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Carnot Thermodynamics Professor Lee Carkner Lecture 12

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PAL # 11 Second Law Refrigerant 134a flowing through a condenser Heat output of condenser is equal to the change in enthalpy of fluid Q H = h 1 = 271.22 kJ/kg (superheat vapor, Table A-13) h 2 = 95.47 kJ/kg (saturated liquid, Table A-12) Q H = (0.018)(271.22-95.47) = COP = Q H /W = 3.164/1.2 = Q L = Q H – W = 3.164 – 1.2 =

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Reversible A reversible process: has a net heat and work exchange for all systems as zero is the theoretical limits for a process

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Irreversible An irreversible process can be due to: Friction Unrestrained expansion of a gas into a vacuum Heat transfer through temperature difference

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Achieving Reversibility Heat transfer through a very small temperature differential dT becomes reversible as dT approaches zero Example Isothermal Work: dT always very small

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The Carnot Cycle The Carnot engine consists of all reversible processes and thus is the most efficient Carnot Cycle An adiabatic fall from T H to T L Adiabatic process is frictionless and isothermal process has very small temperature differentials

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Carnot Cycle

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Carnot Principles All Carnot engines operating between two heat reservoirs have the same efficiency While we cannot build a real Carnot engine, it gives us the upper limit for the efficiency of a real engine

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Carnot Efficiency The efficiency of a reversible engine depends only on the temperatures of the reservoirs th,rev = 1 – (T L /T H ) Maximum efficiency for any real engine Can increase the efficiency of any engine by:

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Kinds of Engines

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Quality of Energy Since work is what we want, we can say that high temperature sources have higher quality energy than low temperature sources Quality is different from quantity

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Efficiency and Temperature

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Carnot Refrigerator We can also make the same determination for the efficiency of the Carnot refrigerator or heat pump COP R = 1 / (T H /T L -1) COP HP = 1 / (1 – T L /T H ) Smaller temperature difference means more efficiency

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Carnot Refrigeration Cycle

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Kinds of Refrigerators

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Heating a House

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Thermodynamic Temperature Scale The efficiency of any engine depends on the ratio of the heats Thus we can determine the temperature of two reservoirs by measuring the heat flow in and out of an ideal engine operating between them

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Kelvin Scale If assign a magnitude to the degree size we get a complete temperature scale, independent of any substance in a thermometer Note that we don’t actually use an engine to find T

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Perpetual Motion 1st kind: Machine that creates energy 2nd kind: Machine that converts heat completely into work 3rd kind: Machine with no dissipation

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Next Time Read: 7.1-7.6 Homework: Ch 6, P: 131, 138, Ch 7, P: 29, 37

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