For the horizontal flow listed below perform the following a

For the horizontal flow listed below, perform the following, assuming it\'s a positive constant:

(a) Construct a simple graph of the flow (sketching by hand is fine);

(b) Compute the vertical vorticity;

(c) Compute the horizontal divergence;

(d) Comment on the result.

u(x,y) = ax; v(x,y) = -ay

Solution

From the equation given we see that the velocity in x direction is ax and in the y direction is -ay

Hence the flow if we look from the top is towards x direction wqith magnitude a and in the ydirection with the magnitude -a

Hence the flow is at 45 degree from the x axis, and inclined to the negative side of y axis.

The vertical voricity is Curl V ie

i [ dw/dy -dv/dz) + j [ du/dz-dv/dx] + k [dv/dx-du/dy] ( all should be read as partial derivatives)

For the vertical component , the last term is non-zero. .

for horizontal component, the first term should be non-zero.

The horizontal divergeence is du/dx +dv/dy ( both partial derivatives)

= a + (-a) = 0 that is there is no change in velocity in magnitude and direction.

The flow do not consider friction. the assumpton is no loss.