Shell-and-tube heat exchanger sizing: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, Undated paper, Q8

Question 8 — Shell-and-tube heat exchanger sizing (Part B, equal value)

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.

Quantity	Symbol	Value
Cold water flow / temps	$\dot m_c$	2.0 kg/s, 30 → 55 °C
Hot water flow / inlet	$\dot m_h,\,T_{h,i}$	3.0 kg/s, 84 °C
Overall coefficient	$U$	1690 W/m²·K
Tube velocity / diameter	$V,\,D$	0.502 m/s, 10.5 mm
Passes	—	4 tube passes

Find. number of tubes per pass and tube length.

Cold water in the tubes gains 209 kW; hot shell-side water leaves at 67.3 °C.

Approach. Get the duty from the cold-stream rise, size the surface area from the corrected LMTD, fix the tubes-per-pass from the required tube-side flow area at the stated velocity, then divide the area among the tubes for the length.

Duty and hot-outlet temperature. With $c_{p}=4.18$ kJ/kg·K, $$\dot Q=\dot m_c c_p(T_{c,o}-T_{c,i})=2.0(4.18)(25)=\boxed{209\ \text{kW}}$$ $$T_{h,o}=T_{h,i}-\frac{\dot Q}{\dot m_h c_p}=84-\frac{209}{3.0(4.18)}=67.3\ ^\circ\text{C}$$
Corrected log-mean temperature difference. Counterflow end differences $\Delta T_1=84-55=29$ K and $\Delta T_2=67.3-30=37.3$ K give $$\Delta T_\text{lm}=\frac{37.3-29}{\ln(37.3/29)}=33.0\ \text{K}$$ With $P=0.463$, $R=0.667$ the multipass factor is $F\approx0.93$, so the required area is $$A=\frac{\dot Q}{U\,F\,\Delta T_\text{lm}}=\frac{209{,}000}{1690(0.932)(33.0)}=\boxed{4.02\ \text{m}^2}$$
Tubes per pass from the tube-side velocity. The total tube-side flow area needed is $A_\text{flow}=\dot m_c/(\rho V)=2.0/(1000\times0.502)=3.98\times10^{-3}$ m². Dividing by one tube's bore $a=\tfrac\pi4D^2=8.66\times10^{-5}$ m², $$N_\text{tubes/pass}=\frac{3.98\times10^{-3}}{8.66\times10^{-5}}=\boxed{46\ \text{tubes per pass}}$$ (which reproduces $V=0.502$ m/s exactly).
Tube length. With four passes the total tube count is $N=46\times4=184$; the heat-transfer area is $A=N\pi D L$, so $$L=\frac{A}{N\pi D}=\frac{4.02}{184\,\pi(0.0105)}=\boxed{0.66\ \text{m per pass}}$$ i.e. about a 2.6 m total flow path through the four passes.

Quantity	Result
Duty $\dot Q$	209 kW
Hot-water outlet	67.3 °C
Corrected LMTD ($F=0.93$)	33.0 K
Surface area	4.02 m²
Tubes per pass / total	46 / 184
Tube length per pass	≈ 0.66 m

Check — pass configuration.

The paper reads "2-pass, 4-tube passes"; the solution adopts a four-tube-pass exchanger and uses the standard multipass correction factor $F\approx0.93$ for $P=0.46$, $R=0.67$. The tubes-per-pass count is pinned by the stated 0.502 m/s velocity and is independent of $F$; only the length scales with $F$.

Concept & Theory — Q8

Topic: Sizing a multipass shell-and-tube exchanger by the corrected-LMTD (F-factor) method with a velocity-constrained tube count.

Key relations: energy balance $\dot Q=\dot m c_p\Delta T$; rate equation $\dot Q=U A\,F\,\Delta T_\text{lm}$; counterflow $\Delta T_\text{lm}$; tube-side continuity $\dot m=\rho V N a$ (Çengel HT, Ch. 11).

Why this works: The duty and both stream temperatures come from simple energy balances; the F-factor corrects the counterflow LMTD for the imperfect temperature profile of a multipass unit, so $A$ follows directly. The velocity requirement is a separate hydraulic constraint that fixes how many parallel tubes share the tube-side flow, and the area then sets their length.

Common pitfall: Using the parallel-flow or uncorrected LMTD, or confusing tubes-per-pass with total tubes — the flow area sets tubes per pass, while the surface area is distributed over all passes.

Question 8 of 8: Shel​l-and-tube heat exchanger sizing

Question 8 — Shel​l-and-tube heat exchanger sizing (Part B, equal value)

Question 8 of 8: Shell-and-tube heat exchanger sizing

Question 8 — Shell-and-tube heat exchanger sizing (Part B, equal value)