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The continuity equation for fluid flow assuming steady-state and constant fluid density, that is: $\rho = \text{constant}$ and $\partial_t=0$ can be written in the two following forms (differential and integral):

  • Differential form:

$${\partial u\over\partial x} = 0 \tag{1}$$

Where $u$ is the $x$ component of velocity.

This equation states that $$u = \text{constant} \tag{2}$$ otherwise the change of $u$ in $x$ direction must be balanced by the change of $v$ in the $y$ direction (this follows from the 2D form of the continuity equation ${\partial u \over \partial x} = -{\partial v \over \partial y}$).

  • Integral form, aka flow rate conservation (correct me if I am wrong):

$$ Q = u\cdot A = u_1\cdot A_1 = u_2\cdot A_2 = \cdots = \text{constant} \tag{3}$$ Where the numbers 1, 2, ... denote sections of streamtube along $x$ axis.

My questions

  • Could you please explain me this ambiguity, that is why the first form states that there is no change in $u$ otherwise the flow will be 2D and in the other hand the integral form there is a change in $u$ along $x$ axis but the flow still 1D.

  • Is it possible to consider flows in diverging (or converging) nozzles as 1D flows?

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  • $\begingroup$ isn't the second one $\dot{Q}=0$? $\endgroup$ – FGSUZ Feb 28 at 23:19
  • $\begingroup$ Think about it this way -- in your differential form, what is the "control volume" when you integrate it? And in your integral form, what is your control volume? How are those different? $\endgroup$ – tpg2114 Feb 28 at 23:34
  • $\begingroup$ The other thing to think about -- when we say a flow is 1D, what are we really saying? That there are no other directions, or just that the change in those other directions is really small compared to the changes in x? $\endgroup$ – tpg2114 Feb 28 at 23:37
  • $\begingroup$ @tpg2114: Sorry I am not sufficiently comfortable with these deep assumptions I have only some basic knowledge about this subject $\endgroup$ – IamNotaMathematician Feb 28 at 23:44
  • $\begingroup$ I'm sorry but I don't understand the equation of your comment. Where does it come from? $\endgroup$ – FGSUZ Mar 1 at 0:49
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Okay now I understand your question, I think.

I'll leave a brief answer because I'm not sure if it will be enough, but I don't have much time now.

You're right, in 1D, there is no $y$, nor $d/dy$, forget about vertical axes.

It is true that $dv/dx=0$, provided that the flow is non-compresible.

And, why does it seem to contradict the other one? Well, you must keep in mind that the divergence is after all sort of a volumetric quantity. It refers to an infinitesimal volume. $div(v)=0$ means that there is no growth of flow from that point, or, in other words: there are as many flow lines coming in as coming out.

That is coherent with the flow conservation.

On the other hand, the integral form has been converted into a surface integral, but not any surface, but the boundary surface. It is now the velocity crossing the surface, not a volume. Of course the flow crossing the entering surface must be the same than the exiting surface.

So, the key is that one: the first one is something related to infinitesimal volumes, whereas the second one is a velocity crossing the surface.

If you want to go deeper, this has to do with Eulerian versus Lagragian points of view. Considering that $v$ is a vector field, versus a stream of particles.

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When you go from the continuity equation in differential form to the integral form, you choose a certain volume (control volume) to integrate over.

The time derivative of the (mass) density $\dot{\rho}$ then turns into the time derivative $\dot{Q}$ of the mass inside the control volume.

The divergence of the (mass) current density then turns into an integral over the surface of the control volume. Since you are integrating (summing up) the vector of the current density over this surface, you have to distinguish flow into and flow out of the control volume. Flow into the control volume will contribute with a negative sign, flow out of the control volume contributes with a positive sign.

If the change in time of density $\rho$ is zero, then also the change of the mass inside the control volume is zero ($\dot{Q}=0$). As a consequence, the out-flow of mass through the surface of the control volume must exactly balance the in-flow. The sum of these two contributions is zero, because in- and out-flow contribute with opposite signs.

In your 1D example, if we consider a control volume of length $\Delta x$ and cross-sectional area $A$ and denote the current density as $j(x)=\rho u$, you will have $$ -j(x)A + j(x+\Delta x)A = 0.$$ The first term is the flux into the control volume from the left, the second term is the out-flux to the right. Note the difference in sign.

If the density $\rho$ is spatially homogeneous, you can rewrite the above relation in terms of the velocities $u(x)$ and $u(x+\Delta x)$. This gives $$ v(x+\Delta x)-v(x) = \frac{\partial v}{\partial x}\Delta x = 0\Rightarrow \frac{\partial u}{\partial x}=0,$$ consistent with $\dot{Q}=0$.

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  • $\begingroup$ I cannot understand how could this answer my question. what do you mean by $\dot\rho$? density is assumed constant and the flow is steady $\endgroup$ – IamNotaMathematician Mar 1 at 11:51
  • $\begingroup$ I cannot understand any sentence in your answer. $\endgroup$ – IamNotaMathematician Mar 1 at 11:53
  • $\begingroup$ @IamNotaMathematician If you can't understand anything written here, then you should revisit the basics/textbook for the equations you are referencing. In every book on fluids/aerodynamics/flow that I am aware of, discussions of control volumes and surface integrals are in the first or second chapter because it's fundamental to deriving the equations. $\endgroup$ – tpg2114 Mar 1 at 12:32
  • $\begingroup$ And so if you need help understanding what control volumes are, and how to convert volume integrals into surface integrals, and so on -- that's a different question than what you've asked here and should be answered first. We probably have questions about those on this site somewhere. $\endgroup$ – tpg2114 Mar 1 at 12:33
  • $\begingroup$ I am not asking about the mathematical steps I know both forms are true (I am not asking how to switch from one to other) but how to clear the ambiguity and to explain in a straitforward manner $\endgroup$ – IamNotaMathematician Mar 1 at 15:32

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