Lecture 13 Heat Engines • Thermodynamic processes and entropy • Thermodynamic cycles • Extracting work from heat - How do we define engine efficiency? - Carnot cycle: the best possible efficiency Reading for this Lecture: Reading for Lecture 14: Elements Ch 4D-F Elements Ch 10 Lecture 13, p 1 A Review of Some Thermodynamic Processes of an Ideal Gas We will assume ideal gases in our treatment of heat engines, because that simplifies the calculations. Isochoric (constant volume) 2 W= pdV = 0 by ∫ p ∆=Uα Nk ∆= T α V ∆ p 1 Q Q= ∆ U Temperature V changes Isobaric (constant pressure) p 1 2 Wby = pdV = pV ∆ ∫ p ∆=Uα NkT ∆= α pV ∆ Q QUW=∆+by =()α +1 pV ∆ Temperature and V volume change Lecture 13, p 2 Review (2) 1 p ∝ V Isothermal (constant temperature) 1 V2 V 2 2 NkT V2 p Wby = pdV = dV = NkT ln Q ∫ ∫ V V V1 V 1 1 Thermal Reservoir ∆U = 0 V T Q= W by Volume and pressure change Adiabatic (no heat flow) 1 p ∝ 1 V γ Wby = − ∆ U p 2 ∆=Uα NkT()()2 −= T 1 α pV 22 − pV 11 Q = 0 Volume, pressure and V temperature change Lecture 13, p 3 Closed Thermodynamic 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 p, V, and T) For example: p 2 3 1 V • U is a state function. Therefore: ∆U = 0 • The net work is the enclosed area. W= pdV ≠ 0 This is the reason that neither W nor Q is a state function. It makes no st ∫ • Energy is conserved (1 Law): Q= W ≠ 0 sense to talk about a state having a certain amount of W. Though of course we could use any curves to make a closed cycle, here we will consider isochoric, isobaric, isothermal, and adiabatic processes. Lecture 13, p 4 Introduction to Heat Engines One of the primary applications of thermodynamics is to Turn heat into work . The standard heat engine works on a cyclic process: 1) extract heat from a hot reservoir, 2) perform work , using some of the extracted heat, 3) dump unused heat into a cold reservoir (often the environment). 4) repeat over and over. We represent this process with a diagram: Hot reservoir at T A “reservoir” is a large body whose h temperature doesn’t change when it Qh absorbs or gives up heat Engine: Wby For heat engines we will define Qh , Q c Qc , and Wby as positive. Cold reservoir at Tc Energy is conserved: Qh = Q c + Wby Lecture 13, p 5 A Simple Heat Engine: the Stirling Cycle p 2 1 Th 3 4 Tc Va Vb V Two reservoirs: Th and Tc. Four processes: Two isotherms and two isochors The net work during one cycle: The area of the “parallelogram”. Lecture 13, p 6 The Stirling Cycle (2) 1) Isochoric Gas temperature increases 2) Isothermal at constant volume (piston can’t move) Gas expands at constant Th Qh1 Th Qh2 Th gas gas W2 T The cycle goes T c around like this. c We don’t describe the mechanical parts that move the gas cylinder back and forth between the reservoirs. 3) Isochoric 4) Isothermal Gas temperature decreases Gas is compressed at constant Tc at constant volume (piston can’t move) Th Th W4 gas gas Qc4 Qc3 Tc Tc Lecture 13, p 7 How Does this Engine Do Work? Look at the two isothermal processes (2 and 4) on the previous slide Process 2: expanding gas does W2 on the piston , as it expands from Va to V b. Process 4: contracting gas is done W4 by the piston , as it contracts from Vb to V a. If W 2 > W 4, the net work is positive. This is true, because the contracting gas is colder ( ⇒lower pressure). During one cycle : • The hot reservoir has lost some energy (Qh=Q h1 +Q h2 ). • The cold reservoir has gained some energy (Q c=Q c3 +Q c4 ). • The engine (the gas cylinder) has neither gained nor lost energy. The energy to do work comes from the hot reservoir, not from the engine itself. The net work done by the engine is: Wby = W 2 -W4 = Qh -Qc= Q h2 -Qc4 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 possible. Lecture 13, p 8 Act 1 On the last slide, why did we write Qh -Qc= Q h2 -Qc4 ? What happened to Q h1 and Q c3 ? (since Qh = Q h1 + Q h2 , and Q c = Q c1 + Q c2 ) a) They both = 0. b) They are not = 0, but they cancel. c) They average to zero over many cycles. Lecture 13, p 9 Solution On the last slide, why did we write Qh -Qc= Q h2 -Qc4 ? What happened to Q h1 and Q c3 ? (since Qh = Q h1 + Q h2 , and Q c = Q c1 + Q c2 ) a) They both = 0. b) They are not = 0, but they cancel. c) They average to zero over many cycles. They cancel, because the amount of heat needed to raise the temperature from Tc to Th at constant volume equals the amount of heat needed to lower the temperature from Th to Tc at constant volume*, even though Va ≠ Vb. *Note: This is only valid for ideal gases. CV is only independent of V for an ideal gas. Lecture 13, p 10 Heat Engine Efficiency We pay for the heat input, Q H, so: Cartoon picture of a heat engine: Define the efficiency Hot reservoir at Th workdone by the engine results ε ≡ = Q heat extracted from reservoir cost h Engine W W Q− Q Q by ε ≡by =h c =−1 c Q Qh Q h Q h c Cold reservoir at T Valid for all heat engines. c (Conservation of energy) Remember: We define Qh and Q c as positive . The arrows define direction of flow. What’s the best we can do? The Second Law will tell us. Lecture 13, p 11 Review Entropy in Macroscopic Systems Traditional thermodynamic entropy: S = k ln Ω = 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 13, p 12 Entropy in Quasi -static Heat Flow When V is constant: dS ≡ dU/T = dQ/T ⇐ W = 0, so dU = dQ In fact, dS = dQ/T during any reversible (quasi-static) process , even if V changes. The reason: In a reversible process, Stot (system plus environment) doesn’t change: 0 =dSsys + dS E ∆Stot = 0 if process is reversible. dU =dS + E The reservoir is supplying (or absorbing) heat. sys T dQ =dS − The reservoir’s energy gain is the system’s heat loss. sys T That’s how they interact. dQfinal dQ d S=, o r ∆ S = ∫ for any reversible process Tin it T Lecture 13, p 13 ∆S in Isothermal Processes Suppose V & p change but T doesn’t. Work is done (dW by = pdV). Heat must enter to keep T constant: dQ = dU+dW by . So: dQdU+ dW dUpdV+ dS = =by = Remember: This holds for quasi-static processes, in which T T T the system remains near thermal equilibrium at all times. Special case, ideal gas: p T=constant For an ideal gas, if dT = 0, then dU = 0. pdV NkTdV NkdV 1 dS = = = T VT V V2 NkdV V ∆S = = Nk ln 2 ∫ V V 2 V1 1 V Lecture 13, p 14 Quasi -static Adiabatic Processes Q = 0 (definition of an adiabatic process) V and T both change as the applied pressure changes. For example, if p increases (compress the system): • V decreases, and the associated S also decreases. • T increases, and the associated S also increases. These two effects must exactly cancel !! system Why? Because: • This is a reversible process, so Stot = 0. Insulated walls • No other entropy is changing. (Q = 0, and Wby just moves the piston.) p 1 So, in a quasi-static adiabatic process, ∆S = 0 . TH Note: We did not assume that the system is an 2 TC ideal gas. This is a general result. V Lecture 13, p 15 Heat Engine Efficiency We pay for the heat input, Q H, so: Cartoon picture of a heat engine: Define the efficiency Hot reservoir at Th workdone by the engine results ε ≡ = Q heat extracted from reservoir cost h Engine W W Q− Q Q by ε ≡by =h c =−1 c Q Qh Q h Q h c Cold reservoir at T Valid for all heat engines. c (Conservation of energy) Remember: We define Qh and Q c as positive . The arrows define direction of flow. What’s the best we can do? The Second Law will tell us. Lecture 13, p 16 The 2 nd Law Sets the Maximum Efficiency (1) ∆Stot ≥ 0 How to calculate ∆Stot ? Over one cycle: ∆Stot = ∆Sengine + ∆Shot +∆Scold = 0 Remember: From the definition of T Qh is the heat taken from the hot reservoir, so ∆Shot = -Qh/T h. Qc is the heat added to the cold reservoir, so ∆Scold = +Q c/Tc.