Question

Heat is escaping at a constant rate [$\frac{\mathbf{d}\mathbf{Q}}{\mathbf{d}\mathbf{t}}$in $\left(1.1\right)$is constant] through the walls of a long cylindrical pipe. Find the temperature T at a distance r from the axis of the cylinder if the inside wall has radius $\mathit{r}\mathbf{=}\mathbf{1}$and temperature $\mathit{T}\mathbf{=}\mathbf{100}$and the outside wall has $\mathit{r}\mathbf{=}\mathbf{2}$and $\mathit{T}\mathbf{=}\mathbf{0}$

Short answer

Answer

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

Step-by-step solution

Given information

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

Definition of differential equation

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

Find temperature 

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

Therefore,

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

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, $\mathrm{\pi Arh}=2$.

Therefore,

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

Use the boundary conditions of the inside and outside cylinder to find two arbitrary constants $Q_0$and $\alpha$

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

$$\begin{gathered} \frac{Q_0}{\mathrm{\pi }2\mathrm{hK}}h\left(1\right)=100+ \\ \Rightarrow \alpha =-100 \end{gathered}$$

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

$$\begin{gathered} \frac{Q_0}{\mathrm{\pi }2\mathrm{hK}}In\left(2\right)=0-100 \\ \Rightarrow Q_0=\frac{\mathrm{\pi hK}200}{In\left(2\right)} \end{gathered}$$

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

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