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Calculate flow rate in parallel pipes

  1. Jul 28, 2014 #1
    So I have a 30" pipe that splits into 3 parallel paths, each 24" in dia. The pipes converge into a 30" outlet and then flows at about 40,000 gallons per minute at full head pressure. Assuming the same head pressure and one path closed off (i.e. now only flowing through 2 parallel paths), how can I determine my new flow rate from the outlet? Testing this would be a real hassle. I'm just looking for a good estimate, not necessarily right on the money. Thanks
     
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  3. Jul 28, 2014 #2

    Simon Bridge

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    Hint: Ohm's Law.

    You want to know the current for two identical resistors in parallel given the current for three and the same voltage across them.
     
  4. Jul 29, 2014 #3
    Good analogy. But do I know the current? I know the "current" for 3 open parallel paths, but if I eliminate one path, I can't exactly assume it's the same flow rate as before - right?
     
  5. Jul 29, 2014 #4

    SteamKing

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    You know the total flow into the parallel pipes must equal the total flow out. This is the continuity relation. If you block one pipe but don't change the flow rate into the network, the the flow velocity must increase.

    You must also have the same pressure drop (or resistance to flow, to continue the electrical analogy) from the entrance of the parallel pipes to their exit, but having all of these pipes the same diameter and length makes it easy to satisfy this requirement. If the flow velocity is the same in each pipe, then the pressure drop thru each pipe should also be the same, as long as the pipes are identical.
     
  6. Jul 31, 2014 #5

    Simon Bridge

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    That is correct - the same pressure difference can produce a different flow rate like the same voltage can produce different currents. Your main advantage is that you know the pipes are identical so the resistance to flow will be identical and that the pressure difference (voltage) is the same.

    Write out the equations - use dummy variables for everything you don't know.
    Just pretend it's an electric circuit.
     
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