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Just one physics question pls....?
A consecutive wall consisting of four layers, with thermal conductivities k1 = 0.0639 W/m·K,k3 = 0.0402 W/m·K and k4 = 0.127 W/m·K (k2 is unknown). The layer thicknesses are L1 = 1.43 cm,L3 = 2.81 cm and L4 = 3.42 cm(L2 is unknown). The known temperatures are T1 = 31.1 ˚C, T12 = 24.0 ˚C and T4 = -11.1 ˚C. Energy transfer through the wall is steady. What is interface temperature T34?
1 Answer
- ?Lv 75 years ago
The heat flow must be continuous, so J = -k grad(T) should be continuous across each boundary. In the steady state, the temperature gradient grad(T) is just delta(T)/L, so
k1 (T1 - T12)/L1 = k2 (T12 - T23)/L2
k2 (T12 - T23)/L2 = k3 (T23 - T34)/L3
k3 (T23 - T34)/L3 = k4 (T34 - T4)
You now all except k2, L2 and T34 and you have three linear equations to solve for these three unknowns. Enjoy! Notice that the first equation tells you what k2/L2 is. That's the only thing you need to find T34 in the second equation...
