How can I predict the flow rate for a real life water hose?

[Roy S Ramirez]

Hello!

I hope you are all doing well! I've been always intrigued by fluid mechanics and decided to try out a quick experiment with my water faucet. After using some equations I found online [1] [2], I plugged a hose to the faucet and let the water run. I filled a 1 quart bowl and recorded the time it took to fill it. Without closing the valve, the experiment was repeated several times.

After comparing the experimental to theoretical results, there was a huuuge discrepancy.

Could you please help me see where the problem is? I feel like this should be a very simple problem...

ref:
[1] https://engineering.stackexchange.c...o-calculate-flow-rate-of-water-through-a-pipe
[2] https://www.lmnoeng.com/moody.php

 

[Tom.G]

Really not my field, but is that pressure measurement made with water flowing thru the hose? Since you mention 'static' pressure, maybe not. Connect the pressure gauge as close as possible to the hose, definitely after the valve! I wouldn't be suprised at a 30% lower pressure.

 

[jrmichler]

The clue is that measured flow is far lower than theoretical flow. That, by itself, tells you that you missed something. The pressure in a city water system is only constant at the water tower. The pressure at your tap when the water is flowing is the pressure at the water tower minus all line losses between the tower and your tap. Most (almost all) of that line loss is in your house. There should be a pressure gauge near your water meter. Watch that when the valve is opened and closed.

Remove the hose and measure flow again. The total line loss is then the supply pressure. The hose just adds a little more line loss to the total system.

 

[Roy S Ramirez]

Really not my field, but is that pressure measurement made with water flowing thru the hose? Since you mention 'static' pressure, maybe not. Connect the pressure gauge as close as possible to the hose, definitely after the valve! I wouldn't be suprised at a 30% lower pressure.

Thanks @Tom.G yes that pressure measurement was made like shown in picture 1. The gauge is supposed to measure total gauge pressure, and since the fluid is static this should be the static pressure.

 

[Roy S Ramirez]

The clue is that measured flow is far lower than theoretical flow. That, by itself, tells you that you missed something. The pressure in a city water system is only constant at the water tower. The pressure at your tap when the water is flowing is the pressure at the water tower minus all line losses between the tower and your tap. Most (almost all) of that line loss is in your house. There should be a pressure gauge near your water meter. Watch that when the valve is opened and closed.

Remove the hose and measure flow again. The total line loss is then the supply pressure. The hose just adds a little more line loss to the total system.

@jrmichler thanks! I see what you are saying. So how would you calculate this theoretically? I used the my tap because that's what I had available, but imagine a similar case where I have a tank pressurized at 60 psi, and I connect a 1/2", 36 in long hose to it through a valve. If I open the valve, what flow should I expect in theory? Are my calculations correct or did I miss something? Also, I started to play with some input parameters and I noticed that if I set delta P to 0.7, the approximation does match the results.

 

[Asymptotic]

Don't know where the calculation is going sideways, but

• the observed flow (0.23 quarts/min; 3.45 GPM) seems about right for a 1/2" line at 60 PSI.
• Calculated velocity of 17.9 meter/sec (58.8 FPS) is well beyond expectation. Design rule of thumb is for a velocity in the 3 to 10 FPS range. 20 FPS would be exceptionally high. Nearly 60 FPM is scary high.
  
• Calculated flow of 2.401 quarts/sec is just over ten times higher than observed 0.23 quarts/sec rate.

Is there anything in the chain of metric to English conversion factors that accounts for a 10:1 difference?

 

[OmCheeto]

Could you please help me see where the problem is?

Probably not, given your 5144 gallons per minute flow rate:

But thank you for pointing out the Hagen–Poiseuille equation. I think I can use that in my research.

ps. My fish tank was a gallon low, and I determined that your flow rate is about 2200 times what my flow rate is. You have very good water pressure.

 

[boneh3ad]

Thanks @Tom.G yes that pressure measurement was made like shown in picture 1. The gauge is supposed to measure total gauge pressure, and since the fluid is static this should be the static pressure.

That's not what static pressure means. The pressure measurement with flow off is both a static pressure and a total pressure measurement since there is no flow. With flow on, those two values are different, and neither of them necessarily equals the pressure you measured with the spigot plugged. The word "static" in this case is used as an adjective for something pertaining to the state of the fluid, not to convey information about whether it is in motion. It is the thermodynamic pressure of the fluid.

I would say you have a lot of pretty huge problems here.

1. Your volumetric flow rate is 100% not correct, as you are not getting 5144 gallons per minute out of that hose. You are at least 3 orders of magnitude too high there. Your velocity and Reynolds number calculations are therefore way off as well. This is probably due to a unit mismatch, as you give no indication of what values you actually plugged into the Hagen-Poiseuille equation. If your unit system wasn't self-consistent (and it doesn't seem to be if you just used the values you have above that), then your answer will be nonsense.
2. You are improperly using the Hagen-Poiseuille equation. Hagen-Poiseuille applies strictly to laminar flows, and your flow is likely turbulent.
3. You are improperly using the Darcy-Weisbach equation. It applies to either laminar or turbulent flows, but for laminar flow reduces identically to the Hagen-Poiseuille equation. You can't use them together like that. The Darcy-Weisbach equation is actually meant as a correction to the Bernoulli equation to account for frictional losses and that is how you use it.

For instance, if you call the outlet of your valve location 1 and the outlet of your system as a whole location 2, then you would set up the following equation:
$$\left(\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + gz_1\right) = \left(\dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + gz_2\right) + f\left(\dfrac{L}{D}\right)\dfrac{V_2^2}{2}.$$
Since this is water, you could also relate the two velocities by noting constant volumetric flow rate. You would still need to determine $Re$ and $f$, which would likely need to be done iteratively.

ps. My fish tank was a gallon low, and I determined that your flow rate is about 2200 times what my flow rate is. You have very good water pressure.

My guess is that his unit system is 2200 times more favorable than yours.

 

[Roy S Ramirez]

Don't know where the calculation is going sideways, but

• the observed flow (0.23 quarts/min; 3.45 GPM) seems about right for a 1/2" line at 60 PSI.
• Calculated velocity of 17.9 meter/sec (58.8 FPS) is well beyond expectation. Design rule of thumb is for a velocity in the 3 to 10 FPS range. 20 FPS would be exceptionally high. Nearly 60 FPM is scary high.
  
• Calculated flow of 2.401 quarts/sec is just over ten times higher than observed 0.23 quarts/sec rate.

Is there anything in the chain of metric to English conversion factors that accounts for a 10:1 difference?

I would love to just find that factor but I'm starting to think that those equations I used are not appropriate/ suitable...

 

[Roy S Ramirez]

That's not what static pressure means. The pressure measurement with flow off is both a static pressure and a total pressure measurement since there is no flow. With flow on, those two values are different, and neither of them necessarily equals the pressure you measured with the spigot plugged. The word "static" in this case is used as an adjective for something pertaining to the state of the fluid, not to convey information about whether it is in motion. It is the thermodynamic pressure of the fluid.

I would say you have a lot of pretty huge problems here.

1. Your volumetric flow rate is 100% not correct, as you are not getting 5144 gallons per minute out of that hose. You are at least 3 orders of magnitude too high there. Your velocity and Reynolds number calculations are therefore way off as well. This is probably due to a unit mismatch, as you give no indication of what values you actually plugged into the Hagen-Poiseuille equation. If your unit system wasn't self-consistent (and it doesn't seem to be if you just used the values you have above that), then your answer will be nonsense.
2. You are improperly using the Hagen-Poiseuille equation. Hagen-Poiseuille applies strictly to laminar flows, and your flow is likely turbulent.
3. You are improperly using the Darcy-Weisbach equation. It applies to either laminar or turbulent flows, but for laminar flow reduces identically to the Hagen-Poiseuille equation. You can't use them together like that. The Darcy-Weisbach equation is actually meant as a correction to the Bernoulli equation to account for frictional losses and that is how you use it.

For instance, if you call the outlet of your valve location 1 and the outlet of your system as a whole location 2, then you would set up the following equation:
$$\left(\dfrac{p_1}{\rho} + \dfrac{V_1^2}{2} + gz_1\right) = \left(\dfrac{p_2}{\rho} + \dfrac{V_2^2}{2} + gz_2\right) + f\left(\dfrac{L}{D}\right)\dfrac{V_2^2}{2}.$$
Since this is water, you could also relate the two velocities by noting constant volumetric flow rate. You would still need to determine $Re$ and $f$, which would likely need to be done iteratively.
My guess is that his unit system is 2200 times more favorable than yours.

@boneh3ad Thank you very much for your reply! I think I was using the Hagen-Poiseuille equation to confirm that the flow was turbulent, but using the resulting Reynolds number for the Darcy-Weisbach is not correct, you are right. Also there should be no unit errors since the software I'm using takes care of them. Then, I noticed that if I put my pressure gauge after the valve and let the water flow through the entire hose, it reads a pressure between 0-1-psi, so I would guess that this is the dynamic pressure (P_dyn), and that the pressure I measured with the spigot plot (p_total) minus p_dyn = static pressure? Interestingly, for any numbers between this range, the equations I used actually do a better job approximating the experimental values.

So is there any way I could theoretically calculate the dynamic pressure from a total pressure known (from a pressure vessel or tap)?

 

[boneh3ad]

Also there should be no unit errors since the software I'm using takes care of them.

This is a very perilous way to do science or engineering. Make sure you understand your unit system. Clearly something is wrong with what you think your software is doing and what it is actually doing.

I noticed that if I put my pressure gauge after the valve and let the water flow through the entire hose, it reads a pressure between 0-1-psi, so I would guess that this is the dynamic pressure (P_dyn)

You guessed incorrectly. You cannot directly measure dynamic pressure. You always have to calculated it by measuring static and total pressure.

the pressure I measured with the spigot plot (p_total)

What you measured at the spigot with no flow has no directly and easily calculable relationship with the pressure you measure with flow on other than the fact that the flow on pressure will be lower. With flow off, that pressure is both the static and the total pressure in that flow-off situation because there is no flow, and therefore no dynamic pressure. With the flow on, it is difficult to relate because there are losses in the system due to motion, so the total pressure of the system in motion is going to be less than your originally-measured pressure. The static pressure will be lower still due to the motion of the fluid. If you want to measure static pressure while the flow is running, then put a T fitting downstream of your valve and attach the gauge there. There will be some additional losses due to the fitting, but it should be a good estimate.

Interestingly, for any numbers between this range, the equations I used actually do a better job approximating the experimental values.

That is purely coincidence. You need to use the equations correctly.

 

[jrmichler]

@jrmichler thanks! I see what you are saying. So how would you calculate this theoretically? I used the my tap because that's what I had available, but imagine a similar case where I have a tank pressurized at 60 psi, and I connect a 1/2", 36 in long hose to it through a valve. If I open the valve, what flow should I expect in theory? Are my calculations correct or did I miss something? Also, I started to play with some input parameters and I noticed that if I set delta P to 0.7, the approximation does match the results.

Start by looking the house water system. Run some flow tests with and without a hose on the tap. Look at the piping in the house, and estimate the size and length. You should be able to plug those numbers into your calculations and match the measured flow without the hose. If your pipe inside diameter is correct, the length where the calculated flow matches the measured flow will be longer than the actual length because of losses in the fittings and the valve. Keep in mind that pipe inside diameter is not always obvious - 1/2" schedule 40 pipe is 0.62" inside diameter and 0.84" outside diameter (both iron and PVC).

When the calculated and measured flows (without the hose) match, compare the measured length of pipe to the length used in your calculations. They should be reasonably close. Then add your hose (length and diameter) to the calculation. You should be able to calculate the new total flow and the pressure between the tap and the hose with small error.

Personal experience: I once designed a heated water system for an entire paper mill. Total flow was about 2,000,000 gallons per day. The main headers were 6" to 12" diameter. The taps for small uses, such as hand wash up hoses, were 2" NPT down to the tap, where they were reduced to the hose size. All hoses were cut to the actual length needed, as opposed to the previous practice of connecting one end of a 50 foot length and letting the excess lay on the floor. The operators were absolutely amazed at how much flow they got out of the new system. They could no longer put a hose into a tank, open the valve, and walk away. My hazy recollection is that the system had over 750 connections.

 

[Mech_Engineer]

A thread I was involved in from 2017 covered this exact topic in some detail: Finding the flow rate through an open-end pipe

Great catch, I got sloppy with my friction factors and wasn't distinguishing between Fanning and Darcy.

Here's the updated equation set:
View attachment 112450

And the results:
View attachment 112452

Looks like this is about as accurate a calculation as I can provide given the parameters known? Thanks @Chestermiller and @JBA , I learned a good deal in this thread!

I thought of one more possible error- the water main's pressure will probably be specified at gauge pressure, meaning the discharge pressure P2 would be zero, not 1 atm (gauge pressure of zero).

That increases the flow a bit, see here:
View attachment 112455

I also tried a quick check by comparing calculated flow rates against a hose discharge chart here: http://www.engineeringtoolbox.com/water-discharge-hose-d_1524.html. The flow rates calculated by my sheet appear to be right in-line with the values from this chart.

Calculated values from my sheet (100 ft length for all):

• 1/8" hose, 100 psi: 0.25 gpm
• 1/2" hose, 40 psi: 6.38 gpm
• 1/2" hose, 100 psi: 10.75 gpm
• 3/4" hose, 40 psi: 19.02 gpm
• 3/4" hose, 100 psi: 32.03 gpm
• 6" hose, 10psi: 1965 gpm
• 6" hose, 60 psi: 5202 gpm

Compare to chart from EngineeringToolBox.com:
View attachment 193336

 

[Mech_Engineer]

For reference, the equation set used in that MathCAD sheet looks like this:

I used a numeric solver and some trial and error to get it working, but in the end the calculations looked to match very closely to the estimates on Engineering Toolbox's site: https://www.engineeringtoolbox.com/water-discharge-hose-d_1524.html

 

[Roy S Ramirez]

For reference, the equation set used in that MathCAD sheet looks like this:
View attachment 239881

I used a numeric solver and some trial and error to get it working, but in the end the calculations looked to match very closely to the estimates on Engineering Toolbox's site: https://www.engineeringtoolbox.com/water-discharge-hose-d_1524.html

View attachment 239882

@Mech_Engineer Thank you very much! I tried using the same equations you used in mathcad but I'm sill getting a similar result which is off by a factor of 10 :/

 

[Mech_Engineer]

The equation set I have gets the same answer you get within 10%. I'm guessing the other aspects of your experimental setup (valve flow rate, tank pressure, etc.) are having an effect on the result. The calculations I was doing were for a 100 ft hose, where the length of the hose dominated the flow characteristics.

 

[Roy S Ramirez]

Ok, I'm going to try with a longer hose this time, and maybe try to get another pressure gauge. When I put my hose with a pressure gauge, it reads like 1 psi which interestingly does kind of match the equations. But yes, when I only put the gauge (without the hose), and open the valve, it reads 60 psi. I'm really confused now... Should I by using the 60 psi value for P_1?

 

[Roy S Ramirez]

Ok, I'm going to try with a longer hose this time, and maybe try to get another pressure gauge. When I put my hose with a pressure gauge, it reads like 1 psi which interestingly does kind of match the equations. But yes, when I only put the gauge (without the hose), and open the valve, it reads 60 psi. I'm really confused now... Should I by using the 60 psi value for P_1?

Ok, I ran the experiment again with a 96 in hose this time, and I got 0.17 qt/s, whereas the mathcad says 1.7! There always seems to be a factor of 10, what's going on here?!

 

[boneh3ad]

If I run this in Matlab using what I know to be consistent units, for a 36 inch hose and supply pressure of 60 psig, I get something similar (at least the correct order of magnitude) to what you got most recently. I get an exit velocity of 27.3 m/s and 0.0035 m³/s, which translates to 89 ft/s and 55.5 gpm. Given, I threw that together in just a few minutes and so I can't guarantee there are no typos. However, I checked it against the curves that @Mech_Engineer posted a few posts back and it matches up well with those values, so I am pretty confident it is giving a good answer.

To match your experimental results, I have to drop the supply pressure down to 0.4 psig.

 

[Mech_Engineer]

Ok, I ran the experiment again with a 96 in hose this time, and I got 0.17 qt/s, whereas the mathcad says 1.7! There always seems to be a factor of 10, what's going on here?!

You might be seeing a large pressure drop across the valve being used such that the apparent pressure at the start of the hose is not in fact 60 psi.

 

[JBA]

Have you run your fill test directly from the faucet without the hose connected. My flow program indicates that the pressure loss for 3 ft of 1/2" hose at your flow rate is essentially zero. In your past testing you are not just testing the pressure loss for the hose but for the combined loss of your house piping system to the faucet, the faucet globe valve and the hose all combined.

 

[Roy S Ramirez]

If I run this in Matlab using what I know to be consistent units, for a 36 inch hose and supply pressure of 60 psig, I get something similar (at least the correct order of magnitude) to what you got most recently. I get an exit velocity of 27.3 m/s and 0.0035 m³/s, which translates to 89 ft/s and 55.5 gpm. Given, I threw that together in just a few minutes and so I can't guarantee there are no typos. However, I checked it against the curves that @Mech_Engineer posted a few posts back and it matches up well with those values, so I am pretty confident it is giving a good answer.

To match your experimental results, I have to drop the supply pressure down to 0.4 psig.

Thank you very much for checking it out with your MATLAB code. If the calculations seem to be right, there must be an error in my measurements I guess. But practically speaking, for a 1/2" hose (36" long) with a 60 psig supply pressure, isn't 55 or 40 gpm way too much?

 

[Mech_Engineer]

Thank you very much for checking it out with your MATLAB code. If the calculations seem to be right, there must be an error in my measurements I guess. But practically speaking, for a 1/2" hose (36" long) with a 60 psig supply pressure, isn't 55 or 40 gpm way too much?

I don't think you have a true 60 psi supply pressure, I suspect your pressure drops significantly when the flow is turned on. I would recommend using a pressure gauge to monitor the supply pressure between the valve and the hose when the valve is turned on; I'm guessing you're seeing a large pressure drop when water is flowing.

 

[boneh3ad]

Thank you very much for checking it out with your MATLAB code. If the calculations seem to be right, there must be an error in my measurements I guess. But practically speaking, for a 1/2" hose (36" long) with a 60 psig supply pressure, isn't 55 or 40 gpm way too much?

Like I said earlier, the static pressure at the outlet of your spigot is not 60 psig. That's the pressure with zero flow going through your house, which means zero losses throughout your entire system. Once your system is running, the actual pressure at the outlet of your spigot is clearly much lower.

 

[Roy S Ramirez]

You might be seeing a large pressure drop across the valve being used such that the apparent pressure at the start of the hose is not in fact 60 psi.

Mmmm that could be it, but as @boneh3ad noticed, I would have to lower the pressure from 60 to around 0.5 psi to make the eqs match, and I doubt that's the valve's pressure drop :/

 

[Mech_Engineer]

I would have to lower the pressure from 60 to around 0.5 psi to make the eqs match, and I doubt that's the valve's pressure drop :/

Why doubt them? Do the science and find out!

 

[boneh3ad]

Mmmm that could be it, but as @boneh3ad noticed, I would have to lower the pressure from 60 to around 0.5 psi to make the eqs match, and I doubt that's the valve's pressure drop :/

That wouldn't be the valve's pressure drop. That would be the pressure drop (59.5 psi) of the entire flow system in your house. Like we said, go check the pressure at your water meter, then take measures to check the pressure while the flow is on at the exit of your valve. I bet you will be surprised.

 

[JBA]

Maybe a bit late to the party, but, I just ran my program including an L/D of 380 for a fully open globe valve (Ref Crane Co Tech Paper 410, Flow of Fluids) plus 3 ft of hose and the result is a flow rate of .233 qt/sec with an inlet pressure to the valve of 59.2 psig

 

[boneh3ad]

Maybe a bit late to the party, but, I just ran my program including an L/D of 380 for a fully open globe valve (Ref Crane Co Tech Paper 410, Flow of Fluids) plus 3 ft of hose and the result is a flow rate of .233 qt/sec with an inlet pressure to the valve of 59.2 psig

Clearly your program takes a lot more factors into account than mine. I went ahead and threw a minor loss from a globe valve into my code for grins. I assumed a loss coefficient of 10. I still get 1.14 qt/s, so it's still missing something that your code has unless you assumed $K=250$ or something.

I need to stop farting around with this. I am supposed to be writing about high-speed wind tunnels right now.

 

[gmax137]

Here's my input.
It sounds like we can believe the 4.33 seconds per quart measured flowrate? That's 3.5 gpm

Can we believe the 1/2 inch diameter measurement? That's a measured inside diameter, right?

If those are correct then for the hose Reynolds number I calculate

Re=50.6 Q rho/ d / mu = 50.6 * 3.5 * 62 / 0.5 / 1.0 = 22,000

and velocity v=5.7 ft/sec

Then for 1/2 inch ID smooth pipe the friction factor is ~ 0.03

For a 36 inch long hose, L/D = 72; Kfric = 0.03 * 72 = 2.2

Add Kexit = 1.0

Ktot = 3.2

Pressure loss in the hose

dP = (rho/144)*Ktot * v^2/2g

dP = (62/144)*3.2*(5.7)^2 /32.2 / 2 = 0.7 psi

As mentioned above, the 60 psig measurement (with zero flow) is irrelevant to the flow condition.