Two-shell-pass, twelve-tube-pass oil cooler: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2019, Q8

Question 8 — Two-shell-pass, twelve-tube-pass oil cooler (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.

Stream / parameter	Value
Water (tubes): $\dot m,\,c_p,\,T_\text{in}$	0.1 kg/s, 4180 J/kg·°C, 18 °C
Oil (shell): $\dot m,\,c_p,\,T_\text{in}$	0.2 kg/s, 2200 J/kg·°C, 160 °C
Geometry	2 shell / 12 tube passes, $D=0.018$ m, $L=3.8$ m/pass
Overall coefficient	$U=340$ W/m²·°C

Find. Heat-transfer rate and both exit temperatures.

Water (the lower heat-capacity rate) is $C_\text{min}$; the multi-pass arrangement permits the water outlet (110.6 °C) to exceed the oil outlet (72 °C) — a temperature cross a single pass could not achieve.

Approach. Compute the two heat-capacity rates and identify $C_\text{min}$; find the area and NTU; apply the $n$-shell-pass effectiveness relation to get $\varepsilon$; then $\dot Q=\varepsilon C_\text{min}\Delta T_\text{max}$ and close the two energy balances for the exit temperatures.

Heat-capacity rates and $C_r$. $$C_w=0.1(4180)=418\ \text{W/K},\quad C_o=0.2(2200)=440\ \text{W/K}$$ $$C_\text{min}=418\ (\text{water}),\ C_\text{max}=440,\ C_r=418/440=0.950$$
Area and NTU. $A=12\,\pi D L=12\pi(0.018)(3.8)=2.578\ \text{m}^2$: $$\text{NTU}=\frac{UA}{C_\text{min}}=\frac{340(2.578)}{418}=2.097$$
Effectiveness (2 shell passes). First one shell pass at $\text{NTU}_1=\text{NTU}/2=1.048$: $$\varepsilon_1=\frac{2}{(1+C_r)+\sqrt{1+C_r^2}\,\dfrac{1+e^{-\text{NTU}_1\sqrt{1+C_r^2}}}{1-e^{-\text{NTU}_1\sqrt{1+C_r^2}}}}=0.478$$ then combine two identical passes: $$\varepsilon=\frac{\left(\dfrac{1-\varepsilon_1 C_r}{1-\varepsilon_1}\right)^{2}-1}{\left(\dfrac{1-\varepsilon_1 C_r}{1-\varepsilon_1}\right)^{2}-C_r}=\boxed{0.652}$$
Heat-transfer rate. $$\dot Q=\varepsilon\,C_\text{min}(T_{h,i}-T_{c,i})=0.652(418)(160-18)=\boxed{38.7\ \text{kW}}$$
Exit temperatures (energy balances). $$T_{w,o}=18+\frac{38\,700}{418}=\boxed{110.6\ ^\circ\text{C}},\qquad T_{o,o}=160-\frac{38\,700}{440}=\boxed{72.0\ ^\circ\text{C}}$$

Check — tube length and water outlet.

The paper prints the tube length as "3.8 cm", which is physically impossible (it gives $A=0.026$ m² and a negligible NTU); it is read as 3.8 m per pass, consistent with a real shell-and-tube unit. Note the computed water outlet, 110.6 °C, exceeds 100 °C — at atmospheric pressure the tube water would begin to boil, so a real installation would run pressurised (or with a higher water flow). The calculation is reported as posed; the boiling caveat is flagged rather than folded into the answer.

Quantity	Result
Effectiveness $\varepsilon$	0.652
Heat-transfer rate	38.7 kW
Water exit temperature	110.6 °C
Oil exit temperature	72.0 °C

Concept & Theory — Q8

Topic: Effectiveness–NTU analysis of a multi-pass shell-and-tube exchanger with both outlet temperatures unknown.

Key relations: $C_r=C_\text{min}/C_\text{max}$; $\text{NTU}=UA/C_\text{min}$; single-shell-pass $\varepsilon_1$ formula; $n$-pass combination $\varepsilon=\big[(\tfrac{1-\varepsilon_1 C_r}{1-\varepsilon_1})^n-1\big]/\big[(\tfrac{1-\varepsilon_1 C_r}{1-\varepsilon_1})^n-C_r\big]$; $\dot Q=\varepsilon C_\text{min}\Delta T_\text{max}$.

Why this works: When both exit temperatures are unknown, the LMTD method needs iteration, but $\varepsilon$–NTU gives the duty directly from $U$, $A$, and the capacity rates alone. The effectiveness is the actual heat transfer divided by the thermodynamic maximum $C_\text{min}\Delta T_\text{max}$; the multi-pass correlation captures the mix of parallel and counter sections in a 2-shell-pass bundle. Because $C_\text{min}$ is the water side, the maximum possible transfer is set by how much the water can be heated.

Common pitfall: Applying the simple counter-flow $\varepsilon$ formula to a 2-shell-pass unit, or mis-identifying $C_\text{min}$ (here the smaller product is the water, not the hotter oil). A temperature cross ($T_{w,o}>T_{o,o}$) is legitimate for multi-pass exchangers and should not be "corrected".

Source: Çengel & Ghajar, Heat and Mass Transfer 6th ed., §11-4 (effectiveness–NTU method, multi-pass relations); Incropera 8th ed., §11.4.

Question 8 of 8: Two-s​hell-pass, twelve-tube-pass oil cooler

Question 8 — Two-s​hell-pass, twelve-tube-pass oil cooler (Part B, equal value)

Question 8 of 8: Two-shell-pass, twelve-tube-pass oil cooler

Question 8 — Two-shell-pass, twelve-tube-pass oil cooler (Part B, equal value)