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 ![= 100 mm = 0.1 m](https://tex.z-dn.net/?f=%3D%20100%20mm%20%3D%200.1%20m)
Surface emissivity ε = 0.8
Temperature of steam
= 150° C = 423K
Atmospheric air temperature ![T_{\infty} = 20^0 \ C = 293 \ K](https://tex.z-dn.net/?f=T_%7B%5Cinfty%7D%20%3D%2020%5E0%20%5C%20C%20%3D%20293%20%5C%20K)
Velocity of wind V = 8 m/s
To calculate average film temperature:
![T_f = \dfrac{T_s+T_{\infty}}{2}](https://tex.z-dn.net/?f=T_f%20%3D%20%5Cdfrac%7BT_s%2BT_%7B%5Cinfty%7D%7D%7B2%7D)
![T_f = \dfrac{423+293}{2}](https://tex.z-dn.net/?f=T_f%20%3D%20%5Cdfrac%7B423%2B293%7D%7B2%7D)
![T_f = \dfrac{716}{2}](https://tex.z-dn.net/?f=T_f%20%3D%20%5Cdfrac%7B716%7D%7B2%7D)
![T_f = 358 \ K](https://tex.z-dn.net/?f=T_f%20%3D%20358%20%5C%20K)
To calculate volume expansion coefficient
![\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}](https://tex.z-dn.net/?f=%5Cbeta%3D%20%5Cdfrac%7B1%7D%7BT_f%7D%20%5C%5C%20%5C%5C%20%5Cbeta%3D%20%5Cdfrac%7B1%7D%7B358%7D%20%5C%5C%20%5C%5C%20%20%5Cbeta%3D%202.79%20%5Ctimes%2010%5E%7B-3%7D%20%5C%20K%5E%7B-1%7D)
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:
![Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}](https://tex.z-dn.net/?f=Ra_%7BD%7D%20%3D%20%5Cdfrac%7Bg%20%5Ctimes%20%5Cbeta%20%28T_s-T_%7B%5Cinfty%7D%29%20%5Ctimes%20D_b%5E3%7D%7B%5Calpha%5Ctimes%20v%7D)
![Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}](https://tex.z-dn.net/?f=Ra_%7BD%7D%20%3D%20%5Cdfrac%7B9.8%20%5Ctimes%20%282.79%20%2A10%5E%7B-3%7D%29%28150-20%29%20%5Ctimes%20%280.1%29%5E3%7D%7B%2831.244%5Ctimes%2010%5E%7B-6%7D%29%20%5Ctimes%20%2821.7984%5Ctimes%2010%5E%7B-6%7D%29%7D)
![Ra_{D} = 5.224 \times 10^6](https://tex.z-dn.net/?f=Ra_%7BD%7D%20%3D%205.224%20%5Ctimes%2010%5E6)
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)
![Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2](https://tex.z-dn.net/?f=Nu_D%20%3D%20%5CBig%20%5C%7B%20%20%20%200.60%20%2B%204.226%20%5CBig%20%5C%7D%5E2)
![Nu_D = 23.29](https://tex.z-dn.net/?f=Nu_D%20%3D%2023.29)
However, for the heat transfer coefficient; we have:
![h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}](https://tex.z-dn.net/?f=h_D%20%3D%20%5Cdfrac%7BNu_D%5Ctimes%20k%7D%7BD_b%7D%20%5C%5C%20%5C%5C%20%20h_D%20%3D%20%5Cdfrac%7B%2823.29%29%20%5Ctimes%20%2830.608%20%5Ctimes%2010%5E%7B-3%7D%20%29%7D%7B0.1%7D)
![h_D = 7.129 \ Wm^2 .K](https://tex.z-dn.net/?f=h_D%20%3D%207.129%20%5C%20Wm%5E2%20.K)
Hence, Stefan-Boltzmann constant ![\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4](https://tex.z-dn.net/?f=%5Csigma%20%3D%205.67%20%5Ctimes%2010%5E%7B-8%7D%20%5C%20W%2Fm%5E2.K%5E4)
Now;
To determine the heat loss using the formula:
![q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)](https://tex.z-dn.net/?f=q%27_b%20%3D%20q%27_%7Bev%7D%20%2B%20q%27_%7Brad%7D%20%5C%5C%20%5C%5C%20q%27_b%20%3D%20h_D%20%28%5Cpi%20D_o%29%20%28T_t-T_%7B%5Cinfty%7D%29%2B%5Cvarepsilon%28%5Cpi%20D_b%29%5Csigma%20%28T_t%5E4-T_%7B%5Cinfty%20%7D%5E4%29)
![q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}](https://tex.z-dn.net/?f=q%27_b%20%3D%20%287.129%29%28%5Cpi%2A0.1%29%20%28423-293%29%20%2B%20%280.8%29%20%28%5Cpi%2A0.1%29%20%285.67%20%2A10%5E%7B-8%7D%29%20%28423%5E4-293%5E4%29%20%5C%5C%20%5C%5C%20%20q%27_b%20%3D%20291.153%20%2B%20351.205%20%5C%5C%20%5C%5C%20%20%5Cmathbf%7Bq%27_b%20%3D%20642.258%20%5C%20W%2Fm%7D)
Now; here we need to determine the Reynold no and the average Nusselt number:
![Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4](https://tex.z-dn.net/?f=Re_D%20%3D%20%5Cdfrac%7BVD_b%7D%7Bv%20%7D%20%5C%5C%20%5C%5C%20%20Re_D%20%3D%20%5Cdfrac%7B8%20%2A0.1%7D%7B21.7984%20%5Ctimes%2010%5E%7B-6%7D%7D%20%5C%5C%20%5C%5C%20%20Re_D%20%3D%203.6699%20%5Ctimes%2010%5E4)
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)
![Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86](https://tex.z-dn.net/?f=Nu_D%20%3D%20%280.3%20%2B%5Cdfrac%7B105.359%7D%7B1.140%7D%5Ctimes%201.218%29%20%5C%5C%20%5C%5C%20Nu_D%20%3D%20112.86)
SO, the heat transfer coefficient for forced convection is determined as follows afterward:
![h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K](https://tex.z-dn.net/?f=h_D%20%3D%20%5Cdfrac%7BNu_%7BD%7D%2A%20k%7D%7BD_b%7D%20%5C%5C%20%5C%20%20h_D%20%3D%20%5Cdfrac%7B112.86%2A30.608%20%2A10%5E%7B-3%7D%7D%7B0.1%7D%20%5C%5C%20%5C%5C%20%20h_D%20%3D%2034.5%20%5C%20W%2Fm%5E2%20.K)
Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
![q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}](https://tex.z-dn.net/?f=q%27b%20%3D%20h_D%20%28%5Cpi%20D_b%29%20%28T_s-T_%7B%5Cinfty%7D%29%20%2B%20%5Cvarepsilon%20%28%5Cpi%20D_b%29%20%5Csigma%20%28T_s%5E4-T_%20%7B%5Cinfty%7D%5E4%29%20%5C%5C%20%5C%5C%20%20q%27b%20%3D%20%2834.5%29%20%28%5Cpi%20%2A0.1%29%20%28423-293%29%20%2B%20%280.8%29%20%28%5Cpi%2A0.1%29%20%285.67%2A10%5E%7B-8%7D%29%20%28423%5E4%20-%20293%5E4%29%20%5C%5C%20%5C%5C%20%3D%201409%20%2B351.205%20%5C%5C%20%5C%5C%20%20%5Cmathbf%7Bq%27b%20%3D%201760.205%20%5C%20W%2Fm%7D)