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Hydrostatic pressure and Gravitational Potential Energy

  1. Aug 24, 2015 #1

    K41

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    So say I had a glass of water on a table ,with the glass filled to the top. The bottom of the glass, the water would have a higher pressure than the top of the glass. This is the concept of hydrostatic pressure etc.

    But how does GPE fit into all of this? Does the water at the top of the glass contain more GPE than the water at the bottom? If that is the case, how can there be equilibrium (assume there is no net movement of water) in the glass of water when the derivation of the hydrostatic pressure only takes into account the pressure of the water and its weight above a certain fixed point?
     
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  3. Aug 24, 2015 #2

    Orodruin

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    Yes, the water at the top is higher and therefore has more potential energy per mass.

    It is unclear why you would think there is a problem here. Changing places of two volumes of water would involve raising the potential energy of the volume going up and vice versa, resulting in a net energy change of zero and thus no reason for the system to do that.
     
  4. Aug 24, 2015 #3

    BvU

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    Hello,

    this hydrostatic pressure expression in fact comes from an equilibrium calculation for a little chunk of liquid: it doesn't move, so the forces on e.g. the bottom square of a little cube must be in equilibrium. On this bottom square of the cube: from below the liquid pressure times the area, from above the liquid pressure from the liquid above the little cube times the area plus the weight of the cube. That gives a very simple equation for the pressure as a function of the height in the liquid.

    The potential mgh is from the constant graviational force mg in an energy balance : ##\Delta ## PE = F x ##\Delta ## h
     
  5. Aug 24, 2015 #4

    K41

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    On that little chunk of liquid, from above, would there not also be some differential potential energy as well as the weight of the fluid and the liquid pressure?
     
  6. Aug 24, 2015 #5

    Orodruin

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    No, this is just the gravitational force which is a volume force and not a surface force, it acts directly on the liquid and not on its surface.
     
  7. Aug 24, 2015 #6

    K41

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    I don't understand this. The chunk of liquid could be anywhere in the fluid. In fact, its easier to assume its far aware from the surface, perhaps in the middle of the glass. I don't know what you mean by volume or surface forces. A surface can be a volume to depending on the definition of its thickness??

    EDIT:

    Or are you saying the GPE involves the ENTIRE liquid inside the glass whereas the hydrostatic pressure (as discussed earlier, refers to differential elements?
     
    Last edited: Aug 24, 2015
  8. Aug 24, 2015 #7

    Orodruin

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    No, a surface is two-dimensional and a volume is three-dimensional. The surface I am talking about is not the surface on top of the glass, it is the boundary of the chunk you are looking at.

    The surface of the chunk is its boundary, the only forces acting on the surface are due to the pressure on the surface. The chunk is also subject to a gravitational force, this force is proportional to the mass contained in the chunk, which is proportional to its volume and not to its surface.
     
  9. Aug 24, 2015 #8

    K41

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    Okay, in another way, what prevents the fluid at the top of the glass from releasing its higher potential energy into kinetic energy (and lets assume the bottom of the glass is the reference zero GPE)?
     
  10. Aug 24, 2015 #9

    Orodruin

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    There is a pressure force from the liquid below it counteracting the gravitational force.
     
  11. Aug 24, 2015 #10

    K41

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    So hence my confusion arises from why, if this pressure force is involved in counteracting gravitational energy, why does it not appear in the derivation for hydrostatic pressure ,unless is the row*g*h term can be considered a "potential" energy of sorts?
     
  12. Aug 24, 2015 #11

    Orodruin

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    You seem to have a confusion relating to energy and forces, they are not the same thing. If there is a difference in total energy between two setups, then the system will generally have a force striving to drive the system to the lower energy state. This is not the situation here. In order for a chunk of the water to go down, another chunk would have to go up and the energy would be the same. Forces do not counteract energies. In equilibrium, forces counteract other forces.

    The hydrostatic pressure comes, as BvU said, from solving the equilibrium equations in a static fluid. This computation is quite trivial, you do not even need calculus to do it.

    Big nono. This expression does not even have the units of an energy (it has units of pressure). Also, not that the Greek letter ##\rho## is transcribed as "rho".
     
  13. Aug 24, 2015 #12
    In fluid mechanics, the gravitational potential energy per unit volume of fluid is often written as ##φ=ρgz##. So in a hydrostatic situation, the variation of pressure with elevation is written as:

    $$\frac{dp}{dz}=-ρg=-\frac{dφ}{dz}$$

    or p + φ = const.

    As others have said, the pressure difference between the bottom and top of a fluid parcel supports the weight of the parcel so that it is in hydrostatic equilibrium. If the parcel is an oddball shape, the pressure variation around the surface of the parcel supports its weight.

    Chet
     
  14. Aug 25, 2015 #13

    K41

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    Thanks. So I can think of the rho * g*z term as a GPE per unit volume, although I should be extremely careful and realize this is a balance of forces and not energy?
     
  15. Aug 25, 2015 #14
    Yes.
     
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