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Home , Carnot heat engine

Lecture 11. (Ch. 4)

A heat – any device that is capable of converting thermal (heating) into mechanical energy (). We will consider an important class of such devices whose operation is cyclic.

Heating – the transfer of energy to a system by thermal contact with a reservoir. Work – the transfer of energy to a system by a change in the external parameters (V, el.-mag. and grav. fields, etc.).

The main question we want to address: what are the limitations imposed by thermodynamic on the performance of heat engines? Machines are Impossible

Perpetual Motion Machines of the first type – these designs seek to violation of the First Law create the energy required for their (energy conservation) operation out of nothing. Perpetual Motion Machines of the second type - these designs extract the energy required for their operation violation of the Second in a manner that decreases the Law of an isolated system.

hot reservoir Word of caution: for non-cyclic processes, T H 100% of heat can be transformed into work without violating the Second Law.

heat Example: an expands isothermally work being in thermal contact with a hot reservoir. Since U = const at T = const, all heat has been transformed into work. impossible cyclic Fundamental Difference between Heating and Work

- is the difference in the entropy transfer!

Transferring purely mechanical energy to or from a system does not (necessarily) change its entropy: ΔS = 0 for reversible processes. For this reason, all forms of work are thermodynamically equivalent to each other - they are freely convertible into each other and, in particular, into mechanical work. An ideal el. motor converts el. work into mech. work, an ideal el. generator converts mech. work into el. work.

Work can be completely converted into heat, but the inverse is not true. The transfer of energy by heating is accompanied with the entropy transfer δ Q Sd = T Thus, entropy enters the system with heating, but does not leave the system with the work. On the other hand, for a continuous operation of a heat engine, the net entropy change during a cycle must be zero! How is it possible??? The Price Should be Paid...

An engine can get rid of all the Essential parts of a heat engine entropy received from the hot (any continuously operating reservoir by transferring only part reversible device generating work from “heat”) of the input thermal energy to the cold reservoir. δ Q hot reservoir, TH d S = T heat Thus, the only way to get rid of the accumulating entropy is to absorb work more in the heating process than the amount converted entropy heat to work, and to “flush” the entropy with the flow of the waste heat out of the system. cold reservoir, TC An essential ingredient: a “Working substance” – the system difference between that absorbs heat, expels waste hot and cold reservoirs. energy, and does work (e.g., water vapor in the ) Perfect Engines (no extra S generated) (to simplify equations, I’ll omit δ in δQ throughout this lecture) The condition of continuous operation: hot reservoir, T Q Q H ΔS = ΔS H = C Sadi Carnot H C T T Q H C S H Δ H = QH T T C H QC = QH TH heat The work generated during one cycle of a work reversible process: W entropy

heat TH −TC W = QH − QC = QH TH Q ΔS = C Q C T C Carnot efficiency: C W T the highest possible e ≡ =1− C <1 max Q T cold reservoir, TC value of the energy H H conversion efficiency Consequences

Any difference TH –TC ≠ 0 can be exploited to generate mechanical energy.

The greater the TH –TC difference, the more efficient the engine. Energy waste is inevitable.

0 Example: In a typical nuclear power plant, TH = 300 C (~570K), TC = 0 40 C (~310K), and the maximum efficiency emax=0.45. If the plant generates 1000 MW of “work”, its waste heat production is at a rate ⎛ 1 ⎞ HC WWQQ ⎜ −=−= ⎟ ≈12201 MW ⎝ e ⎠ - more fuel is needed to get rid of the entropy then to generate useful power. This creates the problem of waste heat - e.g., the waste heat produced by human activities in the LA basin exceeds 7% of the solar energy falling on the basin (~ 1kW/m2). Real Engines

Real heat engines have lower efficiencies because the processes within the devices are hot reservoir, T H not perfectly reversible – the entropy will be generated by irreversible processes: QH ΔSH = QH TH W TC e = ≤1− = emax heat QH TH e = e only in the limit of reversible work max operation.

entropy W Some sources of irreversibility: heat heat may flow directly between reservoirs;

not all temperature difference TH –TC may QC be available (temperature drop across thermal ΔSC = QC TC resistances in the path of the heat flow); part of the work generated may be cold reservoir, TC converted to heat by friction; gas may expand irreversibly without doing work (as gas flow into vacuum). Problem The temperature inside the engine of a helicopter is 20000C, the temperature of the exhaust gases is 9000C. The mass of the helicopter is M = 2⋅103 kg, the heat of 3 combustion of gasoline is Qcomb = 47⋅10 kJ/kg, and the density of gasoline is ρ = 0.8 g/cm3. What is the maximum height that the helicopter can reach by burning V = 1 liter of gasoline?

The work done on lifting the helicopter: = MgHW W T For the ideal (the maximum efficiency): e 1−=≡ C QH TH ⎛ T ⎞ Thus, ⎜ C ⎟ W ⎜1−= ⎟QH ⎝ TH ⎠

The heat released in the gasoline combustion: H = comb ⋅ ρ ⋅VqQ ⎛ T ⎞ ⎜ C ⎟ MgH ⎜1−= ⎟ comb ρ ⋅⋅ Vq ⎝ TH ⎠ ρ ⋅⋅ Vq ⎛ T ⎞ 1047 3 ×⋅ kg/liter0.8kJ/kg ×1liter ⎛ 1173 K ⎞ H = comb ⎜1− C ⎟ = 1− = 928m ⎜ ⎟ 2 ⎜ ⎟ Mg ⎝ TH ⎠ × 8.92000 m/skg ⎝ 2273 K ⎠ Note: if TH and/or TC vary in the process, we still can δ H −δ QQ C δ W introduce an “instanteneous” efficiency: ()Te = = δ QH δ QH

Problem [ TH = f(t) ]

A reversible heat engine operates between two reservoirs, TC and TH.. The cold reservoir can be considered to have infinite mass, i.e., TC = T1 remains constant. However the hot reservoir consists of a finite amount of gas at constant (ν moles with a specific cV), thus TH decreases with time (initially, TH =T2, T2 > T1). After the heat engine has. operated for some long period of time, the temperature TH is lowered to TC =T1 (a) Calculate the heat extracted from the hot reservoir during this period. (b) What is the change of entropy of the hot reservoir during this period? (c) How much work did the engine do during this period?

T1 ∂QH ν dTc HV ν dTc HV T1 (a) ν VH ()−= TTcQ 12 (b) dS = = S =Δ→ =νc ln T T ∫ T V T H H T2 H 2

(c) H ∂−∂ QQ C ∂W T1 ()Te = = 1−= ∂ H = −ν dTcQ HV ∂QH ∂QH TH ⎛ T ⎞ ⎛ T ⎞ ⎜ 1 ⎟ ⎜ 1 ⎟ QW H ⎜1−∂=∂ ⎟ ν dTc HV ⎜1−−= ⎟ ⎝ TH ⎠ ⎝ TH ⎠ T1 ⎛ T ⎞ T W ⎜1−−= 1 ⎟ νν()−−= ν TcTTcdTc ln 2 ∫⎜ T ⎟ VHV 12 V 1 T T2 ⎝ H ⎠ 1 Problem Given 1 kg of water at 1000C and a very large block of ice at 00C. A reversible heat engine absorbs heat from the water and expels heat to the ice until work can no longer be extracted from the system. The heat capacity of water is 4.2 J/g·K. At the completion of the process: a) What is the temperature of the water? b) How much heat has been absorbed by the block of ice in the process? c) How much ice has been melted (the heat of fusion of ice is 333 J/g)? d) How much work has been done by the engine?

(a) Because the block of ice is very large, we can assume its −= ice TTe water = 0/1 temperature to be constant. When work can no longer be 0 extracted from the system, the efficiency of the cycle is zero: water ice ==→ 0 CTT (b) The heat absorbed by the block of ice:

δ H −δ QQ C TI (),TTe IW = δ C []−= (),1 δ HIW []−−= (),1 dTcmTTeQTTeQ WWWIW −= dTcm WWW δ QH TW T f T 273 dT ⎛ 373 ⎞ Q −= I −= cmTdTcm W kgK kJ/kg×××= ln2.41273 ⎜ ⎟ = 9.357 kJ C ∫ T WWW ProblemWWI ∫ T (cont.) 273 Ti W 373 W ⎝ ⎠

QC 9.357 kJ (c) The amount of melted ice: M == = 07.1 kg I L 333 J/g

(d) The work : = − QQW CH = × ⋅ × 9.357-100KKkJ/kg2.4kg1 = 1.62kJ kJ Carnot Cycle - is not very practical (too slow), but operates at the maximum efficiency allowed by the Second Law.

P 1 1 – 2 isothermal expansion (in contact with TH) absorbs 2 – 3 isentropic expansion to T heat C 3 – 4 isothermal compression (in contact with TC) 2 4 – 1 isentropic compression to TH (isentropic ≡ adiabatic+quasistatic) TH 4 3 Efficiency of Carnot T TC e 1−= C rejects heat cycle for an ideal gas: max T (Pr. 4.5) H V S On the S -T diagram, the work done is 3 2 the area of the loop: ∫ dU = 0 = ∫TdS − ∫ PdV

entropy The heat consumed at T (1 – 2) is the area 4 1 H

contained in gas surrounded by the broken line: Q = T (S − S ) S - entropy TC TH T H H H C contained in gas The Carnot heat engine operates at the maximum efficiency allowed by the Second Law. Other heat engines may have a lower efficiency even if the cycle is reversible (no friction, etc.) Problem Problem. Consider a heat engine working in a reversible cycle and using an ideal

gas with constant heat capacity cP as the working substance. The cycle consists of two processes at constant , joined by two adiabatic processes.

(a) Which temperature of TA, TB, TC, and TD is highest, and which is lowest? (b) Find the efficiency of this engine in terms of P1 and P2 . (c) Show that a Carnot engine with the same gas working between the highest and lowest has greater efficiency than this engine.

P (a) From the for an ideal gas (PV=RT), A B we know that P2 > AB > TTTT DC

From the adiabatic equation : > CB > TTTT DA P 1 D C Thus B = ( ,,,max DCBA ) D = ( ,,,min TTTTTTTTTT DCBA )

V Problem (cont.)

= ( −TTCQ ) P (b) The heat absorbed from the hot reservoir ABPAB A B P2 The heat released into the cold reservoir CD = ( −TTCQ DCP ) − QQ −TT Thus, the efficiency e = AB CD 1−= DC P 1 D C QAB −TT AB

From the equation for an : V γ constPV →= / γγ−1 = constPT γ γ γ γ TA TD TB TC γ −1 = γ −1 γ −1 = γ −1 PA PD PB PC

( − )/1 γγ ()− /1 γγ ( / BCB ) − ( / PPTPPT ADA ) ()− /1 γγ e 1−= −= ()/1 PP 21 −TT AB

γ −1 ⎛ P ⎞ γ T T (c) ⎜ 1 ⎟ D D e 1−= ⎜ ⎟ 11 =−<−= emax ⎝ P2 ⎠ TA TB Stirling heat engine

Stirling engine – a simple, practical heat engine using a gas as working substance. It’s more practical than Carnot, though its efficiency is pretty close to the Carnot maximum efficiency. The contains a fixed amount of gas which is transferred back and forth between a "cold" and and a "hot" end of a long cylinder. The "displacer piston" moves T 2 3 the gas between the two ends and the "power piston" T changes the internal volume as the gas expands and 2 contracts.

T 4 1 1 The gasses used inside a Stirling engine never leave the engine. There are no exhaust valves that vent high-pressure gasses, as in a gasoline or , and there are no V1 V2 explosions taking place. Because of this, Stirling engines are very quiet. The uses an external heat source, which could be anything from gasoline to solar energy to the heat produced by decaying plants. Today, Stirling engines are used in some very specialized applications, like in submarines or auxiliary power generators, where quiet operation is important. Efficiency of Stirling Engine

In the Stirling heat engine, a working substance, which may be assumed to be an

ideal monatomic gas, at initial volume V1 and temperature T1 takes in “heat” at constant volume until its temperature is T2, and then expands isothermally until its volume is V2. It gives out “heat” at constant volume until its temperature is again T1 and then returns to its initial state by isothermal contraction. Calculate the efficiency and compare with a Carnot engine operating between the same two temperatures. 3 T 1-2 12 V ()12 ()TTnRTTCQ 12 >−=−= 0 W12 = 0 2 3 2 V V T2 2 2 dV ⎛V ⎞ 2-3 == nRTPdVQ = nRT ⎜ 2 ⎟ 0ln −=> WQ 23 2 ∫∫ V 2 ⎜ V ⎟ 23 23 V1 V1 ⎝ 1 ⎠ T1 4 3 1 3-4 ()()TTnRTTCQ <−=−= 0 W = 0 34 V 21 2 21 34 V1 V1 V dV ⎛ V1 ⎞ V1 2 4-1 == nRTPdVQ = nRT ⎜ ⎟ 0ln −=< WQ 41 1 ∫∫ V 1 ⎜V ⎟ 23 23 V2 V2 ⎝ 2 ⎠ 3 ()()+− /ln VVTTT 1 Q + QQ 12212 T 3 1 = H = 2312 = 2 = 2 + > e − QQ CH − QQ 4123 ()()− /ln VVTT 1212 − 12 ()/ln2 12 eVVTT max Internal Combustion Engines () - engines where the fuel is burned inside the engine cylinder as opposed to that where the fuel is burned outside the cylinder (e.g., the Stirling engine). More economical than ideal-gas engines because a small engine can generate a considerable power. Otto cycle. Working substance – a mixture of air and 1 P vaporized gasoline. No hot reservoir – thermal energy w e is produced by burning fuel. or xp k a do ns ignition n i 0 – 3 intake (fuel+air is pulled into the cylinder e on b y by the retreating piston) 4 ga com s 3 – 4 isentropic compression w press ork d ion 2 one o 4 – 1 isochoric heating n gas 0 Patm 1 – 2 isentropic expansion 3 exhaust intake/exhaust 2 – 3 – 0 exhaust V - maximum cylinder volume V V2 V 2 1 - minimum cylinder volume V1 γ −1 ⎛ ⎞ V2 The efficiency: V1 T2 - the compression ratio e 1−= ⎜ ⎟ 1−= V (Pr. 4.18) ⎜ ⎟ 1 ⎝V2 ⎠ T1 γ = 1+2/f - the adiabatic exponent

For typical numbers V1/V2 ~8 , γ ~ 7/5 → e = 0.56, (in reality, e = 0.2 – 0.3)

(even an “ideal” efficiency is smaller than the second law limit 1-T3/T1) Otto cycle (cont.)

1 P S w e or xp k a do ns 1 Q = 0 ignition n i 2 e on b S1 y 4 ga com s QH w press QC ork d ion 2 one o n gas 0 4 Q = 0 Patm S2 3 exhaust 3 intake/exhaust

V1 V2 V V1 V2 V

QC e 1−= = VH ( −TTCQ 41 ) = VH ()−TTCQ 23 < 0 QH

P3 1− γ −1 −TT V ()− PP PV P PV T ⎛V ⎞ e 1−= 32 1−= 2 32 1−= 22 2 22 2 111 −=−=−= ⎜ 2 ⎟ −TT V ()− PP PV P PV T ⎜ V ⎟ 41 1 41 11 1− 4 11 1 ⎝ 1 ⎠ /CR V P1 R ⎛V2 ⎞ γ 1 =− e 1−= ⎜ ⎟ ⎜ V ⎟ CV ⎝ 1 ⎠ Refrigerators

The purpose of a refrigerator is to make thermal energy flow from cold to hot. The coefficient of hot reservoir, T H performance for a fridge:

QH QC QC 1 ΔSH = QH COP ≡ = = TH W QH − QC QH / QC −1

heat QH TH TC ≥ COP ≤ COPmax = QC TC TH −TC work W entropy COP is the largest when T and T are close

heat H C to each other!

For a typical kitchen fridge TH ~300K, TC~ Q 250K COP ~ 6 (for each J of el. energy, ΔS = C Q ⇒ C C the coolant can suck as much as 6 J of heat TC from the inside of the freezer). cold reservoir, TC A fridge that cools something from RT to LHe temperature (TC~ 4K) would have to be much less efficient. Example:

A “perfect” heat engine with e = 0.4 is used as a refrigerator (the heat reservoirs remain the same). How much heat QC can be transferred in one cycle from the cold reservoir to the hot one if the supplied in one cycle work is W =10 kJ?

QC W − QQ Q Q Q e == CH COP c == c = H Q Q W − QQ Q H H CH 1− C QH

QC 1− e − e 6.01 1−= e COP == = == 5.1 QH −11 + e e 4.0

QC one(in cycle) = ×COPW =15 kJ