A household oven door of 0.6-m height and 0.7-m width reaches an average surface temperature of 32°C during operation. The temperature of ambient air is 22°C. Assuming the door has an emissivity of 0.4 and the surroundings are also at 22°C, determine the heat transfer by free convection and the heat transfer by radiation.

a)  ≈ 11.7 watts

b)  ≈ 21.4 watts

Explanation:

Given data :

Oven door dimensions : 0.6m  by 0.7m

average surface temperature = 32°c

Temperature of ambient air = 22°c

emissivity of door = 0.4

surrounding temperature = 22°c

a) Calculate heat transfer by free convection

q = 3.34 ( 0.5 * 0.7 ) (32-22 ) ≈ 11.7 watts

attached below is the detailed solution

b) calculate heat transfer by radiation

q = ∈.As б ( Ts^4 - Tsun^4 )

   = 1( 0.5*0.7 ) * 5.67 * 10^-8 [( 273+32)^4 - ( 273+22)^4 ]

   ≈ 21.4 watts

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation:

answer:

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