# Homework Help: An electric pump raise 9.1m^3 of water from a reservoir

1. Oct 25, 2011

### lionely

1. An electric pump raise 9.1m^3 of water from a reservoir whose water-level is 4m
below ground level to a storage tank above ground. If the discharge pipe outlet is
32m above ground and the operation takes 1hour,find the minimum power rating of
the pump if it's efficiency is 70%.(1m^3 of water has a mass of 1000kg)

2. Relevant equations

3. The attempt at a solution

An Idea suddenly came to me I need to find the energy but using Potential Energy formula? mgh?

I'm sorry I have no Idea how to attempt it, please help me, give me a hint on what to do.

2. Oct 25, 2011

### edgepflow

Re: Machines-

Yes, potential energy = mgh.

And power = Energy / Time = mgh / time

Or the ideal power is expressed as:

Power (ideal) = (Δm / Δt) gh

And we divide this by the pump efficiency $\eta$ to get the real power:

Power (ideal) = (Δm / Δt) X gh / $\eta$

Notice that Δm can be figured our from your volume and density and Δt is given.

3. Oct 25, 2011

### lionely

Re: Machines-

Sorry I'm not a pro physics student that triangle means change right? so it's change in mass/change in time x gravity x height/ 70%?

4. Oct 25, 2011

### edgepflow

Re: Machines-

Yes ! And you are getting the idea, so maybe one day soon you will be a pro physics student.

5. Oct 25, 2011

### lionely

Re: Machines-

But the thing is I don't see a change in mass or density. :S

6. Oct 25, 2011

### lionely

Re: Machines-

Ohh wait I should find the density by doing 9100KG x 9.1? because Density = Mass x volume?

7. Oct 25, 2011

### edgepflow

Re: Machines-

Think of the change is mass as follows:

First note: mass = density X volume = 1000 kg / m^3 X 9.1 m^3

After 1 hour, all this mass will be pumped out, so this is your "change" in mass.

8. Oct 25, 2011

### edgepflow

Re: Machines-

Close. But, Density = Mass / Volume

9. Oct 25, 2011

### lionely

Re: Machines-

Alright should I do this 9100 x 10 x 32( I believe this is the height) to get the energy then divide it by the efficiency?

10. Oct 25, 2011

### edgepflow

Re: Machines-

Close. Notice that the water is 4 m underground, so work is done over a total height change of: 32m + 4m. And you have to introduce your time of 1 hour in the equation to get power.

11. Oct 25, 2011

### lionely

Re: Machines-

Alright so it's Power= Energy/Time = mgh/time = 9100kg x 10 x 36/ 60 or convert the 36m to km?

12. Oct 25, 2011

### edgepflow

Re: Machines-

With some practice, you can learn to manipulate units: (kg/sec) X (m/s^2) X m = watt. For now, if you stay in base SI units of kg, m, s, then energy will be in joules, and power will be in watts. So your time should be noted as: 1 hr = 3600 sec.

13. Oct 25, 2011

### lionely

Re: Machines-

I can't seem to get the answer in the back of the book it says 1.3kW

14. Oct 25, 2011

### edgepflow

Re: Machines-

Show me the numbers you used.

15. Oct 25, 2011

### lionely

Re: Machines-

9100kg x 10m/s2 x 36m / 60 seconds = 54,600J/70%

16. Oct 25, 2011

### edgepflow

Re: Machines-

The elapsed time is 1 hr:

1 hr = 60 minutes = 3600 sec.

17. Oct 25, 2011

### lionely

Re: Machines-

LOL I didn't even realize LOL I'm STUPID

18. Oct 25, 2011

### edgepflow

Re: Machines-

Honest mistake.

This is a good "real world" problem. Try to understand all we did.

19. Oct 25, 2011

### lionely

Re: Machines-

I will this text book I'm using is full of challenging questions, Physics by Abbott you know of it?

20. Oct 25, 2011

### edgepflow

Re: Machines-

I used Modern Physics by Tipler back in my college days.

21. Oct 30, 2011

### lionely

Re: Machines-