1
$\begingroup$

If A is a property of the fluid, then in a steady flow is $\frac{dA}{dt} = 0$ or $\frac{\partial A}{\partial t} = 0$ ? My professor says it's $\frac{dA}{dt} = 0$ but I don't see how that makes sense. Isn't a flow steady if all of it's properties are independent of time? And if A is independent of time, doesn't that only mean that it isn't a function of time, i.e., that it doesn't explicitly depend on time?

$\endgroup$
  • 3
    $\begingroup$ It's the partial derivative at constant spatial position that is zero, not the material (Lagrangian) derivative. So you are correct and your professor is wrong. $\endgroup$ – Chet Miller Nov 16 '19 at 13:57
1
$\begingroup$

I think you are correct if temperature is constant.

Time Independent Fluids

This class of fluids is characterized by the fact that, provided the temperature of the fluid remains constant, the shear rate depends only on the shear stress and is a single valued function of it. Newtonian fluids which have a viscosity that is independent of the shear rate are described. This behaviour is exhibited by all gases and, in general, liquids, and solutions of materials of low molecular weight. Fluids whose viscosity is a single valued function of shear rate are termed non-Newtonian fluids.

$\endgroup$
0
$\begingroup$

I would argue that $\frac{\partial A}{\partial t}=0$ should be giving you time-independence in all fluid properties.

My idea here is that you can find a system in which the $v \partial_x v$-part of the co-moving time-derivative $\frac{dv}{dt}$, with $$\frac{dv}{dt} = \frac{\partial A}{\partial t} + v\partial_xv$$ plays an important role and makes the system steady-state. Then you can argue for the remainder, a.k.a. $\partial A/\partial t$ being responsible for any time-dependent behaviour.

The system I suggest to look at is the three-way force balance in the 1D-Euler cartesian equations $$ v\partial_xv = -\frac{1}{\rho}\partial_x P + g .$$ This system gives rise to a number of sonic-transition solutions which are all time-independent, i.e. if one would simulate them on a computer with the $\partial v/\partial t$-term, a found solution would not change. This system gives rise to solutions that can be called "solar wind" or "rocket nozzle flow" and others, if you're interested in the literature.

So, because the above system exhibits steady-state solutions, and the first term on the l.h.s is part of the co-moving time-derivative, I would say your notion is the correct one. But I would be curious as to what your prof's rebuttal of that would be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.