CurtisMTK,<br>
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If I understand your problem statement, as the ambient or heat sink temperature increases, the differential temperature between the temperature of the surface of the resistor (T) and ambient temperature (Ta) goes down.<br>
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As Ta increases, what effect does this have on the resistance value of your resistor? If the resistance decreases with increasing temperature, then the power disapated in the resistor must go down. That would explain the reduction in the delta (T-Ta). Q=hA(T-Ta) (assuming the resistor is immersed in air only.) h is the convection coefficient, A is the surface area of the resistor.<br>
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The value of h can be found as follows:<br>
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Determine the product of the Grashof number (Gr) and the Prandt number(Pr) sometimes refered to as the Rayleigh number. This is not a simple process and you can find the methods and equations in any PE review book or heat tranfer text. Based on the value of this product, you can select the appropreate film coefficient equation to determine h. The Rayleigh number is the ratio of the fluid (air in your case) viscous and boyant forces.<br>
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After all this, you may find that the value of h has increased with temperature and therefore a smaller differential temperature is required to drive the heat transfer.<br>
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It could also be a combination of the two discussions above.<br>
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Also, it must be remembered that radiant heat transfer tends to be the dominant form of heat transfer in convection heat transfer cases.<br>
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This is probably clear as the Mississippi. For that I apologize.<br>
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V/R<br>
<br>
Tom<br>
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<p>Tom Worthington<br><a href=mailto

worthi@astro.as.utexas.edu>pworthi@astro.as.utexas.edu</a><br><a href= > </a><br>