Refrigeration Load in Freezers:

Example 1 shows the process for manipulating the refrigeration load in freezers. Additionally to the product load, other loads which have to be considered are the fan heat, heat leakage by insulation, heat load because of internal lighting, containers and conveyers, heat load because of defrosting, infiltration load, etc.

Example:

32,400 kg/day of chilly fish is to be frozen to -30°C in 10 cm thick blocks each of weighing 45 kg. The secondary refrigerant temperature found - 40°C. The evaporating refrigerant temperature found - 47°C. The fish enters at temperature 30°C. The freezing cycle time can be taken as 4 hours. It is given:

Specific heat of thawed fish = 3.77 kJ/kg°C

Latent heat of fusion of fish = 251.2 kJ/kg

Specific heat of frozen fish = 1.67 kJ/kg°C

Compute the number of blocks frozen per cycle & refrigeration duty of the plant for 18 hours running time.

Solution

Number of blocks/day = 32400/45 = 720 blocks/day

Number of freezing cycles/day

= 24 /4= 6

Number of blocks frozen per cycle

= 720 /6= 120 blocks/cycle

Fish loading

 m =32400 /(24)(3600) = 0.375kg/s

 Sensible cooling to a freezing temperature of 0°C

Latent heat of fusion

Sensible cooling to 0-30°C

Total product load

Product load for 18 hours running time