DK44
Mechanical
- Sep 20, 2017
- 196
In a typical Shell & Tube Heat Exchanger, the tubes are of Bi-metal, having different Thickness and different Thermal Conductivity.
Tube OD = 19.05 mm
Outer Tube MOC X Thk = Incoloy 825 X 0.84 mm
Inner Tube MOC X Thk = 70/30 Cu Ni X 1.22 mm
I guess,
If resistance of composite tube = t/K,
Outer Tube resistance = t1 / K1
Inner Tube resistance = t2 / K2
Then adding the resistances serially.
t/K = (t1/K1)+ (t2/K2)
Is my approach correct or is there some other method?
Tube OD = 19.05 mm
Outer Tube MOC X Thk = Incoloy 825 X 0.84 mm
Inner Tube MOC X Thk = 70/30 Cu Ni X 1.22 mm
I guess,
If resistance of composite tube = t/K,
Outer Tube resistance = t1 / K1
Inner Tube resistance = t2 / K2
Then adding the resistances serially.
t/K = (t1/K1)+ (t2/K2)
Is my approach correct or is there some other method?