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Power and Energy - energy in a tidal power plant

  1. Jul 12, 2009 #1
    1. The problem statement, all variables and given/known data
    A tidal power plant generates 10 MW of electricity during a period of 12.5h. Assume that the water is collected behind a dam 6.0m high and is allowed to pass through a turbine twice to generate electricity, once as the tide comes in and once as the tide goes out. What mass of water flows through the turbine if the efficiency is 85%?

    Given:
    P=10MW= 10*10^6W
    t=12.5h= 45 000s
    h= 3m?
    g= 9.8 N/kg

    -the tide comes in twice, so I'll have to multiply the energy produced by two
    -the efficiency is 85%, so power will have to be multiplied by 0.85
    -the max height given is 6m, but there's an example problem in this textbook where the max height of the reservoir is 6m again, but they use an average height of 3m? I tried using both 6m and 3m.

    Required:
    mass of water

    2. Relevant equations
    P=Eg/t
    Eg=P*t

    Eg=m*g*h
    m=Eg/g*h


    3. The attempt at a solution

    The problem seems simple enough but I can't seem to get the answer at the back of the book, which is 9.1*10^6 kg:

    efficiency is 85%, so 85% of the power produced should be

    P= 10*10^6W * 0.85
    P= 8.5*10^6W

    then, solve for the gravitational energy produced

    Eg=P*t
    Eg=8.5*10^6W * 45 000s
    Eg=3.825*10^11J

    the tide pass through the turbine twice, so

    Eg=3.825*10^11J * 2
    Eg=7.65*10^11J

    plug this in into the equation for gravitational energy and solve for mass

    m=Eg/g*h
    m=7.65*10^11J/ (9.8N/kg*3m)
    m=2.6*10^10kg

    so, that didnt work... let's try with h=6m instead

    m=Eg/g*h
    m=7.65*10^11J/ (9.8N/kg*6m)
    m=1.3*10^10kg

    nope :/.
     
  2. jcsd
  3. Jul 12, 2009 #2

    cepheid

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    The first and most obvious problem that I see is your handling of the information given about the efficiency. 10 MW is the final power generated (the power that is usable electricity). What the question is saying is that the turbine WAS only 85% efficient in generating that 10 MW.
     
  4. Jul 12, 2009 #3
    so then the answer i get for my gravitational energy should be divided by two, not multiplied right?

    is there anything else wrong in the process?
     
  5. Jul 12, 2009 #4

    cepheid

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    Umm, no. That has nothing to do with anything. I'm saying that you're making use of the 85% wrong. Read what I said again.

    10 MW is the output of the plant. You know this because it says that the plant produces 10 MW of electricity.
     
  6. Jul 12, 2009 #5
    yes i understand now about the 85% now, thank you, but im asking if there are any more errors.

    you said "first and most obvious problem". if thats the first, what are the rest?
     
    Last edited: Jul 12, 2009
  7. Jul 12, 2009 #6
    E=P*t
    E=10*10^6W * 45 000s
    E=4.5*10^11 J

    so, the tide coming in twice doesnt have anything to do with anything, so ill leave out any dividing/multiplying by 2 that i thought i had to do. ill go straight to solving for mass.

    m=Eg/g*h
    m=4.5*10^11 J/ (9.8N/kg*3m)
    m=1.5*10^10 kg

    i really dont know what im doing wrong :/.
     
  8. Jul 12, 2009 #7

    cepheid

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    No, I did not mean to imply that you had definitely made more errors than that. What I meant, when I said this, in fact, was that I had not looked more deeply into your solution for further problems and that this one was merely the first one that struck me.

    No, I didn't mean to say that the tide going in and coming out twice was irrelevant to the problem. I just meant that it had nothing to do with what I was trying to say in my first post, which is what you seemed to be responding to.

    Sorry for the confusion on both counts. I'll take a closer look at your work now.
     
  9. Jul 12, 2009 #8

    cepheid

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    For some reason, you are still not taking the efficiency into account properly. You went from doing it wrong to just ignoring it completely. Let me state this again. The 10 MW is the output of the power plant (in electrical energy). However, the plant is not 100% efficient, suggesting that the input to the plant was, in fact, greater than 10 MW, and that some of the water energy was lost due to the inefficiency of the plant. Now do you understand?

    What happened to the factor of two? If the water runs through the turbine twice, then that amount of gravitational potential energy is indeed being used twice. Also, isn't the height 6 m?

    EDIT: Nevermind. 3.0 m is the height of the centre of mass of the fluid, which is how the gravitational potential energy is being measured. That's why 3.0 m is used. My bad.
     
  10. Jul 12, 2009 #9
    OH okay.

    thank you!
     
  11. Jul 12, 2009 #10
    and sorry, thank you for your patience lol.
     
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