Heat Loss from Air Flowing in a Square Duct: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, May 2013, Q6

Question 6: Heat Loss from Air Flowing in a Square Duct (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. Inlet air 60 °C at $\dot m=120{,}000\ \text{kg/hr}=33.33$ kg/s; square duct 0.5 m side, 4 m long, thin metal wall (negligible resistance); surroundings 30 °C, outside coefficient $h_o=5\ \text{W/m}^2\text{}\cdot\text{K}$; air properties $k\approx0.028$, $\mu\approx2.0\times10^{-5}$, $Pr\approx0.71$, $c_p\approx1007$.

Find. The rate of heat transfer from the duct air to the surroundings.

Figure 6 — Square duct: hot air is cooled by internal forced convection and outside natural/forced convection in series; the thin metal wall adds negligible resistance so the weak outside coefficient controls.

Approach. Find the internal convection coefficient from a turbulent duct correlation, combine it in series with the outside coefficient for an overall $U$, then apply the constant-surroundings exchanger relation to get the outlet temperature and heat rate.

Internal convection. The hydraulic diameter of the square is $D_h=0.5$ m, so $Re=\dot m D_h/(A_c\mu)=33.33(0.5)/(0.25\cdot2.0\times10^{-5})=3.33\times10^6$ (fully turbulent). With Dittus–Boelter (cooling),$$Nu=0.023\,Re^{0.8}Pr^{0.3}=3430,\qquad h_i=\frac{Nu\,k}{D_h}=192\ \text{W/m}^2\text{}\cdot\text{K}.$$
Overall coefficient. The thin wall is neglected, so$$\frac1U=\frac1{h_i}+\frac1{h_o}=\frac1{192}+\frac15\;\Rightarrow\;U=4.87\ \text{W/m}^2\text{}\cdot\text{K}.$$ The outside film dominates.
Surface area and heat rate. $A_s=(4\times0.5)(4)=8\ \text{m}^2$. With surroundings fixed at 30 °C, the outlet temperature follows$$\frac{T_\text{out}-T_\infty}{T_\text{in}-T_\infty}=\exp\!\left(\frac{-UA_s}{\dot m c_p}\right)=\exp(-1.16\times10^{-3})\approx0.9988,$$ so $T_\text{out}=59.97$ °C (the air barely cools) and$$\dot Q=\dot m c_p(T_\text{in}-T_\text{out})\approx U A_s(T_\text{in}-T_\infty)=\boxed{1.17\ \text{kW}}.$$

Check

The stated 120,000 kg/hr through a 0.5 m duct implies a very high velocity (~130 m/s); the problem is solved with the data as printed. Because the outside coefficient is so much smaller than the inside one, $\dot Q\approx UA_s\,\Delta T$ is essentially independent of the exact internal flow, so the result is robust to that reading.

Quantity	Result
Internal / overall coefficient	h_i = 192, U = 4.87 W/m²·K
Surface area	8.0 m²
Air outlet temperature	≈ 59.97 °C
Heat-transfer rate	≈ 1.17 kW

Concept & Theory

Topic. Combined internal/external convection on a duct; series resistances.

Key relations. Hydraulic diameter $D_h=4A_c/P$; Dittus–Boelter $Nu=0.023Re^{0.8}Pr^n$ ($n=0.3$ cooling); overall $1/U=1/h_i+1/h_o$; single-stream exchanger $T_\text{out}=T_\infty+(T_\text{in}-T_\infty)e^{-UA_s/\dot m c_p}$.

Why this works. Heat crosses two convective films in series, and the smaller coefficient controls the overall $U$; the enormous $\dot m c_p$ makes the temperature drop tiny, so the duct wall stays essentially at the inlet air temperature and the loss is close to $UA_s\,\Delta T$.

Common pitfall. Using the side length rather than the hydraulic diameter (for a square they coincide, but not in general), or ignoring the outside film and hugely over-predicting the loss.

Source. Çengel & Ghajar, Heat and Mass Transfer §8 (internal forced convection) and §11 (overall heat-transfer coefficient).

Question 6 of 8: Heat Loss from Air Flowing in a Square Duct

Question 6: Heat Loss from Air Flowing in a Square Duct (equal value)