23-Chem-A2 Unit Operations and Separation Processes · May 2014
Question 6 of 6: Cross-Flow Heat Exchanger — Overall Coefficient
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exam 04-Chem-A2 Mechanical and Thermal Operations, May 2014 — open-book, 3 hours. Two sections: Section A (Mechanical Operations, A1–A3) and Section B (Thermal Operations, B1–B3); all problems are worth 25 marks. The rubric asks candidates to attempt two problems per section; all six are solved in full below as a study resource.
Reference texts: McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed., McGraw-Hill) — pipe-flow friction and loss coefficients, agitator power correlations (Fig. 9.13, Table 9.3), and filtration theory; de Nevers, Fluid Mechanics for Chemical Engineers (3rd ed.) and Brodkey & Hershey, Transport Phenomena — mechanical-energy balance, fitting equivalent lengths, and the appended Fanning chart; Coulson & Richardson, Chemical Engineering Vol. 2 — rotary-drum filtration; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (7th ed., Wiley) and Lienhard, A Heat Transfer Textbook — composite-wall resistance networks, the Colburn (Chilton–Colburn) analogy, and LMTD/ε–NTU cross-flow exchangers.
Property note. Water properties are as printed in each question; the appended Table A.2 loss coefficients ($k$) and Table A3 turbine constants ($K_T$) are used directly, and pipe-friction factors are taken from Colebrook (identical to the appended Fanning chart, Fig. A1) so the numbers are reproducible without reading a graph.
Given. Cross-flow exchanger, both fluids unmixed; duty is set by the water stream.
Quantity
Value
Tubes: number / $D_i$ / $L$
40 / 5 mm / 0.65 m
Water: $\dot m$ / in / out
36 kg/min / 90 °C / 65 °C
Air: in / out
20 °C / 40 °C
Water $c_p$ (mean $\approx77.5\ \mathrm{^\circ C}$, Table B1)
$4195\ \mathrm{J/kg\,K}$
Find. the overall heat-transfer coefficient $U_i$ on the inner tube area.
Figure B3 — Cross-flow arrangement, both streams unmixed: water through the tubes (90→65 °C), air across the bundle (20→40 °C).
Approach. Compute the duty and the heat-capacity rates, form the effectiveness $\varepsilon$, invert the both-fluids-unmixed $\varepsilon$–NTU relation for NTU, then $U_i=NTU\,C_{\min}/A_i$. Cross-check with the LMTD–$F$ route.
Duty and heat-capacity rates. Water: $\dot m=36/60=0.6\ \mathrm{kg/s}$, so $$q=\dot m c_p\Delta T=0.6(4195)(90-65)=6.29\times10^{4}\ \mathrm{W}.$$ $C_w=\dot m c_p=2517\ \mathrm{W/K}$; from the air balance $C_{\text{air}}=q/\Delta T_{\text{air}}=62{,}925/20=3146\ \mathrm{W/K}$. Thus $C_{\min}=2517$ (water), $C_r=C_{\min}/C_{\max}=0.800$.
Effectiveness. With the maximum possible temperature difference $90-20=70\ \mathrm{K}$, $$\varepsilon=\frac{q}{C_{\min}\Delta T_{\max}}=\frac{62{,}925}{2517(70)}=0.357.$$
Invert the cross-flow (both unmixed) $\varepsilon$–NTU relation. The exact single-pass cross-flow result (both fluids unmixed; the relation behind the appended chart, Fig. B1) is $$\varepsilon=\frac{1}{C_r\,NTU}\sum_{n=0}^{\infty}\Big[1-e^{-NTU}\sum_{m=0}^{n}\frac{NTU^m}{m!}\Big]\Big[1-e^{-C_rNTU}\sum_{m=0}^{n}\frac{(C_rNTU)^m}{m!}\Big].$$ Solving numerically for $\varepsilon=0.357$, $C_r=0.800$ gives $NTU=0.543$. (The popular closed-form approximation $\varepsilon\approx1-\exp[\tfrac{NTU^{0.22}}{C_r}(e^{-C_rNTU^{0.78}}-1)]$ gives 0.565 here, about 4% high, so it is not used for the final answer.)
Overall coefficient. Since $NTU=U_iA_i/C_{\min}$, $$U_i=\frac{NTU\,C_{\min}}{A_i}=\frac{0.543(2517)}{0.4084}=3.35\times10^{3}\ \mathrm{W/m^2K}.$$ $U_i\approx 3350\ \mathrm{W/m^2K}$ (on the inner area)
LMTD–$F$ cross-check (Fig. B1). The counter-flow LMTD is $\dfrac{(90-40)-(65-20)}{\ln(50/45)}=47.46\ \mathrm{^\circ C}$. With water in the tubes, $P=\dfrac{65-90}{20-90}=0.357$ and $R=\dfrac{20-40}{65-90}=0.80$; the $R=0.8$ curve of Fig. B1 gives $F\approx0.97$. Then $U_i=q/(A_iF\,\mathrm{LMTD})=62{,}925/(0.4084\cdot0.970\cdot47.46)=3.35\times10^{3}\ \mathrm{W/m^2K}$, so the two routes agree.
