Flow of Water From Tubing

[RJB32482]

I am trying to calculate the flow rate of water from tubing into an atmospheric vessel from the tube breaking. The tube is about 5/8" ID and 15 ft long with 2 90 degree bends and a vertical drop of about 15 ft. The header pressure is around 80 psig. How would I go about the calculation of flow rate from the exit of the tube? I can't just take delta P as 80 psig right? Any help is appreciated. Thanks

 

[quark]

It is 80+14.7+7-losses in tube. You have to go for a trial and error solution with flowrate and corresponding losses in tube.

 

[RJB32482]

Thanks for the initial help.
Why take in account the 14.7 psi when we are talking about gauge pressures? Do you need to convert gauge into atmospheric?

 

[quark]

Because you are finally letting the water out into atmosphere which is 14.7abs.

 

[Latexman]

quark,

I disagree. It's 94.7 psia - 14.7 psia + 6.5 psi - losses, or 80 psig - 0 psig + 6.5 psi - losses.

Good luck,
Latexman

 

[quark]

oops, perhaps I might be thinking of the available pressure before discharge. You both guys are right.

 

[Latexman]

With 15' of 0.625" ID tubing, 2 short radius elbows, a sharp edge inlet loss, an outlet loss, and dP = 86.5 psi, I got 33-34 U.S. gpm.

Good luck,
Latexman

 

[RJB32482]

Why would the vertical drop of 15 ft add 6.5 psi to the pressure drop? Wouldn't it decrease the pressure drop?

 

[briand2]

For every foot you drop the vertical tube, you've got another foot of liquid (presumably) above the bottom of the tube, so they balance out. However, you've also got the loss due to friction within the vertical tube.

 

[Latexman]

RJB32482,

The governing equation for most common fluid flow problems is the Bernoulli equation, which is a mechanical energy balance involving kinetic energy (velocity), potential energy (pressure and elevation), and frictional losses. If your header pressure was at 0 psig, you would still have flow to the atmosphere because the header has 15 feet more potential energy than the discharge. I converted this potential energy from an elevation basis to a pressure basis by the conversion factor of 2.31 feet of water = 1 psi of pressure. Since the potential energy due to elevation is causing or adding to the flow of fluid I added to the potential energy due to pressure. It also is called "head". You have 80 psi of pressure AND 15 feet of head.

If you had 15 feet of vertical rise, I would have subtracted, but that was not the situation.

Basically, I did the Bernoulli equation in my head and entered the resulting overall potential energy in terms of differential pressure into my handy dandy spreadsheet and trial and error solved for the flow rate which generates losses that equal the dP. I think it took at most 2-3 minutes time.

Good luck,
Latexman

 

[abeltio]

sorry, i do not understand the question...
unless i completely misread the problem...
if the tube is breaking... then the flow at the end of the tube (at the bottom of the 15 ft drop) will be negligible...

If the tube breaks, wouldn't the max flow occurr at the connection with the header?

In US customary units.

Flow will be around Q ~80 gpm

with the basic flow of fluid formula for an open discharge to atmosphere (all the pressure converts to kinetic energy):

v = (2gh)^0.5

then Q = v*A

A = 0.25*pi*(d/12)^2

h = (80/62.4)*144 = 109ft
g = 32.2 ft/s^2
v = 84 ft/s
A = 0.25*3.14*(0.625/12)^2 = 0.002131 sqft
Q = 84 * 0.002131 = 0.1785 cuft/s = 10.71 cuft/m
[Q = 80 gpm]

using the formula for discharge thru a sharp inlet and exit, from CRANE TECHNICAL PAPER 410 - page 3-4 Eq 3-19
Considering K= 0.5 for sharp entrance and K=1.0 for sharp exit (broken pipe) -
L = 0 because the pipe broke at the connection with the header. Worst case.

Q = 236 d^2 (deltaP/(rho*K))^0.5
d = 0.625 in
Q = 236*(0.625)^2*(80/(62.4*(0.5 + 1.0))
[Q = 79 gpm]

pretty much the same result.

What do you think? is this the correct interpretation of the problem?

saludos.
a.

saludos.
a.

 

[quark]

abeltio,

If breaking means physical damage of the tube then your calculations based on total conversion of potential energy to kinetic energy are correct.

But I took it as pressure breaking(as the OP said water flow into the vessel, and I think Latexman took the same meaning). In this case, we consider the flow which creates the frictional drop equivalent to the potential energy. I got it close to 25gpm.

 

[Latexman]

Yes, I interpreted the situation the same as quark.

Good luck,
Latexman

 

[abeltio]

hell... 2nd language hits again...
from the tube breaking <> breaking of the tubing

question lads... breaking of the tubing means:
tube ends and discharges into the atmosphere inside the tank?

saludos.
a.

 

[abeltio]

grrr...
my question should have said:
tube breaking means:...

saludos.
a.

 

[Latexman]

abeltio,

I know that you believe that you understood what you think I said, but I am not sure you realize that what you heard is not what I meant.

Good luck,
Latexman

 

[Latexman]

Let me try that again.

Quote attributed to Robert McCloskey, State Department spokesman:

I know that you believe that you understood what you think I said, but I am not sure you realize that what you heard is not what I meant.

Good luck,
Latexman

 

[abeltio]

i don't know... third base!

saludos.
a.