Estimate of flow in pipe given the pressure?

[ruggerjvd]

Hi
I am trying to estimate the flow in a 1" PVC pipe, given a pressure of about 80 psi, discharging to atmosphere. If I try to do Bernoulli's, I get a very high flow rate. Are there tables or a shortcut method that can be used to better estimate the flow? I assumed length of about 50 feet.

thanks for the help

 

[RWF7437]

I get about 40 gallons per minute ( if it is water). That doesn't seem high to me. Don't see why you would want to use anything except Bernoulli's equation since it is simply a special case of the law of conservation of energy.

 

[JStephen]

Using Bernouli's equation, you need to use the pressure while the pipe is flowing and at a point near the outlet (or else allow for pipe friction loss). If you turn a valve off, measure the pressure, then turn it on, that pressure may drop drastically.

If this is air or gas rather than water, you may have compressible gas effects going on that complicate the situation. In any case, if you calculate a velocity more than about a third the speed of sound, you'll start getting errors due to that.

 

[katmar]

It depends which version of Bernoulli you are using. The original version was for that mythical "frictionless pipe" but in this case about 90% of the energy goes to overcoming friction. It would be better to use the Darcy-Weisbach equation, including an exit loss of K=1.0 to take Bernoulli into account. If you have neglected friction you will get an unrealistically high flow rate.

For what it is worth, my software says that the flowrate will be 81 gpm at a velocity of 33 ft/s (assuming no change in elevation). But I have assumed a pipe ID of exactly 1" and it could be significantly different from that depending on the pipe class.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ruggerjvd]

Thank you for the input

I was using Bernoulli, and then Hazen Williams with a C of 150 for the friction loss. and a K of 1 for the outlet minor loss.

So Bernoulli gave me

P1/gamma + v1^2/2g + Z1 = P2/gamma +v2^2/2g +z2 +hf + Hm

P1 = 180 ft
P2 = 0 (free atmosphere discharge)
Z1 = Z2
V2 = 0 (free atmosphere discharge)
hf = 3.022 (V1^1.85)(50)/150^1.85/0.083^1.165
Hm = (1) v1^2/2g

The 0.083 is the nominal 1" pipe in feet.

I'm not sure how to treat the free atmosphere discharge velocity in this case. If I were discharging to a reservoir it's obvious, but I can't really find a good reference for this case. Would velocities cancel out? If I take Point 2 just inside the end of the pipe, then I don't really have an exit loss, so I assumed that the exit loss is accounting for full dissipation of velocity head in the hm term, so I should set v2 = 0 in the body of Bernoulli.

thanks again.

 

[katmar]

I don't use Hazen Williams, but I suspect that a factor of 150 will be rather conservative for a smooth plastic pipe. Hopefully one of the real water distribution guys can give some advice on this. But your process looks OK - you have included the friction loss.

You have got V1 and V2 confused. This is not surprising because what is called the exit loss does not occur at the exit. In your case V1 would be zero as it is the velocity in the tank that has the 80 psi source. V2 is the velocity at the exit, but the loss of energy occurs where the water is accelerated so in reality it is lost an the entrance and the water velocity remains constant from there onwards.

However, if the discharge were going into a pressurised tank this velocity head could be recovered as pressure as the water decelerates. Because this recovery either occurs, or does not occur, at the exit it has come to be known as an exit loss. Countless engineers have been confused by this (myself included).

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ruggerjvd]

Hi Katmar
actually, point 1 is in the pipe itself, there really is no reservoir, but there is flow, so v1 is the start point in the pipe, v2 is the point of discharge. How then should I cahnge BErnoulli and loss terms?

 

[katmar]

If Point 1 is where you measure the 80 psi, and the water is already flowing at the same velocity as it will be at the exit then the acceleration losses have been suffered before you measure the 80 psi. In this case V1 = V2 and there are no acceleration losses (i.e. no exit losses in conventional terms).

The total head at P1 would be the 80 psi static pressure plus the velocity head (neglecting height effects). The total head at P2 is 0 psi static pressure plus the same velocity head as was present at P1. The only difference then between P1 and P2 is the static pressure (i.e. the 80 psi) and the friction losses along the line.

Theoretically this means your flowrate increases to 85 gpm, but we are splitting hairs at this stage and there are other uncertainties like the actual roughness, pipe ID, pipe length, fittings etc that will have an effect as well.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[bimr]

I agree with Katmar that you should be using the Darcy formula. For schedule 40 PVC pipe:

h = .00259 K (Q)2/(d)4

K = .024 * 50 * 12 / 1.029" + 1 = 14.99

184.6 = .00259 * 14.99 * (Q)2 / (d)4

Q = 73 GPM

Reference Crane Technical Paper No. 410

 

[ruggerjvd]

Hi bimr

I'm not used to seeing darcy expressed this way.

I use

hf=fLv^2/(2Dg)

Is your expression for k from the Crane paper, as well as the Darcy equation?

 

[bimr]

Yes, substitute (f L/D) for (K). K factors are obtained from the Crane Paper. The "Paper" is actually a handbook not a paper.

http://www.tp410.com/

The Darcy formula has been converted to somewhat more convenient terms that are more applicable to liquid flow and are written in terms of flow rate in gallons per minute.

 

[lalvarez]

i need to figure out what the volume for a 1" pvc pipe with a flow of 3ft/sec of water is. Help show math please, also for a 3/4"pvc

 

[cr1973]

lalvarez -
Q=v*a... do the math.

 

[burnt2x]

ruggerjvd,

Try this head loss formula, a variation of the D'Arcy-Weisbach equation in the English units:

0.000216 X(f)X(L)X rho X Q^2
Delta P = ---------------------------------
d^2
where: Delta P = pressure head drop, in psi
f = friction factor from Moody chart
L = pipe length, in feet
rho = density of fluid in lbs per cu. feet
Q = fuid flow rate in gpm
If the drop in pressure is 80 psi to zero, my calculations yielded a flow of 70.9 gpm using a friction factor f = 0.0236

 

[burnt2x]

typo error correction to my previous reply:

the denominator d ^2 should be d^5. with d= 1, that should not be giving wrong values

thanks.

 

[49merc]

If it discharges to atmosphere, you can measure the area of the pipe opening, length and drop of discharge (x and y direction), and calculate the velocity. V X A = flow rate. There are a few handbooks (The Pump Handbook I believe is one) that show this. What kind of accuracy is tolerable?