|Mar5-08, 10:16 PM||#1|
PRV valve pressure drop
I'm in need of some assistance in calculating some flow vs pressure specifications, in order to make some changes to the system.
I have a PRV valve that has a preset setting of:
PRV NO.1 & 2 SET PRESSURE = 1034 KPA (150 PSI) @ 0.63 L/S (10 USGPM)
Meanwhile the design criteria states the output to be:
PRV NO.1 & 2 OUTLET = 683 KPA (99 PSIG) @ 33 L/S (520 USGPM)
= 1034 KPA (150 PSIG) @ 0.63 L/S (10 USGPM)
I'm having difficulty determining or verifying the calculation that will prove that the flow @ 520 USGPM will yield a pressure drop to 99 PSIG, which equals 10 USGPM @ 150 PSIG.
Something just seems odd and I've tried a number of different formula's, obviously not the correct one. I'm hoping to understand what calculation was done, and conduct it again for a flow of 400 USGPM.
The valve is a FORD 4" PRV.
Any help would be appreciated.
|Mar6-08, 07:21 AM||#2|
You're right... something really doesn't add up. I see you've also posted at Eng Tips. BigInch missed a paren. Try this web site for calculating fluid flow through an orifice:
Generally, the relief valve mfg will provide the discharge coefficient and orifice diameter for their device. Do you have this information for the valves?
You say there is design criteria which requires a much higher flow rate at a slightly lower pressure, which doesn't make sense to me. In industry, these 'design criteria' are provided by a Process Department and it is up to a Mechanical Department to come up with valve specs. So can you explain a bit about this design criteria? Where did it come from and what are you doing with it? If you can provide a bit more detail around this system, such as what is creating this overpressure condition (a pump, high pressure source flowing through a valve, heat flux, etc...) it would be very helpful. Sorry if I've totally misunderstood your post, but it seems a bit confusing to me.
|Mar6-08, 10:39 AM||#3|
Unfortunately I have little information from FORD, nor have they been able to provide this detail, my understanding is that they do not necessarily have a chart identifying orifice sizes at given flow rates. The orifice size will fluctuate as the flows change. The PRV is spring loaded and compensates according to flow / pressure downstream of the valve
Apologies, the design criteria is from an old print whereby the designer calculated the approximate outlet pressure of the valve based on the designed flow rate. The PRV valves come factory set at a certain design outlet pressure, in this case 150 PSI at 10 USGPM. In our application, the PRV valves will see a range in flow anywhere from 100-520 usgpm. The application is in an underground mine, whereby the PRV valves are reducing the pressure every 600 ft, in order to feed the lower mine workings with water at a suitable pressure. Now, in this case, the print lists in it's design criteria that a flow of 520 USGPM will be reduced at the outlet of the PRV to 99 PSIG, which equals the factory setting of 150 PSIG at 10 USGPM.
This is what I find confusing. I'm unable to determine how this was calculated.
This example is but one such example on our prints. I have other examples where the PRV set pressure is 120 psi @ 10 usgpm, which is stated to equal 106 psi @ 145 usgpm. Again, I'm not understanding the calculation involved.
|Mar7-08, 07:02 AM||#4|
PRV valve pressure drop
We are talking about Pressure Relief Valves (PRV) aren't we? Or are you refering to a pressure regulator (perhaps pressure reducing valve)? A relief valve for water may not pop open, it may be designed to lift gradually to the full open pressure, but it has a set pressure at which it begins to open, and a pressure at which it is full open, typically 10% above the set pressure. A relief valve that is rated to flow 10 GPM at 150 psig can't flow any more than that unless the pressure differential across the valve is increased. And if it's increased, it will only flow more in proportion to dP^.5, so the valve shouldn't be able to flow 520 GPM unless the dP was on the order of 405,000 psi. So yea, there's something wrong here.
On the other hand, a 4" valve should easily be able to flow 520 GPM. For a 4" valve at 150 psi, 10 GPM sounds like a leakage rate, not a wide open flow rate. I've not heard of a valve made by FORD. Perhaps you can locate their web site and point to the valve you're using, it may help.
|Mar7-08, 03:35 PM||#5|
Apologies, I am talking about a Ford Pressure Regulating Valve as seen here.
They come factory set at 10 USGPM based on the specified PSI value you want, in this case 150 psig.
What I am finding difficult to calculate is that in our system we flow up to 520 usgpm, at XXX pressure based on that setting. Our system is an underground mine where we use the PRV valves to regulate pressure as the water descends down the mine shaft.
I have historical print that states that a flow of 520 USGPM through the valve will yield a pressure drop down to 99 psig. There are other similar calculations as well, but no one (including the manufacturer) seems to be able to tell me how this can be calculated for your desired flow rate.
I am currently looking at retrofitting an installation which will see flows at 400 USGPM and I want to know what the pressure will be based on a factory setting of 150psig @ 10 usgpm. There seems to have been some form of calculation done in the past, but I do not know what it might be.
|Mar7-08, 08:28 PM||#6|
That clears it all up. I'm looking at the cut-away here. This is a pressure reducing valve or pressure regulator. The configuration is for a balanced valve because the inlet sliding seal (#10, Cup Washer) has the same diameter as the seat, #12. The seat also takes the place of a diaphragm or piston in a conventional (under seat) pressure regulator. This valve has flow over the seat.
What you’re asking to calculate is called “droop”. Note that the flow of 10 GPM is just shy of “lockup”. Lockup is the pressure at which there is no flow whatsoever, and that pressure is going to be higher than this 150 psi set pressure. I suspect the 10 GPM is set at 150 psig simply for convenience. That is, the manufacturer knows the lockup is not particularly consistant from valve to valve, and may vary considerably, say from 155 psig to 160 or 170 psig. So rather than say the regulator is set at 155 or 165 psig, they “set” the valve at this very small, 10 GPM flowrate at 150 psig. This gives you a more consistant pressure response. If they went with the lockup pressure, the valve flow rate may vary considerably over the pressure range. Note that this is a fairly inexpensive valve, cheaply made, so you probably can’t expect tremendous accuracy in flow and opening capability.
You could calculate flow through this valve using basic principals. You'd have to do a force balance on the poppet in order to determine valve lift for any given set of inlet and outlet pressure. I've generally done this by setting up a spreadsheet. Works fairly well, but it might give you a headache just to read it all. If you'd like, I can explain how to do that.
The other way is to make a simplifying assumption if you have two data points to go by and it looks as if you do. Assuming valve lift is fairly small, and I think that's a safe bet, then the Cv of the valve is linear between two given set pressures. You have these two, 10 GPM at 150 psi and 520 GPM at 99 psi. You didn't give the inlet pressure but you'll need that also, and I'm assuming it is the same for both cases. For liquid flow,
Q = Cv (( Pi - Po ) / G )^.5
Q = Flow (GPM)
Cv = flow coefficient
Pi = inlet pressure (psi)
Po = outlet pressure (psi)
G = specific gravity with respect to water
Use your known inlet and outlet pressure and flow rate to determine Cv for your 2 cases. Cv will vary linearly between the two outlet pressures. For example, let's say your inlet pressure is 300 psi. Cv at 150 is .812 and Cv at 99 it is 36.5. Those are your 2 points, and Cv varies linearly between these two outlet pressures. Now find the Cv and Po somewhere on this line which correspond to the flow you want (400 GPM).
If your two cases have different inlet pressure, make sure you put that in to determine Cv for each of the two different cases. You'll still make the assumption that Cv varies linearly depending on drop in outlet pressure.
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