What is the heat current in rod 1 and how can it be calculated?

jackson1984
Messages
1
Reaction score
0
Three identical rods are welded together to form a Y-shaped figure. The cross-sectional area of each rod is A, and they have length L and thermal conductivity k.
The free end of rod 1 is maintained at T_1 and the free ends of rods 2 and 3 are maintained at a lower temperature T_0. You may assume that there is no heat loss from the surfaces of the rods.

What is the heat current H_1 in rod 1?
Express the heat current in terms of any or all of k, L, A, and the temperatures T_1 and T_0.

I guess the ans is (2kA/3L)(T_1+T_2), but it is wrong...
Can anyone teach me...
 

Attachments

  • 36907.jpg
    36907.jpg
    5.9 KB · Views: 690
Physics news on Phys.org
What is your T2?

Anyway:
label the junction temperature as T2. Then write the heat flow equations for heat from T1 to T2 and T2 to T0. Solve for T2, then dQ/dt is obvious.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top