Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with the same area. The glass of a single pane is 4.2 mm thick, and the air space between the two panes of the double-pane window is 7.0 mm thick. The glass has thermal conductivity 0.80 W /m ? K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m2 ? K / W.

Solution 109P Given, area A = 0.15 m 2 3 L glass = 4.2 mm = 4.2 × 10 m 3 L air = 7.0 mm = 7.0 × 10 m K glass= 0.80 W/m.K K = 0.024 W/m.K air 2 Combined thermal resistance R = 0.15 m .K/W Now, the rate of heat loss by the single pane window is, A(T hT )c H single= (Lglass glassR) …..(1) Rate of heat loss by the double pane window is, A(T hT )c H double = 2×(L /K ) + (L /K ) + R…..(2) glass glass air air Now, dividing equation (1) by equation (2), H single 2×(L glass glass+ (airKair + R H = L /K + R) double glass glass Now, substituting the given values, we get H single 2×(4.2×103)/0.80 + 7.0×10/0.024 +0.15 = 3 H double 4.2×10 /0.80 + 0.15 H 3 3 singl= 10.5×10 +291×10 +0.15 H double 5.25×103+0.15 H single 0.0105+0.291+0.15 H = 0.0053+0.15 double H single = 0.45 H double 0.155 H single H = 2.90 double Therefore, the required ratio is 2.9.