calculating water pressure

[KurskKnyaz]

I live in an old building and I'm installing a washing machine. The machine will be hooked up to the hot and cold lines that run to my kitchen faucet (not the faucet itself.

The machine requires a water pressure between 30 and 120 psi. The difference between the hot and cold lines cannot be more than 10 psi I need to be sure that I have the right pressure even when the kitchen faucet is on and my mom is doing the dishes ( I know I'm an asshole, but help me anyway).

How do I calculate the water pressure coming into the machine based on how quickly given volume fills up? Basically what I'm going to do is fill up a container to a certain volume and time how long it takes to fill up.

I will know (1) the volume of water i got and (2) the time it took to fill that volume.

What is the equation to determine the water pressure based on these two variables. Thank you for your help (I sold my physics textbooks on eBay).

 

[esun]

I don't believe you can determine it from just those two parameters. If I were to take the same faucet with the same water flow and attach two differently sized hoses on the end, the water pressure would be different depending on the size of the hose (and independent of the rate of water flow). It must be defined for some standard faucet size (i.e. some cross-sectional area) I would think.

 

[KurskKnyaz]

one more variable, hose diameter is 5/16"

in can give you:

hose diameter = 5/16"
total volume
time to achieve that volume

 

[esun]

Hmm, looks like this problem is more complex than I'd guessed. Here are some Google links that may help you:

http://www.askthebuilder.com/B...er_Pressure_Loss.shtml

http://www.engineeringtoolbox....liams-water-d_797.html

http://www.efunda.com/formulae...calc_pipe_friction.cfm

 

[Veramocor]

esun, he doesn't need to calculate the loss across the whole piping system, only the pressure at where he'll be extracting the water. He needs to use the 'orifice equation'. http://www.mcnallyinstitute.com/13-html/13-12.htm

He know the flow rate in gallons/min and he knows the area of the orifice. He can calculate out the ft/s of water speed. then he plugs it into the orifice equation. then he can get the head of the water/dive by 2.31 and he ends up with pressure in psi.

 

[silverpig]

Originally posted by: Veramocor
esun, he doesn't need to calculate the loss across the whole piping system, only the pressure at where he'll be extracting the water. He needs to use the 'orifice equation'. http://www.mcnallyinstitute.com/13-html/13-12.htm

He know the flow rate in gallons/min and he knows the area of the orifice. He can calculate out the ft/s of water speed. then he plugs it into the orifice equation. then he can get the head of the water/dive by 2.31 and he ends up with pressure in psi.

Heh. Orifice.

 

[skull]

Here is the easy way to do this. Get one of those gauges that read refrigerant pressure in cars available cheaply at any auto parts store. Cut an old washer hose in half attach hose to gauge. Turn faucet on upstairs attach hose to water line turn it on you are now reading your water pressure with the faucet on. You might be able to buy a water pressure gauge for cheap. I don't know as I always had a spare gauge laying around so I just made my own. In reality I wouldn't even worry about it I'm sure your water pressure is fine.

 

[KurskKnyaz]

Skull, the building I live in is 90 years old and i'm on the 4th floor. You wouldn't even believe the shit i have to worry about

 

[KurskKnyaz]

OK, so here's the physics word problem:

It takes KurskKnyaz 8 seconds to fill up a 1.8 Liter container with water flowing from a hose that has a 5/16" diameter. What is the water pressure?

 

[KurskKnyaz]

Originally posted by: Veramocor
esun, he doesn't need to calculate the loss across the whole piping system, only the pressure at where he'll be extracting the water. He needs to use the 'orifice equation'. http://www.mcnallyinstitute.com/13-html/13-12.htm

He know the flow rate in gallons/min and he knows the area of the orifice. He can calculate out the ft/s of water speed. then he plugs it into the orifice equation. then he can get the head of the water/dive by 2.31 and he ends up with pressure in psi.

I have no idea what is going on there:

I know that Pressure = force/ diameter

How do i figure out the force based on the mass of fluid i move per second?

 

[PlasmaBomb]

Originally posted by: KurskKnyaz
OK, so here's the physics word problem:

It takes KurskKnyaz 8 seconds to fill up a 1.8 Liter container with water flowing from a hose that has a 5/16" diameter. What is the water pressure?

Originally posted by: KurskKnyaz

Originally posted by: Veramocor
esun, he doesn't need to calculate the loss across the whole piping system, only the pressure at where he'll be extracting the water. He needs to use the 'orifice equation'. http://www.mcnallyinstitute.com/13-html/13-12.htm

He know the flow rate in gallons/min and he knows the area of the orifice. He can calculate out the ft/s of water speed. then he plugs it into the orifice equation. then he can get the head of the water/dive by 2.31 and he ends up with pressure in psi.

I have no idea what is going on there:

I know that Pressure = force/ diameter

How do i figure out the force based on the mass of fluid i move per second?

The first thing to do is convert the litres to cubic feet. 1 litre ~= 1 decimeter cubed.

1.8 decimeters cubed = 0.0635664 cubic feet in 8 seconds.

Now convert that to cubic feet per second (Q) = 0.0079458

Q= AV

Area = Pi*r^2

A = 3.14159*0.013*0.013

We know Q and A solve for V.

V = Q/AK

Assuming K is 1 for the minute...

V = 0.0079/0.0005

V = 14.918 ft/sec

Knowing that we can work out the head

h = V^2/2g

g = 32.2 ft/sec^2

h = 14.918^2/64.4

h = 3.4557

Pressure = head x specific gravity / 2.31

The specific gravity of water varies depending on temp. but is ~62.28 lb per cubic foot @ 20 Celsius.

P = 3.4557*62.28/2.31
P = 93.16 psi

Edit: For simplicities sake if you time the test again with the kitchen faucet running there is a quicker way of determining pressure -

(Original time/ New time) x 93.16 = New pressure

Since the other variables remain constant.

 

[Paperdoc]

With the data you propose, you cannot predict the pressure under projected operating conditions. You could use Skull's suggestion and do an actual measurement with a gauge. Two things to keep in mind doing this:
1. Connect your pressure gauge AFTER the kitchen faucet. For example, you might need to install a tee and small length of pipe, ending in a shut-off (gauge isolation) valve and a threaded nipple in which to mount the gauge. Do this on EACH of the hot and cold lines.
2. On each of these hot and cold sample locations separately, measure the water pressure with the kitchen tap closed, and again when fully open. At this point you'll know whether or not the pressure on each line meets the first criterion: must be 30 to 120 psi at all times.

But you have another criterion: pressures of hot and cold must be within 10 psi of each other under all conditions. The measurements above MAY answer that, but maybe not. The only way to guarantee that match would be to buy and install a water pressure regulator valve in each line, and then use your gauge to set them to some value (say, 60 psi). Unfortunately, such pressure regulators are not cheap - check that out yourself, but you'll need to know the expected flow rate to choose the valve size - so this is probably impractical.

If you're actually concerned about this matching of pressures, there is a cheaper way you could plan to meet the requirement. Where you install the water lines and outlet valves for the washer, add a bit more piping components. Just ahead of the outlet valves you connect hoses to, place another in-line valve in each line. Then, between this and the outlet valve, install a tee, gauge isolation valve and gauge fitting. Once it's all running, put a gauge in one line's fitting and run the washer. With both waters flowing into the machine, measure the real pressures on each line. Go to the one with higher pressure under load and close down the extra throttling valve until the pressure is reduced to match the pressure in the other line. It may not match exactly, but you should be able to get a match within 10 psi under all real conditions. Basically, what this does is establish in each line a variable orfice (the throttling valve) that you can control. Of course, if your first tests demonstrate that you already know which line will have higher pressure always, just mount the throttling valve in that line only.

 

[hanoverphist]

cant you ask the bldg maint guys what the average pressure range is on the 4th floor? ive never seen a system that would run up 4 floors without boosters, unless they had a roof tank. if its boosted, they will have regulators by floor that maintain normal pressure, which should be 35-45 psi. if its a roof tank (seriously doubt it) then your pressure would be decent either way, especially if it were a tall bldg.

 

[KurskKnyaz]

I tried messing around with equations. I basically converter the volume water moved into a cylindrical mass with the diameter of the pipe, calculated its weight based on density. I then squared something and came to ~60 psi. I give up. This is taking too much time and the results don't matter anyhow since I can't do anything to change the pressure. But thank you all for your help. If I knew how fast the water was accelerating i would be able to calculate the pressure.

 

[Paperdoc]

The calculations for what you want are a good deal more complicated than F=ma, etc. We're into fluid dynamics here with the Bernoulli Equation and orfice sizes, etc., most of which you cannot measure anyway. Actual field measurements under real operating conditions will give you better answers than a theoretical calculation. You are right not to spend a lot of time on this, especially if you feel you cannot use the answers to change anything and optimize the system.

 

[KurskKnyaz]

Calculating it was fun. This is what I did:

1. converted 1.8L of water into a cylinder who's diameter is equivalent to the diameter of the pipe.
2. From this, I determined the length of the cylinder.
3. Then I determined the mass of the water from the volume and density.
4. I attempted to determine how much force would be needed to move that mass of water the distance of the cylinders length. However, I only know the velocity and not the acceleration so I can't get the force; only the momentum.
5. Then I can divide the force by the area of the pipe opening and i get the PSI.

I understand that this is only my theoretical pressure but it should give me an idea of what ballpark I'm in.

If someone can tell me how I can calculate the force that a body of a given mass mass with an given momentum exerts, I would be at peace.

Thank you all for your help. Those "head" equations are very strange and I don't think they are applicable to my situation. But thank you anyway.

BTW can I determine the acceleration of water somehow?

 

[KurskKnyaz]

(Original time/ New time) x 93.16 = New pressure

Since the other variables remain constant.

That would mean my pipes are pushing about 10 PSI. That doesn't sound realistic.

 

[KurskKnyaz]

OK this is my theoretical:

Pipe diameter = 0.3125in = 0.79375cm
Pipe orifice area = 0.0767in^2
Volume of water = 1.8L = 1800cm^3
Mass of water = (1800cm^3) X (1g / 0.998cm^3) = 1803g = 3.9749lb

Distance water traveled = height of cylinder (h) where base is pipe diameter
Volume of Cylinder = Pi X R^2 X h
1803cm^3 = pi X (0.79375cm / 2)^2 X h
h = 3643.6756cm = 119.543ft

Time to fill container with 1.8L = time to travel (h) = 8 seconds
Velocity = 119.543ft / 8s= 14.943ft/s

Momentum = Mass X Velocity

Force = (Final Momentum - Initial Momentum ) / (Final time - Initial time)
Force = (3.9749lb X 14.943ft/s - 0) / (8s - 0)
Force = 7.4246lb x ft / s^2
Force = 7.4246lbf (pound-force)

PSI = 1lbf / in^2
PSI = 7.4246lbf / 0.0767in^2
PSI = 98.8

 

[PlasmaBomb]

Originally posted by: KurskKnyaz

(Original time/ New time) x 93.16 = New pressure

Since the other variables remain constant.

That would mean my pipes are pushing about 10 PSI. That doesn't sound realistic.

When the taps are on how long does it take to fill the 1.8 l container?

@ 10 PSI it would take 75-80 seconds to fill.

 

[KurskKnyaz]

It takes me 8 sec to fill 1.8L that's why I think those equations are either wrong or for something else.

 

[KurskKnyaz]

.....Anyway I got a pressure gauge at Home Depot which I just returned. My theoretical pressure was 98.8 PSI but my experimental is.....drum roll please.........29psi-30psi

That is a huge % of error. Any ideas where I went wrong on the calculations?

 

[PlasmaBomb]

Originally posted by: KurskKnyaz
It takes me 8 sec to fill 1.8L that's why I think those equations are either wrong or for something else.

The pressure in the pipes was 93 ish by the equations posted (give or take some rounding and assumptions about water temp and that the aperture is perfect). So a broad agreement with your 98 psi.

One thing that is wrong with your equation is that 1800 cc of water doesn't weigh 1803 g - it weighs 1796.4 g

Which drops your pressure to 96.1 PSI.

I have no idea why there is such a large discrepancy

 

[Paperdoc]

Now we're getting somewhere. Instead of trying to derive a mathematical model (and let's not forget that all models need verification by comparison of predictions with reality), you have a real pressure measurement. No details provided - was that value on one line or both? - and was the water running into the kitchen sink at the time, of a no-flow condition?. But we have real-world direct measurement for one criterion: line pressure must be 30 to 120 psi. Looks like you just make it. The other objective, a pressure match within 10 psi, we have no measured value for yet. But I would speculate that, if one line as only 30 psi, the other can't be different by more than 33%, so you're probably OK.

So, why do the washer makers provide these specs? I can speculate thusly: in any of the tub fill operations (initial or rinse cycle), the filling is stopped by a water level sensor. But often it also is limited by the fact that the timer has a fixed time slice during which fill can happen. It just normally is completed (sensed by level) before that time slot finishes. If you have exceptionally low water pressure there is a chance the tub filling might not complete before the timer advances. There's also the matter of spray rinse operation. With low water pressure you might not get enough water flow in that phase to accomplish all the rinsing needed. As for the pressure match, that has to do with the temperature of the Medium temp wash and rinse operations. On our machine, "Hot" supplies only water from the hot line, "Cold" only form cold, and "Medium" from both lines simultaneously. There is no control of each line's flow beyond ON or OFF with solenoid valves. So the actual balance of water flow rates (and hence temperature) is entirely dependent on the balance of supply pressures.

 

[KurskKnyaz]

Originally posted by: Paperdoc
Now we're getting somewhere. Instead of trying to derive a mathematical model (and let's not forget that all models need verification by comparison of predictions with reality), you have a real pressure measurement. No details provided - was that value on one line or both? - and was the water running into the kitchen sink at the time, of a no-flow condition?. But we have real-world direct measurement for one criterion: line pressure must be 30 to 120 psi. Looks like you just make it. The other objective, a pressure match within 10 psi, we have no measured value for yet. But I would speculate that, if one line as only 30 psi, the other can't be different by more than 33%, so you're probably OK.

The kitchen faucet was NOT running. When it is the pressure drops to about 10 psi. I tested each line separately and got the same results so I did just make it.

So, why do the washer makers provide these specs? I can speculate thusly: in any of the tub fill operations (initial or rinse cycle), the filling is stopped by a water level sensor. But often it also is limited by the fact that the timer has a fixed time slice during which fill can happen. It just normally is completed (sensed by level) before that time slot finishes. If you have exceptionally low water pressure there is a chance the tub filling might not complete before the timer advances. There's also the matter of spray rinse operation. With low water pressure you might not get enough water flow in that phase to accomplish all the rinsing needed. As for the pressure match, that has to do with the temperature of the Medium temp wash and rinse operations. On our machine, "Hot" supplies only water from the hot line, "Cold" only form cold, and "Medium" from both lines simultaneously. There is no control of each line's flow beyond ON or OFF with solenoid valves. So the actual balance of water flow rates (and hence temperature) is entirely dependent on the balance of supply pressures.

Yup, this is why i listen to the manual and the mannufacture support people rather than the hardware store guy.

 

[heyheybooboo]

Originally posted by: Paperdoc
With the data you propose, you cannot predict the pressure under projected operating conditions. You could use Skull's suggestion and do an actual measurement with a gauge. Two things to keep in mind doing this:
1. Connect your pressure gauge AFTER the kitchen faucet. For example, you might need to install a tee and small length of pipe, ending in a shut-off (gauge isolation) valve and a threaded nipple in which to mount the gauge. Do this on EACH of the hot and cold lines.
2. On each of these hot and cold sample locations separately, measure the water pressure with the kitchen tap closed, and again when fully open. At this point you'll know whether or not the pressure on each line meets the first criterion: must be 30 to 120 psi at all times.

But you have another criterion: pressures of hot and cold must be within 10 psi of each other under all conditions. The measurements above MAY answer that, but maybe not. The only way to guarantee that match would be to buy and install a water pressure regulator valve in each line, and then use your gauge to set them to some value (say, 60 psi). Unfortunately, such pressure regulators are not cheap - check that out yourself, but you'll need to know the expected flow rate to choose the valve size - so this is probably impractical.

If you're actually concerned about this matching of pressures, there is a cheaper way you could plan to meet the requirement. Where you install the water lines and outlet valves for the washer, add a bit more piping components. Just ahead of the outlet valves you connect hoses to, place another in-line valve in each line. Then, between this and the outlet valve, install a tee, gauge isolation valve and gauge fitting. Once it's all running, put a gauge in one line's fitting and run the washer. With both waters flowing into the machine, measure the real pressures on each line. Go to the one with higher pressure under load and close down the extra throttling valve until the pressure is reduced to match the pressure in the other line. It may not match exactly, but you should be able to get a match within 10 psi under all real conditions. Basically, what this does is establish in each line a variable orfice (the throttling valve) that you can control. Of course, if your first tests demonstrate that you already know which line will have higher pressure always, just mount the throttling valve in that line only.

Pretty much this.

The discrepancy you are seeing is the drop from static to residual pressure.