Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivities of these rods are: LALB = 12, rArB = 2 and KAKB = 12. The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is K, when equilibrium is established.
Answer
360
Solution
Thermal resistance of a rod: Rth = LKA, with A = r2
AAAB = ( rArB )2 = 4
RARB = LALB KBKA ABAA = 12 2 14 = 14
In the steady state the same heat current flows through both rods (interface temperature T):
400 - TRA = T - 200RB
4(400 - T) = T - 200
5T = 1800
T = 360 K
Answer: 360