Question

A pipe of inner radius R1, outer radius R2 and constant thermal conductivity K is maintained at an inner temperature T1 and outer temperature T2. For a length of pipe L find the rate at which the heat is lost and the temperature inside the pipe in the steady state.

Answer 1

Answer:

Heat lost from the cylindrical pipe is given by the formula,

$d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }$

Temperature distribution inside the cylinder is given by,

$\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}$    

Explanation:

Inner radius = $R_{1}$  

Outer radius = $R_{2}$

Constant thermal conductivity = K

Inner temperature = $T_{1}$

Outer temperature = $T_{2}$

Length of the pipe = L

Heat lost from the cylindrical pipe is given by the formula,

$d Q= \frac{2 \pi K L (T_{1} - T_{2} )}{log_{e}(\frac{R_{2} }{R_{1} } ) }$------------ (1)

Temperature distribution inside the cylinder is given by,

$\frac{T - T_{1} }{T_{2} - T_{1} } = \frac{log_{e} \frac{R}{R_{1} }}{log_{e} \frac{R_{2}}{R_{1} }}$     ------------ (2)