Question 3 of 6: Part A: Pipe Length in a Cross-Flow Tube-Bank Gas Heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: open-book, three-hour exam; Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6); all problems 25 points, at least two attempted per part. Every problem is worked in full below.
Reference texts: J. M. Coulson & J. F. Richardson, Chemical Engineering Vols. 1&2 (Coulson & Richardson) — boiling curves, tube-bank convection, gas absorption and column sizing; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free-convection correlations and the gas-property tables reproduced in the paper; J. R. Welty, C. E. Wicks & R. E. Wilson, Fundamentals of Momentum, Heat and Mass Transfer — the laminar vertical-plate local Nusselt relation; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — stagnant-film diffusion, differential (Rayleigh) distillation and packed-tower flooding; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 3 — Part A: Pipe Length in a Cross-Flow Tube-Bank Gas Heater (25 points)
Given. Air (inside the tubes) is heated $283\to366$ K; flue gas (crossing outside) is cooled $700\to366$ K. $N=20\times20=400$ tubes, $d=0.012$ m, free-flow gas mass velocity $G=10$ kg/m²s, $c_p=1000$ J/kg·K, in-line arrangement factor $C_h=0.95$. Property table ($k$ in W/m·K, $\mu$ in mN·s/m²):
$T$ (K)
250
500
800
$k$ (W/m·K)
0.022
0.044
0.055
$\mu$ (mN·s/m²)
0.0165
0.0276
0.0367
Find. The length $L$ of each of the 400 pipes.
Figure 3 — In-line tube bank: air flows inside the 400 tubes (into the page) while flue gas sweeps across the outside. Inside coefficient from Dittus–Boelter, outside from a cross-flow bank correlation; the two combine into $U$ and the area sets the tube length.
Approach. Get the duty from the air enthalpy rise; the inside coefficient from Dittus–Boelter on the per-tube air flow; the outside coefficient from a cross-flow tube-bank correlation on the flue-gas mass velocity; combine into $U$ (wall neglected, OD = ID), take the counter-current LMTD, and solve $Q=UA\,\Delta T_{lm}$ for area, hence length.
Heat duty. From the air side,
$$Q=\dot m\,c_p\,\Delta T=(0.9)(1000)(366-283)=\boxed{74{,}700\ \text{W}}.$$
Inside (air) coefficient. Air properties at its mean 324.5 K interpolate to $k=0.0286$ W/m·K, $\mu=1.98\times10^{-5}$ Pa·s. Per tube $\dot m_1=0.9/400=2.25\times10^{-3}$ kg/s, so
$$Re_i=\frac{4\dot m_1}{\pi d\,\mu}=\frac{4(2.25\times10^{-3})}{\pi(0.012)(1.98\times10^{-5})}=1.21\times10^{4}\;(\text{turbulent}),\quad Pr=\frac{c_p\mu}{k}=0.694.$$
Dittus–Boelter (heating): $Nu_i=0.023\,Re_i^{0.8}Pr^{0.4}=36.5$, giving
$$h_i=\frac{Nu_i k}{d}=\frac{(36.5)(0.0286)}{0.012}\approx 87\ \text{W/m}^2\text{K}.$$
Outside (flue-gas) coefficient. Gas properties at its mean 533 K interpolate to $k=0.0452$ W/m·K, $\mu=2.86\times10^{-5}$ Pa·s. On the free-flow mass velocity,
$$Re_o=\frac{G\,d}{\mu}=\frac{(10)(0.012)}{2.86\times10^{-5}}=4.20\times10^{3},\quad Pr=0.633.$$
For an in-line bank, $Nu_o=0.33\,C_h\,Re_o^{0.6}Pr^{0.3}=0.33(0.95)(4196)^{0.6}(0.633)^{0.3}=40.8$, so
$$h_o=\frac{Nu_o k}{d}=\frac{(40.8)(0.0452)}{0.012}\approx 154\ \text{W/m}^2\text{K}.$$
Overall coefficient. With OD = ID and wall/scale neglected, the two films add in series:
$$U=\left(\frac1{h_i}+\frac1{h_o}\right)^{-1}=\left(\frac1{87}+\frac1{154}\right)^{-1}=\boxed{55.5\ \text{W/m}^2\text{K}}.$$
Log-mean temperature difference. Counter-current, ends $700\to366$ (gas) against $283\to366$ (air):
$$\Delta T_{lm}=\frac{(700-366)-(366-283)}{\ln\!\frac{334}{83}}=\frac{334-83}{\ln 4.024}=180.3\ \text{K}.$$
Area and pipe length. From $Q=U A\,\Delta T_{lm}$,
$$A=\frac{Q}{U\,\Delta T_{lm}}=\frac{74{,}700}{(55.5)(180.3)}=7.46\ \text{m}^2.$$
Spread over $N=400$ tubes of surface $\pi d L$ each,
$$L=\frac{A}{N\,\pi d}=\frac{7.46}{400\,\pi(0.012)}=\boxed{0.50\ \text{m per pipe}.}$$
Quantity
Result
Duty $Q$
74.7 kW
$h_i$ (air, inside) / $h_o$ (gas, outside)
87 / 154 W/m²K
Overall $U$
55.5 W/m²K
$\Delta T_{lm}$ / total area $A$
180.3 K / 7.46 m²
Length of each pipe
≈ 0.50 m
Check
The counter-current LMTD is the usual textbook simplification for this problem. Strictly, the flue gas crosses the tubes at right angles, so the exchanger is single-pass cross-flow with both streams unmixed. With $P=83/417=0.20$ and $R=334/83=4.0$, the effectiveness–NTU relations give a correction factor $F\approx0.94$. That raises the required area by about 7 %, so each pipe is about $0.50/0.94\approx0.53$ m. Either value is a defensible answer if the flow-arrangement assumption is stated.
