Pipe diameter & flow rate
Forrest@hotmail.com Jim
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
Aidan
Jim wrote:
> 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
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.
Forrest@hotmail.com Jim
> 1gpm ;-)
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
Aidan
Jim wrote:
> Just what I am looking for. thanks
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.
Helen Deborah Vecht
I'm sure they taught me that flow is proportionate to the radius^4 a
long time ago...
--
Helen D. Vecht: helen...@zetnet.co.uk
Edgware.
manat...@hotmail.com
Jim wrote:
> > > 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)
> >
>
> Just keeping the numbers simple. Only looking for a rough idea.
>
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
Matt
fred
Deborah Vecht <helen...@zetnet.co.uk> writes
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
Doctor Drivel
"Matt" aka Lord Hall <pa...@duluxthesh.aggydog.com> wrote in message
news:r8e9n1h5d52ic0pjd...@4ax.com...
Lord hall, leaking via these inferior fitting is a real possibility . Well
spotted. 10/10
Aidan
manatba...@hotmail.com wrote:
> I think you missed the joke.
But it was a very tiny joke and easily overlooked.
James
> flow out of the pipe must be 1gpm. Where else can it go?
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
Aidan
James wrote:
> try this site
> http://www.efunda.com/formulae/fluids/calc_pipe_friction.cfm#calc
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.
Norman Billingham
"fred" <n...@for.mail> wrote in message news:y0nsRJAJFMdDFw6H@y.z...
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
Ed Sirett
Disagree. Based on the OD (or radius) it's 4.6:1 based on the ID (or
radius) it's 5:1
>
>
> 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
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
Gas fitting FAQ http://www.makewrite.demon.co.uk/GasFitting.html
Sealed CH FAQ http://www.makewrite.demon.co.uk/SealedCH.html
fred
Ed Sirett <e...@makewrite.demon.co.uk> writes
>On Fri, 11 Nov 2005 22:51:46 +0000, Norman Billingham wrote:
>
>>
>> "fred" <n...@for.mail> wrote in message news:y0nsRJAJFMdDFw6H@y.z...
>>> In article <3130303037363...@zetnet.co.uk>, Helen
>>> Deborah Vecht <helen...@zetnet.co.uk> writes
>>>>
>>>>long time ago...
>>>
>>> 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.
>>>
>Disagree. Based on the OD (or radius) it's 4.6:1 based on the ID (or
>radius) it's 5:1
>
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.
>>
>> 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
>
>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
>>>> 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.
>>>>
>>Disagree. Based on the OD (or radius) it's 4.6:1 based on the ID (or
>>radius) it's 5:1
>>
> 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.
>
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
fred
Ed Sirett <e...@makewrite.demon.co.uk> writes
>On Sat, 12 Nov 2005 21:46:45 +0000, fred wrote:
>
>
>>>>> 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.
>>>>>
>>>Disagree. Based on the OD (or radius) it's 4.6:1 based on the ID (or
>>>radius) it's 5:1
>>>
>> 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.
>>
>
>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
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 . . .
Aidan
Ed Sirett wrote:
>
> > Poiseuille's equation relates laminar flow through a cylidrical tube to
> 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.
Probably D'Arcy's equation is more relevant
h= 4fl/d v^2/2g
h is head loss due to friction, f is friction factor.
I have always used tables of pressure loss per m at given flow rates
for the pipe material & BS. Probably an equation won't deliver accurate
results over the full range.
Ed Sirett
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.
fred
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.
Andy Wade
> Sorry, a confusing analogy, it was simply there to explain that you cannot
> substitute a simple ratio where terms with powers are present.
Are you trying to say that x^k / y^k != (x/y)^k ?
Shurely shome mishtake...
--
Andy
fred
<spamb...@ajwade.clara.co.uk> writes
these days I will get to the bottom of this and will report back I <weekly
affirm> - sorry, 'promise' just seems too strong a form of words for the effort
I am likely to give it, given the effort involved and the level of satisfaction
likely to be gained from a correct result :-/.
Aidan
fred wrote:
> **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.
CDA site has info for copper tube. N.B., the d dimensions quoted are
outside diameters (15mm, 22mm, 28mm, etc). The fluid equations
mentioned above would use the inside diameters/radii.
I found a fluids textbook whilst looking for another book ( which I
never did find). D'arcy equation uses an empirical factor f to describe
the friction losses in turbulent flow. The difference (to Poseuille's)
is due to the complexity fo turbulent flow, so that a solely analytical
solution is not possible.
> 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.
D'Arcy equation; h= (4fl/d) x (v^2/2g)
v = Q/area = Q/(pi d^2/4)
Substituting gives h = (4fl/d) x 16Q^2/ 2g pi^2 d^4
h = flQ^2/3.03d^5; near enough to h = flQ^2/3d^5
h = rQ^2 where r is a resistance coefficient. r =fl/3d^5.
Above gives head losses for volumetric flow rate. Pressure loss P = hg
x density.
I'm sure you can figure it out further if you want to.