Question

Heat is escaping at a constant rate, (dQ/dt is constant) through the walls of a long cylindrical pipe. Find the temperatureT at a distance r from the axis of the cylinder if the inside wall has radius r=1, and temperatureT=100, and the outside wall has r=2, andT=0.

Answer 1

Answer:

$T = \dfrac{-1}{6.93*10^{-3}}ln(r)+100$

Step-by-step explanation:

Hello,

To solve this, we can use the famous Fourier's Law of Heat Conduction

$\frac{dQ}{dt} = -kA\frac{dT}{dr}$

What do we know so far?

• Cylinder with radii $r_1 = 1 & r_2 = 2$
• Surface of Area of Cylinder is $A = 2\pi rl$
• Inner temperature :$T_1 = 100$

• Outer temperature: $T_2 = 0$
• Heat dissipation rate is constant: $\frac{dQ}{dt} = Q$

When solving keep in mind that the values provided are "boundary conditions" that we'll need when solving this differential equation.

$\frac{dQ}{dt} = -kA\frac{dT}{dr}\\\\Q = -k(2\pi rl)\frac{dT}{dr}\\\frac{1}{r}dr = \dfrac{-2k\pi l }{Q} dt$

$\int\limits^{r_1}_{r_2} {\dfrac{1}{r}} \, dr = \int\limits^{T_2}_{T_1} {\dfrac{-2k\pi l }{Q}} \, dT$

keep in mind that the whole term $\dfrac{2k\pi l }{Q}$ is a constant and will not take part in the integration, so we can just call this whole term $C$

$\int\limits^{r_2}_{r_1} {\dfrac{1}{r}} \, dr = {\dfrac{-2k\pi l }{Q}} \int\limits^{T_2}_{T_1} \, dT \\\int\limits^{r_2}_{r_1} {\dfrac{1}{r}} \, dr = -C \int\limits^{T_2}_{T_1} \, dT$

$ln(r_2) -ln(r_1) = -C(T_2 - T_1)$

We can simplify now,

$ln(\dfrac{r_2}{r_1}) = -C(T_2 - T_1)$

This is our solution to the differential equation, lets call it Eq(1)

Putting all the known values, and forming an equation we can find the value of C

$ln(\dfrac{r_2}{r_1}) = -C(T_2 - T_1)\\ln(\dfrac{2}{1}) = -C(0-100)\\C = \dfrac{ln(2)}{100} \\C = 6.931 * 10^{-3} = 0.00693$

We can put this in value in our original equation Eq(1)

$ln(\dfrac{r_2}{r_1}) = -C(T_2 - T_1)\\T_2 = \dfrac{1}{-C}ln(\dfrac{r_2}{r_1})+T_1\\T_2 = \dfrac{-1}{6.93*10^{-3}}ln(\dfrac{r_2}{r_1})+T_1$

This is our general equation, but the question is asking to find the temperature $T$ at a distance r, i.e $T(r)$, given that $r_1 = 1$ and $T_1 = 100$, this means that the above equation needs to be converted in the form such that $T$ is only a function of $r$

$T_2 = \dfrac{-1}{6.93*10^{-3}}ln(\dfrac{r_2}{r_1})+T_1\\T_2 = \dfrac{-1}{6.93*10^{-3}}ln(\dfrac{r_2}{1})+100$

We can remove the subscripts if we like

$T = -\dfrac{1}{6.93*10^{-3}}ln(r)+100$

Finally, this is our equation for the outer temperature of the cylinder at any distance r