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Heat Transfered from plate to plate

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A514

Mechanical
Feb 16, 2007
8
US
I have a problem with heat from a 1800 degree hot slab
Size 82" X 132" (9" thick)
Stefan - Boltzman Constant .174E-8
2227 Temp in Rankine
Area = 75.167
I used Q=.174E-8 * Area Ft^2 * Temp ^4
Q= 3.217E+6 in BTU / Hr.

Directly above this slab I have an air space of about 4.5" and then I have a heat shield with the same size as the slab
about 1" thick. I figure the slab will radiate heat and assumed it to be a black body. Would anyone know what formula I would use to get a rough estimate of the temperature on the surface of the heat shield? The heat shield is a piece of carbon steel plate.
Thanks in advance for your help,
John
 
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OK, you made the problem much harder. The heat shield will radiate heat back towards the slab. The difference in those two quantities is the net energy flow.

Depending on the material in the heat shield, you should also include a convection/conduction term, for completeness.

You then need to figure out how much heat flow the slab can sustain. It's usually the solution to these two simultaneous constraints that determines the surface temperature.

TTFN

FAQ731-376
 
For the slab problem you really need to know how heat is lost from the heat shield, otherwise the heat shield will be at the same temperature as the slab. As the two are close together and the emissivity of the slab is close to 1 you can assume you have the equivalent of a surface radiating to an ambient temperature of the slab. If on the opposite surface of the heat shield heat is lost by natural convection and radiation to ambient surroundings, say at room temperature, then you have a 1D non-linear problem through the heat shield thickenss which you can solve by finite elements or finite difference methods. If you assume that the temperature is uniform through the thickness then all you need to solve is the equation that relates the radiation to the slab to the heat loss to the surroundings so that f(T^4) = g(T^0.25)+ h(T^4) where g() is the function relating to heat loss by natural convection, h() relates to radiation to the surroundings, and f() is to the slab.

corus
 
T Rankine = T Fahrenheit + 459.67

It looks like he accidentally subtracted 32 degrees when converting from F to R.
 
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