Question

Consider a circular furnace that is 0.5 m long and 0.5 m in diameter. The two ends have diffuse, gray surfaces that are maintained at 400 and 500 K with emissivities of 0.4 and 0.5, respectively. The lateral surface is also diffuse and gray with an emissivity of 0.7 and a temperature of 800 K. Draw and label a schematic for this system. Determine the net radiative heat transfer rate from each of the surfaces.

Answer 1

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)