Radiation heat exchange between two surfaces.

  • Thread starter goldark
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  • #1
goldark
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Homework Statement


Prepairing for a test and this one came up that's confusing me.

A spherical object, water cooled, with a diameter of 10 milimeters and ε = 0,9 is kept at 353 degrees Kelvin, when placed in a very large furnace where a vacuum is formed and which walls are kept at 673 degrees kelvin.

What's the Transfered heat exchange (same as heat flux I think) by the furnace walls into the object?


Homework Equations


q =A ε σ T4
q =A ε σ (Ts4 - To4)
hr = A ε σ (Ts4 + To4)(Ts + To)


Ts - Surface temperature
To - object temperature


The Attempt at a Solution


I know from the solution is 3,04 W. But I don't know how to get there.
At first it seemed very straight forward since I'm given all the data to just replace the values in the formula. I've tried that but the value is always different from the solution.
What confuses me is that what is asked is the transferred heat exchange from the walls to the object not the other way arround.
Would really apreciate some help on this.
 
Last edited:

Answers and Replies

  • #2
LawrenceC
1,198
5
Since the sphere is in a vacuum, the only heat transfer mechanism is radiation.

The equation for determining the rate of heat transfer between two objects is:

q = sigma * epsilon * (T1**4 - T2**4)

where sigma is Stefan Boltzmann and epsilon is emissivity.

Note that the equation can be used to determine heat into or out of an object. I have computed the numbers and the answer drops right out.

If the small sphere is the only item in the furnace, the heat transfer from the walls equals the heat transfer to the sphere.
 
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