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Refrigeration Load Calculations for Cold Storage & Food Processing

Refrigeration Load Calculations for Cold Storage & Food Processing

Accurate refrigeration load calculations determine equipment capacity, operating costs, and temperature control performance. This guide covers all heat gain components for cold storage, food processing, and commercial refrigeration applications, with calculation methodology per ASHRAE Refrigeration Handbook.

Total Refrigeration Load Components

Total heat gain:

$$Q_{total} = Q_{trans} + Q_{prod} + Q_{resp} + Q_{inf} + Q_{equip} + Q_{people} + Q_{other}$$

Where:

  • $Q_{trans}$ = transmission load through walls, floor, ceiling
  • $Q_{prod}$ = product load (cooling from initial to final temperature)
  • $Q_{resp}$ = respiration heat from living produce
  • $Q_{inf}$ = infiltration load (air exchange, door openings)
  • $Q_{equip}$ = equipment heat (fans, lights, forklifts)
  • $Q_{people}$ = occupant heat gain
  • $Q_{other}$ = miscellaneous loads (defrost, piping heat gain)

Transmission Load

Wall, Ceiling, Floor Heat Gain

Heat transfer through envelope:

$$Q_{trans} = U \cdot A \cdot \Delta T$$

Where:

  • $U$ = overall heat transfer coefficient (Btu/hr·ft²·°F)
  • $A$ = surface area (ft²)
  • $\Delta T$ = temperature difference, outdoor - indoor (°F)

U-value calculation:

$$U = \frac{1}{R_{total}} = \frac{1}{R_{out} + R_{wall} + R_{insulation} + R_{in}}$$

Typical U-values for insulated cold storage:

ApplicationU-value (Btu/hr·ft²·°F)R-value (ft²·°F·hr/Btu)
Cooler wall (35-50°F)0.04-0.06R-16 to R-25
Freezer wall (-10 to 0°F)0.025-0.04R-25 to R-40
Cooler ceiling0.03-0.05R-20 to R-33
Freezer ceiling0.02-0.03R-33 to R-50
Floor (insulated slab)0.05-0.08R-12 to R-20

Design temperature differences:

Space TypeIndoor TempOutdoor Design TempΔT
Produce cooler35°F95°F60°F
Meat cooler32°F95°F63°F
Freezer-10°F95°F105°F
Ice cream hardening-20°F95°F115°F

Solar Radiation Load

Roof solar gain (additional to transmission):

$$Q_{solar} = A_{roof} \cdot CLTD$$

Where CLTD = Cooling Load Temperature Difference (includes solar effect)

Typical CLTD for dark roof in summer: 30-50°F (add to ambient ΔT)

Mitigation:

  • White reflective roof coating (reduces solar gain 40-60%)
  • Insulation above roof deck
  • Ventilated air space

Product Load

Sensible Cooling Above Freezing

Heat removal to cool product from initial to final temperature:

$$Q_{prod,sensible} = \frac{m \cdot c_p \cdot (T_i - T_f)}{t}$$

Where:

  • $m$ = product mass (lb)
  • $c_p$ = specific heat above freezing (Btu/lb·°F)
  • $T_i$ = initial product temperature (°F)
  • $T_f$ = final product temperature (°F)
  • $t$ = cooling time (hr)

Typical specific heats (above freezing):

ProductSpecific Heat (Btu/lb·°F)Water Content (%)
Beef0.7775%
Pork0.6860%
Chicken0.8075%
Apples0.9084%
Lettuce0.9695%
Eggs0.7674%
Milk0.9387%

Approximation: $c_p \approx 0.20 + 0.008 \times %_{water}$

Freezing Load

Heat removal during freezing:

$$Q_{freezing} = m \cdot h_{latent}$$

Where $h_{latent}$ = latent heat of fusion (Btu/lb)

Latent heat approximation:

$$h_{latent} = 144 \times %_{water}/100$$

Example: Beef (75% water): $h_{latent} = 144 \times 0.75 = 108$ Btu/lb

Sensible Cooling Below Freezing

Heat removal after freezing:

$$Q_{prod,frozen} = \frac{m \cdot c_{p,frozen} \cdot (T_f - T_{storage})}{t}$$

Where $c_{p,frozen}$ = specific heat below freezing (typically 0.35-0.50 Btu/lb·°F)

Total freezing load:

$$Q_{prod,total} = Q_{above} + Q_{freezing} + Q_{below}$$

Worked Example 1: Product Cooling Load

Given:

  • Product: 10,000 lb beef (initially 70°F)
  • Cooling: 70°F → 28°F (above freezing), then freeze to 0°F
  • Cooling time: 24 hours
  • Specific heat above freezing: 0.77 Btu/lb·°F
  • Latent heat: 108 Btu/lb (75% water content)
  • Specific heat below freezing: 0.42 Btu/lb·°F
  • Freezing point: 28°F

Find: Total product load (Btu/hr)

Solution:

Sensible cooling (70°F → 28°F): $$Q_1 = \frac{10,000 \times 0.77 \times (70 - 28)}{24} = 13,479 \text{ Btu/hr}$$

Latent heat (freezing at 28°F): $$Q_2 = \frac{10,000 \times 108}{24} = 45,000 \text{ Btu/hr}$$

Sensible cooling below freezing (28°F → 0°F): $$Q_3 = \frac{10,000 \times 0.42 \times (28 - 0)}{24} = 4,900 \text{ Btu/hr}$$

Total average product load: $$Q_{prod} = 13,479 + 45,000 + 4,900 = 63,379 \text{ Btu/hr} = 5.3 \text{ tons}$$

Peak load occurs during freezing phase: 45,000 Btu/hr

Answer: Average load: 63,379 Btu/hr (5.3 tons) over 24 hours; Peak: 45,000 Btu/hr during freezing

Respiration Load

Fresh fruits and vegetables generate heat through respiration:

$$Q_{resp} = m \cdot q_{resp}$$

Where:

  • $m$ = product mass (lb or ton)
  • $q_{resp}$ = specific respiration heat (Btu/day·ton or Btu/day·lb)

Respiration heat (at storage temperature):

ProductStorage TempHeat Generation (Btu/day·ton)
Apples32°F600-1,200
Lettuce32°F5,000-8,000
Broccoli32°F10,000-16,000
Tomatoes (ripe)50°F3,000-4,000
Potatoes40°F800-1,200
Carrots32°F2,000-3,000

Important: Respiration increases exponentially with temperature (Q₁₀ ≈ 2-3)

Rate at higher temperature:

$$q_{resp,T2} = q_{resp,T1} \times Q_{10}^{(T_2-T_1)/10}$$

Where $Q_{10}$ = temperature coefficient (typically 2.0-2.5)

Infiltration Load

Air Exchange Load

Heat gain from outdoor air entering space:

$$Q_{inf} = V \cdot \rho \cdot c_p \cdot (T_{out} - T_{in}) \cdot N_{ACH}$$

Or using simplified form:

$$Q_{inf,sensible} = 1.08 \cdot CFM \cdot \Delta T$$

$$Q_{inf,latent} = 4840 \cdot CFM \cdot \Delta W$$

Where:

  • $CFM$ = infiltration airflow rate
  • $\Delta T$ = temperature difference (°F)
  • $\Delta W$ = humidity ratio difference (lb/lb)

Infiltration air changes per day:

SpaceAir Changes per Day
Well-sealed cooler, low traffic1-2
Average cooler3-5
High-traffic cooler8-15
Well-sealed freezer0.5-1.5
Average freezer2-4

Infiltration through door openings (Gosney-Olama equation):

$$CFM_{door} = K \cdot A_{door} \cdot \rho_{avg} \cdot \sqrt{H \cdot (1 - \frac{\rho_{in}}{\rho_{out}})} \cdot F_{usage}$$

Where:

  • $K$ = flow coefficient (0.6-0.8)
  • $A_{door}$ = door opening area (ft²)
  • $H$ = door height (ft)
  • $\rho_{in}, \rho_{out}$ = air density inside/outside (lb/ft³)
  • $F_{usage}$ = usage factor (door open fraction of time)

Mitigation:

  • Air curtains (reduce infiltration 60-80%)
  • Strip curtains (reduce 50-70%)
  • Vestibules/airlocks
  • High-speed doors (minimize open time)

Equipment and People Loads

Internal Equipment

Typical heat gains:

EquipmentHeat Gain
Evaporator fan motors2,545 Btu/hr per hp
Lighting (LED)3.4 Btu/hr per watt
Lighting (fluorescent)4.1 Btu/hr per watt
Electric forklift (during use)8,000-12,000 Btu/hr
Electric forklift (charging)All input power (100% heat)
Conveyor motors2,545 Btu/hr per hp

Forklift load calculation:

$$Q_{forklift} = N \cdot P \cdot U \cdot 2,545$$

Where:

  • $N$ = number of forklifts
  • $P$ = motor power (hp)
  • $U$ = usage factor (0.2-0.4 typical)

People Load

Heat gain per person:

Space TemperatureSensible (Btu/hr)Latent (Btu/hr)Total (Btu/hr)
50°F640140780
40°F730120850
0°F (freezer)920801,000
-20°F1,050501,100

Occupancy diversity: Average number, not peak (use time-weighted average)

Safety Factors and Design Margins

Refrigeration load safety factors:

  1. Equipment capacity: Add 10-20% to calculated load

    • Uncertainty in usage patterns: 10%
    • Future expansion: 10-20%
    • Equipment degradation over time: 5-10%
  2. Pull-down load: Additional capacity for initial cool-down

    • Empty room: 2-3 × steady-state transmission load
    • With product: Calculate pull-down time requirement
  3. Peak vs. average load:

    • Design for peak load, not annual average
    • Consider seasonal variations (higher summer transmission)

Avoid over-sizing:

  • Excessive capacity → short cycling → poor humidity control
  • Target: equipment runs 16-20 hr/day at design conditions

Complete Load Calculation Example

Worked Example 2: Walk-In Cooler Load Calculation

Given:

  • Cooler size: 20 ft × 30 ft × 12 ft high
  • Wall/ceiling U-value: 0.05 Btu/hr·ft²·°F
  • Floor (uninsulated): 0.15 Btu/hr·ft²·°F
  • Indoor temperature: 35°F
  • Outdoor temperature: 95°F (summer design day)
  • Product: 5,000 lb apples per day (70°F → 35°F)
  • Apple specific heat: 0.90 Btu/lb·°F
  • Apple respiration: 1,000 Btu/day·ton (at 35°F)
  • Infiltration: 4 air changes per day
  • Evaporator fans: 2 × 1/2 hp
  • Lighting: 400 watts, 8 hr/day
  • People: 2 workers, 4 hr/day average

Find: Total refrigeration load

Solution:

1. Transmission load:

Wall area: $2 \times (20 + 30) \times 12 = 1,200$ ft² Ceiling area: $20 \times 30 = 600$ ft² Floor area: $20 \times 30 = 600$ ft²

Walls + ceiling: $$Q_{walls} = 0.05 \times (1,200 + 600) \times (95 - 35) = 5,400 \text{ Btu/hr}$$

Floor: $$Q_{floor} = 0.15 \times 600 \times (70 - 35) = 3,150 \text{ Btu/hr}$$

Total transmission: 8,550 Btu/hr

2. Product load:

Daily product: 5,000 lb/day

$$Q_{prod} = \frac{5,000 \times 0.90 \times (70 - 35)}{24} = 6,563 \text{ Btu/hr}$$

3. Respiration load:

Product mass: 5,000 lb = 2.5 tons

$$Q_{resp} = \frac{2.5 \times 1,000}{24} = 104 \text{ Btu/hr}$$

4. Infiltration load:

Room volume: $20 \times 30 \times 12 = 7,200$ ft³

Infiltration rate: $\frac{7,200 \times 4}{24 \times 60} = 20$ CFM average

Sensible: $Q_{inf,s} = 1.08 \times 20 \times (95 - 35) = 1,296$ Btu/hr

Latent (summer 75°F, 60% RH, W=0.0114 vs 35°F sat, W=0.0037): $Q_{inf,l} = 4840 \times 20 \times (0.0114 - 0.0037) = 746$ Btu/hr

Total infiltration: 2,042 Btu/hr

5. Equipment load:

Evaporator fans: $2 \times 0.5 \times 2,545 = 2,545$ Btu/hr (continuous)

Lighting: $400 \times 3.4 \times \frac{8}{24} = 453$ Btu/hr (average)

Equipment total: 2,998 Btu/hr

6. People load:

People: $2 \times 850 \times \frac{4}{24} = 283$ Btu/hr (average)

7. Total load:

$$Q_{total} = 8,550 + 6,563 + 104 + 2,042 + 2,998 + 283 = 20,540 \text{ Btu/hr}$$

8. Safety factor (15%):

$$Q_{design} = 20,540 \times 1.15 = 23,621 \text{ Btu/hr} = 1.97 \text{ tons}$$

Select equipment: 2.5 ton condensing unit (next standard size)

Answer: Design load: 23,621 Btu/hr (1.97 tons); Select 2.5 ton unit

Load breakdown:

  • Transmission: 42%
  • Product: 32%
  • Equipment: 15%
  • Infiltration: 10%
  • People: 1%

Related Technical Guides:

References:

  • ASHRAE Refrigeration Handbook, Chapter 24: Refrigeration Load Calculations
  • ASHRAE Refrigeration Handbook, Chapter 19: Thermal Properties of Foods
  • ASHRAE Fundamentals Handbook, Chapter 14: Climatic Design Information
  • Stoecker, W.F., “Industrial Refrigeration Handbook”
  • USDA: The Commercial Storage of Fruits, Vegetables, and Florist and Nursery Stocks