Head loss - Parallel pipes
Head loss - Parallel pipes
(OP)
Hi,
This question was already debated too much times in thsi forum, but i still doesn't understand one thing that questions all that i've learned until now.
In series pipes, everyone knows that to determine the head loss we have to sum all the head losses of each consumer.
But in parallel flow, i've red every kind of possibilities to solve this problem.
I've red a lot that if we have for example 5 consumer that have the same pressure loss (for example solar collectors), the total head loss is equal to the head loss of 1 collector. But in other books i red that to determine the pressure loss in parallel flow, we have to make some kind of simillarity to electronics. So in this case some books refer:
1/Rtotal = 1/R1 + 1/R2 + 1/r3 + ...
So, how do we solve this simple problem that many books solve them differently.
I've attached an image to show an example.
Thanks!
This question was already debated too much times in thsi forum, but i still doesn't understand one thing that questions all that i've learned until now.
In series pipes, everyone knows that to determine the head loss we have to sum all the head losses of each consumer.
But in parallel flow, i've red every kind of possibilities to solve this problem.
I've red a lot that if we have for example 5 consumer that have the same pressure loss (for example solar collectors), the total head loss is equal to the head loss of 1 collector. But in other books i red that to determine the pressure loss in parallel flow, we have to make some kind of simillarity to electronics. So in this case some books refer:
1/Rtotal = 1/R1 + 1/R2 + 1/r3 + ...
So, how do we solve this simple problem that many books solve them differently.
I've attached an image to show an example.
Thanks!





RE: Head loss - Parallel pipes
When in series, head losses can be different in each component and you can add the head losses, but flow through each component is equal.
When in parallel, flows proportion through all components, such that the head loss between any common points are equal, thus flows change and head loss is equal.
in this diagram you know the values of, Cv1, Cv2, Cv3, Cv4
assume n = 1/2 (or some other appropriate fraction according to your particular head loss vs flow equation)
You also get to assume one value of head, lets say we know the head at B = 10.
The head losses, dH are,
dHab = dHcd+dHdb3 = dHcd+dHdb4
Now just proportion all flows to make the above true.
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
I must refer that this way of calculation is the one that i've always learned. What made me confuse was the appearance of that relation 1/R = sum (1/Ri) in many books.
Can you clarify to me why does this relation appears in many places, has the explanation for parallel flow head loss calculation method?
Thanks for the time and for the good answer!
RE: Head loss - Parallel pipes
I'd look for a better understanding with an explanation of the Hardy-Cross method. Try this,
http
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
Patricia Lougheed
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RE: Head loss - Parallel pipes
flow = sum(flowi)
= V*sum(1/Ri)
The effective resistanec of the paralleled system is the V/flow = 1/sum(1/Ri)
I couch this specifically to be, hopefully, consistent with your fluid flow problem, so that in any problem where flow, or whatever, can be described this way, the equation will look very similar.
TTFN
FAQ731-376: Eng-Tips.com Forum Policies
RE: Head loss - Parallel pipes
RE: Head loss - Parallel pipes
First, thank you all for the answers!
Biginch, your link was very helpful. I've studied in the pas hardy-cross method but i doesn't use it a long time ago.
To confirm all of this, int my example, because the head loss in esch collector is the same, the total head loss would be equal to the head loss of 1 collector...correct?
Thanks once more.
RE: Head loss - Parallel pipes
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
The parallel resistor equation is an analog to the pipe flow equation but it is not the same. The pipe flow equation in terms of resistance, k, is P 1 - P 2 = kQ 2 for liquids and P 1 2 - P 2 2 = kQ 2
For gases.
Parallel resistances are added as follows:
1/√k total = 1/√k 1 + 1/√k 2 + ...
The analysis of k does not lend very much insight to the pipe designer, but if you compare ΔP = kQ 2 to Darcy's liquid flow equation you'll see that k=cfLQ 2 /D 5 , then,
1/√{ cfLQ 2 /D 5 } total = 1/√{ cfLQ 2 /D 5 } 1 + 1/√{ cfLQ 2 /D 5 } 2 + ...
If c, f and L are the same for pipes 1 and 2, the equation becomes
1/√{ Q 2 /D 5 } total = 1/√{ Q 2 /D 5 } 1 + 1/√{ Q 2 /D 5 } 2 + ...
Where:
C =constant
f = friction factor
Q = flow rate
L = length
D = diameter
ΔP = pressure drop
RE: Head loss - Parallel pipes
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
k=cfL/D 5
1/√{ cfL/D 5 } total = 1/√{ cfL/D 5 } 1 + 1/√{ cfL/D 5 } 2 + ...
and
1/√{ 1/D 5 } total = 1/√{ 1/D 5 } 1 + 1/√{ 1/D 5 } 2 + ...
RE: Head loss - Parallel pipes
It is one of SCHAUM'S SOLVED OUTLINE SERIES
You can follow through & see how just about any kind of flow problem is solved.
RE: Head loss - Parallel pipes
I found the attached Google preview of 2500 SOLVED PROBLEMS IN FLUID MECHANICS & HYDRAULICS. It looks like it's worth buying.
http://
RE: Head loss - Parallel pipes
Thak you all! Y've understand everything that you all said!
Thanks for the clarification!!!
I have another post with another title, that you may like to give it a shot, because is one of those big stupid doubts that once in a while come to my head!
THANKS!
RE: Head loss - Parallel pipes
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
Sorrey BigInch, but i had a problem with my internet connection.
I've already posted it.
Please, take a look!
RE: Head loss - Parallel pipes
It is true that the flow resistance of a pipe is like an electrical resistor. But a bigger pipe is not a bigger resistor.
I admire Dianad's attempts to develop an understanding of basic principles behind the equations. In fact eqautions are useless without this understanding.
RE: Head loss - Parallel pipes
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
Your observation that a pipe is not analogous to a resistor is correct. A pipe has a uniformly distributed resistance and is analogous to an electrical wire or conduit. A resistor is more like an orifice plate which has a concentrated resistance.
Nonetheless, you may recall that in the early days of network analysis using analog computers, the McIlroy analyzer used nonlinear resistors, called Fluistors, to represent network pipes. This was done for a good reason. The Fluistors had the equation, ΔE = R I 1.85, which had the same form as the Fritzsche gas flow equation, (P12 - P22) = k Q 1.85. Thus, this was a case where strict adherence to the correct analogy would not be very helpful. I would say that the analogy needs to be applied correctly and intelligently.
RE: Head loss - Parallel pipes
- if the flows in parallel are through equal paths
and
- the head loss though the inlet and the outlet collectors is negligible with respect to the head loss in the branches,
then the flow rates in each branch will be the same.
Please note dianad that the second condition above is unlikely to be true, especially if the dynamic head in the collectors is non negligible.
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: Head loss - Parallel pipes
"When in parallel, flows proportion through all components, such that the head loss between any common points are equal"
Tells me that, if head loss is equal, flows are equal and that doesn't require that flow losses be negligable anywhere.
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com
RE: Head loss - Parallel pipes
BigInch, the sketch from dianad has 6 (supposedly) equal branches and two headers. For the flow through each branch to be the same, what I require is that the loss+dynamic head in the headers is negligible with respect to the loss in the branches.
prex
http://www.xcalcs.com : Online engineering calculations
http://www.megamag.it : Magnetic brakes for fun rides
http://www.levitans.com : Air bearing pads
RE: Head loss - Parallel pipes
"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics
http://virtualpipeline.spaces.msn.com