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Question
Find the rate of heat flow through a cross section of the rod shown in figure (28-E10) (θ2 > θ1). Thermal conductivity of the material of the rod is K.
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Solution
We can infer from the diagram that ΔPQR is similar to ΔPST.
So, by the property of similar triangles,
`x/l = (r - r_1)/(r_2 - r_1)`
⇒ `(r_2 - r_1) x/l = r - r_1`
⇒ `r = r_1 + ( r_2 - r_1 ) x/l`
Let ;
`(r_2 - r_1)/l = a`
⇒ r = ax + r_1................(i)
Thermal resistance is given by
`dR = dx/(K.A)`
⇒ dR = `(dx)/(K.pir^2`]
`dR = (dx)/(K.pi (ax + r_1)^2`
\[\int\limits_0^R\] dR = `1/(kpi)` \[\int\limits_0^l\] `dx/(ax + r_1)`
`R = -(1)/(Kpi) [(1)/(ax + r_1)]_0^6`
`R = -(1)/(aKpi)[(1)/(al + r_1) -1/(r1)]`
`R = -(1)/(aKpi) [ 1/(((r_2 -r_1)/l) l + r_1) - 1/r_1]`
`R = - 1/ (aKpi)[(1)/(r_2)-(1)/(r_1)]`
`R = -(1)/(Kpir_1r_2)`
Rate of flow of heat = `(DeltaQ)/R`
`⇒ q = `( Q_2 - Q_1)/l Kpir_1r_2`
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