Respuesta :
Answer:
[tex]T = \dfrac{-1}{6.93*10^{-3}}ln(r)+100[/tex]
Step-by-step explanation:
Hello,
To solve this, we can use the famous Fourier's Law of Heat Conduction
[tex]\frac{dQ}{dt} = -kA\frac{dT}{dr}[/tex]
What do we know so far?
- Cylinder with radii [tex]r_1 = 1 & r_2 = 2[/tex]
- Surface of Area of Cylinder is [tex]A = 2\pi rl[/tex]
- Inner temperature :[tex]T_1 = 100[/tex]
- Outer temperature: [tex]T_2 = 0[/tex]
- Heat dissipation rate is constant: [tex]\frac{dQ}{dt} = Q[/tex]
When solving keep in mind that the values provided are "boundary conditions" that we'll need when solving this differential equation.
[tex]\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[/tex]
[tex]\int\limits^{r_1}_{r_2} {\dfrac{1}{r}} \, dr = \int\limits^{T_2}_{T_1} {\dfrac{-2k\pi l }{Q}} \, dT [/tex]
keep in mind that the whole term [tex]\dfrac{2k\pi l }{Q}[/tex] is a constant and will not take part in the integration, so we can just call this whole term [tex]C[/tex]
[tex]\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[/tex]
[tex]ln(r_2) -ln(r_1) = -C(T_2 - T_1)[/tex]
We can simplify now,
[tex]ln(\dfrac{r_2}{r_1}) = -C(T_2 - T_1)[/tex]
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
[tex]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[/tex]
We can put this in value in our original equation Eq(1)
[tex]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\\[/tex]
This is our general equation, but the question is asking to find the temperature [tex]T[/tex] at a distance r, i.e [tex]T(r)[/tex], given that [tex]r_1 = 1[/tex] and [tex]T_1 = 100[/tex], this means that the above equation needs to be converted in the form such that [tex]T[/tex] is only a function of [tex]r[/tex]
[tex]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\\[/tex]
We can remove the subscripts if we like
[tex]T = -\dfrac{1}{6.93*10^{-3}}ln(r)+100[/tex]
Finally, this is our equation for the outer temperature of the cylinder at any distance r
