Cross-Flow Heat Exchanger — Required Area: 23-Chem-A2 Unit Operations and Separation Processes, May 2013, Q5

Question B2: Cross-Flow Heat Exchanger — Required Area (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $\dot m_h=\dot m_c=75.6\ \mathrm{kg/min}=1.26\ \mathrm{kg/s}$; hot 94 → 72 °C; cold inlet 38 °C; $U=2270\ \mathrm{W/(m^2K)}$; $c_p\approx4199\ \mathrm{J/(kg\,K)}$ (Table B1, near the mean temperatures); cross-flow, both fluids unmixed.

Find. the heat-transfer surface area $A$.

Figure B2 — Temperature–duty profile. Equal water flows give equal end-approaches ($\Delta T=34,{}^ rc$C at both ends), so the counter-flow LMTD equals 34 °C; the cross-flow correction $F$ and area follow.

Approach. Close the energy balance for the cold outlet and duty, compute the counter-flow LMTD, obtain the cross-flow correction $F$ (equivalently solve the $\varepsilon$–NTU relation for both-unmixed flow), then $A=q/(UF\,\Delta T_{lm})$.

Cold outlet and duty. Equal flows and (near-)equal $c_p$ give equal temperature changes: $T_{c,o}=38+22=60,{}^ rc$C. The duty is $$q=\dot m c_p\,\Delta T_h=1.26(4199)(22)=\boxed{116.4\ \mathrm{kW}}.$$
Log-mean temperature difference (counter-flow basis). $\Delta T_1=94-60=34$, $\Delta T_2=72-38=34$; equal, so $$\Delta T_{lm}=34,{}^ rc\mathrm{C}.$$
Correction factor / effectiveness. With $P=\dfrac{60-38}{94-38}=0.393$ and $R=\dfrac{94-72}{60-38}=1.0$, the both-unmixed cross-flow chart (Fig. B1) gives $F\approx0.90$. Independently, effectiveness $\varepsilon=q/[\,C_{\min}(T_{h,i}-T_{c,i})]=0.393$; solving the both-unmixed relation $\varepsilon=1-\exp\!\big\{N^{0.22}[\,e^{-N^{0.78}}-1]\big\}$ gives $\mathrm{NTU}=0.716$, i.e. $F=0.90$.
Surface area. $$A=\frac{q}{U F\,\Delta T_{lm}}=\frac{116\,400}{2270(0.90)(34)}=\boxed{1.67\ \mathrm{m^2}}.$$ Equivalently $A=\mathrm{NTU}\,C_{\min}/U=0.716(5291)/2270=1.67\ \mathrm{m^2}$ — the two routes agree.

Quantity	Result
Cold outlet temperature	60 °C
Duty $q$	116.4 kW
$\Delta T_{lm}$ (counter-flow) / $F$	34 °C / 0.90
Effectiveness $\varepsilon$ / NTU	0.393 / 0.716
Surface area $A$	1.67 m$^2$

Concept & Theory

Topic. Heat-exchanger sizing — LMTD–$F$ and $\varepsilon$–NTU methods for cross-flow (both fluids unmixed).

Key relations. $q=\dot m c_p\Delta T=UAF\,\Delta T_{lm}$; $\Delta T_{lm}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$; $P,R$ chart parameters; $\varepsilon=q/[C_{\min}(T_{h,i}-T_{c,i})]$, $A=\mathrm{NTU}\,C_{\min}/U$.

Why this works. The correction factor $F$ discounts the ideal counter-flow LMTD for the imperfect temperature crossing in a cross-flow unit; because both streams have identical $C$ ($C_r=1$) and equal end-approaches, the counter-flow $\Delta T_{lm}$ is exactly 34 °C and only $F$ (or NTU) carries the geometry.

Common pitfall. Reading $F$ against the wrong $P$/$R$ definitions, or using $F=1$ (that is counter-flow, not cross-flow). Also note the small $c_p$ variation with temperature makes $T_{c,o}$ 60.1 °C, negligibly different from 60 °C.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer, §11.3–11.4; Lienhard, A Heat Transfer Textbook, Ch. 3 (Fig. 3.17 correction charts).

Question 5 of 6: Cross-Flow Heat Exchanger — Required Area

Section A — Mechanical Operations

Question B2: Cross-Flow Heat Exchanger — Required Area (25 marks)