#### 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.

#### 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`