Engines Physics 202 Professor Lee Carkner Lecture 18
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 = J/K (heat gained) S room = Q/T = /293 S room = J/K (heat lost) S = – = 83.3 J/K Entropy increased, second law holds
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
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
Engines General engine properties: A working substance (usually a gas) An output of heat
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
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
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
Engine Elements
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
Efficiency and Heat = 1 - (Q L /Q H ) Q H = W + Q L Reducing the output heat means improving the efficiency
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.
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
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
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
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
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
Next Time Read: 20.8 Homework: Ch 20, P: 27, 29, 32, 39, 42