Irreversible Processes

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1 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

2 Irreversible Processes Entropy-increasing processes are irreversible, because the reverse processes would reduce entropy. Examples: Free-expansion (actually, any particle flow between regions of different density) Heat flow between two systems with different temperatures. Consider the four processes of interest here: Isothermal: Heat flow but no T difference. Reversible Adiabatic: Q = 0. No heat flow at all. Reversible Isochoric & Isobaric: Heat flow between different T s. Irreversible (Assuming that there are only two reservoirs.) p isotherm p adiabat p isochor p isobar V V reversible reversible irreversible irreversible V V Lecture 15, p 2

3 ACT 1 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 2) Consider instead the free expansion (i.e., not quasi-static) of a gas. What happens to the total entropy during this process? a) S tot < 0 b) S tot = 0 c) S tot > 0 gas vacuum Remove the barrier Lecture 15, p 3

4 Solution 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 Energy (heat) leaves the environment, so its entropy decreases. In fact, since the environment and gas have the same T, the two entropy changes cancel: S tot = 0. This is a reversible process. 2) Consider instead the free expansion (i.e., not quasi-static) of a gas. What happens to the total entropy during this process? a) S tot < 0 b) S tot = 0 c) S tot > 0 gas vacuum Remove the barrier Lecture 15, p 4

5 Solution 1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 Energy (heat) leaves the environment, so its entropy decreases. In fact, since the environment and gas have the same T, the two entropy changes cancel: S tot = 0. This is a reversible process. 2) Consider instead the free expansion (i.e., not quasi-static) of a gas. What happens to the total entropy during this process? a) S tot < 0 b) S tot = 0 c) S tot > 0 There is no work or heat flow, so U gas is constant. T is constant. However, because the volume increases, so does the number of available states, and therefore S gas increases. Nothing is happening to the environment. Therefore S tot > 0. This is not a reversible process. Lecture 15, p 5

6 Review Entropy in Macroscopic Systems Traditional thermodynamic entropy: S = k lnω = kσ 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 σ 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 S T T T V = = = If C V is constant: T T 1 T2 dt T 2 = CV = CV ln T T T 1 1 Lecture 15, p 6

7 ACT 2 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 7

8 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 T = C ln + C ln = C ln = (0.028) C = J/K 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 8

9 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 9

10 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 = Pa p f = p i /2 = 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 10

11 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 11

12 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 = pv γ i i γ f f 7 / 2 γ = = / 2 i pf = pi V f T f V nrt V i i = Vi Vf = Pa x pv f f = = 395 K nr γ γ Equation relating p and V for adiabatic process in α-ideal gas γ is the ratio of C p /C V for our diatomic gas (=(α+1)/α). 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 α i i α f f 1 α V i Tf = Ti = 395 K V f Alternative: Use the equation relating T and V for an adiabatic process to get the final temperature. α = 5/2 Lecture 15, p 12

13 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 Q h T h Q h T 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 = εq h Carnot: ε = 1 - T c /T h This has large ε when T h - T c is large. Refrigerator: We pay for W on, we want Q c. Q c = Q h - W on = ΚW on Carnot: Κ = T c /(T h - T c ) Heat pump: We pay for W on, we want Q h. Q h = Q c + W on = ΚW on Carnot: Κ = T h /(T h - T c ) These both have large Κ when T h - T c is small. Lecture 15, p 13

14 ACT 3: 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? Q c Cold reservoir at T c 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 14

15 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 15

16 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 16

17 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 17

18 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 p b p a b a c d Combustion exhaust / intake adiabats V 1 V 2 V b c and d a are nearly adiabatic processes, because the pistons move too quickly for much heat to flow. Lecture 15, p 18

19 Solution p p b b Calculate the efficiency: c W W Q = C ( T T ) in v b a W = W W by b c d a = U U = C ( T T ) b c b c v b c = 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 b c because Q = 0 d a Wby Cv ( Tb Tc ) Cv ( Ta Td ) ( Tc Td ) ε = = = 1 Q C ( T T ) ( T T ) in v b a b a p a a adiabats d V 1 V 2 V Lecture 15, p 19

20 Solution Write it in terms of volume instead of temperature. We know the volume of the cylinders. ( Tc Td ) ε = 1 ( T T ) T V = T V T = T ( V / V ) α α 1/ α c 2 b 1 c b 1 2 T V = T V T = T ( V / V ) α α 1/ α d 2 a 1 d a 1 2 ( Tc Td ) ( Tb Ta )( V1 / V2 ) = ( T T ) ( T T ) b a b a ε 1/ α 1/ α 1/ α 1 γ b 2 = V1 a V 1 V 2 V 2 = = = V V V V 1 γ p p b p a Compression ratio V 2 /V 1 10 γ = 1.4 (diatomic gas) ε = 60% (in reality about 30%, due to friction etc.) b a c d V 1 V 2 adiabats V Lecture 15, p 20

21 Solution Efficiency of Otto Cycle 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 21

22 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 22

23 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 - T S, where T is the temperature of the environment. U = mc T = 1 kg * 4186 J/kg. K * 80 K = J. S = mc ln(t H2O /T env ) = 1011 J/K F = J Remember to measure temperature in Kelvin. Otherwise, you ll get the entropy wrong. Lecture 15, p 23

24 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 = εq = -εmc dt Thus, the total work obtained as T drops from 100 C to 20 C is: 293 W = mc dt T 373 = ( 4186 J/K) T ( 293 K) ln = K Lecture 15, p 24

This geometry gives us a hot and cold side. These devices aren t as efficient as conventional refrigerators, but are much more compact (and don t have mechanical parts).

25 Non-mechanical Example: Peltier Cooler Electrons in an n-type semiconductor have larger U than electrons in a p-type semiconductor. Therefore, if we push current through a series of n-p and p-n junctions, as shown, the electrons are cooled (because they slow down) at p-n and heated at n-p. This geometry gives us a hot and cold side. These devices aren t as efficient as conventional refrigerators, but are much more compact (and don t have mechanical parts). One can obtain tens of degrees of T. Despite the radically different construction, this heat pump obeys the same limits on efficiency as the gas-based pumps, because these limits are based on the 1 st and 2 nd laws, not on any details. Lecture 15, p 25

26 Peltier Cooler (2) Here s the mechanical equivalent of a Peltier cooler. Raise and lower some gas (i.e., between high and low potential). The raising and lowering is fast, so there is little heat flow, and the pressure at any height is the ambient air pressure (i.e., it decreases with height). T 2 Let the gas cool the upper air. T 2 > T 2 T 2 Raise the gas. You ll do work The gas expands adiabatically. T 2 < T 1 Lower the gas. You ll do work. The gas contracts adiabatically. T 1 > T 2 Note: You must do work when raising and lowering the gas, due to the buoyancy (i.e., the mgh energy), which is not described by the pv diagram. Let the gas warm the lower air. T 1 < T 1 T 1 T 1 p T 1 T 1 T 2 Two adiabats, two isobars T 2 V This refrigerator does not achieve Carnot efficiency, because both of the processes represented by the horizontal arrows involve irreversible heat flow between objects at different temperatures. Lecture 15, p 26

Irreversible Processes

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 Irreversible Processes Entropy-increasing