A semicircular rod is joined at its end to a straight rod of the same material and the same cross -sectional area. The straight rod forms a diameter of the other rod. The junctions are maintained at different temperatures. Find the ratio of the heat transferred through a cross section of the semicircular rod to the heat transferred through a cross section of the straight rod in a given time.

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Solution

We know $Q = \frac{k A (θ_{1} - θ_{2})}{d}$

$Q_{1} = \frac{k A (θ_{1} - θ_{2})}{d_{1}}$

$Q_{2} = \frac{k A (θ_{1} - θ_{2})}{d_{2}}$

$\frac{Q_{1}}{Q_{2}} = \frac{\frac{K A (θ_{1} - θ_{2})}{π r}}{\frac{K A (θ_{1} - θ_{2})}{2 r}} = \frac{2 r}{π r} = \frac{2}{π}$

where $d_{1} = π r$ $d_{2} = 2 r$

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We know Q=kA(θ1−θ2)d Q1=kA(θ1−θ2)d1 Q2=kA(θ1−θ2)d2 Q1Q2=KA(θ1−θ2)πrKA(θ1−θ2)2r=2rπr=2π where d1=πr d2=2r