Plotting Streamlines: Origin at t=0,1,2

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SUMMARY

The discussion focuses on plotting streamlines for the velocity field defined by V = (x^2)*yi + (x^2)*tj at times t = 0, 1, and 2. The streamline equation derived is (y^2)/2 = tx, with the constant C set to 0 for the origin. Participants confirm that plugging in the values of t yields the equations y = 0, y = (2x)^(1/2), and y = 2*(x)^(1/2). The conversation also highlights that streamlines in this context apply to steady-state flow, while pathlines vary with time.

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Homework Statement


The velocity field in a flow is given by V = (x^2)*yi + (x^2)*tj. (a) Plot the streamline through the origin at times t = 0, t = 1, and t = 2. (b) Do the streamlines plotted in part (a) coincide with the path of particles through the origin? Explain.

i&j are directional vectors.

Homework Equations



Providing this just in case you guys haven't heard of stream lines. "A streamline is a line everywhere tangent to the velocity vector at a given instant."

The Attempt at a Solution



dx/u=dy/v (equation from streamline from Fluid mechanics textbook)
where u=x^2 * y
v=(x^2)*t

After plugging those in and differentiating I get (y^2)/2=tx+C
where C is a constant.


My problem is I don't know how to go about plotting "the streamline through the origin at t=0,1,2."

Do I plug the 3 t's into the equation 3 times and plot those 3 equations I get? If so, how do I get C?
 
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Different values of C correspond go different streamlines. You have to choose C so that the equation holds for x=0, y=0.
 
vela said:
Different values of C correspond go different streamlines. You have to choose C so that the equation holds for x=0, y=0.

Oh I see since the question is through the original that would make C=0 right for this equation?

Which would give me (y^2)/2=tx. Then I just plug in t=0,1,2 into this equation and plot the 3 equations?
 
Yup.
 
That would give me
the equations
y=0
y=(2x)^.5
y=2*x^.5

The solutions seem to only have the streamlines plotted in the first quadrant. But doesn't the 2 square root functions exist in both the 1st and 4th quadrant? The first equation y=0 would also exist in all 4 quadrants.
 
In fluid mechanics, the concept of streamlines only applies to steady state flow. This is a non-steadystate problem. In such problems, you can solve for the pathlines of particles, but the pathlines change with time. For your problem, the pathlines are determined by:

dx/dt = vx = x2y

dy/dt = vy= x2t

You need to solve this coupled set of ODEs for sets of initial values of x, y, and t.

Chet
 

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