# Deriving Bernoulli's equation from Euler's equation

I am hoping that some of you can point me in the right direction. I am doing a project regarding Navier-Stokes', Euler's and Bernoulli's equations. I am currently looking for source material that can help me understand the derivation of Bernoulli's equation from Euler's equation of motion. Ideally the source would cover the "transformation" from this version of Euler's equation:

$$\rho\left( \frac{\partial u}{\partial t} + u(u \cdot\nabla)\right)=-\nabla p + \rho g$$

to a version of Bernoulli's equation, eg. $$P_1 + \frac{1}{2} \rho v_1^2 +\rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 +\rho g h_2.$$ I have already looked around on the internet and in previous posts on this forum; however, I have not been able to find anything that describes this derivation in greater detail. Does anyone know of some material, where a derivation like this is described/explained.

Start from Euler (correctly written-- yours is not quite right) $$\rho\left(\frac{\partial u}{\partial t}+ (u\cdot \nabla) u\right)= -\nabla (p+\rho g z)$$ and use $$[(u\cdot \nabla) u]_j=(u_i\partial_i) u_j= u_i \partial_j u_i + u_i(\partial_i u_j-\partial_j u_i)$$ in the form of the vector identity $$(u\cdot \nabla) u=- u\times (\nabla\times u)+\nabla \left(\frac 12 |u|^2\right)$$ to write it as $$\rho\left(\frac{\partial u}{\partial t}- u\times (\nabla\times u)\right)= -\nabla \left(p+\rho g z+\frac 12\rho|u^2|\right)$$ Now consider the steady flow in which $$\partial u/\partial t=0$$ and take a dot product with $$u$$ on both sides. You get $$(v\cdot \nabla)\left(p+\rho g z+\frac 12\rho|u^2|\right)=0$$ which is Bernoulli --- i.e the quanity inside the parentheses is constant along a streamline.

For compressible flow you need to write $$\frac 1 \rho \nabla p= \nabla h$$ where $$h$$ is the specific enthalpy ($$H=E+PV$$ per unit mass) and then Bernoulli becomes $$h+ g z+\frac 12|v^2|= constant.$$

• Thank you; however, I am not quite sure I understand why it is that $$\rho g z$$ must be inside of the parenthesis of the $\nabla$-operator with the pressure, P?
– user292543
Commented Mar 19, 2021 at 12:34
• Because $-\nabla (\rho g z) = \rho {\bf g}$ (where ${\bf g}= -g {\bf e}_z$) is the force due to the weight of the fluid. (assuming incompressible). This is what you have in your Euler eqation. Commented Mar 19, 2021 at 12:40
• What was wrong in your version of Euler is that you have $u(u\cdot \nabla)$ rather than $(u\cdot \nabla)u$. I assume that this was just a typo rather than a misunderstanding. Commented Mar 19, 2021 at 12:43
• Okay! Yes, you are correct. That is just a typo on my part. This is a little advanced for me; however, I think it is beginning to make sense to me.
– user292543
Commented Mar 19, 2021 at 12:54
• Good... Have fun exploring it all! Commented Mar 19, 2021 at 12:55