1 T. Figure 19.16 (Example 19.7) Energy transfer by conduction through two slabs in thermal contact with each other. At steady state, the rate of energy transfer through slab 1 equals the rate of energy transfer through slab 2.
Two slabs of thickness L1 and L2 and thermal conductivities k1 and k2 are in thermal contact with each other as shown. The temperatures of their outer surfaces are Tc and Th, respectively, and Th >Tc. Determine the temperature at the interface and the rate of energy transfer by
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In the question its asks for the rate of energy transfer P after you find what the interface temperature T is.
Upon completion from submitting T in into the intial rate equation the final expansion gives:
P = A(Th - Tc) / (L1/ k1) + (L2/ k2)
I understand how T was achieved from expanding the P1 = P2 substitution, but how is P found from inserting T into P1 = k1A (T - Tc / L1)?
by Bartleby ExpertHello, I am still confused on the second part of this question, how you find the rate of energy transfer by inserting what T is into one of the initial equations:
P1 = k1A (T - Tc / L1)
Could you show the worked out procedure as you did for finding T please?
by Bartleby Expert
