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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[sheiko]

How do you link this average specific volume to the specific volumes of saturated steam and water, respectively vf and vg in steam tables? and which volume to follow on the line of constant specific volume of steam charts/tables to determine the equilibrium temperature?

Reminder:
Initial state is saturated steam at 40 barg (and T=250ºC)
Final state is saturated liquid. What is the final T?

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[zdas04]

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

Think of a 200 mile long gas pipeline that is at 80F and 1,400 psig at the head and 55F and 600 psig at the foot. The density of 0.6 SG gas at the head is 4.2346 lbm/ft^3 and 1.9247 lbm/ft^3 at the foot--less than half.

Mass flow rate is constant for the two points. This is accomplished in the face of declining density by increasing velocity.

David

 

[insult2injury]

zdas04: Now I'm confused. I thought that he was considering a case where the line is blocked in, initially filled with steam, and wanted to determine the final pressure after the temperature equalized. Reading back, I'm not sure that it is ever clearly defined whether this is a blocked in line or a normal operating process line with heat transfer.

sheiko:
1st approach (blocked in line, initially filled with steam)
- look up the specific volume corresponding to the initial state
- follow the line of constant specific volume corresponding to the initial state specific volume
- from this line, you can determine the quality and pressure of the mixture at the temperatures of interest
- how much the temperature changes is a function of the ambient temperature and heat transfer coefficients
- if left sufficiently long, the pipe will equalize in temperature with the surroundings
- to determine the pressure, look up the pressure value corresponding to the initial (and final) state specific volume and ambient temperature

2nd approach (flowing process line with heat transfer)
- the pressure difference will equal the two-phase frictional flow losses, several correlations have been published for this case

I2I

 

[DLANDISSR]

If we are considering a blocked line that is cooling the pressure will follow the saturation line as the steam cools.
The practical answer is: Can the line withstand total vacuum without collapsing, if not, install a vacuum breaker set at a pressure above the collapse pressure.

 

[zdas04]

The blocked line case is trivial. If there is no flow then pressure (within a few seconds of shut in) is the same on both ends of the contraction. As the line cools, pressure drops on both sides of the contraction.

The question only makes sense if the line is flowing.

David

 

[sheiko]

It was dealing with blocked in line.
My conclusion is that, as i want to check the pipe against vacuum, maybe i should account for full vacuum, as even with a final temperature close to room temp, the vapour pressure will be substantially zero.

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[quark]

I am not sure about your formula but you can do this in the following way unless the entire steam is condensed.

Suppose, you have 1cu.mtr of pipe volume filled with steam at 40barg and it started condensing. At 40barg the density is about 20.6078kg/cu.mtr. If 10 kgs of steam is condensed then you have about 10.6078kgs of steam left to occupy a volume of approximately 1cu.mtr (actually, water specific volume at 40barg is 0.00125cu.mtr/kg, so 10kgs of water occupies 0.0125cu.mtr or 10.6078kgs of steam occupies 0.9875cu.mtr and density should be 10.6078/0.9875 = 10.74kg/cu.mtr)

Now, get a steam table program like steam tab and go use the saturation temperature corresponding to 40barg and go on varying the pressure till you end up with 10.74kg/cu.mtr density. This will give you the pressure of 23.75bara.

The important point is to first calculate the heat loss and then use that data to check the rate of condensation and then using the above method, you can check the pressure at intervals.

 

[sheiko]

Quark,
Could you please clarify/explain your last sentence "The important point is to first calculate the heat loss and then use that data to check the rate of condensation and then using the above method, you can check the pressure at intervals. ", because i don't really see how to perform that...

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"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[sheiko]

Moreover,

1/ Why the fact that the entire steam is condensed invalidates your method?

2/ Why did you make all your calcs assuming steam and water at 40 barg while the pressure is suppose to vary?

"We don't believe things because they are true, things are true because we believe them."
"Small people talk about others, average people talk about things, smart people talk about ideas and legends never talk."

 

[Artisi]

I would allow for full vacuum and at the end of the day it probaly make little difference considering the wall thickness you are using for 40 bar - this should have a wall thickness well in excess of requirements for a full vacuum.

 

[quark]

Could you please clarify/explain your last sentence

Suppose, xkw/m²/k is the heat transfer coefficient and y kJ/kg is the latent heat of steam at the given conditions, then the amount of condensate formed in one hour from 1 m²surface area of the pipe and with 1K temperature difference is x*3600/y.

Why the fact that the entire steam is condensed invalidates your method?

When the entire steam condenses then it will be the vapor pressure at the condensate temperature.

Why did you make all your calcs assuming steam and water at 40 barg while the pressure is suppose to vary?

I just showed one example. You can better split the process into no. of intervals and do the calcs as your time permits.

However, like earlier suggested, if the steam quantity trapped is insignificant compared with the exposed area and if it happens for a longer periods of time then full vacuum is a better assumption.