Topic: Heat Transfer
The wall of a chemical plant is modified with two additional layers having half the thermal conductivity of original wall material. The inner layer is of half the thickness and outer layer of same thickness as original wall. What will be the heat flow across the modified wall?
- A. Same as original
- B. Twice of original
- C. 50% of original
- D. 25% of original
Answer: D. 25% of original
Explanation: Original thermal resistance R1 = L/k. New total resistance R_new = R_inner + R1 + R_outer = (L/2)/(k/2) + L/k + L/(k/2) = L/k + L/k + 2L/k = 4L/k. Since the new thermal resistance is 4 times the original, the heat flow (which is inversely proportional to resistance) becomes 1/4th, or 25% of the original.