A Practical Approach to the Carnot Efficiency

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1 Paper ID #9198 A Practical Approach to the Carnot Efficiency Dr. David C Zietlow, Bradley University Professor of Mechanical Engineering c American Society for Engineering Education, 2014 Page

2 Background Practical Approach to the Carnot Efficiency The Carnot cycle is a valuable tool to help students and practitioners understand the theoretical limits of heat engines, refrigeration cycles and heat pumps. Presented here is a practical approach to developing the Carnot efficiency for a heat engine and coefficient of performance for refrigeration and heat pump systems. This practical approach helps students connect actual hardware with the ideal cycle. The material presented here is designed for first semester thermodynamic students before they take a course in heat transfer. Although the course is open to all engineers the vast majority of the students are mechanical engineers. The prerequisite knowledge of the students is algebra and an introduction to conduction and convection heat transfer along with the corresponding constitutive equations. Providing students with an application is a significant aid to their understanding of theory. In one study the authors discovered 68 percent of second-year students tend to agree that having fun activities, interesting assignments, and applying the material to the real world make a course engaging. (1) Another study found that student participation in laboratory or demonstration exercises in conjunction with a lecture produces a more positive learning outcome. (2) Application and demonstration are two key tools of an educator and will be explained further in this paper so students will be able to comprehend the abstract ideas of irreversibility and Carnot efficiency. This paper ties the theoretical Carnot efficiency to the applications to which it is related. One of the more challenging concepts for students is the irreversibility of heat transfer through a finite temperature difference. It is easier to understand the irreversibility of other processes such as friction. To lead into the concept of irreversible heat transfer, a multi-sensory and multimedia demonstration is conducted. Easily understood irreversible processes in the production of tea are used; from the harvesting of the tea plant to the making of a hot cup of tea. This natural progression helps with the more abstract concept of what is necessary to restore the hot tea to its original condition once it cools. It is noted that as the temperature difference between the tea and the room increases it becomes more challenging (greater irreversibility) to restore the temperature of the tea to its original condition. Assuming the students have been exposed to the three basic modes of heat transfer, and the associated constitutive equations, it is possible to make the connection with the conductance form of the heat exchanger equation. At this point in their education, they may be able to understand a qualitative explanation of the conductance form (UA) of the heat exchanger equation. Qualitatively, the conductance represents the combination of Newton s Law of Cooling for the convective heat transfer through the boundary layers of the fluids and Fourier s Law for the conduction through the material used in the heat transfer surface. With the conductance form, the students are equipped to relate the size of the heat exchanger to the temperature difference between the two fluids. Since the Carnot Cycle is ideal, the irreversibility is zero. No irreversibility means the temperature difference driving the heat transfer is zero. For a finite rate of heat transfer, no temperature difference can only happen when the heat transfer surface area is infinite. Therefore, maximizing the efficiency (i.e. Carnot efficiency) of a thermal system is not a realistic objective function. This paper will take you through the process of how Page

3 to present this approach to students. It will also show you how much of this information students retain into the next semester. Irreversible Heat Transfer In thermodynamics, it is critical students understand the abstract terms, heat and irreversibility. The learning process is facilitated when concrete concepts are connected with more abstract concepts. It is useful to begin the discussion of irreversibility with concrete examples that most students comprehend. Therefore, it is recommended to lead students through the tea making process. Begin with cutting the green tea plant from its roots. Next, ask the students to consider the irreversible nature of this process. The plant cannot be reattached at this point; it has died. To make this a multi-sensory and multi-media presentation it is helpful to grow your own tea plants. Spearmint or peppermint is most effective because of the strong aroma produced when a leaf is rubbed between the fingers. The next irreversible step is to dry the plant. Here the moisture has left the leaves and cannot be returned to the plant cells. The color has changed from green to brown and cannot be restored. Dry some of the tea plants by hanging them in a dry place for a few weeks. The results will be an effective visual aid. After drying, comes separation of the leaves from the stems. (The use of a document camera will immensely aid the presentation of this process.) The leaves cannot be reattached without significant input from outside the system (e.g. a human with superglue). Even then, the dried leaves are not restored to their original condition. After the separation from the stems, the leaves are crushed into small pieces. Illustrate this under a document camera to amplify the futility of restoring the leaves to their original condition. Next, a tea bag emerges from the pocket of the instructor and is dropped into a clear glass mug where the diffusion process is easily observed. Note the color and flavor of the tea has left the tea leaves irreversibly. For the motivated teacher who would like to expand the multi-sensory experience to the students senses of smell and taste, a thermos of hot tea and disposable cups are a pleasant way to drive the point home. The irreversibility of absorbing the tea into the human body can be discussed as well. Now with the presentation of several concrete examples of irreversible processes, the more abstract concept of heat can be addressed as shown in Figure 1. T H T L Time (t) Figure 1. Cooling a Cup of Tea Page

4 The cup of tea begins with a highh initial temperature (T H ) and cools towards the ambient temperature (T L ) as the heat flows from the high temperature to the low temperature. The students are asked how this cup of tea can be restored to its original condition. Whether the answer is a microwave, electric heating element, or a flame, the students need to consider how the surroundings are affected. For the microwave and electric heating element the students need to understand the abstract concept of the heat engine (power plant) that is required to supply electrical power to the outlet to which the appliance is connected. T H Q H W net Boiler Pump Turbine Condenser T L Q L Figure 2. Schematic of a Heat Engine The heat engine, like a flame, requires the consumption of a fuel (e.g. gas, oil, coal or wood) which is highly irreversible. To restore the gas, oil or coal to its original state requires biomass and several months at high temperatures and pressures. Restoring the wood requires a seed, sun, soil (fertilized by the ashes) and decades of time. Taking students through these processes, Page

5 clearly and concretely, demonstrates the irreversibility of the heat transfer process. It should also be emphasized that the greater the temperature difference, the greater the irreversibility. Heat Exchangers Unfortunately, heat exchangers are not adequately addressed in most courses of thermodynamics. In most problems, heat transfer is either a given input or is a dependent variable where the conservation of energy is used to solve for this elusive quantity. In this section, a basic understanding of heat exchangers (see Figure 3) is presented. A triangle is used to represent one-side of a heat exchanger. The hypotenuse of the triangle represents the heat transfer surface area while the sides allow for the entering and exiting of the heat transfer fluid. T L Q T H Figure 3. Schematic of a Heat Exchanger Assuming the students have already been exposed to the three modes of heat transfer and their corresponding constitutive equations, the conductance form of the heat exchanger equation (Equation 1) is not difficult for them to grasp in a qualitative sense. For the ambitious instructor, an optional lecture on the development of this equation using Ohm s analogy and a mathematical relationship will give them a firmer understanding of the conductance equation. ( ) Q = UA T T H L (1) The rate of heat transfer,, equals the product of the conductance of the heat exchanger, UA, and the driving potential or the temperature difference between the high-tempera T, ) ature fluid ( ) and the low-temperature fluid ( T L. Qualitatively, the conductance represents a combination of convective heat transfer of the two fluids and conduction through the wall of the heat transfer H Page

6 surface area. For a given application, the key benefit of this equation is it establishes a relationship between the size of the heat exchanger (A) and the temperature difference between the two fluids. The higher the temperature difference the greater the irreversibility. The greater the irreversibility the lower the efficiency and the higher the operating costs for a cycle in a given application. Heat Engines The state points from the heat engine shown in Figure 2 are plotted in Figure 4 on pressure versus specific enthalpy coordinates. The first state point is the inlet to the pump. Each subsequent state-point is numbered sequentially at the inlet to the next component of the heat engine as you follow the working fluid through the cycle. For this particular application, the heat engine cycle is receiving heat from a high temperature thermal reservoir at 370 C and is rejecting heat to a low temperature thermal reservoir at 15 C. The condenser is sized to allow a condensing temperature of 35 C while the boiler is sized to allow a saturation temperature of 320 C Water T H = 370 C T sat,b =320 [C] P [kpa] T sat,c =35 [C] 10 0 T L = 15 C ,000 1,500 2,000 2,500 3,000 3,500 h [kj/kg] Figure 4: Property Plot for the Heat Engine Cycle Next, the definition of efficiency is applied to the heat engine cycle. This produces Equation 2. (2) Page

7 Here the desired result is the net power output from the turbine. The required input is the rate of heat transfer received by the boiler from the high temperature thermal reservoir. Since reversible heat transfer can be calculated with the conductance form of the heat transfer equation it would be useful to replace the net work output with equivalent heat transfer rates. To determine the equivalent rates apply the conservation of energy to the entire heat engine cycle. At steady-state conditions, the rate of change of energy stored within the heat engine cycle is zero, so the rate of energy in equals the rate of energy out. This is shown in Equation 3. (3) Solving for net power output and substituting into Equation 2 yields Equation 4. (4) This expression of efficiency works for all heat engine cycles, real and ideal. If the flow rate of one of the heat transfer fluids for each heat exchanger and the corresponding state-points at both the inlet and outlet of the heat exchanger were known for both the condenser and boiler then the actual efficiency could be determined. Next, the conductance equation will be applied to the two heat exchangers. For the boiler, focusing on the phase change region of the heat exchange and an high mass flow rate of combustion air, the conductance equation can be used to calculate the heat transfer rate as seen in Equation 5., (5) For a given application and set of components, the heat transfer rate is constant and known, along with the flow rates, as well as the heat exchanger material, which fixes U. The flame temperature will fix the temperature of the high temperature thermal reservoir (T H ). As the heat transfer surface area increases the saturation temperature of the water (T sat,b ) in the boiler approaches the temperature of the high temperature thermal reservoir. This decreases the irreversibility of the heat transfer to zero when the area reaches infinity. For the condenser, the conductance equation can be used to determine the heat transfer in the condensing regions of the heat exchanger, when the cooling water flow rate is high. Equation 6 is the result., (6) The same arguments that were used for the boiler can be used for the condenser to fix the heat transfer rate and U. Since the location of the heat engine is fixed for a given application, the temperature of the low temperature thermal reservoir (T L ) is fixed by temperature of the water in the lake, river or cooling tower. As the heat transfer surface area increases, the saturation temperature of the water in the condenser (T sat,c ) approaches the temperature of the low Page

8 temperature thermal reservoir. Similar to the boiler, this decreases the irreversibility of the heat transfer to zero when the area reaches infinity. For the ideal case, an argument is made using mathematical reasoning that the ratio of heat transfer rates is only a function of the ratio of the thermal reservoirs. This reasoning is currently presented in most traditional thermodynamic textbooks. Technically you would need to divide both the numerator and denominator of Equation 4 by the rate of heat transferred from the high temperature thermal reservoir to obtain a ratio of the heat transfer rates. This ratio can then be replaced by the ratio of temperatures for the thermal reservoirs. Practically though, for the Carnot Cycle, where the heat transfer surfaces are infinite, and the irreversibility of the cycle is zero the heat transfer rates can be replaced by the temperature of the corresponding thermal reservoir. This yields Equation 7. TH T ηi = T H L (7) Refrigeration Cycles The basic refrigeration cycle is composed of four primary components as shown in Figure 5. The compressor, condenser, expansion device and evaporator are the primary components. Since the evaporator transfers more heat than the work required to operate the compressor, a new term is necessary to replace the efficiency of this cycle. Therefore, instead of Carnot efficiency the term Carnot coefficient of performance is used to assess the performance of refrigeration and heat pump cycles. Q H Condenser W Q L Compressor Evaporator Figure 5. Schematic of the Refrigeration Cycle The state-points are plotted in Figure 6 on pressure versus specific enthalpy (P-h) coordinates. For this particular application, the refrigeration cycle is cooling a building wheree the indoor environment is being maintained at T L =20 C in a climate where the outdoor temperature T H is 40 C. The condenser is sized to allow a condensing saturation temperature (T sat t,c) of 45 C while the evaporator is sized to allow an evaporating saturation temperature (T sat t,e) of 12.5 C. Page

9 2 x 10 3 R134a Τc Isentropic s=.9 Isotherms 3 T sat,c =45 C 2s P [kpa] 10 3 T L =20 C T H =40 C 4 T sat,e =12.5 C 1 Τe 2 x Figure 6. Property plot of a refrigeration cycle h [kj/kg] Next the definition of Coefficient of Performance is applied to the refrigeration cycle. This produces Equation 8. (8) Here the desired result is the rate of heat transfer removed from the low temperature reservoir. The required input is the power into the compressor. Since reversible heat transfer can be determined with the conductance form of the heat transfer equation, it would be useful to replace the compressor work with equivalent heat transfer rates. Applying the conservation of energy to the refrigeration cycle will accomplish this relationship. At steady-state conditions the rate of change of energy stored within the refrigeration cycle is zero so the rate of energy in equals the rate of energy out. This is shown in Equation 9. (9) Solving for compressor power and substituting into Equation 8 yields Equation 10. Page

10 (10) Next, the conductance equation will be applied to the two heat exchangers. For the evaporator, the conductance equation yields Equation 11. The conductance equation assumes no superheat of the refrigerant at the outlet, and an high mass flow rate of air., (11) For a given application, the heat transfer rate is specified and constant. The flow rates are known and fixed as well as the heat exchanger material which fixes U. The thermostat setting will fix the temperature of the low temperature thermal reservoir. As the heat transfer surface area increases, the saturation temperature of the refrigerant in the evaporator approaches the temperature of the low temperature thermal reservoir. This decreases the irreversibility of the heat transfer to zero as the area approaches infinity. For the condenser, the conductance equation is used to determine the heat transfer in the condensing region alone. This assumes the refrigerant is a saturated vapor at the inlet and a saturated liquid at the outlet and the air flow rate is high. These assumptions are valid if the condenser is divided into three separate heat exchangers, one for de-superheating, one for condensing and one for sub-cooling the refrigerant. Equation 12 is the result of applying the conductance equation to the condensing region., (12) The same arguments can be used to fix the heat transfer rate and U. Since the location is fixed for a given application, the temperature of the high temperature thermal reservoir is fixed by the climate. As the heat transfer surface area increases, the saturation temperature of the refrigerant in the condenser approaches the temperature of the high temperature thermal reservoir. This decreases the irreversibility of the heat transfer to zero as the area approaches infinity. Replacing the heat transfer rates, with the temperature of the corresponding thermal reservoir, yields Equation 13. (13) Heat Pumps Heat pumps use the same cycle as the refrigeration system. The main difference is the desired result of the system is the heat rejected by the condenser rather than the heat removed by the evaporator. This affects the numerator of the COP as given in Equation 14. (14) Page

11 The application of the heat transfer equation to the heat exchangers is the same as the refrigeration cycle. In the ideal case, the heat transfer rates are replaced by the temperature of the corresponding thermal reservoir. (15) Conclusions The multi-media and multi-sensory demonstration of the irreversible processes in the production of tea is a excellent analogy to help students understand irreversible heat transfer. This practical approach to developing the Carnot Cycle was presented to first semester students of thermodynamics on October 22, They were tested, without advance warning, over four months later on February 25, On the test they were asked to derive the Carnot efficiency for a heat engine. Out of forty students 82% were able to apply the definition of efficiency to the cycle, and 75% successfully derived the expression for the Carnot efficiency as a function of the temperatures of the thermal reservoirs. This approach is an effective way for students to understand energy systems and determine the maximum possible performance of these systems. Instructors are encouraged to adopt this practical approach to Carnot efficiency. Expanding the coverage of heat exchangers from the one-sided presentation in traditional thermodynamics to the other side of the exchanger leads to two key concepts. First, it exposes the students to the conductance form of the heat transfer equation. Second, it allows a graphical representation of the irreversibility on the pressure-enthalpy diagram. The conductance equation not only helps students understand the development of the Carnot Cycle but also relates the tradeoffs necessary to optimize energy systems. First, they realize using the efficiency or COP as an objective function leads to unrealistic optimums as the heat exchangers grow to infinity. Second, the conductance form of the heat exchanger equation relates the variables associated with the initial (heat transfer surface area) with the operating costs which are affected by the irreversibilities in the cycle. The higher the irreversibilities the less efficient the cycle and the more energy required for a given application. References 1. Student and Faculty Perception of Engagement in Engineering. Heller, Rachelle S., Beil, Cheryl and Dam, Kim. 3, s.l. : Journal of Engineering Education, 2010, Vol. 99, pp Embedding Laboratory Experiences in Lectures. Morgan, James R., Barroso, Luciana R. and Simpson, Nancy. 4, s.l. : Advances in Engineering Education, 2009, Vol. 1, pp Page

12 Acknowledgements First, I would like to thank my wife, Dixie, Ph.D., for the added load in parenting she has taken. It has permitted me time to write. Next, I appreciate the academic freedom the chair of the Department of Mechanical Engineering, Dr. Desh Paul Mehta, has allowed me. This freedom gave me the opportunities to test these ideas in the crucible of the classroom. Last, I would like to thank my student, Jack Stein, who polished the figures for me so they were consistent in font and format. Page

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