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Water pressure vs pressure drop

  1. Jul 15, 2017 #1
    Hi guys,

    I know the difference between these two things but I'm struggling with something that I hope someone could help me with.

    What dictates the water pressure needed for a "system" to work? I would imagine that if you've calculated your system to have a pressure drop of 1 bar (for example) does that mean that you need the water to start at the pressure of at least 2 bar in the system so that after the pressure drop the water is still at or above atmospheric pressure?

  2. jcsd
  3. Jul 15, 2017 #2
    In regards to the second question: Well, yes. If you want the water at the end of the system to be at least y = 1 [bar], and you know the pressure drops by d = 1 [bar], then of course you need at least y - d = x = 2 [bar]. But I think your first question is more interesting.

    Bernoulli's equation tells us that in the absence of friction, a fluid system contains a pressure head, velocity head, and elevation head, the sum of which must be constant between any two points in the system (provided no external work is introduced). i.e. hp1 + hv1 + hz1 = hp2 + hv2 + hz2. By subtracting one side from the other, this can be restated as the sum of the differences must equal zero. This means that in a system where the pressure head drops by one [bar], it must be the case that the velocity and elevation heads summed increase by one [bar].

    For example, consider a garden hose where the inlet pressure is two [bar] and the outlet pressure is one [bar]. If the outlet is held level with the inlet, then the elevation head does not change, so the velocity head must increase by one [bar], which is equivalent to a velocity difference of about 14 [m s-1]. Alternatively, to have the velocity head remain unchanged, the elevation head must increase by one [bar], which is equivalent to an elevation difference of about 10 [m].

    From the engineering standpoint, to determine the pressure needed for a system to work, you need to know the velocity and/or elevation difference you want to create along with the outlet pressure. From there, you work backwards through Bernoulli's equation to find the necessary inlet pressure.

    Friction introduces additional head losses that can be accounted for using empirical data for the type of pipe or duct that is used. External work also can be accounted for in a similar manner.
  4. Jul 15, 2017 #3
    It seems that your question is referring to frictional pressure drop in a pipe of constant cross section discharging to the atmosphere. is this correct?
  5. Jul 17, 2017 #4

    jim hardy

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    Your intuition is right. If you want water to run out the end of your system it needs to have at least an atmosphere left at that end. That's in an "open" system.

    A closed system is different, for example the condenser of a steam turbine which might be at absolute pressure of only 1/30 atmosphere.
    As a mechanical surely you're familiar with NPSH ?
    The 'minimum' pressure you need there is determined by the saturated pressure-temperature curve of water. You don't want cavitation(micro-boiling) inside the pump that's circulating condensed steam from the condenser hotwell back to the boiler.

    That's why power plant condensate pumps are so tall.


    Similar things go on in in your automobile radiator.
    As stylists dictated lower hoods radiators got smaller. They have to run at higher temperature to maintain same heat transfer to the air with less area, which means higher pressure. They have to prevent boiling at engine's hot spots, probably around the exhaust valves. I see radiator caps up to 17psi nowadays.

    old jim
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