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2 Questions - Heating room air, and fuel combustion efficiency

  1. Jun 3, 2013 #1
    2 Questions -- Heating room air, and fuel combustion efficiency

    1. The problem statement, all variables and given/known data
    2 questions. In advance, i needed to translate it so i am sorry if it is states incorrect. I really did my best to do it right.

    1.
    The density of air at 1 atm and 10.0 ° C is equal to 1.24 kg / m3. (cubic meter)
    A room has the following sizes: 8,0 m x 4,5 m x 2,7 m
    Calculate how long it takes for a heater of 1.5 kW to heat the air from the room to 21.0 ° C.
    Use c = 1005 J / kg / K

    2.
    An internal combustion engine uses a fuel with calorific value/heating value of 38 MJ / kg.
    The density of the fuel is 900 kg / m3. (cubic meter)
    Of each MJ heat only 0.25 MJ is converted to mechanical energy.
    The efficiency is 25%.
    The engine can deliver 30 kW.
    Calculate how many liters of fuel per hour the engine used.

    2. Relevant equations
    http://imgur.com/WQx3BqI

    these are the once i used. I probly made some/alot mistakes so i dont know if i need more.

    3. The attempt at a solution
    1.
    8,0m x 4,5m x 2,7m = 97,2m3 (cubic meters)
    1 atm and 10.0 ° C = 1,24 kg/m3
    Heater 1,5 kW = 1500 W = 1500 J/s
    C = 1005 J/kg/K

    This is where I am atm. I figured I needed to change the heater to J/s hence you need to know how long it takes. But after I gathered all the data I could find I don’t know where to start with the calculation or what formula to use first.

    2.
    Usage = 38 MJ/kg = 0,38 x 10^8 J/kg
    P = 900 kg/m3 (cubic meters)
    Energy usage = x 0.25
    30 kw = 30000 J/s
    30000 J/s x 3600 = 1,08 x 10^8 J/hour (first I figured I needed to set it in hours)
    0,38x10^8 / 1,08x10^8 = 0,3519 J/kg/hour
    900x0,3519 = 316,71 m3 (cubic meters)
    316,71 m3 = 316710 dm3 = 316710 Liters of fuel.

    I doubt there is an engine that uses that much so I think I made some mistakes.


    I hope i can get some help here on 1 or maybe even both questions. Thanks in advance. I tried my best on both and these are the answers and solutions i could produce
     
  2. jcsd
  3. Jun 3, 2013 #2

    CWatters

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    One problem at a time...

    Look up "Specific Heat Capacity" in your text book.

    Write down the equation that relates energy to mass and the temperature rise.

    Then calculate the following in this order..

    a) how many Kg of air you are heating.
    b) What the temperature rise is in Kelvin (ΔT)
    c) The number of joules needed to raise that much mass of air by that ΔT.
    d) Then it might help to know that 1W = 1J/S. If you know how many joules are required and the rate at which the heater produces them you can calculate how many seconds it will take.
     
  4. Jun 3, 2013 #3
    thanks, gonna try that now
     
  5. Jun 3, 2013 #4

    CWatters

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    I'm off to bed as late here.

    For Q2

    Eff(%) = 100 * Power_out/power_in

    rearrange to give

    Power_in = 100 * Power_out/Eff

    I make the power going into the engine 120kW.

    More tomorrow.
     
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