This chapter covers the fundamental principles of thermodynamics applied to refrigeration systems, vapour-compression cycles, heat load calculations, and applicable Canadian standards. By the end of this chapter, you will be able to analyse a refrigeration cycle, calculate performance parameters, and identify relevant regulatory requirements for the Red Seal exam.
Fundamental Principles of Thermodynamics
The Laws of Thermodynamics Applied to Refrigeration
The first law of thermodynamics states that energy is neither created nor destroyed; it is transformed. In a refrigeration system, this translates into the energy balance equation:
Q̇_evap + Ẇ_comp = Q̇_cond + Q̇_losses
Where:
Q̇_evap = heat absorbed at the evaporator (in watts, W)
Ẇ_comp = compressor work (in watts, W)
Q̇_cond = heat rejected at the condenser (in watts, W)
Q̇_losses = parasitic heat losses (in watts, W)
The second law of thermodynamics establishes that heat cannot spontaneously flow from a cold body to a hot body. This is why a compressor must provide mechanical work to enable this "reverse" transfer against the natural direction.
Temperature, Heat, and Pressure
Temperature measures the degree of molecular agitation. It is expressed in degrees Celsius (°C) in the International System, but you will sometimes encounter Fahrenheit (°F) on imported equipment. The conversion is:
°F = (°C × 9/5) + 32
°C = (°F − 32) × 5/9
Heat is a form of energy. We distinguish between:
Sensible heat: heat that causes a temperature change without a change of state. Q = m × c × ΔT, where m is the mass (kg), c is the specific heat capacity (kJ/kg·°C), and ΔT is the temperature difference (°C).
Latent heat: heat that causes a change of state at constant temperature. Q = m × L, where L is the latent heat (kJ/kg).
Pressure is expressed in kilopascals (kPa) or bars (1 bar = 100 kPa). Pressure gauges on refrigeration systems typically indicate gauge pressure (relative), while thermodynamic calculations use absolute pressure:
P_abs = P_gauge + P_atm (101.325 kPa at sea level)
The Pressure-Enthalpy (P-h) Diagram
The P-h diagram is the fundamental tool for analysing refrigeration cycles. It represents enthalpy (h, in kJ/kg) on the x-axis and pressure (P, in kPa) on a logarithmic y-axis. The saturation curves delimit three zones:
Below the saturation curve: liquid-vapour mixture
To the left of the saturated liquid curve: subcooled liquid
To the right of the saturated vapour curve: superheated vapour
For the exam, you must know how to read this diagram to determine:
The enthalpy at each point of the cycle
The saturation temperature corresponding to a given pressure
The vapour quality of a mixture: x = (h − h_liquid) / (h_vapour − h_liquid)
38.Isentropic compression (1→2): the compressor draws in superheated vapour at low pressure and compresses it to high pressure. In the ideal cycle, entropy remains constant. The temperature increases significantly.
39.Isobaric condensation (2→3): the refrigerant passes through the condenser at constant pressure. It first goes from superheated vapour to saturated vapour (desuperheating), then from saturated vapour to saturated liquid (condensation), then from saturated liquid to subcooled liquid (subcooling).
40.Isenthalpic expansion (3→4): the expansion device causes a sudden pressure drop without heat exchange or work. Enthalpy remains constant (h₃ = h₄). A portion of the liquid flashes to vapour, lowering the temperature of the mixture.
41.Isobaric evaporation (4→1): the low-pressure mixture absorbs heat from the medium being cooled. The liquid evaporates completely, then the vapour becomes slightly superheated before being drawn into the compressor.
The Real Cycle and Its Deviations
The real cycle differs from the ideal cycle due to:
Pressure drops in the piping and heat exchangers (pressure loss in the suction and liquid lines)
Non-isentropic compression: the compressor isentropic efficiency η_is = (h₂s − h₁) / (h₂actual − h₁) is typically 0.70 to 0.85
Useful superheat at the evaporator (to protect the compressor from liquid slugging)
Subcooling at the condenser (to improve efficiency and prevent flash gas in the liquid line)
Cycle Performance Calculations
Net refrigerating effect (NRE): heat absorbed per kilogram of refrigerant at the evaporator.
Where v₁ is the specific volume of the suction vapour (m³/kg).
Complete Calculation Example
Given data: A system uses R-134a. At the evaporator, the pressure is 200 kPa (saturation temperature ≈ −10.1 °C). The vapour leaving is superheated to 5 °C (h₁ = 397 kJ/kg). The condenser operates at 800 kPa (saturation temperature ≈ 31.3 °C). The liquid leaving is subcooled to 25 °C (h₃ = 230 kJ/kg). The actual compression gives h₂ = 430 kJ/kg.
Step 1: Determine h₄. The expansion is isenthalpic, so h₄ = h₃ = 230 kJ/kg.
Step 2: Calculate the NRE.
NRE = 397 − 230 = 167 kJ/kg
Step 3: Calculate the compression work.
W_comp = 430 − 397 = 33 kJ/kg
Step 4: Calculate the COP.
COP = 167 / 33 = 5.06
Step 5: If the mass flow rate is 0.05 kg/s, calculate the refrigerating capacity.
Q̇_evap = 0.05 × 167 = 8.35 kW
Step 6: Calculate the heat rejected at the condenser.
In Canada, the Ozone-depleting Substances and Halocarbon Alternatives Regulations (SOR/2016-137) govern the use of refrigerants. Key requirements for the technician:
Mandatory certification to handle refrigerants: certification from Natural Resources Canada's (NRCan) Office of Energy Efficiency
Prohibition on intentionally releasing refrigerants into the atmosphere
Mandatory recovery of refrigerants before any repair or demolition of equipment
Record keeping: quantities of refrigerant purchased, used, recovered, and disposed of
Maximum allowable leak rate: 10% per year for systems with more than 50 kg of charge
Refrigerant Blends
Zeotropic blends (R-404A, R-410A, R-407C) exhibit temperature glide: the saturation temperature varies during the phase change at constant pressure. Practical consequences:
Charging must be done in the liquid phase to avoid fractionation of the blend
Superheat and subcooling are measured using the dew point temperature (vapour) and the bubble point temperature (liquid)
Leaks alter the composition of the remaining blend
Azeotropic blends (R-500, R-502) behave like a pure substance: no temperature glide.
System Components and Their Diagnostics
The Compressor
Types of compressors encountered:
Type
Typical Applications
Characteristics
**Hermetic**
Refrigerators, freezers, small systems
Motor and compressor in a welded shell, not serviceable
**Semi-hermetic**
Medium commercial systems
Motor and compressor in a bolted shell, serviceable
**Open**
Large industrial systems
Compressor driven by an external motor via coupling
Volumetric efficiency η_v = actual volumetric flow rate / theoretical volumetric flow rate. It depends on clearance volume, valve pressure drops, and vapour heating. A typical value is 0.75 to 0.90.
Quick diagnostics:
Compressor runs but does not compress: broken valves, worn reeds, excessive clearance volume
Compressor does not start: defective start relay, failed capacitor, open winding
Air-cooled (natural or forced convection): the most common in appliances
Water-cooled (shell-and-tube, plate): for medium and large capacity systems
Evaporative: combines air and water for very large installations
Subcooling is the difference between the saturation temperature at the condenser and the actual temperature of the liquid leaving. A normal value is 5 to 10 °C. Subcooling that is too low indicates an insufficient refrigerant charge or a dirty condenser.
The Evaporator
The evaporator can be:
Direct expansion: the refrigerant expands directly into the evaporator
Flooded: a liquid accumulator keeps the evaporator full of liquid, with a liquid separator upstream of the compressor
Superheat is the difference between the temperature of the vapour leaving the evaporator and the saturation temperature at the evaporating pressure. A typical value is 5 to 8 °C. Superheat that is too high indicates a refrigerant shortage or an underfeeding expansion valve. Superheat that is too low risks liquid slugging at the compressor.
The Expansion Device
The main types:
Capillary tube: a small-diameter tube with a calibrated length. Used on small hermetic systems. The pressure drop is fixed and does not adapt to load variations.
Thermostatic expansion valve (TXV): maintains constant superheat using a temperature-sensing bulb at the evaporator outlet. It adapts to load variations.
Electronic expansion valve: controlled by an electronic controller with temperature and pressure sensors. Superior precision.
TXV adjustment: superheat is adjusted by turning the adjustment screw (generally a quarter turn at a time). Turning clockwise increases superheat (spring more compressed), counterclockwise decreases it.
Heat Load Calculations
Components of the Heat Load
The total heat load of a refrigerated enclosure includes:
127.Transmission through the walls: Q_walls = U × A × ΔT, where U is the thermal transmission coefficient (W/m²·°C), A is the surface area (m²), and ΔT is the temperature difference between inside and outside.
128.Air infiltration: due to door openings. Q_inf = ṁ_air × c_air × ΔT + ṁ_air × L_condensation (for humidity).
129.Stored products: cooling products from their entry temperature to the storage temperature.
130.Internal loads: lighting, fan motors, defrost, people.
131.Miscellaneous loads: defrost heaters, air circulation.
Simplified Calculation Example
Given data: A cold room measuring 4 m × 3 m × 2.5 m (height) is maintained at 2 °C. The walls have a U = 0.35 W/m²·°C. The outside temperature is 25 °C. You store 500 kg of products at 15 °C with a specific heat of 3.5 kJ/kg·°C, to be cooled over 24 hours. Lighting is 200 W, used 4 hours per day. Evaporator fans consume 150 W continuously.
Q_lighting = 200 × 4 / 24 = 33 W (average over 24 h)
Q_fans = 150 W
Step 5: Total.
Q_total = 475 + 263 + 33 + 150 = 921 W
Step 6: Add a 10% safety margin.
Q_total_adjusted = 921 × 1.10 = 1013 W ≈ 1.0 kW
The system will need to provide approximately 1 kW of refrigerating capacity. In practice, you would select a system of 1.2 to 1.5 kW to account for defrost cycles and peak loads.
Canadian Standards and Codes
The Canadian Electrical Code
The Canadian Electrical Code, Part I (CE Code) is published by the CSA Group under standard CSA C22.1. The relevant rules for refrigeration systems:
Rule 8-200: calculation of circuit loads. Compressor motors are considered continuous loads; the circuit must be sized at 125% of the rated current.
Rule 26-250: requirements for compressor motors and controls. Overload and overcurrent protection.
Rule 28-600: wiring of refrigeration units. Conductors must be sized according to the full-load current of the motor.
The Canadian Electrical Code, Chapter V
The Canadian Electrical Code, Chapter V — Wiring of Refrigeration Equipment (CSA C22.2 No. 120) deals specifically with refrigeration systems. It covers:
Wiring of hermetic and semi-hermetic compressors
Thermal protection devices
Control and safety circuits
Grounding requirements
CSA Standards for Appliances
CSA C22.2 No. 92: household refrigerators and freezers
CSA C22.2 No. 120: wiring of refrigeration equipment
CSA B149.1
CSA B149.1 — Natural Gas and Propane Installation Code applies when refrigeration systems use flammable refrigerants (R-290, R-600a) or are installed in locations where gas is used. Key requirements:
Section 6: ventilation of rooms containing gas appliances
Section 7: venting of combustion products
Flammable refrigerants are classified as gases; rooms must meet ventilation and leak detection requirements
The Refrigeration Appliances Regulations
The Refrigeration Appliances Regulations (SOR/80-120) require that appliances sold in Canada bear an EnerGuide label indicating annual energy consumption. Maximum consumption values are established by this regulation.
Pitfalls to Avoid
175.Confusing gauge pressure and absolute pressure: thermodynamic tables use absolute pressure. If you use the pressure read on the gauge without adding atmospheric pressure, all your enthalpy calculations will be wrong.
176.Forgetting that expansion is isenthalpic: h₄ = h₃. Many candidates calculate h₄ as if there were heat exchange.
177.Using the saturation temperature instead of the actual temperature to calculate superheat or subcooling. Superheat = T_actual_vapour − T_saturation; subcooling = T_saturation − T_actual_liquid.
178.Reversing the direction of the TXV adjustment: clockwise = more superheat, counterclockwise = less superheat.
179.Neglecting the temperature glide of zeotropic blends: for R-410A, the saturation temperature is not a single value at a given pressure.
180.Forgetting units: enthalpies are in kJ/kg, powers in kW. A mass flow rate in kg/s multiplied by an enthalpy in kJ/kg gives kW, not kJ.
181.Confusing COP and energy efficiency ratio (EER): COP is unitless (W/W), EER is in BTU/h per watt. EER = COP × 3.412.
182.Not accounting for isentropic efficiency in real cycle calculations. Actual work is always greater than isentropic work.
183.Ignoring refrigerant recovery requirements: Canadian regulations require recovery before any opening of the circuit, regardless of the type of refrigerant.
184.Forgetting the 125% rule for sizing electrical circuits for compressors (CE Code Rule 8-200).
Summary
The first law governs the energy balance of the cycle: Q_evap + W_comp = Q_cond.
The vapour-compression cycle comprises four processes: isentropic compression, isobaric condensation, isenthalpic expansion, isobaric evaporation.
NRE = h₁ − h₄; COP = NRE / W_comp.
The P-h diagram is the reference tool for all cycle calculations.
Refrigerants are classified according to their environmental impact: CFCs (banned), HCFCs (being phased out), HFCs (high GWP), HFOs (low GWP), natural.
Superheat (typically 5 to 8 °C) protects the compressor; subcooling (typically 5 to 10 °C) improves efficiency.
Zeotropic blends exhibit temperature glide; charging must be done in the liquid phase.
The heat load includes transmission, infiltration, products, internal loads, and miscellaneous loads.
Applicable standards: CE Code Part I (CSA C22.1), CE Code Chapter V (CSA C22.2 No. 120), CSA B149.1 for flammable refrigerants.
Refrigerant recovery is mandatory; NRCan certification is required for handling.
Review Questions
197.An R-134a system operates with an evaporating pressure of 150 kPa and a condensing pressure of 1000 kPa. The vapour leaving the evaporator has an enthalpy of 390 kJ/kg, the liquid leaving the condenser has an enthalpy of 240 kJ/kg, and the vapour leaving the compressor has an enthalpy of 425 kJ/kg. Calculate the NRE, the compression work, and the COP.
198.Explain why superheat at the evaporator is necessary despite the slight efficiency loss it causes.
199.A hermetic compressor has a volumetric efficiency of 0.80. The swept volume is 0.002 m³ per revolution and the rotational speed is 2900 rpm. The specific volume of the suction vapour is 0.10 m³/kg. Calculate the refrigerant mass flow rate.
200.What are the practical differences between a capillary tube and a thermostatic expansion valve in terms of load adaptation?
201.An R-410A system has a leak. Explain why it is imperative to recover all remaining refrigerant, repair the leak, and then recharge with new refrigerant, rather than simply topping up the charge.
202.Calculate the heat load of a refrigerated display case measuring 2 m × 1 m × 1.5 m (height), with a U of 0.8 W/m²·°C, an interior temperature of 4 °C, an ambient temperature of 22 °C, lighting of 150 W operating 10 h/day, and fans of 80 W running continuously. Neglect infiltration and products.
This chapter covers the essential knowledge for the "Refrigeration Systems and Thermodynamics" section of the Red Seal exam. Make sure you master cycle calculations, P-h diagram reading, and Canadian regulatory requirements before moving on to the following chapters.
