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Following is the differential form of continuity equation for a steady in-compressible flow

$$ ∂u/∂x+ ∂v/∂y + ∂w/∂z = 0$$

Now can we obtain the variable area flow equation A1V1=A2V2 by solving this equation. Because usually we use the integral form of the equation

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  • $\begingroup$ Are you familiar with the divergence theorem? $\endgroup$ – Chet Miller Aug 15 '18 at 19:23
  • $\begingroup$ @ChesterMiller Yes I am. It is used to convert the integral form of the continuity equation to differential form $\endgroup$ – Siddharth Prakash Aug 15 '18 at 19:25
  • $\begingroup$ Good. So what was your integral form of the equation? $\endgroup$ – Chet Miller Aug 15 '18 at 19:53
  • $\begingroup$ @Chester Miller can you do it without using the divergence theoram. I mean can you take volume integral of ∂u/∂x over a control volume with varaiable area of cross section and one dimensional flow and reach the result AV=constant $\endgroup$ – Siddharth Prakash Aug 22 '18 at 7:12
  • $\begingroup$ The divergence theorem was derived for an arbitrary volume. $\endgroup$ – Chet Miller Aug 22 '18 at 12:09
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Note that the density is constant in this case i.e.:

$$\partial_t \rho = 0.$$

Using the continuity equation we have:

$\partial_t \rho + \nabla \cdot (\rho u) = 0$. $u$ is the flow velocity vector. This gives $\nabla \cdot (\rho u) =0$. $\rho$ is constant and therefore we get $\nabla \cdot u = 0$. Divergence theorem gives then:

$$ \int_V (\nabla \cdot u )dV = \int_{\partial V} u \cdot \vec{da} =0.$$

I will give the final step also. We have that $$ \int_{\partial V} u da_u = u_1 A_1 - u_2 A_2 = 0 $$ so $$u_1 A_1 = u_2 A_2.$$

Notation: $da_u$ is the projected area element in the direction of $u$ (Where I mean $da_u = cos(\theta) da$.)and I used the fact that at two different surfaces we have different $u$. The - occurs because the normal vector is pointing the other way around for one of the surfaces. For fixed surface perpendicular to some $u$ we have constant $u$ so $u_1,u_2$ can be pulled out of the integral in both cases.

Hope it helps

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  • $\begingroup$ can you do it without using the divergence theoram. I mean can you take volume integral of ∂u/∂x over a control volume with varaiable area of cross section and one dimensional flow and reach the result AV=constant $\endgroup$ – Siddharth Prakash Aug 22 '18 at 7:11

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