If the focal area in Example 9.1 is made of a material with an emissivity of 0.9 and this loses heat only by radiation from the 5 cm diameter area, calculate its
equilibrium temperature

Question

Quizplus · Accepted Answer

The Answer is The area of the dish is

( 3.14 ) imes ( 5 \mathrm {~m} ) ^ { 2 } = 78.5 \mathrm {~m} ^ { 2 }

The area of the focus is

( 3.14 ) imes ( 0.025 \mathrm {~m} ) ^ { 2 } = 1.96 imes 10 ^ { - 3 } \mathrm {~m} ^ { 2 }

So the insolation of the focus area is

\left( 650 \mathrm {~W} / \mathrm { m } ^ { 2 } ight) imes \left( 78.5 \mathrm {~m} ^ { 2 } ight) / \left( 1.96 imes 10 ^ { - 3 } \mathrm {~m} ^ { 2 } ight) = 26.4 \mathrm { MW } / \mathrm { m } ^ { 2 }

The power radiated per area from the focus is and its equilibrium temperature will be

T = \left[ \frac { P / A } { \partial \sigma } ight] ^ { 1 / 4 }

Substituting appropriate values gives

T = \left[ \frac { 2.64 imes 10 ^ { 7 } \mathrm {~W} / \mathrm { m } ^ { 2 } } { ( 0.9 ) imes \left( 5.67 imes 10 ^ { - 8 } \mathrm {~W} \cdot \mathrm { m } ^ { - 2 } \cdot \mathrm { K } ^ { - 4 } ight) } ight] ^ { 1 / 4 } = 4769 \mathrm {~K}

If the Focal Area in Example 9