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

The problem requires the use of the Stefan-Boltzmann law to determine the net radiative heat transfer rate from surfaces with different emissivities and temperatures, considering they are not perfect black bodies (gray surfaces).

Step-by-step explanation:

The concepts of thermal radiation and the Stefan-Boltzmann law to calculate the net radiative heat transfer rate from surfaces with different temperatures and emissivities. The law states that the power radiated from a black body is proportional to the fourth power of its temperature. When dealing with real materials (gray surfaces), we adjust this by the material's emissivity, which accounts for how well it radiates heat compared to a perfect black body.

First, we would determine the radiative heat transfer rate from each individual surface using the formula:
Q = ε•A•σ•(T^4 - T_{env}^4), where ε is the emissivity, A is the surface area, σ is the Stefan-Boltzmann constant, and T and T_{env} are the absolute temperatures of the surface and environment, respectively.

Then, for a non-black surface (a gray surface), we use the emissivity values given in the problem (0.4, 0.5, and 0.7 for the respective surfaces). The net radiative heat transfer rate from each surface is the sum of the heat radiating away from the surface and the heat being absorbed by the surface from the surroundings.

Answer 2

Answer:

Surface 1 = -1999.39 W

Surface 2 = -2086.15 W

Surface 3 = 10076.28 W

Step-by-step 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)