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pressure drop

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mielke

Mechanical
Aug 24, 2009
181
I want to make sure that i am interpretting equations right. The equation for pressure drop across bare tube banks has (pi/po-1) in it for the effects of the change in density. (pi is inlet density, po is outlet density)

Now if pi is greater than po (ie the fluid is being heated) than we have a positive pressure drop. but if pi is less than po than we get a negative term in the equation.

So does this mean if we have a fluid on the outside of a bare tube bank it will lose pressure due to frictional resistance of tube bank but gain pressure due to temperature change (due to change of density).
 
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It seems as if your equation is an emperical equation that cannot be used for reverse engineering.

Without knowing the full equation for dP its a little hard to tell what to do when trying to move outside valid area of the eq.

Best regards

Morten
 
my guess is that you have misinterpreted the definition of p- that should be the pressure, not density.

The second law of thermo can be used to show there must be frictional pressure loss as the process proceeds, and pressure gains would need to be due to either hydrostatic head ( gravity) gains or by adding energy ( via a fan or pump or some external energy addition).

For very high gas velocities ( compressible flow) , one sometimes uses % pressure loss as an indication of the frictional loss across a component; eg, a 2.5% pressure drop is 0.025= (Pi-Po)/Pi = 1- (Po/Pi).
 
dave

Im not reading his post as to suggest that you will geta negative dP - in total - just that you will get a negative contribution from the "density part" of an equation where we have not seen the full equation. Thats why i ask for the full equation

Best regards

Morten
 
The equation is the pressure drop for flow around a tube bank from Kays and London(1984).

dP=G^2/(2pi)[f(At/Amin)(pi/pm)+(1+rho^2)(pi/po-1)]
dP is change in pressure
G is mass velocity
pi,o,m is density at inlet, outlet, and average temp
At,Amin,rho are physical area dimensions.

the (pi/po-1) term in question would imply that if you are cooling down the liquid then there will be a negative term in the pressure drop calc and that it is possible that a flow over a tube bundle may see a pressure rise (if the density term is greater than the frictional terms).
 
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