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Homework Help: Thermodynamics: Designing a hydrocooling unit

  1. Dec 29, 2017 #1
    1. The problem statement, all variables and given/known data
    A hydrocooling unit can cool fruits and vegetables from 30 to 5 C at a rate of 20,000 kg/h under the following conditions: the channel is 3 m wide by 90 cm high. The water is circulated and cooled by the evaporator section of a refrigeration system. The refrigerant temperature inside the evaporator coils is to be -2 C, and the water temperature is not to drop below 1 C and not to exceed 6 C. Assuming reasonable values for the average product density, specific heats, and space taken up by the fruits, recommend reasonable values for (a) the water velocity through the channel and (b) the refrigeration capacity (how much heat is removed) of the refrigeration system
    the figure shows the unit
    2. Relevant equations
    m1*h1(at inlet)+m2*h2(at inlet)=m1*h1(at output)+ m2*h2(at output)
    let's consider that
    m1 is mass rate of water
    m2 is mass rate of the fruits and vegetables
    Cp of water= 4.2
    from the steam tables:
    for water:
    h4=25.32 kJ/kg
    h3=4.28 kJ/kg

    3. this is the attempt at a solution
    I want to know if I approached the problem correctly or not?? and if the answer is reasonable??

    m1*h1(at inlet)+m2*h2(at inlet)=m1*h1(at output)+ m2*h2(at output)
    I assumed that Cp of fruits is 3.7 ( as an average of Cp of fruits and vegetables)
    then,
    m(water)=m(fruits) *(h1-h2) / (h4-h3)
    = m(fruits) *Cp * (T1-T2) / (h4-h3)

    I got that m(water)= 0.9774 kg/s
    Q( that transferred from fruits to water)= m(water) *(h4-h3) = 20.55 kJ/s
    velocity of water = m(water)/ [(density of water) * width of unit* its height ] = 3.62*10^-4 m/s
     

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  2. jcsd
  3. Dec 29, 2017 #2

    BvU

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    Homework Helper
    2017 Award

    Hi,

    Two comments:

    Method looks OK, but the math can't be right. Cooling 20000 kg/h fruit by 25 degrees requires a lot more water if that can only increase 5 degrees in temperature.

    Water speed is not through an otherwise empty channel: the area that the fruit occupies is not available for the water
     
  4. Dec 29, 2017 #3
    Ok,
    we can ignore the effect of the area that the fruits occupy,, Also, as you said we need a lot more water that's why I'm not believing that this is the answer..
    Thanks,
     
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