Answer:
Surface 1 = -1999.39 W
Surface 2 = -2086.15 W
Surface 3 = 10076.28 W
Explanation:
Lenght of furnace = 0.5 m
Diameter d = 0.5 m
Radius r = d/2 = 0.5/2 = 0.25 m
The two ends will have a circular surface area whose area will be
A(ends) = ¶r^2 = 3.142 x 0.25^2 = 0.1964 m^2
Area of lateral surface = ¶dl = 3.142 x 0.5 x 0.5 = 0.7855 m^2
Lets name the surfaces.
Surface 1:
Area = 0.1964 m^2
Emmisivity e = 0.4
Temperature = 400 K
Surface 2:
Area = 0.1964 m^2
Emmisivity e = 0.5
Temperature = 500 K
Surface 3:
Area = 0.7855 m^2
Emmisivity e = 0.7
Temperature = 800 K
Net radiative heat transfer rate for surface 1 is,
E = Ake(T1^4 - T2^4 - T3^4)
where k is Stefan constant = 5.7x10^-8 W/m^2-k^4
T1, T2, and T3 are temperatures of surfaces.
A is the area of the surface
E = 0.1964 x 5.7x10^-8 x 0.4 x (400^4 - 500^4 - 800^4)
E = 0.1964 x 2.28x10^-8(-4.465x10^11)
E = -1999.39 W (the negative sign shows that it receives net heat but doesn't radiate any net heat)
For surface 2:
E = Ake(T2^4 - T1^4 - T3^4)
E = 0.1964 x 5.7x10^-8 x 0.5 x (500^4 - 400^4 - 800^4)
E = -2086.15 W
For surface 3:
E = Ake(T3^4 - T1^4 - T2^4)
E = 0.7855 x 5.7x10^-8 x 0.7 x (800^4 - 400^4 - 500^4)
E = 10076.28 W (it radiates net heat)
