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Given the question:

Calculate the energy transfer rate across a 6 in wall of firebrick with a temperature difference across the wall of 50 °C. The thermal conductivity of the firebrick is $0.65\ \frac{\text{Btu}}{\text{hr ft }^\circ\text{F}}$ at the temperature of interest.

The correct answer is 369 W/m2


I used the following approach:

$${ x = 6in => 0.5ft }$$ $${ \triangle T = 50^oC => 122^oF }$$ $${ k = 0.65Btu/hr-ft-F }$$

$${ \frac{Q}{A} = \frac {k\triangle T}{x} = 158.6Btu/hr*ft^2 => 500W/m^2 }$$

But I did not calculate the correct answer. Is there a step that I am missing?

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Your C to F conversion is correct for a different question.
The Celsius and Fahrenheit scales have their zero points offset by a factor equal to 32 F degrees and this has to be allowed for when converting from one measurement system to the other

However

In this case you wish not to convert between two "scales" but to express the delta difference in degrees F rather than degrees C. So the ratio 9:5 is relevant and the offset of 32 degrees F between the two scales is irrelevant.

Short version of answer is: Doh! :-)

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L=6 inches(2.54cm/(1 inch))=15.24cm

∆T=50℃

k= 0.65 Btu/hr-ft℉((1W/cm℃)/(57.79Btu/hr-ft℉))= 65/5779 W/cm℃

q_k/A=k/L ∆T

q_k/A= ( 65/5779 W/cm℃)/15.24cm × 50℃

q_k/A=0.0369 W/〖cm〗^2 = 369 W/m^2

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    $\begingroup$ Ok, but what step is missing? $\endgroup$ – peterh - Reinstate Monica Dec 13 '20 at 9:46
  • $\begingroup$ the conversion of the thermal conductivity from Btu/hr ft ℉ to W/cm℃ $\endgroup$ – Carl Justine Pogi Dec 14 '20 at 6:26

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