RADIANT HEATING WITH INFRARED - Watlow

E = 1 = 1 = 0.70
1 + 1 - 1 1 + 1 - 1
Eh Ep 0.85 0.8
C. Determine the average product temperature (Tp)
Tp = 300 + 60 = 180°F = 640°R
2
D. Plug into the radiant heat transfer equation
W/in 2 = S(Th 4 - Tp 4 ) x E x F
144 in 2 /ft 2 x 3.412 BTU/watt hr
From above we found:
E = 0.70
F = 0.85
Tp = 640°R
Required W/in 2 = 4.54 W/in 2
Therefore:
4.54 W/in 2 = S(Th 4 - (640) 4 ) x 0.7 x 0.85
144 in 2 /ft 2 x 3.412 BTU/watt hr
7.6 W/in 2 = S(Th 4 - (640) 4 )
144 in 2 /ft 2 x 3.412 BTU/watt hr
S = 0.1714 x 10 -8 BTU/hr ft 2 °R 4
E. Determine the required heater temperature (Th)
All information required to solve the above equation for the heater
temperature (Th) is now available. Note that the equation gives Th in °R.
Th (°F) = Th (°R) - 460
Another option is to use Figure 7
Radiated watts = 7.6 W/in 2
Average Tp = 180°F
From the graph or the calculation, find:
Th = 780°F
To transfer the required watts, the heater must operate at 780°F.
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