Pressure reduction during contraction 3

[sheiko]

Hello,

I was asking how to accurately estimate pressure reduction during contraction. I saw some experienced engineers using the formula P2=(rho1/rho2)*P1 to estimate the pressure after contraction of 40 barg saturated steam to saturated water but i cannot find where it comes from...Is it correct?

Thanks in advance

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[katmar]

Your formula is just a simplification of the ideal gas law - assuming the temperature is constant. It will certainly not be accurate (or even relevant) if the second state involves a liquid. Also, be aware that saturated steam at 40 bar has a compressibility of around 0.82, so the ideal gas law becomes suspect even if you stay with gas phases.

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[sheiko]

Yes Z=0.872 for steam at 40 barg

So?
considering Z and that:
State 1 is saturated steam at 40 barg
state 2 is saturated liquid
what would be the correct approximation for determining P2?


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[katmar]

P2 is determined by the vapour pressure of the liquid at T2.

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[sheiko]

BTW katmar says "Your formula is just a simplification of the ideal gas law - assuming the temperature is constant."
Using the ideal gas law would lead to the formula P2=(rho2/rho1)*P1 and not P2=(rho1/rho2)*P1 ...

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[zdas04]

The formula you should be using for the first cut is Bernoulli's equation. In the contraction, the velocity increases rapidly and the pressure decreases (the equation assumes a constant density).

Then you can use Joule-Thomson coeeficients to get the temperature change to calculate the probable change in density. Finally you can use your equation.

Typically Bernoulli is close enough for this sort of calculation, but if you need more precision this will get you there.

David

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[sheiko]

Thank you zdas04,

Could you please tell me where does the relationship P2=(rho1/rho2)*P1 come from (different from ideal gas law which wuold lead to P2=(rho2/rho1)*P1...)?

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[zdas04]

Well, I don't really know. If I were to bastardize Bernoulli by saying that the density isn't constant and assuming velocity and elevation are constant, you would get P2/rho2=P1/rho1 so I don't have a clue where thay could have come from beyond a long-lived typo.

David

 

[25362]

Steam tables show that below 40 barg, the saturated steam line shows that the products P.v are about constant (with an error of less than 5%) down to, say, 10 Bar.

where P are absolute pressures and v specific volumes = 1/[ρ] of saturated steam.

P.v = P/[ρ]

Is this sheiko's intention ?

 

[sheiko]

Thanks 25362,

However your relationship still leads to P2/rho2=P1/rho1 while some engineers use the relationship P2/rho1=P1/rho2 to estimate the pressure reduction during contraction...

So:
1/ Where does this expression come from?

2/ In the case i mentionned:
State 1 is saturated steam at 40 barg (rho1=19.46 kg/m3)
state 2 is saturated liquid (rho2=799.36 kg/m3)
Is this expression valid?

Please.

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[sheiko]

Also i considered:
T1=Tsat=250°C, Z1=0.872 and
rho2 is obtained from specific volume in steam tables at same temperature T2=T1=Tsat=250°C (maybe it is wrong...)

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[katmar]

Guys, we are talking at cross purposes here. I took sheiko's "contraction" to be "contraction during cooling" and David has assumed it is a flow process through a reducer. These are two very different situations and require different analyses. Even with the information now presented I cannot say that I know what the problem actually is. I must admit that I did not check carefully enough to see that the formula was upside down compared with the ideal gas law.

Sheiko, what actually are you trying to do?

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[sheiko]

Thanks,

I'm trying to check if the minimum design pressure of a 40 barg steam pipe is low enough. So i considered the formula that i presented you in order to figure out the pressure reduction during condensation.

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[katmar]

If you fill your pipeline with 40 barg steam and isolate it and let it cool down the pressure will be determined by the vapour pressure of water at the temperature reached. The temperature reached will depend on the time and the external weather conditions, but if you left it long enough in freezing conditions the pressure in the pipe would get down to below 1 kPa absolute. Effectively full vacuum.

You may well find that your steam traps will work as vacuum breakers. Depends what type you are using and what your condensate piping looks like.

Off topic - but if you are talking saturated 40 barg steam your density is wrong.

Harvey

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[sheiko]

For saturated steam at 40 barg:
rho1=4100000*18/(0.872*8314.5*(250+273.15))=19.46 with real gas law and rho1=19.95 using steam tables at 250°C.

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[sheiko]

In addition, Katmar, what about using the Joule-Thomson coeeficients to get the temperature change to calculate the probable change in density?

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[insult2injury]

The process is constant volume; therefore, just follow a line of constant specific volume on your steam charts/tables until you reach the equilibrium temperature.

I2I

 

[insult2injury]

In addition, Katmar, what about using the Joule-Thomson coeeficients to get the temperature change to calculate the probable change in density?[/sheiko]

If it's a closed system, there is no change in density.

I2I

 

[sheiko]

How can the volume be constant during a contraction (which is by definition a volume reduction)? Also does condensation always produce a contration?

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[insult2injury]

You have a two-phase vapor-liquid mixture. The portion that condenses to a liquid takes up less space; however, the amount that remains vapor takes up more space as it expands with the corresponding decrease in pressure. The average specific volume of the complete mixture remains the same.

I2I