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What is the power input to the plant?

  1. May 25, 2003 #1
    I have a few questions that i need help understanding how to answer:

    1. At a power plant, the operating temperatures are 700 C and 310 C. The efficiency of the plant is 65% of its Carnot efficiency. The output of the plant is 500 MW.
    a. Determine the efficiency of the plant.
    b. What is the power input to the plant?
    c. The heat released by the plant is carried away by water. How many kg/sec must pass through the
    plant if the maximum allowed temperature increase is 20C?
    d. The heat of combustion of coal is 2.8 x 109 J/kg. How many kg of coal must be burned per hour to
    supply the plant?

    2. An electron initially moves in a horizontal direction midway between two charged parallel plates with a kinetic energy of 2000 eV. The plate separation d = 0.02 m and the length of the plates L=0.040m. The potential difference between the plates is 250 volts.
    a. Determine the initial velocity, v, in the position shown.
    b. What are the magnitude and direction of the electric field between the
    c. What are the magnitude and direction of the acceleration of the electron
    between the plates.
    d. What are the magnitude and direction of a magnetic field that, if applied
    between the plates, would cause the electron to travel in a straight line?

    For the first problem, I have determined that the efficiency of the engine is 26.1%. I can't figure out how to solve question 1b and the rest of the problem. I also don't know how to do problem number 2. Any help that is provided will be greatly appreciated, I need to know how to correctly solve these two problems before my final.
  2. jcsd
  3. May 25, 2003 #2
    No HW!

    We don't do your homework in this forum but rather we discuss any ideas and questions. Problems are welcome but don't give us your homework....
  4. May 25, 2003 #3


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    Forum rules state that for us to help you'll need to at least try the problem and show us where you got stuck.

    It doesn't help you if we do your homework for you.
  5. May 25, 2003 #4

    Tom Mattson

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    I deleted your other thread, because it was identical to this one. As the others said, tell us your thoughts on the problems and we will help you through your difficulties.
  6. May 25, 2003 #5

    I'd love to help with #1 but they're all gonna jump on me if I give it away too easily, so you have to try harder.

    I'll tell you this:
    I agree with your answer to part a, so you seem to have the basic idea.

    Now, forget about applying a formula and ask yourself, exactly what does efficiency mean?
  7. May 25, 2003 #6
    Sorry about the Misinterpretation

    I was not asking if someone could do the hw for me, I was simply just asking if someone could explain to me how i should go about solving the problems. I have used these forums before and never once asked for someone to "do my homework for me". I just felt that if I were to put the entire question in the post that I would be giving the person that helped me a full sense of what I was working on. In the earlier post it may have appeared that what I was asking for was for someone to do my HW for me but I am aware that this isnt the purpose of these forums. All I would like to know is what steps I should follow to answewr question 1b.
  8. May 25, 2003 #7

    Tom Mattson

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    How did you get that? I got about 36%.

    This is what I mean by "show me your work". That way, I can point your mistake out to you, instead of posting my solution.

    Once you get part 1a, this should be dead easy. The efficiency is the power output divided by the power input.
  9. May 25, 2003 #8

    Tom Mattson

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    Re: Re: Please Help

    And, of course, I forgot to convert to Kelvin, like a dumbass. 26% is correct.
  10. May 25, 2003 #9

    JM's answer of 26.1% for part a is correct.
    The Carnot efficiency EC = 1 - Tc/Th
    = 1 - 583K/973K = 40.1%

    It is given that the engine's efficiency is 65% of that:
    40.1% * 65% = 26.1%


    The next step is to use the general definition of efficiency which (as Tom pointed out) is
    E = output/input (expressed either in terms of energy or power)

    see if you can make any progress with that...
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