Q.63 An evaporator is insulated using glass wool material of 0.15 m thickness. The inner most surface and the outer surface of the insulation are at 700 °C and 80 °C, respectively. The mean thermal conductivity of the glass wool under these conditions is 0.29 W.m−1.K−1. The rate of heat loss (in W) through 1.2 m2 of the evaporator wall surface (rounded off to the nearest integer) is ____________.
Problem Overview
The rate of heat loss through the evaporator wall is calculated using Fourier’s law for steady-state conduction through a plane wall. Given the parameters, the heat transfer occurs purely by conduction across the glass wool insulation, with no convection or radiation specified.
Problem Parameters
Temperature difference: ΔT = 700 – 80 = 620 K (or °C, as the difference is equivalent).
Calculation Steps
Q = k⋅A⋅ΔT/L
Key Assumptions
- Steady-state conditions: No heat accumulation; temperatures fixed
- One-dimensional conduction: Heat flows perpendicular to surfaces only
- Constant mean k: 0.29 W/m·K valid despite high temperatures (glass wool stable up to ~600°C)
- Plane wall: Flat evaporator wall approximation
Common Errors to Avoid
- Forgetting ΔT conversion: Wrong units → invalid Q
- Using ambient k (0.04 W/m·K): Underestimates at mean ~390°C
- Area misread (12 m²): Yields 10x error (14,384 W)
- No rounding: Exam demands nearest integer
GATE/IIT JAM Exam Tips
- Memorize Fourier’s law variants: Plane wall, cylinder, sphere
- Practice numericals: 80% weightage in BT/CH papers
- Verify units: W/m·K, m², K, m → W output guaranteed
- Check rounding: Always to nearest integer for numerical answers
Why Glass Wool Excels for High-Temp Applications?
- Low thermal conductivity (0.030-0.045 W/m·K typical) traps air effectively
- High temperature resilience up to 600°C+
- Cost-effective vs. ceramic fiber alternatives
- Excellent for evaporator insulation applications
