Cross-Flow Finned-Tube Heat Exchanger (ε–NTU): 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2017, Q8

Question 8: Cross-Flow Finned-Tube Heat Exchanger (ε–NTU) (equal value)

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Stream	Data
Air (hot, cooled)	$\dot m_a=0.80$ kg/s, 30 °C → 7 °C, $c_p=1007$ J/kg·°C
Water (cold)	$\dot m_w=0.76$ kg/s, in at 3 °C, $c_p=4180$ J/kg·°C
Overall coefficient	$U=55$ W/m²·°C
Geometry	Cross-flow, both fluids unmixed (finned tube)

Given. The two stream conditions and $U$ above. Find. (a) the heat-exchanger area required; (b) the percentage reduction in heat-transfer rate when the water flow is halved.

Figure 8 — Cross-flow finned-tube unit: air (hot) sweeps across tubes carrying water (cold). Air, with the smaller capacity rate, is $C_{min}$.

Approach. Compute the duty from the air side, identify $C_{min}$, form the effectiveness and hence NTU from the cross-flow (both-unmixed) relation to size the area; for part (b) the area and $U$ are fixed so NTU is unchanged (air stays $C_{min}$), and the higher capacity ratio lowers the effectiveness and thus the duty.

Heat duty and water outlet. $\dot Q=\dot m_ac_{p,a}(30-7)=0.80(1007)(23)=18.53$ kW. Water rise: $T_{w,o}=3+\dfrac{\dot Q}{\dot m_wc_{p,w}}=3+\dfrac{18528}{3176.8}=8.83$ °C.
Capacity rates and effectiveness. $C_a=805.6$, $C_w=3176.8$ W/K, so $C_{min}=C_a$, $C_r=805.6/3176.8=0.254$. With $\Delta T_{max}=30-3=27$ °C, $\;\varepsilon=\dfrac{\dot Q}{C_{min}\Delta T_{max}}=\dfrac{18528}{805.6(27)}=0.852$.
NTU (cross-flow, both unmixed) and area. Inverting $\varepsilon=1-\exp\!\left\{\tfrac1{C_r}NTU^{0.22}\!\left[e^{-C_rNTU^{0.78}}-1\right]\right\}$ for $\varepsilon=0.852$, $C_r=0.254$ gives $NTU=2.44$, so $$A=\frac{NTU\,C_{min}}{U}=\frac{2.44(805.6)}{55}$$ $A\approx35.7$ m²
Part (b): water flow halved. Now $\dot m_w=0.38$ kg/s, $C_w=1588.4$ W/K; air is still $C_{min}$ so $NTU=UA/C_a=2.44$ is unchanged, but $C_r=805.6/1588.4=0.507$.
New effectiveness and duty. The same NTU at the higher $C_r$ gives $\varepsilon'=0.783$, so $\dot Q'=\varepsilon'C_{min}\Delta T_{max}=0.783(805.6)(27)=17.04$ kW. $$\%\text{ reduction}=\frac{\dot Q-\dot Q'}{\dot Q}\times100=\frac{18.53-17.04}{18.53}\times100$$ ≈ 8.0 % reduction in heat-transfer rate

Quantity	Result
Heat duty	18.53 kW (water exit 8.83 °C)
Effectiveness / NTU ($C_r=0.254$)	0.852 / 2.44
Required area	≈ 35.7 m²
Effectiveness with half water flow ($C_r=0.507$)	0.783
Reduction in heat-transfer rate	≈ 8.0 %

Concept & Theory

Topic. Cross-flow (both fluids unmixed) heat-exchanger sizing and off-design performance by the ε–NTU method.

Key relations. $\dot Q=\dot m c_p\Delta T$; $\varepsilon=\dot Q/[C_{min}(T_{h,i}-T_{c,i})]$; cross-flow both-unmixed $\varepsilon=1-\exp\{\tfrac1{C_r}NTU^{0.22}[e^{-C_rNTU^{0.78}}-1]\}$; $NTU=UA/C_{min}$.

Why this works. The ε–NTU method sizes an exchanger from inlet temperatures alone (no iterative LMTD guess of the outlet). Since the air has the smaller capacity rate it sets $C_{min}$; with the area and $U$ frozen, cutting the water flow does not change $NTU$ (which is referenced to the unchanged air capacity) but raises $C_r$, and a larger capacity ratio always lowers effectiveness — the warmer water outlet reduces the mean temperature difference, so the duty falls modestly.

Common pitfall. Recomputing NTU with the reduced water capacity (NTU is based on $C_{min}=C_a$, which is unchanged); or assuming halving the water flow halves the heat transfer — the effect is muted (~8 %) because water is the over-capacity stream.

Source. Çengel & Ghajar, Heat and Mass Transfer §11 (heat exchangers — the effectiveness–NTU method and cross-flow relations).

Question 8 of 8: Cross-Flow Finned-Tube Heat Exchanger (ε–NTU)

Question 8: Cross-Flow Finned-Tube Heat Exchanger (ε–NTU) (equal value)