Hi all. I design large septic systems, and we have an application where I am trying to "dose" several tanks an equal rate, or as close to equal as possible. There will be one tank with a pump, a ~1" sch 40 pvc pipe exiting that pump and out of the tank that leads to a ball valve, and after the ball valve, splits off to each tank. The pipe will then go into a tank, and a cap will be placed at the end. The contractor will then drill a ~5/16" inch hole into that cap. This essentially puts me in a position where the flow into each tank will be virtually identical, so all the contractor has to do is adjust the pressure to what I specify and set the timer of the pump (x times per day for x minutes each time). I've included a very rough image of that here. My question is, how the heck to I determine flow out of this 5/16" hole? My original thought was thought I could add an imaginary ~1" stick of pipe at 5/16" diameter in my calculations to determine head loss, but it's looking like that may not work. I'm also not quite sure how to account for multiple streams going into multiple tanks at once, not just 1 orifice (find out overall Q, and divide that by the numbers of orifice's?). I found articles like this and this calculator that specifically mentions orifice. I have no clue how to obtain the pressure difference across the orifice. Would any of y'all have any idea have this is done? Thanks a bunch.
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Calculating flowrate out of orifice 1
- Thread starter AHickman
- Start date
-
1
- #2
LittleInch
Petroleum
AHickman,
What you're looking at is a header and branch flow distribution issue.
The usual ROT is that the header (your 1" pipe) should be a short as possible and also the sum of the area of each branch should equal the header area. At 5/16" you're at about max 10 exits off the 1" header.
If you need exactly equal flow into each tank you're better off installing a set of needle valves and start with each at the same opening and then measure flow at each location and adjust as required, then lock it down....
But if you want to do it by orifice then your pressure is going to be based on flow as your d/s pressure is 0 psig. You will just need to play around with flow to find out the pressure needed.
This is where it gets complicated though as this would need to match exactly your pump.
Hence why some sort of variable flow valve is much better. IMHO.
so how did you light on 5/16"?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
What you're looking at is a header and branch flow distribution issue.
The usual ROT is that the header (your 1" pipe) should be a short as possible and also the sum of the area of each branch should equal the header area. At 5/16" you're at about max 10 exits off the 1" header.
If you need exactly equal flow into each tank you're better off installing a set of needle valves and start with each at the same opening and then measure flow at each location and adjust as required, then lock it down....
But if you want to do it by orifice then your pressure is going to be based on flow as your d/s pressure is 0 psig. You will just need to play around with flow to find out the pressure needed.
This is where it gets complicated though as this would need to match exactly your pump.
Hence why some sort of variable flow valve is much better. IMHO.
so how did you light on 5/16"?
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #3
LittleInch,
Thanks for the response! I'll be honest though, a lot of this is going over my head, so I hope it's okay if I ask a couple question:
1 - What does ROT mean? I'm sure is an acronym for something obvious, but it's my first time reading that
2 - And by header, this would be the length of pipe before it splits, correct? Why does it need to be as short as possible?
3 - Would you have any recommendation on what needle valve exactly? The wastewater will be pretreated (aerobic), but post settled. Basically, it will still be contaminated, but there wont be any large solids. Will this be a problem for the needle valves? I'm worried that they would get clogged fairly quickly.
4 - By d/s, I assume you mean the differential pressure? I'm assuming I'm wrong, because that would make my Q = 0.
5 - No reason I specified 1" and 5/16". My contractor told me he drills a 5/16", so I was going to run with that until I understood the math and could spec out a size that may be more appropriate, for both the supply line and the orifice sizing.
Thanks for the response! I'll be honest though, a lot of this is going over my head, so I hope it's okay if I ask a couple question:
1 - What does ROT mean? I'm sure is an acronym for something obvious, but it's my first time reading that
2 - And by header, this would be the length of pipe before it splits, correct? Why does it need to be as short as possible?
3 - Would you have any recommendation on what needle valve exactly? The wastewater will be pretreated (aerobic), but post settled. Basically, it will still be contaminated, but there wont be any large solids. Will this be a problem for the needle valves? I'm worried that they would get clogged fairly quickly.
4 - By d/s, I assume you mean the differential pressure? I'm assuming I'm wrong, because that would make my Q = 0.
5 - No reason I specified 1" and 5/16". My contractor told me he drills a 5/16", so I was going to run with that until I understood the math and could spec out a size that may be more appropriate, for both the supply line and the orifice sizing.
LittleInch
Petroleum
Sorry - Rule Of Thumb - basically a rough guide which mostly works
Yes the pipe before it splits. As short as you can to reduce the impact of frictional losses along it. If you have say two tanks 1m apart then 20m before the next tank, that 20m will impact the pressure at the branch and hence have less flow into that tank despite the same branch size or orifice.
Ah - that's not so good - but the same issue will come up with an orifice? You might be better with a globe valve. One problem with a hole type is that the hole should be drilled in the bottom of the cap or orifice plate so that you don't end up with half a header full of dirt...
D/s is Downstream
5/16" could easily work, but getting the maths to work might not be so easy. Your pump will need to be fairly low pressure. What sort of pressure drop are you getting for whatever your required flowrate is ( you don't specify the flow rate into the tanks??
Another alternative might be to have a short ish header and fit a set of equal length smaller hoses / tubes ( say 1/4" or 1/2"?) coming off the bottom of the header and feed them into each tank. Even if you need to coil some up you should get nearly equal flow. At least you could clear them out more easily if they get blocked. If you want more flow just shorten all the hoses the same amount. Less flow add some hose.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Yes the pipe before it splits. As short as you can to reduce the impact of frictional losses along it. If you have say two tanks 1m apart then 20m before the next tank, that 20m will impact the pressure at the branch and hence have less flow into that tank despite the same branch size or orifice.
Ah - that's not so good - but the same issue will come up with an orifice? You might be better with a globe valve. One problem with a hole type is that the hole should be drilled in the bottom of the cap or orifice plate so that you don't end up with half a header full of dirt...
D/s is Downstream
5/16" could easily work, but getting the maths to work might not be so easy. Your pump will need to be fairly low pressure. What sort of pressure drop are you getting for whatever your required flowrate is ( you don't specify the flow rate into the tanks??
Another alternative might be to have a short ish header and fit a set of equal length smaller hoses / tubes ( say 1/4" or 1/2"?) coming off the bottom of the header and feed them into each tank. Even if you need to coil some up you should get nearly equal flow. At least you could clear them out more easily if they get blocked. If you want more flow just shorten all the hoses the same amount. Less flow add some hose.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
EdStainless
Materials
When I have set up similar systems I started by drilling smaller holes, maybe 3/16".
Holes are easy to make larger.
Then run the system with plain water and see what flow I get.
You can use header/branch and orifice plate calculations, but you will only end up with approximate answers.
= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
Holes are easy to make larger.
Then run the system with plain water and see what flow I get.
You can use header/branch and orifice plate calculations, but you will only end up with approximate answers.
= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed
The flow through a liquid flow orifice is a pretty straight forward calculation. Attached is Cameron Hydraulic Data handbook equation for flow through an orifice. The calculation is for water but if your liquid specific gravity and viscosity is close to or does not vary significantly from water then it would give you similar results. Your case would resemble the square edge orifice with Cd = 0.61. So then the only remaining unknown in the equation is "h" which is the difential head across the orifice = head upstream minus head downstream = (pressure upstream minus pressure downstream)*144/62.4 for water with SG of 62.4 #/cu ft.
This is the little tricky part since you are in control of how much pressure is on the upstream side based your pump selection and how much head your selected pump outputs for a given flowrate.
So assuming that your 1" header pipe is size for a very negligible pressure drop and you continue the 1" size on each branch to each tank then you will be assured that the output pressure of the pump will equally equal the pressure upstream of each orifice if there is no pressure drop in the piping.
Then from the pump curve selected you can determine the head/pressure at the orifices. For your given total dousing flowrate you would select a pump with a pump curve which develops the required "h" at the discharge which without friction is the head required upstream of the orifice such that the head/pressure at the orifice inlet = pump output head at flow (if no pressure drop in piping). Then this pump head minus the head in the tank must equal the diffential head you used in the orifice sizing calculation.
So in esscence you size the orifice at a given diffential pressure to give the flow you require and then you pick a pump to give you the head output at that flowrate that when subtracting the downstream head in the tank (which is just the static liquid height in the tank) you get the differential head "h" you used to size the orifice. Visa Versa if you already have a existing pump with known head at flow you then size the orifice at that head (minus head in tank) to get the flow you require.
This is the little tricky part since you are in control of how much pressure is on the upstream side based your pump selection and how much head your selected pump outputs for a given flowrate.
So assuming that your 1" header pipe is size for a very negligible pressure drop and you continue the 1" size on each branch to each tank then you will be assured that the output pressure of the pump will equally equal the pressure upstream of each orifice if there is no pressure drop in the piping.
Then from the pump curve selected you can determine the head/pressure at the orifices. For your given total dousing flowrate you would select a pump with a pump curve which develops the required "h" at the discharge which without friction is the head required upstream of the orifice such that the head/pressure at the orifice inlet = pump output head at flow (if no pressure drop in piping). Then this pump head minus the head in the tank must equal the diffential head you used in the orifice sizing calculation.
So in esscence you size the orifice at a given diffential pressure to give the flow you require and then you pick a pump to give you the head output at that flowrate that when subtracting the downstream head in the tank (which is just the static liquid height in the tank) you get the differential head "h" you used to size the orifice. Visa Versa if you already have a existing pump with known head at flow you then size the orifice at that head (minus head in tank) to get the flow you require.
This is like the Aqilibrium competitions we used to run at high school science fairs if everything is equal after the split flow would be equal. same size diameter pipes, same lengths of pipe, same elevations etc
you could counteract the additional pipe losses for the longer pipe runs (to the tanks further away) by using sections of larger pipe.
If you are doing this as a mental challenge go ahead, if you just want it to work then with your 1" pipes you could probably just throttle the flow in each pipe with a 1/4-turn valve and just mark the position it should stay at. I have seen this done in irrigation settings. Setting would be different on each branch
you could counteract the additional pipe losses for the longer pipe runs (to the tanks further away) by using sections of larger pipe.
If you are doing this as a mental challenge go ahead, if you just want it to work then with your 1" pipes you could probably just throttle the flow in each pipe with a 1/4-turn valve and just mark the position it should stay at. I have seen this done in irrigation settings. Setting would be different on each branch
- Thread starter
- #8
LittleInch - I'm trying to get 5 GPM per tank. I have 2 tanks, so 10 GPM overall. I will have a tee at the tom of the pump, with a ball valve and a return directly back to the tank, that way the contractor can increase/decrease the flow back into the tank to alter the pressure in the line.
Snickster - Awesome image, thanks for sharing that. This will be post-settled wastewater, so the characteristics would be virtually identical. I don't quite understand the differences between the 2 square edged, would you mind explaining why I would use the c 0.61 square edged and not the other? Honestly, all of these seem somewhat viable, minus sharp edge and well rounded. I'm just trying to understand a little better the differences.
Yeah, the differential pressure has been my point of confusion. I assumed head upstream will be whatever pressure I have at the orifice, which I will determine myself, so this would be the variable that I would use to determine my Q. My downstream pressure, I assume, would be 0?
I will use the setup described above, so I can easily specify "20 psi at the orifice" or whatever is needed and the contractor can easily obtain that by allowing more of less wastewater to pump back into the pump tank.
Yeah, the main head loss will come from pumping out of the tank, which will be around 7' or 3 psi.
Seems like the formula gets a little trickier (from an algebraic perspective obviously) if the orifice is larger, so keeping it smaller simplifies things. Just curious though, any idea why the formula changes based on d1/d2? Doesn't seems like sqrt(1/(1-....)) really changes much at all regardless. I've solved for h, and I end up with around 9, and I assume "feet of liquid" is just total head in feet. If I had 9 feet of head at that orifice, this pump would end up producing about 105 GPM, and I need 10. Really, I need to have about 19 feet of head at the orifice, so that means I need to change my orifice size. I am starting to see how this is a sliding scale, or multiple sliding scales! I think the solution here is to keep resizing the orifice until I end up with 10 GPM at ~19' of head. Idk, this is making my head spin a little and I've been at this all day, so I might need to rest my brain on this overnight lol. I've included my calcs in the attached excel sheet if you're interested in how I came up with it
Thank y'all so much for the help by the way, this is incredible! I would be lost otherwise.
Edit: I've edited this so many times now, but I can't stop thinking about this. I'm losing about 7' of head to the orifice currently, but most of that is loss pumping out of the tank, and my starting point is basically right past that point. There's also virtually 0 loss to the orifice, so I can assume my only loss is that at the orifice. If I were to put a pressure gauge directly before the orifice, it would need to read 9 feet (I should convert this to psi, but I'm just keeping it simple for now). So all the contractor would have to do is adjust the main ball valve that controls how much flow goes back into the tank into the gauge reads 9'. I think I've cracked the case! What do you think?
Snickster - Awesome image, thanks for sharing that. This will be post-settled wastewater, so the characteristics would be virtually identical. I don't quite understand the differences between the 2 square edged, would you mind explaining why I would use the c 0.61 square edged and not the other? Honestly, all of these seem somewhat viable, minus sharp edge and well rounded. I'm just trying to understand a little better the differences.
Yeah, the differential pressure has been my point of confusion. I assumed head upstream will be whatever pressure I have at the orifice, which I will determine myself, so this would be the variable that I would use to determine my Q. My downstream pressure, I assume, would be 0?
I will use the setup described above, so I can easily specify "20 psi at the orifice" or whatever is needed and the contractor can easily obtain that by allowing more of less wastewater to pump back into the pump tank.
Yeah, the main head loss will come from pumping out of the tank, which will be around 7' or 3 psi.
Seems like the formula gets a little trickier (from an algebraic perspective obviously) if the orifice is larger, so keeping it smaller simplifies things. Just curious though, any idea why the formula changes based on d1/d2? Doesn't seems like sqrt(1/(1-....)) really changes much at all regardless. I've solved for h, and I end up with around 9, and I assume "feet of liquid" is just total head in feet. If I had 9 feet of head at that orifice, this pump would end up producing about 105 GPM, and I need 10. Really, I need to have about 19 feet of head at the orifice, so that means I need to change my orifice size. I am starting to see how this is a sliding scale, or multiple sliding scales! I think the solution here is to keep resizing the orifice until I end up with 10 GPM at ~19' of head. Idk, this is making my head spin a little and I've been at this all day, so I might need to rest my brain on this overnight lol. I've included my calcs in the attached excel sheet if you're interested in how I came up with it
Thank y'all so much for the help by the way, this is incredible! I would be lost otherwise.
Edit: I've edited this so many times now, but I can't stop thinking about this. I'm losing about 7' of head to the orifice currently, but most of that is loss pumping out of the tank, and my starting point is basically right past that point. There's also virtually 0 loss to the orifice, so I can assume my only loss is that at the orifice. If I were to put a pressure gauge directly before the orifice, it would need to read 9 feet (I should convert this to psi, but I'm just keeping it simple for now). So all the contractor would have to do is adjust the main ball valve that controls how much flow goes back into the tank into the gauge reads 9'. I think I've cracked the case! What do you think?
- Thread starter
- #9
Tired of editing the previous post, I made a rookie mistake and screwed up the math. Turns out I need 32 psi, or 73' of head at the orifice. That is quite a lot. There's no way I could produce that, so I'd need to increase my orifice size. Increasing it to 1/2" makes a HUGE difference. Updated calculation attached.
Compositepro
Chemical
Constant flow valves use the principle of using a pressure regulator to maintain a constant pressure differential across an orifice. You can do the same by using a dome loaded pressure regulator before each of your orifices and one pneumatic pressure regulator.
LittleInch
Petroleum
Honestly, you're complicating this beyond all requirements.
I would buy a PD metering pump with variable speed or stroke and set it at 10 GPM and then split it into two equal branches in length, height , bends etc and you should be fixed. If you need to add some sort of device to adjust flow, but equal resistance should get you to within 5% equal flow.
An AODD or similar dirty water complaint PD pump would do it.
Any orifice where you have more than about 5 psi difference will result in a high velocity flow and end up needing to be quite small which could more easily clog up.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
I would buy a PD metering pump with variable speed or stroke and set it at 10 GPM and then split it into two equal branches in length, height , bends etc and you should be fixed. If you need to add some sort of device to adjust flow, but equal resistance should get you to within 5% equal flow.
An AODD or similar dirty water complaint PD pump would do it.
Any orifice where you have more than about 5 psi difference will result in a high velocity flow and end up needing to be quite small which could more easily clog up.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #12
LittleInch
Petroleum
You probably need something a bit more robust for "dirty water". Try looking up AODD or even electrically driven.
Maybe peristaltic pump?
here's info for you
Diaphragm pumps are often air operated, but do come in electric plunger form as well
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Maybe peristaltic pump?
here's info for you
Diaphragm pumps are often air operated, but do come in electric plunger form as well
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Electric operated diaphragm metering pumps are typically used for chemical dousing in the waste water treatment industry. You can get high pressure with relatively low flows with constant flow output regardless of the variation in pressure. Then you would not use orifice arrangement but just pipe to each tank wirh equal runs. The centrifugal pump arrangement with orifices I have never seen in WWT. Not only do varitions in pressure occur and therefore flow but if fluids are shear sensitive then centrifugal pumps would be the worst to use. PD metering pumps pump provide constant flow regardless of pressures and do not shear the fluid.
Here is link to PD metering pump manufacturer:
Here is link to PD metering pump manufacturer:
Here is response to your questions from previous post. If you really want to go with the centrifugal pump arrangement with orifices. Just a question of how precise you want to be with the dousing. But now you say you are not dousing but pumping post-settled wastewater?
Snickster - Awesome image, thanks for sharing that. This will be post-settled wastewater, so the characteristics would be virtually identical. I don't quite understand the differences between the 2 square edged, would you mind explaining why I would use the c 0.61 square edged and not the other? Honestly, all of these seem somewhat viable, minus sharp edge and well rounded. I'm just trying to understand a little better the differences.
I said the square edge Cd 0.61 because if you drill a hole in a cap on the end of a PCV pipe then what you will get will mostly resemble a square edge orifice. Now if you purchase an orifice and then install in the line then you can purchance any type you like.
Yeah, the differential pressure has been my point of confusion. I assumed head upstream will be whatever pressure I have at the orifice, which I will determine myself, so this would be the variable that I would use to determine my Q. My downstream pressure, I assume, would be 0?
Your pressure at the orifice upstream will be your pump discharge pressure minus friction loss in the piping. If the pipe size is 1" throughout the combined header flowing 10 gpm will have a pressure drop of 4 psi per 100 ft. So if less than 10 feet will be less than 0.4 psi which is negligible. When flow splits in two 1" branches the pressure drop will really be negligible. Therefore using 1" pipe you can consider the pressure upstream of your orifices the same as the discharge of the pump at given flowrate (assuming your net suction loss is negligible = suction tank head minus suction friction losses).
Your downstream pressure will be based on the liquid level in the tank you are pumping into. So the differential head used in the orifice equation is pump head minus tank liquid level.
I will use the setup described above, so I can easily specify "20 psi at the orifice" or whatever is needed and the contractor can easily obtain that by allowing more of less wastewater to pump back into the pump tank.
I am not sure but it seems that you are saying you will adjust the pressure by recirculation some flow back to the suction as necessary? A centrifugal pump does not work like that. If you recirculate some flow then you will not have it availabe to flow to the tank unless you oversize your pump for the recirculated flow.
Yeah, the main head loss will come from pumping out of the tank, which will be around 7' or 3 psi.
So you are saying there is 3 psi loss in the suction line? Seems high. You should have a 1 1/2" suction to reduce the suction head loss and limit suction velocity to below 3 ft/sec.
Seems like the formula gets a little trickier (from an algebraic perspective obviously) if the orifice is larger, so keeping it smaller simplifies things. Just curious though, any idea why the formula changes based on d1/d2? Doesn't seems like sqrt(1/(1-....)) really changes much at all regardless. I've solved for h, and I end up with around 9, and I assume "feet of liquid" is just total head in feet. If I had 9 feet of head at that orifice, this pump would end up producing about 105 GPM, and I need 10. Really, I need to have about 19 feet of head at the orifice, so that means I need to change my orifice size. I am starting to see how this is a sliding scale, or multiple sliding scales! I think the solution here is to keep resizing the orifice until I end up with 10 GPM at ~19' of head. Idk, this is making my head spin a little and I've been at this all day, so I might need to rest my brain on this overnight lol. I've included my calcs in the attached excel sheet if you're interested in how I came up with it
So you are saying that you already have a pump with a known pump curve that produces 105 gpm at 9 ft of discharge head and that the two 5/16" orifices produce 10 gpm at 9 ft upstream? Then you need to size the orifices so that the head at flow of the orifices is the same as the head at flow for the pump you have. That will be the operating point of the system. So if you already have the pump with known pump curve and you want 10 gpm flow then you would consider the discharge head of the pump equal to what the pump curve says it is a 10 gpm, and you will size the orifices based on this discharge head at 10 gpm. Actually the pump head shown on the pump curve is the differntial pressure between the pump suction and discharge so you have to add or subtract the suction pressure to the pump head output to get the discharge pressure.
Snickster - Awesome image, thanks for sharing that. This will be post-settled wastewater, so the characteristics would be virtually identical. I don't quite understand the differences between the 2 square edged, would you mind explaining why I would use the c 0.61 square edged and not the other? Honestly, all of these seem somewhat viable, minus sharp edge and well rounded. I'm just trying to understand a little better the differences.
I said the square edge Cd 0.61 because if you drill a hole in a cap on the end of a PCV pipe then what you will get will mostly resemble a square edge orifice. Now if you purchase an orifice and then install in the line then you can purchance any type you like.
Yeah, the differential pressure has been my point of confusion. I assumed head upstream will be whatever pressure I have at the orifice, which I will determine myself, so this would be the variable that I would use to determine my Q. My downstream pressure, I assume, would be 0?
Your pressure at the orifice upstream will be your pump discharge pressure minus friction loss in the piping. If the pipe size is 1" throughout the combined header flowing 10 gpm will have a pressure drop of 4 psi per 100 ft. So if less than 10 feet will be less than 0.4 psi which is negligible. When flow splits in two 1" branches the pressure drop will really be negligible. Therefore using 1" pipe you can consider the pressure upstream of your orifices the same as the discharge of the pump at given flowrate (assuming your net suction loss is negligible = suction tank head minus suction friction losses).
Your downstream pressure will be based on the liquid level in the tank you are pumping into. So the differential head used in the orifice equation is pump head minus tank liquid level.
I will use the setup described above, so I can easily specify "20 psi at the orifice" or whatever is needed and the contractor can easily obtain that by allowing more of less wastewater to pump back into the pump tank.
I am not sure but it seems that you are saying you will adjust the pressure by recirculation some flow back to the suction as necessary? A centrifugal pump does not work like that. If you recirculate some flow then you will not have it availabe to flow to the tank unless you oversize your pump for the recirculated flow.
Yeah, the main head loss will come from pumping out of the tank, which will be around 7' or 3 psi.
So you are saying there is 3 psi loss in the suction line? Seems high. You should have a 1 1/2" suction to reduce the suction head loss and limit suction velocity to below 3 ft/sec.
Seems like the formula gets a little trickier (from an algebraic perspective obviously) if the orifice is larger, so keeping it smaller simplifies things. Just curious though, any idea why the formula changes based on d1/d2? Doesn't seems like sqrt(1/(1-....)) really changes much at all regardless. I've solved for h, and I end up with around 9, and I assume "feet of liquid" is just total head in feet. If I had 9 feet of head at that orifice, this pump would end up producing about 105 GPM, and I need 10. Really, I need to have about 19 feet of head at the orifice, so that means I need to change my orifice size. I am starting to see how this is a sliding scale, or multiple sliding scales! I think the solution here is to keep resizing the orifice until I end up with 10 GPM at ~19' of head. Idk, this is making my head spin a little and I've been at this all day, so I might need to rest my brain on this overnight lol. I've included my calcs in the attached excel sheet if you're interested in how I came up with it
So you are saying that you already have a pump with a known pump curve that produces 105 gpm at 9 ft of discharge head and that the two 5/16" orifices produce 10 gpm at 9 ft upstream? Then you need to size the orifices so that the head at flow of the orifices is the same as the head at flow for the pump you have. That will be the operating point of the system. So if you already have the pump with known pump curve and you want 10 gpm flow then you would consider the discharge head of the pump equal to what the pump curve says it is a 10 gpm, and you will size the orifices based on this discharge head at 10 gpm. Actually the pump head shown on the pump curve is the differntial pressure between the pump suction and discharge so you have to add or subtract the suction pressure to the pump head output to get the discharge pressure.
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