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Gas cooling in a pipe and pressure drop

Gas cooling in a pipe and pressure drop

Gas cooling in a pipe and pressure drop

(OP)
I'm working on a design in which a cooler is used to bring down the temperature of a gas in a pipe from 80C to 50C. Is that going to cause a significant pressure drop in the pipe?

RE: Gas cooling in a pipe and pressure drop

It depends.  It's an economic issue.  Buy a big cooler and the pressure drop will be low.  Buy a small cooler and the pressure drop will be high.  You need to optimize the design.

Good luck,
Latexman

RE: Gas cooling in a pipe and pressure drop

For compressible fluids, heat loss reduces the temperature of the fluid and thus increases the fluid density, which reduces the pressure drop of the fluid.

"We don't believe things because they are true, things are true because we believe them."

RE: Gas cooling in a pipe and pressure drop

compressible, non compressible, it doesn't matter.  The cooler will have a pressure drop that you can specify in the design. As Latexman says, you have a choice each joice has a different consequence in pressre drop and cost.

RE: Gas cooling in a pipe and pressure drop

(OP)
Thank you for your replies. I must clarify something here: I was not talking about the pressure drop due to friction in the cooler but wondering if the cooling process itself could trigger a pressure drop.
From an ideal gas point of view, if PV = nRT and T decreases, then P or V must decrease. What will happen?

RE: Gas cooling in a pipe and pressure drop

the result (80->50C) would be a reduced density, velocity and pressure drop   

RE: Gas cooling in a pipe and pressure drop

Quote (Whelp):

if PV = nRT and T decreases, then P or V must decrease. What will happen?
Yes, V (volume) will decrease.  This is the same thing as an increase in density - which was explained by sheiko.  I suspect this is what drilling53 meant to say as well because an increase in density will bring about a decreased velocity.

Katmar Software
Engineering & Risk Analysis Software
http://katmarsoftware.com

RE: Gas cooling in a pipe and pressure drop

In a closed system (gas cylinder), I'd expect the pressure to change with temperature as katmar indicates.

However, for an open system (flowing), I'd expect the pressure to be a function of the source pressure, destination pressure, system pressure drop, fluid physical properties, and bernouilli's law.  I don't see how cooling or heating a flowing gas is going to make a pressure change by itself.  Yes, the physical properties of the gas will change and that will change pressure drop somewhat, but there is not going to be a step change in pressure across a cooler or heater solely because of the temperature change.

If you blow through a straw that runs into and out of a pot of boiling water to exit to atmosphere, it will NOT come out at a pressure higher than atmospheric pressure.

Good luck,
Latexman

RE: Gas cooling in a pipe and pressure drop

Latexman,
I think you have a feel for the subject.  The arithmetic supports it.

If we start with an ideal gas (say air) then as said above PV=nRT, so P1V1/(n1RT1)=1=P2V2/(n2RT2).  Now since the continuity equation says that mass flow rate at any point in a closed system is the same as any other point in the system, we can choose a control volume (V2) so that n1=n2.  Let's say that V1 equals a unit volume so:

P2V2=P1V1*(T2/T1)

It is important to note that we don't have any idea what size V2 has to be to hold the same "n" as V1.  So while the product of P2V2 will have to be about 9% lower than P1V1, we have absolutely no way of knowing the magnitude of either variable (P2 could be higher than P1 and V2 a lot smaller than V1, or vice versa, or any combination).

I don't think that there is any analytical way to determine either P2 or V2, but my observations have been that pressure at the outlet of a cooler is very close to inlet pressure minus friction within the cooler.

David

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