# Homework Help: Power output of a pump

1. Oct 10, 2015

### amanda.ka

1. The problem statement, all variables and given/known data
A pump draws 500 kg of water/min from a 12 m deep well. The water is ejected with a speed of 20 m/s.
a) What is the power output of the pump?
b) How high is the fountain (ignoring air resistance)

2. Relevant equations

3. The attempt at a solution
For part a I found the potential energy and kinetic energy:
PE: m*g*delta h = (500)(9.81)(12) = 58860 J
KE: 1/2mv^2(final) - 1/2mv^2(initial) = 1/2(500)(20)^2 - 0 = 100, 000 J
Then I used Power = change in Work/change in time
= KE+PE/60 seconds
= 2647.7 W
Is this correct?

And I'm stuck on part b, here is what I did but I am not sure if it is correct:
[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]
-1/2mv^2(initial) = mgh(final)
-1/2*20^2/9.81 = h (final)
so height final = 20.4 m

Any help is appreciated :)

2. Oct 10, 2015

### TSny

Looks correct. "change in work" should just be "work". Does your instructor care about significant figures?

Is this supposed to be an equation? I don't see an equal sign.

You can see that something went wrong here. Your next to last equation implies that h is negative. Did you just arbitrarily drop the negative sign to get your final answer?

3. Oct 10, 2015

### SteamKing

Staff Emeritus
You need to examine the units of your calculations.

The pump draws water at the rate of 500 kg/min, which is not the same as 500 kg.
You are then multiplying 500 kg/min by v2 in (m/s)2. I don't think the end result of this calculation is Joules. Ditto for the calculation of PE.

4. Oct 10, 2015

### amanda.ka

For my equation i tried to do -1/2v^2 = mgh then solve for h but I think that is incorrect because I did get a negative :/

5. Oct 10, 2015

### TSny

Go back to your starting expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)]. What does this represent? What should it be equal to?

6. Oct 10, 2015

### amanda.ka

It represents the total energy Emech = K + U, should it be equal to zero (conservation of energy)?

7. Oct 10, 2015

### TSny

It does not represent the total energy. Let's take part of your expression: [1/2mv^2(final)-1/2mv^2(initial)] . What does this represent?

8. Oct 10, 2015

### amanda.ka

It represents the change in kinetic energy

9. Oct 10, 2015

### TSny

Yes. And the second part represents the change in potential energy. So, what does the entire expression represent?

10. Oct 10, 2015

### amanda.ka

Is it the change in work?

11. Oct 10, 2015

### TSny

No. As you noted, total mechanical energy is the sum of kinetic energy and potential energy: Emech = KE + PE. So, if you add together the change in KE and the change in PE, what do you get?

12. Oct 10, 2015

### amanda.ka

K(final) + U (final) = K(initial) + U (initial) ?
Sorry I don't think I am understanding the concept :(

13. Oct 10, 2015

### TSny

OK. This equation is correct. Note that this is equivalent to saying [K(final) - K(initial)] + [U (final) - U (initial)] = 0.

When you wrote the expression [1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] in your first post, you were writing [K(final) - K(initial)] + [U (final) - U (initial)] . So, I was trying to prod you into seeing that the expression you wrote down represents the total change in mechanical energy, ΔEmech.

[1/2mv^2(final)-1/2mv^2(initial)] + [mgh(final) -mgh(initial)] = ΔEmech.

The law of conservation of energy tells you that the change in total energy is zero. ΔEmech = 0. Therefore

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

You can then rearrange this as K(final) + U (final) = K(initial) + U (initial). This just says that the final total energy equals the initial total energy. You can solve the problem by starting with either this equation

[K(final) - K(initial)] + [U (final) - U (initial)] = 0

or, this equation

K(final) + U (final) = K(initial) + U (initial).

They represent the same idea of conservation of energy.

So, using either equation, try applying it to your situation and see if you can get a positive value for h.

14. Oct 10, 2015

### amanda.ka

Kf + Uf = Ki + Ui
= 1/2mv^2 + mgh = 1/2mv^2 + mgh
= 100,000 + (500)(9.81)h = 0 + (500)(9.81)(12)
= 100, 000 + 4905h = 58,860
41140 = - 4905 h
h = -8.387 m
My answer is still negative, is something wrong with the initial/final values ?

15. Oct 10, 2015

### TSny

Let's be clear on where you are taking the initial position and where you are taking the final position.

16. Oct 10, 2015

### amanda.ka

Would the initial U be negative 12 since it is below the well? If I take the top of the well as a reference point (zero) then the water would initially start -12 m below that

17. Oct 11, 2015

### TSny

My interpretation of the problem is that the water is projected upward with a speed of 20 m/s at ground level. So, if you are going to let your initial point be the point where the water has a speed of 20 m/s, then what would be your initial height?

18. Oct 11, 2015

### amanda.ka

Kf + Uf = Ki + Ui
0 + (500)(9.81)h = 1/2(500)(20^2) + 0
Now I get the same answer as I originally got but with a positive sign

19. Oct 11, 2015

### TSny

Yes, that looks good.

20. Oct 11, 2015