# Substantial derivative of density in the derivation of mass conservation equation.

We are all taught the derivation of the mass conservation using a fixed Eulerian control volume in a typical fluid dynamics course.

That is, first we think about the total rate of change of mass in that control volume. $$\frac{d}{dt}\int_V\rho dV$$ Next, we say that in the absence of sinks or sources, the fluid that enters this volume contributes to the increase. The expression is : $$-\int_{\partial V}\rho\vec{u}\cdot \hat{n}dA$$

Now these expressions are equal, $$\frac{d}{dt}\int_V\rho dV = -\int_{\partial V}\rho\vec{u}\cdot \hat{n}dA$$

to give us the integral form of the mass equation.

Now, to get the differential form, we use the gauss-divergence theorem and do some strange manipulation to the total derivative on the LHS :

$$\int_{V}\left[\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho \vec{u})\right]dV = 0$$

So, my question is how did the substantial derivative for density turn into a partial derivative?

I understand that if you consider the total mass of the volume, it's essentially changing only with time. i.e. $$\frac{dM}{dT} = \frac{\partial M}{\partial T}$$

where M is the mass of the control volume.

But how do you explain the use material derivative with density above?

Shouldn't this be true :

$$\frac{d}{dt}\int_V\rho dV = \int_V\frac{d\rho}{dt}dV = \int_V\left[\frac{\partial\rho}{\partial t} + \vec{u}\cdot\nabla\rho\right]dV$$

Can this only be explained by $$\vec{u}\cdot\nabla\rho=0$$?

## 1 Answer

I figured it out. Understand that the substantial derivative exists for only field variables i.e. functions that vary with position vector x and time t.

Thus, $$\frac{DM}{DT}$$

doesn't really make sense.

For a fixed control volume the mass can only vary with time.