- #1
member 428835
Homework Statement
A cylindrical wire with radius ##r_i## carries a current ##I \frac{A}{cm^2}## with a resistance of ##\Omega \: ohms\times cm##. It is insulated with a material with radius ##r_o## whose thermal conductivity is ##k \frac{W}{cm \times K}##. The insulator is exposed to air that convects heat at ##h \frac{W}{cm^2 \times K}##. Find the temperature at the wire\insulator interface.
Homework Equations
Heat equation with fourier's law using constant thermal conductivity and steady state conditions: $$k\nabla ^2 T = \dot{g}$$
where ##\dot{g}## is the heat generation term.
The Attempt at a Solution
I will use cylindrical coordinates. i first want to conduct a relationship with the temperature passing through the insulator. Thus, i think the first thing I can do is state that: $$k \nabla ^2 T_1 = AI^2$$ where my units are ##\frac{W}{cm^3}##. I do need boundary conditions, however. I should state, before proceeding, that I assume heat only transfers radially. now, to work on those boundary conditions, isn't the following relationship among heat fluxes true: $$\underbrace{AI^2 \times r_i}_{\text{heat flux from generation term at r=r_i}} = \underbrace{-k \nabla T_1}_{\text{heat flux from conductive term at } r=r_i}$$ and $$ \underbrace{-k\nabla T_2}_{\text{heat flux from conductive term at } r=r_o} = \underbrace{h(T_{r_o}-T_{air})}_{\text{heat flux of convective term at }r=r_o}$$
although I do not know ##T_{air}##. ##T_1## and ##T_2## are the heat profiles throughout the copper wire and then the insulator respectively.
Now, solving ##k \nabla^2 T_1 = AI^2## to solve yields: $$T_1(r) = \frac{kAI^2 r^2}{4} + C_1 \ln r + C_2$$
and $$\nabla T_1 = \frac{kAI^2r}{2} + \frac{C_1}{r}$$
It seems now we can solve for ##C_1## using the first boundary condition.
Assuming we find ##C_1## (this is easy enough) i really don't know how to proceed? can someone please help?
thanks!