Understanding heat transfer processes is fundamental for thermal engineers and students of physics. When we talk about the operation of a refrigeration machine, we are considering a cyclic process in which the working fluid does work using external energy. The key point here is not only the cooling of the internal chamber, but also the removal of energy to the external environment, which often raises questions when solving problems.
In the everyday understanding, a refrigerator “produces cold”, but from the point of view of physics, it transfers thermal energy from a less heated body to a more heated one. Amount of heatwhich is given to the surrounding air (heater) always exceeds the energy taken from refrigerated products. This inequality is due to the cost of electricity required to operate the compressor.
To accurately determine the required value, it is necessary to operate with the concepts of internal energy, gas work and cycle efficiency. In this article we will analyze calculation algorithms for ideal and real cycles, consider the influence of efficiency and analyze typical errors when substituting data into formulas. Thermodynamic analysis will allow you to accurately solve problems of any complexity.
The physical essence of the operation of a refrigeration machine
The refrigeration machine is a heat pump operating in a reverse cycle. Unlike a heat engine, which converts heat into mechanical work, here mechanical work is spent on transferring heat. The working fluid (refrigerant) circulates in a closed loop, alternately evaporating and condensing. In the evaporator, the refrigerant boils at low pressure, taking energy $Q_2$ from the cooled chamber.
Then the vaporous refrigerant is compressed by a compressor. In this case, external forces perform work $A$, the temperature and pressure of the gas increase sharply. In the condenser, usually located on the rear wall of the unit (or built into the side panels), the hot gas releases $Q_1$ energy to the environment, turning into a liquid state. It is this value $Q_1$ that is the required amount of heat transferred to the heater.
It is important to understand that the process is not absolutely efficient. Some of the energy is inevitably dissipated in the form of heat from the motor-compressor itself and friction in the mechanical parts. Therefore, the real cycle always differs from the ideal cycle Carnot, although the basic laws of conservation of energy remain unchanged. To calculate, we need knowledge of the parameters at each stage of the cycle.
⚠️ Attention: When solving problems, the directions of energy flows are often confused. Remember: $Q_2$ is the heat taken from the refrigerator compartment (useful effect), and $Q_1$ is the heat released into the room (heater). The amount of $Q_1$ is always greater than $Q_2$ by the amount of work expended.
The first law of thermodynamics in calculations
The basis for all calculations is the law of conservation of energy. For a closed cycle, the change in the internal energy of the working fluid is zero, since the system returns to its original state. Therefore, the algebraic sum of the heat received by the system and the work done on the system must be zero. In the context of a refrigerator, this equation takes the form of a power balance.
The amount of heat given to the heater ($Q_1$) consists of two components: the heat taken away from the refrigerator ($Q_2$), and the work of external forces ($A$) spent on compressing the steam. Mathematically, this is expressed by the formula:
Q_1 = Q_2 + A
where all quantities are measured in Joules (J). If the problem statement gives the compressor power $P$ and operating time $t$, then the work can be found as the product $A = P \cdot t$.
Often in problems it is required to find exactly $Q_1$, knowing the other parameters. For example, if it is known that the refrigerator took 500 kJ of energy from the food, and the compressor did 150 kJ of work, then 650 kJ of heat will be released into the room. The principle of superposition is not used here, a simple arithmetic sum of energies works.
- 🔹 Q1 —module of the amount of heat given to the heater (environment).
- 🔹 Q2 —module of the amount of heat received from the refrigerator (cooled body).
- 🔹 A —work done by the compressor on the working fluid.
When using this formula, it is critical to monitor the units of measurement. If the power is given in Watts and the time is in hours, you need to convert the time to seconds to get the work done in Joules. An error in dimension is the most common reason for an incorrect answer.
Ideal Carnot cycle and its efficiency
In theoretical physics, an ideal reversible Carnot cycle is often considered, which has the highest possible efficiency for given temperature limits. Although real refrigerators do not achieve these figures, Carnot cycle calculations provide insight into efficiency limits and allow comparisons between different installations. In this case, the ratios of heat and temperatures are related by a strict proportion.
For an ideal cycle, the ratio of the amounts of heat is equal to the ratio of the absolute temperatures of the heater and refrigerator:
Q_1 / Q_2 = T_1 / T_2
where $T_1$ is the absolute temperature of the heater (ambient), and $T_2$ is the absolute temperature of the refrigerator (inside the chamber). Temperatures must be converted to Kelvin ($T = t^\circ C + 273$).
Using this relationship together with the first law of thermodynamics, it is possible to express the required heat $Q_1$ in terms of temperatures and perfect work. This allows you to solve problems where direct values of heat flows are not given, but the temperature conditions of the unit's operation are known. Carnot cycle assumes the absence of losses due to friction and heat transfer at a finite temperature difference.
Why is the real efficiency always less than the ideal?
In real conditions it is impossible to ensure infinitely slow heat exchange (necessary for reversibility) and completely eliminate friction in the moving parts of the compressor. In addition, throttling (sharp expansion of the refrigerant) is an irreversible process, which reduces the efficiency of the cycle compared to Carnot.
Consider an example: if the temperature in the room is +27°C ($300K), and in the freezer -23°C ($250K), then the temperature ratio is $1.2$. This means that for every Joule of cold taken into the room, $1.2$ Joule of heat will go into the room, of which $0.2$ Joule is the work of the compressor.
Calculation through the coefficient of performance (COP)
In engineering practice, instead of the ideal cycle, the concept of refrigeration coefficient ($\varepsilon$ or $COP$ - Coefficient of Performance) is more often used. It shows the efficiency of the machine and is defined as the ratio of the cold received to the work expended: $\varepsilon = Q_2 / A$. Knowing this coefficient, you can easily find the desired value $Q_1$.
If the problem statement contains the cooling coefficient, the calculation algorithm is as follows. First we express the work in terms of $Q_2$ and $\varepsilon$: $A = Q_2 / \varepsilon$. Then we substitute into the balance equation: $Q_1 = Q_2 + (Q_2 / \varepsilon) = Q_2 \cdot (1 + 1/\varepsilon)$. This method is especially convenient when analyzing the energy efficiency of modern models.
Modern refrigerators of the A++ or A+++ class have a high coefficient of refrigeration, which means less electricity consumption with the same amount of heat removed. However, even the most efficient models will always have $Q_1$ greater than $Q_2$. The higher the efficiency, the smaller the difference between them, but it will never become zero.
| Parameter | Designation | Unit of measurement | Physical meaning |
|---|---|---|---|
| Heat to the heater | Q1 | J (Joule) | Energy given to the room |
| Heat to the refrigerator | Q2 | J (Joule) | Energy taken from the chamber |
| Work compressor | A | J (Joule) | Consumed electricity |
| Coefficient of performance | ε (COP) | Dimensionless | Ratio of Q2 to A |
Practical examples of solving problems
Let's analyze a typical problem to fix the material. Condition: A refrigeration engine with an efficiency of 30% (here we mean the ratio of work to energy expended in the context of the engine, but for a refrigerator it is more convenient to use the concept of cycle efficiency, let’s say the ratio $A/Q_1 = 0.3$ is given). Let the work expended be $A = 30$ kJ. Find $Q_1$.
Solution: If $A/Q_1 = 0.3$, then $Q_1 = A / 0.3 = 30 / 0.3 = 100$ kJ. In this case, the heat transferred to the heater is 100 kJ. The heat taken from the refrigerator will be equal to $Q_2 = Q_1 - A = 100 - 30 = 70$ kJ. It is important to carefully read the definition of efficiency in the problem statement, since in different textbooks it can be understood as different ratios of values.
Another example: compressor power 200 W, operating time 10 minutes. How much heat is given to the room if $Q_2 = 800$ kJ?
First, let's find the work: $t = 600$ s, $A = 200 \cdot 600 = 120\,000$ J $= 120$ kJ.
Then $Q_1 = 800 + 120 = $920 kJ.
Such problems are often encountered in exams and require a clear understanding of the physical essence of the processes.
- 🔹 Always convert the time to seconds before calculating the work.
- 🔹 Check the dimensions: kJ and J are often mixed in one condition.
- 🔹 The temperature must be in Kelvin for thermodynamic proportions.
Factors influencing heat transfer in real conditions
In real operation, the amount of heat given off by the refrigerator can vary depending on many factors. Dust contamination of the condenser (rear grille) impairs heat transfer, causing the compressor to run longer and consume more energy, which ultimately increases $Q_1$. The temperature in the room also affects: the hotter the room, the worse the heat is removed, and the higher the condensation temperature.
The frequency of door opening and the number of loaded products also play a role. Loading warm products requires significant energy expenditure to cool them. During this period, the refrigerator operates in intensive mode, and the amount of heat released into the kitchen increases significantly. Thermal inertia Thermal inertia
⚠️ Attention: Installing the refrigerator in a niche without proper ventilation leads to overheating of the condenser. This not only increases energy consumption, but can also lead to compressor failure due to overload.
☑️ Heat transfer diagnostics
Frequently asked questions (FAQ)
Why is the amount of heat given to the heater always greater than that taken away from the refrigerator?
This follows from the first law of thermodynamics. To transfer heat from a cold body to a hot one, external work (electricity) must be expended. This work does not disappear without a trace, but turns into thermal energy, which is added to the transferred heat and is also released into the heater.
Is it possible to cool a room by opening the door of a working refrigerator?
No, the room will heat up. Although the open door will blow cold air, the back grill of the refrigerator will emit more heat than cool the inside. The total heat balance of the room will increase by the amount of work done by the compressor electric motor.
How to convert Celsius to Kelvin for calculations?
To convert, you need to add the number 273.15 to the temperature in Celsius (in school problems it is often rounded to 273). For example, 0°C = 273 K, -20°C = 253 K.
Does the amount of heat depend on the type of refrigerant?
Yes, indirectly. Different refrigerants have different thermodynamic properties, which affects the efficiency of the cycle and the required work of the compressor to transfer the same amount of heat. However, the law of conservation of energy ($Q_1 = Q_2 + A$) is true for any substance.