Lecture 3 Heat Engines hermodynamic rocesses and entroy hermodynamic cycles Extracting work from heat - How do we define engine efficiency? - Carnot cycle: the best ossible efficiency Reading for this Lecture: Elements Ch 4D-F Reading for Lecture 4: Elements Ch 0 Lecture 3,
A Review of Some hermodynamic Processes of an Ideal Gas We will assume ideal gases in our treatment of heat engines, because that simlifies the calculations. Isochoric (constant volume) Wby = d = 0 U = αnk = α Q = U 2 Q emerature changes Isobaric (constant ressure) by W = d = U = αnk = α by ( α ) Q = U + W = + 2 Q emerature and volume change Lecture 3, 2
Review (2) Isothermal (constant temerature) by 2 2 Nk 2 = = = ln W d d Nk U = 0 Q = W by 2 Q hermal Reservoir olume and ressure change Adiabatic (no heat flow) Wby = U Q = 0 ( ) α ( ) U = αnk = 2 2 2 2 γ olume, ressure and temerature change Lecture 3, 3
Review Entroy in Macroscoic Systems raditional thermodynamic entroy: S = k lnω = kσ We want to calculate S from macrostate information (,,, U, N, etc.) Start with the definition of temerature in terms of entroy: σ S, or k U U, N, N he entroy changes when changes: (We re keeing and N fixed.) ds 2 du C d C d S = = = If C is constant: 2 d 2 = C = C ln Lecture 3, 4
Entroy in Quasi-static Heat Flow When is constant: ds du/ = dq/ W = 0, so du = dq In fact, ds = dq/ during any reversible (quasi-static) rocess, even if changes. he reason: In a reversible rocess, S tot (system lus environment) doesn t change: 0 = dssys + dse du = dssys + dq = dssys E S tot = 0 if rocess is reversible. he reservoir is sulying (or absorbing) heat. he reservoir s energy gain is the system s heat loss. hat s how they interact. d S fin a l d Q =, o r S = in it d Q for any reversible rocess Lecture 3, 5
S in Isothermal Processes Suose & change but doesn t. Work is done (dw by = d). Heat must enter to kee constant: dq = du+dw by. So: dq du + dw by du + d ds = = = Remember: his holds for quasi-static rocesses, in which the system remains near thermal equilibrium at all times. Secial case, ideal gas: For an ideal gas, if d = 0, then du = 0. =constant d Nkd Nkd ds = = = 2 Nkd 2 S = = Nk ln 2 Lecture 3, 6
AC ) We just saw that the entroy of a gas increases during a quasi-static isothermal exansion. What haens to the entroy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 2) Consider instead the free exansion (i.e., not quasi-static) of a gas. What haens to the total entroy during this rocess? a) S tot < 0 b) S tot = 0 c) S tot > 0 gas vacuum Remove the barrier Lecture 3, 7
Solution ) We just saw that the entroy of a gas increases during a quasi-static isothermal exansion. What haens to the entroy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 Energy (heat) leaves the environment, so its entroy decreases. In fact, since the environment and gas have the same, the two entroy changes cancel: S tot = 0. his is a reversible rocess. 2) Consider instead the free exansion (i.e., not quasi-static) of a gas. What haens to the total entroy during this rocess? a) S tot < 0 b) S tot = 0 c) S tot > 0 gas vacuum Remove the barrier Lecture 3, 8
Solution ) We just saw that the entroy of a gas increases during a quasi-static isothermal exansion. What haens to the entroy of the environment? a) S env < 0 b) S env = 0 c) S env > 0 Energy (heat) leaves the environment, so its entroy decreases. In fact, since the environment and gas have the same, the two entroy changes cancel: S tot = 0. his is a reversible rocess. 2) Consider instead the free exansion (i.e., not quasi-static) of a gas. What haens to the total entroy during this rocess? a) S tot < 0 b) S tot = 0 c) S tot > 0 here is no work or heat flow, so U gas is constant. is constant. However, because the volume increases, so does the number of available states, and therefore S gas increases. Nothing is haening to the environment. herefore S tot > 0. his is not a reversible rocess. Lecture 3, 9
Quasi-static Adiabatic Processes Q = 0 (definition of an adiabatic rocess) and both change as the alied ressure changes. For examle, if increases (comress the system): decreases, and the associated S also decreases. increases, and the associated S also increases. hese two effects must exactly cancel!! system Why? Because: his is a reversible rocess, so S tot = 0. No other entroy is changing. (Q = 0, and W by just moves the iston.) So, in a quasi-static adiabatic rocess, S = 0. Insulated walls H Note: We did not assume that the system is an ideal gas. his is a general result. 2 C Lecture 3, 0
Closed hermodynamic Cycles Closed cycles will form the basis of our heat engine discussion. A closed cycle is one in which the system returns to the initial state. (same,, and ) For examle: 2 3 U is a state function. herefore: he net work is the enclosed area. Energy is conserved ( st Law): U = 0 W = d Q = W 0 0 his is the reason that neither W nor Q is a state function. It makes no sense to talk about a state having a certain amount of W. hough of course we could use any curves to make a closed cycle, here we will consider isochoric, isobaric, isothermal, and adiabatic rocesses. Lecture 3,
Introduction to Heat Engines One of the rimary alications of thermodynamics is to urn heat into work. he standard heat engine works on a cyclic rocess: ) extract heat from a hot reservoir, 2) erform work, using some of the extracted heat, 3) dum unused heat into a cold reservoir (often the environment). 4) reeat over and over. We reresent this rocess with a diagram: Hot reservoir at h Q h A reservoir is a large body whose temerature doesn t change when it absorbs or gives u heat Engine: W by Q c For heat engines we will define Q h, Q c, and W by as ositive. Cold reservoir at c Energy is conserved: Q h = Q c + W by Lecture 3, 2
A Simle Heat Engine: the Stirling Cycle 2 4 3 h c a b wo reservoirs: h and c. Four rocesses: wo isotherms and two isochors he net work during one cycle: he area of the arallelogram. Lecture 3, 3
he Stirling Cycle (2) ) Isochoric Gas temerature increases at constant volume (iston can t move) 2) Isothermal Gas exands at constant h Q h h Q h2 h gas gas W 2 c 4) Isothermal Gas is comressed at constant c he cycle goes around like this. We don t describe the mechanical arts that move the gas cylinder back and forth between the reservoirs. c 3) Isochoric Gas temerature decreases at constant volume (iston can t move) h h gas W 4 gas Q c4 c Q c3 c Lecture 3, 4
How Does this Engine Do Work? Look at the two isothermal rocesses (2 and 4) on the revious slide Process 2: exanding gas does W 2 on the iston, as it exands from a to b. Process 4: contracting gas is done W 4 by the iston, as it contracts from b to a. If W 2 > W 4, the net work is ositive. his is true, because the contracting gas is colder ( lower ressure). During one cycle: he hot reservoir has lost some energy (Q h =Q h +Q h2 ). he cold reservoir has gained some energy (Q c =Q c3 +Q c4 ). he engine (the gas cylinder) has neither gained nor lost energy. he energy to do work comes from the hot reservoir, not from the engine itself. he net work done by the engine is: W by = W 2 - W 4 = Q h - Q c = Q h2 - Q c4 Not all of the energy taken from the hot reservoir becomes useful work. Some is lost into the cold reservoir. We would like to make Q c as small as ossible. Lecture 3, 5
Act 2 On the last slide, why did I write Q h - Q c = Q h2 - Q c4? What haened to Q h and Q c3? (since Q h = Q h + Q h2, and Q c = Q c + Q c2 ) a) hey both = 0. b) hey are not = 0, but they cancel. c) hey average to zero over many cycles. Lecture 3, 6
Solution On the last slide, why did I write Q h - Q c = Q h2 - Q c4? What haened to Q h and Q c3? (since Q h = Q h + Q h2, and Q c = Q c + Q c2 ) a) hey both = 0. b) hey are not = 0, but they cancel. c) hey average to zero over many cycles. hey cancel, because the amount of heat needed to raise the temerature from c to h at constant volume equals the amount of heat needed to lower the temerature from h to c at constant volume*, even though a b. *Note: his is only valid for ideal gases. C is only indeendent of for an ideal gas. Lecture 3, 7
Heat Engine Efficiency We ay for the heat inut, Q H, so: Define the efficiency work done by the engine results ε = heat extracted from reservoir cost W Q Q Q ε = = Q Q Q by h c c h h h alid for all heat engines. (Conservation of energy) Cartoon icture of a heat engine: Engine Remember: Hot reservoir at h Q h Q c Cold reservoir at c W by We define Q h and Q c as ositive. he arrows define direction of flow. What s the best we can do? he Second Law will tell us. Lecture 3, 8
he 2 nd Law Sets the Maximum Efficiency () How to calculate S tot? Over one cycle: S tot 0 S tot = S engine + S hot + S cold Remember: From the definition of Q h is the heat taken from the hot reservoir, so S hot = -Q h / h. Q c is the heat added to the cold reservoir, so S cold = +Q c / c. H C Stot = + H C = 0 Q Q Q C H Q C H 0 2 nd Law Q c cannot be zero. Some energy is always lost. Lecture 3, 9
he 2 nd Law Sets the Maximum Efficiency (2) efficiency W Q = ε = Q Q Q Q C H C H ε H C H C H nd from the 2 Law his is a universal law!! (equivalent to the 2 nd law) It is valid for any rocedure that converts thermal energy into work. We did not assume any secial roerties (e.g., ideal gas) of the material in the derivation. he maximum ossible efficiency, ε Carnot = - c / h, is called the Carnot efficiency. he statement that heat engines have a maximum efficiency was the first exression of the 2 nd law, by Sadi Carnot in 824. Lecture 3, 20
How Efficient Can an Engine be? Consider an engine that uses steam ( h = 00 C ) as the hot reservoir and ice ( c = 0 C ) as the cold reservoir. How efficient can this engine be? he Carnot efficiency is C 273 K ε Carnot = = = 0.27 373 K herefore, an engine that oerates between these two temeratures can, at best, turn only 27% of the steam s heat energy into useful work. H Question: How might we design an engine that has higher efficiency? Answer: By increasing h. (hat s more ractical than lowering c.) Electrical ower lants and race cars obtain better erformance by oerating at a much higher h. Lecture 3, 2
When Is ε Less than ε Carnot? We can write the efficiency loss in terms of the change of total entroy: QH QC QH QH W Stot = + = + H C H C C W = Q S H H C tot Engine Hot reservoir at h Q h W W S ε = QH H QH C C tot Q c Cold reservoir at c ε Carnot Lesson: Avoid irreversible rocesses. (ones that increase S tot ). direct heat flow from hot to cold free exansion (far from equilibrium) sliding friction Lecture 3, 22
Irreversible Processes Entroy-increasing rocesses are irreversible, because the reverse rocesses would reduce entroy. Examles: Free-exansion (actually, any article flow between regions of different density) Heat flow between two systems with different temeratures. Consider the four rocesses of interest here: Isothermal: Heat flow but no difference. Reversible Adiabatic: Q = 0. No heat flow at all. Reversible Isochoric & Isobaric: Heat flow between different s. Irreversible (Assuming that there are only two reservoirs.) isotherm adiabat isochor isobar reversible reversible irreversible irreversible Lecture 3, 23
Act 3: Stirling Efficiency Will our Stirling engine achieve Carnot efficiency? a) Yes b) No 2 3 h 4 c a b Lecture 3, 24
Solution Will our Stirling engine achieve Carnot efficiency? a) Yes b) No Processes and 3 are irreversible. (isochoric heating and cooling) 2 : Cold gas touches hot reservoir. 3: Hot gas touches cold reservoir. 4 3 h c a b Lecture 3, 25
Examle: Efficiency of Stirling Cycle 2 4 3 c h Area enclosed: W by = d a b otal work done by the gas is the sum of stes 2 and 4: b b d b W2 = d = Nkh = Nkh ln a a a a d b W4 = d = Nkc = Nkc ln a b b Wby = W2 + W4 = Nk ( h c ) ln a a b We need a temerature difference if we want to get work out of the engine. Lecture 3, 26
Solution 2 4 3 c h Area enclosed: W by = d a b Heat extracted from the hot reservoir, exhausted to cold reservoir: b Qh = Q + Q2 = αnk ( h c ) + Nkh ln a α b Qc = Q3 + Q4 = Nk ( h c ) Nkc ln a Lecture 3, 27
Solution Let s ut in some numbers: b = 2 a α = 3/2 (monatomic gas) h = 373K (boiling water) c = 273K (ice water) b ( h c )ln Wby ε = a ( ln 2 = 0.69) Qh 3 ( ) + ln b h c h 2 a 00(0.69) = = 6.9% 50 + 373(0.69) For comarison: ε carnot = 273/373 = 26.8% Lecture 3, 28
How to Achieve Carnot Efficiency o achieve Carnot efficiency, we must relace the isochors (irreversible) with reversible rocesses. Let s use adiabatic rocesses, as shown: Processes and 3 are now adiabatic. Processes 2 and 4 are still isothermal. his cycle is reversible, which means: S tot remains constant: ε = ε Carnot. 2 his thermal cycle is called the Carnot cycle, and an engine that imlements it is called a Carnot heat engine. 4 3 h c b a c d Lecture 3, 29
Heat Engine Summary For all cycles: ε = Q c Q h Some energy is dumed into the cold reservoir. For the Carnot cycle: Q Q c h = c h Q c cannot be reduced to zero. Carnot (best) efficiency: ε = Carnot c h Only for reversible cycles. Carnot engines are an idealization - imossible to realize. hey require very slow rocesses, and erfect insulation. When there s a net entroy increase, the efficiency is reduced: ε = ε C a rn o t C S Q H to t Lecture 3, 30
Next ime Heat Pums Refrigerators, Available Work and Free Energy Lecture 3, 3