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Homework Help: Calculating heat req. by fuel in boiler, interpolation needed.

  1. Nov 25, 2011 #1
    1) Feed water enters a boiler at 172.52 degrees Celsius. The boiler produces super heated steam at 5600kpa and 472 degrees Celsius. The boiler efficiency is 82%. Find the Heat required by fuel.

    I interpolated the value given for feed water and super heated steam.

    h@5600kpa 472 degrees Celsius = 3359.0128 kj/kg
    hf@ 172.52 degrees Celsius = 730.30 kj/kg


    delta h = 3359.0128 kj/kg - 730.30kj/kg
    delta h = 2628.7128 kj/kg (this is the energy that the boiler would produce running at 100% to make the given steam right?)


    Qfuel = (2628.7128 kj/kg)(1.18) (I'm multiplying by 1.18 to make up for the 18% efficiency difference)
    Qfuel = 3101.88 kj/kg

    efficiency = Qsteam/Qfuel
    .82 = 2628.7128/Qfuel

    Qfuel = 3205.07 kj/kg

    3) I get two different answers for quantity of heat required by fuel. Can someone please explain why? Thank you.
    Last edited: Nov 25, 2011
  2. jcsd
  3. Nov 25, 2011 #2


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    The temperatures that you are given are based on the 82% efficiency.

    82% will give dh as 2628.7128 kJ/kg
    100% will give 2628.718/0.82

    which is the same as your last method.
  4. Nov 25, 2011 #3
    hmm sorry, I still am not following

    dh = 2628.718 (this is the amount of heat made by the boiler to make 5600kpa and 472 degrees Celsius of steam) boiler running at a 82% efficiency.

    18% of heat from the fuel is lost correct?

    So why couldn't I take 2628.718 and multiply that by 1.18 to make up the missing 18% of heat lost by the fuel thus giving me the total heat of the fuel required.

    I just don't understand why I can't do this.
  5. Nov 26, 2011 #4


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    Because what you'd be doing is this 2628.718(1)+2628.718(0.18), you'd be adding the value at the 82% efficiency to 18% of the 82% efficiency value.Which is incorrect.
  6. Nov 26, 2011 #5
    ah yes that makes sense. Thanks for the help!
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