Chapter 9: Applications of the Laws of hermodynamics Goals of Period 9 Section 9.1: Section 9.2: Section 9.3: o review the first law of thermodynamics o discuss heat engines and their efficiency o discuss air conditioners, refrigerators, and heat pumps and their efficiency 9.1 Conservation of Energy and the First Law of hermodynamics In Chapter 8 we applied the law of conservation of energy to thermodynamic systems: a change in internal energy of a system U equals the heat added to or subtracted from the system minus the work done by the system. Equation 9.1 describes this relationship. he law of conservation of energy tells us that the total amount of energy in this system does not change. U (Equation 9.1) U = total internal energy (thermal energy) (joules) U = the change in internal energy (joules) = heat added to or subtracted from the system (joules) = work done by the system (joules) A system moving towards equilibrium has the ability to do mechanical work. As long as the system is not in equilibrium with its environment, the entropy it has compared to the entropy it will have when it reaches equilibrium is a measure of the amount of work that can be obtained from the system as it goes to equilibrium. As shown in equation 9.1, not all of the energy can be recovered as work. In Chapter 9, we consider applications of the first law of thermodynamics to heat engines, heat pumps, refrigerators, and air conditioners. 9.2 eat Engines As mentioned previously, work can be obtained from a system as it progresses from a non-equilibrium to an equilibrium situation. eat engines and motors are examples of devices that utilize this principle. Both are used to convert other forms of energy into kinetic energy. In an engine, thermal energy at a high temperature is produced by the combustion of some sort of fuel, and then transformed into kinetic energy. In a heat engine, thermal energy from the combustion process is transferred to what is called the working substance. For example, coal may be burned to produce steam in the boiler of a steam engine; the steam is the working substance. he steam expanding in the cylinders of the steam engine moves the pistons to produce kinetic energy. 73
In a motor, the primary transformation process is the direct conversion of some other form of energy into kinetic energy. In an electric motor, for example, electrical energy is converted directly into kinetic energy. An electric motor will not run without voltage across its terminals, and a steam engine will not operate if the steam reaching the pistons has the same temperature as the external environment. A system will run only as long as a non-equilibrium situation exists. Figure 9.1 illustrates a heat engine that runs because the system is not in equilibrium. Figure 9.1: A eat Engine eat engines may be classified as either internal or external combustion engines. An aircraft gas turbine engine and the automobile gasoline engine are examples of internal combustion engines, engines in which the fuel is actually burned inside the engine. he steam turbine engine is an example of an external combustion engine. An external combustion engine is an engine in which the fuel is burned outside the engine in a furnace. he furnace heats water in a boiler producing steam to run the steam turbine. Ideal eat Engines and heir Efficiency Just how efficient can we hope to make a device? his question has great importance in view of our limited fuel reserves. e clearly wish to attempt to use these reserves in an efficient manner. Engineers must concern themselves with calculating efficiencies of processes that are very complex or performing an experiment in order to find the efficiency. For heat engines the second law of thermodynamics provides an absolute upper limit to the efficiency. he efficiency of an actual heat engine is the ratio of the work obtained from the engine divided by the heat energy of the system as shown in Equation 9.2. 74
Eff (Equation 9.2) = work obtained from the system (joules) = heat energy of the system (joules) In an ideal situation, all of the heat energy would be converted into work. Since some energy is wasted in every process, the work obtained is the difference in the high and low heat energies, that is = - L. Equation 9.2 can be written as Eff L (Equation 9.3) In an ideal situation the heat transferred is proportional to the temperature. For an ideal heat engine, that is, one that operates with reversible processes, the maximum efficiency is given by Equation 9.4. Eff (Equation 9.4) = the high temperature (in Kelvin) of the working substance (water, steam, hot gas) reaching the piston or turbine after combustion in the engine L = the low temperature (in Kelvin) of the working substance leaving the engine. L he efficiency of even an ideal heat engine always results in a number less than one because we cannot reach absolute zero for the low temperature. Of course, no such engine exists or can exist, but it does provide an upper limit on what we can expect. (Example 9.1) hat is the efficiency of an ideal engine that uses working substances at 400 K and 300 K? Eff L = 400 K 300K = 100 K = 0.25 = 25% 400 K 400 K Steam turbines are used in electric power plants to provide the mechanical energy required to run the electric generators. ypically the steam turbines are run with an efficiency of about 50% for converting thermal energy to mechanical energy. o obtain the overall efficiency of the power plant, we must multiply this 50% by the 75
efficiencies of the other energy conversions taking place in the chain from fuel to electricity. Modern boilers can convert about 88% of the chemical energy of the fuel to thermal energy of the working substance. Electric generators can convert up to 99% of the mechanical energy produced by the steam turbine into electricity. he overall efficiency of a device is equal to the product of the efficiencies of each energy conversion step. hus the overall efficiency is less than the least efficient process. (Example 9.2) hat is the overall efficiency of the process described in the paragraph above for generating electricity? Eff overall = Eff 1 x Eff 2 x Eff 3 = 0.50 x 0.88 x 0.99 = 0.44 = 44% In the U.S. nuclear power plants operate at a lower efficiency than 44% because the present reactors cannot, for reasons of safety, run at temperatures as high as those used with conventional boilers. For the complete cycle of conversion from the stored energy in the nuclear fuel to electrical energy, the nuclear power reactors now in use have an efficiency of about 30%. 9.3 Refrigerators, Air Conditioners, and eat Pumps Refrigerators and Air Conditioners e next consider devices that use energy to transfer thermal energy from a cold body to a hot body. A window air conditioner is an example of such a device. he air conditioner uses electrical energy to pump heat from a room and exhausts that heat into the outside air. A refrigerator pumps heat from inside the refrigerator into the room. An input of electrical energy is required to move heat from a cool environment inside the refrigerator into a warmer environment into the room. Air conditioners and refrigerators make use of expansion cooling. hen a fluid, such as freon, is compressed to a higher pressure and the fluid is allowed to expand, the result is a cooling of the fluid. Cooling occurs in the air conditioner coils when the pressure on the freon (or other liquid) is quickly reduced as the freon is allowed to expand through a small opening. As the liquid freon expands, it changes phase from a liquid to a gas. Freon s low boiling point makes it useful as the fluid in air conditioners and refrigerators. o change phase from a liquid to a gas, heat must have been added to the freon. hat heat comes from the surroundings, and the surroundings are cooled. Fans blowing across the air conditioner coils send cool air into the room. Fans on the opposite side of the air conditioner blow warm air outside the room. Refrigerators operate in a similar way, blowing cold air into the refrigerator and warm air into the room in which the refrigerator sits. An input of electrical energy is required to move heat from a cool environment inside the refrigerator into a warmer environment into the room. Most of this electrical energy is used to run the compressor; a very small amount is used by the fans. he electrical energy used by the air conditioner or refrigerator is not converted into thermal 76
energy except in frictional processes. Rather, the electrical energy is used to pump thermal energy from one place to another. In this case the amount of thermal energy transferred can be greater than the electrical energy required to run the device. his is not a violation of conservation of energy. he second law of thermodynamics does not say that you cannot transfer thermal energy from a cold body to a hot body; it only says that it takes energy to accomplish such a transfer. eat Pumps A window air conditioner sends cold air into the room to be cooled and pumps hot air outside. Although an air conditioner is normally used for cooling, it could be used either to cool or heat a room. Used as an air conditioner, it pumps thermal energy out of the room into the warmer air outside the house. Used as a heat pump, it pumps thermal energy from the colder air outside the house into the house. hus, a heat pump can be thought of as a window air conditioner that has been turned 180 degrees so that warm air is blown into the room and cold air is blown outside. Figure 9.2 illustrates a heat pump and Figure 9.3 illustrates an air conditioner or a refrigerator. Figure 9.2: A eat Pump Figure 9.3: An Air Conditioner or Refrigerator eating and Cooling with Geothermal Energy he heating and cooling devices described above, air conditioners and heat pumps, use electricity to move thermal energy into or out of a closed structure, such as a building. During hot weather, an air conditioner extracts heat from the air inside a building and pumps this heat outside, adding heat to the outside air. In cold weather, a heat pump extracts heat from the cooler outside air and pumps this heat inside. 77
A geothermal heating and cooling system operates in a similar manner but pumps heat into and out of ground water rather than air. In moderate climates, ground water at least 10 feet below the Earth s surface remains at a constant temperature of about 55 O F (13 O C) year round. During summer, fans blowing across coils of this cool water can help cool a room. During winter, heat pumped from this 55 O F water can help warm a room. o extract heat from ground water, the ground water itself can be pumped to the surface, heat extracted from the water, and the water returned underground. Alternatively, fluid can be circulated through a system of vertical pipes buried underground. he fluid absorbs heat as it circulates through the underground pipes. A heat pump extracts heat from the fluid and pumps the heat into a room. he fluid is then returned through the pipes into the ground to again absorb heat. o cool a room using this method, a heat pump removes heat from inside a room and transfers that heat to the fluid. he fluid is pumped underground it is chilled to 55 O F. he cool fluid is returned to the surface it absorbs heat. he process is similar to that of a window air conditioner with the waste heat transferred from a room into a fluid pumped underground instead of to outside air. Coefficient of Performance A measure of how well heat pumps, air conditioners or refrigerators perform is called the coefficient of performance of the device. For example, the coefficient of performance of a heat pump is the ratio of the amount of thermal energy transferred from outside the house, which is at a low temperature, to inside the house, which is at a higher temperature. he coefficient of performance in this case is the ratio of the heat transferred to the house to the work required to transfer it. (Equation 9.5) COP for heat pumps COP = coefficient of performance = heat transferred into the home (joules) = work required to transfer the heat (joules) (Example 9.3) hat is the coefficient of performance of a heat pump that requires 2,500 joules of energy to pump 7,500 joules extracted from outside air into a house? COP = 7,500 J = 3 2,500 J If you have an ideal heat pump, you follow what was done for the maximum efficiency of the heat engine because and L are proportional to and C. Equation 9.5 can be written as 78
COP (Equation 9.6) for heat pumps COP = coefficient of performance = heat transferred into the home (joules) = work required to transfer the heat (joules) = the high temperature (in Kelvin) = the low temperature (in Kelvin) L C e know that efficiencies must be less than one, since some energy always goes into friction. owever, the coefficient of performance ratio may be greater than one. his is because the coefficient of performance is the inverse of the efficiency of the device as seen by comparing Equations 9.3 and 9.6. For actual devices the coefficient of performance is usually between two and six. For air conditioners and refrigerators the coefficient of performance is COP L for air conditioners and refrigerators (Equation 9.7) COP = coefficient of performance L = heat taken out of the refrigerator or the home (joules) = work required to transfer the heat (joules) If you have an ideal refrigerator or air conditioner, the coefficient of performance is (Equation 9.8) COP COP = coefficient of performance L = heat taken out of the refrigerator or the home (joules) = work required to transfer the heat (joules) = the high temperature (in Kelvin) = the low temperature (in Kelvin) L L C C for air conditioners and refrigerators his ratio is the heat transferred from inside the refrigerator or inside the house L divided by the work, = C. 79
Period 9 Summary 9.1: he first law of thermodynamics: the change in internal energy of a system equals the heat added to the system minus the work done by the system. U = As systems move toward equilibrium, they can give off energy or do work or do both. ork can be obtained from a system as it moves from a non-equilibrium situation to an equilibrium situation. 9.2 eat engines do work using the non-equilibrium situation between differences in temperature. eat engines convert thermal energy into mechanical energy. he efficiency of an ideal heat engine = Eff = - L with measured in Kelvin hermal energy flows from warmer environments to cooler environments. 9.3 eat pumps, refrigerators, and air conditioners use energy to move heat from cooler environments to warmer environments. eat can be moved to and from outside air. Using geothermal energy, heat is moved to and from underground water or fluid flowing through underground pipes. Geothermal heating and cooling makes use of the year-round constant underground temperature of 55 O F. he efficiency of heat transfer is measured by the coefficient of performance. eat pump COP = / = heat transferred from a lower temperature environment to a higher temperature environment and = work required to transfer the energy Air Conditioner or refrigerator COP = L / L = heat transferred from a lower temperature environment to a higher temperature environment. = work required to transfer the energy 80