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- 2069 Bhadra · 8 marks
Rock wool to reduce wall heat transfer 50%
Short answer
The existing wall resistance is 0.205 m²·K/W. Halving the heat loss requires doubling the resistance, so the additional rock wool resistance must be 0.205 m²·K/W. Thickness = 0.065 × 0.205 = 0.0133 m (1.33 cm).
An exterior wall of a house consists of 10cm of common brick (k = 0.8 W/mK) followed by a 4cm layer of gypsum plaster (k = 0.5 W/mK). What thickness of rock wool insulation (k = 0.065 W/mK) should be added to reduce the heat transfer through the wall by 50%?
Assumptions: one-dimensional steady conduction; same temperature difference before and after; surface film resistances not given, so neglected. Per m² of wall.
Resistance of the existing wall
Brick: R_b = L/k = 0.10/0.8 = 0.1250 m²·K/W Gypsum plaster: R_g = 0.04/0.5 = 0.0800 m²·K/W R_wall = 0.2050 m²·K/W
Required resistance
Heat transfer is reduced by 50 %, so q_new = 0.5 q_old. With the same ΔT:
R_new = R_wall/0.5 = 0.4100 m²·K/W
Rock wool thickness
R_wool = R_new − R_wall = 0.2050 m²·K/W
L_wool = k R_wool = 0.065 × 0.2050 = 0.01333 m
Answer: rock wool thickness ≈ 1.33 cm (about 13.3 mm). (Halving the heat flow means adding a resistance equal to the existing one.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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