Hard Momentum problem (Force on blades for turbine)

[rohanlol7]

Homework Statement

Here it is http://imgur.com/a/Wltb0

Homework Equations

F=dP/dt
Cosine rule

The Attempt at a Solution

For the first part of the question I'm not exactly sure what the resultant of these vectors represent. First i thought they represented the relative speed of water to the blades but that's only for the first diagram.
For the second Part i got the change in momentum to be (let m= mass per unit time here) = mt(Vi*cos(alpha)+Vo*cos(beta)). Now i assumed that collision is elastic and so the relative speed of approach and relative speed of separation should be the same. this gave me Vi^2 + Vb^2-2Vb*Vi Cos(alpha)=Vo^2+Vb^2-2Vo*Vb*cos(B). from these I could not get the required result.
However now that I'm thinking about it i think that the first vector sum represents the relative speed of approach and the second one represents the relative speed of sepration and i think the left hand side of the equation should thus be the same thing except Cos(alpha) should be replaced by Cos(pi-alpha) that would give me the result. But was that my mistake ?

 

[haruspex]

In part b), each diagram shows the absolute velocity of the blades and a velocity of a water stream. in each case, you have to decide whether the given water stream velocity is absolute or relative to the blades, then use a triangle of vectors to find... what?

You should certainly not assume the collision is elastic. In general, never assume conservation of work without good cause.
Use momentum conservation wherever that can apply (i.e., no external forces in the direction considered).

 

[rohanlol7]

In part b), each diagram shows the absolute velocity of the blades and a velocity of a water stream. in each case, you have to decide whether the given water stream velocity is absolute or relative to the blades, then use a triangle of vectors to find... what?

You should certainly not assume the collision is elastic. In general, never assume conservation of work without good cause.
Use momentum conservation wherever that can apply (i.e., no external forces in the direction considered).

In the question doesn't it say that Vin is the absolute velocity ?
the first triangle is basically Vin-Vb and the second one is Vout+vb. Doesnt the first triangle represent the relative velocity of water coming in relative to the blades and the second triangle seems to represent the resultant of Vin and Vb which i don't see why this is relevant
I can assume conservation of momentum in direction perpendicular to X.

 

[haruspex]

In the question doesn't it say that Vin is the absolute velocity ?
the first triangle is basically Vin-Vb and the second one is Vout+vb. Doesnt the first triangle represent the relative velocity of water coming in relative to the blades and the second triangle seems to represent the resultant of Vin and Vb

Yes, that is all correct. It does not have to be relevant to the next part, but it is what is asked for in part b).

I can assume conservation of momentum in direction perpendicular to X.

No. There is an unknown force from the axle supporting the blades in the direction perpendicular to X.

 

[rohanlol7]

So the change of momentum of the water in a time t is : m*t*(Vi*cos(alpha)+Vo*cos(beta)) ( only the change in direction of x is relevant since the force is only horizontal)
Next from the geometry: Vi*sin(alpha)=Vo*sin(beta) ( equating 'virtical' momentum )
Force = m*(Vi*cos(alpha)+Vo*cos(beta))
From there I could not eliminate alpha and beta to get the required equation, and i don't get why the Vb is relevant.

 

[rohanlol7]

Please ignore the above comment its wrong

 

[rohanlol7]

Please ignore the above comment its wrong

 

[rohanlol7]

Yes, that is all correct. It does not have to be relevant to the next part, but it is what is asked for in part b).

No. There is an unknown force from the axle supporting the blades in the direction perpendicular to X.

But since momentum is not conserved in a direction perpendicular to x in the direction of x, doesn't this imply that momentum is not conserved in any direction ?

 

[haruspex]

But since momentum is not conserved in a direction perpendicular to x in the direction of x, doesn't this imply that momentum is not conserved in any direction ?

No, why?
Edit: just realized you meant "nor in the direction of X". Why is it not conserved in that direction?

Edit 2:
Sorry, I was wrong to be discussing conservation of momentum. The point is that the water undergoes a change in momentum, and that is what applies the force to the blades.
You can do the calculation for both directions, but only the component in the direction X is interesting.

 

[rohanlol7]

No, why?

Sorry i mistyped, i meant if momentum is not conserved both in the direction of x and perpendicular to x.
Unless momentum is conserved in the direction of x ?

 

[rohanlol7]

No, why?
Edit: just realized you meant "nor in the direction of X". Why is it not conserved in that direction?

Edit 2:
Sorry, I was wrong to be discussing conservation of momentum. The point is that the water undergoes a change in momentum, and that is what applies the force to the blades.
You can do the calculation for both directions, but only the component in the direction X is interesting.

It is not conserved because the blades continue moving in direction X with the same speed.( unless I'm wrong here )
Okay, computed the change in momentum of the water in the direction X. I got m*t*(Vi*cos(alpha)+Vo*cos(beta))
From there i need to get rid of the trigonometric parts, maybe from some kind of relationship that I'm not seeing

 

[haruspex]

So the change of momentum of the water in a time t is : m*t*(Vi*cos(alpha)+Vo*cos(beta))

Two things...
You are told to assume that the water speed does not change; on the other hand, angle beta is relative to the blade motion. You need to find the X component of the absolute exit velocity.