# Bernoulli's Principle equation help

## Main Question or Discussion Point

Hi everyone and thank you in advance. My first problem is: if v1=constan in a section S1, it should not be ∑ F = 0 ? Or maybe Σ M = 0 ? Because all forces that i recognize are gravity and pressure, but i never found the equation mg - PS = 0 or something similar.
I was thinking about something like F = d/dt (p) = (dm/dt * v + dv/dt*m) = (dm/dt * v + 0 ) = dm/dt * v but i don't know.

Can someone help me please? I'm going crazy.

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DEvens
Gold Member
Some context would help. It is pretty difficult to tell what you are referring to.

DEvens
Gold Member
Did you read the derivation of that equation? And did you read the text immediately following that equation? It reads as follows. "In other words, if the speed of a fluid decreases and it is not due to an elevation difference, we know it must be due to an increase in the static pressure that is resisting the flow."

So, who says v is a constant?

Gold Member
Sorry, you're right. I'm considering the equation 1/2 ρv^2 + ρgz + P = constant. I'm wondering why that "v" in the equation is a constant (in the related section S) if this equation is obtained considering pressure forces and gravity force ≠ 0.

for example here:
https://en.wikipedia.org/wiki/Bernoulli's_principle#Derivations_of_Bernoulli_equation
The ##v## in Bernoulli's equation need not be constant. Only that whole ##\rho v^2/2 + \rho g z + p## term needs to remain constant.

We are measuring the volume V = S * v * Δt, so v is a constant, infact we are talking about a stationary fluid flow.

DEvens
Gold Member
If A1v1 = A2v2 and A1,A2 are constants with A1≠A2 ⇒ v1 is constant.
If A1v1 = A2v2 and A1,A2 are constants with A1≠A2 ⇒ v1 is ##not## constant. I fixed it for you. In the example you linked to, A is the area of the flow, and A changes. So v has to change to conserve fluid.

Yes, but my problem is that in the section A1, v1 is constant. I know that v1 ≠ v2, but in their regions (A1*v1*Δt and A2*v2*Δt) they are constants, with v1 ≠ v2.

In a stationary flow, d/dt (v1) = 0, where v1 is the fluid velocity through the volume V1. I can't understand why, because we have 2 kinds of forces acting on our mass, and i can't find the equation Σ F = 0.

Gold Member
In a stationary flow, d/dt (v1) = 0, where v1 is the fluid velocity through the volume V1. I can't understand why, because we have 2 kinds of forces acting on our mass, and i can't find the equation Σ F = 0.
What do you mean you don't understand why? I am struggling to grasp at the source of confusion here. If ##\frac{dv_1}{dT} = 0## then by definition, ##v_1## is constant. You can't look at it quite the same way as a rigid body because in this case, you are dealing with a number of particles passing through a point, not a single particle with its own time history.

I know that d/dt (x) = 0 ⇔ x = constant (of time), it's obvious. I can't understand why we have a constant (v1) when i don't see the equation ∑ F = 0 or ∑ = r x F = 0. This v1 is the velocity of a mass (for example of water) through an entire volume of the tube and it's a constant of time, but the definition of force (pression) used by wikipedia in this case is F = m d/dt (v). I was thinking about the total force as the sum of gravity and pressure forces: 1) F = d/dt(p) = d/dt (m*v) = dm/dt * v + m * dv/dt = dm/dt * v because dv/dt = 0 but i don't know if it's correct to suppose this.

For example we have a mass of water flowing through an horizontal tube with a constant section for 10 cm, " 1/2 ρv^2 + ρgz + P = constant " where "ρ, g, z, P" are constants of time in that region, and so "v" is a constant of time too. But the definition of pressure itself is "P = F/S", and we have also the gravity involved in, so i can't realize why that "v" is a constant of time in that 10 cm, unless 1) works.

The force that will accelerate a small section (like a cylindrical slice) of fluid is given by the pressure difference between the two ends of the cylinder.
If the pressure is constant along the pipe there is no force and the speed is constant. If the pressure changes you have a force that accelerates the fluid.

Gold Member
You can't just apply a force balance to a single particle based on Bernoulli's equation. That isn't how it works. Bernoulli's equation is valid along a streamline (or in a control volume if the conditions in the volume are suitable) where ##v_1## and ##v_2## are some beginning and end states. The (average) force involved is the different between the pressures at the two states. If you want to start going smaller and smaller so that you have something that more resembles a force balance, you are better of going with a different model, such as the Euler equations (inviscid) or Navier-Stokes equations (viscous).

Force is not just
$$F = m\dfrac{dv}{dt}.$$
Newton actually said that
$$F = \dfrac{d(mv)}{dt} = m\dfrac{dv}{dt} + v\dfrac{dm}{dt}.$$
Since you have already assumed that ##\frac{dv}{dt} = 0## (which is required for Bernoulli's equation to be valid in the first place), then that reduces to
$$F = v\dfrac{dm}{dt},$$
which you might recognize as simply the basic equation for thrust.

Basically, though, the force balance in a fluid control volume is most conveniently handled as a balance of momentum, i.e. the net momentum going into a control volume has to equal the net momentum out of it, and when you take this to the infinitesimal scale, you get the Navier-Stokes or Euler equations depending on whether or not you include viscosity. If you don't take it to the infinitesimal scale, one possible result is the Bernoulli equation.

Perfect, thank you!