Carnot heat engine and carnot cycle
To find out how to increase the efficiency of the heat engine, a French scientist named Sadi Carnot (1796-1832) examined an ideal theoretical caloric machine in 1824. At that time, the first law of thermodynamics had not been formulated, nor the second law of thermodynamics. The first law has not been formulated because scientists do not yet know that heat is energy. After Joule and his colleagues experimented in the 1830s, scientists discovered heat is energy that moves due to temperature differences. So first law of thermodynamics was formulated after 1830. Sadi Carnot had been researching the theoretical ideal caloric engine in 1824. His research was actually to increase the efficiency of the steam engine. Most steam engines of that time were less efficient.
The cycle on the ideal heat engine of Sadi Carnot’s thought is named Carnot cycle. Before learning the Carnot cycle, we understood the irreversible process again. Every process of energy change and energy transfer occurs irreversibly. For example, we rub hands so that hands warm. The heat is generated through the work we do. The process is irreversible. The heat can not rub hands.
The purpose of a heat engine is to reverse some of this process, where heat can be utilized to perform work with the highest possible efficiency. For the heat engine to have maximum efficiency, then we must avoid all irreversible processes. Heat transfer is naturally irreversible so that the movement of heat should not occur.
When engine takes the heat (QH) at a high temperature (TH), the working fluid in the machine must also be at TH temperature. Similarly, if the engine exhausts the QL heat at a low temperature (TL), the working fluid in the machine must also be at TL temperature. Thus any process involving heat transfer must be isothermal (same temperature). Conversely, if the temperature of the workpiece in the machine is between TH and TL, no heat transfer occurs between the machine with the place having the TH temperature (heat supplier) and the area having the TL temperature (discharge). For the heat does not move then the process must be done adiabatically.
The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes.
The image on the side is The pistonCarnot cycle for an ideal gas. First, the heat is absorbed during the isothermal expansion (a-b). During isothermal expansion, the gas temperature in the cylinder is kept constant.
Next, the gas expands adiabatically so that the temperature drops from TH to TL (b-c). TH = high temperature, TL = low temperature. During adiabatic expansion, no heat enters or exits the cylinder. After that gas is isothermally pressed (c-d). During isothermal pressing, the gas temperature is kept constant. Gas is then pressed adiabatically (d-a). Because it is adiabatically suppressed, gas temperature increases. The temperature of the gas increases so that gas pressure also increases. The whole process in the Carnot cycle is reversible.
During isothermal expansion and isothermal presses, the gas temperature is kept constant. The goal is to avoid any temperature differences. The existence of temperature difference can cause heat transfer (irreversible process). For the isothermal process to occur (gas temperature is always constant) the gas must be gradually expanded or pressed. In reality, expansion or gas suppression occurs more quickly. This is due to turbulence (remember dynamic fluid material), friction, viscosity, etc. As a result, the perfect isothermal process will never exist. Conversely, adiabatic expansion and suppression are rapid.
The goal is to keep the heat from flowing into the cylinder or out of the cylinder. The presence of friction, viscosity, etc. causes no complete adiabatic expansion and emphasis. Keep in mind that the Carnot engine is only theoretical. The Carnot machine does not exist in our lives. Although only theoretical the existence of Carnot engine is the beneficial development of thermodynamics. It can at least be known of any irreversible processes that may occur during the process and attempt to minimize them so that the efficiency of the engine is maximum.
The significant result of the Carnot engine is for the perfect heat engine, the received heat (QH) is proportional to the high temperature (TH), and the thawed exhaust (QL) is proportional to the low temperature (TL). Thus, the efficiency of the perfect heat engine is:
This is the equation of the perfect heat engine’s efficiency.
Question 1 :
A steam engine works between 500 oC and 300 oC. Determine the ideal efficiency (Carnot efficiency) of the steam engine.
The temperature should be converted into the Kelvin scale.
TH (high temperature) = 500 oC = 500 + 273 = 773 K
TL (low temperature) = 300 oC = 300 + 273 = 573 K
The efficiency of a perfect heat engine that works between 500 °C and 300 °C is 26%. When the machines we use in our daily life work between 500 °C and 300 °C, a maximum efficiency that machines can achieve is about 0.7 times ideal efficiency (18.2%).
Question 2 :
A heat engine receives 600 J at a temperature of 300 ° C, performs work (W) 100 Joules and discharges 500 J at 100 ° C. Determine the actual efficiency and ideal efficiency (Carnot efficiency) of this machine.
Temperature should be converted into Kelvin scale
TH (high temperature) = 300 oC — 300 + 273 = 573 K
TL (low temperature) = 100 oC — 100 + 273 = 373 K
QH = 600 J
QL = 500 J
The ideal efficiency of this machine:
Efficiency of a perfect heat engine working between 300 °C and 100 °C is 35%. Maximum efficiency that the machine can achieve is usually about 0.7 times the ideal efficiency = 0.7 x 35% = 24.5% (24.5% x 600 J = 147 J heat that used to do work).
The actual efficiency of this machine is 17% (only 100 J heat are used to do work). Still around 147 J – 100 J = 47 J heat that can be used to do work.
Question 3 :
A machine receives 1000 Joule calories and produces 400 Joules of work on each cycle. This machine works between 500 oC and 200 oC. Calculate the actual efficiency and ideal efficiency of this machine.
TH (high temperature) = 500 oC — 500 + 273 = 773 K
TL (low temperature) = 200 oC — 200 + 273 = 473 K
QH = 1000 J
QL = 400 J
The ideal efficiency of this machine:
Carnot Efficiency = 40%. The actual engine efficiency = 60%. This machine does not exist. The efficiency of a machine could not have been greater than Carnot’s efficiency.
Question 4 :
For Carnot engine efficiency to reach 100% (1), what is the required discharge temperature (TL)?
For perfect engine efficiency to reach 100% (all heat can be used to do work), then low temperature (TL) should be 0 Kelvin.
Reach 0 Kelvin’s temperature is impossible. This statement is the third law of thermodynamics.
Since reach 0 Kelvin is not possible, then perfect heat engine may not have 100% efficiency. Since 100% efficiency cannot be achieved by the heat engine, can be concluded that is not possible to use all the heat (QH) to do work. There must be some wasted heat (QL).
There can be no heat engine (which works in a cycle) that can turn all the heat into work (Second law of thermodynamics, Kelvin-Planck statement).
This graph shows the isothermal process (isothermal expansion) occurring in one stage only. In this process, all heat (Q) can be converted into work (W). The amount of work done = shaded area.
Heat engine must work continuously (process must occur repeatedly, can not happen in just one step). For example an alternating type steam engine. The piston on an alternating steam engine must move to the right and left continuously so wheels can spin (can be used to move something). Wheels can not rotate when the piston moves to stop right then (process happens in one stage only). If the process occurs repeatedly, it is impossible that all heat can be turned into work (Kelvin-Planck statement).
The graph on the left shows an isothermal expansion (down arrow) and isothermal press (up arrow). The process occurs continuously isothermally (No work is produced). The graph on the right is an isothermal expansion process (down arrow), isobaric pressing process (arrow to the left) and isochoric process (up arrow). From these two graphs, it appears that for a continuous process, there is always a wasted heat.