Help: Flow in parallel pipe formulas
Help: Flow in parallel pipe formulas
(OP)
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?
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?





RE: Help: Flow in parallel pipe formulas
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.
RE: Help: Flow in parallel pipe formulas
Good luck....
RE: Help: Flow in parallel pipe formulas
Sorry for the confusion... All pipe lengths are 1000ft.
The 20ft, 10ft, and 0ft were for the elevations.
RE: Help: Flow in parallel pipe formulas
Look at this tread
Thread378-25256 posting should be possible to expand to 3 unknown (and three equations)
Best Regards
Morten
RE: Help: Flow in parallel pipe formulas
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.
RE: Help: Flow in parallel pipe formulas
Hardy-Cross is a tedious calculation, thats why people always think there is an easier solution...
Bob
RE: Help: Flow in parallel pipe formulas
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
RE: Help: Flow in parallel pipe formulas
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
RE: Help: Flow in parallel pipe formulas
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.
RE: Help: Flow in parallel pipe formulas
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
RE: Help: Flow in parallel pipe formulas
Perhaps Lohms law will help? try this web site:http://www.microhydraulics.com/EFSWEB/213a.htm
RE: Help: Flow in parallel pipe formulas
Best Regards
Morten
RE: Help: Flow in parallel pipe formulas
That is what I've assumed too...
Hope kevinsst5 can elucidate this imbroglio...
Regards
fvincent
fvincent
Figener S/A
RE: Help: Flow in parallel pipe formulas
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.
RE: Help: Flow in parallel pipe formulas
Grtz,
RE: Help: Flow in parallel pipe formulas
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.