Deriving the variable area flow equation from the differential form of continuity equation

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

• Are you familiar with the divergence theorem? – Chet Miller Aug 15 '18 at 19:23
• @ChesterMiller Yes I am. It is used to convert the integral form of the continuity equation to differential form – Siddharth Prakash Aug 15 '18 at 19:25
• Good. So what was your integral form of the equation? – Chet Miller Aug 15 '18 at 19:53
• @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 – Siddharth Prakash Aug 22 '18 at 7:12
• The divergence theorem was derived for an arbitrary volume. – Chet Miller Aug 22 '18 at 12:09

1 Answer

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

• 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 – Siddharth Prakash Aug 22 '18 at 7:11