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Find amount of H20 needed to cool Steel

  1. Jul 31, 2013 #1
    Hello people, thanks for reading. I am glad you are checking out my problem. Basically I need to confirm a "theory/formula" that my boss had written down from a decade ago declaring how much water (roughly) would be needed to cool Slag from steel making process to below boiling.

    PROBLEM: Figure out how much H20 is needed to cool Slag down to 212°C
    GIVEN:
    MASS: 2000 lbs. or 1 ton or 907.185 kg
    HEAT CAPACITY
    Carbon Steel: 0.12 (Kcal/Kg°)C
    Heat of H20 Vaporization (Enthalpy): 2260 J/g°C
    TEMPERATURE
    T1: 2200°F or 1104.4°C
    T2: 212°F or 100°C
    ΔT: 1104.44 °C
    CONVERSIONS
    1 Kcal = 4.184 Kilajoules
    1 Gallon = 3.785 Kg

    FORMULA
    (MASS) (HEAT CAPACITY) (ΔTEMPERATURE) = Energy Removed From Slag
    (907.168 Kg) (0.12 Kcal/Kg°)C (1104.44°)C
    (907.168 Kg) (0.12 Kcal/Kg°)C (1104.44°)C
    (120229.5151 Kcal)

    (120229.5151 Kcal) 4.184 Kj
    1 1 Kcal

    503,040.2912 Kj = Energy Needed to be Removed from Slag
    503,040.2912 Kj 503,040,291 Joules

    503,040,291 J 1
    1 2.260 J/g°C

    (222584199.6 Grams) in H20 222,584 Kg

    222,584 Kg 1 gallon
    1 3.785 Kg

    58806.86 Gallons of Water


    Does anyone think I am close to the right solution? Yet again this need not be an extremly precise calculation. I really just want a ballpark that has been found using good math
     
  2. jcsd
  3. Jul 31, 2013 #2

    SteamKing

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    Check your ΔT calculation for the slag again.
     
  4. Jul 31, 2013 #3
    Got it, I just had the wrong number written down for the conversion of 2200F to C. I believe my ΔT is ok because I just had a recording error there. Thanks!
     
  5. Jul 31, 2013 #4
    ΔT is fine, the error is T1, which should have been 1204 C.
     
  6. Jul 31, 2013 #5
    I think the calculation is fine. What I would normally complain about is that your data in converted units retains absurdly many "significant" digits, with none of them most likely correct; but since you need a ballpark estimate, then you should be fine.
     
  7. Jul 31, 2013 #6

    SteamKing

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    T1 = 2200 F should be 1204 C

    Once you get the energy to be removed (503000 kJ) you use 2.26 J/g-C for the heat of vaporization of water. The correct figure is 2270 kJ/kg-C, which puts your amount of water off by a factor of 1000.
     
  8. Jul 31, 2013 #7

    SteamKing

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    Basically, you need to evaporate one 55-gallon drum of water to cool each ton of slag.
     
  9. Jul 31, 2013 #8
    Indeed. 2260 J/g°C was correct. How did that become 2.260 J/g°C?
     
  10. Jul 31, 2013 #9
    Thanks everyone! It means a lot to have people helping out. I see where I went wrong. I got ahead of myself and tried to conver 2260 J/g°C (for reasons I dont even remember). I am honestly not a very strong in physics or basic math. I have not taken a course in 5 years and do not keep up enough... I am far removed from even basic mathmatical equations but I am seeing with my new job, I better start remembering. Thanks again everyone. It means a lot
     
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