LSN 16 Heat Engines.pdf

12/07/2024 ENGG431 – THERMODYNAMICS AND FLUID MECHANICS The Second Law of Thermodynamics Physics & Engineering LESSON OUTLINE » Heat Engines and the...

12/07/2024 ENGG431 – THERMODYNAMICS AND FLUID MECHANICS The Second Law of Thermodynamics Physics & Engineering LESSON OUTLINE » Heat Engines and the Second Law of Thermodynamics » Heat Pumps and Refrigerators Physics & Engineering 1  12/07/2024 INTENDED LEARNING OBJECTIVES 1. State the second law of thermodynamics. 2. Describe the operation of a heat engine. 3. Describe the operation of a refrigerator and a heat pump. Physics & Engineering THE LAWS OF THERMODYNAMICS The First Law of Thermodynamics » The first law is a statement of Conservation of Energy » The first law states that a change in internal energy in a system can occur as a result of energy transfer by heat, by work, or by both ∆𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊 » The first law makes no distinction between processes that occur spontaneously and those that do not » Only certain types of energy transformation and transfer processes actually take place in nature Physics & Engineering 4 2  12/07/2024 THE LAWS OF THERMODYNAMICS The Second Law of Thermodynamics » The second law establishes which processes do and do not occur » Most NATURAL processes are observed to occur only in one direction ▪ Energy always transfers from the hot object to the cold object We expect to see energy transfer by heat from a hot object to a cold object with which it is in contact ▪ This directionality is governed by the second law » These types of processes are irreversible ▪ An irreversible process is one that occurs naturally in one direction only ▪ No irreversible process has been observed to run backwards ▪ An important engineering implication is the limited efficiency of heat engines Physics & Engineering 5 LORD KELVIN » William Thomson, Lord Kelvin » 1824 – 1907 » British physicist and mathematician » First to propose the use of an absolute scale of temperature » His work in thermodynamics led to the idea that energy cannot pass spontaneously from a colder object to a hotter object Physics & Engineering 6 3  12/07/2024 Physics & Engineering Heat Engines and the Second Law of Thermodynamics HEAT ENGINE » A heat engine is a device that takes in energy by heat and, operating in a cyclic process, expels a fraction of that energy by means of work » Examples: ▪ Power plant producing electricity ▪ Internal combustion engine of an automobile » A heat engine carries some working substance through a cyclic process during which: ▪ The working substance absorbs energy by heat from a high- temperature energy reservoir (Qh) ▪ Work is done by the engine (Weng) ▪ Energy is expelled as heat to a lower-temperature reservoir (Qc) Physics & Engineering 8 4  12/07/2024 HEAT ENGINE Physics & Engineering Components of a simplified steam engine 9 HEAT ENGINE Physics & Engineering A steam engine may be used to turn a turbine 10 5  12/07/2024 HEAT ENGINE » Since it is a cyclical process: ∆𝐸𝑖𝑛𝑡 = 0 ▪ Its initial and final internal energies are the same. » From the 1st Law of Thermodynamics: ∆𝐸𝑖𝑛𝑡 = 𝑄 + 𝑊 = 𝑄𝑛𝑒𝑡 − 𝑊𝑒𝑛𝑔 = 0 » Therefore: 𝑊𝑒𝑛𝑔 = 𝑄𝑛𝑒𝑡 = 𝑄ℎ − 𝑄𝑐 ▪ The net work done by a heat engine equals the net energy transferred to it Physics & Engineering 11 HEAT ENGINE Thermal Efficiency of a Heat Engine » Thermal efficiency is defined as the ratio of the net work done by the engine during one cycle to the energy input at the higher temperature: 𝑊𝑒𝑛𝑔 𝑄ℎ − 𝑄𝑐 𝑄𝑐 𝑒= = =1− 𝑄ℎ 𝑄ℎ 𝑄ℎ » We can think of the efficiency as the ratio of what you gain (W) to what you give (Qh) » In practice, all heat engines expel only a fraction of the input energy by mechanical work ▪ Their efficiency is always less than 100% ▪ For example, diesel engines have efficiencies ranging from 35% to 40% Physics & Engineering 12 6  12/07/2024 HEAT ENGINE 𝑊𝑒𝑛𝑔 𝑄𝑐 𝑒= =1− 𝑄ℎ 𝑄ℎ Perfect Heat Engine » A heat engine has 100% efficiency (e = 1) only if |Qc| = 0 ▪ That is, no energy is expelled to the cold reservoir » A heat engine with perfect efficiency would have to expel all the input energy by work ▪ It takes in some amount of energy and does an equal amount of work ▪ It is impossible to construct such an engine Physics & Engineering 13 HEAT ENGINE The Second Law of Thermodynamics (Kelvin-Planck Form) » It is impossible to construct a heat engine that, operating in a cycle, produces no effect other than the input of energy by heat from a reservoir and the performance of an equal amount of work » This statement of the second law means that during the operation of a heat engine: ▪ Weng can never be equal to |Qh| ▪ Means that Qc cannot equal 0 ▪ Some energy |Qc| must be expelled to the environment ▪ Means that e cannot equal 100% Physics & Engineering 14 7  12/07/2024 SAMPLE PROBLEM 1 An engine transfers 2.00103 J of energy from a hot reservoir during a cycle and transfers 1.50103 J as exhaust to a cold reservoir. (a) Find the efficiency of the engine. (b) How much work does this engine do in one cycle? 𝑊𝑒𝑛𝑔 𝑄𝑐 𝑊𝑒𝑛𝑔 = 𝑄ℎ − 𝑄𝑐 𝑒= =1− 𝑄ℎ 𝑄ℎ Physics & Engineering 15 Physics & Engineering Heat Pumps and Refrigerators 8  12/07/2024 HEAT PUMPS AND REFRIGERATORS » Heat engines can run in reverse (transfer energy from the cold reservoir to the hot reservoir) ▪ This is not a natural direction of energy transfer ▪ We must put some energy into a device to do this ▪ Devices that do this are called heat pumps or refrigerators » Example: Air conditioner ▪ An air conditioner transfers energy from the cool room in the home to the warm air outside Physics & Engineering 17 HEAT PUMPS AND REFRIGERATORS Heat Pump Process » Energy is extracted from the cold reservoir, |QC| » Energy is transferred to the hot reservoir, |Qh| ▪ Work must be done on the engine, W » From the first law: energy given up to the hot reservoir must equal the sum of the work done and the energy taken in from the cold reservoir ▪ The refrigerator or heat pump transfer energy from a colder body to a hotter body ▪ It is desirable to carry out this process with a minimum amount of work Physics & Engineering 18 9  12/07/2024 HEAT PUMPS AND REFRIGERATORS » A heat pump includes a circulating fluid that passes through two sets of metal coils that can exchange energy with the surroundings ▪ The fluid is cold and at low pressure when it is in the coils located in a cool environment, where it absorbs energy by heat ▪ The resulting warm fluid is then compressed and enters the other coils as a hot, high-pressure fluid ▪ There it releases its stored energy to the warm surroundings » In an air conditioner, energy is absorbed into the fluid in coils located in a building’s interior ▪ After the fluid is compressed, energy leaves the fluid through coils located outdoors » In a refrigerator, the external coils are behind the unit or underneath the unit ▪ The internal coils are in the walls of the refrigerator and absorb energy from the food Physics & Engineering 19 HEAT PUMPS AND REFRIGERATORS 5 This liquid expands adiabatically at a rate controlled by the expansion valve so that the liquid is greatly cooled. 4 Vapor entering the condenser gives off heat and 1 Liquid refrigerant becomes a hot, passing through the high-pressure liquid. evaporator absorbs heat from the air and becomes a vapor. 3 Next the vapor enters the compressor, whose piston compresses it adiabatically. 2 Vapor entering the food compartment Positive work Wmotor is done on the vapor is colder than the air in the compartment, and it leaves the compressor at a higher Physics & so heat is transferred to the refrigerant. pressure and higher temperature. Engineering 20 10  12/07/2024 HEAT PUMPS AND REFRIGERATORS Perfect Heat Pump » Takes energy from the cold reservoir » Expels an equal amount of energy to the hot reservoir » No work is done » This is an impossible heat pump The Second Law of Thermodynamics (Clausius Form) » It is impossible to construct a cyclical machine whose sole effect is to transfer energy continuously by heat from one object to another object at a higher temperature without the input of energy by work ▪ Energy does not transfer spontaneously by heat from a cold object to a hot object ▪ Work input is required to run a refrigerator Physics & Engineering 21 HEAT PUMPS AND REFRIGERATORS Coefficient of Performance » The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) ▪ Similar to thermal efficiency for a heat engine ▪ It is the ratio of what you gain (energy transferred to or from a reservoir) to what you give (work input) » In cooling mode, you “gain” energy removed from a cold temperature reservoir: 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑎𝑡 𝑙𝑜𝑤 𝑡𝑒𝑚𝑝 𝑄𝑐 𝐶𝑂𝑃 𝐶𝑜𝑜𝑙𝑖𝑛𝑔 𝑀𝑜𝑑𝑒 = = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑝𝑢𝑚𝑝 𝑊 » A good refrigerator should have a high COP ▪ Typical values are 5 or 6 Physics & Engineering 22 11  12/07/2024 HEAT PUMPS AND REFRIGERATORS Coefficient of Performance » In addition to cooling applications, heat pumps are becoming increasingly popular for heating purposes ▪ In heating mode, energy is absorbed from the cool air outside a building and warm air is released inside the building » In heating mode, the COP is the ratio of the heat transferred to the work required: 𝑒𝑛𝑒𝑟𝑔𝑦 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟𝑟𝑒𝑑 𝑎𝑡 ℎ𝑖𝑔ℎ 𝑡𝑒𝑚𝑝 𝑄ℎ 𝐶𝑂𝑃 𝐻𝑒𝑎𝑡𝑖𝑛𝑔 𝑀𝑜𝑑𝑒 = = 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑏𝑦 𝑡ℎ𝑒 ℎ𝑒𝑎𝑡 𝑝𝑢𝑚𝑝 𝑊 » Qh is typically higher than W ▪ For outside temperature of about 25°F (–4°C ) or higher, values of COP are generally about 4 The amount of energy transferred to the building is about four times greater than the work done by the motor in the heat pump ▪ As the outside temperature decreases, however, it becomes more difficult for the heat pump to extract sufficient energy from the air outside and so the COP decreases The use of heat pumps that extract energy from the air is most satisfactory in moderate climates but Physics & Engineering not appropriate during very low winter temperatures 23 SAMPLE PROBLEM 2 A certain refrigerator has a COP of 5.00. When the refrigerator is running, its power input is 500 W. A sample of water of mass 0.500 kg and temperature 20.0°C is placed in the freezer compartment. How long does it take to freeze the water to ice at 0°C? Assume all other parts of the refrigerator stay at the same temperature and there is no leakage of energy from the exterior, so the operation of the refrigerator results only in energy being extracted from the water. 𝑄𝑐 𝑊 𝐶𝑂𝑃 = 𝑄𝑐 = 𝑚𝑐∆𝑇 + −𝑚𝐿𝑓 𝑃= 𝑊 ∆𝑡 Physics & Engineering 24 12  12/07/2024 NEXT MEETING » Reading Assignment – Reversible and Irreversible Processes, The Carnot Engine ▪ Chapter 21 of Physics for Scientists and Engineers by Serway & Jewett ▪ Chapter 7 of Fundamentals of Thermal-Fluid Sciences by Cengel, Cimbala & Turner » Practice Problems ▪ Problem Set 08 – Second Law of Thermodynamics » Starting next week ▪ Shifting ▪ Schedule for ENGG431 is TTh ☺ Physics & Engineering 25 13