A small sphere (2.5 cm-diam) is placed in a heating oven. The oven cavity is a 30 cm cube filled with air at 101 kPa(abs); it contains 3 % water vapor at 810 K, and its walls are at 1370 K. The emissivity of the sphere is equal to 0.44 – 0.00018 T, where T is the surface temperature in K. When the surface temperature of the sphere is 810 K determine (a) the total irradiation received by the walls of the oven from the sphere, (b) the net heat transfer by radiation between the sphere and the walls of the oven, and (c) the radiant heat transfer coefficient.

GIVEN

A small sphere in a 30 cm cubic heat oven filled with air

Sphere diameter (D) = 30 cm – 2.5 cm= 27.5 cm

Air pressure = 101 kPa(abs) = 1.01*105 Pa

Air contains 3% water vapor

Air temperature (Tm) = 810 K

Oven wall temperature (T2) = 1370 K

Sphere emissivity (?1) = 0.44 – 0.00018 T (T in K)

Sphere surface temperature (T1) = 810 K

FIND

(a) Total irradiation received by the walls from the sphere (q2)

(b) The net heat transfer by radiation between the sphere and the walls (q12)

(c) The radiant heat transfer coefficient (hr)

ASSUMPTIONS

The gas is a gray body

The oven walls are black (?2 = 1)

The sphere is near the center of the oven

SKETCH



PROPERTIES AND CONSTANTS

the Stephan-Boltzmann constant


(a)



The surface area of the sphere is



Since the air and the oven completely enclose the sphere, F12 = 1.0 and F1g = 1.0

the portion of the total radiation leaving the sphere that is received by the walls



where tm is the transmissivity of the air and the radio sity of the sphere, J1, is



Since the air temperature is the same as the sphere temperature, there will be no net heat transfer between the air and the sphere and the heat transfer from the sphere (q1) will be the same as the net heat transfer between the sphere and the walls (q12). Simplifying with the shape factors above and



Since the sphere is near the center of the cube, one half the beam length for a cube will be used



The partial pressure of the water vapor is



The emissivity of the water vapor (?m) can be calculated where Tm = 810 K



By Kirchoff’s radiation law



(b) The net radiation between the sphere and walls is 146.5 W from the walls to the sphere.



(c) The radiative heat transfer coefficient must satisfy the following equation

Physics & Space Science

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