00:01
So we're asked to find the force exerted by the jet on the buckets as well as the work done per kilogram of water, the power developed by the runner, and the hydraulic efficiency of the turbine.
00:16
So to get started, we'll first look at the force exerted by the jet of the buckets.
00:23
Let's first start out by writing out our given data in the problem.
00:28
We know that the tan velocity here is equal to our 20 meters per second.
00:46
Our head race to tell race level difference is going to be our h here.
00:55
And this is going to be our 60 meters.
01:02
So we have our head loss due to friction, which is our h, subscript f, and this is going to be 10 meters.
01:22
Our discharge through the nozzle is going to be our q, which this is going to be equal to 0 .03 meters cubed per second.
01:40
We have the coefficient of velocity, which is 0 .97.
01:52
And then we have our acceleration due to gravity, which this is our already given, 9 .81 meters per second squared.
02:09
Okay, so now we're going to first find our jet velocity at the nozzle exit.
02:45
Actually, we'll put this on the bottom.
02:51
There you go.
02:52
Nozzle exit.
02:54
All right.
03:01
So to calculate this, we're going to write out our h effective.
03:14
Is going to be equal to our head, which we were given, minus the head loss, which is the lowercase h.
03:31
Let me read this.
03:34
H, script, f.
03:39
All right, and now we can just add what we have here.
03:44
So up here, we have 60 minus 10.
03:48
So let's put this in.
03:51
60.
03:53
Minus 10 and this is going to be equal to which we'll write this on a new line actually 50 meters all right now next we need to calculate the theoretical velocity which this is v the theoretical velocity is going to be equal to the square root of 2g times h effective.
04:40
All right, now we can add in our given.
04:43
So v -t -h is going to be equal to 2 times 9 .81 times the 50.
04:59
All right.
05:04
Now we simplify this, taking the square root of everything.
05:12
So the skirt of 981, which gives us 31 .32 meters per second.
05:31
And now we find the actual jet velocity, considering the coefficient of the velocity.
05:39
Velocity.
05:42
So our last step is to set v is equal to c our coefficient of velocity times our theoretical velocity.
05:58
So up here our coefficient velocity velocity is 0 .97 .0 .97 times our theoretical velocity which is here 3 .97 times 31 .32 meters per second.
06:27
All right.
06:34
Now we can just simplify, which will give us 30 .38 meters per second.
06:49
And this will be our actual jets velocity.
06:58
All right.
07:02
So now we're going to to find the force exerted by the jet on the buckets.
07:11
So we're going to write the force.
07:18
It's going to be equal to row.
07:26
All right, hold on.
07:27
Let me rewrite this.
07:31
Row times q times v minus you, and then times one plus the cosine of theta.
07:58
All right.
08:00
So now we need to first write out our given.
08:07
So our row, we'll write it over here.
08:13
Our row is going to be equal to 1 ,000 kilograms per meters cubed.
08:30
And then we have our q from above, which is equal to 0 .03 meters cubed per second.
08:56
And we have our velocity, which we got from before, 30 .38 meters per second.
09:07
And then we have our u, which is equal to 20 .38 meters per second.
09:09
And then we have our u, which is equal to 20.
09:17
Meters per second.
09:22
And then we have our theta, which is equal to 15 degrees.
09:31
And this is our given.
09:37
Let's put this in a box right here.
09:46
All right.
09:48
So now let's first calculate the cosine of theta.
09:54
So our theta down here will use a different color.
09:59
So our co -sign.
10:00
Which our theta is 15 degrees and simplified we're going to get 0 .9 659.
10:20
All right now we can substitute everything from what we've written above for the force exerted here.
10:34
So f is going to be equal to our 1 ,000 times our 0 .03 times our 30 .38 minus 20 times 1 plus our cosine of 15 degrees is 0...