AP Physics B (Heat Engines)

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18. (II) At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate ehat input of the second. Th eoperating temeprautres of the first are 680 degrees C and 430 degrees C, and of the second 415 degrees C and 280 degrees C. If the heat of combuston of coal is 2.8 E8 J/kg, at what rate msut coal be burned if the plant is to put out 900 MW of power. Assume the efficiency of the engines is 65 percent of the ideal (Carnot) efficiency.

Ok I calculated the efficiency of the first engine to be about 17.05 % efficient and the second one to be about 14.64 % efficient.

I do not know were to go from here. Please show me the formulas you use. Thanks You!

  • AP Physics B (Heat Engines) -

    I know that efficiency is

    e = |W|/|QH|

    were the H is the subscript

    I know that 900 MW is a value of power which is not work

    I know that |QH| is 2.8 E8 J/kg but I do not like the units for this, J/kg, as I thought Q was measured in joules only...

    How do I find the work done?

  • PLEASE SOMEONE -

    I need help with this, thanks!

  • AP Physics B (Heat Engines) -

    the work that engine one does is (using your numbers)

    .65*.17Heatinput
    Then the waste heat is
    Heatinput- work doneinengine1
    But this is used as input to engine2, so the work done in engine 2 is
    Heatinput*massrate(1-.65*.17)(.65*.14)

    But the total work done is

    Heat input*massrate(.65*.17)+Heatinput*massrate(1-.65*.17)(.65*.14)

    but we want that to be 900Mwatts,
    900M=
    Heat input*massrate(.65*.17)+Heatinput*massrate(1-.65*.17)(.65*.14)

    Solve for mass rate in kg/sec

  • AP Physics B (Heat Engines) -

    Do not multiply the percentages you calculated by .65, because you already multiplied it to get 17% and 14.6% :)

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