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Carnot. Thermodynamics Professor Lee Carkner Lecture 12. 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)
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Carnot Thermodynamics Professor Lee Carkner Lecture 12
PAL # 11 Second Law • Refrigerant 134a flowing through a condenser • Heat output of condenser is equal to the change in enthalpy of fluid • QH = • h1 = 271.22 kJ/kg (superheat vapor, Table A-13) • h2 = 95.47 kJ/kg (saturated liquid, Table A-12) • QH = (0.018)(271.22-95.47) = • COP = QH/W = 3.164/1.2 = • QL = QH – W = 3.164 – 1.2 =
Reversible • A reversible process: • has a net heat and work exchange for all systems as zero • is the theoretical limits for a process
Irreversible • An irreversible process can be due to: • Friction • Unrestrained expansion of a gas into a vacuum • Heat transfer through temperature difference
Achieving Reversibility • Heat transfer through a very small temperature differential dT becomes reversible as dT approaches zero • Example Isothermal Work: • dT always very small
The Carnot Cycle • The Carnot engine consists of all reversible processes and thus is the most efficient • Carnot Cycle • An adiabatic fall from TH to TL • Adiabatic process is frictionless and isothermal process has very small temperature differentials
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
Carnot Efficiency • The efficiency of a reversible engine depends only on the temperatures of the reservoirs hth,rev = 1 – (TL/TH) • Maximum efficiency for any real engine • Can increase the efficiency of any engine by:
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
Carnot Refrigerator • We can also make the same determination for the efficiency of the Carnot refrigerator or heat pump COPR = 1 / (TH/TL -1) COPHP = 1 / (1 – TL/TH) • Smaller temperature difference means more efficiency
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
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
Perpetual Motion • 1st kind: • Machine that creates energy • 2nd kind: • Machine that converts heat completely into work • 3rd kind: • Machine with no dissipation
Next Time • Read: 7.1-7.6 • Homework: Ch 6, P: 131, 138, Ch 7, P: 29, 37