Physics 207: Lecture 26, Pg 1 Dec. 1

Physics 207: Lecture 26, Pg 2 Lecture 26, Dec. 1 Goals: Chapter 19 Chapter 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 Friday, Dec. 5 th  HW12, Due Friday, Dec. 12 th  For Wednesday, Read through all of Chapter 20

Physics 207: Lecture 26, Pg 3 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.

Physics 207: Lecture 26, Pg 4 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

Physics 207: Lecture 26, Pg 5 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

Physics 207: Lecture 26, Pg 6 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

Physics 207: Lecture 26, Pg 7 Refrigerator (Heat pump) l Device that uses work to transfer heat from a colder object to a hotter object.

Physics 207: Lecture 26, Pg 8 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.

Physics 207: Lecture 26, Pg 9 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:

Physics 207: Lecture 26, Pg 10 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

Physics 207: Lecture 26, Pg 11 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

Physics 207: Lecture 26, Pg 12  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

Physics 207: Lecture 26, Pg 13 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

Physics 207: Lecture 26, Pg 14 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?

Physics 207: Lecture 26, Pg 15 The Sterling Cycle l Return of a 1800’s thermodynamic cycle Isothermal expansion Isothermal compression SRS Solar System (~27% eff.)

Physics 207: Lecture 26, Pg 16 Sterling cycles Gas T=T H Gas T=T H Gas T=T C Gas T=T C V P TCTC THTH VaVa VbVb 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

Physics 207: Lecture 26, Pg 17  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:

Physics 207: Lecture 26, Pg 18 Ocean Conversion Efficiency  Carnot = 1 - T c /T h = 1 – 275 K/300 K = (even before internal losses and assuming a REAL cycle) Still: “This potential is estimated to be about 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: So  =1-Q c /Q h is always correct but  Carnot =1-T c /T h only reflects a Carnot cycle

Physics 207: Lecture 26, Pg 19 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

Physics 207: Lecture 26, Pg 20 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.

Physics 207: Lecture 26, Pg 21 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

Physics 207: Lecture 26, Pg 22 Fuel Cell Structure

Physics 207: Lecture 26, Pg 23 Problem-Solving Strategy: Heat-Engine Problems

Physics 207: Lecture 26, Pg 24 Going full cycle l 1 mole of an ideal gas and PV= nRT  T = PV/nR T 1 = / 8.3 = 100 K T 2 = / 8.3 = 300 K T 3 = / 8.3 = 600 K T 4 = / 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 = J W by =0Q in =-4980 J Q C =4980 J 4  1  E th = 1.5 nR  T = -1.5x8.3x100 = 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 N/m N/m 2

Physics 207: Lecture 26, Pg 25 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 )

Physics 207: Lecture 26, Pg 26 Lecture 26, Dec. 1 Assignment Assignment  HW11, Due Friday, Dec. 5 th  HW12, Due Friday, Dec. 12 th  For Wednesday, Read through all of Chapter 20