Given the following velocity field u Bx1 Atx Cy y with A

Given the following velocity field u = Bx{1 + At)x + Cy y, with A = 0.5 s^-1, B = C = 1 s^-1 and coordinates measured in meters, plot the pathline of the particle that passed through the point (1, 1, 0)at t = 0. Compare with the streamlines through the same point at the instants t = 0, 1 and 2s.

Solution

solution:

1)here velocity along x and y direction is given by

u=Bx(1+At)=x(1+.5t)

v=Cy=y

2)equation of streamline is given by

dx/u=dy/v

for t=0

u=x and v=y

dx/x=dy/y

on integrating

logx=logy+logc

for point (1,1,0) x=1 and y=1

so we get logc=0

logx=logy

y=x

3)for t=1 sec

dx/u=dy/v

for t=1

u=x+.5 and v=y

dx/x+.5=dy/y

on integrating

log(x+.5)=logy+logc

for point (1,1,0) x=1 and y=1

so we get logc=log1.5

logx=logy+log1.5

(x+.5)=1.5y

y=.666x+.333

4) for t=2 sec

dx/u=dy/v

for t=2 sec

u=x+1 and v=y

dx/x+1=dy/y

on integrating

log(x+1)=logy+logc

for point (1,1,0) x=1 and y=1

so we get logc=log2

log(x+2)=logy+log2

x+2=2y

y=.5x+.5

7)in this way it is observed eqiation of streamline has slope of 1 for t=0 means horizontal flow ,

at t=1,slope is m=.666 means inclined flow at same point

at t=2 sec,slope m=.5,slope decreasing from .666 to .5 means flow line shifting downward

and by summarising all it is proved that flow is unsteady at that point