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Can I calculate how much air is released from water due to pressure change?

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MechEng92

Mechanical
May 12, 2015
61
This might be a stretch but is there a way of calculating how much air is released from water due to pressure change such as a valve opening and closing? If so, what is the method and what do I need to carry out the calculation?
I am trying to find out if there's a tangible number I can associate to how much air is being released in a closed system by the pressure changing over a moving valve.

Thanks in advance.
 
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SNAP!

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Took you 2 minutes too long.

A black swan to a turkey is a white swan to the butcher ... and to Boeing.
 
Thanks both, I assume I can use the table if I know the water temperature and the pressure difference over the valve once opened to determine solubility ratios, can I use this to work out how much air is released over time with a water flow rate?
 
Transient and time dependant flows can be quite difficult as air solubility takes time and depends on the surface area to volume ratio and also how volatile the water air interface is.

I'm not rally understanding what this means "a closed system by the pressure changing over a moving valve."

Is there a diagram or schematic of what it is you're doing here?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
I'll try to provide as much detail as I can. We have a product that consists of pipework inlet and outlet connections with a chilled water cooling coil inside the product and a control valve on the outlet of the coil. The product would be piped up to a cooling unit providing the chilled water.

We are being asked why the control valve can't go on to the supply side of the cooling coil in the product and being asked to provide some sort of tangible benefit for having the valve on the outlet of the coil.

My only thoughts in doing this is trying to calculate an amount of air released in the water from the control valve that would get trapped in the coil over time if the valve was on the supply side of the coil.

Hope this helps.
 
OK - Is the chilled water loop a closed loop?

If so I can't really see where the air is going to come from.

If the velocity in the cooling pipes is about 1m/sec or even as low as 0.7, then this should be enough to flush out any air that does escape.

Valve on the inlet or outlet? I would tend to go for the outlet to keep the coil at more pressure, but if your closed loop is at sufficient pressure you shouldn't get any gas here. Having trouble seeing any disadvantage to having it on the outlet.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
It is a closed loop. And I would imagine the installer would fit devices to remove air from the system if there was any, but we have made our product to most recommended practices in fitting the valve to the outlet.

However they are being very stubborn in demanding tangible evidence to support fitting it to the outlet so I'm grasping at straws to trying put numbers to paper.
 
I think you're going to struggle to find "tangible benefits".

The only other one I can think of is if the pressurisation bellows or tank fails then the coolant won't flash off across the valve because it is held at a higher pressure. that alone is good enough for most people.

But if they want it on the inlet side then point out in writing it is their decision and you take no responsibility for anything going wrong in the future.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
Suggest criterion to use is to ask what will happen if there were to be pinhole leak or rupture of the cooling coil immersed in this product vis a vis the location of the control valve.
Pressure of the coolant is usually lower downstream of the control valve.
What would happen : Will coolant leak into product or will product leak into coolant ?
And what does that do to either product quality or the rest of the coolant system.
Which is the less of the 2 evils.
In some cases, coolant contamination of product is completely unacceptable if coolant is toxic or carcinogenic etc. In this case, you may be persuaded to (a)change coolant type or (b) ask for completely corrosion resistant coils plus 100% radiography of welds in the cooling coil. Dont know if (b) is a good idea - what if the coil sheared off the internal nozzle support due to high vibration ( due to poor coil supports or transient high flowing velocity) or some mechanical damage.
 
So I assume that you are thinking of trapped air in the water. This is my way to look at this experiment. Hook up a vacuum pump to an insulated container of water and have the vacuum pump outlet connected to an insulated recipient container that can measure the increase in volume caused by the removed air and water vapor from the supply container. Pressure and temperature sensing devices will need to be connected to both containers. Obviously the pressure sensors readings speak for themselves. On the other hand the temperature sensors will indicate by using the Mollier diagram how much water vapor is included on the discharge side of the vacuum pump. Knowing the initial and final mass of the water in the supply container will tell you how much water vapor was removed and added to the recipient container. The volume change of the recipient container will be from the water vapor and from the removed trapped air. Since you have the pressure and temperature readings of the mixture within the recipient container, you can figure out the mass of the trapped air and of the water vapor. Your OP requires a knowledge of the Mollier diagram noted in thermodynamics, ideal gas equations that were introduced in chemistry and in thermodynamics, and mass transfer, a particular topic introduced in heat transfer.









spor water andr, ,samass o
 
Note that the solubility part of the problem is a static state, i.e., it doesn't necessarily mean that any pressure change over a transient period will result in the change in dissolved gas. There is a finite time for diffusion of the gas through the solution, so any brief transient pressure will result in a much smaller change than the change in solubility will suggest.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! faq731-376 forum1529 Entire Forum list
 
Sakshi is just hijacking a thread to get his link / advert. This post closed four months ago.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
 
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