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B The pressure term in Bernoulli's equation

  1. Jun 30, 2016 #1
    Seeing as P(F/A) + KE + PE = Constant, if the fluid's flowing through a constriction (So the area's decreasing(Which would net an overall larger pressure)), How does the pressure term decrease? Does the Force decrease even more greatly than the Area, to net an overall lower pressure?

    Assuming we're going from a larger area, through a constriction, into a smaller area.

    Just to note: I know it's decreasing because either the KE or PE are increasing, i understand that. I'm just confused as to if it's actually the Force(P=F/A) increasing/decreasing significantly to make up for the Area increase/decrease at points.
     
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  3. Jun 30, 2016 #2

    FactChecker

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    Remember that the pressure decrease is where the velocity has already increased after the area decrease has already happened. So the pressure there is lower than the pressure that was pushing it into the constriction. The trade-off between average molecular velocity in the direction of flow and the average random molecular velocity that causes non-directional pressure can be imagined as a simple application of the Pythagorean Theorem.
     
  4. Jun 30, 2016 #3
    Say we want to represent Pressure as Force/Area though, and the cross-sectional area of a (horizontal) pipe decreases(in this hypothetical situation), we will have a decrease in pressure (F/A in this situation), and an increase in velocity. So is force decreasing by a larger amount, to make up for the area that is also decreasing? (Otherwise, a smaller area would just net an even larger pressure(F/A))

    Also thank you for your reply, i really appreciate the help
     
  5. Jun 30, 2016 #4
    The converging walls are also exerting a force on the fluid in the direction opposite to its direction of flow. So the upstream pressure has to overcome this too.
     
  6. Jun 30, 2016 #5

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    You should be careful about your terminology. The term "pressure" is used in different ways. There is dynamic pressure, d, and static pressure, s, which correspond to the kinetic and potential energy, respectively. They total to the total pressure, p = d + s. The connection between dynamic pressure, d, and velocity, v, is d = ½ρv2. In a steady-state condition, p is constant and the rate of material flowing through the tube at every point is equal. So the product of the area, A, and v is constant. You should be able to piece those facts together to get your equation.
     
  7. Jun 30, 2016 #6
    So the P term in bernoulli's equation P+½ρv^2+ρgh=constant isn't the same type of pressure usually represented as P=F/A?
     
  8. Jun 30, 2016 #7
    Actually, that is exactly what it is.
     
  9. Jun 30, 2016 #8
    So when the pressure term increases/decreases from a change in velocity, the force(in the term P=F/A) would be considered to increase/decrease to overcome the new smaller/larger area, right?

    My example thought is below:
    If we were going from an area of low velocity, to high velocity, the pipe's cross-sectional area would be smaller, so for pressure to decrease the force term would have to decrease to overcome the new, smaller area. Otherwise the force over the new, smaller area, would just cause pressure to increase.

    I'm just confused as to how the P term physically looks as it changes, as it's represented as F/A, if it's as simple as Force overcoming the change in area, or if i'm missing something
     
    Last edited: Jun 30, 2016
  10. Jun 30, 2016 #9
    No. This is not the whole story. The fluid is accelerating as it flows from the larger cross section region to the lower cross section region. In order of this to happen, the upstream force must be higher than the downstream force. You know, Fnet = ma, right? But, the converging wall is pushing back on the fluid. This is part of the force balance too. However, the net effect is still that the upstream pressure must be higher than the downstream pressure. The combination of these factors is captured by the Bernoulli equation.
     
  11. Jun 30, 2016 #10
    This is exactly what i was looking for, thank you.
     
  12. Jun 30, 2016 #11

    rcgldr

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    In case you're wondering about the effect of the converging wall pushing back on the flow, it's taken care of in the frame invariant version of Bernoulli's equation for constricted flow:

    ## p1 - p2 = \frac{\rho}{2} \ (v2 - v1)^2 \ ( \frac{a1 + a2}{a1 - a2} )##

    http://www.loreto.unican.es/Carpeta2012/EJP%28Mungan%29Bernoulli%282011%29.pdf [Broken]
     
    Last edited by a moderator: May 8, 2017
  13. Jul 1, 2016 #12
    Actually, one more question. If we mathematically represent the P=(F/A) term as P=Mass*Acceleration/Area, what exactly would be increasing/decreasing? Does Mass increase/decrease at all? Or is it only acceleration increasing/decreasing?
     
  14. Jul 1, 2016 #13
    You can't represent P = F/A = ma/A because F is not equal to ma. Fnet = ma, where Fnet is the NET force acting on a body. F = ma does not apply to each and every individual force acting on a body. How could a set of different forces all give the same acceleration?
     
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