# Finding the Density Change of a Fluid

Consider the motion of a fluid with velocity field defined in Eulerian variables by the following equations $$u=kx,\,\,v=-ky,\,\,w=0$$ where $k$ is a constant. Also assume that the density is given by $$\rho = \rho_0 + Aye^{kt}$$ What is the rate of change of density for each individual fluid particle? ($\rho_0$, $A$ are constant)

I am pretty unsure what to do with the information that I'm given. I know that the $\textbf{Conservation of Mass}$ states that the rate-of-increase of mass inside a region $\Sigma$ must equal to the mass flux into $\Sigma$ across the surface $S$. Thus

$$\iiint_{\Sigma}\frac{\partial\rho}{\partial t}\,dV = \iint_{S}(\rho\underline{u})\cdot\underline{\hat{n}}\,dA$$

From $\textbf{Gauss's Divergence Theorem}$ I know that

$$\iint_{S}(\rho\underline{u})\cdot\underline{\hat{n}}\,dA = \iiint_{\Sigma}\underline{\nabla}\cdot(\rho\underline{u})\,dV$$

$$\iiint_{\Sigma}\Big[\frac{\partial\rho}{\partial t} + \underline{\nabla}\cdot(\rho\underline{u})\Big]\,dV = 0$$

So the $\textbf{mass-conservation equation}$ is

$$\frac{\partial\rho}{\partial t} + \underline{\nabla}\cdot(\rho\underline{u})$$

So in tensor notation

$$\frac{\partial\rho}{\partial t} + \frac{\partial}{\partial x_j}(\rho u_j) = 0$$

Now am I right to think that

$$\frac{\partial \rho}{\partial t} = Ayke^{kt},\,\,\frac{\partial}{\partial x_j}(\rho u_j) = \frac{\partial\rho}{\partial x_j}u_j + \rho\frac{\partial u_j}{\partial x_j}$$ But I don't know what to do next...

The rate of change of density $\rho$ for each individual fluid particle is the time derivative of $\rho$ in Lagrangian frame. Since the velocity field is given in Eulerian variables, this corresponds to the material derivative of $\rho$ which is given by $$\frac{D\rho}{Dt} = \frac{\partial\rho}{\partial t} + u\cdot\nabla\rho.$$
• Oh yes! I have seen this in my notes. So then if $\rho = \rho_0 + Aye^{kt}$ then $$\underline{u}\cdot\underline{\nabla}\rho = kx\cdot\frac{\partial\rho}{\partial x}\underline{\hat\imath} - ky\cdot\frac{\partial\rho}{\partial y}\underline{\hat\jmath} + 0\cdot\frac{\partial\rho}{\partial z}\underline{\hat k} = kx\cdot(0+0) - ky\cdot(0+Ae^{kt})\underline{\hat\jmath} + 0\cdot(0+0)\\ = -kyAe^{kt}\underline{\hat\jmath}$$ Now if I differentiate $\rho$ wrt $t$ I get $$\frac{\partial\rho}{\partial t} = 0 + Ayke^{kt}\,\,\Rightarrow\,\, \frac{D\rho}{Dt} = Ayke^{kt} - kyAe^{kt} = 0.$$ – MRT Feb 18 '18 at 12:39
• @Michael It is consistent with the conversation of mass equation, since $\frac{\partial\rho}{\partial t} + \nabla\cdot\rho(\underline{u}) = \frac{D\rho}{Dt} + \rho\nabla\cdot\underline{u} = 0$ and the given flow field is incompressible. – Chee Han Feb 18 '18 at 12:44