HVACR technicians, refrigeration engineers and cold storage managers in the commissioning, often due to expansion valve selection, cold storage tonnage calculation and other formulas to find the difficulties and fumbling, a missing parameter may affect the efficiency of the system, and even cause equipment failure.
Refrigerant charge error, chiller load calculation error, will also lead to compressor overload, cold storage temperature does not meet the standard and other issues.
This article summarizes the core refrigeration and air conditioning formula, analyze the applicable scenarios and operational points, to help you avoid calculation errors and improve the efficiency of the system cycle.
Core Basic Formula of Refrigeration System
The stable operation of the refrigeration system, inseparable from the expansion valve, compressor, refrigerant, the three core components of the accurate matching. The following formula is the “cornerstone” of building the system, which can help you avoid the problem of over or under sizing of components.
Expansion Valve Sizing Formula (with margin)
Expansion valve is the “regulating valve” of refrigerant circulation, and a certain margin must be reserved to cope with the peak load when sizing.
Selection formula: cold tons × (1 + 1.25% margin)
- Parameter interpretation:
- Cold tons: refers to the actual cooling capacity required by the system, as the basic calculation value;
- 25% margin: Used to cope with sudden load fluctuations caused by extreme high temperature and frequent door opening and closing in summer to ensure the stability of system operation.
Engineering case: Take 5 refrigerating tons cold storage system as an example, the expansion valve selection should be calculated by substituting into the formula: 5 × (1 + 0.0125) = 5.0625 refrigerating tons, which can effectively avoid the shortage of refrigerant supply in the peak refrigeration period by reserving the load margin.
Compressor Power Conversion
Compressor as the core of the system “power source”, accurate mastery of the power conversion formula is the key to select the right model.
Power conversion formula: 1P (electric power) = 0.735KW
Pit avoidance guide: Special attention should be paid to the “P” here on behalf of the electric power, and the concept of cooling capacity is different (1P cooling capacity of about 2.56KW), which is a common cognitive misunderstanding of the novice.
Take 3P compressor as an example, its rated power is 3×0.735=2.205KW, when designing the circuit configuration, make sure to choose the appropriate wire specifications based on this power parameter.
Refrigerant Charge Calculation
The refrigerant charge is directly related to the operating efficiency of the air-conditioning system and the equipment life: overcharging or undercharging will not only significantly reduce the refrigeration effect, but may also lead to compressor damage. Specifically, the following formula can be used to accurately calculate the refrigerant charge:
Calculation formula: Refrigeration capacity (KW) ÷ 3.516 × 0.58
Application example: Take the air-conditioning system with 10KW cooling capacity as an example, the theoretical charge is 10÷3.516×0.58≈1.65KG. It should be noted that the ambient temperature will have a slight influence on the calculation result.
so in the actual charging process, it is recommended to adopt the dynamic operation method of “monitoring pressure while charging” to avoid relying on the theoretical calculation alone and ensure that the system is in the best operating condition. Ensure that the system reaches the optimal operating condition.
Air Cooling and Heat Load Calculation Formulas
Whether it’s a shopping mall air conditioner or a cold storage unit, you need to first calculate the “heat load”, which is the amount of heat that needs to be removed from the system. This part of the formula helps you to accurately determine the cooling requirements and avoid “big horse, small horse” or “small horse, big horse”.
Air Cooling Total Heat (QT) Calculation
Total heat contains sensible heat (temperature change) and latent heat (humidity change), which is the core parameter of air conditioning and cold storage ventilation design.
Formula: QT = 0.24 × ∝ × L × (h₁ – h₂)
- Variable interpretation: ∝ is the air density (usually take 1.2kg/m³), L is the air volume flow (m³/h), h₁, h₂ are the enthalpy of the inlet and outlet air (kJ/kg, can check the enthalpy and humidity diagram).
- Actual case: a supermarket freezer area, air flow 2000m³/h, inlet enthalpy 45kJ/kg, outlet enthalpy 30kJ/kg, the total heat QT = 0.24 × 1.2 × 2000 × (45-30) = 8640W = 8.64KW, the selection of the chiller needs to meet the cooling capacity of at least 8.64KW.
Calculation of air cooling sensible heat (QS)
The sensible heat is directly related to the temperature change, which is crucial in the temperature control of dry environment, and is often used in the server room, archive room and other places with heavy temperature precision adjustment and light humidity control. In the design of air-conditioning systems, a reasonable calculation of the apparent heat can ensure that the cooling capacity of the equipment to match the environmental needs.
Calculation formula: QS = Cp×∝×L×(T₁ – T₂)
- Variable description:
- C_p: constant pressure specific heat capacity of air, often taken as the value of 1.01 \text{kJ/kg}\cdot°C in engineering calculations;
- \rho: air density, about 1.2 \text{kg/m}^3 in standard condition;
- L: air flow rate (unit: \text{m}^3/\text{h});
- T_1, T_2: inlet and outlet air temperature (unit: ℃), respectively.
Application example: Take a server room as an example, it is known that the server room needs to reduce the ambient temperature from 30℃ to 22℃, and the air flow is 1500 \text{m}^3/\text{h}. Calculated by the formula, the required sensible heat QS = 1.01 \times 1.2 \times 1500 \times (30 – 22) = 14544 \text{W} = 14.54 \text{kW}.
Therefore, when selecting air-conditioning equipment for this server room, the focus should be on its sensible heat and cold index to ensure that the sensible heat and cold of the selected equipment is not lower than the value obtained from the calculation, so as to ensure that the equipment in the server room can operate stably in a suitable temperature environment.
Calculation of air cooling latent heat (QL)
Latent heat is closely related to humidity. In typical high humidity environment scenarios such as fruit and vegetable cold storage and food processing rooms, accurate calculation of latent heat is the key to maintaining environmental stability.
Calculation formula: QL = 600×∝×L×(W₁ – W₂)
- Variable interpretation:
- W_1, W_2: moisture content of the incoming and outgoing air respectively (unit: \text{kg/kg}dry air)
- 600: approximate latent heat of vaporization of water (unit: \text{kJ/kg})
- \alpha: air density (usually taken as 1.2 \text{kg/m}^3)
- L: air flow (unit: \text{m}^3/\text{h})
Practical application: Take fruit and vegetable cold storage as an example, if the moisture content of the incoming air is 0.015 \text{kg/kg}dry air, the moisture content of the outgoing air is 0.008 \text{kg/kg}dry air, and the air flow rate is 1,000 \text{m}^3/\text{h}, then the latent heat can be calculated by the formula:
Therefore, this kind of scenario needs to be Therefore, this type of scenario needs to be equipped with a chiller with dehumidification function in order to effectively control the ambient humidity and ensure the quality and safety of the stored items.
Temperature Difference Flow Method
When the system uses chilled water for cooling, the following formula is the most intuitive and convenient for calculating the heat load, and it can also be used to reverse the flow rate of the water required.
Calculation formula: Q = Cp·r·Vs·∆T
- Description of parameters:
- Q: heat load (KW);
- Cp: specific heat at constant pressure (KJ/kg-°C), e.g.: 4.1868KJ/Kg-°C;
- r: specific weight (Kg/m³), such as: 1000Kg/m³;
- Vs: water flow rate (m³/h), e.g.: 1.5m³/h;
- ∆T: water temperature difference (℃), ∆T = T₂ (outlet water temperature) – T₁ (inlet water temperature) = 10℃.
Application example: Take a water-cooled air-conditioning system as an example, its water flow rate is known to be 1.5 \text{ m}^3/\text{h}, and the temperature difference between inlet and outlet water is set to be 10^{\circ}\text{C}.
The heat load Q = 4.1868 \times 1000 \times 1.5 \times 10 \div 3600 \approx 17.45 \text{ KW} can be obtained from the formula. This indicates that the system needs to remove 17.45 kW of heat per hour to maintain the set operating conditions.
Timed Temperature Rise Method to Calculate Heat Load
If you can’t measure the water flow rate, such as for tank cooling in a small cooler, it’s easier to use the Timed Temperature Rise Method.
Calculation formula: Q = Cp·r·V·∆T/H
- Parameter interpretation:
- Q: heat load (KW);
- Cp: specific heat at constant pressure (KJ/Kg-°C), e.g.: 4.1868KJ/Kg-°C;
- r: specific weight (kg/m³), e.g.: 1000kg/m³;
- V: total water volume (m³), e.g.: 0.5m³;
- ∆T: water temperature difference (℃), ∆T = T₂ – T₁ = 5℃;
- H: time (h), e.g.: 1h.
The meaning of the rest of the parameters is the same as that of the temperature difference flow method.
Application example: Take a 0.5m^3 water tank as an example, when the water temperature rises from 15℃ to 20℃ (\Delta T = 5℃) in 1 hour, the heat load Q = 4.1868 \times 1000 \times 0.5 \times 5 \div 1 \approx 10467W = 10.47kW.
Therefore, it is not possible to calculate the heat load Q by the formula. Therefore, when selecting a refrigeration unit, its cooling capacity should at least reach this heat load value to ensure that the actual demand is met.
Exclusive Calculation Formula for Cold Storage
The core demand of cold storage is “to store and freeze”, the three major formulas of tonnage calculation, incoming volume control and chiller matching directly determine the practicality and economy of cold storage.
Cold Storage Tonnage Calculation
Calculate the tonnage in order to avoid waste of space or insufficient stock, the key is to consider the “volume utilization coefficient” – not all space can be put goods.
- Core formula: Cold storage tonnage = Internal volume of cold storage room × Volume utilization coefficient × Food unit weight
- Base volume: Cold room content area = Length × Width × Height (must use the inside dimensions of the warehouse, excluding the thickness of the walls)
- Volume utilization factor (by warehouse capacity):
- 500-1000 cubic meters: 0.4
- 1001-2000 cubic meters: 0.5
- 2001-10000 cubic meters: 0.55
- 10001-15000 m3 : 0.6
A cold storage warehouse has a length of 10m, a width of 8m and a height of 5m, with an internal volume of 10×8×5=400 m3 (<500 m3, coefficient 0.4), storing pork (unit weight of 800kg/m³), with a tonnage of 400×0.4×800=128,000kg=128 tons.
Calculation of The Maximum Capacity of Cold Storage
The storage capacity cannot be full, it is necessary to leave space for cold air circulation, otherwise it will be partially warmed up.
- Standard cold storage (fixed):
Effective volume = total internal volume × 0.9 (10% of passage and ventilation space is reserved)
Maximum storage capacity (tons) = total internal volume ÷ 2.5 (1 ton of goods per 2.5 cubic meters, general estimation value)
- Active cold storage (mobile):
Effective volume = total internal volume × 0.9
Maximum storage capacity (tons) = total internal volume × (0.4-0.6)÷2.5 (0.4-0.6 is adjusted according to the way of stacking goods, for example, 0.6 for carton goods and 0.4 for bulky goods)
Reminder: a 100 cubic meters of active cold storage, storage of cartons of fruits, the amount of storage = 100 × 0.6 ÷ 2.5 = 24 tons, do not be greedy to pile up to 30 tons, otherwise the middle of the goods will be smothered.
Calculate Matching of Chiller for Cold Storage
The chiller will be slow to cool down if it is small, and it will cost electricity if it is big, and it is necessary to calculate the load according to the type of cold storage (refrigerating/freezing/processing room).
- Refrigerated cold storage:
base load W₀=75W/m³, coefficient A (by volume): V＜30m³ take 1.2, 30≤V＜100m³ take 1.1, V≥100m³ take 1.0; single cold storage plus coefficient B=1.1.
final load W=A×B×W₀, evaporation temperature match by -10℃. The evaporating temperature is matched at -10℃.
Example: 25 cubic meters single cold storage, W=1.2×1.1×75×25=2475W=2.48KW, choose 2.5KW chiller.
- Freezer:
base load W₀=70W/m³, coefficients A and B are the same as freezer; evaporation temperature: -30℃ for individual freezer, -35℃ for common use with low-temperature cabinet.
Example: 80 cubic meters single freezer, W=1.1×1.1×70×80=6776W≈6.8KW, choose 7KW chiller, evaporation temperature – 30℃.
- Processing room of cold storage:
Base load W₀=110W/m³, coefficient A: 1.1 for V＜50m³, 1.0 for V≥50m³.
Final load W=A×W₀, evaporation temperature: 0℃ for separate processing room, -10℃ for shared use with medium temperature cabinet.
Example: 40 cubic meters processing room, W=1.1×110×40=4840W≈4.8KW, choose 5KW chiller.
Fluid (water / air) Flow Calculation Formula
Water and air are the “heat mover” of the refrigeration system, the flow rate calculation error will lead to pipe blockage, fan noise, and even system paralysis.
Cooling Water / Chilled Water Flow Calculation
Water flow algorithms for different types of units (air-cooled / water-cooled screw) are different, don’t mix the formulas.
- Water flow rate of air-cooled machine: cooling capacity (KW) ÷ temperature difference ÷ 1.163 (the temperature difference is usually taken as 5℃)
Example: 20KW air-cooled machine, the flow rate = 20 ÷ 5 ÷ 1.163 ≈ 3.44m³/h.
- Water-cooled screw machine chilled water flow (frozen water, to the end of the cooling): refrigeration capacity (KW) × 0.86 ÷ temperature difference (temperature difference of 5-7 ℃)
Example: 100KW water-cooled screw machine, temperature difference of 6 ℃, the flow rate = 100 × 0.86 ÷ 6 ≈ 14.33m³ / h
- Water-cooled screw machine cooling water flow(to take away the heat of the compressor): (refrigeration capacity KW + press power) × 0.86 ÷ temperature difference (temperature difference of 8-10 ℃)
Example: 100KW refrigeration capacity, the press power of 25KW, the temperature difference of 10 ℃, the flow rate = (100 + 25) × 0.86 ÷ 10 = 10.75m³ / h
Reminder: When selecting the pump, the flow rate should be 5%-10% larger than the calculated value to prevent piping resistance resulting in insufficient flow.
Calculation of Air Supply Volume
The air supply volume determines the speed and uniformity of space cooling, especially suitable for large places (such as workshops, shopping malls).
- Formula: L = Qs/(Cp×∝×(T₁ – T₂))
- Variable description: L is the air volume (m³/h), Qs is the sensible heat load (KW, need to be calculated first), other parameters as before.
Example: a workshop apparent heat load of 50KW, the need to reduce the temperature from 35 ℃ to 25 ℃ (T ₁ – T ₂ = 10 ℃), the amount of air supply L = 50,000 / (1.01 × 1.2 × 10) ≈ 4,132m ³ / h, choose 4200m ³ / h fan.
Calculating Water Pipe Diameter
Pipe diameter too small will have “water hammer” noise, too big will be costly, the most accurate calculation is based on flow rate and flow velocity.
- Formula: D = √(4×1000L₂/(π×v))
- Variable interpretation: D is the diameter of the pipe (mm), L₂ is the water flow (m ³ / h), v is the water flow rate (m / s, suggested: cold water v = 1.0-1.5m / s, cooling water v = 1.2-2.0m / s).
Example: 10m³/h cold water flow, v=1.2m/s, D=√(4×1000×10/(3.14×1.2))≈103mm, choose DN100 water pipe.
Calculation of Duct Area
The duct area should match the air supply volume to avoid noise generated by too fast wind speed (general wind speed ≤ 8m/s).
Formula: F = a × b × L₁/(1000u)
Explanation of variables: F is the cross-sectional area of the duct (m²), a, b is the side length of the duct (m, such as rectangular duct), L ₁ is the volume of air supply (m³/h), u is the wind speed in the duct (m / s).
Example: 4000m³/h air volume, choose 1m×0.5m rectangular duct, u=6m/s, F=1×0.5×4000/(1000×6)≈0.33m² (1m×0.5m=0.5m², enough to be used, the wind speed will be lower and quieter).
Key Unit Conversion Formula
HVACR equipment often involves Chinese and foreign parameters (such as the United States with cold tons, Europe with KW), the conversion of the wrong directly choose the wrong equipment.
Temperature Conversion (℃/℉/K)
- Degrees Fahrenheit (℉) = 32 + Degrees Celsius (℃) x 1.8
- Celsius (°C) = (Fahrenheit (℉) – 32) ÷ 1.8
- Kelvin (K) = Celsius (℃) + 273.15
Practical Scenario: Imported compressor labeled “Evaporation Temperature 5℉”, converted to ℃ is (5-32)÷1.8 ≈ -14℃, match the chiller should be selected according to – 14℃.
Pressure Conversion (MPa/KPa/bar/psi)
There are all kinds of pressure gauge units, remember these core conversion relationships:
- 1MPa = 10bar = 1000KPa
- 1bar ≈ 1 atmospheric pressure = 14.5 psi (pounds per square inch)
- 1kg/cm² = 1bar = 10mH₂O (meters of water column)
Reminder: domestic air conditioning pressure is usually 0.4-0.5MPa (4-5bar), if the table shows 60psi, converted to bar is 60 ÷ 14.5 ≈ 4.14bar, in the normal range.
Refrigeration Capacity and Power Conversion
This is the most confusing part, remember “cooling capacity” and “electric power” are two different things:
- Cooling capacity: 1KW = 3.517 BTU/h (British Thermal Units/hour) = 0.284 Refrigerated Tons (RT); 1 U.S. Refrigerated Tons (USTR) = 3517W = 3024 Calories/hour (Kcal/h)
- Electric power: 1P (electric) = 0.75KW; 1KW = 1.34P (electric)
Example: The businessman said “5P air conditioner”, if it refers to the cooling capacity, it is 5 × 2.56KW ≈ 12.8KW; if it refers to the electric power, it is 5 × 0.75KW = 3.75KW, when you buy it, you should ask clearly.
Energy Efficiency and Performance Calculation Formulas
The system should not only “work”, but also “save money”, these formulas help you determine whether the equipment is energy efficient.
Energy Efficiency Ratio (EER)
Mainly used for fixed-frequency air conditioners, the higher the value, the more energy-saving.
- Formula: EER = cooling capacity (Mbtu/h) ÷ power consumption (KW)
- Standard: domestic air conditioner EER ≥ 3.2 for the first grade of energy efficiency, for example, an air conditioner cooling capacity of 12Mbtu / h, power consumption of 3.5KW, EER = 12 ÷ 3.5 ≈ 3.43, belonging to the first grade of energy efficiency.
Coefficient of Performance (COP)
applies to inverter air conditioners and refrigeration units, which is more comprehensive than EER (covering partial load).
Formula: COP = cooling capacity (KW) ÷ power consumption (KW)
Reference value: the unit with COP＞3.5 is considered high efficiency, for example, the unit with 100KW cooling capacity and 28KW electricity consumption, COP=100÷28≈3.57, and the operation cost is low.
Part-load performance value (NPLV)
Large units (e.g. chiller units) run at part-load most of the time, and NPLV can reflect the real energy efficiency.
- Formula: NPLV = 1/(0.01/A + 0.42/B + 0.45/C + 0.12/D)
- Variable Description: A is 100% load COP, B is 75% load COP, C is 50% load COP, D is 25% load COP.
Reminder: Units with NPLV > 4.0 are more suitable for shopping malls and factories (long operating hours and high percentage of part load).
Three-phase System Full Load Current (FLA)
Calculate the correct current in order to choose the right circuit breaker and wire to avoid electrical fire.
- Formula: FLA = N/(√3×U×COSφ)
Variable interpretation: N is the power (KW), U is the voltage (V, three-phase usually 380V), COSφ is the power factor (motor equipment to take 0.8-0.9).
Example:15KWcompressor,U=380V,COSφ=0.85, FLA=15000/(1.732×380×0.85)≈27.2A, choose 32A circuit breaker.
