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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 appriciate 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?

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# Flow problem with a two diemsional water flow divider

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