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Heat Transfer Refrigeration Problem

  1. Jul 23, 2012 #1
    Estimate how long a time it will take to cool a trailer filled with coke ?
    Given the ambient temperature is 22 degrees C and heat transfer provided by the refrigeration unit is 10 kW and the drinkable temperature is 4 degrees C. Make reasonable estimates of how much insulation is included on the walls of the truck. Assume an air flow velocity inside the truck of 0.1 m/s. Assume the sun is incident on the external surface of the truck for part of the time. Indicate any assumptions you make in your analysis. Take the properties of coke as: specific heat capacity 4200 J/kgK, density 1000 kg/m^3 and thermal conductivity of 0.6 W/mK.

    6.1m X 1.6m X 1.5m or 20ft X 6ft X 5ft
     
  2. jcsd
  3. Jul 24, 2012 #2
    The actual heat transfer given out by your refrigeration unit will be different than your input power (10KW) if that's the case?. Even with insulation your gonna get some loss. I think the actual time to cool it down to steady state temperature wont be long given your power and convection.
     
    Last edited: Jul 24, 2012
  4. Jul 24, 2012 #3
    mellowoout. Good day ( why the extra o )

    I guess your first guestimate is how full is the trailer of coke - a range of a minimum of one bottle or packed full to the brim - which you would estiamte by the type of the coke container ( bottles or cans ), size of container ( 950 ml, 1 L, 2 L ) , cans/bottles per case, and the number of cases per pallet, and the number of pallets per trailer ( single or double or triple stacked ). That is if you use pallets or just load the cases directly into the truck. I guess you will have to go the store and see what size case(s) coke comes in.
    So you have to decide how much coke is in the trailer is and that would be your first assumption.
     
  5. Jul 24, 2012 #4
    Ok so starting outside and working in, you have:
    (A) Heat of radiation, depending on the reflective/absorptive characteristics of the outer panels.
    (B) Heat of convection outside, based on the 22C and, say, 100kph wind speed (highway), assume its fairly turbulent, over a smooth flat plate... I'm not gonna look up the Reynolds numbers and Nusselt numbers but yeah.
    (C) Heat of conduction through the outer panel (steel?).
    (D) Heat of conduction through the insulation (this is the unknown in the equation)
    (E) Heat of convection between inside wall and inside air, based on 4C and that .1m/s air speed pulling heat from the wall into the room air.
    (F) Heat of convection between inside air and coke can, again using 4C and .1m/s.
    (G) Heat of conduction through coke can wall.
    (H) Heat of conduction through coke liquid (neglect convection here if the coke is sittin still.
    (I) Heat of convection between inside air and cooling coil wall. (given as 10kW of thermal energy)

    Items (A) through (H) can either be calculated the hard way, or looked up in an ASHRAE table somewhere. There's some equivalent thermal resistance. Invert that, multiply by the surface area of the truck walls and the temperature difference (18C). You'll get something like (U_insulation*Area_of_walls + U_otherstuff*Area_of_otherstuff )*dT = 10kW, where U_otherstuff is found by math wizardry, and U_insulation is unknown. We're assuming that 10kW is enough to get us down to 4C on the inside, and solving to figure out what value of U_insulation makes that true. Then simply invert U_insulation to figure out the R-value of the insulation. As for the amount of time, t = mass_of_coke * heat_capacity * delta_T / 10kW. As far as time goes, it doesn't really matter *how* that heat is transferred since you already know how *much* is transferred.... (mass of coke is total volume of coke (you pick!) times density.


    (side-note: upon further review, the above is anything but simple... it made far more sense to me when typing it than I'm sure it will to anyone reading it. sorry!)
     
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