Head loss - Parallel pipes

[dianad]

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!

 

[BigInch]

No no no.

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

 

[dianad]

So, according to you, the example that i've attached that is related to solar pannels, the head loss is equal to the head loss of 1 collector and not that relation that i've mentioned, that has simillarities to electronics...

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!

 

[BigInch]

Maybe you're reading too many electrical books.

I'd look for a better understanding with an explanation of the Hardy-Cross method. Try this,

http://www.rpi.edu/dept/chem-eng/Biotech-Environ/NETWORKS/termproject.html

"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[vpl]

or heat transfer books

Patricia Lougheed

Please see FAQ731-376: Eng-Tips.com Forum Policies for tips on how to make the best use of the Eng-Tips Forums.

 

[IRstuff]

The parallel resistor equation is NOT an electrical equation, per se. It comes from a very simple paradigm; the flow in each leg is inversely proportional to resistance of that leg and proportional to the driving potential, hence,

flow = sum(flow_i)
= V*sum(1/R_i)

The effective resistanec of the paralleled system is the V/flow = 1/sum(1/R_i)

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

 

[stanier]

Download Epanet and model it. Epanet is free ware and will enable you to model networks.

 

[dianad]

Hi,

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.

 

[BigInch]

Yes, that would be correct.


"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[vzeos]

IRstuff

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 [Δ]P = 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
[Δ]P = pressure drop

 

[BigInch]

Ya the 1/K thing just gets people confussed when they try to use it with pipeflow.

"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[vzeos]

Correction to my previous post.

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 + …

 

[DIRTYDAVE]

Check out 2500 SOLVED PROBLEMS IN FLUID MECHANICS & HYDRAULICS
It is one of SCHAUM'S SOLVED OUTLINE SERIES
You can follow through & see how just about any kind of flow problem is solved.

 

[vzeos]

I found the attached Google preview of 2500 SOLVED PROBLEMS IN FLUID MECHANICS & HYDRAULICS. It looks like it's worth buying.

http://books.google.com/books?id=oB...&+HYDRAULICS&source=gbs_toc_s&cad=1#PPA280,M1

 

[dianad]

Hi there,

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!

 

[BigInch]

Where is it?

"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[dianad]

Hi,

Sorrey BigInch, but i had a problem with my internet connection.
I've already posted it.

Please, take a look!

 

[Compositepro]

A similar topic came up in one of the electrical forums about the usefulness the hydraulic analogy in understanding electricity. The analogy can be very useful but has to be applied correctly. The biggest mistake is equate the wrong quantities like: a pipe is a resistor.

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.

 

[BigInch]

A bigger pipe diameter could have a bigger resistance, if it were longer.

"If everything seems under control, you're just not moving fast enough."
- Mario Andretti- When asked about transient hydraulics

http://virtualpipeline.spaces.msn.com

 

[vzeos]

Compositepro,

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, [Δ]E = 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.