Vertical foce of fluid jet split in two

[Andy Salter]

Homework Statement

A water jet hits a wall and splits into two streams (see attachment). The jet carries 3kg of water per second. The cross-sectional area of the flow at (2) is A = 2*10^-3 m^2. The average velocities at the outlets are V2 = V3 = 1m/s

What is the net vertical force exerted by the jet on the combined wall and path surfaces?

The Attempt at a Solution

So, I gather the vertical force on the wall is zero because the first turning of the two streams cancel in the y-direction. But stream (3) turns 90 degrees again. This should create a net force in the y-direction on the path because the y component of the momentum is changing. If the stream stayed constant cross-sectional area this could be worked out, but it isn't constant. Then there's also the weight force on the path, but without knowing the total volume how can this be worked out? I'm totally at a loss on this one

 

[Chestermiller]

Are you currently studying the application of macroscopic momentum balances?

(You don't need to include weight because the implication is that they are neglecting gravity)

 

[Andy Salter]

Yes. I think I have the solution:

ΣFy = ΣρQ(outlet) (because V_1y = 0, inlet term of momentum equation is zero)

for mass flow rate at outlet we only need to consider outlet 2 because velocity y-component of 3 is also zero. The mass flow rate at 2 is 2 kg/s and V₂ = 1 so we have:

ΣFy = 2 * 1 = 2N

correct?

 

[Chestermiller]

It should be $(\rho v A)v$

 

[Andy Salter]

yep, a typo haha