1. Physics 207: Lecture 27, Pg 1 Lecture 27 Goals: Ch. 18 Ch. 18 Qualitatively understand 2 nd Law of Thermodynamics Qualitatively understand 2 nd Law of Thermodynamics Ch. 19 Ch. 19  Understand the relationship between work and heat in a cycling process  Follow the physics of basic heat engines and refrigerators.  Recognize some practical applications in real devices.  Know the limits of efficiency in a heat engine. Assignment Assignment  HW11, Due Tues., May 5 th  HW12, Due Friday, May 9 th  For Thursday, Read through all of Chapter 20

2. Physics 207: Lecture 27, Pg 2 The need for something else: Entropy You have an ideal gas in a box of volume V 1. Suddenly you remove the partition and the gas now occupies a larger volume V 2. (1)How much work was done by the system? (2) What is the final temperature (T 2 )? (3) Can the partition be reinstalled with all of the gas molecules back in V 1 ? P V1V1 P V2V2

3. Physics 207: Lecture 27, Pg 3 Free Expansion and Entropy You have an ideal gas in a box of volume V 1. Suddenly you remove the partition and the gas now occupies a larger volume V 2. (3) Can the partition be reinstalled with all of the gas molecules back in V 1 (4) What is the minimum process necessary to put it back? P V1V1 P V2V2

4. Physics 207: Lecture 27, Pg 4 Free Expansion and Entropy You have an ideal gas in a box of volume V 1. Suddenly you remove the partition and the gas now occupies a larger volume V 2. (4) What is the minimum energy process necessary to put it back? Example processes: A. Adiabatic Compression followed by Thermal Energy Transfer B. Cooling to 0 K, Compression, Heating back to original T P V1V1 P V2V2

5. Physics 207: Lecture 27, Pg 5 Exercises Free Expansion and the 2 nd Law What is the minimum energy process necessary to put it back? Try: B. Cooling to 0 K, Compression, Heating back to original T Q 1 = n C v  T out and put it where…??? Need to store it in a low T reservoir and 0 K doesn’t exist Need to extract it later…from where??? Key point: Where Q goes & where it comes from are important as well. P V1V1 P V2V2

6. Physics 207: Lecture 27, Pg 6 Modeling entropy l I have a two boxes. One with fifty pennies. The other has none. I flip each penny and, if the coin toss yields heads it stays put. If the toss is “tails” the penny moves to the next box. l On average how many pennies will move to the empty box?

7. Physics 207: Lecture 27, Pg 7 Modeling entropy l I have a two boxes, with 25 pennies in each. I flip each penny and, if the coin toss yields heads it stays put. If the toss is “tails” the penny moves to the next box.  On average how many pennies will move to the other box?  What are the chances that all of the pennies will wind up in one box?

8. Physics 207: Lecture 27, Pg 8 2 nd Law of Thermodynamics l Second law: “The entropy of an isolated system never decreases. It can only increase, or, in equilibrium, remain constant.” l The 2 nd Law tells us how collisions move a system toward equilibrium. l Order turns into disorder and randomness. l With time thermal energy will always transfer from the hotter to the colder system, never from colder to hotter. l The laws of probability dictate that a system will evolve towards the most probable and most random macroscopic state Entropy measures the probability that a macroscopic state will occur or, equivalently, it measures the amount of disorder in a system Increasing Entropy

9. Physics 207: Lecture 27, Pg 9 Home exercise: Entropy l Two identical boxes each contain 1,000,000 molecules. In box A, 750,000 molecules happen to be in the left half of the box while 250,000 are in the right half. In box B, 499,900 molecules happen to be in the left half of the box while 500,100 are in the right half. l At this instant of time:  The entropy of box A is larger than the entropy of box B.  The entropy of box A is equal to the entropy of box B.  The entropy of box A is smaller than the entropy of box B.

10. Physics 207: Lecture 27, Pg 10 Home Exercise: Entropy l Two identical boxes each contain 1,000,000 molecules. In box A, 750,000 molecules happen to be in the left half of the box while 250,000 are in the right half. In box B, 499,900 molecules happen to be in the left half of the box while 500,100 are in the right half. l At this instant of time:  The entropy of box A is larger than the entropy of box B.  The entropy of box A is equal to the entropy of box B.  The entropy of box A is smaller than the entropy of box B.

11. Physics 207: Lecture 27, Pg 11 Reversible vs Irreversible l The following conditions should be met to make a process perfectly reversible: 1. Any mechanical interactions taking place in the process should be frictionless. 2. Any thermal interactions taking place in the process should occur across infinitesimal temperature or pressure gradients (i.e. the system should always be close to equilibrium.) l Based on the above answers, which of the following processes are not reversible? 1. Melting of ice in an insulated (adiabatic) ice-water mixture at 0°C. 2. Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston. 3. Lifting the piston described in the previous statement by removing one grain of sand at a time. 4. Freezing water originally at 5°C.

12. Physics 207: Lecture 27, Pg 12 Reversible vs Irreversible l The following conditions should be met to make a process perfectly reversible: 1. Any mechanical interactions taking place in the process should be frictionless. 2. Any thermal interactions taking place in the process should occur across infinitesimal temperature or pressure gradients (i.e. the system should always be close to equilibrium.) l Based on the above answers, which of the following processes are not reversible? 1. Melting of ice in an insulated (adiabatic) ice-water mixture at 0°C. 2. Lowering a frictionless piston in a cylinder by placing a bag of sand on top of the piston. 3. Lifting the piston described in the previous statement by removing one grain of sand at a time. 4. Freezing water originally at 5°C.

13. Physics 207: Lecture 27, Pg 13 Home Exercise l A piston contains two chambers with an impermeable but movable barrier between them. On the left is 1 mole of an ideal gas at 200 K and 1 atm of pressure. On the right is 2 moles of another ideal gas at 400 K and 2 atm of pressure. The barrier is free to move and heat can be conducted through the barrier. If this system is well insulated (isolated from the outside world) what will the temperature and pressure be at equilibrium? p,T,V R p,T,V L

14. Physics 207: Lecture 27, Pg 14 Home Exercise l If this system is well insulated (isolated from the outside world) what will the temperature and pressure be at equilibrium? At equilibrium both temperature and pressure are the same on both sides.  E Th(Left) +  E Th(Right) = 0 1 x 3/2 R (T-200 K) + 2 x 3/2 R (T-400 K) = 0 (T-200 K) + 2 (T-400 K) = 0 3T = 1000 K T=333 K Now for p….notice p/T = const. = n R / V n L R / V L = n R R / V R n L V R = n R V L V R = 2 V L

15. Physics 207: Lecture 27, Pg 15 Exercise l If this a system is well insulated (isolated from the outside world) what will the temperature and pressure be at equilibrium? V R = 2 V L and V R + V L = V initial = (1 x 8.3 x 200 / 10 5 + 2 x 8.3 x 400 / 2x10 5 ) V initial = 0.050 m 3 V R =0.033 m 3 V L = 0.017 m 3 P R = n R RT / V R = 2 x 8.3 x 333 / 0.033 = 1.7 atm P l = n L RT / V l = 1 x 8.3 x 333 / 0.017 = 1.7 atm

16. Physics 207: Lecture 27, Pg 16 Heat Engines and Refrigerators l Heat Engine: Device that transforms heat into work ( Q  W) l It requires two energy reservoirs at different temperatures  An thermal energy reservoir is a part of the environment so large with respect to the system that its temperature doesn’t change as the system exchanges heat with the reservoir.  All heat engines and refrigerators operate between two energy reservoirs at different temperatures T H and T C.

17. Physics 207: Lecture 27, Pg 17 For practical reasons, we would like an engine to do the maximum amount of work with the minimum amount of fuel. We can measure the performance of a heat engine in terms of its thermal efficiency η (lowercase Greek eta), defined as We can also write the thermal efficiency as Heat Engines

18. Physics 207: Lecture 27, Pg 18 l Consider two heat engines:  Engine I:  Requires Q in = 100 J of heat added to system to get W=10 J of work (done on world in cycle)  Engine II:  To get W=10 J of work, Q out = 100 J of heat is exhausted to the environment Compare  I, the efficiency of engine I, to  II, the efficiency of engine II. Exercise Efficiency

19. Physics 207: Lecture 27, Pg 19 Compare  I, the efficiency of engine I, to  II, the efficiency of engine II.  Engine I:  Requires Q in = 100 J of heat added to system to get W=10 J of work (done on world in cycle)   = 10 / 100 = 0.10  Engine II:  To get W=10 J of work, Q out = 100 J of heat is exhausted to the environment  Q in = W+ Q out = 100 J + 10 J = 110 J   = 10 / 110 = 0.09 Exercise Efficiency

20. Physics 207: Lecture 27, Pg 20 Refrigerator (Heat pump) l Device that uses work to transfer heat from a colder object to a hotter object.

21. Physics 207: Lecture 27, Pg 21 The best thermal engine ever, the Carnot engine l A perfectly reversible engine (a Carnot engine) can be operated either as a heat engine or a refrigerator between the same two energy reservoirs, by reversing the cycle and with no other changes. l A Carnot cycle for a gas engine consists of two isothermal processes and two adiabatic processes l A Carnot engine has max. thermal efficiency, compared with any other engine operating between T H and T C l A Carnot refrigerator has a maximum coefficient of performance, compared with any other refrigerator operating between T H and T C.

22. Physics 207: Lecture 27, Pg 22 The Carnot Engine l All real engines are less efficient than the Carnot engine because they operate irreversibly due to the path and friction as they complete a cycle in a brief time period. l Carnot showed that the thermal efficiency of a Carnot engine is:

23. Physics 207: Lecture 27, Pg 23 Problem l You can vary the efficiency of a Carnot engine by varying the temperature of the cold reservoir while maintaining the hot reservoir at constant temperature. Which curve that best represents the efficiency of such an engine as a function of the temperature of the cold reservoir? Temp of cold reservoir

24. Physics 207: Lecture 27, Pg 24 The Carnot Engine (the best you can do) l No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs. A.A  B, the gas expands isothermally while in contact with a reservoir at T h B.B  C, the gas expands adiabatically (Q=0,  U=W B  C,T h  T c ), PV  =constant C.C  D, the gas is compressed isothermally while in contact with a reservoir at T c D.D  A, the gas compresses adiabatically (Q=0,  U=W D  A,T c  T h ) V P QhQh QcQc A B C D W cycle

25. Physics 207: Lecture 27, Pg 25  Carnot = 1 - Q c /Q h Q A  B = Q h = W AB = nRT h ln(V B /V A ) Q C  D = Q c = W CD = nRT c ln(V D /V C ) (here we reference work done by gas, dU = 0 = Q – P dV) But P A V A =P B V B =nRT h and P C V C =P D V D =nRT c so P B /P A =V A /V B and P C /P D =V D /V \C as well as P B V B  =P C V C  and P D V D  =P A V A  with P B V B  /P A V A  =P C V C  /P D V D  thus  ( V B /V A )=( V D /V C ) Q c /Q h =T c /T h Finally  Carnot = 1 - T c / T h Carnot Cycle Efficiency A B C D W cycle Q=0 QhQh QcQc

26. Physics 207: Lecture 27, Pg 26 Other cyclic processes: Turbines l A turbine is a mechanical device that extracts thermal energy from pressurized steam or gas, and converts it into useful mechanical work. 90% of the world electricity is produced by steam turbines. l Steam turbines &jet engines use a Brayton cycle

27. Physics 207: Lecture 27, Pg 27 Steam Turbine in Madison l MG&E, the electric power plan in Madison, boils water to produce high pressure steam at 400°C. The steam spins the turbine as it expands, and the turbine spins the generator. The steam is then condensed back to water in a Monona-lake-water- cooled heat exchanger, down to 20°C. l Carnot Efficiency?

28. Physics 207: Lecture 27, Pg 28 The Sterling Cycle l Return of a 1800’s thermodynamic cycle Isothermal expansion Isothermal compression SRS Solar System (~27% eff.)

29. Physics 207: Lecture 27, Pg 29 Sterling cycles Gas T=T H Gas T=T H Gas T=T C Gas T=T C 1 1 2 3 4 V P TCTC THTH VaVa VbVb 12 3 4 x start l 1 Q, V constant  2 Isothermal expansion ( W on system < 0 )  3 Q, V constant  4 Q out, Isothermal compression ( W on sys > 0) 1 Q 1 = nR C V (T H - T C ) 2 W on2 = -nR T H ln (V b / V a )= -Q 2 3 Q 3 = nR C V (T C - T H ) 4 W on4 = -nR T L ln (V a / V b )= -Q 4 Q Cold = - (Q 3 + Q 4 ) Q Hot = (Q 1 + Q 2 )  = 1 – Q Cold / Q Hot

30. Physics 207: Lecture 27, Pg 30  Carnot = 1 - Q c /Q h = 1 - T c /T h Carnot Cycle Efficiency Power from ocean thermal gradients… oceans contain large amounts of energy See: http://www.nrel.gov/otec/what.html

31. Physics 207: Lecture 27, Pg 31 Ocean Conversion Efficiency  Carnot = 1 - T c /T h = 1 – 275 K/300 K = 0.083 (even before internal losses and assuming a REAL cycle) Still: “This potential is estimated to be about 10 13 watts of base load power generation, according to some experts. The cold, deep seawater used in the OTEC process is also rich in nutrients, and it can be used to culture both marine organisms and plant life near the shore or on land.” “Energy conversion efficiencies as high as 97% were achieved.” See: http://www.nrel.gov/otec/what.htmlhttp://www.nrel.gov/otec/what.html So  =1-Q c /Q h is always correct but  Carnot =1-T c /T h only reflects a Carnot cycle

32. Physics 207: Lecture 27, Pg 32 Internal combustion engine: gasoline engine (Adiabats) l A gasoline engine utilizes the Otto cycle, in which fuel and air are mixed before entering the combustion chamber and are then ignited by a spark plug. Otto Cycle

33. Physics 207: Lecture 27, Pg 33 Internal combustion engine: Diesel engine l A Diesel engine uses compression ignition, a process by which fuel is injected after the air is compressed in the combustion chamber causing the fuel to self-ignite.

34. Physics 207: Lecture 27, Pg 34 Thermal cycle alternatives l Fuel Cell Efficiency (from wikipedia) Fuel cells do not operate on a thermal cycle. As such, they are not constrained, as combustion engines are, in the same way by thermodynamic limits, such as Carnot cycle efficiency. The laws of thermodynamics also hold for chemical processes (Gibbs free energy) like fuel cells, but the maximum theoretical efficiency is higher (83% efficient at 298K ) than the Otto cycle thermal efficiency (60% for compression ratio of 10 and specific heat ratio of 1.4).Carnot cycleGibbs free energyOtto cycle l Comparing limits imposed by thermodynamics is not a good predictor of practically achievable efficiencies l The tank-to-wheel efficiency of a fuel cell vehicle is about 45% at low loads and shows average values of about 36%. The comparable value for a Diesel vehicle is 22%.fuel cell vehicle l Honda Clarity (now leased in CA and gets ~70 mpg equivalent) This does not include H 2 production & distribution

35. Physics 207: Lecture 27, Pg 35 Fuel Cell Structure

36. Physics 207: Lecture 27, Pg 36 Problem-Solving Strategy: Heat-Engine Problems

37. Physics 207: Lecture 27, Pg 37 Going full cycle l 1 mole of an ideal gas and PV= nRT  T = PV/nR T 1 = 8300 0.100 / 8.3 = 100 K T 2 = 24900 0.100 / 8.3 = 300 K T 3 = 24900 0.200 / 8.3 = 600 K T 4 = 8300 0.200 / 8.3 = 200 K (W net = 16600*0.100 = 1660 J) 1  2  E th = 1.5 nR  T = 1.5x8.3x200 = 2490 J W by =0Q in =2490 J Q H =2490 J 2  3  E th = 1.5 nR  T = 1.5x8.3x300 = 3740 J W by =2490 J Q in =3740 J Q H = 6230 J 3  4  E th = 1.5 nR  T = -1.5x8.3x400 = -4980 J W by =0Q in =-4980 J Q C =4980 J 4  1  E th = 1.5 nR  T = -1.5x8.3x100 = -1250 J W by =-830 J Q in =-1240 J Q C = 2070 J Q H(total) = 8720 J Q C(total) = 7060 J  =1660 / 8720 =0.19 (very low) V P 100 liters 200 liters 1 2 3 4 24900 N/m 2 8300 N/m 2

38. Physics 207: Lecture 27, Pg 38 Home Exercise If an engine operates at half of its theoretical maximum efficiency (  max ) and does work at the rate of W J/s, then, in terms of these quantities, how much heat must be discharged per second. l This problem is about process (Q and W), specifically Q C ?  max  = 1- Q C /Q H and  = ½  max  = ½(1- Q C /Q H ) also W =  Q H =  ½  max Q H  2W /  max = Q H -Q H (  max  -1) = Q C  Q C = 2W /  max (1 -  max )

39. Physics 207: Lecture 27, Pg 39 Lecture 27 Assignment Assignment  HW11, Due Tues., May 5 th  HW12, Due Friday, May 9 th  For Thursday, Read through all of Chapter 20