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Heat loss from a cylinder in cross flow

Short answer

Using Churchill-Bernstein correlation, Re = 2.39×10^4, Nu = 88.5, average heat transfer coefficient h = 52.3 W/m^2·K, heat loss per metre length = 821 W/m.

Air at 25 °C flows at 10 m/s across a long horizontal cylinder of outer diameter 50 mm whose surface is at 125 °C. Properties of air at the film temperature of 75 °C: nu = 20.92 x 10^-6 m^2/s, k = 0.02953 W/m K, Pr = 0.7202. Using the Churchill-Bernstein correlation, find the average heat transfer coefficient and the heat loss per metre length. Nu = 0.3 + [0.62 Re^0.5 Pr^(1/3)] / [1 + (0.4/Pr)^(2/3)]^0.25 x [1 + (Re/282000)^(5/8)]^(4/5)

Reynolds number

ReD=VDν=10×0.0520.92×10−6=2.39×104Re_D = \frac{VD}{\nu} = \frac{10\times0.05}{20.92\times10^{-6}} = 2.39\times10^4

ReD Pr=1.72×104>0.2Re_D\,Pr = 1.72\times10^4 > 0.2, so the Churchill-Bernstein equation is valid.

Nusselt number

Compute the parts:

Re0.5=154.6,Pr1/3=0.896numerator=0.62×154.6×0.896=85.9[1+(0.4/0.7202)2/3]1/4=(1.676)0.25=1.138[1+(Re/282000)5/8]4/5=(1.2138)0.8=1.168\begin{aligned} Re^{0.5} &= 154.6, \qquad Pr^{1/3} = 0.896\\ \text{numerator} &= 0.62\times154.6\times0.896 = 85.9\\ \left[1+(0.4/0.7202)^{2/3}\right]^{1/4} &= (1.676)^{0.25} = 1.138\\ \left[1+(Re/282000)^{5/8}\right]^{4/5} &= (1.2138)^{0.8} = 1.168 \end{aligned} Nu=0.3+85.91.138×1.168=0.3+88.2≈88.5Nu = 0.3 + \frac{85.9}{1.138}\times1.168 = 0.3 + 88.2 \approx 88.5

Heat transfer coefficient

h=Nu kD=88.5×0.029530.05=52.3 W/m2Kh = \frac{Nu\,k}{D} = \frac{88.5\times0.02953}{0.05} = 52.3\ \text{W/m}^2\text{K}

Heat loss per metre

QL=h (πD)(Ts−T∞)=52.3×π×0.05×100=821 W/m\frac{Q}{L} = h\,(\pi D)(T_s-T_\infty) = 52.3\times\pi\times0.05\times100 = 821\ \text{W/m}

Answer: h≈52.3 W/m2h \approx 52.3\ \text{W/m}^2K; Q/L≈821 W/mQ/L \approx 821\ \text{W/m}.

✓ Answer checked by Claude Sonnet 5.5

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