Flow problem with a two diemsional water flow divider

[psuaero]

Question Details:


A two-dimensional water flow divider is shown in the sketch. The flow in through opening A varies linearly from 3 to 6 m/s and is inclined with angle α=45 degrees to the opening itself. The area of the A opening is .4m2 per unit depth, that of B is .2m2 per unit depth, and that of C is .15 m2 per unit depth. the flow at B, normal to the opening, varies linearly from 3 to 5 m/s; and the flow at C, normal to the opening is approximately uniform. Find velocity at C. I'm having difficult setting up the relation needed to solve the problem.

I've attached a figure for the problem.

I would appreciate a step by step solution. This is what I was thinking:


Min=Mb out +Mc out.

what expression would I use for Min and Mb out
I'm thinking Mc out would equal V*.45m2*cos(0)

would Mb out= integral from 0 to 4 of (x/2)+3 dx and
Min= integral from 0 to7 of (3x/7)+3dx

than just sove for V correct?

 

[Navin A S]

Yes, that is correct. You can use the equations you wrote for Mb out and Min. To solve for V, you would set them equal to each other and solve for V.