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Finding significant differences in heat transfer formulas. Which method should be used?

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andy_oohhh

Chemical
Jul 11, 2019
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Hi everyone. I had to solve a simple problem for losses in pipe. Used two different methods and got significant differences.

Givens:
Pipe Surface temp, 450F
Ambient temp, 65F
Convection coeff, 5 Btu/(hrft2F)
Pipe Length, 20 feet
Emissivity, 0.5

Method 1:
Qtot = Qrad + Qcon = hA(Ts-Ta)+eaA(Ts4-Ta4) = 76,900 Btu/hr
back solving for equivalent hrad when radiation and convection is in parallel gives...
hrad = 0.1623

Method 2: Same method used here ( ). With formula for hrad here (
Qtot = UA*deltaT
1/UA = sum of resistance in parallel
Qtot = 11,352 Btu/hr
hrad = 1.35

The differences comes from the hrad formula. I don't understand why there is such a huge difference. It would be regrettable to use the wrong method in industry. Wondering if anyone can make sense of which one of these methods should not be used.
 
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What formula does method 1 use for the rad compliant?
The last time that I checked radiant heat transfer was T to the fourth power.
What you use for Tamb can be tricky if there is a radiation absorption factor also.
The are also corrections for curved surfaces, but you likely don't need those.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
 
Total hc+hr of 5.2 btu/hr/ft2/degF in method 1 is within the ballpark as that published in Perry Chem Engg Handbook of approx 4.5 btu/hr/ft/degF for horizontal bare pipe - see table 5-2 on page 5-14 in the 7th edition. But value for hr of 0.16 is too low when compared with that from fig 5-7 in the same chapter in Perry ( approx 2.5 after correcting for gray body emissitivity - assume to be 0.8).
 
hi,
From my notes :
hr= ε*τ*(Tskin^4-Tamb^4)/(Tskin-Tamb) w/(m2*K)
τ stefan bolzman constant
5.67037E-08 W/(m2*K4)
hr= 7.689 w/(m2.K)or 1.35411 Btu/(ft2.h.F)
hr +hco = 6.354 Btu/(ft2.h.F)

hope this is helping you
Pierre


 
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