Answer:
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Explanation:
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The required thickness of insulation is 3.38 mm
Rate of heat transfer using the combined heat transfer co-efficient,[tex]Q=h_{combined}A_{3}\left ( T_{\infty } -T_{i}\right )[/tex]
[tex]=20\left ( \Pi \times D_{3}\times L \right ) \left [ \left ( 20+273 \right ) -\left ( -200+273 \right )\right ][/tex]
[tex]=13823D_{3} L[/tex]
Total thermal resistance from liquid oxygen to the outer surface of the insulation,
[tex]R_{total}=\frac{1}{h_{1}A_{1}}+\frac{In\left ( \frac{r_{2}}{r_{1}} \right )}{2\Pi k_{e}L}+\frac{In\left ( \frac{r_{3}}{r_{2}} \right )}{2\Pi k_{i}L}[/tex]
[tex]R_{total}=\frac{1}{h_{i}\left ( \Pi \times D_{1} \times L\right )}+\frac{In\left ( \frac{D_{2}}{D_{1}} \right )}{2\Pi k_{e}L}+\frac{In\left ( \frac{D_{3}}{D_{2}} \right )}{2\Pi k_{i}L}[/tex]
[tex]R_{total}=\frac{1}{120\left ( \Pi \times 20\times 10^{-3} \right )}+\frac{In\left ( \frac{25\times 10^{-3}}{20\times 10^{-3}} \right )}{2\times \Pi \times \left ( 400 \right )}+\frac{In\left ( \frac{D_{3}}{25\times 10^{-3}} \right )}{2\Pi \left ( 0.05 \right )}[/tex]
[tex]=0.1326+\left ( 8.88\times 10^{-5} \right )+3.18In\left ( \frac{D_{3}}{0.025} \right )[/tex]
[tex]=0.1327+3.18In\left ( \frac{D_{3}}{0.025} \right )[/tex]
Condensation of air occurs when the surface reaches dew point temperature.
[tex]T_{3}=10^{\circ}C=10+273=283K[/tex]
The rate of heat transfer can also be expressed as,
[tex]Q=\frac{T_{3}-T_{i}}{R_{total}}[/tex]
[tex]13823D_{3} L=\frac{283-\left ( -200+273 \right )}{R_{total}}[/tex]
[tex]R_{total}L=\frac{210}{13823D_{3}}[/tex]
[tex]0.1327+3.18In\left ( \frac{D_{3}}0.025{} \right )=\frac{210}{13823D_{3}}[/tex]
[tex]8.735+209.32In\left ( \frac{D_{3}}{0.025} \right )=\frac{1}{D_{3}}[/tex]
Solve the equation using an online calculator/EES/by trial and error to obtain [tex]{D_{3}[/tex].
[tex]{D_{3}=0.028375m\approx 28.38mm[/tex]
The thickness of the insulation,
[tex]t=D_{3}-D_{2}[/tex]
[tex]=28.38-25[/tex]
[tex]=3.38mm[/tex]
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