Head loss and Volumetric flow in piping?

Question

How do you calculate flow rates in parallel piping with different head losses steming from the same initial piping stream and pump?

Question is for my husband. Thanks! :)

Answer 1

i suggest you to get an assume calculation in the same head loss condition.. Then the result giving tolerance up to 10% if the deferentiation up to 5 %. By the way if you or a friend of you interesting for packaging & printing engineering, visit my blog : www.interpak.blogspot.com,  I'm a chemical engineer, my next blog about process control in Industry.  Interesting ? wanna joint or share your capability, contact in [e-mail]

Answer 2

For liquid flowing thru a pipe, the equation for calculating pressure drop can be given as follows:

Pin - Pout = 0.000216*f*L*d*(Q^2/D^5)

where:

Pin - Pout = Pressure drop in the pipe

f = Friction Factor in the pipe

L = Equivalent Pipe length in feet

d = Density of the liquid in lbs per cubic foot = 62.4 for water

Q = Volumetric Flow Rate in GPM

D = Pipe inside diameter in inches.

* = times sign

Q^2 = Flow rate in GPM squared.

D^5 = Inside diameter to the 5th power

If the inlet pressures of the two pipes are the same, and the outlet pressures are at atmospheric, then the pressure drop in the two lines has to be the same, and with a lot of work, you can solve for the flow rate in each line. This would be the maximum flow rate assuming your pump had sufficient output to furnish the maximum flow to each pipe.

Usually we know what flow rate we want in each line and we install a flow control valve in each so that we can restrict the flow to what we want it to be.

The flow control valves make the calculation much easier.

I'm sure you are going to have to install some sort of shutoff valve in each line that could be used to throttle the flow.

Answer 3

Use a spreadsheet. Set up pressure drop calculations for each leg, preferably using the Colebrook equation. Use the Excel goal seek feature to numerically iterate volume in each leg so that the pressures are identical. When the volumes match the pump capability, you are done.

BTW, the Colebrook equation  has a complex form that can not be explicitly solved. However, subtract one side so the sum is equal to zero, and use a macro with the goal seeker to set equal to zero. The equation closes quickly, so not too many iterations should be required.