# How to derive Bernoulli's equation in steady irrotational flow?

Condsider an incompressible, inviscid, irrotational fluid with constant density $$\rho$$. Let $$\overrightarrow u$$ be its velocity field, $$p$$ its pressure field and $$\overrightarrow F$$ be an external body force given by some potential $$\chi$$ so that $$-\nabla \chi = \overrightarrow F$$. The momentum equation reads $$\frac{ \partial \overrightarrow u}{\partial t}= -\nabla(\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi).$$ I was asked to show that $$\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi$$ is constant in steady flow, and the impled line of reasoning is to say that it simply follows from $$0= -\nabla(\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi).$$ But I can only see why we may conclude that $$\frac{p}{\rho}+\frac{1}{2}|\overrightarrow u|^2+\chi$$ is a function of time. And we could have this be any function of time, by absorbing it into $$\chi$$. Am I missing something?

• In steady flow you have $\partial/\partial t=0$ so the right hand side of your equations cannot change as a function of time. Jan 21 at 17:02
• @Newbie Here by steady I mean $\partial \overrightarrow u / \partial t = 0$, what does you $\partial / \partial t$ act on? I see that all the gradients in my equations are time independent, but that wouldn't rule out some time dependence in $p/ \rho+u^2/2+\chi$? Jan 21 at 17:12
• Are you worried that your right hand side cannot change as a function of space but can change as a function of time? Jan 21 at 17:15
• @Newbie Yes exactly. Jan 21 at 17:16
• Hint: The term involving $|\vec u|^{2}$ cannot change as a function of time since you already have set $\partial\vec u/\partial t=0$ on the left hand side. Jan 21 at 17:18

## 1 Answer

Your observation is correct. For steady irrotational flow of an incompressible, inviscid fluid with constant density, the Euler equation of momentum reduces to

$$\nabla \left(\frac{p}{\rho} + \frac{1}{2} |\mathbf{u}|^2 + \chi \right) = 0$$

Integrating the components with respect to the spatial variables, we get the general solution

$$\frac{p}{\rho} + \frac{1}{2} |\mathbf{u}|^2 + \chi = c(t),$$

where the arbitrary function $$t \mapsto c(t)$$ changes nothing about the flow field and can be absorbed into the pressure. Recall that the pressure field corresponding to a particular velocity field is never uniquely determined since an arbitrary reference pressure can be added. The pressure and body-force potential only affect the velocity through their spatial gradients and adding an arbitrary function of time has no impact.

More generally for irrotational flow where $$\nabla \times \mathbf{u} =0$$, the velocity field is the gradient of a potential, $$\mathbf{u} = \nabla\phi$$, and the momentum equation reduces to

$$\frac{\partial\nabla \phi}{\partial t}= -\nabla \left(\frac{p}{\rho} + \frac{1}{2} |\nabla \phi|^2 + \chi \right),$$

which upon integration yields

$$\frac{\partial \phi}{\partial t}+\frac{p}{\rho} + \frac{1}{2} |\nabla \phi|^2 + \chi = c(t)$$

Again, in steady flow where $$\frac{\partial \phi}{\partial t} = 0$$, the appearance of the time-dependent $$c(t)$$ is not excluded.

• Thank you very much for your reply! I think this settles it for me, and I also happened to have a chat to the relevant professor who gave me a similar reply. Jan 27 at 23:19
• @LobLikelyhood: You're welcome.
– RRL
Jan 28 at 3:38