Engines Physics 202 Professor Lee Carkner Lecture 16
PAL #15 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 = S room = Q/T = /293 S room = S = – = Entropy increased, second law holds
PAL #15 Entropy Entropy of isobaric process, 60 moles compressed from 2.6 to 1.7 ms at 100 kPa S = nRln(V f /V i ) + nC V ln(T f /T i ) T i = (100000)(2.6)/(60)(8.31) =521 K T f = (100000)(1.7)/(60)(8.31) = 341 K S = -530 J/K Something else must increase in entropy by more than 530 J/K
Engines General engine properties: A working substance (usually a gas) A net output of work, W Note that we write Q and W as absolute values
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 cooling fins hot
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
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 Efficiency C = 1 - (T L / 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 You cannot get out more than you put in 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 Net output Q is Q L Find p, V and T at these points to find W and Q
Carnot Engine A Carnot engine has two isothermal processes and two adiabatic processes Heat is only transferred at the highest and lowest temperatures
Next Time Read: 20.8
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