How Do You Calculate the Streamlines and Pathlines of an Incompressible Fluid?

[mathmari]

Hey!

Construct and draw the streamlines of the velocity field $u=az-bt, v=\frac{b}{4}z-cy, w=2(a-1)$. Calculate $c$ (as a function of the constants $a$, $b$) such that the flow field $\overrightarrow{u}=(u, v, w)$ represents the flow of an incompressible fluid. Find also the pathlines of the element of the fluid that at the time $t=0$ it is at the position $(x_0, y_0, z_0)$.

I have done the following:

$$\frac{dx}{u}=\frac{dy}{v}=\frac{dz}{w}$$

$$\frac{dx}{az-bt}=\frac{dy}{\frac{b}{4}z-cy}=\frac{dz}{2(a-1)}$$

$$\frac{dx}{az-bt}=\frac{dz}{2(a-1)} \Rightarrow 2(a-1)dx=(az-bt)dz \Rightarrow 2(a-1)x=a\frac{z^2}{2}-btz+d$$

$$\frac{dx}{az-bt}=\frac{dy}{\frac{b}{4}z-cy} \Rightarrow \frac{x}{az-bt}=-\frac{\ln \left (\frac{b}{4}z-cy\right )}{c}+m$$

Is this correct so far?? (Wondering)

So, which are the streamlines?? (Wondering)
From the fact that the fluid is incompressible we have the following:

$$div \overrightarrow{u}=0 \Rightarrow (u, v, w)=(0, 0, 0) \\ \Rightarrow az-bt=0 \text{ and } \frac{b}{4}z-cy=0 \text{ and } 2(a-1)=0 \\ \Rightarrow az=bt \text{ and } c=\frac{bz}{4y} \text{ and } a=1$$

Is this correct?? (Wondering)
To find the pathlines we do the following:

• $$\frac{dx(t)}{dt}=u \Rightarrow \frac{dx(t)}{dt}=az-bt \Rightarrow \frac{dx(t)}{dt}=a(2(a-1)+z_0)-bt \\ \Rightarrow \frac{dx(t)}{dt}=2a(a-1)+az_0-bt \Rightarrow x(t)=(2a(a-1)+az_0)t-\frac{b}{2}t^2+x_0 $$
  
• $$ \frac{dy(t)}{dt}=v \Rightarrow \frac{dy(t)}{dt}=\frac{b}{4}z-cy(t) \Rightarrow \frac{dy(t)}{dt}=\frac{b}{4}(2a(a-1)t+z_0)-cy(t) \\ \Rightarrow \frac{dy(t)}{dt}+cy(t)=\frac{b}{2}a(a-1)t+\frac{bz_0}{4} \\ \Rightarrow y(t)=\frac{(y_0c^2-c\frac{bz_0}{4}+\frac{b}{2}a(a-1))}{c^2}e^{-ct}+\frac{\frac{b}{2}a(a-1)t}{c}+\frac{c\frac{bz_0}{4}-\frac{b}{2}a(a-1)}{c^2}$$
  
• $$ \frac{dz(t)}{dt}=w \Rightarrow \frac{dz(t)}{dt}=2(a-1) \Rightarrow z(t)=2(a-1)t+z_0$$

Is this correct?? (Wondering)

 

[jvicens]

As for the streamlines, we can plot them by using the equations we found earlier. The streamlines will be curves that satisfy the following equations:

$$2(a-1)x=a\frac{z^2}{2}-btz+d \text{ and } \frac{x}{az-bt}=-\frac{\ln \left (\frac{b}{4}z-cy\right )}{c}+m$$

To simplify the equations, we can set $d=m=0$. Then, the streamlines will have the form:

$$2(a-1)x=a\frac{z^2}{2}-btz$$

and

$$\frac{x}{az-bt}=-\frac{\ln \left (\frac{b}{4}z-cy\right )}{c}$$

We can plot these equations using a graphing software or by hand. The resulting curves will be the streamlines of the velocity field.

As for the pathlines, we have found the equations for $x(t)$, $y(t)$, and $z(t)$ in terms of time. To find the pathline of a specific element at time $t=0$, we simply plug in the initial position $(x_0, y_0, z_0)$ into these equations. The resulting parametric equations will give us the pathline of that element.