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2 steam engines

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 718∘C and 427∘C, and of the second 419∘C and 270∘C.
    A) If the heat of combustion of coal is 2.8×107J/kg, at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65 % of the ideal (Carnot) efficiency.
    B)Water is used to cool the power plant. If the water temperature is allowed to increase by no more than 5.5 C∘ , estimate how much water must pass through the plant per hour.

    2. Relevant equations
    e = 1-QL/QH
    eideal = 1-TL/TH
    e = W/QH

    3. The attempt at a solution
    for part A
    second engine
    e2 = W/(QH2) = (950000000J/s)/(QH2)

    e2ideal =1-(TL2/TH2) = 1-(427+273)/(718+273) = 1 - 543/692 = .215
    e2 = .65(e2ideal) = .65(.215) = .140

    .140 =(950000000J/s)/(QH2)
    QH2 = (950000000J/s)/(.140) = 6785714286 J/s

    QH2 = QL1

    e1ideal = 1 - 700/991 = .294
    e1 = .65(.294) = .191
    .191 = 1 - (QL1)/(QH1)
    (QL1)/(QH1) = 1 - .191 = .809
    QL1 = (.809)(QH1)

    so QH2 = .809(QH1)
    QH2 = 6785714286 J/s
    .809(QH1) = 6785714286 J/s
    QH1 = 8387780329 J/s - heat input of first engine

    (8387780329 J/s) (kg/(2.8x10^2 J))
    = 299.6 kg/s
    but it came up wrong. I cant figure out what I did wrong
     
  2. jcsd
  3. Oct 25, 2014 #2

    haruspex

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    The given power output is the sum of the outputs of the two engines, not the output of the second engine alone.
    You seem to have used the first engine's temperatures in your calculation for the second engine.
    Haven't checked past there.
     
  4. Oct 25, 2014 #3
    ahh ok thanks I got it right now
     
  5. Oct 25, 2014 #4
    im a little confused about the set up for part B
    would it work if i used
    q = mc(deltaT) where q = 950000000 J/s
    c is the specific heat of water and deltaT is 5.5 and solve for m?
     
  6. Oct 26, 2014 #5

    haruspex

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    The 950MW is the output power. It is not heat. What is the heat that must be removed?
     
  7. Oct 26, 2014 #6
    the sum of QL for each engine?
     
  8. Oct 26, 2014 #7

    haruspex

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    No. Some of the QL from the first engine was turned into useful work by the second.
     
  9. Oct 26, 2014 #8
    so only the QL for the second?
     
  10. Oct 26, 2014 #9
    yeah i got it. thanks!
     
  11. Oct 26, 2014 #10

    haruspex

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    Yes.
     
  12. Oct 26, 2014 #11

    NTW

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    You have written, toothpaste666, in your calcs above:

    e2ideal =1-(TL2/TH2) = 1-(427+273)/(718+273) = 1 - 543/692 = .215

    And that's wrong. It should be:

    e2ideal =1-(TL2/TH2) = 1-(427+273)/(718+273) = 1 - 700/991 = .2936

    There may be other mistakes (or not). I've spotted just this one...
     
  13. Oct 26, 2014 #12

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    I would prefer a global approach to solve the problem, haruspex permitting... The power plant can be considered as a 'black box' that ingests coal at a given rate, produces a given power of electricity and rejects some quantity of heat per unit of time. We are told that the the efficiencies of the engines are 65% of the Carnot efficiency. Well, the efficiency of the first engine, working between 718 ºC and 427 ºC is 0,29360. That of the second, working between 419 ºC and 270 ºC is 0,21532. Their compounded efficiency is thus 0,29360 + 0,21532 = 0,50892. As we are told that we are left with 65% of that, then we have a global, real efficiency of 0,33080.

    Now that we know the efficiency of that 'black box', we can reckon that, from 100 units of coal energy input, 33,08 units become electric energy, and 66,92 units go to heat a mass of water by 5,5ºC .

    Little is left, but doing a few sums. I believe that toothpaste666 was doing quite well, but had confused some data in his calculations. I believe that my suggestions are not excessive, in the sense that I'm not violating the rules...
     
  14. Oct 26, 2014 #13

    haruspex

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    Looks like a typo in copying into PF. The 543 and 692 are right, it's the 427 and 718 that are wrong.
     
  15. Oct 26, 2014 #14

    haruspex

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    Is that valid? Maybe 0,29360 + (1-0,29360)*0,21532?
     
  16. Oct 26, 2014 #15

    NTW

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    You are right, I believe...

    The efficiency along the all the fall from 718 ºC to 270 ºC would be 1- (543/991) = 0.45207. You calculate a lower one, 0,4457, that is somewhat lower than the 'extreme' 718 ºC/270 ºC efficiency because of the eight degrees lost in the 'transfer'...
     
  17. Oct 26, 2014 #16

    NTW

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    Which is quite logic, because -in order to obtain total, 'chained' efficiency, you calculate the efficiency of the second machine from the heat rejected from the first machine...
     
  18. Oct 26, 2014 #17

    NTW

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    Accepting, of course, haruspex' correction concerning the chained efficiency, I re-write the global approach to solve the problem. The power plant can be considered as a 'black box' that ingests coal at a given rate, produces a given power of electricity and rejects some quantity of heat per unit of time. We are told that the the efficiencies of the engines are 65% of the Carnot efficiency. Well, the efficiency of the first engine, working between 718 ºC and 427 ºC is 0,29360. That of the second, working between 419 ºC and 270 ºC is 0,21532. Their compounded efficiency is thus 0,29360 + (1-0,29360) * 0,21532 = 0,44570. As we are told that we are left with 65% of that, then we have a global, real efficiency of 0,28971.
    Now that we know the efficiency of that 'black box', we can reckon that, from 1000 units of coal energy input, 289,71 units become electric energy, and 710,29 units go to heat a mass of water by 5,5ºC .
     
  19. Oct 26, 2014 #18
    yeah it looks like i got a little sloppy and confused myself =\. So I can treat this problem as one engine with the combined efficiency instead of two separate engines? thanks for all your help guys
     
  20. Oct 26, 2014 #19

    NTW

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    Yes, but not simply adding the individual efficiencies, as I mistakenly did. Read the posts of haruspex...
     
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