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(B) The heat current flows radially outward through the cylindrical pipe.For a cylindrical shell of radius $x$ and thickness $dx$,the thermal resistan

Vedclass · Accepted Answer

$3445$

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(B) The heat current flows radially outward through the cylindrical pipe.For a cylindrical shell of radius $x$ and thickness $dx$,the thermal resistance $dR_T$ is given by $dR_T = \frac{dx}{k(2\pi x \ell)}$.Integrating from inner radius $R_1$ to outer radius $R_2$,the total thermal resistance $R_T$ is:$R_T = \int_{R_1}^{R_2} \frac{dx}{2\pi k \ell x} = \frac{1}{2\pi k \ell} \ln\left(\frac{R_2}{R_1}ight)$.The rate of heat flow is $\frac{dQ}{dt} = \frac{\Delta T}{R_T} = \frac{2\pi k \ell (T_1 - T_2)}{\ln(R_2/R_1)}$.Given: $T_1 = 110^{\circ} C$,$T_2 = 10^{\circ} C$,$\Delta T = 100^{\circ} C$,$k = 0.38 \,kW/m/^{\circ} C$,$\ell = 10 \,m$,$R_1 = 2 \,cm$,$R_2 = 4 \,cm$.Substituting the values:$\frac{dQ}{dt} = \frac{2 	imes 3.14159 	imes 0.38 	imes 10 	imes 100}{\ln(4/2)} = \frac{2387.6}{0.6931} \approx 3444.6 \,kW$.Thus,the rate of heat flow is approximately $3445 \,kW$.

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