Lecture 15 Heat Engines Review & Examples

p b Hot reservoir at Th pb Q Heat h c leak p Heat W a pump a adiabats Qc

d Cold reservoir at Tc

V1 V2 V

Lecture 15, p 1 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 p p isotherm adiabat isochor p isobar

V V V V reversible reversible irreversible irreversible

Lecture 15, p 2 ACT 1

1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment?

a) Senv < 0 b) Senv = 0 c) Senv > 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) Stot < 0 b) Stot = 0 c) Stot > 0

gas vacuum

Remove the barrier

Lecture 15, p 3 Solution

1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment?

a) Senv < 0 b) Senv = 0 c) Senv > 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: Stot = 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) Stot < 0 b) Stot = 0 c) Stot > 0

gas vacuum

Remove the barrier

Lecture 15, p 4 Solution

1) The entropy of a gas increases during a quasi-static isothermal expansion. What happens to the entropy of the environment?

a) Senv < 0 b) Senv = 0 c) Senv > 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: Stot = 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) Stot < 0 b) Stot = 0 c) Stot > 0

There is no work or heat flow, so Ugas is constant. ⇒ T is constant. However, because the volume increases, so does the number of available states,

and therefore Sgas increases. Nothing is happening to the environment. Therefore Stot > 0. This is not a reversible process.

Lecture 15, p 5 Review Entropy in Macroscopic Systems

Traditional thermodynamic entropy: S = k ln = 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∂ UVN, T  ∂ U VN ,

The entropy changes when T changes : (We’re keeping V and N fixed.)

dU CdTT2 CdT dS==V⇒ = S V T T∫ T T1

T2 dT T2  If C V is constant: =CV = C V ln   ∫ T T T1 1 

Lecture 15, p 6 ACT 2

Two blocks, each with heat capacity* T1 = 250K C = 1 J/K are initially at different T1 T2 T2 = 350K temperatures, T1 = 250 K , T2 = 350 K . They are then placed into contact, and

eventually reach a final temperature of Tf Tf Tf = 300 K 300 K. (Why?) What can you say about the total change in entropy S ? Two masses each tot with heat capacity C = 1J/K* a) Stot < 0 b) Stot = 0 c) Stot > 0

*For a solid, C = C V = C p, to good approximation, since V ≈≈≈ 0. Lecture 15, p 7 Solution

Two blocks, each with heat capacity* T1 = 250K C = 1 J/K are initially at different T1 T2 T2 = 350K temperatures, T1 = 250 K , T2 = 350 K . They are then placed into contact, and eventually reach a final temperature of Tf Tf Tf = 300 K 300 K. (Why?) What can you say about the total change in entropy S ? Two masses each tot with heat capacity C = 1J/K* a) Stot < 0 b) Stot = 0 c) Stot > 0

This is an irreversible process, so there must be a net increase in entropy. Let’s calculate S: Tf  T f Stot = Cln + C ln  T1  T 2 300   300  The positive term is slightly =Cln  + C ln   250   350  bigger than the negative term. 300 2  =Cln  = (0.028) C = 0.028 J/K 250× 350 

*For a solid, C = C V = C p, to good approximation, because V ≈ 0. Lecture 15, p 8 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 Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures?

Lecture 15, p 9 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 Vi = 10 liters to Vf = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures?

st Q = Wby The 1 law. For an ideal gas, T = 0 → U = 0. Positive Q means heat flows into the gas.

Wby = nRT ln(V f/V i) = 6048 J An expanding gas does work. 5 pi = nRT/V i = 8.72 ×10 Pa Use the ideal gas law, pV = nRT 5 pf = p i/2 = 4.36 ×10 Pa

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 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 Vi = 1 liter , Ti = 300 K . What are the final temperature and pressure?

Lecture 15, p 11 Solution

γ γ pVii= pV ff Equation relating p and V for adiabatic process in α-ideal gas 7 / 2 γ = = 1.4 γ is the ratio of C /C for our diatomic gas ( =( α+1)/ α). 5 / 2 p V γ Vi  pf= p i   Solve for p f. Vf  γ nRT V  = i i  We need to express it in terms of things we know. Vi V f  = 6.57x 106 Pa p V T =f f = 395 K Use the ideal gas law to calculate the final temperature. f nR

αα TVii= TV ff Alternative: Use the equation relating T and V for an adiabatic 1 process to get the final temperature. α = 5/2 α Vi  Tf= T i   = 395 K Vf 

Lecture 15, p 12 Helpful Hints in Dealing with Engines and Fridges Sketch the process (see figures below).

Define Qh and Q c and Wby (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 T T h h h Q = Q Qh Qh Qh leak h

Wby Won Won Q Q Q Tc c Tc c Tc c Qleak = Q C Engine: Refrigerator: Heat pump:

We pay for Qh, We pay for W on , We pay for W on , we want Wby . we want Q c. we want Qh. Wby = Qh -Qc = εQh Qc = Qh -Won = ΚWon Qh = Q c + W on = ΚWon Carnot: ε = 1 - Tc/T h Carnot: Κ = Tc/(T h -Tc) Carnot: Κ = Th/(T h -Tc) This has large ε These both have large Κ when T -T is small. when T -T is large. h c h c Lecture 15, p 13 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 at Th hot reservoir during one cycle? Qh a) Sh < 0 b) Sh = 0 c) Sh > 0 Engine W

Qc 2) What is the sign of the entropy change of the Cold reservoir at Tc cold reservoir ?

a) Sc < 0 b) Sc = 0 c) Sc > 0

3) Compare the magnitudes of the two changes.

a) | Sc| < | Sh| b) | Sc| = | Sh| c) | Sc| > | Sh|

Lecture 15, p 14 Solution

Consider a Carnot heat pump.

1) What is the sign of the entropy change of the Hot reservoir at Th hot reservoir during one cycle? Qh a) Sh < 0 b) Sh = 0 c) Sh > 0 Engine Energy (heat) is entering the hot reservoir, so the W number of available microstates is increasing. Qc 2) What is the sign of the entropy change of the Cold reservoir at Tc cold reservoir ?

a) Sc < 0 b) Sc = 0 c) Sc > 0

3) Compare the magnitudes of the two changes.

a) | Sc| < | Sh| b) | Sc| = | Sh| c) | Sc| > | Sh|

Lecture 15, p 15 Solution

Consider a Carnot heat pump.

a) Sc < 0 b) Sc = 0 c) Sc > 0 Energy (heat) is leaving the cold reservoir.

3) Compare the magnitudes of the two changes.

a) | Sc| < | Sh| b) | Sc| = | Sh| c) | Sc| > | Sh|

Lecture 15, p 16 Solution

Consider a Carnot heat pump.

a) Sc < 0 b) Sc = 0 c) Sc > 0 Energy (heat) is leaving the cold reservoir.

3) Compare the magnitudes of the two changes.

a) | Sc| < | Sh| b) | Sc| = | Sh| c) | Sc| > | Sh|

It’s a reversible cycle, so Stot = 0. Therefore, the two entropy changes must cancel. Remember that the entropy of the “engine” itself does not change.

Lecture 15, p 17 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

pb exhaust / intake c pa a adiabats d

V V1 V2

b→c and d→a are nearly adiabatic processes, because the pistons move too quickly for much heat to flow. Lecture 15, p 18 Solution

p b pb

Calculate the efficiency: c pa a adiabats

Qin= CT v( b − T a ) d W= W − W by b→ c d → a V1 V2 V

Wbc→ =−= UU b c CTT vb( − c ) because Qb→ c = 0

Wda→ =−= UU d a CTT vd( − a ) because Qd→ a = 0

Wby= CT vb( −− T c )( CT va − T d ) W CTT(−−− )( CTT )( TT − ) ε ==by vbc vad =−1 cd Qin CTT vba()− () TT ba −

Lecture 15, p 19 Solution

Write it in terms of volume instead of temperature. We know the volume of the cylinders.

(T− T ) ε =1 − c d p (T− T ) b b a p α α α 1/ b TVc2= TV b 1⇒ T cb = TVV( 12 / ) α α α 1/ TVd2= TV a 1⇒ T da = TVV( 12 / ) c 1/ α pa (TTcd− )( T ba − TVV )(/)1 2 ∴ = a adiabats (TTba− ) ( TT ba − ) γ α 1/α − 1/ 1 − V  V  V d =1 =  2 =  2 V V V V2  V 1  V 1 1 2 1−γ V  ε =1 − 2  V 1  Compression ratio ≡ V2/V 1 ≈ 10 γ = 1.4 (diatomic gas) → ε = 60% (in reality about 30%, due to friction etc.)

Lecture 15, p 20 Solution

1 0.9 0.8 0.7 0.6 Monatomic 0.5 Diatomic 0.4 Nonlinear' 0.3 0.2 0.1 Efficiency of Otto Cycle Otto of Efficiency 0 1 5 9 13 17

Conmpression Ratio (V 2/V 1)

Why not simply use a higher compression ratio?

• If V 2 big, we need a huge, heavy engine (OK for fixed installations).

• If V1 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 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 cH2O = 4186 J/kg K.

Lecture 15, p 22 Solution

F = U - TS, where T is the temperature of the environment. U = mc T = 1 kg * 4186 J/kg .K * 80 K = 3.349 ×10 5 J.

S = mc ln(T H2O /T env ) = 1011 J/K F = 3.87 ×10 4 J

Remember to measure temperature in Kelvin. Otherwise, you’ll get the entropy wrong.

Lecture 15, p 23 Solution

If we run a Carnot engine, the efficiency at a given water temperature is:

E(T) = 1 - T/Tenv . 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 293  W= − mc∫ 1 −  dT 373 T  373 373   =()()4186 J/KT − 293 K ln    293   293 =3.87 × 104 K

Lecture 15, p 24 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 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).

Let the gas cool the upper air. T’ > T T2 2 2 T’2

Note: You must do work when Raise the gas. The gas expands Lower the gas. The gas contracts raising and lowering the gas, You’ll do work adiabatically. T < T You’ll do work. adiabatically. T’1 > T’2 due to the buoyancy ( i.e. , the 2 1 mgh energy), which is not described by the pV diagram.

Let the gas warm the lower air. T 1 < T’1 T1 T’1

p Two adiabats, T’1 two isobars T1 This refrigerator does not achieve Carnot efficiency, T’2

T2 because both of the processes represented by the horizontal arrows involve irreversible heat flow between objects at different temperatures. V Lecture 15, p 26