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
∆Qin = T+ ∆S
ηmax =
heat
engine
T+ −T−
T+
Today: closed-cycle heat engines
• working fluid in container
• no fluid leaves or enters
work out
∆W = ∆Qin − ∆Qout
“waste” heat output
? ∆Qout ≥ T− ∆S
piston
working
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
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
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 = NkB T
8.21 Lecture 9: Heat Engines
4 / 16
Thermodynamic Processes
Carnot engine
Stirling engine
expanding fluid
does work on piston
dW
fluid kept at
temperature T
Isothermal expansion
dQ
reservoir at temperature T
Constant temperature ⇒ dU = 0 ⇒ dQ = p dV
R
Note: W = pdV = area under p-V curve!
p1
T1 = T2
T
p
pV = NkB T
= constant
p2
V1
V
V2
S1
8.21 Lecture 9: Heat Engines
S
S2
5 / 16
Thermodynamic Processes
Carnot engine
Stirling engine
Quasiequilibrium processes
In real process,
gradient ⇒ heat flow (q = −k∇T)
expanding fluid
does work on piston
dW
fluid kept at
temperature T
dQ
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 = ĉv NkB dT
= ĉv (pdV + Vdp) = −pdV
dW
Adiabatic
expansion
(reversible)
fluid cools as
expands
⇒ (ĉv + 1)pdV + ĉv Vdp = 0
⇒ pV γ = constant, γ = ĉp /ĉv
dQ = 0, no heat added
No heat added ⇒ dQ = 0 ⇒ dU = −p dV, dS = 0
T1
pV γ = constant
p1
p
T
(γair = 1.4)
p2
T2
V1
V
V2
S1 = S2
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
p1
2→3: Adiabatic expansion,
∆Q = ∆S = 0, work W2→3
3→4: Isothermal compression (TC )
Heat out, work in ∆Qout = W3→4
1
p
pV = const
p4
p2
p3
4→1: Adiabatic compression,
∆Q = ∆S = 0, work in W4→1
pV γ =const
4
2
pV = const
V1 V4
V
V2
pV γ = const
3
V3
Total work: area
W = W12 + W23 − W34 − W41
= Qin − Qout = (TH − TC )∆S
η = W/Qin = (TH − TC )/TH = ηC
8.21 Lecture 9: Heat Engines
8 / 16
Thermodynamic Processes
Carnot engine
Stirling engine
Compare: pV and TS plots
p1
1
heat in
1
T1 = T2
2
p4
p2
4
2
V1 V4
R
T3 = T4
V
p dV = ∆Q =
V2
V3
R
T dS
3
4
?
heat out
3
p3
W=
T
p
?
S1 = S4
S
S2 = S3
Net work out = net heat in = area on both graphs
Cycle analysis: ∼ Sudoku–4 relations p1 V1 = p2 V2 , p2 V2γ = p3 V3γ , . . .
⇒ 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
T
p
p4
4
p3
V1
V2
1
T1 = T2
2
p2
V
V4
T3 = T4
2
3
4
3
V3
S1 = S4
8.21 Lecture 9: Heat Engines
S
S2 = S3
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
p
T
V = constant
p1
T1
V1 = V2
V
S = constant + CV ln T
S1
8.21 Lecture 9: Heat Engines
S
S2
11 / 16
Thermodynamic Processes
Carnot engine
Stirling engine
Stirling cycle (note different [engineering] labeling 1-4!)
3
p3
T1 = T2
2
p
V2 = V3
2:
3:
4:
1:
1
2
4
1
p4
p1
1→
2→
3→
4→
4
T
p2
3
T3 = T4
V
V1 = V4
S2S3
S
S1S4
Isothermal (TC ) compression, heat out Qout , work in Win .
Isometric heating TC → TH , heat in Qh , no work.
Isothermal (TH ) expansion, heat in Qin , work Wout .
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 = NkB T ⇒ p’s
T
p3 = p2 H ∼
3.18 Atm
TC =
3
p3
p2
p
p2 =
R
2. Work = ∆Q = p dV
R4
p dV = p3 V3 ln VV34
3
2
p1 V1
= 2.5 Atm
V2
Qin ∼
= 118 J
= 129 J (ln 2.5) ∼
p4 =∼
= 1.27 Atm
4
1
p4
p1 = 1 Atm
Qout ∼
= 101 J (ln 2.5) ∼
= 92.5 J
W = Qout − Qin ∼
= 25.5 J
Compare Carnot: W ∼
= 8.6 J
V2 = V3 = 0.4L
V
V1 = V4 = 1L
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. pV γ = 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