How does the angle of the plate affect the fraction of liquid that flows up?

[alpine_steer]

Homework Statement

A horizontal water jet produces a wide sheet of parallel-flowing liquid that strikes an inclined plate and then divides, with a fraction of the liquid flowing up the plate and the remainder flowing down the plate. The figure shows a section of the flow. All surfaces of the liquid are exposed to atmospheric pressure and both the gravity forces and frictional losses in the flow may be neglected.
Shear stress between the plate surface and the liquid may be neglected in comparison with pressure forces there and, therefore, the force between the plate and the liquid is perpendicular to the plate surface with zero component tangent to the plate surface.

With these facts, use mass and momentum balances to:

a) Determine the relation between the fraction of the liquid that travels up along the plate and the angle θ.

Homework Equations

Bernoulli's equation { u₁=u₂=u₃}
Conservation of mass {m₁=m₂+m₃
Conservation of momentum {um₁=um₂+um₃

The Attempt at a Solution

F_x:
um₂cos(θ)-um₃cos(-θ)-um₁=0
F_y:
um₂sin(θ)-um₃sin(-θ)=0

m₂=-m₃--> can not be true

 

[256bits]

Fx:
um2cos(θ)-um3cos(-θ)-um1=0
Fy:
um2sin(θ)-um3sin(-θ)=0

Are you sure that ƩFx and ƩFy equal 0.
There is a plate that the jet strikes.

 

[alpine_steer]

Fx:
um₂cos(θ)-um₃cos(-θ)-um₁=F_x
[(m₂+m₃)cos(θ)-m₁]u=F_x
[m₁cos(θ)-m₁]u=F_x
[cos(θ)-1]m₁u=F_x
Fy:
um₂sin(θ)-um₃sin(-θ)=F_y

 

[rude man]

Gravity is neglected? So the image is looking down on the experiment?

 

[256bits]

Not necessarily.
Gravity can be neglected if the change in energy of the water stream due to pgh is small in comparison to other energy changes. Note also that frictional forces have been neglected also so there is no shear force on the plate.

 

[256bits]

Alpine Steer
To make this problem conceptually more understandable, you could rotate the picture so that the plane of the plate is either vertical or horizontal - in other words line up your axis parallel and perpendicular to the plate.

Making the axis vertical:
In that case, the water jet approaches the plate at an angle θ measured from the plate.
Since friction is neglected the plate will experience a reaction force of F only in the x-direction.
Your equations are set up in a similar manner as before but in this case there is no reaction parallel to the plate.

And since the velocity of the jet after it hits the plate is zero your momentum equationlooks much neater.

ie F = $\rho$Q ( initial velocity - final velocity )
or Fx = $\rho$Q( u1x - u2x ) , where u2x = 0
and u1x = Vcos(θ) or sin θ depending whether you measure the angle from the plate or the normal to the plate.