A reversible heat engine operates between a hot reservoir at T1 and a radiating surface at T2. Heat radiated from the surface is proportional to the surface area and temperature of surface raised to power 4. Determine the condition for minimum surface area for a given work output.

Question

A reversible heat engine operates between a hot reservoir at [latex] extbf{T}_{1}[/latex] and a radiating surface at [latex] extbf{T}_{2}[/latex]. Heat radiated from the surface is proportional to the surface area and temperature of surface raised to power 4. Determine the condition for minimum surface area for a given work output.

Accepted Answer

[latex] ext{Heat rejected = Heat radiated from surface at }T_{2} = K . A . T^{4}_{2} , ext{where}\ ext{A is surface area and K is proportionality constant.}\\begin{aligned}\frac{Q_{2}}{W}&= \frac{T_{2}}{T_{1} - T_{2}}\\frac{K . A . T^{4}_{2}}{W}&= \frac{T_{2}}{T_{1} - T_{2}}\⇒ \qquad \qquad \qquad \qquad A&= \frac{W}{T_{2}^{3}(T_{1} - T_{2}) . K}\end{aligned}\ ext{In order to have minimum surface area the denominator in above expression of A should be maximum}\ ext{i.e. }T_{2}^{3}(T_{1} - T_{2}) ext{ should be maximum. Differentiating with respect to }T_{2}.\\begin{aligned}\frac{d}{dT_{2}}\left\{(T_{2}^{3}(T_{1} - T_{2}) ) ight\} &= 0\3T_{1} . T_{2}^{2} - 4T^{3}_{2}&= 0\⇒ \qquad \qquad \qquad \qquad \qquad \frac{T_{2}}{T_{1}}&= \frac{3}{4} ⇒ T_{2} = T_{1} . \frac{3}{4}\ ext{Taking second differential}\qquad \qquad &\\frac{d^{2}}{dT^{2}_{2}}\left\{T_{2}^{3} . (T_{1} - T_{2}) ight\} &= 6T_{1} . T_{2} – 12 T_{2}^{2}\end{aligned}\ ext{Upon substitution it is –ive so}\left\lgroup\frac{T_{2}}{T_{1}}=\frac{3}{4} ight group ext{ is the condition for }\left\{T_{2}^{3} . (T_{1} - T_{2}) ight\} ext{ to be maximum and so}\ ext{the minimum surface area}\ \large{\frac{T_{2}}{T_{1}} = \frac{3}{4}}[/latex]

Question 4.16: A reversible heat engine operates between a hot reservoir at...

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