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Second Law Thermodynamics Professor Lee Carkner Lecture 11
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PAL # 10 Flow Devices Argon gas in a turbine, want temperature of exit gas Energy balance between enthalpy, ke and work Input = Output Mass flow rate P v = RT v = RT 1 /P 1 = (0.2081)(723) / (900) = m’ = 1 A 1 V 1 = (1/ v 1 )A 1 V 1 = (1/0.167)(0.006)(80) =
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PAL # 10 Flow Devices Want to find h = c p T h = h = (250/2.874) – [(80 2 /2) – (150 2 /2)] / 1000 h = T = h/c p = 95.04 / 0.5203 = Input T is 450 C so output T is 267.3 C
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The Second Law of Thermodynamics Processes have preferred directions Example: melting an ice cube Energy has a quality as well as a quantity A tank of gasoline contains highly useful energy, a slightly warmer atmosphere contains fairly useless energy
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Heat Reservoir Something with a large mass or specific heat Allows for heat transfer at constant T
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Heat Engine Work can be converted to heat directly e.g. Heat can only be converted to work with special devices In general: Need a fluid called the working substance
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Engine Notation Four basic engine properties (all positive): Output heat to cold reservoir, Q L Work needed for compression, W in Some of the output work is needed to run the cycle and thus the net work is: A cycle has U = 0, so: W net,out = Q H – Q L
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Efficiency We want to maximize net output work for a given input heat Called the thermal efficiency Can express as:
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Refrigerators Instead of using a temperature difference to create work, uses work to create a temperature difference
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Refrigerator Cycle Expanding a gas obviously does work Need a refrigerant undergoing a vapor- compression refrigeration cycle: Fluid is compressed to high pressure Expansion valve returns the fluid to low pressure and is moved to the cold chamber
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Coefficient of Performance We want to move the most heat out of the cold chamber for the least work Since energy is conserved: W net,in + Q L = Q H Thus: Can be greater than one
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Heat Pump A heat pump is just a refrigerator with the inside of your house as the hot chamber and the outside as the cold chamber We want a lot of heat added to the house for a little work Note that, COP HP = COP R + 1
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Air Conditioner Uses same COP as refrigerator For every 2-3 joules of heat removed from cold chamber, need about 1 joule of work
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Kelvin – Planck Statement Why does an engine have to have Q out ? We need to compress the gas back to the initial state and that requires an output of heat Kelvin – Planck Statement: Means that th is always less then one
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Clausius Statement It is impossible for a cyclic device to move heat from cold to hot with no input work
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Equivalence A perfect engine powering a refrigerator will make it perfect as well
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Next Time Read: 6.5-6.11 Homework: Ch 6, P: 28, 78, 91, 97
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