Bernoulli Equation for Irrotational Flows

Recall that the angular velocity can be defined in terms of vorticity:

$$\omega =\frac{1}{2}\zeta =\frac{1}{2}\nabla \times \mathbf{V}$$

An irrotational flow is a special case of inviscid flow where

$$\omega =\vec{0}:{\omega }_x={\omega }_y={\omega }_z=0$$

This happens obviously when vorticity is zero.

Recall the Euler’s Equations of Motion for Inviscid Flows:

$$\rho \mathbf{g}-\nabla p=\rho \left[\frac{\partial \mathbf{V}}{\partial t}+(\mathbf{V}\cdot \nabla )\mathbf{V}\right]$$

For a steady flow (no changes with respect to time), the time derivative disappears:

$$\rho \mathbf{g}-\nabla p=\rho (\mathbf{V}\cdot \nabla )\mathbf{V}$$

The acceleration term simplifies:

$$(\mathbf{V}\cdot \nabla )\mathbf{V}=\frac{1}{2}\nabla (V^2)-\mathbf{V}\times (\nabla \times \mathbf{V})$$

and the gravitational force can be expressed as a gradient:

$$\mathbf{g}=-g\mathbf{k}=-\nabla (gz)$$

Thus Euler’s Equation becomes:

$$-\nabla p-\rho \nabla (gz)=\rho \left[\frac{1}{2}\nabla (V^2)-\mathbf{V}\times (\nabla \times \mathbf{V})\right]$$

Re-arranging:

$$\nabla p+\frac{1}{2}\rho \nabla (V^2)+\rho \nabla (gz)=\rho \mathbf{V}\times (\nabla \times \mathbf{V})$$

Dividing through with $\rho$:

$$\frac{\nabla p}{\rho }+\frac{1}{2}\nabla (V^2)+\nabla (gz)=\mathbf{V}\times (\nabla \times \mathbf{V})$$

To derive Bernoulli’s equation, we integrate this expression along a streamline. Recall that a streamline is represented by a path vector $ds$:

$$d\mathbf{s}=dx\mathbf{i}+dy\mathbf{j}+dz\mathbf{k}$$

Dotting our previous expression with $ds$:

$$\frac{1}{\rho }\nabla p\cdot d\mathbf{s}+\frac{1}{2}\nabla (V^2)\cdot d\mathbf{s}+\nabla (gz)\cdot d\mathbf{s}=(\mathbf{V}\times (\nabla \times \mathbf{V}))\cdot d\mathbf{s}$$

which simplifies to

$$\frac{1}{\rho }dp+\frac{1}{2}d(V^2)+gdz=(\mathbf{V}\times (\nabla \times \mathbf{V}))\cdot d\mathbf{s}$$

If the flow is irrotational, the right-hand side is zero, since $\nabla \times \mathbf{V}=0$:

$$\begin{array}{r}(\mathbf{V}\times (\nabla \times \mathbf{V}))\cdot d\mathbf{s}=0\\ \therefore \frac{1}{\rho }dp+\frac{1}{2}d(V^2)+gdz=0\end{array}$$

Now we integrate:

$$\begin{array}{rl}{\int }_1^2\frac{1}{\rho }dp+{\int }_1^{2}d\left(\frac{1}{2}{V}^2\right)+{\int }_1^{2}gdz& =0\\ \frac{p}{\rho }+\frac{1}{2}{V}^2+gz& =c\end{array}$$

If the fluid is incompressible, the equation applies across different streamlines:

$$p_1+\frac{1}{2}\rho V_1^2+\rho g{z}_1=p_2+\frac{1}{2}\rho V_2^2+\rho g{z}_2$$

Under what conditions is $(\mathbf{V}\times (\nabla \times \mathbf{V}))\cdot d\mathbf{s}=0$?

• Along a streamline:

$$\mathbf{V}\parallel d\mathbf{s}⟹(\mathbf{V}\times (\nabla \times \mathbf{V}))\perp \mathbf{V}⟹(\mathbf{V}\times (\nabla \times \mathbf{V}))\perp d\mathbf{S}$$

• Irrotational flows:

$$\nabla \times \mathbf{V}=0$$

For irrotational flows, Bernoulli Equation can be applied between any two points in the flow field.