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

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

    Quiz4Physics.jpg

    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. jcsd
  3. Feb 17, 2009 #2

    LowlyPion

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    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

    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.
     
  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

    LowlyPion

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    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

    LowlyPion

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    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

    LowlyPion

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    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.
     
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