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

In fluid mechanics what does pressure energy mean?

  1. Apr 11, 2016 #1
    In Bernoulli equation we use pressure energy... Pressure* volume gives the unit of energy... But what does this mean..... If someone says it is potential energy per unit volume, then what does datum head refers to???... Please help... Can't go ahead without having thorough knowledge about this..
  2. jcsd
  3. Apr 11, 2016 #2
    Pressure energy is simply pressure. Force per unit area, or energy per unit volume are just two ways to express the same thing.

    Pressure and potential energy per unit volume are related by Bernoulli's equation, but they are not the same thing.
    [itex]P + 1/2*\rho*v^2 + \rho*g*h = constant[/itex]
    [itex]P[/itex] is the pressure (energy/volume)
    [itex]1/2*\rho*v^2[/itex] is the kinetic energy per unit volume
    [itex]\rho*g*h[/itex] is the potential energy per unit volume

    Have you dove to the bottom of a pool before? If so, you've experienced the exchange of potential energy for pressure energy. The water at the bottom of the pool doesn't have the same potential energy as the water at the top, but it is at a higher pressure.

    The head form of Bernoulli's equation simply divides through by density * force of gravity. This leaves the equation in units of length. For why that's called the head form, I'll defer to someone more knowledgeable.
  4. Apr 11, 2016 #3


    User Avatar
    Science Advisor
    Gold Member

    Pressure is still, in a certain sense, a form of potential energy. If you want to view it that way, then it is more akin to the potential energy in a compressed spring than it is to gravitational potential energy. Also like potential energy, it must be "expended" in order to increase the kinetic energy (in the form of dynamic pressure) of a flow.
  5. Apr 11, 2016 #4
    I think you are referring to the thermodynamic pressure, due to the movement of air molecules. The kinetic and potential pressure energies are just due to the net velocity and relative height of a fluid respectively.
  6. Apr 12, 2016 #5
    U can see that in the equation u have written....the pressure energy (P) and potential energy (ro*g*h) are the same...isnt it???.....really confused
  7. Apr 12, 2016 #6
    please elaborate more...with the spring as eg.
  8. Apr 12, 2016 #7
  9. Apr 12, 2016 #8


    User Avatar
    Science Advisor
    Gold Member

    I am not really sure how much more elaborate I can make this. Bernoulli's equation is a conservation of energy statement. The ##\rho v^2/2## term (dynamic pressure) is kinetic energy, the ##\rho g z## term (hydrostatic pressure) is the gravitational potential energy, and ##p## (static pressure) is sort of like spring potential energy in the sense that it comes from a bunch of moving molecules being in some space together and as you relax how much it is compressed, the pressure goes down and is transferred into some other sort of energy. In the case of a flow that is subject to Bernoulli's equation, that other form of energy is either kinetic or gravitational depending on the system.
  10. Apr 12, 2016 #9
    I understand your confusion. No, they are not the same. However, they are closely related.

    Imagine a column of water 10 meters tall open to the atmosphere. Assume zero velocity and incompressible (water is very nearly incompressible.)
    With no velocity, the kinetic energy portion will drop out and leave:
    [itex]P + \rho*g*h = constant[/itex]

    Now image two points in the water; point one at the top and point two at the bottom. You can use the reduced Bernoulli's equation to describe their relationship.
    [itex]P1 + \rho*g*h1 = P2 + \rho*g*h2[/itex]
    For the h terms, h1 is 10m and h2 is 0m (at the bottom of the column.)
    This leaves you with:
    [itex]P1 + \rho*g*10 m = P2[/itex]
    Now, because point one is open to the atmosphere we know that P1 (the static pressure at point one) is equal to about 101 kPa (atmospheric pressure) or more usefully, zero gauge. Taking the second option (0 gauge pressure) leaves this:
    [itex]\rho*g*10 m = P2[/itex]
    So you can see that the static pressure at the bottom of the column is equal to the potential at the top of the column.
    For the sake of completeness:
    [itex]P2 = \rho*g*10m = 1000 kg/m^3 * 9.81 m/s^2 * 10 m = 100kPa (gauge) [/itex]
  11. Apr 12, 2016 #10
    If you are calling "pressure energy" P*V in the Bernoulli equation, then you should be using the form of the Bernoulli equation expressed in terms of ##P/\rho##. This quantity has units of energy per unit mass, and is equal to the pressure times the volume per unit mass. If you have a fixed specified control volume, then, from the open system version of the first law of thermodynamics, ##P/\rho## at the exit of the control volume represents the amount of work required to push a unit mass of fluid out of the control volume against the back pressure of the downstream fluid. ##P/\rho## at the entrance to the control volume represents the amount of work required to push a unit mass of fluid into the control volume by the forward pressure of the upstream fluid. If you want to call these quantities "pressure energy," knock yourself out.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted