By what factor is the heat flow increased if a window 0.550 mm on a side is inserted in the door? The glass is 0.450 cmcm, and the glass has a thermal conductivity of 0.80 W/(m⋅K)W/(m⋅K). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cmcm of glass.

This question is incomplete, the complete question is;

A carpenter builds a solid wood door with dimensions 1.95 m × 0.99 m × 4.5 cm . Its thermal conductivity is k=0.120W/(m.K). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional 1.6 cm thickness of solid wood. The inside air temperature is 19.0°C , and the outside air temperature is -6.50°C .

a) What is the rate of heat flow through the door?

b) By what factor is the heat flow increased if a window 0.550 m on a side is inserted in the door? The glass is 0.450 cm , and the glass has a thermal conductivity of 0.80 W/(m.K). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 cm of glass.

a) the rate of heat flow through the door is 97 watts

b) The factor of increased heat flow is 1.353

Explanation:

Given that;

room dimension = 1.95m × 0.99m × 4.5cm,

thermal conductivity = 0.120 w/m.k

additional thickness of solid wood Δt = 1.6 cm

a)

first we determine the effective thickness of the door;

t = 4.5cm + 1.6 cm = 61 cm ≈ 0.061 m

Now rate of heat flow is given by the relation

Q = KA( (TH -TC)/L)

= 0.12 × (1.95 m × 0.99 m) × ( (19°C - (-6.50°C)) / 0.061m)

= 0.23166 × 418.0327

= 96.8414 watts

Q = 97 watts

therefore the rate of heat flow through the door is 97 watts

b)

by intensity the glass of thickness 0.450 cm

the effective thickness is

L = 0.45cm + 12 cm = 12.45 cm = 0.1245 m

additionally area of glass A = (0.550 m)²

A = 0.3025 m²

Now

Qglass = KA ((TH-TC)/L)  

= 0.80 w/m.k × 0.3025 m² × (19°C - (-6.50°C)) / 0.1245m)

= 0.242 × 204.819

Qglass = 49.57 watt  

Qwood = KA ((TH-TC)/L)  

area of wooden door = (1.95×0.99) - 0.3025 m² = 1.628m²

so Qwood = 0.12 × 1.628 × (19°C - (-6.50°C)) / 0.061m)

= 0.19536 × 418.0327

Qwood = 81.67 watt

Q = Qglass + Qwood

Q = 49.57 watt  + 81.67 watt

Q = 131.24 watt

The factor of increased heat flow is;

f = 131.24 watt / 97 watts

f = 1.353

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