Find the rate of heat flow through a cross section of the rod shown in figure -2-1- Thermal conductivity of the material of the rod is K-

Tan Φ = r_2 r_1/L = y r_1/x Differentiating w. r to x r_2 r_1 = Ldy/dx 0 ⇒dy/dx = r_2 r_1/L ⇒r_2 r_1/L dx = dyL/r_2 r_1 ...(i) Q/ ...

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Standard XII

Physics

Conduction Law: Differential Form

Find the rate...

Question

Find the rate of heat flow through a cross section of the rod shown in figure $(θ_{2} > θ_{1})$ . Thermal conductivity of the material of the rod is K.

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Solution

$T a n Φ = \frac{r_{2} - r_{1}}{L} = \frac{y - r_{1}}{x}$

Differentiating w. r to x

$r_{2} - r_{1} = \frac{L d y}{d x} - 0$

$\Rightarrow \frac{d y}{d x} = \frac{r_{2} - r_{1}}{L}$

$\Rightarrow \frac{r_{2} - r_{1}}{L} d x = \frac{d y L}{r_{2} - r_{1}}$ ...(i)

$\frac{Q}{t} = \frac{k π y^{2} d θ}{d x}$

Now $\frac{Q d x}{t} = \frac{k π y^{2} d θ}{d x}$

$\Rightarrow \frac{Q L d y}{t (r_{2} - r_{1})} = k π y^{2} Δ θ$ [From equation (i)]

$\Rightarrow d θ = \frac{Q L d y}{t (r_{2} - r_{1}) k π y^{2}}$

Integrating both side

$\Rightarrow \int_{θ_{1}}^{θ_{2}} d θ = \frac{Q L}{t (r_{2} - r_{1}) k π} \int_{r_{1}}^{r_{2}} \frac{d y}{y^{2}}$

$\Rightarrow (θ_{2} - θ_{1}) = \frac{Q L}{t (r_{2} - r_{1}) k π} \times {[\frac{- 1}{y}]}_{r_{1}}^{r_{2}}$

$\Rightarrow (θ_{2} - θ_{1}) = \frac{Q L}{t (r_{2} - r_{1}) k π} \times [\frac{1}{r_{1}} - \frac{1}{r_{2}}]$

$\Rightarrow (θ_{2} - θ_{1}) = \frac{Q L}{t (r_{2} - r_{1}) k π} \frac{(r_{2} - r_{1})}{r_{2} r_{1}}$

$\Rightarrow \frac{Q}{t} = \frac{k π r_{1} r_{2} (θ_{1} - θ_{2})}{L}$

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Find the rate of heat flow through a cross section of the rod shown in figure (θ2>θ1). Thermal conductivity of the material of the rod is K.

Find the rate of heat flow through a cross section of the rod shown in figure $(θ_{2} > θ_{1})$ . Thermal conductivity of the material of the rod is K.