AME 436
Energy and Propulsion
Lecture 7
Unsteady-flow (reciprocating) engines 2:
Using P-V and T-s diagrams
Outline
 Air cycles
 What are they?
 Why use P-V and T-s diagrams?
 Using P-V and T-s diagrams for air cycles






Seeing heat, work and KE
Constant P and V processes
Inferring efficiencies
Compression & expansion component efficiencies
Correspondence between processes on P-V and T-s
Hints & tricks
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Air-cycles - what are they?
 In this course we will work primarily with “air cycles” in which
the working fluid is treated as just air (or some other ideal gas)
 In air cycles, changes in gas properties (CP, Mi, , etc.) due to
changes in composition, temperature, etc. are neglected; this
greatly simplifies the analysis and leads to simple analytic
expressions for efficiency, power, etc.
 Later we’ll examine “fuel-air cycles” (using GASEQ) where the
“real” gases are considered and the properties change as
composition, temperature, etc. change (but still can’t account
for slow burn, heat loss, etc. since it’s still a thermodynamic
analysis that tells us nothing about reaction rates, burning
velocities, etc.)
 In addition to the analytical results, P-V and T-s diagrams will be
used extensively to provide a visual representation of cycles
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Why use P-V diagrams?
 Pressure vs. time and cylinder volume vs. time are easily
measured in reciprocating-piston engines
 ∫ PdV = work
 ∫ PdV over whole cycle = net work transfer + net change in KE +
net change in PE = net heat transfer
 Heat addition is usually modeled as constant pressure or constant
volume, so show as straight lines on P-V diagram
 Note that V on P-V diagram is cylinder volume (m3) which is NOT a
property of the gas, whereas specific volume (v) (m3/kg) IS a
property of the gas (we need to use V since cycle work = ∫ PdV;
can’t use v since mass changes during intake/blowdown/exhaust)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Why use T-s diagrams?
 Idealized compression & expansion processes are constant S
since dS ≥ Q/T; for adiabatic process Q = 0, for reversible =
(not >) sign applies, thus dS = 0 (note that dS = 0 still allows for
any amount of work transfer to occur in or out of the system,
which is what compression & expansion processes are for)
 For reversible process, ∫TdS = Q, thus area under T-s curves
show amount of heat transferred
 ∫ TdS over whole cycle = net heat transfer = net work transfer +
net change in KE + net change in PE
 T-s diagrams show the consequences of non-ideal compression
or expansion (dS > 0)
 For ideal gases, T ~ heat xfer or work xfer or KE (see next 2
slides)
 Efficiency can be determined by breaking any cycle into Carnotcycle “strips,” each strip (i) having th,i = 1 - TL,i/TH,i
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v for control mass: work, heat & KE
 For an ideal gas with constant CP & Cv:
h = CpT, u = CvT
and the 1st Law says, for a control mass with PE = 0 (in internal
combustion engines we can almost always neglect PE)
dE = d(U + KE) = d[m(u + KE)] = q - w
(note the use of “KE” rather than “u2/2” to avoid confusion
between u {internal energy} and u {velocity})
q = Q/m (heat transfer per unit mass)
w = W/m (work transfer per unit mass)
 If no work transfer (dw = 0) or KE change (dKE = 0), du = CvdT = dq
 q12 = Cv(T2 - T1)
 If no heat transfer (dq = 0) or KE, du = CvdT = dw
 w12 = -Cv(T2 - T1)
 If no work or heat transfer
 KE = KE2 – KE1 = -Cv(T2 - T1)
 For a control mass containing an ideal gas with constant Cv,
T ~ heat transfer - work transfer - KE
 Note that the 2nd law was not invoked, thus the above statements
are true for any process, reversible or irreversible
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v for control volume: work, heat & KE
 1st Law says, for a control volume, steady flow, with PE = 0
0 = Q -W + méë(hin - hout ) + ( KEin - KEout )ùû
 For an ideal gas with constant CP, dh = CPdT  h2 - h1 = CP(T2 - T1)
 If 2 = outlet, 1 = inlet, and noting
h2 - h1 = q1®2 - w1®2 - ( KE2 - KE1 )
 If no work transfer (dw = 0) or KE change (dKE = 0), du = CvdT = dq
 q12 = CP(T2 - T1)
 If no heat transfer (dq = 0) or KE, du = CvdT = dw
 w12 = -CP(T2 - T1)
 If no work or heat transfer
 KE = KE2 – KE1 = -CP(T2 - T1)
 For a control volume containing an ideal gas with constant CP,
T ~ heat transfer - work transfer - KE
(Same statement as control mass with CP replacing Cv)
 Again true for any process, reversible or irreversible
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v diagrams: work, heat & KE
3
w34 + (KE4 – KE3) = -Cv(T4-T3) (control mass)
w34 + (KE4 – KE3) = -CP(T4-T3) (control vol.)
q23 = Cv(T3-T2) (const. V)
q23 = CP(T3-T2) (const. P)
4
2
1
w12 + (KE2 – KE1) = -Cv(T2-T1) (control mass)
w12 + (KE2 – KE1) = -CP(T2-T1) (control vol.)
Case shown: cons. vol. heat in, rc = re = 3,  = 1.4, Tcomb = fQR/Cv = 628K, P1 = 0.5 atm
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v diagrams: work, heat, KE & PE
 How do I know that work shown on the P-v (via ∫PdV) diagram is
the same as that shown (via CvT) on the T-s diagram? As an
example, for isentropic compression
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v diagrams: work, heat, KE & PE
 Going back to the 1st Law again
dE = dQ - dW Þ
ò dE = ò dQ - ò dW
 Around a closed path, since E = U + KE + PE; since U is a property of the
system, ò dU = 0 , thus around a closed path, i.e. a complete thermodynamic
cycle (neglecting PE again)
But wait - does this mean that the thermal efficiency
No, the definition of thermal efficiency is
 For a reversible process, Q = TdS, thus q = Tds and
 Thus, for a reversible process, the area inside a cycle on a T-s diagram is
equal to (net work transfer + net KE) and the net heat transfer
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s & P-v diagrams: work, heat & KE
 Animation: using T-s diagram to determine heat & work
Heat
Heat transfer
transfer in
out
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Constant P and V curves
 Recall for ideal gas with constant specific heats (1st lecture)
æ T2 ö
æ V2 ö
æ T2 ö
æ P2 ö
S2 - S1 = CP lnç ÷ - Rlnç ÷ S2 - S1 = CV lnç ÷ + Rlnç ÷
è T1 ø
è V1 ø
è T1 ø
è P1 ø
 If P = constant, ln(P2/P1) = 0  T2 = T1exp[(S2-S1)/CP]
 If V = constant, ln(V2/V1) = 0  T2 = T1exp[(S2-S1)/Cv]
  constant P or V curves are growing exponentials on a T-s
diagram
 Since constant P or V curves are exponentials, as s increases, the
T between two constant-P or constant-V curves increases; as
shown later, this ensures that compression work is less than
expansion work for ideal Otto or Brayton cycles
 Since CP = Cv + R, CP > Cv or 1/CP < 1/Cv, constant v curves rise
faster than constant P curves on a T-s diagram
 Constant P or constant V lines cannot cross (unless they
correspond to cycles with different CP or CV)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Constant P and V curves
3
T(s) = T2exp[(s-s2)/Cv] (const vol.)
T(s) = T2exp[(s-s2)/CP] (const press.)
4
2
Constant v or P curves spread out as
s increases  T3 - T4 > T2 - T1
1
T(s) = T1exp[(s-s1)/Cv] (const vol.)
T(s) = T1exp[(s-s1)/CP] (const press.)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Constant P and V curves
“Payback” for compression work and KE decrease
3
(T3 - T4) - (T2 - T1) ~ net work + KE increase
4
2
1
T2 - T1 ~ Work input + KE decrease during compression
Net work + KE decrease ~ (T3 - T4) - (T2 - T1)
= (T2 - T1)[exp(s/Cv) - 1] > 0 or (T2 - T1)[exp(s/CP) - 1] > 0
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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T-s diagram
Constant P and V curves
1200
Constant-v heat addition &
rejection (Otto cycle)
Temperature (K)
1000
800
600
400
200
Double-click chart to
open Excel spreadsheet
0
-100
0
100
200
300
Constant-P heat addition &
rejection (Brayton cycle)
400
500
600
700
800
Entropy (J/kg-K)
Constant-v curves are steeper than constant-P curves on the T-s
Both cases: T2/T1 = 1.552,  = 1.4, fQR = 4.5 x 105 J/kg, P1 = 0.5 atm
The two cycles shown also have the same thermal efficiency (th)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Inferring efficienciesT-s diagram
1200
Temperature (K)
1000
Constant-v heat addition &
rejection (Otto cycle)
800
TH,i
600
400
200
0
-100
Carnot cycle “strip”
th,i = 1 - TL,i/TH,i
TL,i
Double-click chart to
open Excel spreadsheet
0
100
200
300
400
500
600
700
Entropy (J/kg-K)
Carnot cycles appear as rectangles on the T-s diagram; any cycle
can be broken into a large number of tall skinny Carnot cycle
“strips,” each strip (i) having th,i = 1 - TL,i/TH,i
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Compression & expansion efficiency
 If irreversible compression or expansion, dS > Q/T; if still adiabatic
(Q = 0) then dS > 0
 Causes more work input (more T) during compression, less work
output (less T) during expansion
 Define compression efficiency comp & expansion efficiency exp
hcomp º
Reversible adiabatic work input for given V or P ratio
Actual work input required for same V or P ratio
-CV [T1 (V1 /V2 )g -1 - T1 ] (V1 /V2 )g -1 -1
=
=
-CV (T2 - T1 )
T2 /T1 -1
(control mass, specified volume ratio)
hexp º
Actual work output for given V or P ratio
Reversible adiabatic work output for same V or P ratio
-CV (T2 - T1 )
T2 /T1 -1
=
=
g -1
-CV [T1 (V1 /V2 ) - T1] (V1 /V2 )g -1 -1
(control mass, specified volume ratio)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Compression & expansion efficiency
-CV [T1 (P2 /P1 )(g -1) / g - T1 ] (P2 /P1 )(g -1)/ g -1
hcomp =
=
-CV (T2 - T1)
T2 /T1 -1
(control mass, specified pressure ratio)
-CV (T2 - T1)
T2 /T1 -1
hexp =
=
-CV [T1(P2 /P1)(g -1)/ g - T1 ] (P2 /P1 )(g -1) / g -1
(control mass, specified pressure ratio)
 Control volume: replace Cv with CP, but it cancels out so definitions
are same
 These relations give us a means to quantify the efficiency of an
engine component (e.g. compressor, turbine, …) or process
(compression, expansion) as opposed to the whole cycle
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Compression & expansion efficiency
 Animation: comparison of ideal Otto cycle with non-ideal
compression & expansion
 Same parameters as before but with comp = exp =0.9
1200
Temperature (K)
1000
800
600
400
2 charts on top of each other;
double-click each to open Excel
spreadsheets
200
0
-200
0
200
400
600
800
Entropy (J/kg-K)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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4.5
4.5
4.0
4.0
Temperature (T/To)
Pressure (P/Po)
Correspondence between P-V & T-s
3.5
3.0
2.5
2.0
1.5
1.0
0.5
3.5
3.0
Pressure increasing
2.5
2.0
1.5
1.0
0.5
0.0
0.0
0
1
2
3
4
5
0
1
2
3
4
5
6
Entropy ((s-so)/R)
4.5
4.5
4.0
4.0
Temperature (T/To)
Pressure (P/Po)
Volume (V/Vo)
3.5
3.0
2.5
2.0
1.5
1.0
3.5
3.0
2.5
Volume increasing
2.0
1.5
1.0
0.5
0.5
0.0
0.0
0
1
2
3
Volume (V/Vo)
4
5
0
1
2
3
4
Entropy ((s-so)/R)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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6
20
4.5
4.5
4.0
4.0
Temperature (T/To)
Pressure (P/Po)
Correspondence between P-V & T-s
3.5
3.0
Temperature increasing
2.5
2.0
1.5
1.0
3.5
3.0
2.5
1.5
1.0
0.5
0.5
0.0
0.0
0
1
2
3
4
Constant temperature
2.0
0
5
1
3
4
5
6
Entropy ((s-so)/R)
Volume (V/Vo)
2.5
3.5
2.0
Temperature (T/To)
Pressure (P/Po)
2
1.5
Entropy increasing
1.0
0.5
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0
1
2
3
Volume (V/Vo)
4
5
Constant entropy
0.0
0
1
2
3
4
5
Entropy ((s-so)/R)
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Example
 Why do internal combustion engines compress before burning? Is it
possible to produce work or thrust without compression?
To generate positive area (thus net work) on a T-s diagram (left), the pre-heataddition process must be more nearly vertical than the heat addition process,
otherwise there is no area thus no work (middle). The best way to do this is with
isentropic compression followed by heat addition (left) - but it’s not required. You
could have (for example) constant-V heat addition followed by isentropic
expansion back to ambient P (right) and work would be generated without a
“real” compressor (although with lower efficiency.)
T
T
T
No area,
no work!
Net work
V
S
P
S
S
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Example
 Why is it necessary to add heat to generate work or thrust?
Without heat transfer, dq = 0 and thus (for reversible cycles) TdS = 0, thus
T dS = P dV = Net work = 0 (left figure). If the process is irreversible,
TdS < 0, thus T dS < 0 and thus = P dV = Net work < 0 (right figure).
ò
ò
ò
ò
T
T
Negative
work
No area,
no work!
S
S
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Example
Consider the “Lenoir cycle” engine shown on the P-V diagram
a) Sketch the T-s diagram corresponding to the P-V diagram
2
1  2 const. V, increasing P 
heat addition, s increases
2
P
T
2  3 const. s, decreasing P 
expansion, T decreases
V
3
P
1
3  1 const. P, decreasing v 
heat rejection, s decreases
1
3
V
s
b) Sketch modified P-V and T-s diagrams if the initial temperature is
increased by 10% (same P1, V1 and P2)
2’
2
2
P
T
P
P - V plot unchanged (but higher
T means less mass processed)
3’
V
P
1
3
V
1 P 1’
3
Higher T, same P  s1, s2
increases
Still need 3  1 const. P
s
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Example
c) Sketch modified P-V and T-s diagrams if the “compression ratio” V3/V1 is
increased by 20% (same P1, V1)
2’
1  2 const. V, P2 larger than
base cycle
2’
2
2
P
T
V
3’
3
P
1, 1’
Still need 3  1 const. P
1,1’
3 3’
V
On P-V, curves of const. T
slope less steeply than const. s,
thus T3’ > T3
s
d) Sketch modified P-V and T-s diagrams if a gas with higher  is used
(same P1, V1, P2, R)
PV curves steeper if  increases
2’
2, 2’
P
2
V
T
3’ 3
P
1,1’
1, 1’
3’
V
CV = R/(-1); if R = const. and 
increases, CV decreases, thus
const.-v curves on T – s are
steeper
T2 doesn’t change since V1 = V2,
P1, P2 and R don’t change
3
s
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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Using T-s and P-v diagrams - summary
 Thermodynamic cycles as they occur in IC engines are often approximated
as a series of processes occurring in an ideal gas
 T-s and P-v diagrams are very useful for inferring how changes in a cycle
affect efficiency, power, peak P & T, etc.
 The T (on T-s diagrams) and areas (both T-s & P-v) are very useful for
inferring heat & work transfers
 Each process (curve or straight line) on T-s or P-v diagram has of 3 parts
 An initial state
 A process (const. P, v, T, s, as shown in the previous slides), constant area
(Rayleigh, Fanno or shock flow, discussed in propulsion section), etc.
 A final state, which is usually
» For compression and expansion processes in reciprocating piston engines, a
specified volume relative to the initial state (i.e. a particular compression or
expansion ratio)
» For compressors in propulsion cycles, a specified pressure ratio
» For turbines in propulsion cycles, a specified temperature that makes the work
output from the turbine equal the work required to drive compressor and/or fan
» For diffusers in propulsion cycles, a specified Mach number (usually zero)
» For nozzles in propulsion cycles, the pressure after expansion (usually the
ambient pressure)
» For heat addition processes, either a specified heat input = ∫ Tds (i.e. a mixture
having a specified FAR and QR) thus a given area on the T-s diagram, or a
specified temperature (i.e. for temperature limited turbines in propulsion
cycles)
» The constant P and constant v exponential curves on the T-s diagram are very
useful for determining end states
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AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
Using T-s and P-V diagrams - summary
 Three or more processes combine to make a complete cycle
 When drawing P-V or T-s diagrams, ask yourself
 What is the P, V, T and s of the initial state? Is it different from the
baseline case?
 For each subsequent process
» What is the process? Is it the same as the baseline cycle, or does it change
from (for example) reversible to irreversible compression or expansion?
Does it change from (for example) constant pressure heat addition to heat
addition with pressure losses?
» When the process is over? Is the target a specified pressure, volume,
temperature, heat input, work output, etc.?
» Is a new process (afterburner, extra turbine work for fan, etc.) being added or
is existing one being removed?
» In gas turbine cycles, be sure to make work output of turbines = work input
to compressors and fans in gas
 Be sure to close the cycle by having (for reciprocating piston
cycles) the final volume = initial volume or (for propulsion cycles)
(usually) the final pressure = ambient pressure
AME 436 - Spring 2013 - Lecture 7 - Using P-V & T-s diagrams
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