# Derivation of the energy equation in fluid dynamics

I'm working through Acheson's 'Elementary Fluid Dynamics' and i'm having trouble deriving the conservation of energy equation (exercise 1.4)

$\frac{d}{dt} \int_V \frac{\rho \vert u\vert^2}{2} dV = -\int_{\partial V}(p +\rho \phi + \frac{\rho \vert u\vert^2}{2})\mathbf{u} \cdot \mathbf{n}dS$

where $\mathbf{u}$ is the velocity of the fluid, $\rho$ is the density of the fluid, $\phi$ is the gravitational potential and $p$ is the atmospheric pressure.

No matter what I do I can't get the factor of $\frac{1}{2}$ on the left hand side. Here's my attempt:

Start from Eulers equation, $\frac{\partial \mathbf{u}}{\partial t} + \mathbf{\omega} \times \mathbf{u} = - \nabla (\frac{p}{\rho} +\phi + \frac{1}{2}\vert u\vert ^2)$, and multiply by $\rho$ and dot both sides with $\mathbf{u}$ to get

$\frac{\partial \rho \vert u\vert^2}{\partial t} = - \nabla \tau \cdot \mathbf{u}$, where $\tau = \frac{p}{\rho} +\phi + \frac{1}{2}\vert u\vert ^2$.

Now, using the identity $\nabla \cdot (\tau \mathbf{u}) = \nabla \tau \cdot \mathbf{u} + \tau \nabla \cdot \mathbf{u}$ and the incompresibility condition, this becomes $\frac{\partial \rho \vert u\vert ^2}{\partial t} = - \nabla \cdot (\tau \mathbf{u})$. Integrating over a closed volume $V$ and using the divergence theorem gives

$\frac{d}{dt} \int_V \rho \vert u\vert^2 dV = -\int_{\partial V}(p +\rho \phi + \frac{\rho \vert u\vert^2}{2})\mathbf{u} \cdot \mathbf{n}dS$.

Can anyone see where the missing 2 went? Sorry about the poor latex, i'm learning that too!

• Does $u = \Vert u \Vert$? Commented Jun 28, 2014 at 22:15
• yes, thanks for pointing that out! i've edited it now. Commented Jun 28, 2014 at 22:19

The square of the norm of the velocity field is

$$|\mathbf{u}|^2=u_x^2+u_y^2+u_z^2.$$

Taking the partial derivative with respect to time we get

$$\frac{\partial}{\partial t}|\mathbf{u}|^2=2u_x\frac{\partial u_x}{\partial t}+2u_y\frac{\partial u_y}{\partial t}+2u_z\frac{\partial u_z}{\partial t}= 2\mathbf{u} \cdot\frac{\partial \mathbf{u}} {\partial t}.$$

Hence,

$$\mathbf{u} \cdot\frac{\partial \mathbf{u}} {\partial t}=\frac1{2}\frac{\partial}{\partial t}|\mathbf{u}|^2$$

• It should be:$\frac{\partial}{\partial t}|\mathbf{u}|^2= 2\mathbf{u} \cdot\frac{\partial} {\partial t}\mathbf{u}$.
– mike
Commented Jun 28, 2014 at 22:19
• Yes -- that's what I have -- last term of second equation. Thanks
– RRL
Commented Jun 28, 2014 at 22:21
• Brilliant! I can't believe it was something so simple! Thank you Commented Jun 28, 2014 at 22:40
• You're very welcome
– RRL
Commented Jun 28, 2014 at 22:52