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Engines Physics 202 Professor Lee Carkner Lecture 18
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PAL #17 Entropy 1 kg block of ice (at 0 C) melts in a 20 C room S = S ice + S room S ice = Q/T = mL/T = [(1)(330000)]/(273) S ice = +1219.8 J/K (heat gained) S room = Q/T = 330000/293 S room = -1136.5 J/K (heat lost) S = +1219.8 – 1136.5 = 83.3 J/K Entropy increased, second law holds
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When water condenses out of the air onto a cold surface the entropy of the water, A)Increases, since entropy always decreases B)Decreases, but that is OK since the 2 nd law does not apply to phase changes C)Decreases, but that is OK since the entropy of the air increases D)Increases, since phases changes always increase entropy E)Remains the same
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Water is heated on a stove. Which of the following temperature changes involve the greatest entropy change of the water? A)T increased from 20 to 25 C B)T increased from 40 to 45 C C)T increased from 80 to 85 C D)T increased from 90 to 95 C E)All are equal
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Engines General engine properties: A working substance (usually a gas) An output of heat
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The Stirling Engine The Stirling engine is useful for illustrating the engine properties: The input of heat is from the flame The output of heat makes the fins hot
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Parts of the Cycle Cycle can be broken down into specific parts In general: One involves compression One involves the output of heat Q L Change in internal energy is zero
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Heat and Work How does the work compare to the heat? Since the net heat is Q H -Q L, from the first law of thermodynamics: E int =(Q H -Q L )-W =0 W = Q H - Q L
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Engine Elements
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Efficiency In order for the engine to work we need a source of heat for Q H = W/Q H An efficient engine converts as much of the input heat as possible into work The rest is output as Q L
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Efficiency and Heat = 1 - (Q L /Q H ) Q H = W + Q L Reducing the output heat means improving the efficiency
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The Second Law of Thermodynamics (Engines) This is one way of stating the second law: It is impossible to build an engine that converts heat completely into work Engines get hot, they produce waste heat (Q L ) You cannot completely eliminate friction, turbulence etc.
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Carnot Engine C = 1 - (T C / T H ) This is the Carnot efficiency Any engine operating between two temperatures is less efficient than the Carnot efficiency < C There is a limit as to how efficient you can make your engine
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The First and Second Laws The first law of thermodynamics says: The second law of thermodynamics says: You cannot break even The two laws imply: W < Q H W Q H
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Dealing With Engines W = Q H - Q L = W/Q H = (Q H - Q L )/Q H = 1 - (Q L /Q H ) C = 1 - (T L /T H ) If you know T L and T H you can find an upper limit for (=W/Q H ) For individual parts of the cycle you can often use the ideal gas law: PV = nRT
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Engine Processes We can find the heat and work for each process Net input Q is Q H Four processes are bordered by 4 points Find p, V and T at these points to find W and Q
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Carnot Processes Isothermal W = Q = nRTln(V f /V i ) Need volume at a, b, c and d and T H and T L Adiabatic Q = 0 W = E int = nC V T Just need T H and T L
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Next Time Read: 20.8 Homework: Ch 20, P: 27, 29, 32, 39, 42
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