1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question re. the Pressure Energy of a Fluid

  1. Apr 27, 2013 #1
    I’m confused as to how energy per unit weight can be quantified as a head of fluid. I have tried to understand this from a couple of angles but am getting confused. I’ll go through my methodology to see where you can spot the cracks.

    If energy = work done = (force * length over which it acts) then pressure energy of a fluid is Pressure*Area*Length

    A column of water exerts a downward force, being its weight, and so the pressure required to support a column of water is equal to the weight of the column divided by the base area of the column. So, if the pressure exerted by the column of water is equal to

    mass density*gravity*height

    then the pressure at the bottom is equal to

    mass density*gravity*height

    but acts vertically upward. The energy of the fluid supporting this head would then be

    pressure*area*length = mass density*gravity*height*area*length

    However, what is confusing me is what the final term length is. From my reading it seems to suggest that the column of water is the governing factor, in which case the length would be the height of the column so the pressure energy would be:

    mass density*gravity *area* height^2

    However, on looking at a column of water I just can’t see how this is correct. If the column was split into incremental parts then surely the pressure which acted upward would only move 1 increment by a distance of the height (to the top) whereas another increment would only be moved halfway up the column? Does this not mean that the pressure energy at the base of the column would be more like force*average distance over which it acts, i.e. height/2?

    Cheers for the help
     
  2. jcsd
  3. Apr 28, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Welcome to PF!

    Hello Sir James! Welcome to PF! :smile:
    No, work done = force "dot" displacement of the point of application of the force.

    (it has nothing to do with the length of the column)
     
  4. Apr 29, 2013 #3
    Hello Tiny Tim, thanks very much.

    I'm still a little lost with this one though. In Understanding Hydraulics (Hamill, 2011) it states that the pressure energy per unit weight of a fluid is equal to the height of the head.

    Then total energy indicated by a head is:

    head * weight

    which can also be written

    head * (mass density*gravity*area*head)

    As the (mass density*gravity*area*head) is a force, it would suggest that the head is, in this case at least, equal to the displacement from the point of application of the force?

    What I don't understand is how we can regard this force as constant over this displacement.
     
  5. Apr 29, 2013 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    exactly correct :smile:
    a head of water assumes that the height stays constant

    if you remove water from one end, you have to replace it at the other end …

    you're calculating the work done as if the column collapses :wink:

    (but examining pressure in terms of energy really isn't appropriate …

    pressure is a force, and you can easily find the pressure of a column of water using an ordinary force diagram)
     
  6. Apr 29, 2013 #5
    To me, this seems to be a combination of two separate, but related, questions. How can pressure be thought of as energy per unit weight? and Why is this considered head (height of a column)?

    The first question is related to units. But pressure cannot really be considered energy per unit weight. It is really energy per unit volume. Here is how all this works out:

    p = F/l2 = (F)(l)/l3=energy per unit volume

    Now let's divide pressure by density:
    p/ρ = ((F)(l)/l3)[itex]\div[/itex](m/l3)=(F)(l)/m=energy per unit mass

    Now let's divide by gravitational acceleration g:

    p/(ρg)=(F)(l)/(m)(g)=energy per unit weight

    But p/(ρg) represents the height that the pressure p can raise a column of fluid of density ρ. This height is call the "head" of fluid. So, the energy per unit weight (as interpreted above) is also equal to the heat of fluid.
     
  7. Apr 29, 2013 #6
    Tiny-Tim and Chestermiller thanks very much, your answers have helped me to clear up what I was confused over. I'm looking at this in relation to permeability values of soil so need to get it right before I bring in the Darcy Coefficient.
     
  8. Apr 29, 2013 #7
    I can help you in this area too. I've had considerable experience with groundwater flow in confined aquifers, single phase.
     
  9. Apr 30, 2013 #8
    In that case Chestermiller I do have another question. I am trying to get my head around the equation for the velocity of flow through a soil.
    The equation is:

    V = Ki

    Where:
    V = velocity
    K = Darcy coefficient of permeability
    i = Hydraulic Gradient = (head at point 1-head at point 0/horizontal length between points 1 and 0)

    I guess my main question is how the hydraulic gradient in this equation should be interpreted. Does the component “i” in the equation for velocity mean that we are effectively multiplying by a ratio that gives the pressure energy that has been converted to kinetic energy, per horizontal unit distance?
     
  10. Apr 30, 2013 #9
    Forget about kinetic energy. Kinetic energy does not apply to groundwater flow. The magnitude of the kinetic energy and its gradient are too small to be important. What we are dealing with here is viscous flow through the tiny pores of rock, and the viscous effects dominate.

    In your equation above, V is what they call the superficial velocity. This is the flow rate averaged over the cross sectional area of both rock and pores. This differs from the pore velocity which is averaged only over the cross sectional area of the pores, and is therefore higher.

    Also, in your equation above, K is what I am accustomed to calling the hydraulic conductivity, not the permeability. The hydraulic conductivity has units of velocity, and is expressed in terms of the permeability by:

    [tex]K=\frac{kρg}{μ}[/tex]

    where k is the Darcy permeability, and, if I remember correctly, has units of length squared.

    The hydraulic gradient is the derivative of head with respect to distance. (You have had calculus, correct?) The velocity V is a vector that is pointing in the same direction as the head gradient vector. Incidentally, there should be a minus sign in your equation on the right hand side. The flow is from high heat to low head.

    Get yourself a copy of the book Groundwater by Freeze and Cherry.

    Chet
     
  11. May 2, 2013 #10
    Much appreciated Chet, I was going wrong by not looking at the hydraulic gradient as a vector. This meant I didn't understand why the length L was sometimes measured horizontally, sometimes vertically or sometimes in the direction of flow. I now see that this just depends on the direction of the velocity you are trying to obtain.
    Also noted the minus sign I missed off.
    Cheers again!
     
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: Question re. the Pressure Energy of a Fluid
  1. Fluid Pressure (Replies: 3)

Loading...