MIT OpenCourseWare http://ocw.mit.edu 8.21 The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. Thermodynamic Processes Carnot engine Stirling engine 8.21 Lecture 9 Heat Engines September 28, 2009 8.21 Lecture 9: Heat Engines 1 / 16 Thermodynamic Processes Carnot engine Stirling engine Today: heat engines (cyclic) heat input work out - heat - ∆Q = T ∆S ∆W = ∆Q − ∆Q in + engine in out T+−T− “waste” heat output ηmax = T+ ? ∆Qout ≥ T−∆S piston Today: closed-cycle heat engines • working fluid in container working • no fluid leaves or enters fluid Friday: internal combustion engines 8.21 Lecture 9: Heat Engines 2 / 16 Thermodynamic Processes Carnot engine Stirling engine Components of a closed-cycle heat engine 1. Heat in 2. Expand ⇒ do 3. Release heat 4. Contract work ⇒ cycle Processes need not be distinct (e.g. isothermal expansion combines 1 + 2) 8.21 Lecture 9: Heat Engines 3 / 16 Thermodynamic Processes Carnot engine Stirling engine What we need from thermodynamics for heat engine analysis: We will analyze thermodynamic processes on working fluid —Follow processes in terms of state variables p, V, S, T —Assume near equilibrium at all times (quasiequilibrium) —Combine thermodynamic processes ⇒ heat engine cycles First law⇒ dQ = dU + p dV (quasieq. → no free expansion) Definition of temperature⇒ dQ = TdS (quasieq.) Heat capacity: dU = Cv dT (assume CV independent of T) Ideal gas law: pV = NkBT 8.21 Lecture 9: Heat Engines 4 / 16 Thermodynamic Processes Carnot engine Stirling engine expanding fluid does work on piston dW fluid kept at Isothermal expansion temperature T dQ reservoir at temperature T Constant temperature ⇒ dU = 0 ⇒ dQ = p dV Note: W = R pdV = area under p-V curve! p1 pV = NkBT = constant T1 = T2 p T p2 V1 V2 S1 S2 V S 8.21 Lecture 9: Heat Engines 5 / 16 Thermodynamic Processes Carnot engine Stirling engine expanding fluid Quasiequilibrium processes does work on piston dW fluid kept at In real process, temperature T gradient ⇒ heatdQ flow (q = −k∇T) reservoir at temperature T But for slow expansion, temperature approximately constant at each time. Quasiequilibrium assumption: No gradients, system in equilibrium at each time t. With quasiequilibrium assumption: isothermal expansion reversible Process reversible ⇔ dStotal = 0! So desire reversible processes for efficient engines. 8.21 Lecture 9: Heat Engines 6 / 16 Thermodynamic Processes Carnot engine Stirling engine expanding fluid does work on piston dU = ˆc Nk dT Adiabatic dW v B = ˆc (pdV + Vdp) = −pdV expansion fluid cools as v expands (reversible) ⇒ (ˆcv + 1)pdV + ˆcvVdp = 0 γ ⇒ pV = constant, γ = ˆcp/ˆcv dQ = 0, no heat added No heat added ⇒ dQ = 0 ⇒ dU = −p dV, dS = 0 T1 γ p1 pV = constant (γair = 1.4) p T p2 T2 V1 V2 S1 = S2 V S 8.21 Lecture 9: Heat Engines 7 / 16 Thermodynamic Processes Carnot engine Stirling engine 1→2: Isothermal expansion at TH Heat in, work out ∆Qin = W1→2 Carnot Heat Engine 2→3: Adiabatic expansion, ∆Q = ∆S = 0, work W2→3 3→4: Isothermal compression (TC) p1 1 Heat out, work in ∆Qout = W3→4 pV = const 4→1: Adiabatic compression, γ pV =const ∆Q = ∆S = 0, work in W4→1 p p4 4 p 2 2 γ pV = const Total work: area W = W + W − W − W p3 pV = const 3 12 23 34 41 = Qin − Qout = (TH − TC)∆S η = W/Qin = (TH − TC)/TH = ηC V1 V4 V2 V3 V 8.21 Lecture 9: Heat Engines 8 / 16 Thermodynamic Processes Carnot engine Stirling engine Compare: pV and TS plots p1 1 heat in 1 2 T1 = T2 ? 3 p p 4 T T3 = T4 4 4 p2 2 ? heat out p3 3 V1 V4 V2 V3 S1 = S4 S2 = S3 V S W = R p dV = ∆Q = R T dS Net work out = net heat in = area on both graphs γ γ Cycle analysis: ∼ Sudoku–4 relations p1V1 = p2V2, p2V2 = p3V3 ,... ⇒ solve given 4 independent variables (5 including T’s or N) 8.21 Lecture 9: Heat Engines 9 / 16 Thermodynamic Processes Carnot engine Stirling engine Carnot engines achieve optimal efficiency But generally very little work/cycle Previous example had unrealistic γ = 2.5. With γ = 1.4 p1 1 1 2 T = T p 2 1 2 2 3 p T T3 = T4 4 p4 4 p3 3 V1 V2 V4 V3 S1 = S4 S2 = S3 V S 8.21 Lecture 9: Heat Engines 10 / 16 Thermodynamic Processes Carnot engine Stirling engine Better: Stirling engine Need new process — Isometric heating Easy to do irreversibly, more tricky reversibly No change in volume ⇒ dQ = dU = CV dT ⇒ dS = CV dT/T p2 T2 V = constant p T p1 T1 S = constant + CV ln T V1 = V2 S1 S2 V S 8.21 Lecture 9: Heat Engines 11 / 16 Thermodynamic Processes Carnot engine Stirling engine Stirling cycle (note different [engineering] labeling 1-4!) 3 4 p3 3 T3 = T4 1 p2 2 T1 = T2 2 p T p4 4 p1 1 V2 = V3 V1 = V4 S2S3 S1S4 V S 1 → 2: Isothermal (TC ) compression, heat out Qout, work in Win. 2 → 3: Isometric heating TC → TH , heat in Qh, no work. 3 → 4: Isothermal (TH ) expansion, heat in Qin, work Wout. 4 → 1: Isometric cooling TH → TC , heat out Qc, no work. Key: heat output Qc stored in regenerator, returned as Qh Achieves Carnot efficiency– but greater work output than Carnot! 8.21 Lecture 9: Heat Engines 12 / 16 Thermodynamic Processes Carnot engine Stirling engine Stirling engine implementation (example) Cool Heat 8.21 Lecture 9: Heat Engines 13 / 16 Thermodynamic Processes Carnot engine Stirling engine Stirling cycle: example ◦ ◦ UseTH = 100 C, TC = 20 C, V1 = 1 L, V2 = 0.4 L, p1 = 1 Atm 1. Use pV = NkBT ⇒ p’s TH ∼ p3 = p2 = 3.18 Atm TC 3 p3 2. Work = ∆Q = R p dV 4 p R V4 2 2 3 p dV = p3V3 ln V p1V1 3 p2 = = 2.5 Atm V2 p ∼ ∼ Qin = 129 J (ln 2.5) = 118 J ∼ p4 == 1.27 Atm ∼ ∼ Qout = 101 J (ln 2.5) = 92.5 J p4 4 p1 = 1 Atm 1 ∼ W = Qout − Qin = 25.5 J Compare Carnot: W =∼ 8.6 J V2 = V3 = 0.4L V1 = V4 = 1L V 8.21 Lecture 9: Heat Engines 14 / 16 Thermodynamic Processes Carnot engine Stirling engine Stirling engines Maximum (Carnot) efficiency Higher work/cycle than Carnot for given p, V extremes Theoretically very promising Variant: Ericsson cycle–constant pressure regenerative process External combustion, arbitrary fuel source (solar, nuclear, rice. ) Currently used in niche apps (space, mini cryocoolers. ) Technically challenging! —Hard to get true isothermal expansion —Seal problems at high pressure (Beale: free piston SE) —Materials exposed to high T for long time Maybe widely used in future? (vehicles, solar thermal?) 8.21 Lecture 9: Heat Engines 15 / 16 Thermodynamic Processes Carnot engine Stirling engine SUMMARY • Thermodynamic processes/cycles: use state variables p, V, T, S, U quasiequilibrium assumption, reversible processes. • Isothermal expansion/compression: add heat, do work dQ = dW, dU = 0. pV = constant, T = constant. • Adiabatic expansion/compression: no heat added dU = −pdV, dQ = 0. pVγ = constant, S = constant. • Isometric heating/cooling: reversible with regenerator, no work dQ = dU. V = constant, S = CV ln T+ constant. • Carnot cycle: IT exp., Ad exp., IT compr., Ad compr. Maximum (Carnot) efficiency, but small work done per cycle. • Stirling cycle: IT compr., IM heat, IT exp., IM cool. Maximum efficiency, more work/cycle than Carnot 8.21 Lecture 9: Heat Engines 16 / 16.