Engineering Heat Transfer and Combustion Homework

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The discussion focuses on verifying the calculations for heat transfer and combustion homework, specifically part D, which estimates the heat loss by natural convection from a pipe. The key equation used is Q = hA(DeltaT), where Q represents the heat transfer rate. The calculations provided yield a heat loss of approximately 167.9855 W/m, which aligns closely with another participant's result of 167.1 W/m. The conversation emphasizes the importance of consistent units and accurate arithmetic in these calculations. Overall, the participants confirm that the working out appears correct with minor discrepancies attributed to rounding.
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Homework Statement
The outer surface of the insulation on a horizontal steam pipe has a radius of 50mm and is at a temperature of 90Deg C. The atmospheric air surrounding the pipe is at a temperature of 14 Deg C, and has the property values listed in part (c) above. Estimate the rate of heat loss by natural convection to the atmosphere by each metre length of pipe.

Only require help with part (D).
Relevant Equations
Nu = 0.53Gr^0.25Pr^0.25
Q3.png
Q3.1.png


Does this working out look okay? Thanks in advance.
 

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I'm not going to check your arithmetic. I assume you used consistent units so that the dimensionless groups actually came out dimensionless..

I don't see any part D here. I assume part D requires you to calculate the heat transfer rate per meter of pipe, given the heat transfer coefficient, the pipe diameter, and the temperature difference. Let's see your equations for doing this.
 
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Correct part D is the question sorry - Estimate the rate of heat loss by natural convection to the atmosphere by each metre length of pipe.

The equation for doing this is

Q=hA(DeltaT)

therefore..

Q=h(2Pi*rL)(DeltaT)

Q=1.34*(90-14/0.1)^0.25
*0.1*Pi * 1 * (90-14)

Q = 167.9855 w/m
 
MaisieMitchell said:
Correct part D is the question sorry - Estimate the rate of heat loss by natural convection to the atmosphere by each metre length of pipe.

The equation for doing this is

Q=hA(DeltaT)

therefore..

Q=h(2Pi*rL)(DeltaT)

Q=1.34*(90-14/0.1)^0.25
*0.1*Pi * 1 * (90-14)

Q = 167.9855 w/m
Based on the heat transfer coefficient you calculated in part C, I get about the same result.
 
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Thanks, when you say about the same result - is there something that I have missed?
 
MaisieMitchell said:
Thanks, when you say about the same result - is there something that I have missed?
I don't think so (other than roundoff). I get 167.1
 
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Thanks again!
 

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