The heating surface is gray (emissivity = 0.8) while the receiver has a black surface. The lower plate is heated uniformly over its surface with a power input of 300 W. Assuming that heat losses from the backs of the radiating surface and the receiver are negligible and that the surroundings are at a temperature of 27°C, calculate the following

Question

A 10 cm square, electrically heated plate is placed in a horizontal position 5 cm below a second plate of the same size as shown schematically.



(a) The temperature of the receiver

(b) The temperature of the heated plate.

(c) The net radiation heat transfer between the two surfaces.

(d) The net radiation loss to the surroundings.

(e) Estimate the effect of natural convection between the two surfaces on the rate of

heat transfer.

GIVEN

A square heated plate below a second plate of equal size as shown above

Plate size = 10 cm x 10 cm = 0.1 m x 0.1 m

Distance between plates (L) = 5cm = 0.05 m

Heated surface (A1) is gray with an emissivity (?1) = 0.8

Receiver (A2) is black (?2 = 1.0)

Heater power input



FIND

(a) The temperature of the receiver (T2)

(b) The temperature of the heated transfer (T1)

(c) The net radiation heat transfer (q12)

(d) The net radiation loss to the surroundings (qs)

(e) Estimate the effect of natural convection

ASSUMPTIONS

Steady state

Heat losses from the back of each plate are negligible

Temperature of surroundings (T3) = 27°C = 300 K

The surroundings behave as a blackbody enclosure

PROPERTIES AND CONSTANTS

the Stephan-Boltzmann constant

Answer 1

In steady state, the net rate of heat transfer from the heater must be equal to the power input



The net rate of heat transfer to the receiver in steady state must be zero: q2 = 0

Also, since the receiver and the surroundings are black



The shape factor F12 can be read



Since A1 = A2, F21 = F12 (This is also clear from the symmetry of the problem).

Since neither A1 nor A2 can see itself, F11 = F22 = 0

The shape factors from any given surface must sum to unity





The net rate transfer from a gray surface is





Combining Equations [2] and [4]



Substituting



Solving for T2



(b)





Applying







(c) The net rate of heat transfer between A1 and A2 is



(d) Applying the conservation of energy of both plates and the surroundings yields



The surroundings gain 300 watts from the plates.

(e) An estimate of the natural convection heat transfer will be made by treating the heater and receiver as a horizontal enclosed space, heated from below with the surface temperature calculated above.

for dry air at the average temperature of (459°C + 807°C)/2 = 633°C



The Rayleigh number is





where

[]

indicates that the quantity in the brackets should be taken to be zero if it is negative.



The rate of heat transfer by convection is given by



COMMENTS

The natural convection heat transfer rate is only about 4% of the total heat transfer rate. Therefore, the estimate of natural convection is probably adequate.