Consider a circular tapered shape as shown in figure where heat i

SOLUTION

Concept:

From given figure, it can be seen that diameter is varying linearly and function of x.

D = ax, a is constant

Area \(= \frac{\pi }{4}{a^2}{x^2}\)

Using Fourier law

\(Q = - kA\frac{{\partial T}}{{\partial x}}\)

Calculation:

\(Q = - k\;\left( {\frac{\pi }{4}{a^2}{x^2}} \right)\frac{{\partial T}}{{\partial x}}\;\)

\(Q\mathop \smallint \limits_{{x_1}}^x \frac{{dx}}{{{x^2}}} = \frac{{k\pi {a^2}}}{4}\mathop \smallint \limits_{{T_1}}^T dT\)

\(\theta \left[ {\frac{{ - 1}}{x} + \frac{1}{{{x_1}}}} \right] = \frac{{ - k\pi {a^2}}}{4}\;\left[ {T - {T_1}} \right]\)      ---(1)

Similarly, put x = x₂ and we get

\(Q = \frac{{ - k\pi }}{4}{a^2}\frac{{\left( {{T_2} - {T_1}} \right)}}{{\left[ {\frac{1}{{{x_1}}} - \frac{1}{{{x_2}}}} \right]}}\)

\(Q = \frac{{T - {T_1}}}{{{T_2} - {T_1}}} = \frac{{\left( {\frac{1}{{{x_1}}}} \right) - \left( {\frac{1}{x}} \right)}}{{\left( {\frac{1}{{{x_1}}}} \right) - \left( {\frac{1}{{{x_2}}}} \right)}}\)

\(\frac{{T - 35}}{{45 - 35}} = \frac{{\left( {\frac{1}{{0.2}}} \right) - \left( {\frac{1}{{0.4}}} \right)}}{{\left( {\frac{1}{{0.2}}} \right) - \left( {\frac{1}{{0.5}}} \right)}}\)

∴ T = 43.33°C