Streamlines

Streamlines: Streamlines represents field lines in a fluid flow. It is a path traced out by massless fluid particles moving with the flow, which is tangential to the instantaneous velocity direction. Different streamlines at same instant in a flow do not intersect or flow across it, because a fluid particle cannot have two different velocities at the same point.

For a unsteady flow, streamline pattern is different at different times because the velocity vectors are fluctuating with time both in magnitude and direction.
In a fluid flow, a bundle of streamlines is called a streamtube. Walls of an ordinary garden hose form a streamtube for the water flowing through the hose.
Differential equation for a streamline in two dimensions is  $$vdx - udy = 0$$ Question: x and y components of a velocity field are given as $u = \frac{{4x}}{{\left( {{x^2} + {y^2}} \right)}}$ and $v = \frac{{4y}}{{\left( {{x^2} + {y^2}} \right)}}$, what is the equation of streamlines.  

Solution: Equation of streamline is given as 

$\left( {\frac{{dy}}{{dx}}} \right) = \left( {\frac{v}{u}} \right)$ therefore,

$\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{{4y}}{{\left( {{x^2} + {y^2}} \right)}}}}{{\frac{{4x}}{{\left( {{x^2} + {y^2}} \right)}}}}$

$\left( {\frac{{dy}}{{dx}}} \right) = \frac{{4y}}{{\left( {{x^2} + {y^2}} \right)}} \times \frac{{\left( {{x^2} + {y^2}} \right)}}{{4x}}$

$= \left( {\frac{y}{x}} \right)$

$\frac{{dy}}{{dx}} = \frac{y}{x}$

 $\frac{{dy}}{y} = \frac{{dx}}{x}$ 

$\ln y = \ln x + c$

$\ln y - \ln x = c$

$\ln \left( {\frac{y}{x}} \right) = \ln c$

$\frac{y}{x} = \ln c$

$y = {c_1}x$

The streamlines are straight lines from a source.

Question: A velocity field has radial and tangential components of velocity as ${V_r} = 0$ and ${V_\theta } = 4r$.Obtain equation of streamlines.

Solution: Since  ${V_r} = 0,$,so there is no radial component of velocity.So the streamlines will be circular, with centres at origin.

$$\begin{array}{l}u =  - {v_\theta }\sin \theta \\ =  - 4r\sin \theta \\ =  - 4r\frac{y}{r} =  - 4y\\v = {v_{_\theta }}\cos \theta  = 4r\cos \theta  = 4r\frac{x}{r} = 4x\end{array}$$ therefore equation of streamline $$\begin{array}{l}\left( {\frac{{dy}}{{dx}}} \right) = \frac{v}{u} = \left( {\frac{{ - x}}{y}} \right)\\ \Rightarrow ydy =  - xdx\\ \Rightarrow \frac{{{y^2}}}{2} + \frac{{{x^2}}}{2} = c\\ \Rightarrow {x^2} + {y^2} = {\rm{constant}}\end{array}$$ This is equation of circle,with center at origin.

Question: Consider a velocity field where x and y components of velocity are given by u=4x and v=-4y. Obtain the equation of streamlines.

Solution: $$\begin{array}{l}\left( {\frac{{dy}}{{dx}}} \right) = \frac{v}{u} = \left( {\frac{{ - 4y}}{4x}} \right) = \left( {\frac{{ - y}}{x}} \right)\\ \Rightarrow \left( {\frac{{dy}}{y}} \right) = \left( {\frac{{ - dx}}{x}} \right)\\ \Rightarrow \ln y = x\ln x + c\\ \Rightarrow y = \frac{{{c_1}}}{x}\end{array}$$ So,the streamlines are hyperbolas.