Shell-and-Tube Heat Exchanger (Water Heated by Air): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013, Q8

Question 8: Shell-and-Tube Heat Exchanger (Water Heated by Air) (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. $A=47.5\ \text{m}^2$, $U=200\ \text{W/m}^2\text{}\cdot\text{K}$, 1 shell / 2 tube passes; water $c_p=4180$, air $c_p\approx1010\ \text{J/kg}\cdot\text{K}$.

Stream	Part (a)	Part (b)
Water flow	6.5 kg/s, in 15 °C	10 kg/s, in 15 °C, out 42 °C
Air flow	5.0 kg/s, in 200 °C	5.0 kg/s, inlet = unknown
Capacity rates	C_w = 27,170; C_a = 5050 W/K	C_w = 41,800; C_a = 5050 W/K

Find. (a) water and air outlet temperatures at the stated conditions; (b) the air inlet temperature needed to deliver 10 kg/s of water heated to 42 °C.

Figure 8 — 1 shell-pass / 2 tube-pass exchanger: water in the tubes (cold) is heated by air in the shell (hot). The air stream has the smaller capacity rate, so it limits the effectiveness.

Approach. Use the ε–NTU method with the 1-shell-pass effectiveness relation. For (a) compute ε from NTU and solve the outlet temperatures; for (b) the required duty fixes ε and hence the air inlet temperature.

Part (a): NTU and effectiveness. The air is $C_\text{min}$: $C_a=5.0(1010)=5050$, $C_w=6.5(4180)=27{,}170$ W/K, $C_r=0.186$. Then $NTU=UA/C_\text{min}=200(47.5)/5050=1.881$ and, with $\varepsilon=\dfrac{2}{(1+C_r)+\sqrt{1+C_r^2}\,\coth(NTU\sqrt{1+C_r^2}/1)}$ (1-shell-pass),$$\varepsilon=0.783.$$
Part (a): outlet temperatures. The duty is $\dot Q=\varepsilon C_\text{min}(T_{a,i}-T_{w,i})=0.783(5050)(185)=731\ \text{kW}$, so$$T_{w,o}=15+\frac{\dot Q}{C_w}=\boxed{41.9\ ^{\circ}\text{C}},\qquad T_{a,o}=200-\frac{\dot Q}{C_a}=\boxed{55.2\ ^{\circ}\text{C}}.$$
Part (b): required air inlet. Now $C_w=10(4180)=41{,}800$ W/K; the air is still $C_\text{min}$ so $NTU$ is unchanged and $\varepsilon=0.805$. The required duty is $\dot Q=C_w(42-15)=1129\ \text{kW}$, and since $\dot Q=\varepsilon C_\text{min}(T_{a,i}-15)$,$$T_{a,i}=15+\frac{1129{,}000}{0.805(5050)}=\boxed{293\ ^{\circ}\text{C}}\quad(T_{a,o}\approx69\ ^{\circ}\text{C}).$$

Check

Part (a) with 6.5 kg/s already delivers water at ≈41.9 °C, so part (b) is read as raising the demand to 10 kg/s of water heated to 42 °C (the "as before" refers to the 15 °C inlet and 5.0 kg/s air). The exchanger geometry, $U$ and area are unchanged between the two parts.

Quantity	Result
(a) NTU / effectiveness	1.881 / 0.783
(a) Water outlet / air outlet	41.9 °C / 55.2 °C (duty 731 kW)
(b) Effectiveness / duty	0.805 / 1129 kW
(b) Required air inlet	≈ 293 °C

Concept & Theory

Topic. Shell-and-tube exchanger analysis by the ε–NTU method.

Key relations. $C_r=C_\text{min}/C_\text{max}$; $NTU=UA/C_\text{min}$; 1-shell-pass $\varepsilon=2\big/\big[(1+C_r)+\sqrt{1+C_r^2}\,\tfrac{1+e^{-NTU\sqrt{1+C_r^2}}}{1-e^{-NTU\sqrt{1+C_r^2}}}\big]$; duty $\dot Q=\varepsilon C_\text{min}\Delta T_\text{max}$.

Why this works. With inlet temperatures known but outlets not, the ε–NTU method gives the duty directly without iterating on an LMTD correction factor. The air, having the smaller capacity rate, undergoes the larger temperature change and sets the maximum possible heat transfer $C_\text{min}\Delta T_\text{max}$.

Common pitfall. Using the counter-flow ε relation for a shell-and-tube unit, or mis-identifying $C_\text{min}$ (it is the air here, not the water).

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §11.4 (ε–NTU relations, one shell pass).

Question 8 of 8: S​hell-and-Tube Heat Exchanger (Water Heated by Air)

Question 8: S​hell-and-Tube Heat Exchanger (Water Heated by Air) (equal value)

Question 8 of 8: Shell-and-Tube Heat Exchanger (Water Heated by Air)

Question 8: Shell-and-Tube Heat Exchanger (Water Heated by Air) (equal value)