Lecture 15 Heat Engines Review & Examples p p b b Hot reservoir at T h p a a c adiabats Heat leak Heat pump Q h Q c W d V 1 V 2 V Cold reservoir at T c Lecture 15, p 1
Review Entropy in Macroscopic Systems Traditional thermodynamic entropy: S = k lnw = ks We want to calculate S from macrostate information (p, V, T, U, N, etc.) Start with the definition of temperature in terms of entropy: 1 s 1 S, or kt U T U V, N V, N The entropy changes when T changes: (We re keeping V and N fixed.) ds 2 du C dt CVdT T T T V S If C V is constant: T T 1 T2 dt T 2 CV CV ln T T T 1 1 Lecture 15, p 2
ACT 1: An Irreversible Process Two blocks, each with heat capacity* C = 1 J/K are initially at different temperatures, T 1 = 250 K, T 2 = 350 K. They are then placed into contact, and eventually reach a final temperature of 300 K. (Why?) What can you say about the total change in entropy S tot? a) S tot < 0 b) S tot = 0 c) S tot > 0 T 1 T 2 T f T f T 1 = 250K T 2 = 350K T f = 300 K Two masses each with heat capacity C = 1J/K* *For a solid, C = C V = C p, to good approximation, since V 0. Lecture 15, p 3
Solution Two blocks, each with heat capacity* C = 1 J/K are initially at different temperatures, T 1 = 250 K, T 2 = 350 K. They are then placed into contact, and eventually reach a final temperature of 300 K. (Why?) What can you say about the total change in entropy S tot? a) S tot < 0 b) S tot = 0 c) S tot > 0 T 1 T 2 T f T f T 1 = 250K T 2 = 350K T f = 300 K Two masses each with heat capacity C = 1J/K* This is an irreversible process, so there must be a net increase in entropy. Let s calculate S: T f T f Stot C ln C ln T1 T2 300 300 Cln Cln 250 350 2 300 Cln (0.028) C 0.028 J/K 250 350 The positive term is slightly bigger than the negative term. *For a solid, C = C V = C p, to good approximation, because V 0. Lecture 15, p 4
Example Process To analyze heat engines, we need to be able to calculate U, T, W, Q, etc. for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from V i = 10 liters to V f = 20 liters and the temperature is kept at a constant 350 K? What are the initial and final pressures? Lecture 15, p 5
Solution To analyze heat engines, we need to be able to calculate U, T, W, Q, etc. for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from V i = 10 liters to V f = 20 liters and the temperature is kept at a constant 350 K? What are the initial and final pressures? Q = W by The 1 st law. For an ideal gas, T = 0 U = 0. Positive Q means heat flows into the gas. W by = nrt ln(v f /V i ) = 6048 J p i = nrt/v i = 8.7210 5 Pa p f = p i /2 = 4.3610 5 Pa An expanding gas does work. Use the ideal gas law, pv = nrt Where is the heat coming from? In order to keep the gas at a constant temperature, it must be put in contact with a large object (a heat reservoir ) having that temperature. The reservoir supplies heat to the gas (or absorbs heat, if necessary) in order to keep the gas temperature constant. Very often, we will not show the reservoir in the diagram. However, whenever we talk about a system being kept at a specific temperature, a reservoir is implied. Lecture 15, p 6
Example Process (2) Suppose a mole of a diatomic gas, such as O 2, is compressed adiabatically so the final volume is half the initial volume. The starting state is V i = 1 liter, T i = 300 K. What are the final temperature and pressure? Lecture 15, p 7
Solution Suppose a mole of a diatomic gas, such as O 2, is compressed adiabatically so the final volume is half the initial volume. The starting state is V i = 1 liter, T i = 300 K. What are the final temperature and pressure? pv i i f f 7 / 2 1.4 5 / 2 i pf pi V f T f p V V nrt V i i Vi Vf 6 6.57x10 Pa pv f f 395 K nr Equation relating p and V for adiabatic process in -ideal gas is the ratio of C p /C V for our diatomic gas (=(+1)/). Solve for p f. We need to express it in terms of things we know. Use the ideal gas law to calculate the final temperature. T V T V i i f f V i Tf Ti V f 1 395 K Alternative: Use the equation relating T and V for an adiabatic process to get the final temperature. = 5/2 Lecture 15, p 8
Helpful Hints in Dealing with Engines and Fridges Sketch the process (see figures below). Define Q h and Q c and W by (or W on ) as positive and show directions of flow. Determine which Q is given. Write the First Law of Thermodynamics (FLT). We considered three configurations of Carnot cycles: T h T Q h T h Q h h Q h Q leak = Q h W by W on W on T c Q c T c Q c T c Q c Q leak = Q C Engine: We pay for Q h, we want W by. W by = Q h - Q c = eq h Carnot: e = 1 - T c /T h This has large e when T h - T c is large. Refrigerator: We pay for W on, we want Q c. Q c = Q h - W on = KW on Carnot: K = T c /(T h - T c ) Heat pump: We pay for W on, we want Q h. Q h = Q c + W on = KW on Carnot: K = T h /(T h - T c ) These both have large K when T h - T c is small. Lecture 15, p 9
ACT 2: Entropy Change in Heat Pump Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) S h < 0 b) S h = 0 c) S h > 0 Engine Hot reservoir at T h Q h W 2) What is the sign of the entropy change of the cold reservoir? a) S c < 0 b) S c = 0 c) S c > 0 Q c Cold reservoir at T c 3) Compare the magnitudes of the two changes. a) S c < S h b) S c = S h c) S c > S h Lecture 15, p 10
Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) S h < 0 b) S h = 0 c) S h > 0 Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing. 2) What is the sign of the entropy change of the cold reservoir? Engine Hot reservoir at T h Q h Q c Cold reservoir at T c W a) S c < 0 b) S c = 0 c) S c > 0 3) Compare the magnitudes of the two changes. a) S c < S h b) S c = S h c) S c > S h Lecture 15, p 11
Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) S h < 0 b) S h = 0 c) S h > 0 Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing. 2) What is the sign of the entropy change of the cold reservoir? Engine Hot reservoir at T h Q h Q c Cold reservoir at T c W a) S c < 0 b) S c = 0 c) S c > 0 Energy (heat) is leaving the cold reservoir. 3) Compare the magnitudes of the two changes. a) S c < S h b) S c = S h c) S c > S h Lecture 15, p 12
Solution Consider a Carnot heat pump. 1) What is the sign of the entropy change of the hot reservoir during one cycle? a) S h < 0 b) S h = 0 c) S h > 0 Energy (heat) is entering the hot reservoir, so the number of available microstates is increasing. 2) What is the sign of the entropy change of the cold reservoir? Engine Hot reservoir at T h Q h Q c Cold reservoir at T c W a) S c < 0 b) S c = 0 c) S c > 0 Energy (heat) is leaving the cold reservoir. 3) Compare the magnitudes of the two changes. a) S c < S h b) S c = S h c) S c > S h It s a reversible cycle, so S tot = 0. Therefore, the two entropy changes must cancel. Remember that the entropy of the engine itself does not change. Lecture 15, p 13
Example: Gasoline Engine It s not really a heat engine because the input energy is via fuel injected directly into the engine, not via heat flow. There is no obvious hot reservoir. However, one can still calculate work and energy input for particular gas types. We can treat the gasoline engine as an Otto cycle: p b Combustion p b exhaust / intake p a a c adiabats d V 1 V 2 V bc and da are nearly adiabatic processes, because the pistons move too quickly for much heat to flow. Lecture 15, p 14
Solution p p b b Calculate the efficiency: Q C ( T T ) in v b a W W W by bc d a W U U C ( T T ) bc b c v b c W U U C ( T T ) d a d a v d a W C ( T T ) C ( T T ) by v b c v a d because Q 0 bc because Q 0 da Wby Cv ( Tb Tc ) Cv ( Ta Td ) ( Tc Td ) e 1 Q C ( T T ) ( T T ) in v b a b a p a c a adiabats d V 1 V 2 V Lecture 15, p 15
Solution Write it in terms of volume instead of temperature. We know the volume of the cylinders. ( Tc Td) e 1 ( T T ) T V T V T T ( V / V ) 1/ c 2 b 1 c b 1 2 T V T V T T ( V / V ) 1/ d 2 a 1 d a 1 2 ( Tc Td ) ( Tb Ta )( V1/ V2) ( T T ) ( T T ) b a b a b 1/ 1/ 1/ 1 2 1 1 V e 1 V 2 1 a V 1 V 2 V 2 V V V 1 p p b p a Compression ratio V 2 /V 1 10 = 1.4 (diatomic gas) e = 60% (in reality about 30%, due to friction etc.) b a c d V 1 V 2 adiabats V Lecture 15, p 16
Efficiency of Otto Cycle Solution 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 1 5 9 13 17 Conmpression Ratio (V 2 /V 1 ) Monatomic Diatomic Nonlinear' Why not simply use a higher compression ratio? If V 2 big, we need a huge, heavy engine (OK for fixed installations). If V 1 small, the temperature gets too high, causing premature ignition. High compression engines need high octane gas, which has a higher combustion temperature. Lecture 15, p 17
Free Energy Example Suppose we have a liter of water at T = 100 C. What is its free energy, if the environment is T = 20 C? Verify the result by calculating the amount of work we could obtain. Remember that c H2O = 4186 J/kg. K. Lecture 15, p 18
Solution Suppose we have a liter of water at T = 100 C. What is its free energy, if the environment is T = 20 C? Verify the result by calculating the amount of work we could obtain. Remember that c H2O = 4186 J/kg. K. F = U - TS, where T is the temperature of the environment. U = mct = 1 kg * 4186 J/kg. K * 80 K = 3.34910 5 J. S = mc ln(t H2O /T env ) = 1011 J/K F = 3.8710 4 J Remember to measure temperature in Kelvin. Otherwise, you ll get the entropy wrong. Lecture 15, p 19
Solution Suppose we have a liter of water at T = 100 C. What is its free energy, if the environment is T = 20 C? Verify the result by calculating the amount of work we could obtain. Remember that c H2O = 4186 J/kg. K. If we run a Carnot engine, the efficiency at a given water temperature is: E(T) = 1 - T/T env. So, for each small decrease in water temperature, we get this much work out of the engine: dw = eq = -emc dt Thus, the total work obtained as T drops from 100 C to 20 C is: 293 293 W mc 1 dt T 373 373 4186 J/K T 293 Kln 293 4 3.88 10 K 373 293 Lecture 15, p 20
Beware!!! Maximum S tot Does Not Always Mean Minimum F sys When we introduced the Helmholtz free energy, F, last lecture (see the hot brick discussion), we assumed that the volume of the brick was constant. If the volume weren t constant, then the brick could gain or lose energy (and entropy) by contracting or expanding. That would change the analysis. It is very common that the pressure, not the volume, is constant (e.g., if our system is a gas at constant atmospheric pressure). In this situation, we use a different form of free energy, called Gibbs free energy : G = U+pV-TS. The pv term takes into account the work that is done during volume changes. We won t use Gibbs free energy in this course, but it important to be aware that the calculation of free energy depends on the situation. Lecture 15, p 21
Supplement: Gibbs Free Energy Most phase transitions are observed under constant p,t conditions, not constant-v,t. Unless the stuff is in a closed vessel, it s open to the air (thus, at atmospheric pressure). In this case, the entropy of the environment changes not only when the system energy changes but also when its volume changes. The reservoir is (by assumption) in equilibrium at fixed T and p. So, as heat flows, the change of the reservoir s entropy is: Q U pv U pv T T T R R R SR V R = -V and U R = -U because total volume and total energy are constant. The change in the total entropy is thus: U pv ( U pv TS) G Stot S SR S T T T T where: G U pv TS Variables U, V, and S are of the system. Fixed p and T are of the reservoir. In these conditions, maximizing S tot means minimizing the system s G. Lecture 15, p 22