Module 3, Lesson 2 Efficiency of the Carnot Cycle at Maximum Power Output Objective: Be the end of this lesson you will be able to identify and describe some of the basic thermodynamic processes. To facilitate this we will look at the individual steps of the Carnot process, which as we learned last module has the best possible efficiency for an engine. We will use this to introduce a new expression for efficiency for a Carnot cycle running at maximum power that was derived by two UBC physics faculty in the 1970 s. Introduction In the previous module we described how the second law of thermodynamics limits the maximum possible efficiency a heat engine can obtain, which has the form, The specific engine that achieves this maximum efficiency is called Carnot engine. The thermodynamic cycle of the engine, shown in Figure 1, exists entirely on adiabats and isotherms. A process that follows an adiabat has zero heat transfer between the system and the surrounding. A process that follows an isotherm, is one in which the temperature of the working medium doesn t change.
Figure 1. This plots shows the four steps of the Carnot Cycle, which follows adiabats and! isotherms to complete a cycle. An adiabatic process is one in which the heat transfer is! zero, Q=0, while an isothermal process is one in which all the heat transferred is equal to! the work done during the process, Q=W. Review of Thermodynamic Processes The first law of thermodynamics is the statement that the change in internal energy of a system U is equal to the heat supplied to the system Q plus the work done on the system W, The sign of the work term is often the cause of some confusion. Many texts write the first law in terms of the work done by the system, which results in a negative sign. You can think of work either way, but be sure you get the sign right! The two thermodynamics processes involved in the Carnot cycle are called isothermal and adiabatic expansion and compression. We phrase the processes in terms of expansion and compression because the working medium (the stuff that does the work) of the engine is usually a gas that expands or is being compressed. Because of this we will use the ideal gas law for many of our manipulations,
For any ideal gas process the change in internal energy can be also be related to the specific heat of the working medium CV and the change in temperature T, We are now ready to talk about some basic thermodynamic processes. Adiabatic We ll first look at adiabatic processes. Adiabatic processes are ones in which no heat is exchanged Q=0. In terms of the first law this means that U=W, or the change in the internal energy of the system is equal to the work done on or by the system. The work done by the system is simply given by An example of and adiabatic process is deflating a tire by releasing the valve. The air expands quickly as it leaves the tire and cools quite dramatically. This process happens quickly, before any heat transfer can occur and the work done by the system to expand the gas drops the temperate of the system. Another example is the expansion of hot air as it rises in the atmosphere. The path that an adiabatic process follows on a PV diagram is called and adiabat. The specific curve is described by where γ is the adiabatic constant and takes values 5/3 for a monatomic gas and 7/3 for a diatomic gas. The derivation of this formula isn t critical, and could be found in most textbooks. In Figure 1 the paths 2->3 and 4->1 are the adiabats. Because this product is constant we can simply write Isothermal The second type of process in the Carnot cycle is the isothermal compression and expansion. For these processes the working medium doesn t change temperature. For an ideal gas this means that the change in internal energy is zero, which means that The work done by the system (the negative in front of W indicates it is the system doing work) is equal to the heat that enters the system. The system does work by extracting heat from the surroundings. This is in stark contrast to an adiabatic process where the temperature change of the system is what contributes to the work and there is no heat transfer.
The path that an isothermal process follows on a PV diagram is called an isotherm. From the ideal gas law we can see that for a gas to remain at the same temperature the product nrt must be constant. On a PV diagram the curves look like which looks similar to the curve that adiabatic processes follow, but with out the adiabatic constant. In Figure 1 the paths 1->2 and 3->4 are isotherms. Question: Plotted on the graph below are an isothermal process and an adiabatic process. Each takes 0.10 mols of nitrogen gas from V1 to V2. 1) Which curve is the isothermal process? A 2) Which curve transfers more heat? B 3) Which curve does more work? A 4) If both processes start at P=3 Pa and V = 1 m^3 and end at 3 m^3, what is the... Isochoric and Isobaric There are more processes than the two used in the heat engine. Two of the simplest are the isochoric and isobaric processes. An isochoric process is one in which the volume doesn t change and an isobaric process is one in which the pressure remains constant. As seen in Figure 2, only the isobaric process has a change in volume, and thus is the only one that does work. The work done by it is simply
Figure 2. An illustration of isochoric and isobaric processes. Notice that only the isobaric! process has area under its curve (in blue), so it s the only one of the two that does work. Question: Label the following processes as adiabatic (A), isothermal (T), isobaric (P), or isochoric (V). 1) the melting of an ice cube (phase transition): T, V 2) a piston being slowly compressed: T 3) a piston being quickly compressed: A 4) blowing up a balloon: P 5) letting go of a tennis ball at the bottom of a pool: V The Carnot Engine Isn t So Useful One might assume that because of their efficiency all engines are Carnot engines -- why would you make anything but the most efficient engine. It turns out that the Carnot engine only achieves the theoretical maximum efficiency if the process is truly isothermal, which means that the working substance (the gas inside the engine), is in thermal equilibrium with the heat reservoirs. For this happen the process must be carried out very slowly, infinitely slowly in fact. Engines are great because they do work for us, but only if they do that work in a reasonable time. We want engines with power. The Carnot engine, efficient as it is, doesn t produce power -- it takes an infinite amount of time to do a finite amount of work.
If we change the cycle so it doesn t take an infinite amount of time to perform the isothermal part s, the process the engine will produce power, but not at the maximum theoretical efficiency. In fact, we could perform the isothermal parts of the process so quickly that no heat would be transferred from either the hot or cold reservoirs to the working substance. This engine would take a finite amount of time to run through a cycle, but it would produce no work, which leaves us again with an engine that produces zero power. There must be a happy medium, which leads us to our question: what is the efficiency of a Carnot cycle run at maximum power? This question was addressed in the paper Efficiency of a Carnot Engine at Maximum Power Output by Curzon and Ahlborn. Both are emeritus Professors of the UBC physics department, and for a while this was the most cited paper out of the department. The success of the paper is two-fold. The formula they derived is simple, as elegant as the original expression for the Carnot efficiency, In addition to its compact form, the formula also matched the best observed efficiency of many operational power plants. Question: The CANDU nuclear reactor has a hot reservoir of 300 C and and cold reservoir of about 25 C. Calculate the Carnot efficiency and the Curzen-Ahlborn efficiency of of the reactor. Look up the operating efficiency of the reactor in the Curzen-Ahlborn paper. Is it close? Answer: Carnot: 0.48 CA: 0.28. Looks close to me. Summary