4-T_c=\frac{\dot Q_x}{h_cA}.$$

Each has the form Δ𝑇 =˙𝑄𝑥𝑅$\Delta T=\dot Q_xR$, so the factor multiplying ˙𝑄𝑥$\dot Q_x$ is the thermal resistance:

𝑅ℎ=1ℎℎ𝐴,𝑅𝑖=𝐿𝑖𝑘𝑖𝐴,𝑅𝑐=1ℎ𝑐𝐴$$R_h=\frac{1}{h_hA},\qquad R_i=\frac{L_i}{k_iA},\qquad R_c=\frac{1}{h_cA}$$(1)

The convection resistances 𝑅ℎ$R_h$ and 𝑅𝑐$R_c$ are often called film resistances. They represent heat transfer between each fluid and the adjacent wall surface.

Now add the temperature drops from the hot fluid to the cold fluid:

(𝑇ℎ−𝑇1)+(𝑇1−𝑇2)+(𝑇2−𝑇3)+(𝑇3−𝑇4)+(𝑇4−𝑇𝑐)=˙𝑄𝑥(𝑅ℎ+𝑅1+𝑅2+𝑅3+𝑅𝑐)$$(T_h-T_1)+ (T_1-T_2)+ (T_2-T_3)+ (T_3-T_4)+ (T_4-T_c) =\dot Q_x (R_h+R_1+R_2+R_3+R_c)$$

Each intermediate temperature appears once with a positive sign and once with a negative sign. These terms cancel, leaving the overall temperature difference 𝑇ℎ −𝑇𝑐$T_h-T_c$. Therefore,

˙𝑄𝑥=𝑇ℎ−𝑇𝑐∑𝑅=𝑈𝐴(𝑇ℎ−𝑇𝑐)$$\dot Q_x=\frac{T_h-T_c}{\sum R}=UA(T_h-T_c)$$(2)

The overall heat transfer coefficient is

𝑈=1𝐴∑𝑅$$U=\frac{1}{A\sum R}$$

Here

∑𝑅=𝑅ℎ+𝑅1+𝑅2+𝑅3+𝑅𝑐.$$\sum R=R_h+R_1+R_2+R_3+R_c.$$

The overall heat transfer coefficient 𝑈$U$, in W m−2 K−1$\mathrm{W\,m^{-2}\,K^{-1}}$, combines the effects of both convection processes and all three solid layers.

We can compare this wall to an electrical circuit with resistors in series. A temperature difference drives heat transfer, just as a voltage difference drives electric current. The same heat rate passes through both fluid films and all three layers, so their thermal resistances add to give the total resistance.

Once we know the heat rate, we can calculate the surface and interface temperatures. Starting at the hot fluid, subtract the temperature drop across each resistance in turn. This gives 𝑇1 =𝑇ℎ −˙𝑄𝑥𝑅ℎ$T_1=T_h-\dot Q_xR_h$ at the hot wall surface, followed by 𝑇2 =𝑇1 −˙𝑄𝑥𝑅1$T_2=T_1-\dot Q_xR_1$ at the first interface. Since each drop is ˙𝑄𝑥𝑅$\dot Q_xR$ and the heat rate is the same throughout, the largest temperature drop occurs across the largest resistance.

At fixed fluid temperatures, reducing the total resistance increases the heat rate. At fixed area, the film resistance is inversely proportional to the heat transfer coefficient ℎ$h$. Increasing the fluid speed generally increases ℎ$h$ and therefore reduces the film resistance. A thinner solid layer or a material with higher conductivity reduces the conduction resistance. The effect depends on how much that resistance contributes to the total. Halving a small resistance changes the total much less than halving the largest one. For insulation, the aim is to increase the total resistance and reduce heat loss.

4.2 Series and parallel composite wall

Let’s consider the wall in Figure 3. Layers E and H extend across the full area 𝐴$A$. Between them, materials F and G each occupy half the area and have the same thickness, 𝐿𝐹 =𝐿𝐺$L_F=L_G$. The outer faces are held at uniform temperatures 𝑇1$T_1$ and 𝑇2$T_2$, with 𝑇1 >𝑇2$T_1>T_2$.

Assume steady conduction, constant thermal conductivities, no heat generation and perfect contact between materials. The top and bottom edges are insulated.

Figure 3: Composite wall with thicknesses, conductivities and areas labelled. Network (a) assumes uniform temperatures on the internal interfaces; network (b) prevents heat flow across the horizontal midline.

Heat passes through E, then through F or G, and finally through H. To represent this wall by a resistance network, we need an assumption about how temperature varies across the internal interfaces. The figure shows two possible models.

In model (a), we assume that each internal interface has a uniform temperature. F and G therefore have the same temperature difference Δ𝑇$\Delta T$ across them. Their heat rates depend on their individual resistances and add to give the total heat rate:

˙𝑄=˙𝑄𝐹+˙𝑄𝐺=Δ𝑇𝑅𝐹+Δ𝑇𝑅𝐺=Δ𝑇(1𝑅𝐹+1𝑅𝐺)$$\dot Q=\dot Q_F+\dot Q_G =\frac{\Delta T}{R_F}+\frac{\Delta T}{R_G} =\Delta T\left(\frac{1}{R_F}+\frac{1}{R_G}\right)$$

F and G are in parallel. We can replace them with an equivalent resistance 𝑅𝐹𝐺$R_{FG}$ such that ˙𝑄 =Δ𝑇/𝑅𝐹𝐺$\dot Q=\Delta T/R_{FG}$. Comparing this expression with the equation above gives

1𝑅𝐹𝐺=1𝑅𝐹+1𝑅𝐺,𝑅𝐹𝐺=𝑅𝐹𝑅𝐺𝑅𝐹+𝑅𝐺$$\frac{1}{R_{FG}}=\frac{1}{R_F}+\frac{1}{R_G},\qquad R_{FG}=\frac{R_FR_G}{R_F+R_G}$$(3)

Each resistance must use the area through which heat passes. E and H occupy the full area, whereas F and G each occupy 𝐴/2$A/2$:

𝑅𝐸=𝐿𝐸𝑘𝐸𝐴,𝑅𝐻=𝐿𝐻𝑘𝐻𝐴$$R_E=\frac{L_E}{k_EA},\qquad R_H=\frac{L_H}{k_HA}$$ 𝑅𝐹=𝐿𝐹𝑘𝐹(𝐴2),𝑅𝐺=𝐿𝐺𝑘𝐺(𝐴2)$$R_F=\frac{L_F}{k_F\left(\frac{A}{2}\right)},\qquad R_G=\frac{L_G}{k_G\left(\frac{A}{2}\right)}$$

The total heat rate passes through E and H, so these two resistances are in series with the parallel combination of F and G. The equivalent resistance for model (a) is therefore

𝑅𝑎=𝑅𝐸+𝑅𝐹𝐺+𝑅𝐻$$R_a=R_E+R_{FG}+R_H$$

In model (b), we instead assume that no heat crosses the horizontal midline. Heat then follows two separate paths through the wall. The upper path passes through E, F and H; the lower path passes through E, G and H. Each path has three resistances in series, and the two complete paths are in parallel between 𝑇1$T_1$ and 𝑇2$T_2$.

Each path uses half the area of E and H. Since conduction resistance is inversely proportional to area, their resistances in each path are 2𝑅𝐸$2R_E$ and 2𝑅𝐻$2R_H$. The definitions of 𝑅𝐹$R_F$ and 𝑅𝐺$R_G$ already use 𝐴/2$A/2$. The equivalent resistance for model (b) is therefore

1𝑅𝑏=12𝑅𝐸+𝑅𝐹+2𝑅𝐻+12𝑅𝐸+𝑅𝐺+2𝑅𝐻$$\frac{1}{R_b}= \frac{1}{2R_E+R_F+2R_H}+ \frac{1}{2R_E+R_G+2R_H}$$(4)

The actual wall generally has two-dimensional heat flow. Heat can pass between its upper and lower parts, and the internal interface temperatures need not be uniform. Under the stated assumptions, the two models give lower and upper bounds on its thermal resistance:

𝑅𝑎≤𝑅≤𝑅𝑏$$R_a\le R\le R_b$$

If F and G have the same conductivity, the two models give the same result because the temperature varies only through the wall thickness. If the estimates differ substantially, a two-dimensional conduction calculation is needed to determine the resistance more precisely.

For either model, use its equivalent resistance to calculate the total heat rate from

˙𝑄=𝑇1−𝑇2𝑅.$$\dot Q=\frac{T_1-T_2}{R}.$$

Dividing this heat rate by the full area 𝐴$A$ gives the average heat flux through the wall.

4.3 Thermal contact resistance

So far, we have assumed that adjacent layers are in perfect thermal contact, with no temperature difference at their interface. Real surfaces have small irregularities. When two solids are pressed together, contact occurs at a limited number of points, leaving gaps elsewhere, as shown in Figure 4. Heat must pass through these contact regions and any material in the gaps, adding a thermal resistance at the interface.

Figure 4: Two rough solid surfaces touch at small contact areas, with air gaps between them. Heat flows from the left solid to the right solid.

We describe heat transfer across the interface using the thermal contact conductance ℎcontact$h_{\mathrm{contact}}$. Let 𝑇−$T_-$ and 𝑇+$T_+$ be the temperatures of the two contacting surfaces. The heat rate and contact resistance are

˙𝑄=ℎcontact𝐴(𝑇−−𝑇+)$$\dot Q=h_{\mathrm{contact}}A(T_--T_+)$$

The corresponding contact resistance is

𝑅contact=𝑇−−𝑇+˙𝑄=1ℎcontact𝐴$$R_{\mathrm{contact}}=\frac{T_--T_+}{\dot Q} =\frac{1}{h_{\mathrm{contact}}A}$$

Here 𝐴$A$ is the nominal contact area, rather than the sum of the small areas where the solids actually touch. The contact conductance has units of W m−2 K−1$\mathrm{W\,m^{-2}\,K^{-1}}$, and the contact resistance has units of K W−1$\mathrm{K\,W^{-1}}$. We add this resistance to the other resistances in series. The resulting temperature drop at the interface is 𝑇− −𝑇+ =˙𝑄𝑅contact$T_--T_+=\dot Q R_{\mathrm{contact}}$.

A cold-room panel has two continuous linings with insulation between them. A solid metal support passes through part of the insulation, connecting the two linings. Treat each surface between a lining and the core as having a uniform temperature.

Figure 5: Heat passes through an outer lining, splits between insulation and a metal support, then passes through the inner lining. The two core paths connect the same interface temperatures.

Heat flows steadily from the warmer surroundings into the cold room, with no heat generation in the panel. Which statement must be true?

• The two linings have equal temperature drops, and the insulation and support have equal heat rates.
• The same heat rate passes through each lining, the insulation and the support.
• Both linings, the insulation and the support all have the same temperature drop.
• The same total heat rate passes through each lining. The insulation and metal support have the same temperature drop. ✓

Explanation

Heat passes through the outer lining, divides between the insulation and metal support, then passes through the inner lining. The heat rate through each lining is therefore the sum of the rates through the two middle paths. The insulation and support connect the same two uniform-temperature surfaces, so they have the same temperature drop. Their heat rates depend on their resistances and need not be equal.

An engineer is checking the temperature profile through a furnace wall. For a 1 m2$1\ \mathrm{m^2}$ section, the thermal resistances from the hot gas to the room air are listed below. Assume steady heat transfer with no heat generation. All five resistances are in series.

Part	Thermal resistance 𝑅$R$ in K W−1$\mathrm{K\,W^{-1}}$
Hot-side gas film	0.040$0.040$
Firebrick	0.100$0.100$
Insulation	1.200$1.200$
Steel casing	0.000125$0.000125$
Cold-side air film	0.100$0.100$

Across which part should the engineer expect the largest temperature drop?

• The insulation layer. ✓
• The steel casing, because steel has a high thermal conductivity.
• The hot-side gas film, because it is closest to the furnace.
• All five parts should have the same temperature drop.

Explanation

The same heat rate passes through all five parts. Across each part,

∣Δ𝑇𝑖∣=∣˙𝑄∣𝑅𝑖$$|\Delta T_i|=|\dot Q|R_i$$

The insulation has the largest resistance, so it has the largest temperature drop. The steel has a very small resistance and therefore a small temperature drop.

4.4 Series of cylinders

Let’s apply the resistance method to the wall of a long pipe. The wall has length 𝐿$L$, inner radius 𝑟1$r_1$, outer radius 𝑟2$r_2$ and thermal conductivity 𝑘$k$. Unlike a plane wall, its area normal to heat flow changes through the thickness.

Assume steady conduction, constant conductivity, no heat generation and negligible heat loss through the ends. The surface conditions are uniform along the pipe and around its circumference. Temperature therefore varies only with radius, 𝑇 =𝑇(𝑟)$T=T(r)$.

At steady state, heat enters and leaves each part of the wall at the same rate. The heat rate ˙𝑄𝑟$\dot Q_r$ is therefore independent of radius. At radius 𝑟$r$, heat passes through the cylindrical area 2𝜋𝑟𝐿$2\pi rL$, so Fourier’s law gives

˙𝑄𝑟=−2𝜋𝑟𝑘𝐿𝑑𝑇𝑑𝑟$$\dot Q_r=-2\pi rkL\frac{dT}{dr}$$

Let the inner and outer surface temperatures be 𝑇1 =𝑇(𝑟1)$T_1=T(r_1)$ and 𝑇2 =𝑇(𝑟2)$T_2=T(r_2)$. Separating the temperature and radius terms, then integrating between these surfaces, gives

˙𝑄𝑟∫𝑟2𝑟1𝑑𝑟𝑟=−2𝜋𝑘𝐿∫𝑇2𝑇1𝑑𝑇$$\dot Q_r\int_{r_1}^{r_2}\frac{dr}{r} =-2\pi kL\int_{T_1}^{T_2}dT$$

Evaluating the integrals gives

˙𝑄𝑟ln⁡(𝑟2𝑟1)=2𝜋𝑘𝐿(𝑇1−𝑇2)$$\dot Q_r\ln\left(\frac{r_2}{r_1}\right)=2\pi kL(T_1-T_2)$$

We can now write the result as

˙𝑄𝑟=𝑇1−𝑇2𝑅.$$\dot Q_r=\frac{T_1-T_2}{R}.$$

Comparing this form with the integrated equation gives the cylindrical resistance:

𝑅cyl=ln⁡(𝑟2𝑟1)2𝜋𝑘𝐿$$R_{\mathrm{cyl}}=\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi kL}$$(5)

The heat rate is then

˙𝑄𝑟=𝑇1−𝑇2𝑅cyl$$\dot Q_r=\frac{T_1-T_2}{R_{\mathrm{cyl}}}$$

Integrating between 𝑟1$r_1$ and any radius 𝑟$r$ within the wall gives

𝑇(𝑟)=𝑇1−(𝑇1−𝑇2)ln⁡(𝑟𝑟1)ln⁡(𝑟2𝑟1)$$T(r)=T_1-(T_1-T_2)\frac{\ln\left(\frac{r}{r_1}\right)}{\ln\left(\frac{r_2}{r_1}\right)}$$

Temperature varies logarithmically with radius because the area changes through the wall. It therefore does not have the straight-line profile obtained for a plane wall with constant conductivity.

The local heat flux is

𝑞𝑟=˙𝑄𝑟2𝜋𝑟𝐿.$$q_r=\frac{\dot Q_r}{2\pi rL}.$$

For outward heat flow, the same heat rate passes through progressively larger areas. The heat flux therefore decreases as 1/𝑟$1/r$, even though the heat rate remains constant.

Now consider three concentric layers, A, B and C, with successive radii 𝑟1,𝑟2,𝑟3,𝑟4$r_1,r_2,r_3,r_4$. Heat passes from the inner fluid through its film, through all three layers, and then through the outer film to the outer fluid. These five resistances are in series, as shown in Figure 6.

Figure 6: Layered cylindrical wall: cross-section, pipe geometry and series resistance network. The same heat rate passes through both films and all three layers.

The total resistance is

𝑅total=1ℎ12𝜋𝑟1𝐿+ln⁡(𝑟2𝑟1)2𝜋𝑘𝐴𝐿+ln⁡(𝑟3𝑟2)2𝜋𝑘𝐵𝐿+ln⁡(𝑟4𝑟3)2𝜋𝑘𝐶𝐿+1ℎ42𝜋𝑟4𝐿$$R_{\mathrm{total}}=\frac{1}{h_1 2\pi r_1L} +\frac{\ln\left(\frac{r_2}{r_1}\right)}{2\pi k_AL} +\frac{\ln\left(\frac{r_3}{r_2}\right)}{2\pi k_BL} +\frac{\ln\left(\frac{r_4}{r_3}\right)}{2\pi k_CL} +\frac{1}{h_4 2\pi r_4L}$$(6)

The overall temperature difference is between the two fluids. With inner and outer fluid temperatures 𝑇∞,1$T_{\infty,1}$ and 𝑇∞,4$T_{\infty,4}$, the heat rate is

˙𝑄𝑟=𝑇∞,1−𝑇∞,4𝑅total.$$\dot Q_r=\frac{T_{\infty,1}-T_{\infty,4}}{R_{\mathrm{total}}}.$$

The film resistances use different areas: 𝐴𝑖 =2𝜋𝑟1𝐿$A_i=2\pi r_1L$ on the inside and 𝐴𝑜 =2𝜋𝑟4𝐿$A_o=2\pi r_4L$ on the outside. We must therefore state which area is used when defining an overall heat transfer coefficient. Either choice gives the same heat rate, provided that

𝑈𝑖𝐴𝑖=𝑈𝑜𝐴𝑜=1𝑅total.$$U_iA_i=U_oA_o=\frac{1}{R_{\mathrm{total}}}.$$

The values of 𝑈𝑖$U_i$ and 𝑈𝑜$U_o$ themselves are generally different.

For a thin wall, 𝑟2 −𝑟1 ≪𝑟1$r_2-r_1\ll r_1$, the change in area is small. Using

ln⁡(𝑟2𝑟1)≃𝑟2−𝑟1𝑟1$$\ln\left(\frac{r_2}{r_1}\right)\simeq\frac{r_2-r_1}{r_1}$$

in Eq. 5 gives the plane-wall expression with thickness 𝑟2 −𝑟1$r_2-r_1$ and area 2𝜋𝑟1𝐿$2\pi r_1L$.

𝑅cyl≃𝑟2−𝑟12𝜋𝑟1𝑘𝐿=Δ𝑥𝑘𝐴$$R_{\mathrm{cyl}}\simeq\frac{r_2-r_1}{2\pi r_1kL} =\frac{\Delta x}{kA}$$

Here Δ𝑥 =𝑟2 −𝑟1$\Delta x=r_2-r_1$ and 𝐴 =2𝜋𝑟1𝐿$A=2\pi r_1L$.

4.5 Series of spheres

The same resistance method applies to a hollow spherical wall, such as insulation around a spherical process vessel. Let its inner and outer radii be 𝑟1$r_1$ and 𝑟2$r_2$, and its thermal conductivity be 𝑘$k$.

Assume steady conduction, constant conductivity and no heat generation. The surface conditions are uniform over each spherical surface, so temperature varies only with radius, 𝑇 =𝑇(𝑟)$T=T(r)$.

At steady state, the same heat rate passes through every spherical surface. At radius 𝑟$r$, the area normal to heat flow is 4𝜋𝑟2$4\pi r^2$. Fourier’s law therefore gives

˙𝑄𝑟=−4𝜋𝑟2𝑘𝑑𝑇𝑑𝑟$$\dot Q_r=-4\pi r^2k\frac{dT}{dr}$$

Let the inner and outer surface temperatures be 𝑇1 =𝑇(𝑟1)$T_1=T(r_1)$ and 𝑇2 =𝑇(𝑟2)$T_2=T(r_2)$. Separating the temperature and radius terms, then integrating between the surfaces, gives

˙𝑄𝑟∫𝑟2𝑟1𝑑𝑟𝑟2=−4𝜋𝑘∫𝑇2𝑇1𝑑𝑇$$\dot Q_r\int_{r_1}^{r_2}\frac{dr}{r^2} =-4\pi k\int_{T_1}^{T_2}dT$$

Evaluating the integrals gives

˙𝑄𝑟(1𝑟1−1𝑟2)=4𝜋𝑘(𝑇1−𝑇2)$$\dot Q_r\left(\frac{1}{r_1}-\frac{1}{r_2}\right) =4\pi k(T_1-T_2)$$

We can now write the result as

˙𝑄𝑟=𝑇1−𝑇2𝑅.$$\dot Q_r=\frac{T_1-T_2}{R}.$$

Comparing this form with the integrated equation gives the spherical resistance:

𝑅sph=1𝑟1−1𝑟24𝜋𝑘$$R_{\mathrm{sph}}=\frac{\frac{1}{r_1}-\frac{1}{r_2}}{4\pi k}$$(7)

The heat rate is then

˙𝑄𝑟=𝑇1−𝑇2𝑅sph$$\dot Q_r=\frac{T_1-T_2}{R_{\mathrm{sph}}}$$

Integrating between 𝑟1$r_1$ and any radius 𝑟$r$ within the wall gives

𝑇(𝑟)=𝑇1−(𝑇1−𝑇2)1𝑟1−1𝑟1𝑟1−1𝑟2$$T(r)=T_1-(T_1-T_2)\frac{\frac{1}{r_1}-\frac{1}{r}}{\frac{1}{r_1}-\frac{1}{r_2}}$$

Temperature varies linearly with 1/𝑟$1/r$, rather than with 𝑟$r$. The local heat flux is

𝑞𝑟=˙𝑄𝑟4𝜋𝑟2.$$q_r=\frac{\dot Q_r}{4\pi r^2}.$$

For outward heat flow, it decreases as 1/𝑟2$1/r^2$ because the same heat rate passes through progressively larger areas.

Now consider three concentric spherical layers, A, B and C, with radii 𝑟1,𝑟2,𝑟3,𝑟4$r_1,r_2,r_3,r_4$. Heat passes through the inner fluid film, all three solid layers and the outer fluid film. Their resistances add in series, as shown in Figure 7.

Figure 7: Layered spherical wall: central section, spherical geometry and series resistance network. Each layer has a different area as the heat flows outwards.

The total resistance is

𝑅total=1ℎ14𝜋𝑟21+1𝑟1−1𝑟24𝜋𝑘𝐴+1𝑟2−1𝑟34𝜋𝑘𝐵+1𝑟3−1𝑟44𝜋𝑘𝐶+1ℎ44𝜋𝑟24$$R_{\mathrm{total}}=\frac{1}{h_1 4\pi r_1^2} +\frac{\frac{1}{r_1}-\frac{1}{r_2}}{4\pi k_A} +\frac{\frac{1}{r_2}-\frac{1}{r_3}}{4\pi k_B} +\frac{\frac{1}{r_3}-\frac{1}{r_4}}{4\pi k_C} +\frac{1}{h_4 4\pi r_4^2}$$(8)

With inner and outer fluid temperatures 𝑇∞,1$T_{\infty,1}$ and 𝑇∞,4$T_{\infty,4}$, the heat rate is

˙𝑄𝑟=𝑇∞,1−𝑇∞,4𝑅total$$\dot Q_r=\frac{T_{\infty,1}-T_{\infty,4}}{R_{\mathrm{total}}}$$

Each film uses the area of its own surface, 𝐴𝑖 =4𝜋𝑟21$A_i=4\pi r_1^2$ or 𝐴𝑜 =4𝜋𝑟24$A_o=4\pi r_4^2$. Either area can define an overall heat transfer coefficient, provided that

𝑈𝑖𝐴𝑖=𝑈𝑜𝐴𝑜=1𝑅total.$$U_iA_i=U_oA_o=\frac{1}{R_{\mathrm{total}}}.$$

For a thin spherical wall, 𝑟2 −𝑟1 ≪𝑟1$r_2-r_1\ll r_1$, the two surface areas are nearly equal. Writing

1𝑟1−1𝑟2=𝑟2−𝑟1𝑟1𝑟2$$\frac{1}{r_1}-\frac{1}{r_2}=\frac{r_2-r_1}{r_1r_2}$$

in Eq. 7 and taking 𝑟2 ≃𝑟1$r_2\simeq r_1$ gives the plane-wall limit:

𝑅sph≃𝑟2−𝑟14𝜋𝑘𝑟21=Δ𝑥𝑘𝐴$$R_{\mathrm{sph}}\simeq\frac{r_2-r_1}{4\pi kr_1^2} =\frac{\Delta x}{kA}$$

Here Δ𝑥 =𝑟2 −𝑟1$\Delta x=r_2-r_1$ and 𝐴 =4𝜋𝑟21$A=4\pi r_1^2$.

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.

4.5.1 Conduction into an infinite surrounding fluid

Consider a sphere surrounded by an infinite, stationary fluid, with heat transfer through the fluid by conduction alone. Taking 𝑟2 →∞$r_2\to\infty$ in Eq. 7 gives

𝑅=14𝜋𝑘𝑟1,$$R=\frac{1}{4\pi kr_1},$$

where 𝑘$k$ is the fluid’s thermal conductivity. Equating this to

1ℎ(4𝜋𝑟21)$$\frac{1}{h\left(4\pi r_1^2\right)}$$

gives ℎ =𝑘/𝑟1$h=k/r_1$.

Conduction limit for a sphere

For a sphere in an infinite, stationary fluid, pure conduction gives

𝑁𝑢𝐷=ℎ𝐷𝑘fluid=2$$Nu_D=\frac{hD}{k_{\mathrm{fluid}}}=2$$

Here 𝑁𝑢𝐷$Nu_D$ is the dimensionless Nusselt number based on the sphere diameter 𝐷 =2𝑟1$D=2r_1$, and 𝑘fluid$k_{\mathrm{fluid}}$ is the fluid’s thermal conductivity. Heat transfer persists even without bulk fluid motion. This is the conduction limit approached by sphere convection correlations as fluid motion vanishes.

A long steam pipe is insulated. Heat leaks outwards through the insulation at steady state. Temperature varies only with radius, and there is no heat generation in the insulation or heat loss through the ends.

Moving from the pipe surface to the outside of the insulation, what happens to the total heat rate ˙𝑄𝑟$\dot Q_r$ and local heat flux 𝑞𝑟$q_r$?

• Both the heat rate and heat flux increase.
• The heat rate decreases, while the heat flux stays the same.
• The heat rate stays the same, while the heat flux decreases. ✓
• Both the heat rate and heat flux stay the same.

Explanation

At steady state, the heat entering the insulation must leave it. The same heat rate passes through every cylindrical surface:

˙𝑄𝑟=2𝜋𝑟𝐿𝑞𝑟=constant$$\dot Q_r=2\pi rLq_r=\text{constant}$$

Thus,

𝑞𝑟=˙𝑄𝑟2𝜋𝑟𝐿$$q_r=\frac{\dot Q_r}{2\pi rL}$$

As radius increases, the heat crosses a larger area. The heat flux therefore decreases even though the total heat rate stays the same.

4.6 Classroom example 1: glass window

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.

5 Case 2: 1D steady conduction with heat generation

In the previous examples, no heat was generated within the wall. We now consider a solid with internal heat generation. At steady state, each region must lose both the heat entering it and the heat generated within it. We will use this balance to find the temperature distribution in an electrically heated wire.

5.1 Electric wire and shell energy balance

Consider a long wire of radius 𝑅$R$ and length 𝐿$L$. A uniform electrical current density 𝐼$I$ passes through it. Here 𝐼$I$ is current per unit area, in A m−2$\mathrm{A\,m^{-2}}$. Let 𝑘𝑒$k_e$ be the electrical conductivity, in S m−1$\mathrm{S\,m^{-1}}$, and 𝑘$k$ the thermal conductivity. Electrical resistance produces heat throughout the wire at the volumetric rate 𝑆𝑒 =𝐼2/𝑘𝑒$S_e=I^2/k_e$, in W m−3$\mathrm{W\,m^{-3}}$.

Assume steady state, constant thermal and electrical conductivities, uniform heat generation and negligible heat loss through the ends. The wire is stationary, and its surface is held at a uniform temperature 𝑇0$T_0$. Conditions do not vary around the circumference.

Temperature then depends only on radius, 𝑇 =𝑇(𝑟)$T=T(r)$. This makes the conduction problem one-dimensional, although the wire itself has three dimensions. Variation along the wire or around its circumference would require a model with additional spatial coordinates.

Figure 9: Longitudinal cutaway of a solid wire of radius R and length L. The narrow hatched strips are cuts through the annular control volume, with inner radius r and thickness delta r. Arrows show the radial heat flux at its inner and outer faces.

The shell balance follows Shah (n.d.), slides 22 and 23, and Bird et al. (2006), section 10.2.

Choose the thin cylindrical shell between 𝑟$r$ and 𝑟 +Δ𝑟$r+\Delta r$ in Figure 9. Heat enters its inner face at the rate 2𝜋𝑟𝐿𝑞𝑟(𝑟)$2\pi rLq_r(r)$ and leaves its outer face at the rate

2𝜋(𝑟+Δ𝑟)𝐿𝑞𝑟(𝑟+Δ𝑟).$$2\pi(r+\Delta r)Lq_r(r+\Delta r).$$

The areas differ because the two faces are at different radii.

The shell volume is 𝜋[(𝑟 +Δ𝑟)2 −𝑟2]𝐿$\pi[(r+\Delta r)^2-r^2]L$. Multiplying it by 𝑆𝑒$S_e$ gives the heat generated within the shell. With no accumulation, the balance is heat in minus heat out plus heat generated equals zero:

0=2𝜋𝑟𝐿𝑞𝑟(𝑟)−2𝜋(𝑟+Δ𝑟)𝐿𝑞𝑟(𝑟+Δ𝑟)+𝑆𝑒𝜋[(𝑟+Δ𝑟)2−𝑟2]𝐿$$0=2\pi rLq_r(r)-2\pi(r+\Delta r)Lq_r(r+\Delta r) +S_e\pi[(r+\Delta r)^2-r^2]L$$(9)

Dividing this balance by 2𝜋𝐿Δ𝑟$2\pi L\Delta r$ and moving the heat-flow terms to the left gives

(𝑟+Δ𝑟)𝑞𝑟(𝑟+Δ𝑟)−𝑟𝑞𝑟(𝑟)Δ𝑟=𝑆𝑒(𝑟+Δ𝑟2)$$\frac{(r+\Delta r)q_r(r+\Delta r)-rq_r(r)}{\Delta r} =S_e\left(r+\frac{\Delta r}{2}\right)$$

The left side is the change in 𝑟𝑞𝑟$rq_r$ divided by the shell thickness. Taking the limit as Δ𝑟 →0$\Delta r\to0$ gives

𝑑(𝑟𝑞𝑟)𝑑𝑟=𝑟𝑆𝑒$$\frac{d(rq_r)}{dr}=rS_e$$(10)

The radius remains inside the derivative because the heat-flow area changes with 𝑟$r$. Treating the two shell faces as having the same area would give an incorrect radial balance. In a finite-volume calculation, we can instead apply Eq. 9 to each cell, retaining its actual inner and outer face areas.

5.2 Integrate and apply the boundary conditions

To find the heat flux, integrate Eq. 10 with respect to radius:

𝑟𝑞𝑟=𝑆𝑒𝑟22+𝐶1𝑞𝑟=𝑆𝑒𝑟2+𝐶1𝑟$$rq_r=\frac{S_er^2}{2}+C_1 \qquad q_r=\frac{S_er}{2}+\frac{C_1}{r}$$

At the axis, the flux must be finite. A nonzero 𝐶1$C_1$ would make 𝐶1/𝑟$C_1/r$ unbounded as 𝑟 →0$r\to0$, so 𝐶1 =0$C_1=0$. The resulting flux, 𝑞𝑟 =𝑆𝑒𝑟/2$q_r=S_er/2$, is zero at the axis, consistent with symmetry.

Substituting 𝑞𝑟 = −𝑘 𝑑𝑇/𝑑𝑟$q_r=-k\,dT/dr$ gives the temperature gradient. A second integration gives the temperature:

𝑑𝑇𝑑𝑟=−𝑆𝑒𝑟2𝑘𝑇=−𝑆𝑒𝑟24𝑘+𝐶2$$\frac{dT}{dr}=-\frac{S_er}{2k} \qquad T=-\frac{S_er^2}{4k}+C_2$$

We determine 𝐶2$C_2$ from the prescribed surface temperature 𝑇(𝑅) =𝑇0$T(R)=T_0$. Substitution gives

𝐶2=𝑇0+𝑆𝑒𝑅24𝑘,$$C_2=T_0+\frac{S_eR^2}{4k},$$

and hence

𝑇(𝑟)−𝑇0=𝑆𝑒4𝑘(𝑅2−𝑟2)$$T(r)-T_0=\frac{S_e}{4k}(R^2-r^2)$$(11)

The radial heat flux is

𝑞𝑟=𝑆𝑒𝑟2$$q_r=\frac{S_er}{2}$$

Temperature decreases parabolically from its maximum at the axis to 𝑇0$T_0$ at the surface. The radial heat flux behaves differently: it is zero at the axis and increases linearly with radius. Each cylindrical surface must release all the heat generated inside it. At the axis, the temperature gradient is zero, so Fourier’s law gives zero flux even though the temperature is highest there.

The centre-to-surface temperature difference is 𝑆𝑒𝑅2/(4𝑘)$S_eR^2/(4k)$. Doubling 𝑅$R$ therefore increases this difference by a factor of four when 𝑆𝑒$S_e$ and 𝑘$k$ are fixed. The condition of fixed 𝑆𝑒$S_e$ matters. If the total electrical current were held fixed instead, changing the radius would also change the current density and the heat generation per unit volume.

A long solid wire generates heat uniformly, and its surface is held at 𝑇0$T_0$. At steady state, temperature varies only with radius. What must be true at the axis, 𝑟 =0$r=0$?

• The temperature equals the surface temperature 𝑇0$T_0$.
• The temperature gradient 𝑑𝑇/𝑑𝑟$dT/dr$ and radial heat flux are both zero. ✓
• The radial heat flux is infinite because the cylindrical area is zero.
• The temperature gradient has its largest magnitude.

Explanation

There is no preferred radial direction at the axis. Symmetry requires

[𝑑𝑇𝑑𝑟]𝑟=0=0$$\left[\frac{dT}{dr}\right]_{r=0}=0$$

Fourier’s law then gives

𝑞𝑟(0)=−𝑘[𝑑𝑇𝑑𝑟]𝑟=0=0$$q_r(0)=-k\left[\frac{dT}{dr}\right]_{r=0}=0$$

The centre can still be hotter than the surface. Its temperature is a maximum, where the gradient is zero.

A long wire generates heat uniformly at a rate 𝑆𝑒$S_e$ per unit volume. At steady state, its centre is 12 K$12\ \mathrm{K}$ hotter than its surface. Suppose we increase the radius from 2 mm$2\ \mathrm{mm}$ to 4 mm$4\ \mathrm{mm}$, keeping 𝑆𝑒$S_e$, thermal conductivity 𝑘$k$ and surface temperature 𝑇0$T_0$ unchanged. Neglect heat loss through the ends. What is the new centre-to-surface temperature difference?

• 6 K$6\ \mathrm{K}$.
• 12 K$12\ \mathrm{K}$.
• 24 K$24\ \mathrm{K}$.
• 48 K$48\ \mathrm{K}$. ✓

Explanation

At the axis, Eq. 11 gives

𝑇max⁡−𝑇0=𝑆𝑒𝑅24𝑘$$T_{\max}-T_0=\frac{S_eR^2}{4k}$$

With heat generation and conductivity unchanged, doubling the radius makes this temperature difference four times as large:

(𝑇max⁡−𝑇0)new=22(12)=48 K$$(T_{\max}-T_0)_{\mathrm{new}}=2^2(12)=48\ \mathrm K$$

5.3 Average temperature and conservation

To find the mean temperature, we must weight each temperature by the volume of material at that radius. A ring of thickness 𝑑𝑟$dr$ has area 2𝜋𝑟 𝑑𝑟$2\pi r\,dr$, so equal radial intervals contain more material near the surface than near the axis.

Temperature is uniform along the wire, so the length cancels from the volume average. We can therefore average over the cross-section. Using Eq. 11 gives

¯¯¯¯𝑇−𝑇0=∫𝑅0[𝑆𝑒(𝑅2−𝑟2)4𝑘] 2𝜋𝑟 𝑑𝑟𝜋𝑅2=𝑆𝑒𝑅28𝑘$$\overline T-T_0= \frac{\int_0^R \left[\frac{S_e(R^2-r^2)}{4k}\right]\,2\pi r\,dr} {\pi R^2}=\frac{S_eR^2}{8k}$$(12)

Evaluating the integrals uses

∫𝑅0𝑟 𝑑𝑟=𝑅22$$\int_0^R r\,dr=\frac{R^2}{2}$$

and

∫𝑅0(𝑅2𝑟−𝑟3) 𝑑𝑟=𝑅44.$$\int_0^R(R^2r-r^3)\,dr=\frac{R^4}{4}.$$

The mean temperature rise is half the centre-to-surface rise. Averaging temperatures at equally spaced radii without the factor 𝑟$r$ would overestimate the mean by giving too much weight to the hotter material near the axis.

We can check conservation of energy using the flux at 𝑟 =𝑅$r=R$. Multiplying by the curved surface area gives the total heat loss:

˙𝑄𝑅=𝑞𝑟(𝑅) 2𝜋𝑅𝐿=𝑆𝑒𝑅2 2𝜋𝑅𝐿=𝑆𝑒𝜋𝑅2𝐿$$\dot Q_R=q_r(R)\,2\pi RL =\frac{S_eR}{2}\,2\pi RL =S_e\pi R^2L$$

The result is exactly the heat generation per unit volume multiplied by the wire volume. All the heat generated leaves through the curved surface, as required by the assumptions. A numerical solution should satisfy this overall balance as well as the local energy equation.

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.

6 Reducing the general energy equation

The shell balance gave the energy equation for a particular geometry. We can also obtain it by simplifying the general energy equation. For an incompressible Newtonian fluid with constant properties and negligible compression work, this equation can be written as

𝜌̂𝐶𝑝(∂𝑇∂𝑡+𝒖⋅∇𝑇)=𝑘∇2𝑇+𝑆$$\rho\widehat C_p\left(\frac{\partial T}{\partial t} +\boldsymbol u\cdot\nabla T\right)=k\nabla^2T+S$$(13)

Here 𝜌$\rho$ is density, ̂𝐶𝑝$\widehat C_p$ is specific heat capacity per unit mass and 𝒖$\boldsymbol u$ is fluid velocity. The source 𝑆$S$ represents heat generation per unit volume, in W m−3$\mathrm{W\,m^{-3}}$. Viscous dissipation must either be negligible or be included in 𝑆$S$.

The left side accounts for the local temperature change and heat transport by fluid motion. The terms on the right account for conduction and internal heat generation. In the stationary wire, 𝒖 =0$\boldsymbol u=0$ and 𝑆 =𝑆𝑒$S=S_e$.

6.1 Cartesian and cylindrical forms

The material derivative 𝐷𝑇/𝐷𝑡$DT/Dt$ describes the rate of temperature change of a fluid element as it moves. It combines the local time derivative with the change due to motion through a temperature field. In Cartesian coordinates, the material derivative is

𝐷𝑇𝐷𝑡=∂𝑇∂𝑡+𝑢𝑥∂𝑇∂𝑥+𝑢𝑦∂𝑇∂𝑦+𝑢𝑧∂𝑇∂𝑧$$\frac{DT}{Dt}=\frac{\partial T}{\partial t} +u_x\frac{\partial T}{\partial x} +u_y\frac{\partial T}{\partial y} +u_z\frac{\partial T}{\partial z}$$

The Laplacian is

∇2𝑇=∂2𝑇∂𝑥2+∂2𝑇∂𝑦2+∂2𝑇∂𝑧2$$\nabla^2T=\frac{\partial^2T}{\partial x^2} +\frac{\partial^2T}{\partial y^2} +\frac{\partial^2T}{\partial z^2}$$

For cylindrical coordinates, we use radius 𝑟$r$, angle 𝜃$\theta$ and axial position 𝑧$z$. The distances are measured in metres and the angle in radians. The material derivative is

𝐷𝑇𝐷𝑡=∂𝑇∂𝑡+𝑢𝑟∂𝑇∂𝑟+𝑢𝜃𝑟∂𝑇∂𝜃+𝑢𝑧∂𝑇∂𝑧$$\frac{DT}{Dt}=\frac{\partial T}{\partial t} +u_r\frac{\partial T}{\partial r} +\frac{u_\theta}{r}\frac{\partial T}{\partial\theta} +u_z\frac{\partial T}{\partial z}$$(14)

The Laplacian is

∇2𝑇=1𝑟∂∂𝑟(𝑟∂𝑇∂𝑟)+1𝑟2∂2𝑇∂𝜃2+∂2𝑇∂𝑧2$$\nabla^2T=\frac{1}{r}\frac{\partial}{\partial r} \left(r\frac{\partial T}{\partial r}\right) +\frac{1}{r^2}\frac{\partial^2T}{\partial\theta^2} +\frac{\partial^2T}{\partial z^2}$$(15)

Substituting these expressions into Eq. 13 gives its cylindrical form. The radial conduction term contains the same factor 𝑟$r$ as the shell balance because the area normal to radial heat flow increases with radius.

6.2 Thermal diffusivity

Thermal diffusivity

Thermal diffusivity, 𝛼$\alpha$, is the ratio of thermal conductivity to volumetric heat capacity, 𝛼 =𝑘/(𝜌̂𝐶𝑝)$\alpha=k/(\rho\widehat C_p)$. It characterises the rate at which temperature disturbances spread through a material by conduction. Its SI unit is m2 s−1$\mathrm{m^2\,s^{-1}}$.

The volumetric heat capacity 𝜌̂𝐶𝑝$\rho\widehat C_p$ is the energy required per unit volume for a unit temperature rise. Dividing Eq. 13 by this quantity gives

𝐷𝑇𝐷𝑡=𝛼∇2𝑇+𝑆𝜌̂𝐶𝑝$$\frac{DT}{Dt}=\alpha\nabla^2T+\frac{S}{\rho\widehat C_p}$$

Here the thermal diffusivity is

𝛼=𝑘𝜌̂𝐶𝑝$$\alpha=\frac{k}{\rho\widehat C_p}$$(16)

Each term in the temperature equation has units of K s−1$\mathrm{K\,s^{-1}}$. Thermal diffusivity combines two effects. A higher conductivity allows heat to redistribute more readily, while a higher volumetric heat capacity requires more energy for the same temperature change. Their ratio determines the rate at which temperature differences diminish by conduction.

6.3 Example: estimating a diffusion time

We can use thermal diffusivity to estimate the time required for conduction over a distance 𝐿$L$. With no fluid motion or heat generation, the energy equation is

∂𝑇∂𝑡=𝛼∇2𝑇.$$\frac{\partial T}{\partial t}=\alpha\nabla^2T.$$

For a temperature change of order Δ𝑇$\Delta T$ over distance 𝐿$L$ and time 𝑡$t$, the two terms have the approximate magnitudes

∂𝑇∂𝑡∼Δ𝑇𝑡𝛼∂2𝑇∂𝑦2∼𝛼Δ𝑇𝐿2$$\frac{\partial T}{\partial t}\sim\frac{\Delta T}{t} \qquad \alpha\frac{\partial^2T}{\partial y^2}\sim\alpha\frac{\Delta T}{L^2}$$

Balancing these magnitudes and cancelling Δ𝑇$\Delta T$ gives

𝑡diff∼𝐿2𝛼$$t_{\mathrm{diff}}\sim\frac{L^2}{\alpha}$$(17)

The diffusion time scale therefore increases with the square of the distance. For the same material, doubling 𝐿$L$ increases the estimated time by a factor of four.

For example, consider a piece of beef 2 cm$2\ \mathrm{cm}$ thick, heated from both sides. Conduction must act over 𝐿 =0.01 m$L=0.01\ \mathrm{m}$ to reach the centre. Taking 𝛼 =1.4 ×10−7 m2 s−1$\alpha=1.4\times10^{-7}\ \mathrm{m^2\,s^{-1}}$ gives 𝑡diff ∼714 s$t_{\mathrm{diff}}\sim714\ \mathrm{s}$, or about twelve minutes.

This is a characteristic time for temperature changes to extend through the thickness. Finding the time for the centre to reach a specified temperature requires the initial and boundary conditions. The estimate also does not predict surface browning, which involves chemical reactions.

A mould is repeatedly heated and cooled. A designer proposes using an alloy with a higher thermal conductivity so that temperature changes spread through the mould faster. Is conductivity alone enough to tell whether the proposed alloy has a higher thermal diffusivity?

• Yes. Thermal diffusivity depends only on thermal conductivity.
• Yes. Density and heat capacity affect steady conduction only.
• No. We also need the density and specific heat capacity of each alloy. ✓
• No. A higher conductivity always gives a lower diffusivity.

Explanation

Thermal diffusivity depends on conductivity, density and specific heat capacity:

𝛼=𝑘𝜌̂𝐶𝑝$$\alpha=\frac{k}{\rho\widehat C_p}$$

A higher conductivity increases this ratio, while a higher volumetric heat capacity reduces it. We need all three properties to compare the two alloys.

In a moulding process, a polymer sheet is heated from both sides. For a 10 mm$10\ \mathrm{mm}$ thick sheet, the estimated diffusion time from either surface to the centre is 50 s$50\ \mathrm{s}$.

What is the estimated diffusion time for a 30 mm$30\ \mathrm{mm}$ thick sheet of the same polymer? Assume the thermal diffusivity is unchanged and use 𝑡diff ∼𝐿2/𝛼$t_{\mathrm{diff}}\sim L^2/\alpha$, where 𝐿$L$ is the distance from a heated surface to the centre.

• 150 s$150\ \mathrm{s}$.
• 450 s$450\ \mathrm{s}$. ✓
• 50 s$50\ \mathrm{s}$.
• 50/9 s$50/9\ \mathrm{s}$.

Explanation

The conduction distance is half the sheet thickness. It increases from 5 mm to 15 mm. For the same diffusivity,

𝑡new𝑡old=(155)2=9$$\frac{t_{\mathrm{new}}}{t_{\mathrm{old}}}=\left(\frac{15}{5}\right)^2=9$$

Thus,

𝑡new=9(50)=450 s$$t_{\mathrm{new}}=9(50)=450\ \mathrm s$$

This is a diffusion time estimate. Finding the time to reach a particular temperature also requires the initial and surface conditions.

6.4 Apply the general equation to the wire

Let’s recover the wire equation from the general energy equation. Substitute the material derivative in Eq. 14 and the Laplacian in Eq. 15 into Eq. 13, then apply the wire assumptions:

1. At steady state, ∂𝑇/∂𝑡 =0$\partial T/\partial t=0$.
2. The wire is stationary, so 𝑢𝑟 =𝑢𝜃 =𝑢𝑧 =0$u_r=u_\theta=u_z=0$ and there is no heat transport by bulk motion.
3. Conditions are uniform around the circumference, so derivatives with respect to 𝜃$\theta$ are zero.
4. End effects are negligible, so derivatives with respect to 𝑧$z$ are zero in the region considered.

Only radial conduction and heat generation remain. Taking 𝑆 =𝑆𝑒$S=S_e$ and constant 𝑘$k$ gives

0=𝑘𝑟𝑑𝑑𝑟(𝑟𝑑𝑇𝑑𝑟)+𝑆𝑒$$0=\frac{k}{r}\frac{d}{dr}\left(r\frac{dT}{dr}\right)+S_e$$(18)

Rearranging after multiplication by 𝑟/𝑘$r/k$, then integrating, gives

𝑑𝑑𝑟(𝑟𝑑𝑇𝑑𝑟)=−𝑆𝑒𝑟𝑘𝑑𝑇𝑑𝑟=−𝑆𝑒𝑟2𝑘+𝐶𝑟$$\frac{d}{dr}\left(r\frac{dT}{dr}\right)=-\frac{S_er}{k} \qquad \frac{dT}{dr}=-\frac{S_er}{2k}+\frac{C}{r}$$

A nonzero 𝐶$C$ would produce a logarithmic temperature term that diverges at 𝑟 =0$r=0$. The temperature must remain finite, so 𝐶 =0$C=0$. A second integration, followed by the boundary condition 𝑇(𝑅) =𝑇0$T(R)=T_0$, recovers Eq. 11.

We can also compare the differential equations. Substituting Fourier’s law, 𝑞𝑟 = −𝑘 𝑑𝑇/𝑑𝑟$q_r=-k\,dT/dr$, into Eq. 18 gives

𝑑(𝑟𝑞𝑟)𝑑𝑟=𝑟𝑆𝑒,$$\frac{d(rq_r)}{dr}=rS_e,$$

exactly as in Eq. 10. The shell balance and the general energy equation therefore give the same model. The first derives it from heat entering, leaving and being generated in a control volume; the second obtains it by simplifying an existing energy equation.

We want the steady temperature distribution in a long, stationary wire that generates heat uniformly. Temperature does not vary along the wire or around its circumference, and heat loss through the ends is negligible. Which terms remain in the energy equation after applying these assumptions?

• Accumulation and radial conduction.
• Advection and internal heat generation.
• Radial and axial conduction.
• Radial conduction and internal heat generation. ✓

Explanation

Steady state removes accumulation. The wire is stationary, so there is no advection. Temperature is uniform around the circumference, so there is no angular conduction. Negligible end effects allow us to neglect axial conduction. Only radial conduction and internal heat generation remain, and they balance at steady state.

7 Additional examples

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.

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$123\ \mathrm{kg\,h^{-1}}$.

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.

8 What you should now be able to calculate

Calculate heat loss and interface temperatures in plane and radial networks, and find a maximum temperature when heat is generated inside the solid.

Practice: Workshop 5 for resistance networks, heat generation and variable conductivity.

9 Report an error

To report an error, email Ranjeet Utikar. Include the title Heat transfer modes and steady conduction, the URL /heat-transport/steady-conduction/, and the equation or passage that needs correction.

References

Bird, R. Byron, Warren E. Stewart, and Edwin N. Lightfoot. 2006. Transport Phenomena. Revised 2nd. John Wiley & Sons. https://www.wiley-vch.de/en/areas-interest/engineering/transport-phenomena-978-0-470-11539-8.

Çengel, Yunus A., and Afshin J. Ghajar. 2015. Heat and Mass Transfer: Fundamentals and Applications. 5th ed. McGraw-Hill Education. https://glencoe.mheducation.com/sites/0073398187/information_center_view0/application_chapters.html.

Shah, Milinkumar. n.d. “CHEN3001-TP: Heat Transport 1.” Curtin University. Unpublished.