Tools › Building Systems › HVAC Loads › Cold Room Refrigeration Load › Example

Pear cold storage: refrigeration load during cooling and filling by the daily heat balance (USDA Agriculture Handbook 66)

Tool	Cold Room Refrigeration Load
Source	Gross, K. C., Wang, C. Y., and Saltveit, M. (eds.) (2016). The Commercial Storage of Fruits, Vegetables, and Florist and Nursery Stocks. Agriculture Handbook 66, revised, U.S. Department of Agriculture, Agricultural Research Service, Washington, DC. Heat Load Calculation, pp. 19-21, example A.
Level	Professional

Problem

A 15 × 15 × 4.5 m pear storage held at −1.1 °C is filled at 200 bins (100,000 kg of fruit) per day and holds 300,000 kg. Each day's fruit is cooled from 21 to 4.5 °C on its first day and from 4.5 to −1.1 °C on its second. The outside air is at 30 °C and 50 % relative humidity. Find the refrigeration needed on the peak day, when one lot is on its first day of cooling, the lot before it on its second, and the first 100 tonnes are already in storage, with a 10 % allowance and the plant running 18 hours a day.

Inputs to Enter

Room −1.1 °C, 90 % RH; outside 30 °C, 50 % RH; outside surface 720 m² (walls 270, ceiling 225, floor 225) with U = 1.1 kJ/h·m²·°C (0.3055556 W/m²·K) and 31.1 K on every surface
Room volume 908 m³, 6 air changes per day, heat removed from the air 74.5 kJ/m³
Lots: pears 100,000 kg, 21 to 4.5 °C; bins 12,700 kg, 21 to 4.5 °C; pears 100,000 kg, 4.5 to −1.1 °C; bins 12,700 kg, 4.5 to −1.1 °C; pears 100,000 kg stored at −1.1 °C; specific heat 0.86 kcal/kg·°C for pears (3.59996 kJ/kg·K) and 0.5 for wood bins (2.093)
Respiration 12,206, 1,741 and 812 kJ per tonne per day at 13, 1.7 and −1.1 °C (141.27, 20.15 and 9.398 W per tonne)
Lights 2,400 W for 8 h; three fan motors at 3,112 kJ per hp-hour (864.4 W each) for 24 h; two fork trucks at 36,920 kJ per 8 h (1,281.9 W each); two workers at 1,000 kJ/h (277.8 W) for 8 h
Safety factor 10 %, run time 18 h per day; evaporator TD 5 K, COP 2.5, 0.12 $/kWh (the last three are not in the handbook)

Open the Cold Room Refrigeration Load tool (loaded with this example)

Answers

Transmission 720 × 1.1 × 31.1 × 24 = 591,149 kJ per 24 h (book 591,149; the book's summary misprints it as 519,149)
Air change 908 × 74.5 × 6 = 405,876 kJ (book 405,876); the psychrometric heat of the air at these states is 74.34 kJ/m³
Product cooling 5,939,934 + 438,588 + 2,015,978 + 148,854 = 8,543,354 kJ (book 5,939,934, 438,588, 2,015,977, 148,854; 8,543,353)
Respiration 1,220,600 + 174,100 + 81,200 = 1,475,900 kJ (book the same)
Lights 69,120, fans 224,064, fork trucks 73,840, workers 16,000: 383,024 kJ (book the same)
Subtotal 11,399,303 kJ, 10 % allowance 1,139,930 kJ, total 12,539,233 kJ per 24 h (book 11,399,302; 1,139,930; 12,539,232)
Capacity 12,539,233 / 18 h = 696,624 kJ/h = 193.51 kW = 660,272 Btu/h = 55.02 tons of refrigeration (book 696,624 kJ/h, 55 tons)
Product cooling is 74.9 % of the load, respiration 12.9 %, transmission 5.2 %, infiltration 3.6 %, internal 3.4 %
Tool only: evaporating temperature −6.1 °C, compressor 77.40 kW at COP 2.5, 1,266.6 kWh per day, $151.99 per day; peak demand while both lots cool 131.9 kW, so the capacity is 1.47 times it

Notes

The tool's default inputs are this case; press Submit. The one-unit differences from the book are the book's truncation of each line. The handbook's storage case (7.2 °C and 70 % outside, 1.8 air changes at 15.3 kJ/m³, 300 tonnes respiring, lights and one worker 4 h) gives 757,895 kJ per day and 3.3 tons in the test script. The handbook writes the ton of refrigeration as 12,660 kJ per 24 h; it is 12,660 kJ per hour, which its division uses.