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Equating all the mass flow rates into and out of the differential control volume gives

\[ \begin{eqnarray} \rho v_1 (dx_2 dx_3) + \rho v_2 (dx_1 dx_3) + \rho v_3 (dx_1 dx_2) - \left( \rho v_1 + {\partial (\rho v_1) \over \partial x_1} dx_1 \right) dx_2 dx_3 \\ - \left( \rho v_2 + {\partial (\rho v_2) \over \partial x_2} dx_2 \right) dx_1 dx_3 - \left( \rho v_3 + {\partial (\rho v_3) \over \partial x_3} dx_3 \right) dx_1 dx_2 = {\partial \over \partial t} (\rho dx_1 dx_2 dx_3) \end{eqnarray} \]

Cancelling terms and dividing through by \(dx_1 dx_2 dx_3\) gives

\[ \begin{eqnarray} - {\partial (\rho v_1) \over \partial x_1} - {\partial (\rho v_2) \over \partial x_2} - {\partial (\rho v_3) \over \partial x_3} = {\partial \rho \over \partial t} \end{eqnarray} \]

\[ {\partial \rho \over \partial t} + {\partial (\rho v_1) \over \partial x_1} + {\partial (\rho v_2) \over \partial x_2} + {\partial (\rho v_3) \over \partial x_3} = 0 \]

This can be written concisely as

\[ {\partial \rho \over \partial t} + \nabla \cdot (\rho {\bf v}) = 0 \qquad \qquad { \partial \rho \over \partial t} + (\rho \, v_i),_{\,i} = 0 \]

One might wonder if there is a Lagrangian counterpart to the Eulerian form of this. There is. It is usually written as

\[ \rho \, dV = \rho_o dV_o \]

This simply states that the differential chunk of mass in the deformed state \(\rho \, dV\) must be equal to its original value \(\rho_o dV_o\) in the undeformed state.

\[ \nabla \cdot (\rho {\bf v}) = 0 \qquad \qquad (\rho \, v_i),_{\,i} = 0 \qquad \qquad \text{(steady state)} \]

The second special case is that of incompressibility, i.e., \(\rho = \)constant. In this case the derivative with respect to time is zero and \(\rho\) can be factored out of the equation leaving only

\[ \nabla \cdot {\bf v} = 0 \qquad \qquad \quad v_{i,i} = 0 \qquad \qquad \quad \text{(incompressible)} \]

This result is simple enough that it is often expanded out.

\[ {\partial v_1 \over \partial x_1} + {\partial v_2 \over \partial x_2} + {\partial v_3 \over \partial x_3} = 0 \qquad \qquad \qquad \text{(incompressible)} \]

Note that this is nothing more than \(\text{tr}({\bf D}) = 0\) for the case of incompressible materials.

\[ {\partial \rho \over \partial t} + {\bf v} \cdot \nabla \rho + \rho \, (\nabla \cdot {\bf v}) = 0 \]

And then note that \({\partial \rho \over \partial t} + {\bf v} \cdot \nabla \rho\) is just the material derivative of the density, \({D \rho \over D t}\).

So the continuity equation can also be written as

If the material is incompressible, then \(\rho\) cannot change, so \( {D \rho \over D t} \) must be zero, leaving

\[ \rho \nabla \cdot {\bf v} = 0 \]

And then divide through by \(\rho\) (since it's not zero) to get

\[ \nabla \cdot {\bf v} = 0 \]

\[ {\partial v_1 \over \partial x_1} + {\partial v_2 \over \partial x_2} = 0 \]

Start by looking at the y-component of the flow, \(v_2\). The geometry of the converging nozzle forces the \(v_2\) component to flow upward when \(y \lt 0\) and flow downward when \(y \gt 0\). So \(v_2 \gt 0\) when \(y \lt 0\) and \(v_2 \lt 0\) when \(y \gt 0\).

The net effect of this is that \( {\partial v_2 \over \partial x_2} \lt 0 \) in the converging nozzle.

But the continuity equation dictates that the sum of the two partial derivatives must equal zero. So if the second is less than zero, then

\[ {\partial v_1 \over \partial x_1} \gt 0 \]

And this means that the fluid flow must be accelerating.

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Bob McGinty

bmcginty@gmail.com

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