# Conflicts between Bernoulli's Equation and Momentum Conservation?

The well known Bernoulli's equation states that $P+\frac{\rho V^2}{2}=c$

However, a simple momentum conservation considering $P_1$ and $P_2$ acting on two sides, and velocity changes from $V_1$ to $V_2$, yields $P_1+\rho_1 V_1^2=P_2+\rho_2 V_2^2$, which differs from Bernoulli's by a coefficient $\frac{1}{2}$.

What is going on here? I understand the derivation of both, just want to know how to explain the conflict.

• In what way does a $\rho V^2$ term represent a momentum? It's a specific kinetic energy term. – Gert Nov 23 '15 at 2:44
• The term is a result of the momentum conservation. $(P_2-P_1) \Delta t A=\rho_2 V_2\Delta t V_2-\rho_1 V_1\Delta t V_1$. – David Nov 23 '15 at 2:50
• $(P_2-P_1) A=\rho_2 V_2 V_2-\rho_1 V_1 V_1$ is not dimensionally consistent. Left is $\mathrm{N}$, right $\mathrm{kg m^{-1} s^{-2}}$. – Gert Nov 23 '15 at 3:01
• Sorry I forgot A on the right side. – David Nov 23 '15 at 3:19
• If $A$ and $\rho$ are constant then $v$ has to be constant, due conservation of mass flow. – fibonatic Nov 24 '15 at 9:18

## 3 Answers

The factor $\frac 12$ comes from the relation $\vec v \cdot\nabla \vec v = \nabla \frac{\vec v^2}{2} + (\nabla\times\vec v)\times \vec v$ in the momentum conservation equation $$\rho \left(\frac{\partial \vec v}{\partial t}+\vec v \cdot\nabla \vec v\right)=\vec g-\nabla p$$

(Sorry to post this as an answer, but I can't comment your post yet because of reputation)

EDIT: formalism as follows

• $\vec u \equiv \vec v$ is the velocity field
• $p$ is the pressure field
• $\vec g$ is the gravity field

If you develop the equation, assuming that the gravity field derives from a potential such as $\vec g=-\nabla \phi$, and that the flow is in steady state i.e. $\frac{\partial \vec v}{\partial t}=0$ then: $$\nabla\left(\frac{\vec v^2}{2}+\frac p\rho+\phi\right)+\vec \omega\times\vec v=0$$, where $\vec{\omega}\equiv \nabla \times \vec{v}$ is the vorticity operator

At equilibrium, $W=\Delta E_k + \Delta E_p$ (work of the pressure forces, kinetic energy and potential energy)

• $W=p_{1}A_{1}(v_{1}\Delta t) - p_{2}A_{2}(v_{2}\Delta t)$
• $\Delta E_k=\Delta m(v^{2}_{2}- v^{2}_{1})/2$
• $\Delta E_p=\Delta mgh_{2}- \Delta mgh_{1}$
• Please specify what $u$ stands for. – Gert Nov 23 '15 at 3:13
• Thanks for the answer. I know where that 1/2 comes from. I just want to know why from momentum conservation, I got a different equation which 1/2 is not present. – David Nov 23 '15 at 3:23
• The difference is that the equation without the 1/2 is in integral form while the derivation by @neok is in differential form. If you apply divergence theorem to the equation above (and ignore the external forces), you will get a surface integral of the form $\int (\vec{v} \cdot \hat{n})\vec{v} \,\mathrm{d}A = - \int p \hat{n} \,\mathrm{d}A$, which then reduces to $P_1+\rho_1 V_1^2=P_2+\rho_2 V_2^2$ when the area of the two sides are equal and the velocities are uniform along the cross section. – Darwin Dec 30 '15 at 6:31

I think I figured it out. Bernoulli's assumption is incompressible flow. The equation yielding from momentum conservation always holds. When velocity is low (incompressibility holds), the two equations yields similar results.

The distinction between these two equations is that: $$p+ \rho u^2=constant$$, is valid only for 1D compressible flow while, $$p+(1/2) \rho u^2=constant$$, is valid for incompressible flow.

The difference arises because of the coupling of continuity and momentum equation in compressible flow. This coupling is absent for incompressible flow. You can see this by a simple derivation from 1D Euler's equation. Euler equation is basically the momentum equation where the viscous forces are neglected. $$\frac{dP}{\rho}+ u du + gdz = 0$$

Let's also neglect the body forces. Now this becomes: $$dP=- (\rho u) du$$

Where P is the pressure, $$\rho$$ is density and u is the 1D velocity. This is valid for both compressible and incompressible flows. For deriving the Bernoulli's equation, you simply integrate both sides.

$$\int{dP}=- \int{(\rho u) du}$$

If the density, $$\rho$$ is constant, flow is incompressible, and you can take $$\rho$$ out of the integral sign to get your Bernoulli's equation.

$$P= -\rho \int{u du} = -\rho \frac{u^2}{2}+constant$$ $$P+\frac{\rho u^2}{2}=constant$$

Now, consider the case of compressible flows. Here you cannot take the density $$\rho$$ out of the integral. Instead, you can use continuity equation for compressible flows, which says: $$\rho u=constant$$, to take $$(\rho u)$$ out of the integral. So, we have

$$P= -(\rho u) \int{du} = -\rho u * u+constant$$ $$P+\rho u^2=constant$$

You see? We had to use the continuity equation in order to get this. This is the coupling I'm talking about. Even if you derive the 1D compressible flow equation using a control volume, using $$P_1 , V_1 ; P_2 , V_2$$ at entry and exits, you'll still need to use the continuity equation $$P_1 V_1 =P_2 V_2$$ to get $$P_1 + \rho {V_1}^2=P_2 + \rho {V_2}^2$$. The derivation from Euler equation just helps to see the distinction very easily.