Topic: Heat Transfer
Three metal walls of the same cross sectional area having thermal conductivities in the ratio 1:2:4 transfer heat at the rate of 6000 kJ/hr. For the same wall thickness, the temperature drops will be in the ratio
- A. 1:2:4
- B. 1:1/2:1/4
- C. 1/4:1/2:1
- D. 1:1:1
Answer: B. 1:1/2:1/4
Explanation: For steady state one-dimensional heat conduction in a series, the heat transfer rate Q is constant. According to Fourier's law, Q = k * A * (delta_T / t). Since A, t, and Q are constant across all three walls, delta_T is inversely proportional to k. Therefore, delta_T1 : delta_T2 : delta_T3 = (1/k1) : (1/k2) : (1/k3) = 1 : 1/2 : 1/4.