LECTURE 9

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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 84 LECTURE 9 HEAT ENGINES SECOND LAW OF THERMODYNAMICS 2 DEMONSTRATIONS 4 QUIZ QUESTIONS 12 SUGGESTED PROBLEMS READING ASSIGNMENT: Chapter22

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 85 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 • A heat engine carries some working substance through a cyclical process • Since it is a cyclical process, Δ E int = 0 – Its initial and final internal energies are the same • Therefore, Q net = W eng • The work done by the engine equals the net energy absorbed by the engine • The work is equal to the area enclosed by the curve of the PV diagram – If the working substance is a gas "࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?" = ࠵?ℎ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?ℎ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵? PERFORMANCE COFFICIENT à efficiency or COP Thermal Efficiency of a Heat Engine ࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵? ࠵? = |࠵?| |࠵? ! | • For Heat engine the 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. In practice, all heat engines expel only a fraction of the input energy by mechanical work. • Therefore, their efficiency is always less than 100%. To have e = 100%, Q C must be 0 HOT RESERVOIR AT T h ENGINE Q h COLD RESERVOIR AT T c Q c W eng Area=W eng p V

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 86 Second Law: Kelvin-Planck Form • It is impossible to construct a heat engine that, • operating in a cycle, produces no other effect • than the absorption of energy from a reservoir • and the performance of an equal amount of work – Means that Qc cannot equal 0 • Some Qc must be expelled to the • environment – Means that e cannot equal 100% Heat Pumps and Refrigerators • Heat engines can run in reverse. – This is not a natural direction of energy transfer. – Must put some energy into a device to do this. – Devices that do this are called heat pumps or refrigerators. • Examples – A refrigerator is a common type of heat pump – An air conditioner is another example of a heat pump • Energy is extracted from the cold reservoir, QC • Energy is transferred to the hot reservoir, Qh • Work must be done on the engine, W HEAT PUMPS: Coefficient of Performance (COP) • The effectiveness of a heat pump is described by a number called the coefficient of performance (COP) • In heating mode , the COP is the ratio of the heat transferred into the work required. COP, Heating Mode ࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? = |࠵? ! | |࠵?| • Qh is typically higher than W – Values of COP are generally greater than 1 – It is possible for them to be less than 1 • We would like the COP to be as high as possible! HOT RESERVOIR AT T h Q h W eng ENGINE COLD RESERVOIR AT T c HOT RESERVOIR AT T h HEAT PUMP Q h COLD RESERVOIR AT T c Q c W eng

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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 87 COP, Cooling Mode • In cooling mode , you want fast removal of the energy from a cold temperature reservoir ࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?࠵? = |࠵? " | |࠵?| • A good refrigerator should have a high COP – Typical values are 5 or 6 Second Law – 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 • Or – energy does not transfer spontaneously by heat from a cold object to a hot object. • 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 FIRST AND SECOND LAWS OF THERMODYNAMICS • First Law is an expression of Conservation of Energy • There is class of processes which do not violate Conservation of Energy, yet they never occur. (Boiled Egg is difficult to be un-boiled, shattered glass very rarely becomes whole again, the stone which has been dropped down from certain height, on a soft ground, very seldom cools down an rises back to its original position) • WHY? The Second Law of Thermodynamics • Establishes which processes do occur spontaneously and which do not occur • Some processes can occur in either direction according to the first law • They are observed to occur only in one direction • This directionality is governed by the second law HOT RESERVOIR AT T h HEAT PUMP Q h Q c COLD RESERVOIR AT T c

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 88 Irreversible Processes • 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 Reversible and Irreversible Processes • A reversible process is one in which every point along some path is an equilibrium state, and one for which the system can be returned to its initial state along the same path • An irreversible process does not meet these requirements – All natural processes are known to be irreversible – Reversible processes are an idealization, but some real processes are good approximations • A real process that is a good approximation of a reversible one will occur very slowly- the system is always very nearly in an equilibrium state • A general characteristic of a reversible process is that there are no dissipative effects that convert mechanical energy to internal energy present – No friction or turbulence, for example Summary: The reversible process is an idealization : all real processes on Earth are irreversible

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 89 Carnot Engine • A theoretical engine developed by Sadi Carnot • A heat engine operating in an ideal, reversible cycle (now called a Carnot cycle ) between two reservoirs is the most efficient engine possible – This sets an upper limit on the efficiencies of all other engines Carnot’s Theorem • • No real heat engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs – All real engines are less efficient than a Carnot engine because they do not operate through a reversible cycle – The efficiency of a real engine is further reduced by friction, energy losses through conduction, etc. Carnot Cycle Overview of the processes in a Carnot cycle

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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 90 • A -> B is an isothermal expansion • The gas is placed in contact with the high temperature reservoir, Th • The gas absorbs heat | Qh | • The gas does work WAB in raising the piston • B -> C is an adiabatic expansion • The base of the cylinder is replaced by a thermally nonconducting wall • No heat enters or leaves the system • The temperature falls from Th to Tc • The gas does work WBC • The gas is placed in contact with the cold temperature reservoir • C -> D is an isothermal compression • The gas expels energy Qc • Work WCD is done on the gas • D -> A is an adiabatic compression • The gas is again placed against a thermally nonconducting wall, so no heat is exchanged with the surroundings • The temperature of the gas increases from Tc to Th • The work done on the gas is WDA

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 91 Efficiency of a Carnot Engine • Carnot showed that the efficiency of the engine depends on the temperatures of the reservoirs Temperatures must be in Kelvins. All Carnot engines operating between the same two temperatures will have the same efficiency • Efficiency is 0 if Th = Tc • Efficiency is 100% only if Tc = 0 K – Such reservoirs are not available – Efficiency is always less than 100% • The efficiency increases as Tc is lowered and as Th is raised • In most practical cases, Tc is near room temperature, 300 K – So generally Th is raised to increase efficiency. Carnot Cycle in Reverse • Theoretically, a Carnot-cycle heat engine can run in reverse • This would constitute the most effective heat pump available • This would determine the maximum possible COPs for a given combination of hot and cold reservoirs • • In heating mode: • In cooling mode: Carnot Cycle, PV Diagram • The work done by the engine is shown by the area enclosed by the curve, W eng • The net work is equal to | Q h | – | Q c | ° Δ E int = 0 for the entire cycle h c Carnot h c h c T T e Q Q T T - = = 1 C h h h c Q T COP W T T = = - c c C h c Q T COP W T T = = -

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 92 Exercise: Obtain the efficiency formula for Carnot Engine OTHER TYPES OF ENGINES ( READING ASSIGNMENT: SELF STUDY) STIRLING ENGINE In 1816 Robert Stirling, a Scottish clergyman, patented the Stirling engine , which has found a wide variety of applications ever since. Fuel is burned externally to warm one of the engine’s two cylinders. A fixed quantity of inert gas moves cyclically between the cylinders, expanding in the hot one and contracting in the cold one. A Stirling engine is easier to manufacture than an internal combustion engine or a turbine. It can run on burning garbage. It can run on the energy of sunlight and produce no material exhaust. OTTO’S ENGINE Four stroke combustion engine (This the ideal version of Gasoline combustion engine used today) Diesel Engine: Create the short note (less than a page on Diesel Engine)

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PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 93 SUGGESTED PROBLEMS 1 Given is Starling Engine represented by the diagram Consider n mol of an ideal monatomic gas being taken once through the cycle, consisting of two isothermal processes at temperatures 3 T i and T i and two constant-volume processes. Determine, in terms of n , R , and T i , (a) the net energy transferred by heat to the gas and (b) the efficiency of the engine. 2 Given is Otto Engine The compression ratio of an Otto cycle, as shown in Figure below, is V A / V B = 8.00. At the beginning A of the compression process, 500 cm 3 of gas is at 100 kPa and 20.0 ° C. At the beginning of the adiabatic expansion the temperature is T C = 750 ° C. Model the working fluid as an ideal gas with E int = nC V T = 2.50 nRT and = 1.40. (a) Fill in this table to follow the states of the gas: T (K) P (kPa) V (cm 3 ) E int A 293 100 500 B C 1 023 D A (b) Fill in this table to follow the processes: Q (input) W (output) E int A ® B B ® C C ® D D ® A ABCDA (c) Identify the energy input Q h , the energy exhaust Q c , and the net output work W eng . (d) Calculate the thermal efficiency. (e) Find the number of crankshaft revolutions per minute required for a one-cylinder engine to have output power 1.00 kW = 1.34 hp. Note that the thermodynamic cycle involves four piston strokes. g D

PHY 1321/PHY1331 Principles of Physics I Fall 2023 Dr. Andrzej Czajkowski 94 3 An ideal gas is taken through a Carnot cycle. The isothermal expansion occurs at 250°C, and the isothermal compression takes place at 50.0°C. The gas takes in 1 200 J of energy from the hot reservoir during the isothermal expansion. Find (a) the energy expelled to the cold reservoir in each cycle and (b) the net work done by the gas in each cycle. 4. The exhaust temperature of a Carnot heat engine is 300°C. What is the intake temperature if the efficiency of the engine is 30.0%? 5. A Carnot heat engine uses a steam boiler at 100 ° C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 20.0 ° C. Energy is exhausted to the low-temperature reservoir at the rate of 15.4 W. (a) Determine the useful power output of the heat engine. (b) How much steam will it cause to condense in the high-temperature reservoir in 1.00 h? 6. A heat engine takes in 360 J of energy from a hot reservoir and performs 25.0 J of work in each cycle. Find (a) the efficiency of the engine and (b) the energy expelled to the cold reservoir in each cycle. 7. A heat engine performs 200 J of work in each cycle and has an efficiency of 30.0%. For each cycle, how much energy is (a) taken in and (b) expelled as heat? 8. A particular heat engine has a useful power output of 5.00 kW and an efficiency of 25.0%. The engine expels 8 000 J of exhaust energy in each cycle. Find (a) the energy taken in during each cycle and (b) the time interval for each cycle. 9. Heat engine X takes in four times more energy by heat from the hot reservoir than heat engine Y . Engine X delivers two times more work, and it rejects seven times more energy by heat to the cold reservoir than heat engine Y . Find the efficiency of (a) heat engine X and (b) heat engine Y . 10. Suppose a heat engine is connected to two energy reservoirs, one: a pool of molten , freezing 1.00 g of aluminum and two: melting 15.0 g of mercury during each cycle. The heat of fusion of aluminum is 3.97 ´ 10 5 J/kg; the heat of fusion of mercury is 1.18 ´ 10 4 J/kg. What is the efficiency of this engine? 11 A heat engine operating between 200°C and 80.0°C achieves 20.0% of the maximum possible efficiency. What energy input will enable the engine to perform 10.0 kJ of work? 12. A heat pump, is essentially an air conditioner installed backward. It extracts energy from colder air outside and deposits it in a warmer room. Suppose that the ratio of the actual energy entering the room to the work done by the device’s motor is 10.0% of the theoretical maximum ratio. Determine the energy entering the room per joule of work done by the motor, given that the inside temperature is 20.0°C and the outside temperature is –5.00°C