8.21 The Physics of Energy Fall 2009

Transcription

1 MIT OpenCourseWare The Physics of Energy Fall 2009 For information about citing these materials or our Terms of Use, visit:

2 8.21 Lecture 9 Heat Engines September 28, Lecture 9: Heat Engines 1 / 16

3 Today: heat engines heat input Q in = T + S η max = T+ T T + (cyclic) heat engine work out W = Q in Q out waste heat output Q out T S Today: closed-cycle heat engines working fluid in container no fluid leaves or enters Friday: internal combustion engines working fluid piston 8.21 Lecture 9: Heat Engines 2 / 16

4 Components of a closed-cycle heat engine 1. Heat in 2. Expand do work 3. Release heat 4. Contract cycle Processes need not be distinct (e.g. isothermal expansion combines 1 + 2) 8.21 Lecture 9: Heat Engines 3 / 16

5 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 = C v dt (assume C V independent of T) Ideal gas law: pv = Nk B T 8.21 Lecture 9: Heat Engines 4 / 16

6 expanding fluid does work on piston Isothermal expansion dw dq fluid kept at temperature T reservoir at temperature T Constant temperature du = 0 dq = p dv Note: W = pdv = area under p-v curve! p p 1 pv = Nk B T = constant T T 1 = T 2 p 2 V 1 V 2 V S 1 S 2 S 8.21 Lecture 9: Heat Engines 5 / 16

7 Quasiequilibrium processes In real process, gradient heat flow (q = k T) dw dq reservoir at temperature T expanding fluid does work on piston fluid kept 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 ds total = 0! So desire reversible processes for efficient engines Lecture 9: Heat Engines 6 / 16

8 Adiabatic expansion (reversible) dw expanding fluid does work on piston fluid cools as expands dq = 0, no heat added du = ĉ v Nk B dt = ĉ v (pdv + Vdp) = pdv (ĉ v + 1)pdV + ĉ v Vdp = 0 pv γ = constant, γ = ĉ p /ĉ v No heat added dq = 0 du = p dv, ds = 0 p 1 pv γ = constant (γ air = 1.4) T 1 p T p 2 T 2 V 1 V 2 V S 1 = S 2 S 8.21 Lecture 9: Heat Engines 7 / 16

9 p p 1 p 4 p 2 p 3 Carnot Heat Engine 1 pv = const pv γ =const 4 2 pv = const pv γ = const 3 1 2: Isothermal expansion at T H Heat in, work out Q in = W : Adiabatic expansion, Q = S = 0, work W : Isothermal compression (T C ) Heat out, work in Q out = W : Adiabatic compression, Q = S = 0, work in W 4 1 Total work: area W = W 12 + W 23 W 34 W 41 = Q in Q out = (T H T C ) S η = W/Q in = (T H T C )/T H = η C V 1 V 4 V 2 V 3 V 8.21 Lecture 9: Heat Engines 8 / 16

10 Compare: pv and TS plots p 1 1 heat in T 1 = T p p 4 p 2 p T T 3 = T 4 heat out 4 3 V 1 V 4 V 2 V 3 V S 1 = S 4 S 2 = S 3 S W = p dv = Q = T ds Net work out = net heat in = area on both graphs Cycle analysis: Sudoku 4 relations p 1 V 1 = p 2 V 2, p 2 V γ 2 = p 3V γ 3,... solve given 4 independent variables (5 including T s or N) 8.21 Lecture 9: Heat Engines 9 / 16

11 s achieve optimal efficiency But generally very little work/cycle Previous example had unrealistic γ = 2.5. With γ = 1.4 p 1 1 p p 2 2 T T 1 = T 2 T 3 = T p 4 p V 1 V 2 V 4 V 3 V S 1 = S 4 S 2 = S 3 S 8.21 Lecture 9: Heat Engines 10 / 16

12 Better: Need new process Isometric heating Easy to do irreversibly, more tricky reversibly No change in volume dq = du = C V dt ds = C V dt/t p 2 V = constant T 2 p T p 1 T 1 S = constant + C V ln T V 1 = V 2 V S 1 S 2 S 8.21 Lecture 9: Heat Engines 11 / 16

13 Stirling cycle (note different [engineering] labeling 1-4!) p 3 3 T 3 = T p 2 2 T 1 = T p T p 4 p V 2 = V 3 V 1 = V 4 V S 2 S 3 S 1 S 4 S 1 2: Isothermal (T C ) compression, heat out Q out, work in W in. 2 3: Isometric heating T C T H, heat in Q h, no work. 3 4: Isothermal (T H ) expansion, heat in Q in, work W out. 4 1: Isometric cooling T H T C, heat out Q c, no work. Key: heat output Q c stored in regenerator, returned as Q h Achieves Carnot efficiency but greater work output than Carnot! 8.21 Lecture 9: Heat Engines 12 / 16

14 implementation (example) Cool Heat 8.21 Lecture 9: Heat Engines 13 / 16

15 Stirling cycle: example UseT H = 100 C, T C = 20 C, V 1 = 1 L, V 2 = 0.4 L, p 1 = 1 Atm 1. Use pv = Nk B T p s T p 3 = p H 2 TC = 3.18 Atm p Work = Q = p dv p p 4 p 1 = 1 Atm p 2 2 p 2 = p 1 V 1 = 2.5 Atm V 2 p 4 = = 1.27 Atm p dv = p 3V 3 ln V4 V 3 Q in = 129 J (ln 2.5) = 118 J Q out = 101 J (ln 2.5) = 92.5 J W = Q out Q in = 25.5 J Compare Carnot: W = 8.6 J V 2 = V 3 = 0.4L V V 1 = V 4 = 1L 8.21 Lecture 9: Heat Engines 14 / 16

16 s 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

17 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. pv γ = constant, S = constant. Isometric heating/cooling: reversible with regenerator, no work dq = du. V = constant, S = C V 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