Exercise 1

Try to calculate the mean refrigeration system capacity by considering:

  • Case 1: Freezing time of 6 hours and all the other parameters unchanged;
  • Case 2: Mass of meat equal to 240 kg and all the other parameters unchanged.

Solutions:

  • Case 1: A mean refrigeration system capacity of 1004.5 W is obtained halving the freezing time.
  • Case 2: A mean refrigeration system capacity of 1004.5 W is obtained doubling the mass of meat to be refrigerated.
    Halving the time or doubling the mass have the same influence in determining the refrigeration system capacity.

Exercise 2

Modify the exercise presented in the lesson by changing the following parameters:

  • Case 1: Electric power required for cooling: 140 W, Time required for cooling: 12 h, Electric power required for temperature maintenance: 80 W, Time required for temperature maintenance: 12 h, Location: Kinshasa;
  • Case 2: Electric power required for cooling: 280 W, Time required for cooling: 12 h, Electric power required for temperature maintenance: 160 W, Time required for temperature maintenance: 12 h, Location: Milano;
  • Questions: Check both (1) how many PV modules have to be installed and (2) how many kWh you need from the batteries, compare the obtained results with the one presented.

Solutions:

  • Case 1: (1) 2 PV modules have to be installed and (2) a battery of minimum 1.6 kWh is required. Comparison: reducing the energy required leads to a reduction in both PV modules and batteries.
  • Case 2: (1) 12 PV modules have to be installed and (2) a battery of minimum 3.7 kWh is required. Comparison: for the case considered changing location implies an increase in PV modules (three times more).