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??
So, which are the streamlines??
From the fact that the fluid is incompressible we have the following:
$$div \overrightarrow{u}=0 \Rightarrow \partial_xu + \partial_yv + \partial_zw=0\\ \Rightarrow -cy=0 \Rightarrow c=0$$
Is this correct??
To find the pathlines we do the following:
- $$\frac{dx(t)}{dt}=u \Rightarrow \frac{dx(t)}{dt}=az-bt \Rightarrow \frac{dx(t)}{dt}=a(2(a-1)+z_0)-bt \\ \Rightarrow \frac{dx(t)}{dt}=2a(a-1)+az_0-bt \Rightarrow x(t)=(2a(a-1)+az_0)t-\frac{b}{2}t^2+x_0 $$
- $$ \frac{dy(t)}{dt}=v \Rightarrow \frac{dy(t)}{dt}=\frac{b}{4}z \Rightarrow \frac{dy(t)}{dt}=\frac{b}{4}(2a(a-1)t+z_0) \\ \Rightarrow \frac{dy(t)}{dt}=\frac{b}{2}a(a-1)t+\frac{bz_0}{4} \Rightarrow y(t)=\frac{b}{4}a(a-1)t^2+\frac{bz_0}{4}t+y_0$$
- $$ \frac{dz(t)}{dt}=w \Rightarrow \frac{dz(t)}{dt}=2(a-1) \Rightarrow z(t)=2(a-1)t+z_0$$
Is this correct??