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# Calculate jet force and angle on moving water vane

## Calculate jet force and angle on moving water vane

(OP)
I did a Civil Engineering course some years ago and from my textbook I had a question on this below that I am would like to solve. I am trying to understand what it means when water leaves the vane with no velocity in the direction of motion of the vane. Does this mean it leaves the vane at a 90 degrees angle ?

A jet of water of 100 mm diameter is moving at 36 m/s and is deflected by a vane moving at 15 m/s in a direction at 30 degrees to the direction of the jet. The water leaves the vane with no velocity in the direction of motion of the vane. Determine
(a) The inlet and outlet angles of the vane for no shock at entry or exit,
(b) The force on the vane in the direction of motion
and (c) The force on the vane at right angles to the direction of motion.

Take the outlet velocity of the water relative to the vane to be 0.85 of the relative velocity at entry.

This is what I have so far.

If the water enters without shock the relative velocity at inlet V1r must be tangential to the blade at at an angle (alpha symbol) to the direction of motion.

V1 = absolute velocity of incoming jet.
u = velocity of the vanes
V1r = relative velocity of jet over the blade at inlet
V2r = relative velocity of water at outlet
V2 = absolute velocity of water at outlet

My photo below shows the inlet and outlet velocity triangle for this vane setup

So far I have calculated blade angle at inlet of 48 degrees 3 minutes 12 seconds,

But now I need to get an understanding what it means, when water leaves vane with no velocity in direction of motion as mentioned above.
This will help with understanding how to get outlet angle of vane and horizontal and vertical forces on the vane.

### RE: Calculate jet force and angle on moving water vane

It means that the maximum amount of energy possible has been extracted by the blade - no residual kinetic energy in the water.

### RE: Calculate jet force and angle on moving water vane

Sounds like a typical over-simplified textbook problem which allows you to calculate an answer assuming all the energy of the water jet goes into turning the vane. You seem to be expecting to be able to calculate a real-world answer to a fictional problem.

Rod Smith, P.E., The artist formerly known as HotRod10

### RE: Calculate jet force and angle on moving water vane

(OP)
@compositepro
So with "water leaving the vane with no velocity in the direction of motion" you are saying that, that means there is no residual kinetic energy in the water and maximum amount of energy possible has been extracted or taken out by the blade.

Does this mean that V2r (relative velocity of water at outlet) leaves the vane at a right angle to the vane setup ?

How would I go about calculating the outlet angle of the vane for no shock at exit with this question ?

### RE: Calculate jet force and angle on moving water vane

(OP)
@compositepro
Hi Compositepro

With water leaving the vane with no velocity in the direction of motion of the vane, does this mean that V2r (relative velocity of water at outlet) leaves the vane at a right angle to it ?

How would I go about calculating outlet angle of vane for no shock at exit with this question ?

### RE: Calculate jet force and angle on moving water vane

(OP)
@bridgesmith
Hi BridgeSmith

For this question, I think you are referring to the fact that all of the energy of the water jet goes into turning the vane, hence no shock at entry and exit.
So in real world I imagine that would not happen and you would have to take into account some shock loss, I wonder how that would be done ?

### RE: Calculate jet force and angle on moving water vane

(OP)
Seeing no one is getting back to me on my replies to help me with this question, I am going to see how it would work with water leaving vane at a right angle.

We now want to calculate the outlet angle for Vane setup

Find Beta angle using the outlet triangle:

V2r = 0.8 V1r = 0.8 f1 / (sin * Alpha angle) = (0.8 * 18) / (sin * 48 degrees 3 minutes 12 seconds)
V2r = 19.36 m/s

f2 = V2r * sin 90 degrees = 19.36 * sin 90 degrees = 19.36 m/s

tan Beta = f2 / [ 15 - ( * cos 90 degrees)] = m/s
tan Beta = 19.36 / [ 15 - 0 ] = 19.36 / 15 = 1.290726178092654

Beta = 52 degrees 13 minutes 58.77 seconds (Outlet angle of vane)
This is not the answer of 43 degrees 11 seconds from the textbook

This is why I am putting this post up to figure why I am not getting the correct answer for outlet angle of vane
Appreciate some help understanding this thanks

### RE: Calculate jet force and angle on moving water vane

I'm not sure, but it seems you are not keeping your frame of reference constant. In one frame the blade is moving and in the other it is stationary.

### RE: Calculate jet force and angle on moving water vane

(OP)
@compositepro

It seems like you are not completely sure how it works with the outlet angle of the vane.

I have gone by an example in my textbook.
In that example V2r was at an angle of 150 degrees to horizontal and that angle was taken when the vane was stationary. So I think this relates to what you are saying with one frame the blade is moving and in the other it is stationary.

So just confirming do you actually know how this outlet angle should be calculated ?

If you do, then we can discuss how this should be calculated !

### RE: Calculate jet force and angle on moving water vane

(OP)
Hi @Compositepro and @BridgeSmith

I have some really awesome news you would love to hear.

After typing force on vane at right angles on google search, I came across this link https://holooly.com/solutions/a-jet-of-water-100-m...

This looks to be the same question, it does not calculate the force on the vane at right angles to the direction of motion , but I think this and my textbook, would give me a good idea as to how to work that out.

It confirms that when water leaves the vane with no velocity in the direction of motion of the vane, then the water leaves the vane with an absolute velocity at right angles to the direction of the vane.

The outlet angle of 43 degrees agrees with my textbook answer.
The link has force on vane in direction of motion as 8.814 kN, my textbook has it as 5.63 kN. I will have to check that with the calculations as maybe my textbook may not have that correct.

Overall though I think this is an excellent find to give an understanding on how these vane questions work.

BridgeSmith in the real world though, I think the no shock at entry or exit would not be possible and there must be a way to take this into account with calculations, I would have to research that to figure that out.

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