Answer to Question #260640 in Differential Equations for Tanvir

Question

A steam pipe 20 cm in diameter contains steam at 180°C & is covered with a material 6 cm thick. If the temperature of the surface of the covering is 30°C, find the temperature halfway through the covering under steady state conditions.

Answer 1

Solution;

Let Q be the constant quantity of heat flowing radially through a surface of a Pipe having a radius R,of x cm.

The area of the lateral surface,A=2πx.

By Fourier Law;

$Q=-kA\frac{dT}{dx}=-k(2πx)\frac{dT}{dx}$ .....(1)

Hence;

$dT=\frac{-Q}{2πk}\frac{dx}{x}$ .....(2)

Integrating,we get;

$T=\frac{-Q}{2πk}log (x)+c$ .......(3)

Now,

At $T_1=180°c$ ,$R=10cm$

By direct substitution into equation (3);

$180=\frac{-Q}{2πk}log(10)+c$ .....(3a)

Also;

At $T_2=30°c,R=16cm$

By direct substitution into equation (3);

$30=\frac{-Q}{2πk}log(16)+c$ .....(3b)

Subtract equation (3b) from (3a)

$150=\frac{Q}{2πk}log(\frac{16}{10})$

Hence;

$\frac{Q}{2πk}=\frac{150}{log(1.6)}$ .....(4)

Now let T₃ be the temperature at underway the covering,that is,R=13cm.

By substitution into equation (3);

$T_3=\frac{-Q}{2πk}log(13)+C$ .....(3c).

Subtract (3c) from (3a);

$180-T_3=\frac{Q}{2πk}log(1.3)$ ..….(5)

Substitute (4) into (5);

$180-T_3=\frac{150}{log(1.6)}log(1.3)$

$180-T_3=83.73$

$T_3=180-83.73=96.27°c$

Ans;

96.27°c