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Anonymous Poster

Flow Rate

11/15/2007 4:39 PM

What will be the water flow rate for the pipe of 1" dia. having the pressure of 7 bar or 0.7 MPa

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Guru

Join Date: Sep 2007
Location: Reno, NV (USA)
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#1

Re: Flow Rate

11/15/2007 5:43 PM

The short answer is "depends". Based on the information given it could be anywhere from zero to free discharge. Ignoring losses, I believe you want the Bernoulli Equation. It will allow you to calculate the free discharge condition. The general form:

dP/rho + (v22-v12)/2 + g * dy = 0

where:

dP = change in pressure

rho = fluid density

v2 = final velocity

v1 = initial velocity

g = acceleration due to gravity

dy = change in height

Standardize your units and apply accordingly. I'm assuming v1 is zero and you'll be solving for v2. Knowing v2 and your pipe diameter, you can solve for flow rate.

If what you're really looking for is full-pipe flow, let me know and I'll write that equation out. But beware, it's not particularly nice...

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Anonymous Poster
#2
In reply to #1

Re: Flow Rate

11/15/2007 6:27 PM

Thank you , lets open the actual problem: in the plant I am working, there is a supply water pipiline of 1" dia. having pressure of 7 bar. I need to have a flow rate of 10 liter/min and pressure of 2.5 bar for my valve stand, cooling system purpose and I don't know the existing flow rate. Without using Flow meter & Pressure control valve etc., how do I find ( calculate) the required values . Plant water supply pipeline of 1" having 7 bar pressure available when it runs, now I need to get a supply line of 2.5 bar & 10 Liter/min flow.

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Guru

Join Date: Sep 2007
Location: Reno, NV (USA)
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#3
In reply to #2

Re: Flow Rate

11/16/2007 12:34 AM

Okay, it's been forever since I've done this, but here goes:

Use this form:

V12/2 + P1/rho = V22/2 + P2/rho

Flow(Q) in = Flow out; 10L/min = 10.17 in3/s

Q=AV => V1=10.17/0.864 (True ID of 1" Sch.40 pipe = 1.049" => A1=0.864 in2) = 11.77 in/s

P1=101.53 psi, P2=36.26 psi, rho=0.03613 lb/in3(water)

Therefore: 11.772/2 + 101.53/0.03613 = V22/2 + 36.26/0.0361 => V2=61.25 in/s

A2=10.17/61.25=0.167 in2 => d2=0.46" (3/8" Sch.40 has true ID of 0.493")

Now, for the bad news. This assumes you have the flow you specified. If for some reason your supply does not deliver the 10 L/min you mentioned, none of this works. Long story short, you'll probably need to install a flow regulator.

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Power-User

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#4
In reply to #2

Re: Flow Rate

11/16/2007 4:03 AM

So what happens when the upstream pressure changes? Are you happy with a change in flow, or a change in pressure or both?

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Active Contributor

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#5
In reply to #1

Re: Flow Rate

11/16/2007 8:22 AM

What does Aequam memento rebus in arduis servare mentem mean ?

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Guru

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#6
In reply to #5

Re: Flow Rate

11/16/2007 10:42 AM

Loosely - "keep your wits when things are tough". It has some more eloquent translations, but that's the blue-collar version. It reminds me not to be overwhelmed when problems start coming rapidly and to focus on facts and solutions, not preconceptions and unknowns. A problem is only a problem until it's solved.

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Aequam memento rebus in arduis servare mentem.
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Anonymous Poster
#7
In reply to #6

Re: Flow Rate

11/17/2007 12:12 AM

To : CSM Engineer,

Thanks for your translation.

I did learn it in latin but I have since forgotten it , but your closing statement on the problem reminded me of it :

"Problems are nothing but solutions in disguise"

Labor Omnia Vincit

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Guru

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#8
In reply to #7

Re: Flow Rate

11/17/2007 12:16 AM

"Labor Omnia Vincit"

Also from Virgil, the author generally credited for mine.

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