Recent content by candii

  1. C

    New here Need help with Thermodynamics question

    Ohhhhh haha thanku! I get so lost our teacher is good but he has a really hard way of teaching that everyone in our class has trouble understanding so we have to teach ourselves lol thank you so much :)
  2. C

    New here Need help with Thermodynamics question

    Ohhhhh I get it! 1085600kj/s / 10.4kj/s = 104384.62s / 3600s = 29hrs?
  3. C

    New here Need help with Thermodynamics question

    Ok so after I subtract the loss of 9600w from the 20000w I get 10400w and the steel needs to reach 1200'C to reach forging temperature
  4. C

    New here Need help with Thermodynamics question

    Is the value the constant energy input of 20kw? So in joules per second is it 20000? Lol I'm pretty stuck here, :/
  5. C

    New here Need help with Thermodynamics question

    lmfao >__< I'm doing this at work soo I'm not fully focused but yes furnaces do have 6 sides lol whoops =P your probably right in your values, lol these stupid books are full of mistakes! ok so let's do that again: a = 2(3x2) + 2(2x2) + 2(3x2) a = 32m² 300 x 32 = 9600W so...
  6. C

    New here Need help with Thermodynamics question

    In my book it states in the Appendix the following: Cast Iron = 0.42kJ/kgK Steel = 0.46kJ/kgK I just followed what was in there =) ok, let me see so the environment is the furnace itself and the losses stated is 300W/m² so I should work out the area of the furnace and times that buy...
  7. C

    New here Need help with Thermodynamics question

    Hi thank you for your respone! HaHa cool, I feel like I'm in school again =P am I on the right track?: Q = mc(T2 - T1) Q = 2000 x 0.46(1200 - 20) Q = 1085.6 MJ
  8. C

    New here Need help with Thermodynamics question

    Homework Statement An electric furnace has dimensions 2m x 2m x 3m. When it reaches forging temperature of 1200°C, a 2t block of steel at 20°C is placed in the furnace. If there is a heat loss of 300W/m² from all surfaces, and a constant energy input of 20kW, calculate how long it will...
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