- 2070 Magh · 6 marks
Polystyrene insulation to cut wall heat loss 55%
Short answer
The existing wall resistance is 0.4405 m²·K/W. To reduce the heat loss by 55%, the new total resistance must be 0.9788 m²·K/W, so additional insulation resistance is 0.5384 m²·K/W. Polystyrene thickness = 0.035 × 0.5384 = 0.0188 m ≈ 1.9 cm.
An exterior wall of a residential building of 25 cm thick brick [k = 0.7 W/m.°C] followed by layers of 2cm thick cement plaster [k = 0.48 W/m.°C] on both sides. What thickness of extruded polystyrene insulation [k = 0.035 W/m.°C] should be added to reduce the heat loss (or gain) through the wall by 55 percent?
Assumptions: steady one-dimensional conduction; the same inside and outside temperatures before and after insulation; film resistances are not given and are neglected. Per m² of wall.
Resistance of the existing wall
Brick: R_b = L/k = 0.25/0.7 = 0.3571 m²·K/W Two plaster layers: R_p = 2 × 0.02/0.48 = 0.0833 m²·K/W R_wall = 0.3571 + 0.0833 = 0.4405 m²·K/W
Required resistance
The heat flow is reduced by 55 %, so the new flow is 45 % of the old: q_new = 0.45 q_old.
With the same ΔT: R_new = R_wall/0.45 = 0.9788 m²·K/W
Polystyrene thickness
R_ins = R_new − R_wall = 0.9788 − 0.4405 = 0.5384 m²·K/W
L_ins = k R_ins = 0.035 × 0.5384 = 0.01884 m
Answer: extruded polystyrene thickness ≈ 1.9 cm (about 19 mm).
✓ Answer checked by Claude Opus 5.5 (high effort)
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