Answer:
Heat loss per unit length = 642.358 W/m
The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m
Explanation:
From the information given:
Diameter D 
Surface emissivity ε = 0.8
Temperature of steam
= 150° C = 423K
Atmospheric air temperature 
Velocity of wind V = 8 m/s
To calculate average film temperature:




To calculate volume expansion coefficient

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;
Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s
Thermal conductivity k = 30.608 × 10⁻³ W/m.K
Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s
Prandtl no. Pr = 0.698
Rayleigh No. for the steam line is determined as follows:



The average Nusselt number is:
![Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2](https://tex.z-dn.net/?f=Nu_D%20%3D%20%5CBig%20%5C%7B%20%20%20%200.60%20%2B%20%5Cdfrac%7B0.387%28Ra_D%29%5E%7B1%2F6%7D%7D%7B%5B%201%2B%20%280.559%2FPr%29%5E%7B9%2F16%7D%5D%5E%7B8%2F27%7D%7D%20%5CBig%20%5C%7D%5E2)
![Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2](https://tex.z-dn.net/?f=Nu_D%20%3D%20%5CBig%20%5C%7B%20%20%20%200.60%20%2B%20%5Cdfrac%7B0.387%285.224%5Ctimes%2010%5E6%29%5E%7B1%2F6%7D%7D%7B%5B%201%2B%20%280.559%2F0.698%29%5E%7B9%2F16%7D%5D%5E%7B8%2F27%7D%7D%20%5CBig%20%5C%7D%5E2)
![Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2](https://tex.z-dn.net/?f=Nu_D%20%3D%20%5CBig%20%5C%7B%20%20%20%200.60%20%2B%20%5Cdfrac%7B5.0977%7D%7B%5B%201.8826%5D%5E%7B8%2F27%7D%7D%5CBig%20%5C%7D%5E2)


However, for the heat transfer coefficient; we have:


Hence, Stefan-Boltzmann constant 
Now;
To determine the heat loss using the formula:


Now; here we need to determine the Reynold no and the average Nusselt number:

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;
![Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}](https://tex.z-dn.net/?f=Nu_D%20%3D%200.3%20%2B%20%5Cdfrac%7B0.62%20%5Ctimes%20Re_D%5E%7B1%2F2%7D%2A%20Pr%5E%7B1%2F3%7D%7D%7B%5B1%2B%280.4%2FPr%29%5E%7B2%2F3%7D%5D%5E%7B1%2F4%7D%7D%20%5B1%2B%20%28%5Cdfrac%7BRe_D%7D%7B282000%7D%29%5E%7B5%2F8%7D%5D%5E%7B4%2F5%7D)
![Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}](https://tex.z-dn.net/?f=Nu_D%20%3D%200.3%20%2B%20%5Cdfrac%7B0.62%20%5Ctimes%20%283.6699%2A10%5E4%29%5E%7B1%2F2%7D%2A%20%280.698%29%5E%7B1%2F3%7D%7D%7B%5B1%2B%280.4%2F0.698%29%5E%7B2%2F3%7D%5D%5E%7B1%2F4%7D%7D%20%5B1%2B%20%28%5Cdfrac%7B3.669%2A10%5E4%7D%7B282000%7D%29%5E%7B5%2F8%7D%5D%5E%7B4%2F5%7D)

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
