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Lecture 27

Lecture 27 . Goals:. Ch. 18 Qualitatively understand 2 nd Law of Thermodynamics 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.

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Lecture 27

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  1. Lecture 27 Goals: • Ch. 18 • Qualitatively understand 2nd Law of Thermodynamics • 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 • HW12, Due Tuesday, May 4th • HW13, Due Friday, May 7th • For Thursday, Read through all of Chapter 20

  2. 2nd Law of Thermodynamics • Second law: “The entropy of an isolated system never decreases. It can only increase, or, in equilibrium, remain constant.” • The 2nd Law tells us how collisions move a system toward equilibrium. • Order turns into disorder and randomness. • With time thermal energy will always transfer from the hotter to the colder system, never from colder to hotter. • 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

  3. Reversible vs Irreversible • 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.)

  4. Reversible vs Irreversible • Based on the above comments, which of the following processes do you expect 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 tiny grain of sand at a time. 4. Freezing water originally at 5°C.

  5. Reversible vs Irreversible • 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.

  6. Heat Engines and Refrigerators • Heat Engine: Device that transforms heat into work ( Q  W) • 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 THand TC.

  7. Heat Engines 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 efficiencyη(lowercase Greek eta), defined as We can also write the thermal efficiency as

  8. Exercise Efficiency • Consider two heat engines: • Engine I: • Requires Qin = 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, Qout = 100 J of heat is exhausted to the environment • Compare hI, the efficiency of engine I, to hII, the efficiency of engine II.

  9. Exercise Efficiency • Compare hI, the efficiency of engine I, to hII, the efficiency of engine II. • Engine I: • Requires Qin = 100 J of heat added to system to get W=10 J of work (done on world in cycle) • h = 10 / 100 = 0.10 • Engine II: • To get W=10 J of work, Qout = 100 J of heat is exhausted to the environment • Qin = W+ Qout = 100 J + 10 J = 110 J • h = 10 / 110 = 0.09

  10. Refrigerator (Heat pump) • Device that uses work to transfer heat from a colder object to a hotter object.

  11. The best thermal engine ever, the Carnot engine • 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. • A Carnot cycle for a gas engine consists of two isothermal processes and two adiabatic processes • A Carnot engine has max. thermal efficiency, compared with any other engine operating between THand TC • A Carnot refrigerator has a maximum coefficient of performance, compared with any other refrigerator operating between THand TC.

  12. The Carnot Engine • Carnot showed that the thermal efficiency of a Carnot engine is: • 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.

  13. Problem • 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

  14. Other cyclic processes: Turbines • 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. • Steam turbines & jet engines use aBrayton cycle

  15. Steam Turbine in Madison • 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. • Carnot Efficiency?

  16. P 1 1 2 1 2 Gas Gas Gas Gas T=TC x T=TH T=TH T=TC TH 3 4 start TC V Va Vb 3 4 Sterling cycles • 1 Q, V constant  2 Isothermal expansion ( Won system < 0 )  3 Q, V constant  4 Q out, Isothermal compression ( Won sys> 0) • Q1 = n CV (TH - TC) • Won2 = -nR THln (Vb / Va)= -Q2 • Q3 = n CV (TC - TH) • Won4 = -nR TLln (Va / Vb)= -Q4 QCold = - (Q3 + Q4 ) QHot = (Q1 + Q2 ) h = 1 – QCold / QHot

  17. Power from ocean thermal gradients… oceans contain large amounts of energy Carnot Cycle Efficiency hCarnot = 1 - Qc/Qh = 1 - Tc/Th See: http://www.nrel.gov/otec/what.html

  18. hCarnot = 1 - Tc/Th = 1 – 275 K/300 K = 0.083 (even before internal losses and assuming a REAL cycle) Still: “This potential is estimated to be about 1013watts 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.html So h=1-Qc/Qh is always correct but hCarnot =1-Tc/Th only reflects a Carnot cycle Ocean Conversion Efficiency

  19. Internal combustion engine: gasoline engine • 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. (Adiabats) Otto Cycle

  20. Internal combustion engine: Diesel engine • 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.

  21. Thermal cycle alternatives • 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). • Comparing limits imposed by thermodynamics is not a good predictor of practically achievable efficiencies • 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%. • Honda Clarity (now leased in CA and gets ~70 mpg equivalent) This does not include H2 production & distribution

  22. P 2 3 24900 N/m2 4 1 8300 N/m2 V 100 liters 200 liters Going full cycle • 1 mole of an ideal gas and PV= nRT  T = PV/nR T1 = 8300 0.100 / 8.3 = 100 K T2 = 24900 0.100 / 8.3 = 300 K T3 = 24900 0.200 / 8.3 = 600 K T4 = 8300 0.200 / 8.3 = 200 K (Wnet = 16600*0.100 = 1660 J) 12 DEth= 1.5 nR DT = 1.5x8.3x200 = 2490 J Wby=0 DEth =2490 J QH=2490 J 23 DEth= 1.5 nR DT = 1.5x8.3x300 = 3740 J Wby=2490 J DEth =3740 J QH= 6230 J 34 DEth = 1.5 nR DT = -1.5x8.3x400 = -4980 J Wby=0 DEth =-4980 J QC=4980 J 41 DEth = 1.5 nR DT = -1.5x8.3x100 = -1250 J Wby=-830 J DEth =-1240 J QC= 2070 J QH(total)= 8720 J QC(total)= 7060 J h =1660 / 8720 =0.19 (very low)

  23. Lecture 27 • Assignment • HW12, Due Tuesday, May 4th • HW13, Due Friday, May 7th • For Thursday, Read through all of Chapter 20

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