andy_oohhh
Chemical
- Jul 11, 2019
- 15
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.
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.