1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Potential Energy, Tank on Hill, water power generator downhill.

  1. Feb 17, 2009 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations



    3. The attempt at a solution

    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
    -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. jcsd
  3. Feb 17, 2009 #2


    User Avatar
    Homework Helper

    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?
  4. Feb 17, 2009 #3
    Yeah, I got that far. What about part B though?
  5. Feb 17, 2009 #4
    Your PE seems to be okay. As for b) what is the equation for power?

    Edit: LowlyPion was faster...
  6. Feb 17, 2009 #5
    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


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

    That number seems really small.
  7. Feb 17, 2009 #6
    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
  8. Feb 17, 2009 #7


    User Avatar
    Homework Helper

    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
  9. Feb 17, 2009 #8
    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?
  10. Feb 17, 2009 #9


    User Avatar
    Homework Helper

    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.
  11. Feb 17, 2009 #10
    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
  12. Feb 17, 2009 #11


    User Avatar
    Homework Helper

    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.
  13. Feb 17, 2009 #12
    Ugh, this is rough. I much prefer calc 2 to this stuff.

    Thank you for your help.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook