91Ó°ÊÓ

It is given that the heat is escaping at a constant rate$\frac{dQ}{dt}$ through the walls of a long cylindrical pipe.

A differential equation contains at least onederivative of an unknown function, either an ordinary derivative or a partial derivative.

It is given that the heat is escaping at a constant rate $\frac{dQ}{dt}$through the walls of a long cylindrical pipe.

Therefore,

$$\begin{gathered} \frac{dQ}{dt}=KA\frac{dT}{dr} \\ QKA=\frac{dT}{dr} \\ ∫\frac{Q_0}{KA}dr=∫dT \end{gathered}$$,

Since $\frac{dQ}{dt}$is constant , $\frac{dQ}{dt}$is replaced with $Q_0$above.

The pipe is very long, so the area of the circles at the top and bottom is ignored.

Thus, $\mathrm{Ï€´¡°ù³ó}=2$.

Therefore,

$$\begin{gathered} \frac{Q_0}{\mathrm{π}2\mathrm{hk}}{∫}_r\frac{dr}{}=∫dT \\ \frac{Q_0}{\mathrm{π}2\mathrm{hk}}h\left(r\right)=T+ \end{gathered}$$.

Use the boundary conditions of the inside and outside cylinder to find two arbitrary constants $Q_0$and $\mathrm{Î\pm }$

Now, inside the cylinder, $r=1$and $T=100$.

$$\begin{gathered} \frac{Q_0}{\mathrm{Ï€}2\mathrm{hK}}h\left(1\right)=100+ \\ ⇒\mathrm{Î\pm }=-100 \end{gathered}$$

Now, outside the cylinder, $r=2$and $T=0$.

$$\begin{gathered} \frac{Q_0}{\mathrm{Ï€}2\mathrm{hK}}In\left(2\right)=0-100 \\ ⇒Q_0=\frac{\mathrm{Ï€³ó°­}200}{In\left(2\right)} \end{gathered}$$

Put the values of $Q_0$and $\mathrm{Î\pm }$into the general solution $\frac{Q_0}{\mathrm{Ï€}2\mathrm{hK}}In\left(r\right)=T+$

Thus, $T=100\left[1-\frac{In\left(r\right)}{In\left(2\right)}\right]$.

Therefore, the temperature T at a distance r from the axis of the cylinder is $T=100\left[1-\frac{In\left(r\right)}{In\left(2\right)}\right]$