Answer:
Equation of the streamline V = 4i - 4j m/s
Unit Vector n = (4i+4j)/4√2
Explanation:
Parameters
x=2 and y=-1
V=(2x)i + (yt)j m/s
Substituting (x=2) and (y=-1) into the velocity field V
Therefore V = 4i - tj where t=4s
Equation of the streamline
V = 4i - 4j m/s
Unit vector normal to the streamline n
Note V.n=0 and the velocity only have x- and y - components
Therefore
V.n=(4i - 4j).(nₓi+nyj)=0 or 4nₓ - 4ny = 0
The unit vector requires that nₓ^2+ny^2=1
Therefore n = (4i+4j)/4√2
Answer:
a) jyt + jt + 2xi = 4i
b)2iy + 2i + 8j = 4xj
Explanation:
V = (2x)i + (yt)j
By implicit differentiation:
With V = 0: 2i + dyjt/dy = 0
- dyj/dx = 2i
∴ dy/dx = - 2i/jt
The equation of the streamline passing through (2,-1), using, y - y1 = m(x - x1), where m = dy/dx
y + 1 = -2i/jt(x - 2)
jt(y + 1) = -2xi +4i
jyt + jt + 2xi = 4i
b) Using y - y1 = -1/m(x - x1), where at unit normal, dy/dx = -1/m
y + 1 = 4j/2i(x -2)
2i(y + 1) = 4j(x - 2)
2iy + 2i + 8j = 4xj
