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A steady two-dimensional flow (pure straining) is given by $u = k x$, $v = -k y$, for $k$ constant.

  1. Find the equation for a general streamline of the flow.
  2. At $t = 0$,the fluid on the curve $x^2 + y^2 = a^2$ is marked. Find the equation for this material fluid curve for $t > 0$.
  3. Does the area within the curve change in time, and why?

My Attempt: I have solved a differential equation to find the streamlines by eliminating the parameter $s$; I get $xy = A$ where $A$ is some constant. I am stuck on the second part; I know that the initial position lies on the circle given, but how do I find the equation of the material curve? Will it be a function of $x$, $y$ and $t$?

Any help would be greatly appreciated!

MathematicianP
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    How do you defined $s$? Do I need some background information about fluid-dynamics? – Jónás Balázs Jan 22 '19 at 19:58
  • It is the parameter that was used in my lectures – MathematicianP Jan 22 '19 at 20:25
  • Where does $s$ fit into your equations for $u$ and $v$? We can't see how you've found the streamlines by eliminating $s$ without knowing where it is! – postmortes Jan 22 '19 at 21:14
  • Your answer for part (1) is correct. The streamlines are hyperbolas including the (degenerate) axes. The origin is a stagnation point. – RRL Jan 22 '19 at 21:25
  • Streamlines are curves where the stream function is contstant, and the stream function $\psi$ satisfies differential equations $\psi_y = u = kx, \psi_x = - v = ky$ – RRL Jan 22 '19 at 21:27

1 Answers1

2

Part (2)

In the Lagrangian description of the flow, a fluid particle at a point $\mathbf{x}_0=(x_0,y_0)$ at time $t = 0$ is located at a point $\mathbf{x} =(x,y)$ at time $t > 0$ with coordinates given by

$$x= X(t;\mathbf{x}_0), \,\, y = Y(t;\mathbf{x}_0),$$

where the Eulerian velocity field is related to the Lagrangian mapping by

$$\tag{*}\frac{d}{dt}X(t;\mathbf{x}_0) = u[X(t;\mathbf{x}_0),Y(t;\mathbf{x}_0),t],\\ \frac{d}{dt}Y(t;\mathbf{x}_0) = v[X(t;\mathbf{x}_0),Y(t;\mathbf{x}_0),t] $$

In steady, 2D straining flow the velocity field has the form $$u(x,y,t) = kx,\,\, v(x,y,t) = -ky$$ and (*) simplifies to

$$\tag{**}\frac{dX}{dt} = kX,\\ \frac{dY}{dt} =- kY$$

Solving (**) subject to initial conditions $X(0,\mathbf{x_0})= x_0, \,Y(0,\mathbf{x}_0) = y_0$ we get

$$X(t;\mathbf{x}_0) = x_0e^{kt}, \,\,\, Y(t;\mathbf{x}_0) = y_0 e^{-kt},$$

and for points $(x_0,y_0)$ on the circle of radius $a$ centered at the origin we have the evolution of the material curve given by

$$X(t;\theta) = a\cos(\theta)e^{kt}, \,\,\, Y(t;\theta) = a \sin (\theta) e^{-kt}\quad (0 \leqslant \theta < 2\pi)$$

Part (3)

Intuitively you can see that the circle of marked fluid particles is deformed into an eccentric elliptical shape as it is elongated in the positive and negative $x-$directions and squashed in the $y-$ directions. The enclosed area remains the same for incompressible flow with $\nabla \cdot \mathbf{u} = 0$ -- which holds in this case.

In incompressible flow the density $\rho$ remains constant (within any fluid parcel that moves with the flow). The total mass of fluid particles at $t= 0$ within a cylindrical region with cross-section $\Omega(0)$ in the plane and bounded by the circle $C(0)$ of radius $a$ must remain constant through time as the region $\Omega(t)$ is convected and deformed with the flow. Since the flow is two-dimensional, there is no motion in the direction normal to the plan so we can say that the mass per unit length in the $z-$direction remains constant. Thus,

$$\frac{d}{dt}\int_{\Omega(t)} \rho \, dA = 0,$$

and since the density does not vary with time the area $\int_{\Omega(t)} \, dA$ is constant.

This can be justified rigorously by the Reynolds transport theorem.

RRL
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  • Ah thank you for the detailed answer. I like the parametrisation for the second part1 – MathematicianP Jan 26 '19 at 14:41
  • For the 3rd part, I know that incompressible flow implies that the divergence of the vector field is 0, which it is here but I don't know how to proceed? – MathematicianP Jan 26 '19 at 14:42
  • @MathematicianP: You're welcome. I added to the answer. – RRL Jan 26 '19 at 20:21
  • Hi, trivial question, but in the final part of (ii) could you explain how did you get from $X(t;x_0)=x_0\exp^{kt}$ to $X(t;\theta)=a\cos(\theta)\exp^{kt}$ and similarly for the $Y$ equation as well? – mathnoob Sep 27 '20 at 22:24
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    @mathnoob: I'm just using the polar coordinate representation of the point $(x_0,y_0)$ on the circle of radius $a$ and center $(0,0)$ where a particle is located initially. Better notation would have been $x_0 = a \cos \theta_0$ and $y_0 = a \sin \theta_0$. – RRL Sep 28 '20 at 00:10
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    Probably all that matters here is the initially circular material curve where $x^2 + y^2 = a^2$ is deformed into an ellipse at time $t$ with equation $\frac{x^2}{a^2e^{2kt}}+\frac{y^2}{a^2e^{-2kt}} = 1$ – RRL Sep 28 '20 at 00:21
  • Ah, thank you for clarifying. – mathnoob Sep 28 '20 at 00:36