Cross-Flow Tube-Bank Heat Exchanger: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2016, Q8

Question 8: Cross-Flow Tube-Bank Heat Exchanger (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.

Quantity	Value
Tubes	40 × 1 cm diameter, thin-walled, 1 m long
Water side	18 °C in, $V=3$ m/s, $c_p=4180$
Air side	130 °C, 105 kPa in, $V=12$ m/s, $c_p=1010$
Overall coefficient	$U=80$ W/m²·°C

Figure 8 — Cross-flow arrangement: cold water flows through the 40 tubes (into the page) while hot air sweeps across them along the duct. Both streams are unmixed.

Approach. Compute both heat-capacity rates from the flow areas and inlet densities, identify $C_\text{min}$, evaluate NTU from $UA_s/C_\text{min}$, and apply the both-fluids-unmixed cross-flow effectiveness to get the duty and outlet temperatures.

Capacity rates. Water: $\dot m_w=\rho_w A_\text{tubes}V=998.6\,(40\cdot\tfrac{\pi}{4}0.01^2)(3)=9.41$ kg/s, $C_w=39{,}340$ W/°C. Air: $\rho_a=105000/(287\cdot403)=0.908$ kg/m³, $\dot m_a=0.908(1)(12)=10.89$ kg/s, $C_a=11{,}000$ W/°C. Thus $C_\text{min}=C_a$, $C_r=0.280$.
NTU. Heat-transfer area $A_s=40\,\pi(0.01)(1)=1.257$ m², so $$\text{NTU}=\frac{UA_s}{C_\text{min}}=\frac{80(1.257)}{11000}=0.00914$$
Effectiveness (both unmixed). $$\varepsilon=1-\exp\!\left\{\tfrac{1}{C_r}\text{NTU}^{0.22}\big[e^{-C_r\,\text{NTU}^{0.78}}-1\big]\right\}=0.0091$$
Duty and outlet temperatures. $$\dot Q=\varepsilon\,C_\text{min}(T_{a,i}-T_{w,i})=0.0091(11000)(130-18)$$ ==**$\dot Q\approx11.2$ kW.**== $$T_{a,o}=130-\frac{\dot Q}{C_a}=129.0\ \text{°C},\qquad T_{w,o}=18+\frac{\dot Q}{C_w}=18.3\ \text{°C}$$

Check — the exchanger is drastically undersized.

With only 40 one-metre tubes the surface is $A_s=1.26$ m² and $UA_s=100$ W/°C — barely 1 % of $C_\text{min}$, so $\varepsilon<1\%$ and both streams leave essentially at their inlet temperatures ($\Delta T_\text{air}\approx1$ °C, $\Delta T_\text{water}\approx0.3$ °C). The numbers are solved exactly as printed; a real unit would need vastly more area. (This question is identical to the 2016-May sitting's Q8.)

Quantity	Result
Water / air capacity rate	39.3 / 11.0 kW/°C ($C_\text{min}=$ air)
NTU / effectiveness	0.0091 / 0.91 %
Heat-transfer rate	≈ 11.2 kW
Air outlet temperature	≈ 129 °C
Water outlet temperature	≈ 18.3 °C

Concept & Theory

Topic. Effectiveness–NTU analysis of a single-pass cross-flow (both fluids unmixed) heat exchanger.

Key relations. $C=\dot m c_p$; $C_r=C_\text{min}/C_\text{max}$; $\text{NTU}=UA_s/C_\text{min}$; both-unmixed $\varepsilon=1-\exp\{\frac1{C_r}\text{NTU}^{0.22}[e^{-C_r\text{NTU}^{0.78}}-1]\}$; $\dot Q=\varepsilon C_\text{min}\Delta T_\text{max}$.

Why this works. With both inlet temperatures known but neither outlet, the $\varepsilon$–NTU method gives the duty directly without iterating on an LMTD. The heat-capacity rates come from the volumetric flow and inlet density on each side; whichever is smaller sets $C_\text{min}$ and hence the maximum possible duty $C_\text{min}\Delta T_\text{max}$.

Common pitfall. Using the duct's full 1 m² frontal area as the heat-transfer area (the actual transfer area is the tube surface $N\pi DL$), or mis-assigning $C_\text{min}$ — here the air, despite the larger mass flow, has the smaller $C$ because of its low specific heat.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §11.4 (the effectiveness–NTU method; cross-flow relations).

Question 8 of 8: Cross-Flow Tube-Bank Heat Exchanger

Question 8: Cross-Flow Tube-Bank Heat Exchanger (equal value)