Workshop 7: Forced and natural convection

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Attempt these problems before opening the worked solutions.

1 Water cooling a reactor tube

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.

2 Oil in a cold lake

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.

3 Water pipe in moving or stagnant air

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.