Help: Flow in parallel pipe formulas

I'm looking for help with a flow in parallel pipe problem:

3 pipes in parallel, all 1000ft in length.
Pipe 1=1" dia at 20ft
Pipe 2=3" dia at 10ft
Pipe 3=2" dia at 0ft
A 6" dia pipe with 1000 gpm flow feeds into the 3 parallel pipes. I need flow in each pipe. All pipes are sch 40.

I've tried using: h=(f(L/D))* V2/2g but the values seem too small.
According to Qtot= Q1+Q2+Q3, The sum of the 3 pipes should equal 1000 gpm. I'm not even close. What am I doing wrong?

 

[TD2K]

It's a trial and error solution. Draw a sketch of the system.

I'd do it on a spreadsheet with an overall mechanical energy balance. Fix the outlet pressure due to the outlet pressure and static head/elevation you have on each line. Adjust the inlet pressure until you have 1000 gpm total through all 3 lines.

 

From your information you only have two pipes, the third has no length. You want to use an equation called Hardy-Cross. Like TD2K said, it's an iterative solution.

Good luck....

 

BobPE,
Sorry for the confusion... All pipe lengths are 1000ft.
The 20ft, 10ft, and 0ft were for the elevations.

 

[MortenA]

I dont think its trial and error

Look at this tread

thread378-25256 posting should be possible to expand to 3 unknown (and three equations)

Best Regards

Morten

 

Mr.Kevinsst5,

Let,
Head at,
Branching Point be, Ho (metre)
First branch Pipe Outlet,H1=(6.097 + 10.3)metre
Second branch Pipe Outlet,H2=(3.048 + 10.3)metre
Third branch Pipe Outlet,H3 = 10.3 metre

Pressure at,
Branching Point,Po = (rho*g*Ho) N/sq.m
First Branch Outlet,P1=(rho*g*H1) N/sq.m
Second Branch Outlet,P2=(rho*g*H2) N/sq.m
Third Branch Outlet,P3=(rho*g*H3) N/sq.m

Calculate R1, R2, & R3 frictional resistance of pipe.
such that discharge^2 * R = Frictional Head Loss in Pipe.

Discharge through,
Pipe 1, Q1 = (P1-Po) / R1 --- 1
Pipe 2, Q2 = (P2-Po) / R2 ---- 2
Pipe 3, Q3 = (P3-Po) / R3 ---- 3

Equating Q1+Q2+Q3 = Qtotal,
The only unknown will be Ho in above equation, find the same. Substitute in equation 1, 2 & 3 and find discharge in individual pipes.

with Thanks,
Sachi.

* Always use elctrical network like approach for easy solution.

Note: Please comment whether my approach is correct.

 

Sure it's trial and error, or more appropriately iteration. Some variables have to be assumed since not all are know, or you wouldnt need to calculate in the first place. Usually velocity is assumed, then iterated until it converges to the real velocity.

Hardy-Cross is a tedious calculation, thats why people always think there is an easier solution...

Bob

 

[fvincent]

Maybe a general formula for multi parallel pipes would help:

Consider the following, for incompressible fluid:

f= frictional coefficient
L = pipe lenght
D = inner diameter
S = section area
ks = singularity coefficient
rho = specific mass
m = mass flowrate
Q= volume flowrate

Each pipe has its pressure drop calculated as:

Delta p = (SUM(f*L/D)+ SUM(ks))/(2*S^2*rho) *m^2 ,or
Delta p = (SUM(f*L/D)+ SUM(ks))/(2*S^2) *rho* Q^2

Let us use the second expression.

Delta p along each pipe i can alternatively be represented as:

Delta p = Ki * Q^2, where

Ki = (SUM(f*L/D)+ SUM(ks))/(2*S^2)* rho

Now consider:

n = number of parallel pipes
Keq = equivalent resistance of the multi parallel pipes
Ki = resistance of pipe i (any one and all from 1 to n), calculated as above
Kj = resistance of pipe j (any one and all from 1 to n)
PI for product
SUM for sum
< > instruction for counting of i or j
SQ square
SQR square root

Then

PI<j=1;n> (Kj)
Keq = ------------------
SQ(SUM<j=1;n>(SQR(PI<i=1;i=!j;n>Ki)))

For instance,

For n= 2

Keq = k1*k2 /(k2^.5 + k1^.5)^2

For n = 3

Keq = k1*k2*k3 / ((k2*k3)^.5 + (k1*k3)^.5 + (k2*k3)^.5)^2

And so on...


After having obtained Keq, you can obtain the model for the complete system as a circuit in serie.
After solving the complete system (and so total flow), you just calculate pressure drop along Keq as:
Delta p = Keq*Q^2

With Delta p, go back to each Ki and calculate the flow along pipe i as:
Qi = SQR(Delta p / Ki)

The sum of all Qi <i=1;n) must be equal to total flow, as verification.

As to the different elevations, disconsider them because it is a closed loop and so the initial and final extremities are the same for all pipes.

Obs: branching singularities are assumed to be either very low or constant disregarding (vi /v main branch)

Good luck

 

[fvincent]

Please correct


For n = 3

Keq = k1*k2*k3 / ((k2*k3)^.5 + (k1*k3)^.5 + (k1*k2)^.5)^2

instead of

Keq = k1*k2*k3 / ((k2*k3)^.5 + (k1*k3)^.5 + (k2*k3)^.5)^2

Sorry

 

Mr. fvincent,

You cannot disconcider the Elevation, because it is the pressure against which water is discharged.

And pressure difference is the main factor which decides discharge.

For Example:
Concider a Single Pipe.
If,Available Absolute Head at Inlet = H
Pipe Outlet is at an Elevation of K metres
Frictional Resistance in Pipe = R
Then,
Discharge = rho*g* ( H - (K+10.3) ) / R,
So you cannot disconcider elevation.


I feel my approach narrated in my previous posting will be
correct. Please comment your suggesions.

with Thanks,
Sachi.

Note: 10.3 metre is added, for atmospheric Pressure.

 

[fvincent]

Dear Mr Sachidhanandhan

I guess I dealed with the wrong problem...

I thought all pipes joined again, and so the circuit could have been treated the way I've done...

If the three pipes reach the same tank whose water level is above 30 ft, my approach is still valid... (please comment)

If not, it is necessary to recur to your solution which is of course valid for all cases

I am sorry I haven't paid due attention...

fvincent
Figener S/A

 

[jmw]

Nice thread.
Perhaps Lohms law will help? try this web site:

http://www.microhydraulics.com/EFSWEB/213a.htm

 

[MortenA]

How can it be parallel flow when the dont have the same (total) elevation change. I assumed (incurrecly?) that the would join together again after the 1000 ft so that the dP for each pipe would be equal.

Best Regards

Morten

 

[fvincent]

Dear Morten

That is what I've assumed too...

Hope kevinsst5 can elucidate this imbroglio...

Regards
fvincent fvincent
Figener S/A

 

[abeltio]

After going through all the answers nobody suggested the good old graphical method...
Try plotting the 3 friction losses on a deltaH (ft) vs. Q (gpm)with their corresponding static heads.
Add all three plots in parallel (i.e. Qtotal = Q1+Q2+Q3 at each static head).
To the resulting plot add in series the deltaH for the inlet pipe (6in)(i.e. at each Q add the static heads)
For Qtotal = 1000 gpm, intersect the resulting plot then draw a horizontal line the Q where this line intersects the original 3 friction losses plots are the flows through each branch.
This has some peculiar results... there will be no flow through the top branch until the deltaH of the system goes above 20ft, same for the 10ft branch...there will be no flow until the deltaH reaches 10ft.

Perhaps you would like to consult: Pump Handbook
by Igor Karassik, William C. Krutzsch, Warren H.Fraser and Joseph P. Messina (Editors) 2nd Edition
Mc-Graw Hill Book Co.
ISBN 0-07-033302-5
Section 8.2 Branch Line Pumping Systems - Page 8.79 Fig 4 seems to match your problem statement.
Page/Fig References correspond to the 2nd Edition.
Hope this helps.

 

The formula posted by fvincent does not make sense after rearranging it there should be the formula delta p= Req*Q^2 which is not the case you get a formula that looks like Delta p= Req*(Q+2*SQRT(Q1)*SQRT(Q2)) which is obviously not correct...

Grtz,

 

[parsec]

Actually, it appears as though you have four unknowns. The three flows (Q1,Q2,Q3) and the common head at the junction, call it h1. Note that the lines probably do not connect at the end otherwise the elevations would be immaterial.

Fortunately, you have four equations namely:

Q1^2/2g*K1=h1-0
Q2^2/2g*K2=h1-10
Q3^2/2g*K3=h1-20
Q1+Q2+Q3=1000

The K in this instance is fL/d times the conversion for Q to velocity which acounts for the size of the piping. You should be able to substitute and solve the resulting quadratic form in the normal fashion.

Let me know if this helps.