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Heat absorption by apples

  1. Nov 30, 2011 #1
    I have a problem here im trying to solve for work. Does not need to be 100% accurate.
    Apple slices are pushed through a bath of hot water to heat the outside of them for 30 seconds.

    - Apple slice has an area of 3500mm² = 0.035m²
    - Apple temperature is 3 degree cel
    - Hot water temp = 53 deg cel. (kept constance from heat exchanger)
    - Apple slices are heated for 30 seconds
    - 82,300 slices per hour are processed.
    - Thermal Conductivity of Apple = 0.5 W/(m K)

    Im trying to find the Heat (energy transfered to the apples) - in other words, heat required by the heat exchanger to keep the bath at 53 degrees.

    Here is my workings.

    82,300 slices / (60*60) = 22.8 slices per second.
    30 seconds bath time * 22.8 slices = 686 slices in bath at any one time.

    686 * 0.035 = 24m2 of surface area in the bath.

    Q/t = K*A*(T1-T2)

    Q/t= 0.5*24*50 = 600 J/s = 600W.

    After looking on google, the formula I have above sometimes get divided with a distance. mainly when looking at heat losses through windows. distance = thickness. This makes sense as the units leaves me a with meter on the top. Q/t = W/(m K) * m2 * K,

    So should I now divide the 600W by the average depth of a slice/2 (say 8mm)

    giving me 75kW of heat being absorbed by the slices. Thats a lot of heat! (the Heat exchanger is capable of having upto 96kW provided to it, i think)

  2. jcsd
  3. Nov 30, 2011 #2
    In your formula: Q/t = K*A*(T1-T2)

    Note the T2 increases and Q varies with time.

    SO, you could assume in 30 seconds the apples reach 50 C. if you want to perform a transient heat transfer analyis on the applies let me know!

    BUT, we can get an upper limit of the energy required as follows:

    The energy to heat one apple slice:

    Q = mass_apple_slice X specific heat_apple X temperature rise of apple.

    THEN, with the rate you pass apples, you can figure the total input power to maintain 50C !

    Let me know if you need any help.
  4. Dec 1, 2011 #3
    I would follow edgepflow's idea. If you assume apples are more or less the same as water (SHC 4200J/kgC) and you know the mass of a slice you should get a reasonable upper limit for the power required.
  5. Dec 1, 2011 #4
    750 kg of apples if processed per hour
    750 (kg/hr) / 82,300 (slc/hr) = 9.11 grams per slice

    Q = 0.00911 * 4,200 (J/kgC) * 50 (C) = 1913 J

    22.8 slices per second;
    1913 *22.8 = 43.62 KJ/s = 43.62 KW

    This 43.62 kW would only be reached if the apples managed to reached 53 degree, correct? But is a maximum possible consumption.
  6. Dec 1, 2011 #5
    I agree with your calculation.....43kW to heat this quantity of apples to 53C in 30secs.
    I don't know but I imagine that in practice you would make sure that your heater could supply about 60kW to give some lee way.
  7. Dec 1, 2011 #6
    Is there a way for me to determine a more accurate final temperature of the apples?
    I don't think the whole slice heats up to 53 degrees.
  8. Dec 1, 2011 #7
    I came up with 38 kW - so I think your numbers are good. That is a lot of power, but that is a lot of apples.
  9. Dec 1, 2011 #8
    Yes, you could start with a simple lumped capacity model if the Biot number < 0.1. Otherwise, we have to move to some 1D models.
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