Question

The end of a cylindrical liquid cryogenic propellant tank in free space is to be protected from external (solar) radiation by placing a thin metallic shield in front of the tank. Assume the view factor Fts between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K. It is given that Tt= 100 K, ε1=ε2= 0. 10, εt= 0. 20, and GS= 1250 W/m2.

Answer 1

If the temperature of the shield is 338 kelvin. Then the heat flux through the tank will be 25.3 Watt per square meter.

What is heat flux?

The increase in heat energy movement through a particular surface is known as heat flux, and the heat flux density is the absolute temperature per unit area.

Assume the view factor between the tank and the shield is unity; all surfaces are diffuse and gray, and the surroundings are at 0 K.

It is given that T= 100 K, ε₁ = ε₂ = 0. 10, $\varepsilon_t$ = 0.20, and GS= 1250 W/m².

Then we have

The temperature of the shield will be

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dot{q}_{ST} = 0$ ...1

and

$\rm q''_{12}=\dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}}$ ...2

Then from equations 1 and 2, we have

$\rm \alpha _sG_s - \varepsilon _1 E_b (T_s) - \dfrac{ \sigma (T_{1}^{4} - T_{2}^{2})}{\frac{1}{\varepsilon _1 }+ \frac{1}{\varepsilon _2} -1}} = 0$

Then the value of $\rm T_s$ will be

$\rm T_s =\left [ \dfrac{\alpha _sGs+\left ( \dfrac{\sigma T_1^4}{\frac{1}{\varepsilon _1}+\frac{1}{\varepsilon _2} - 1} \right )}{\sigma \left ( \varepsilon _1 + \dfrac{1}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2}-1} \right )} \right ] ^{\dfrac{1}{4}}$

Put all the values, then we have

$\rm T_s = \left [ \dfrac{0.05 \times + \left ( \dfrac{\sigma (100)^4}{\frac{1}{0.1}+\frac{1}{0.05}-1} \right )}{\sigma \left ( 0.05 + \dfrac{1}{\frac{1}{0.1}+\frac{1}{0.05} - 1} \right )} \right ]^{\dfrac{1}{4}} \\\\\\T_s = 338 \ K$

Then the heat flux will be

$\rm q"_{ST}=\dfrac{\sigma (T_S^4 - T_t^4)}{\frac{1}{\varepsilon _1} + \frac{1}{\varepsilon _2} - 1} \\\\\\q"_{ST}=\dfrac{5.67 \times 10^{-8}(388^4-100^4)}{\frac{1}{0.1}+\frac{1}{0.05}-1}\\\\\\q"_{ST} = 25.3 \ W/m^2$

More about the heat flux link is given below.

brainly.com/question/12913016

#SPJ1