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Heat Exchanger Vessel Acceptable Insulation Thickness Calculation

Heat Exchanger Vessel Acceptable Insulation Thickness Calculation

Heat Exchanger Vessel Acceptable Insulation Thickness Calculation

Hi All,

I have a Heat Exchanger vessel which is 82.25" OD and 133.25" long shell. I want to calculate the acceptable insulation thickness. The shell side has Hot Boiler feed water at average of 130 Deg C. The worst ambient temperature can be -10 Deg C. Can I calculate the heat loss for this vessel just as we do for pipes as below, assuming no resistance to heat transfer from the vessel wall itself.

q= k. AL. (T1-T2)/(r2-r1)

k is the thermal conductivity of the insulation. (k = 0.029 Btu/hr-ft Deg F for mineral wool at 100 Deg F)
AL is the log mean area = 2.PI()*(r2-2r1)/ln(r2/r1)
T1 is the temperature at the vessel outside surface T1 = 130 Deg C = 266 Deg F
T2 is the temperature at the outside surface of insulation. T2= -10 Deg C = 14 Deg F

r1 and r2 are the radius at the vessel outside and the outside of the insulation

Based on my calcs, I get heat loss as function of insulation thickness as follows:

I know there is a balance on insulation cost vs. cost of heat loss, but I learnt somewhere that there increasing thickness beyond a certain thickness would result in increase in heat loss. I know I have not taken in consideration the effect of air resistance.

My questions :

1. Can I calculate the optimum insulation thickness for a vessel just like a pipe?.
2. Can I ignore for simplicity sake the resistance of air in optimum insulation thickness calculation?.

Thanks and Regards,
Pavan Kumar

RE: Heat Exchanger Vessel Acceptable Insulation Thickness Calculation


I believe the increase in insulation adding to heat loss is usually only a factor for small diameter piping. To answer your questions:

1. Yes, provided you provide a definition of "optimum". This is contingent on correct heat loss calculations and on your criteria for the "optimum" capital expenditure versus operating expenditure.
2. No, you cannot ignore the resistance of air. You will need to determine the natural or forced convection coefficients of the air (depending on if this is inside or outside and what the wind speed is). Air has a significant resistance to heat transfer, so your equation, as written, will over-estimate heat loss by a wide margin.

As written, your equation is only valid if you can measure the skin temperature of the insulation. You'll need to use the combination of resistances method to determine overall heat transfer coefficient (U = 1/(SUM OF RESISTANCES)), where the resistance of natural convection for air can be found by your geometry and the corresponding Nusselt number equation. If outside and needing to calculate forced convection, you can assume a wind speed and calculation the forced convection coefficient using a modified Re number that is designed for flow around the outside of a pipe.

Lastly, the easy way would be to look up generic overall heat transfer coefficients with air and get pretty close. I don't know how accurate you need to be.

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