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Heat Exchanger Pressure Drop

Heat Exchanger Pressure Drop

Heat Exchanger Pressure Drop

(OP)
We have a tube and shell heat exchanger that is made in-house.

It has one face that is ~45" x ~30" with 150 2" pipes open to the gas stream.  The gas goes through some equipment where it is cooled and then crosses back over so it perpendicular to the flow of the first pass.  The flow on the second pass is across the path of the 2" pipes.  The second opening is ~45" x 24" wide.

I'm trying to get some rough numbers for pressure drop.  I have a feeling that the number we initially expected are way too low.   

RE: Heat Exchanger Pressure Drop

Refer to Process Heat Transfer by DQ Kern. Can you attach a drawing of the HX?

RE: Heat Exchanger Pressure Drop

I tried to do the calculation making some assumptions and got the value of 27.85 mmWC (though I am not sure about it). This is excluding entry, exit losses and loss due to directional change. The Kern equation has a constant in denominator which must be a conversion factor and I couldn't balance the equation dimensionally. Moreover, I feel that the equation is for liquids but they used the term fluids instead.

Quoting the equation for your reference.

dP = fG2Ln/(5.22x1010Dsφt)

f is dimensional friction factor, sq.ft/sq.in (I have attached the graph here)
G is mass velocity in lb/hr sq.ft
L is length of tube path in ft
n is no. of tube passes
dP is pressure drop in psi
φ is a factor (of viscosities) to compensate the diffference in heat transfer coefficients at wall temperature and bulk temperature (I think you can omit this as the viscosity change will be not significant for gases)
s is specific gravity (Kern might have considered s value against water)

PS: 1. Heat Exchanger Design by Sadic Kakac may be of help to you. 2. You may get better response from Heat Transfer or Chemical Fora

RE: Heat Exchanger Pressure Drop

(OP)
Quark, I appreciate your help.  The figure you came up with is about what we're measuring across the HX now which would include the exit/entrance losses etc.  The problem is that our fan can not pull enough air to get up to the 15,000acfm to see what the operating pressure drop will be.

Thanks again.

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