Heat transfer and combustion problem review

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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
Handwritten working for a heat transfer question 3c on lined paper, calculating Grashof, Prandtl and Nusselt numbers using listed property values and the formulas Gr = αρ²d³(Ts−Tf)g/μ² and Nu = 0.53 Gr^0.25 Pr^0.25.
Handwritten working for part 3c: Nu = hd/k, so hd/k = 24.46, giving h = 24.46k/d, then h = (24.46 × 2.861×10⁻⁵)/0.1 = 6.998006×10⁻³ kW m⁻² K⁻¹ = 6.998 W m⁻² K⁻¹, boxed.


Does this working out look okay? Thanks in advance.
 
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  • Textbook problem part (d) on natural convection from a horizontal steam pipe, with the Nusselt, Grashof and Prandtl correlations and the simplified form h = 1.34(ΔT/d)^0.25 with air property values listed.
    When air has the values listed below.png
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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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