Understanding the Carnot Cycle: Maximum Heat Engine Efficiency Explained

[00:00:00]A lot of the time, a key question that comes  up for engineers is how efficient something is. Thinking about just heat engines, what  is the maximum efficiency a heat engine can have? We know it can’t be 100 percent  from the second law of thermodynamics, so what is the maximum we can get out of  one? To figure that out, we need to think about reversable and irreversible processes. If we have a process that can be reversed, in other words, back to the original state,  without leaving any trace on the surroundings, then that’s a reversible process. In simple  terms, the system and the surroundings must come back to their initial states for a process  to be reversible. Anything that doesn’t check off both of those requirements are irreversible  processes. Interesting to note that every single process that happens in nature are irreversible  processes. This is due to a bunch of factors called irreversibilities. A few of them include  friction, electric resistance, chemical reactions, inelastic deformations and more. Any single  irreversibility factor will cause a process to be irreversible. Be careful to not misunderstand  that a system can come back to its original state whether a process is irreversible or reversible,  it’s just in irreversible processes, surroundings will do some work on the system to bring it  back to its original state. By thinking about idealized reversible processes, we can figure out  the theoretical maximums for different systems.

[00:01:35]Remember that for a heat-engine cycle, the net  work can be maximized by processes that require the least amount of work and deliver the most.  This can be done by using a reversible process.

[00:01:48]In fact, the most efficient cycles  are always reversible cycles. Now we know we can’t actually achieve a reversible  cycle in practice due to irreversibilities, but we can compare our existing heat engines to  this hypothetical reversible heat engine. One such reversible cycle is called the Carnot cycle  proposed by a French engineer named Sadi Carnot.

[00:02:13]A heat engine that operates on the Carnot cycle  is called a Carnot heat engine. It has four reversible processes and it can work in closed or  steady flow systems. Let’s go through this cycle.

[00:02:27]Imagine we have a closed system with  a gas inside a piston cylinder device.

[00:02:33]The first step of the cycle is the  reversible Isothermal expansion.

[00:02:38]Let’s label the temperature of the gas TH. The  cylinder head, which is this part right here, is close in contact with an energy source.  This source also has a temperature of TH. So, in simple terms, the temperature of the gas and  the energy source temperature is the same. Now the gas inside is allowed to expand. When the  gas expands, it does work on the surroundings, so here, the piston moves to the right. As the  gas is expanding, the temperature of the gas begins to decrease but heat is transferred from  our energy source back into the gas. This means the gas temperature stays constant at TH.  Because the temperature difference between our energy source and the gas never changes more  than an infinitesimal amount, this is a reversible heat transfer process. The total amount of heat  transferred during this step; we can label QH.

[00:03:37]The second step is called the Reversible Adiabatic  Expansion. At this step, we remove the energy source and replace it with insulation. The system  then becomes adiabatic. Adiabatic just means there is no heat transfer to or from the system to  the surroundings. The gas still expands and drops in temperature. We will label the new  temperature TL. In this hypothetical cycle, there is no friction between the piston  and cylinder, so the process is reversible.

[00:04:09]The third step is called the Reversible Isothermal  Compression. Now the insulation is removed and the cylinder is now in contact with an energy sink  at temperature TL. In simple terms, the sink we place is at the same temperature as the gas.  Now an external force pushes on the piston, moving it to the left. This causes the gas to  compress, and the temperature begins to rise, but as soon as it’s rising, the heat is transferred  to the energy sink. So the temperature of the gas remains constant at TL. Because the temperature  difference between our energy sink and the gas never changes more than an infinitesimal amount,  this is a reversible heat transfer process. The total amount of heat rejected from the  gas during this step, we can label QL.

[00:04:59]The fourth and last step is called the Reversible  Adiabatic Compression. Now the low energy sink is removed and insulation is placed back on the  cylinder head. The gas continues to be compressed and it goes back to its initial state. In  other words, the temperature rises from TL to TH which means the cycle is now complete. Let’s look at a pressure volume diagram of a Carnot cycle. Starting from step one to two, the  pressure drops and volume increases as the gas expands. The temperature stays constant at TH.  For the temperature to stay constant at TH, heat needs to be transferred so we have QH which is  the heat transferred from our energy source.

[00:05:43]From step 2 to 3, the pressure drops more and the  volume continues to increase. Because our cylinder is now insulated, the temperature drops to TL.  From step 3 to 4, the piston is pushed inwards, so the gas is compressed. The pressure increases  and volume decreases. There is an energy sink so the temperature is kept constant at TL. For that  to happen, heat needs to be rejected into the energy sink, so that’s QL. From step 4 to step  1, we insulate the cylinder while the piston keeps pushing inwards. The pressure continues to  increase and volume continues to decrease arriving at the same state as step 1 and the temperature  rises to TH. The area inside the path 1,2,3,4,1 shows us the net work done during the cycle. Keep in mind that the Carnot heat-engine cycle is a reversible cycle, so every step can be  reversed, which would result in the Carnot refrigeration cycle. Here is the pressure  volume diagram for a reversed Carnot cycle.

[00:06:50]The biggest difference lies in the direction  along with QL being the heat absorbed from a low-temperature reservoir and QH being the  heat rejected to a high-temperature reservoir.

[00:07:02]Now let’s dive into the efficiency of a Carnot  engine. In the previous heat engines video, we talked about this equation which gives  us the efficiency of any heat engine. Here, QH is the heat transferred to the engine from  a high temperature reservoir and QL is the heat rejected to a low temperature reservoir. When it comes to reversible heat engines, like the Carnot engine, we can go one step further  and write our efficiency equation like this.

[00:07:33]Remember that TH is the temperature of the  high temperature energy source and TL is the temperature of the energy sink. This relation is  called the Carnot efficiency. In simple terms, this equation says the efficiency of reversible  engines is a function of reservoir temperature only. You should note that TL and TH must be in  kelvin for this equation to work because they are absolute temperatures. Do not use Celsius or  Fahrenheit when solving problems, or you will end up with incorrect answers. A relationship between  our Carnot efficiency equation and the normal efficiency equation can be written like this  for reversible heat engines. Again, the values must be in kelvin for this equation to work. If we compare the thermal efficiency equation to the Carnot efficiency equation, we can  determine what type of heat engine it is.

[00:08:27]If the thermal efficiency is less than the Carnot  efficiency, then it’s an irreversible heat engine.

[00:08:33]If it’s equal to the Carnot efficiency, then it’s  a reversible heat engine, and if it’s greater, then it’s an impossible heat engine. Now let’s go through some problems to see how we can apply what we learned. Let’s take a look at this question where we have a Carnot heat engine that receives  heat from a source where we don’t know the temperature. We need to figure out the temperature  of the source and the thermal efficiency of the heat engine. Let’s start by writing down what  we know. The engine receives 650 kJ of energy, and rejects 250 kJ. So that’s our QH and  QL. The sink where heat is rejected to is at a temperature of 24 degrees Celsius. We can use this equation to figure out the temperature of the energy source  since this is a Carnot engine. Let’s plug in the values we know. Solving gives  us the temperature of the energy source.

[00:09:28]We can find the efficiency using the Carnot  efficiency equation. And those are our answers.

[00:09:39]In this problem, we have a heat engine with  heat supplied to it at a steady rate. We need to figure out the maximum power output.  Let’s write down what we know. The source temperature is 477 degrees Celsius and the  sink temperature is 25 degrees Celsius.

[00:09:56]Heat is supplied at a rate of 65000 kJ/min. We have TH and TL, so we can figure out the efficiency using the Carnot efficiency equation. To figure out the maximum power output, we need to use this equation. You should remember this from  the heat engines video. Here, our Qin is the heat supplied. Let’s rearrange the equation and write  it in time rate form. Now we can plug our values in and solving tells us the maximum power output. Let’s take a look at this problem where we have a heat engine that does maximum work equal to 500  kJ. We need to figure out the heat supplied, the heat rejected, and the temperature of  the heat sink. Let’s start by writing down what we know. The heat engine receives heat  from a source at 1200 degrees Celsius. The efficiency of the engine is 40 percent. The  maximum work the heat engine does is 500 kJ.

[00:11:02]We can first figure out the heat supplied  using this equation. Qin is our QH, which is the heat supplied. To figure out QL, remember that net work is equal to QH minus QL.  Solving gives us the heat rejected.

[00:11:20]To figure out the temperature of the heat sink,  so that’s TL, we can use this equation. Let’s plug in our values. Solving gives us our answers. That should cover the types of problems you will face when it comes to Carnot heat engines. Thanks  for watching and best wishes with your studies.