Question 8 of 8: Cross-Flow Tube-Bank Heat Exchanger — Effectiveness–NTU

Question 8: Cross-Flow Tube-Bank Heat Exchanger — Effectiveness–NTU

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. 40 tubes, $D=1$ cm, in a 1 m × 1 m duct. Water (tube-side, hence UNMIXED — each tube's stream never contacts another): $c_p=4180$ J/kg°C, $T_{c,in}=18\,{}^{\circ}\text{C}$, $V=3$ m/s. Air (duct-side, open cross-section, hence MIXED): $c_p=1010$ J/kg°C, $T_{h,in}=130\,{}^{\circ}\text{C}$, 105 kPa, $V=12$ m/s. $U=80$ W/m²°C.

Find. The water and air outlet temperatures and the total heat-transfer rate.

Approach. Compute both capacity rates $C=\dot{m}c_p$ from the given velocities and cross-sections, identify $C_{\min}/C_{\max}$, form $\text{NTU}=UA/C_{\min}$ with $A=N\pi DL$, apply the effectiveness–NTU relation for the one-fluid-mixed/one-unmixed cross-flow configuration (air mixed, water unmixed), then back out both outlet temperatures.

1. Water-side capacity rate (40 parallel tubes). $$\dot{m}_w=\rho_w V_w N\left(\frac{\pi}{4}D^2\right)=(998.6)(3)(40)\left(\frac{\pi}{4}(0.01)^2\right)=9.41\text{ kg/s}$$ $$C_w=\dot{m}_w c_{p,w}=9.41(4180)=39{,}340\text{ W/K}$$
2. Air-side capacity rate (duct approach flow). $$\rho_a=\frac{P_a}{RT_{h,in}}=\frac{105{,}000}{(287)(403.15)}=0.908\text{ kg/m}^3$$ $$\dot{m}_a=\rho_a V_a(1\text{ m}\times1\text{ m})=0.908(12)=10.89\text{ kg/s}$$ $$C_a=\dot{m}_a c_{p,a}=10.89(1010)=11{,}000\text{ W/K}$$
3. $C_{\min}$, $C_{\max}$, and NTU. $C_a
4. Effectiveness ($C_{\min}$ fluid mixed, $C_{\max}$ fluid unmixed): $$\varepsilon=\frac{1}{C_r}\left(1-\exp\!\big[-C_r\big(1-e^{-\text{NTU}}\big)\big]\right)=\boxed{0.00909}$$ (essentially $\varepsilon\approx\text{NTU}$, as expected in the small-NTU limit for any flow arrangement — a useful self-check.)
5. Heat-transfer rate and outlet temperatures. $$q_{\max}=C_{\min}(T_{h,in}-T_{c,in})=11{,}000(130-18)=1{,}232{,}000\text{ W}$$ $$q=\varepsilon\, q_{\max}=0.00909(1{,}232{,}000)=\boxed{11.19\text{ kW}}$$ $$T_{c,out}=T_{c,in}+\frac{q}{C_w}=18+\frac{11{,}190}{39{,}340}=\boxed{18.29\,{}^{\circ}\text{C}}$$ $$T_{h,out}=T_{h,in}-\frac{q}{C_a}=130-\frac{11{,}190}{11{,}000}=\boxed{128.98\,{}^{\circ}\text{C}}$$

The temperature changes are small because the given $U$ (80 W/m²°C) and area (1.26 m²) together give a very small NTU relative to the large capacity rates on both sides — this is a genuinely undersized/lightly-loaded exchanger for the flow rates specified, not a computational error (confirmed by the $\varepsilon\approx\text{NTU}$ self-check above).