Length of a water tube cooled by cross-flowing air: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q6

Question 6 — Length of a water tube cooled by cross-flowing air (Part B, 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. Internal water flow ($V=6$ m/s, in at 77 °C, out at 55 °C) in a $D=2$ cm horizontal tube; external air cross-flow ($V_\infty=30$ m/s, $T_\infty=27$ °C). Thin tube wall. Find. the tube length $L$.

Quantity	Value
Tube diameter, $D$	0.02 m
Water velocity / inlet / outlet	6 m/s / 77 °C / 55 °C
Air velocity / temperature	30 m/s / 27 °C
Water props at 66 °C (bulk mean)	$\rho=980,\ k=0.66,\ Pr=2.7$
Air props at ~47 °C (film)	$\nu=1.76\times10^{-5},\ k=0.0279,\ Pr=0.72$

Figure 6 — Internal water stream (bulk temperature falling 77 → 55 °C) cooled by an external air cross-flow. The outside air film dominates the overall resistance.

Approach. Find the inside (turbulent internal-flow) and outside (cross-flow cylinder) coefficients, combine into an overall $U$, then apply the LMTD single-stream balance $\dot m c_p\,\Delta T = U A_s\,\Delta T_\text{lm}$ to solve for the area and hence the length.

Water mass flow. $\dot m = \rho V (\pi D^2/4) = (980)(6)\,\tfrac{\pi}{4}(0.02)^2 = 1.85\ \text{kg/s}$.
Inside coefficient (Dittus–Boelter, cooling). $Re = \rho V D/\mu \approx 2.76\times10^{5}$ (turbulent); with $Pr=2.7$, $$Nu_i = 0.023\,Re^{0.8}Pr^{0.3} \approx 691,\qquad h_i = Nu_i\,k_w/D \approx 22{,}900\ \text{W/m}^2\text{}\cdot\text{°C}$$
Outside coefficient (cross-flow over a cylinder). $Re_\infty = V_\infty D/\nu \approx 3.40\times10^{4}$; the Churchill–Bernstein correlation gives $Nu_o \approx 108$, so $$h_o = Nu_o\,k_\text{air}/D \approx 151\ \text{W/m}^2\text{}\cdot\text{°C}$$
Overall coefficient (thin wall). The tiny air-side coefficient controls: $$\frac{1}{U} = \frac{1}{h_i}+\frac{1}{h_o} \Rightarrow U \approx 150\ \text{W/m}^2\text{}\cdot\text{°C}$$
Log-mean temperature difference (air ≈ constant). $$\Delta T_\text{lm} = \frac{(77-27)-(55-27)}{\ln\frac{77-27}{55-27}} = \frac{50-28}{\ln(50/28)} = 37.9\ \text{°C}$$
Area and length. The heat removed is $\dot Q = \dot m c_p(77-55) = (1.85)(4185)(22) = 1.70\times10^{5}$ W. Then $$A_s = \frac{\dot Q}{U\,\Delta T_\text{lm}} = \frac{1.70\times10^{5}}{(150)(37.9)} \approx 30.0\ \text{m}^2,\qquad L = \frac{A_s}{\pi D} = \frac{30.0}{\pi(0.02)}$$ $L \approx 4.8\times10^{2}\ \text{m}\ (\approx 477\ \text{m})$

Check — physically large length.

The required length (~0.48 km) is enormous because the fast internal water carries a huge thermal capacity (1.85 kg/s) while the air-side film coefficient is only ~150 W/m²·°C. The number is correct for the stated data; in practice one would use many tubes in parallel or finned/extended surface. The result is dominated by $h_o$, so a ±15% change in the cross-flow correlation moves $L$ proportionally.

Quantity	Result
Water mass flow, $\dot m$	1.85 kg/s
Inside / outside coefficient	≈ 22,900 / 151 W/m²·°C
Overall $U$	≈ 150 W/m²·°C
$\Delta T_\text{lm}$	37.9 °C
Heat duty $\dot Q$	170 kW
Tube length $L$	≈ 477 m

Concept & Theory

Topic. Combined internal-flow / external cross-flow convection with an overall heat-transfer coefficient and the LMTD method for a single stream against a constant-temperature environment.

Key relations. Internal turbulent $Nu=0.023Re^{0.8}Pr^{0.3}$ (cooling); external cylinder Churchill–Bernstein $Nu_o$; series resistance $1/U=1/h_i+1/h_o$; $\dot Q=\dot m c_p\Delta T = U A_s\Delta T_\text{lm}$.

Why this works. Because the air stream is effectively infinite (constant 27 °C) while the water cools, the exchanger behaves like a single stream against a fixed sink, so the LMTD collapses to the two end differences. The overall coefficient is set almost entirely by the weakest film — here the air side — so the internal coefficient barely matters and the tube must be very long to reject the large water-side duty.

Common pitfall. Using $Pr^{0.4}$ (heating) instead of $Pr^{0.3}$ (cooling) for the water, or evaluating air properties at the bulk 27 °C rather than the film temperature; both are minor here because $h_o$ dominates.

Source. Incropera et al., Fundamentals of Heat and Mass Transfer §7 (external flow — cylinder in cross-flow) and §8 (internal flow), with §11 (LMTD).

Question 6 of 8: Length of a water tube cooled by cross-flowing air

Question 6 — Length of a water tube cooled by cross-flowing air (Part B, equal value)