# Pipe diameter & flow rate

Can you calculate how pipe diameter affects flow rate?
What is needed to calculate? Flow rate (gpm) and pressure?
As a guide how does the flow rate decrease as length increases? Does pipe material make a difference (i.e copper / speedfit)
i.e if water pressure is say 2bar, flow into pipe is 1gpm what would be flow be at the end of say (assuming no elbows, etc)
10,20,30 m of LLDPE tubing of inside diameter of 2.5mm,4mm,6mm,7mm,9mm,11.5mm
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Jim wrote:

1gpm ;-)
There's a pressure loss due to friction in the pipe; also pressure loss due to fittings, which is usually expressed as an equivalent length of pipe (le); velocity pressure/head at the oulet is lost. The total head loss @ the design flowrate is equal to the differential pressure (inlet - outlet pressure; pump differential pressure for a circulating system).
Too detailed to explain, you'll find books on how you'd calculate it in reference libraries & I'm not going to try to write a book in reply here. You'll also get tables of pressure losses @ flow rates on manufacturers' websites. The cda does a useful pdf design guide for copper pipes.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Just keeping the numbers simple. Only looking for a rough idea.
The cda does a useful pdf design guide for copper pipes.
Just what I am looking for. thanks
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Jim wrote:

Think it's this one, if not look at their publications list. http://www.cda.org.uk/megab2/build/pub33/33_2000.pdf
Tables in the appendices.
You should be able to get a rule-of-thumb design guide from the makers of your tubes.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
Jim wrote:

I think you missed the joke. If the flow into the pipe is 1gpm then the flow out of the pipe must be 1gpm. Where else can it go?
MBQ
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
snipped-for-privacy@hotmail.com wrote:

Leak - via a dribble fitted pushfit coupling ?
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
wrote:

would be flow

Lord hall, leaking via these inferior fitting is a real possibility . Well spotted. 10/10
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
snipped-for-privacy@hotmail.com wrote:

But it was a very tiny joke and easily overlooked.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

I missed the joke as well.
If you have say, a copper pipe with tee and isolator - the flow at this point is xxx gpm
If you shove on 20m pipe onto the isolator - how much comes out of the end? This will be limited due to resistance through the pipe.
try this site http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm#calc
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
James wrote:

A quick scroll through that link; very interesting. It seems they've assumed all bends are of large radius and have negligible effect. In buildings the fittings (bends, elbows, tees, reducers, valves ) usually have a very significant effect, but you could add on their equivalent length to the pipe length.
It sounds like the OP just wants to estimate flow rate through some blue plastic, so it looks useful. He would still need a value of the friction factor for his material.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

I'm sure they taught me that flow is proportionate to the radius^4 a long time ago...
--
Helen D. Vecht: snipped-for-privacy@zetnet.co.uk
Edgware.
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Someone else has suggested the same here in the past but I had my doubts. My gut feel was that the fourth power was such a high factor that water would flow through microbore like treacle. I looked at equivalent length tables from the CDA and it suggested that _resistance_ to flow would be inverse proportional to the cube of the radius. That brings out the sort of sensible result that 1m of 15mm copper tube will have the same sort of restriction as about 5m of 22mm copper. If it was 4th power of radius then the equivalence would be 12m.
Anyone know better?
--
fred
Plusnet - I hope you like vanilla
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Poiseuille's equation relates laminar flow through a cylidrical tube to pressure drop:
P = (8 F h L)/(pi R^4)
where F = flow rate, P = pressure drop, h = viscosity, L =length and R radius
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Fri, 11 Nov 2005 22:51:46 +0000, Norman Billingham wrote:

Disagree. Based on the OD (or radius) it's 4.6:1 based on the ID (or radius) it's 5:1

But at normal domestic flow rates for gas and water the Reynold's numbers are high, the flow is turbulent. So pressure drop is proportional to the square of velocity and also related to the density of the fluid rather than it's viscosity.
--
Ed Sirett - Property maintainer and registered gas fitter.
The FAQ for uk.diy is at http://www.diyfaq.org.uk
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

More words required Ed, are you saying that 4th power leads to 5:1 or something else? I got 3rd power of radius leading to about 5:1 but I haven't had a chance to re-address my sums.

--
fred
Plusnet - I hope you like vanilla
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Sat, 12 Nov 2005 21:46:45 +0000, fred wrote:

It makes no difference whether you use the radius or the diameter.
Based on OD you have (22/15)^4 approx 4.6 Based on OR you have (11/7.5)^4 approx 4.6 Based on ID you have (0.75"/0.5")^4 approx 5.1 Based on IR you have (0.375/0.25) ^4 approx 5.1
--
Ed Sirett - Property maintainer and registered gas fitter.
The FAQ for uk.diy is at http://www.diyfaq.org.uk
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

If flow is based on equations that contain r**4 (or d**4) then it's not possible to take ratios of the radii (or diameters) and simply raise them to the 4th power to obtain the relative effect, the results are entirely different. The equations need to be balanced with r1**4 on one side and r2**4 (or their reciprocals or whatever) on the other.
Compare with parallel resistor equations (if you're one of those sorts of peeps):
1/Rpara**2 = 1/R1**2 + 1/R2**2
the result cannot be obtained by combining the resistor ratios before any squaring takes place, the result is more complex.
I _will_ look at this in more detail, one of these days . . .
--
fred
Plusnet - I hope you like vanilla
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
On Sun, 13 Nov 2005 12:38:13 +0000, fred wrote:

We are not adding together differing lengths of pipe to find the total resistance but rather comparing the relative resistance of larger and smaller pipes.
--
Ed Sirett - Property maintainer and registered gas fitter.
The FAQ for uk.diy is at http://www.diyfaq.org.uk
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

Sorry, a confusing analogy, it was simply there to explain that you cannot substitute a simple ratio where terms with powers are present.
I am aware that we are not adding different pipe lengths together, I am just curious to find out equivalent lengths at which, for a given (mass) flow rate, the pressure drop on 2 pipes of differing diameters will be the same. That involves the balancing of two pressure equations, apparently having radius **4 terms in each. The confusion that I am experiencing is that equivalent length tables published on the CDA site appear to be proportional to cube of the radius, not the fourth power.
--
fred
Plusnet - I hope you like vanilla
<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>
fred wrote:

Are you trying to say that x^k / y^k != (x/y)^k ?
Shurely shome mishtake...
--
Andy

<% if( /^image/.test(type) ){ %>
<% } %>
<%-name%>

## Site Timeline

• ### Old Wilson Wallflame oil boiler - where can I get/rent a combustion test kit/flue gas test...

• - previous thread in UK Do-It-Yourself Forum
• Share To

HomeOwnersHub.com is a website for homeowners and building and maintenance pros. It is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.