When we talk about the reversibility of processes, it should be taken into account that this is some kind of idealization. All real processes are irreversible, therefore, the cycles by which heat engines operate are also irreversible, and therefore non-equilibrium. However, to simplify the quantitative estimates of such cycles, it is necessary to consider them as equilibrium, that is, as if they consisted only of equilibrium processes. This is required by the well-developed apparatus of classical thermodynamics.
The famous cycle of the ideal Carnot engine is considered to be an equilibrium inverse circular process. In real conditions, any cycle cannot be ideal, since there are losses. It takes place between two heat sources constant temperatures at the heat sink T 1 and heat receiver T 2, as well as the working fluid, which is taken as an ideal gas (Fig. 3.1).
Rice. 3.1. Cycle heat engine
We believe that T 1 > T 2 and heat removal from the heat sink and heat supply to the heat sink do not affect their temperatures, T1 And T2 remain constant. Let us denote the parameters of the gas at the left extreme position heat engine piston: pressure - R 1 volume - V 1, temperature T 1 . This is point 1 on the graph on the axes P-V. At this moment, the gas (working fluid) interacts with the heat source, the temperature of which is also T 1 . As the piston moves to the right, the gas pressure in the cylinder decreases and the volume increases. This will continue until the piston arrives at the position determined by point 2, where the parameters of the working fluid (gas) will take on the values P 2 , V 2 , T2. The temperature at this point remains unchanged, since the temperature of the gas and the heat sink is the same during the transition of the piston from point 1 to point 2 (expansion). Such a process in which T does not change, is called isothermal, and curve 1–2 is called isotherm. In this process, heat is transferred from the heat source to the working fluid. Q1.
At point 2, the cylinder is completely isolated from the external environment (there is no heat exchange) and at further movement piston to the right, a decrease in pressure and an increase in volume occurs along a curve 2-3, which is called adiabatic(process without heat exchange with external environment). When the piston moves to the extreme right position (point 3), the expansion process will end and the parameters will have the values P 3 , V 3 , and the temperature will become equal to the temperature of the heat sink T 2. With this position of the piston, the insulation of the working fluid is reduced and it interacts with the heat sink. If we now increase the pressure on the piston, then it will move to the left at a constant temperature T 2(compression). Hence, this compression process will be isothermal. In this process, heat Q2 will pass from the working fluid to the heat sink. The piston, moving to the left, will come to point 4 with the parameters P4, V4 and T 2 where the working fluid is again isolated from the environment. Further compression occurs along a 4–1 adiabat with an increase in temperature. At point 1, compression ends at the parameters of the working fluid P 1 , V 1 , T 1. The piston returned to its original state. At point 1, the isolation of the working fluid from the external environment is removed and the cycle is repeated.
Coefficient useful action ideal Carnot engine.
The work done by the engine is:
This process was first considered by the French engineer and scientist N. L. S. Carnot in 1824 in the book Reflections on the driving force of fire and on machines capable of developing this force.
The purpose of Carnot's research was to find out the reasons for the imperfection of heat engines of that time (they had an efficiency of ≤ 5%) and to find ways to improve them.
The Carnot cycle is the most efficient of all. Its efficiency is maximum.
The figure shows the thermodynamic processes of the cycle. In the process of isothermal expansion (1-2) at a temperature T 1 , the work is done by changing the internal energy of the heater, i.e., by supplying the amount of heat to the gas Q:
A 12 = Q 1 ,
Cooling of the gas before compression (3-4) occurs during adiabatic expansion (2-3). Change in internal energy ΔU 23 in an adiabatic process ( Q=0) is completely converted into mechanical work:
A 23 = -ΔU 23 ,
The temperature of the gas as a result of adiabatic expansion (2-3) decreases to the temperature of the refrigerator T 2 < T 1 . In the process (3-4), the gas is isothermally compressed, transferring the amount of heat to the refrigerator Q2:
A 34 = Q 2,
The cycle is completed by the process of adiabatic compression (4-1), in which the gas is heated to a temperature T 1.
The maximum value of the efficiency of heat engines operating on ideal gas, according to the Carnot cycle:
.
The essence of the formula is expressed in the proven WITH. Carnot's theorem that the efficiency of any heat engine cannot exceed the efficiency of the Carnot cycle carried out at the same temperature of the heater and refrigerator.
6.3. Second law of thermodynamics
6.3.1. Efficiency thermal engines. Carnot cycle
The second law of thermodynamics arose from the analysis of the operation of heat engines (machines). In Kelvin's formulation, it looks like this: a circular process is impossible, the only result of which is the conversion of the heat received from the heater into work equivalent to it.
The scheme of operation of a heat engine (heat engine) is shown in fig. 6.3.
Rice. 6.3
Heat engine cycle consists of three stages:
1) the heater transfers the amount of heat Q 1 to the gas;
2) the expanding gas does work A ;
3) to return the gas to its original state, heat Q 2 is transferred to the refrigerator.
From the first law of thermodynamics for a cyclic process
Q=A
where Q is the amount of heat received by the gas per cycle, Q \u003d Q 1 - Q 2; Q 1 - the amount of heat transferred to the gas from the heater; Q 2 - the amount of heat given off by the gas to the refrigerator.
Therefore, for an ideal heat engine, the equality
Q 1 − Q 2 = A.
When energy loss (due to friction and its dissipation in environment) are absent, during the operation of heat engines, law of energy conservation
Q 1 \u003d A + Q 2,
where Q 1 is the heat transferred from the heater to the working fluid (gas); A is the work done by the gas; Q 2 is the heat transferred by the gas to the refrigerator.
Efficiency heat engine is calculated by one of the formulas:
η = A Q 1 ⋅ 100% , η = Q 1 − Q 2 Q 1 ⋅ 100% , η = (1 − Q 2 Q 1) ⋅ 100% ,
where A is the work done by the gas; Q 1 - heat transferred from the heater to the working fluid (gas); Q 2 is the heat transferred by the gas to the refrigerator.
The most commonly used in heat engines is the Carnot cycle, as it is the most economical.
The Carnot cycle consists of two isotherms and two adiabats shown in Fig. 6.4.
Rice. 6.4
Section 1–2 corresponds to the contact of the working substance (gas) with the heater. In this case, the heater transfers heat Q 1 to the gas and isothermal expansion of the gas occurs at the temperature of the heater T 1 . The gas does positive work (A 12 > 0), its internal energy does not change (∆U 12 = 0).
Segment 2–3 corresponds to the adiabatic expansion of the gas. In this case, heat exchange with the external environment does not occur, performed positive work A 23 leads to a decrease in the internal energy of the gas: ∆U 23 = −A 23, the gas is cooled to the refrigerator temperature T 2 .
Section 3–4 corresponds to the contact of the working substance (gas) with the cooler. In this case, heat Q 2 is supplied to the refrigerator from the gas and isothermal compression of the gas occurs at the temperature of the refrigerator T 2 . The gas does negative work (A 34< 0), его внутренняя энергия не изменяется (∆U 34 = 0).
Segment 4–1 corresponds to the adiabatic compression of the gas. In this case, there is no heat exchange with the external environment, the negative work A 41 performed leads to an increase in the internal energy of the gas: ∆U 41 = −A 41, the gas is heated to the temperature of the heater T 1, i.e. returns to its original state.
The efficiency of a heat engine operating according to the Carnot cycle is calculated using one of the formulas:
η = T 1 − T 2 T 1 ⋅ 100% , η = (1 − T 2 T 1) ⋅ 100% ,
where T 1 - heater temperature; T 2 - refrigerator temperature.
Example 9. Ideal heat engine performs 400 J of work per cycle. How much heat is transferred to the refrigerator in this case, if the efficiency of the machine is 40%?
Solution . The efficiency of a heat engine is determined by the formula
η = A Q 1 ⋅ 100% ,
where A is the work done by the gas per cycle; Q 1 - the amount of heat that is transferred from the heater to the working fluid (gas).
The desired value is the amount of heat Q 2 transferred from the working fluid (gas) to the refrigerator, which is not included in the written formula.
The relationship between work A, heat Q 1 transferred from the heater to gas, and the desired value Q 2 is established using the law of conservation of energy for an ideal heat engine
Q 1 \u003d A + Q 2.
The equations form a system
η = A Q 1 ⋅ 100% , Q 1 = A + Q 2 , )
which needs to be solved with respect to Q 2 .
To do this, we exclude from the system Q 1 , expressing from each equation
Q 1 \u003d A η ⋅ 100%, Q 1 \u003d A + Q 2)
and writing the equality of the right parts of the resulting expressions:
A η ⋅ 100% = A + Q 2 .
The desired value is determined by the equality
Q 2 \u003d A η ⋅ 100% - A \u003d A (100% η - 1) .
The calculation gives the value:
Q 2 \u003d 400 ⋅ (100% 40% - 1) \u003d 600 J.
The amount of heat transferred per cycle from the gas to the refrigerator of an ideal heat engine is 600 J.
Example 10. In an ideal heat engine, 122 kJ / min is supplied from the heater to the gas, and 30.5 kJ / min is transferred from the gas to the refrigerator. Calculate the efficiency of this ideal heat engine.
Solution . To calculate the efficiency, we use the formula
η = (1 − Q 2 Q 1) ⋅ 100% ,
where Q 2 - the amount of heat that is transferred per cycle from the gas to the refrigerator; Q 1 - the amount of heat that is transferred per cycle from the heater to the working fluid (gas).
Let's transform the formula by dividing the numerator and denominator of the fraction by the time t:
η = (1 − Q 2 / t Q 1 / t) ⋅ 100% ,
where Q 2 /t is the rate of heat transfer from the gas to the refrigerator (the amount of heat that is transferred by the gas to the refrigerator per second); Q 1 /t - the rate of heat transfer from the heater to the working fluid (the amount of heat that is transferred from the heater to the gas per second).
In the condition of the problem, the rate of heat transfer is given in joules per minute; convert it to joules per second:
- from the gas heater -
Q 1 t \u003d 122 kJ / min \u003d 122 ⋅ 10 3 60 J / s;
- from gas to the refrigerator -
Q 2 t \u003d 30.5 kJ / min \u003d 30.5 ⋅ 10 3 60 J / s.
Calculate the efficiency of this ideal heat engine:
η = (1 − 30.5 ⋅ 10 3 60 ⋅ 60 122 ⋅ 10 3) ⋅ 100% = 75%.
Example 11. The efficiency of a heat engine operating according to the Carnot cycle is 25%. How many times will the efficiency increase if the temperature of the heater is increased and the temperature of the refrigerator is reduced by 20%?
Solution . The efficiency of an ideal heat engine operating according to the Carnot cycle is determined by the following formulas:
- before changing the temperature of the heater and refrigerator -
η 1 = (1 − T 2 T 1) ⋅ 100% ,
where T 1 is the initial temperature of the heater; T 2 - the initial temperature of the refrigerator;
- after changing the temperature of the heater and refrigerator -
η 2 = (1 − T ′ 2 T ′ 1) ⋅ 100% ,
where T′ 1 - new temperature heater, T ′ 1 = 1.2 T 1; T ′ 2 - new refrigerator temperature, T ′ 2 = 0.8 T 2 .
Equations for efficiency factors form a system
η 1 = (1 − T 2 T 1) ⋅ 100% , η 2 = (1 − 0.8 T 2 1.2 T 1) ⋅ 100% , )
which needs to be solved with respect to η 2 .
From the first equation of the system, taking into account the value of η 1 = 25%, we find the temperature ratio
T 2 T 1 \u003d 1 - η 1 100% \u003d 1 - 25% 100% \u003d 0.75
and substitute into the second equation
η 2 \u003d (1 - 0.8 1.2 ⋅ 0.75) ⋅ 100% \u003d 50%.
The desired efficiency ratio is equal to:
η 2 η 1 \u003d 50% 25% \u003d 2.0.
Consequently, the indicated change in the temperatures of the heater and refrigerator of the heat engine will lead to an increase in the efficiency factor by a factor of 2.
Modern realities involve the widespread operation of heat engines. Numerous attempts to replace them with electric motors have so far failed. Problems associated with the accumulation of electricity in autonomous systems are solved with great difficulty.
Still relevant are the problems of technology for the manufacture of electric power accumulators, taking into account their long-term use. Speed characteristics electric vehicles are far from those of cars on engines internal combustion.
First steps to create hybrid engines can significantly reduce harmful emissions in megacities, solving environmental problems.
A bit of history
The possibility of converting steam energy into motion energy was known in antiquity. 130 BC: Philosopher Heron of Alexandria presented to the audience a steam toy - aeolipil. A sphere filled with steam began to rotate under the action of jets emanating from it. This prototype of modern steam turbines did not find application at that time.
For many years and centuries, the development of the philosopher was considered only a fun toy. In 1629, the Italian D. Branchi created an active turbine. Steam set in motion a disk equipped with blades.
From that moment began the rapid development steam engines.
heat engine
The conversion of fuel into energy for the movement of parts of machines and mechanisms is used in heat engines.
The main parts of machines: a heater (a system for obtaining energy from outside), a working fluid (performs a useful action), a refrigerator.
The heater is designed to ensure that the working fluid has accumulated a sufficient supply of internal energy to perform useful work. The refrigerator removes excess energy.
The main characteristic of efficiency is called the efficiency of heat engines. This value shows what part of the energy spent on heating is spent on doing useful work. The higher the efficiency, the more profitable job machine, but this value cannot exceed 100%.
Efficiency calculation
Let the heater acquire from outside the energy equal to Q 1 . working body did work A, while the energy given to the refrigerator was Q 2 .
Based on the definition, we calculate the efficiency:
η= A / Q 1 . We take into account that A \u003d Q 1 - Q 2.
From here, the efficiency of the heat engine, the formula of which has the form η = (Q 1 - Q 2) / Q 1 = 1 - Q 2 / Q 1, allows us to draw the following conclusions:
- Efficiency cannot exceed 1 (or 100%);
- to maximize this value, either an increase in the energy received from the heater or a decrease in the energy given to the refrigerator is necessary;
- an increase in the energy of the heater is achieved by changing the quality of the fuel;
- reducing the energy given to the refrigerator, allow you to achieve design features engines.
Ideal heat engine
Is it possible to create such an engine, the efficiency of which would be maximum (ideally, equal to 100%)? The French theoretical physicist and talented engineer Sadi Carnot tried to find the answer to this question. In 1824, his theoretical calculations about the processes occurring in gases were made public.
The main idea behind perfect car, we can consider the carrying out of reversible processes with ideal gas. We start with the expansion of the gas isothermally at a temperature T 1 . The amount of heat required for this is Q 1. After the gas expands without heat exchange. Having reached the temperature T 2, the gas is compressed isothermally, transferring the energy Q 2 to the refrigerator. The return of the gas to its original state is adiabatic.
The efficiency of an ideal Carnot heat engine, when accurately calculated, is equal to the ratio of the temperature difference between the heating and cooling devices to the temperature that the heater has. It looks like this: η=(T 1 - T 2)/ T 1.
The possible efficiency of a heat engine, the formula of which is: η= 1 - T 2 / T 1 , depends only on the temperature of the heater and cooler and cannot be more than 100%.
Moreover, this ratio allows us to prove that the efficiency of heat engines can be equal to unity only when the refrigerator reaches temperatures. As you know, this value is unattainable.
Carnot's theoretical calculations make it possible to determine maximum efficiency heat engine of any design.
The theorem proved by Carnot is as follows. An arbitrary heat engine under no circumstances is capable of having a coefficient of efficiency greater than the similar value of the efficiency of an ideal heat engine.
Example of problem solving
Example 1 What is the efficiency of an ideal heat engine if the heater temperature is 800°C and the refrigerator temperature is 500°C lower?
T 1 \u003d 800 o C \u003d 1073 K, ∆T \u003d 500 o C \u003d 500 K, η -?
By definition: η=(T 1 - T 2)/ T 1.
We are not given the temperature of the refrigerator, but ∆T = (T 1 - T 2), from here:
η \u003d ∆T / T 1 \u003d 500 K / 1073 K \u003d 0.46.
Answer: efficiency = 46%.
Example 2 Determine the efficiency of an ideal heat engine if, due to the acquired one kilojoule of heater energy, useful work 650 J. What is the temperature of the heat engine heater if the coolant temperature is 400 K?
Q 1 \u003d 1 kJ \u003d 1000 J, A \u003d 650 J, T 2 \u003d 400 K, η -?, T 1 \u003d?
In this problem, we are talking about a thermal installation, the efficiency of which can be calculated by the formula:
To determine the temperature of the heater, we use the formula for the efficiency of an ideal heat engine:
η \u003d (T 1 - T 2) / T 1 \u003d 1 - T 2 / T 1.
After performing mathematical transformations, we get:
T 1 \u003d T 2 / (1- η).
T 1 \u003d T 2 / (1- A / Q 1).
Let's calculate:
η= 650 J / 1000 J = 0.65.
T 1 \u003d 400 K / (1- 650 J / 1000 J) \u003d 1142.8 K.
Answer: η \u003d 65%, T 1 \u003d 1142.8 K.
Real conditions
The ideal heat engine is designed with ideal processes in mind. Work is done only in isothermal processes, its value is defined as the area bounded by the Carnot cycle graph.
In fact, it is impossible to create conditions for the process of changing the state of a gas without accompanying changes in temperature. There are no materials that would exclude heat exchange with surrounding objects. The adiabatic process is no longer possible. In the case of heat transfer, the temperature of the gas must necessarily change.
The efficiency of heat engines created in real conditions differ significantly from the efficiency of ideal engines. Note that the processes in real engines occurs so rapidly that the variation in the internal thermal energy of the working substance in the process of changing its volume cannot be compensated by the influx of heat from the heater and return to the cooler.
Other heat engines
Real engines operate on different cycles:
- Otto cycle: the process at a constant volume changes adiabatically, creating a closed cycle;
- Diesel cycle: isobar, adiabat, isochor, adiabat;
- the process occurring at constant pressure is replaced by an adiabatic one, closing the cycle.
Create equilibrium processes in real engines (to bring them closer to ideal ones) under conditions modern technology does not seem possible. The efficiency of heat engines is much lower, even taking into account the same temperature conditions, as in an ideal thermal installation.
But do not reduce the role of the calculated efficiency formulas because it becomes the starting point in the process of working on increasing the efficiency of real engines.
Ways to change efficiency
When comparing ideal and real heat engines, it is worth noting that the temperature of the refrigerator of the latter cannot be any. Usually the atmosphere is considered to be a refrigerator. The temperature of the atmosphere can be taken only in approximate calculations. Experience shows that the temperature of the coolant is equal to the temperature of the exhaust gases in the engines, as is the case in internal combustion engines (abbreviated internal combustion engines).
ICE is the most common heat engine in our world. The efficiency of a heat engine in this case depends on the temperature created by the burning fuel. The essential difference ICE from steam engines is a fusion of the functions of the heater and the working fluid of the device into air-fuel mixture. Burning, the mixture creates pressure on the moving parts of the engine.
An increase in the temperature of the working gases is achieved by significantly changing the properties of the fuel. Unfortunately, it is not possible to do this indefinitely. Any material from which the combustion chamber of an engine is made has its own melting point. The heat resistance of such materials is the main characteristic of the engine, as well as the ability to significantly affect the efficiency.
Motor efficiency values
If we consider the temperature of the working steam at the inlet of which is 800 K, and the exhaust gas is 300 K, then the efficiency of this machine is 62%. In reality, this value does not exceed 40%. Such a decrease occurs due to heat losses during heating of the turbine housing.
The highest value of internal combustion does not exceed 44%. Increasing this value is a matter of the near future. Changing the properties of materials, fuels is a problem that is being worked on the best minds humanity.
