# Potential Energy, Tank on Hill, water power generator downhill.

1. Feb 17, 2009

### Wobble

1. The problem statement, all variables and given/known data

2. Relevant equations

KE=(0.5)mv^2

PE=mgh

3. The attempt at a solution

a.)
The tanks volume is 20m^3. I then multiply that by the density to get the mass of the water, which is 20,000 kg.

Apply that into PE=mgh for gravitational potention
(20,000)(9.8)(50)=9,800,000
-Is everything in the right units?-

b.) I have no idea how to do this part.

If I can get some help with part b, that would be greatly appreciated.

2. Feb 17, 2009

### LowlyPion

m3 = 1000 L
1 L = 1 kg H2O

20m3 = 20,000 kg times g and 50 m is your PE. So far so good.

But what does that give you? Joules?

And they want Watts? Which are what? Joules/sec?

3. Feb 17, 2009

### Wobble

Yeah, I got that far. What about part B though?

4. Feb 17, 2009

### Kruum

Your PE seems to be okay. As for b) what is the equation for power?

Edit: LowlyPion was faster...

5. Feb 17, 2009

### Wobble

Hm, I've been in class all day.

Power is Energy/time. I'm looking for the volume of water to produce 1000W

P=E/t, and I get 1000J of water per second for 1000 watts of power in one second.

Then I plug that into the GPE equation GPE=mgh where I'm looking for mass

1000=m(9.8)50
m=2.04kg

Then divide that by the density to get the cubic meters of water per second, .00204m^3

That number seems really small.

6. Feb 17, 2009

### Wobble

I'll attempt to solve the rest while waiting for a response to my answer for part b.

c.) 86,400 seconds in one day. 86,400 * .00204 = 176.256 m^3 per day.

d.) 1m^3=35.3ft^3

so .00204*35.3=.072cfs

e.) my theoretical creek would have 0.3 cfs
I'm not sure where to go from here

7. Feb 17, 2009

### LowlyPion

Part b looks ok at about 2.04 L/s

1 L = .0353 cf

OK to e) so figure what they want.

It takes 2.04 L a sec to generate what you need - i.e 1000 w.

If the flow of 6 is reduced to a 1/20 that makes it 6/20 cfs *1/.0353 cfs/L 8.5 L and then 1/2.04 kw/L = 4.16 kw

8. Feb 17, 2009

### Wobble

Ok. So then for part F

I got 4.17(.065)(86,400)(365.25)=$8,520,552 as the cost for the year. That doesn't make any sense to me. Does it generate 4.16 kW every second? 9. Feb 17, 2009 ### LowlyPion You mean Joules / s don't you? That's watts. Then you must be real careful in reading your problem. The price they quote is kw-hour = 3600 *J/s = 3600 w. 10. Feb 17, 2009 ### Wobble Oh crap. So should I use 4.16 times 3600 (seconds in an hour), then multiply that by rate of .065? (4.16)(3600)(.065) to get the cost for 1 hour. Then that number times 24, then 365.25 to get the amount in a year? That gave me$8,553,687.48

Last edited: Feb 17, 2009
11. Feb 17, 2009

### LowlyPion

1000j/s is 1 kw.

You need 3600 of those to make a kw-h

At the flow rate they ask about though you have 4.16 j available every second for an hour so that means the stream provides 4.16 kw-hour every hour.

How many hours in a year?

24*365 now times 4.16 kw-h and times again the price of \$.065.

12. Feb 17, 2009

### Wobble

Ugh, this is rough. I much prefer calc 2 to this stuff.

Thank you for your help.