Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Controlling static pressure?

  1. Sep 7, 2016 #1
    Hi everyone,
    Is there any mechanism/method/device to control static pressure?
    Noting that, total pressure = static + dynamic pressure.
    The scenario I am thinking is just water flowing through a pipe (assume incompressible), and I want to control the static pressure at two points along the pipe. I don't want to set the flow rate upstream in the pipe.

    Any thoughts? Physically possible?
    Thanks!
     
  2. jcsd
  3. Sep 7, 2016 #2
    Add downstream pumps.
     
  4. Sep 7, 2016 #3

    russ_watters

    User Avatar

    Staff: Mentor

    You are going to be a lot more detailed about this in order for us to help you. Can you provide a drawing of what you are trying to do?
     
  5. Sep 7, 2016 #4
    Well, consider a simple network of pipes. Two inlets and one outlet. The outlet is into atmosphere.
    One way to drive flow through network is to hook up the two inlets to the same reservoir. Now, the pressure of the reservoir should be treated as a total pressure at the inlets when solving for the flow through the network. In the steady state (assuming the pressure of the reservoir doesn't change) the two inlets may have different static pressures but they will have the same total pressures.

    I am curious to know if there is an experimental set-up where you control the static pressure at the two inlets. Say, for example, I want the static pressure at both inlets to be the same (the total pressures at each inlet could be different) and I would like to be able to "turn a knob" to change the static pressure at the inlets (while making sure the two inlets have the same static pressure).
     
  6. Sep 7, 2016 #5

    russ_watters

    User Avatar

    Staff: Mentor

    How about valves?
     
  7. Sep 7, 2016 #6
    Yea, pressure regulator valve

    Edit: I guess for static, might need to be an adjustable pressure relief valve
     
    Last edited: Sep 7, 2016
  8. Sep 7, 2016 #7
    Okay great, thanks guys. I'll look more into the regulators.
     
  9. Sep 7, 2016 #8

    JBA

    User Avatar

    Stated another way, it appears you are simply saying you want an equal flow velocity through both inlets, that is the only way to achieve an equal static pressure at those two points. The flow rates will still be different if the two inlets are of different sizes or flowing backpressures.
     
  10. Sep 8, 2016 #9
    I agree that if the total pressure is the same at the two inlets, then in order to have the same static pressure at the two inlets as well, the flow velocity would also need to be the same. Flow rate could be different.

    I'll go into more detail about what I am considering, to make sure I understand regulator valves correctly.

    Suppose I am analyzing a pipe network, same as above, two inlets and one (or more) outlets into atmosphere. And I disregard minor losses at the junctions of pipes in the network. Also, I assume fully developed laminar flow at the inlets. And I treat the network like a circuit by writing down the Poiseuille flow equation (https://en.wikipedia.org/wiki/Hagen–Poiseuille_equation) for each pipe segment, which describes the static pressure loss over a straight pipe for a given flow rate. Now I set the static pressure at both inlets to be the same, and I write down conservation of flow rate equations for the junctions. Now I have a system of equations that I can solve, which will tell me the flow rate through each segment of the network. I then can use the calculated flow rate at the inlets along with the static pressure I chose for the inlets and calculate the total pressure at the inlets. I will likely find different total pressures at the inlets, particularly if the network is asymmetric.

    Now I want to test my calculations. Hooking up my network to a reservoir is a different boundary condition than what I used in my calculations (it sets total pressure the same at the inlets, not static pressure). So what if I take a reservoir that is pressurized to a very high pressure. Now I attach a pressure regulator (like the single-stage one on this page, https://en.wikipedia.org/wiki/Pressure_regulator) to each inlet of my network and then attach the regulators to the reservoir, will this achieve the boundary conditions I assumed in my calculations? That is, can I use the valves to set the static pressure at each inlet the same, while allowing the total pressure at the inlets to be different from each other, assuming neither total pressure exceeds the very high pressure that the reservoir is at?

    Thanks again for the discussion everyone.
     
  11. Sep 8, 2016 #10

    JBA

    User Avatar

    On my first reading I was having a bit of a problem following your above and just now I started to read it again and saw something that I think is important to understand. That is that the Poiseuille calculation you used calculates the loss in both the Total Pressure and the Static Pressure in the pipe.

    The pressure losses in a pipe represent losses in energy as the fluid encounters the flow resisting element in the pipe and therefore a loss in Total Pressure.

    If you take a horizontal piece of pipe of constant inside diameter with some amount of flowing friction loss and use a pitot type tube sensor facing upstream to measure the total pressure at the inlet and at the outlet you will find that the total pressure at the outlet is equal to the total pressure at the inlet minus the calculated flowing pressure drop through the pipe and since the flow velocity will be constant through the pipe then the change in static pressure will likewise equal the inlet static pressure minus the calculated flowing pressure.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Controlling static pressure?
  1. Static pressure (Replies: 6)

  2. Static Pressure (Replies: 31)

Loading...