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Fluid mechanics force balance question

  1. Oct 25, 2004 #1
    Please see the attached image to see what I'm talking about.

    I understand how the forces (in terms of the pressures and areas) are equated, but then it says 'for small delta x' and from there on I'm confused.

    It's something to do with differentiation, but I don't get it.

    Can someone explain in simple terms what this all means?

    Carrying out a force balance in the x and z direction.


    p(x). ∆z − p(x+∆x). ∆z = 0

    For small ∆x,

    p(x+∆x) = P(x) + (∂p/∂x). ∆x

    Hence, ∂p/∂x = 0

    Which confirms that the hydrostatic pressure does not change in the horizontal plane; i.e., p ≠ f(x,y).


    p(z). ∆x − p(z+∆z). ∆x − ∆W = 0

    Where ∆W = ρ g ∆x∆z

    assuming ρ is constant over the small element.

    For small ∆z, and as p = f(z)

    p(z+∆z) = p(z) + (dp/dz).∆z

    Hence dp/dz = -ρg

    Stuff in brackets after p is a subscript of p i.e. not a function p of...

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    Last edited: Oct 25, 2004
  2. jcsd
  3. Oct 25, 2004 #2


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    Hi Fritz,

    Welcome to PF!

    For a small delta x, the pressure at the right hand side is the pressure at the left hand side + the slope of the pressure gradient (in the x direction) times the displacement delta x.

    Basically, it's saying that delta x is so small that you can assume that the slope of the pressure is approximated as a linear dp/dx. Since the pressure on the two sides is equal, the slope needs to be zero (which is what the third line says).

    Hope that helps.
  4. Oct 25, 2004 #3
    I'm still confused.

    So the pressure is a function of x on the left and right side; p of x and p of (x + delta x) respectively.

    The rate of change of pressure on the RHS w.r.t. x multiplied by the displacement delta x is approximately zero, since delta x is negligible.

    Is this right?

    What exactly is the pressure function on the RHS?

    I'm confused about the exact meaning of px+∆x (RHS) and delta p by delta x.
  5. Oct 25, 2004 #4
    Is this fluid statics or mechanics? Or is fluid statics a sub-division of fluid mechanics (with fluid dynamics being the other sub-division)?
  6. Oct 25, 2004 #5


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    Pressure is a function of x (and z... but it can be considered seperately), and has the values of px on the left side and px+delta x on the right side. The subscripts are just designating the point in space which is being looked at.

    No, since delta x is negligible, the rate of change of p can be approximated as a straight line with slope dp/dx. Furthermore, since delta x is negligible, the value of p on the right hand side is

    [tex]p(x) + \frac{\partial p}{\partial x} \times \Delta x [/tex]

    (the value on the left hand side) plus (the slope as x changes) times (the distance between the sides). Basically, all the equations are doing is turning a complex slope into a simple (linear) one by looking at a really small piece. If you take a look back in your MV Calculus book, all this is doing is taking a differential.

    It's a bit tricky the first time you see it, but once you get it once, you'll get it every time you see it (and you'll see it a bunch in engineering classes).
    Last edited: Oct 25, 2004
  7. Oct 25, 2004 #6
    If dp/dx is the rate of change of the pressure w.r.t. x, then surely if delta x is negligible (zero), dp/dx would also be zero (if you multiply anything by zero, you get zero.
  8. Oct 25, 2004 #7


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    Really really small does not mean zero.

    If you're looking at two points on a curve to find the slope of the line connecting them and bringing the points closer and closer together, as the distance between them approaches zero, the slope of the line connecting the two points doesn't go to zero... it goes to the derivative.
  9. Oct 25, 2004 #8
    I don't really understand where dp/dx came from in the first place.
  10. Oct 25, 2004 #9


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    -->Look up differential approximations in your calculus text.<--

    They're using the fact that the pressure on the right and the left are equal to prove that the slope is zero for the entire z=constant line. You can't prove that unless you take a differentially small piece and find that the slope is zero. Otherwise, it would be possible that the pressure increases and then decreases between the two points.
  11. Oct 25, 2004 #10
    When you say z = constant line, you mean that it's a horizontal line (with the value on the z-axis being constant)?
  12. Oct 25, 2004 #11
    Maybe I'm too stupid for engineering.
  13. Oct 25, 2004 #12


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    If I had a dollar every time that I said that... I'd have a lot of dollars. :tongue2:
  14. Oct 25, 2004 #13
    You're an aerospace engineer! Is it possible to do a masters in aerospace engineering if you've got a degree in mechanical engineering?
  15. Oct 25, 2004 #14
    The slope of the z=constant line is the same as dp/dx?
  16. Oct 25, 2004 #15


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    Yes. How easy it is depends on what you're focusing on, though.

    If you're going into materials, you won't have any problem. I've actually got several ME classmates, and we only give them a little bit of a hard time about it :tongue2: . If you're shooting for propulsion or something like that, you may have to take some junior/senior classes first.

    The slope of z=constant is dz/dx - how the z value changes as a function of x.

    dp/dx is how the pressure is changing as a function of x.
  17. Oct 26, 2004 #16
    It's just clicked.
  18. Oct 27, 2004 #17
    This is what I love about math/engineering. You get so frustrated and depressed, because you don't understand some relatively simple concept. Then, all of a sudden, it clicks and you understand it.

    That's the best feeling in the world.
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