AME 514
Applications of
Combustion
Lecture 12: Hypersonic Propulsion
Applications
Advanced propulsion systems (3 lectures)
 Hypersonic propulsion background (Lecture 1)
 Why hypersonic propulsion?
 What's different at hypersonic conditions?
 Real gas effects (non-constant CP, dissociation)
 Aircraft range
 How to compute thrust?
 Idealized compressible flow (Lecture 2)
 Isentropic, shock, friction (Fanno)
 Heat addition at constant area (Rayleigh), T, P
 Hypersonic propulsion applications (Lecture 3)
 Crash course in T-s diagrams
 Ramjet/scramjets
 Pulse detonation engines
AME 514 - Spring 2015 - Lecture 12
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Crash course in using T-s diagrams
 T-s diagrams provide a “useful” visual representation of cycles
 Primarily work with “air cycles” in which the working fluid is
treated as just air (or some other ideal gas) - changes in gas
properties (CP, Mi, , etc.) due to changes in composition,
temperature, etc. are neglected; this greatly simplifies the
analysis
 Later study “fuel-air cycles” (using GASEQ) where 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.)
AME 514 - Spring 2015 - Lecture 12
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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 transfer or work transfer or KE
(see next slide)
 Efficiency can be determined by breaking any cycle into
Carnot-cycle “strips,” each strip (i) having th,i = 1 - TL,i/TH,i
AME 514 - Spring 2015 - Lecture 12
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T-s for control volume: work, heat & KE
 1st Law for a control volume, steady flow, with PE = 0
0 = Q -W + méë(hin - hout ) + ( KEin - KEout )ùû
 For 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 or KE change (dw = KE = 0), dh = CPdT = dq
 q12 = CP(T2 - T1)
 If no heat transfer (dq = 0) or KE, du = CPdT = 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
 True for any process, reversible or irreversible
AME 514 - Spring 2015 - Lecture 12
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T-s & P-v diagrams: work, heat & KE for CV
3
w34 + (KE4 – KE3) = -CP(T4-T3)
q23 = CP(T3-T2) (if const. P)
4
2
1
w12 + (KE2 – KE1) = -CP(T2-T1)
Case shown: cons. vol. heat in, rc = re = 3,  = 1.4, Tcomb = fQR/Cv = 628K, P1 = 0.5 atm
AME 514 - Spring 2015 - Lecture 12
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T-s diagrams: work, heat, KE
 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 514 - Spring 2015 - Lecture 12
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T-s diagrams: work, heat & KE
 Animation: using T-s diagram to determine heat & work
Heat
Heat transfer
transfer in
out
AME 514 - Spring 2015 - Lecture 12
8
Constant P curves
 Recall for ideal gas with constant specific heats
æ T2 ö
æ P2 ö
S2 - S1 = CP lnç ÷ - Rlnç ÷
è T1 ø
è P1 ø
 If P = constant, ln(P2/P1) = 0  T2 = T1exp[(S2-S1)/CP]
  constant P curves are growing exponentials on a T-s diagram
 Since constant P curves are exponentials, as s increases, the T
between two constant-P curves increases; as shown later, this
ensures that compression work is less than expansion work for
ideal Brayton cycles
 Constant P curves cannot cross (unless they correspond to
cycles with different CP)
AME 514 - Spring 2015 - Lecture 12
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Constant P and V curves
3
T(s) = T2exp[(s-s2)/CP] (const press.)
4
2
Constant P curves spread out as s
increases  T3 - T4 > T2 - T1
1
T(s) = T1exp[(s-s1)/CP] (const press.)
AME 514 - Spring 2015 - Lecture 12
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Constant P 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/CP) - 1] > 0
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Inferring efficienciesT-s diagram
1200
Temperature (K)
1000
Constant-P heat addition &
rejection (Brayton cycle)
800
TH,i
600
200
0
-100
Carnot cycle “strip”
th,i = 1 - TL,i/TH,i
TL,i
400
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 514 - Spring 2015 - Lecture 12
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Compression & expansion efficiency
 Animation: comparison of ideal Brayton 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 514 - Spring 2015 - Lecture 12
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"Conventional" ramjet






Incoming air decelerated isentropically to M = 0 - high T, P
No compressor needed
Heat addition at M = 0 - no loss of Pt - to max. allowable T (called T)
Expand to P9 = P1
Doesn't work well at low M - Pt/P1 & Tt/T1 low - Carnot efficiency low
As M increases, Pt/P1 and Tt/T1 increases, cycle efficiency increases,
but if M too high, limited ability to add heat (Tt close to Tmax) - good
efficiency but less thrust
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"Conventional" ramjet example
 Example: M1 = 5, T/T1 = 12,  = 1.4
 Initial state (1): M1 = 5
 State 2: decelerate to M2 = 0
æ g -1 2 ö
æ 1.4 -1 2 ö
T2 = T1 ç1+
M1 ÷ = T1 ç1+
5 ÷ = 6T1
è
øg è
ø
2
2
1.4
æ g -1 2 ö g -1
æ 1.4 -1 2 ö 1.4-1
P2 = P1 ç1+
M1 ÷ = P1 ç1+
5 ÷
= 529.1P1
è
ø
è
ø
2
2
 State 4: add at heat const. P; M4 = 0, P4 = P3 = P2 = 529.1P1, T4 =
T = 12T1
 State 9: expand to P1
g
1.4
æ g -1 2 ö
æ 1.4 -1 2 ö 1.4-1
P9t = 529.1P1 = P9 ç1+
M 9 ÷ = P1 ç1+
M9 ÷
Þ M 9 = 5.00
è
ø
è
ø
2
2
æ g -1 2 ö
æ 1.4 -1 2 ö
T9t = T4t =12T1 = T9 ç1+
M 9 ÷ = T9 ç1+
5 ÷ Þ T9 = 2T1
è
ø
è
ø
2
2
g -1
AME 514 - Spring 2015 - Lecture 12
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"Conventional" ramjet example
 Specific thrust (ST) (assume FAR << 1)
Thrust = m˙ a [(1+ FAR) u9 - u1 ] + ( P9 - P1) A9;FAR << 1,P9 = P1
Thrust u9 u1 u9 c 9
T9
Þ ST =
= - =
- M1 = M 9
- M1 = 5 2 - 5 = 2.07
m˙ a c1
c1 c1 c 9 c1
T1
 TSFC and overall efficiency
Heat input m˙ a CP (T4 t - T3t ) m˙ a a1 CP
TSFC º
=
=
T - T3t )
2 ( 4t
Thrust × c1
Thrust × c1
Thrust c1
1 [g /(g -1)]R
1
1
=
12T
6T
=
( 1 1)
(12 - 6) = 7.24;
ST
gRT1
2.07 1.4 -1
M1
5
ho =
=
= 0.691
TSFC 7.24
AME 514 - Spring 2015 - Lecture 12
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"Conventional" ramjet - effect of M1
 "Banana" shaped cycles for low M1, tall skinny cycles for high M1,
"fat" cycles for intermediate M1
Diffuser
Turbine (comp)
Nozzle
2
5
9
2500
Basic ramjet
T/T1 = 7
Temperature (K)
2000
Compressor
Turbine (fan)
Close the cycle
3
6
Combustor
Afterburner
1
4
7
T-s diagram
T-s diagram
T-s diagram
Intermediate
M1
high M1
Low M1
1500
1000
500
0
-250
0
250
500
750
1000 1250 1500 1750 2000
Entropy (J/kg-K)
AME 514 - Spring 2015 - Lecture 12
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"Conventional" ramjet - effect of M1
 Low M1: low compression, low thermal efficiency, low thrust
 High M1: high compression, high T after compression, can't add
much heat before reaching  limit, low thrust but high o
 Highest thrust for intermediate M1
Basic ramjet
 = 7
AME 514 - Spring 2015 - Lecture 12
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Scramjet (Supersonic Combustion RAMjet)
 What if Tt > Tmax allowed by materials? Can't decelerate flow to M = 0
before adding heat
 Need to mix fuel & burn supersonically, never allowing air to decelerate
to M = 0
X-43
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Scramjet (Supersonic Combustion RAMjet)
 Many laboratory studies, only a few (apparently) successful test
flight in last few seconds of sub-orbital rocket flight, e.g.
 Australian project: http://www.uq.edu.au/news/index.html?article=3414
 NASA X-43: http://www1.nasa.gov/missions/research/x43-main.html
» Steady flight (thrust ≈ drag) achieved at M1 ≈ 9.65 at 110,000 ft altitude
(u1 ≈ 2934 m/s = 6562 mi/hr)
» 3.8 lbs. H2 burned during 10 - 12 second test
» Rich H2-air mixtures ( ≈ 1.2 - 1.3), ignition with silane (SiH4, ignites
spontaneously in air)
» …but no information about the conditions at the combustor inlet, or the
conditions during combustion (constant P, T, area, …?)
» Real-gas stagnation temperatures 3300K (my model, lecture 10, slide 6:
3500K), surface temperatures up to 2250K (!)
 USAF X-51: http://www.wpafb.af.mil/news/story.asp?id=123346970
» Acceleration to steady flight achieved at M1 ≈ 5 at 60,000 ft for 240
seconds using hydrocarbon fuel
AME 514 - Spring 2015 - Lecture 12
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X43
AME 514 - Spring 2015 - Lecture 12
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Component performance
 Diffusers (1  2) & nozzles (7  9) subject to stagnation pressure
losses due to friction, shocks, flow separation
 More severe for diffusers since dP/dx > 0 - unfavorable pressure
gradient, aggravates tendency for flow separation
 If adiabatic, Tt = constant, thus d  T2t/T1t = 1 and n  T9t/T7t = 1
 If also reversible, d  P2t/P1t = 1 and n  P9t/P7t = 1, but d or n
may may be < 1 due to losses
 If reversible, d = d(-1)/ = 1, n = n(-1)/ = 1, but if irreversible
d(1)/ < 1 or  (-1)/ < 1 even though  = 1,  = 1
n
d
n
 Define diffuser efficiency
d = d(-1)/ = 1 if ideal
 Define nozzle efficiency
n = n(-1)/ = 1 if ideal
 For nozzle - also need to consider what happens if Pexit ≠ Pambient, i.e.
P9 ≠ P1
AME 514 - Spring 2015 - Lecture 12
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AirCycles4Hypersonics.xls
 Thermodynamic model is exact, but heat loss, stagnation
pressure losses etc. models are qualitative
 Constant  is not realistic (changes from 1.4 to ≈ 1.25 during
process) but only affects results quantitatively (not qualitatively)
 Heat transfer model
 T ~ h(Twall -Tgas,t), where heat transfer coefficient h and
component temperature Twall are specified - physically reasonable
Ti,b,t = Ti,a,t + h(Tw - Ti,a,t );Pi,b,t = Pi,a,t
 Wall temperature Tw is specified separately for each component
(diffuser, compressor, combustor, etc.) since each component
feels a constant T, not the T averaged over the cycle
 Increments each cell not each time step
 Doesn't include effects of varying area, varying turbulence,
varying time scale through Mach number, etc. on h
 Work or heat in or out from step i to step i+1 computed with dQ
= CP(Ti+1,t - Ti,t) or dW = - CP(Ti+1,t - Ti,t) (heat positive if into
system, work positive if out of system)
AME 514 - Spring 2015 - Lecture 12
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AirCycles4Hypersonics.xls
 Diffuser
 Mach number decrements from flight Mach number (M1) to the
specified value after the diffuser in 25 equal steps
 Stagnation pressure decrements from its value at M1 (P1t) to dP1t =
d/(-1)P1t in 25 equal steps
 Static P and T are calculated from M and Pt, Tt
 Sound speed c is calculated from T, then u is calculated from c and
M
 No heat input or work output in diffuser, but may have wall heat
transfer
 Shocks not implemented (usually one would have a series of oblique
shocks in an inlet, not a single normal shock)
AME 514 - Spring 2015 - Lecture 12
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AirCycles4Hypersonics.xls
 Combustor
 Heat addition may be at constant area (Rayleigh flow), P or T
 Mach number after diffuser is a specified quantity (not necessarily
zero) - Mach number after diffuser sets compression ratio since
there is no mechanical compressor
 Rayleigh curves starting at states 1 and 2 included to show
constant area / no friction on T-s
AME 514 - Spring 2015 - Lecture 12
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AirCycles4Hypersonics.xls
 Nozzle
 Static (not stagnation) pressure decrements from value after
afterburner to specified exhaust pressure in 25 equal steps
 Stagnation pressure decrements from its value after afterburner (P7t)
to nP7t = n/(-1)P7t in 25 equal steps
 Static P and T are calculated from M and Pt, Tt
 Sound speed c is calculated from T, then u is calculated from c and
M
 No heat input or work output in diffuser, but may have wall heat
transfer
 Heat transfer occurs according to usual law
AME 514 - Spring 2015 - Lecture 12
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AirCycles4Hypersonics.xls
 Combustion parameter  = T4t/T1 (specifies stagnation
temperature, not static temperature, after combustion)
 Caution on choosing 
 If T1 < rT1 (r = 1 + (-1)/2 M12) (maximum allowable
temperature after heat addition > temperature after deceleration)
then no heat can be added (actually, spreadsheet will try to
refrigerate the gas…)
 For constant-area heat addition, if T1 is too large, you can't add
that much heat (beyond thermal choking point) & spreadsheet
"chokes"
 For constant-T heat addition, if T1 is too large, pressure after
heat addition < ambient pressure - overexpanded jet - still works
but performance suffers
 For constant-P heat addition, no limits!  But temperatures go
sky-high 
 All cases: f (fuel mass fraction) needed to obtain specified  is
calculated - make sure this doesn't exceed fstoichiometric!
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - T-s diagrams
 With const. T combustion, max. temperature stays within sane
limits but as more heat is added, P decreases; eventually P4 < P9
 Also, latter part of cycle has low Carnot-strip efficiency since
constant T and P lines will converge
Diffuser
Close the cycle
1
9
5000
4500
Combustor
Rayleigh from 1
2
Nozzle
Rayleigh from 2
4
Diffuser
2 Area
Combustor
4 Mach
1.E+00
T-s diagram
Nozzle
9 Area
1 Mach
1 Area
Diffuser
2 Mach
12
Area and Mach number
4000
10
1.E-01
Area (m^2)
3500
3000
2500
2000
1500
1000
8
1.E-02
6
1.E-03
4
1.E-04
2
500
0
-500
0
500
1000
1500
Entropy (J/kg-K)
2000
2500
3000
Mach number
Temperature (K)
Combustor
4 Area
Nozzle
9 Mach
1.E-05
0
0
20
40
60
80
Step number
Const. T combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
Stoich. H2-air (f = 0.0283, QR = 1.2 x 108 J/kg   = 35.6)
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - effect of M1
 Minimum M1 = 6.28 - below that T2 < 2000 even if M2 = 0
 No maximum M2
 overall improves slightly at high M1 due to higher thermal (lower T9)
4000
4.5
3500
4
3000
3.5
2500
3
2.5
2000
Tau_lamda/10
Mach No after diffuser
Specific Thrust
Overall efficiency
Specific Impulse
2
1.5
1
1500
1000
500
0.5
Specific impulse (sec)
M2, Tau_lamda, ST & OE
5
0
0
6
7
8
9
10
11
12
13
Flight Mach No. (M1)
Const. T combustion, M1 = varies; M2 adjusted so that T2 = 2000K;
H2-air (QR = 1.2 x 108 J/kg),  adjusted so that f = fstoichiometric
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - effect of 
M1 = 10, T2 = 2000K specified  M2 = 2.61
At  = 21.1 no heat can be added
At  = 35.6, f = 0.0283 (stoichiometric H2-air)
At  = 40.3 (if one had a fuel with higher QR than H2), P4 = P9
f & Specific Thrust increase as more fuel is added ( increasing), overall
& ISP decrease due to low thermal at high heat addition (see T-s
diagram)
4
4500
4000
3.5
3000
2.5
2500
2
Fuel mass fraction (%)
Stoich. fuel mass fraction
Specific Thrust
Overall efficiency
Specific Impulse
1.5
1
2000
1500
1000
0.5
500
Specific impulse (sec)
3500
3
f, ST & OE





0
0
20
25
30
35
40
45
Tau lamda
Const. T combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - effect of M2
4
4000
3.5
3500
3
3000
2.5
2500
Fuel mass fraction (%)
2
2000
Specific Thrust
Overall efficiency
Specific Impulse
1.5
1500
1000
1
500
0.5
Specific impulse (sec)
f, ST & OE
 Maximum M2 = 3.01 - above that P4 < P9 after combustion (you could go
have higher M2 but heat addition past P4 = P9 would reduce thrust!)
 No minimum M2, but lower M2 means higher T2 - maybe beyond materials
limits (after all, high T1t is the reason we want to burn at high M)
 overall decreases at higher M2 due to lower thermal (lower T2)
0
0
0
0.5
1
1.5
2
2.5
3
3.5
Mach No. after diffuser (M2)
Const. T combustion, M1 = 10; M2 varies;
H2-air (QR = 1.2 x 108 J/kg),  adjusted so that f = fstoichiometric
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - effect of d
 Obviously as d decreases, all performance parameters decrease
 If d too low, pressure after stoichiometric heat addition < P1, so
need to decrease heat addition (thus )
 Diffuser can be pretty bad (d ≈ 0.25) before no thrust
3
3500
Fuel mass fraction (%)
Specific Thrust
3000
Overall efficiency
2500
f, ST & OE
Specific Impulse
2
2000
1500
1.5
1000
1
500
0.5
0
Specific impulse (sec)
2.5
-500
0
0
0.2
0.4
0.6
0.8
1
1.2
Diffuser efficiency
Const. T combustion, M1 = 10; M2 = 2.611; H2-air (QR = 1.2 x 108 J/kg),
 adjusted so that f = fstoichiometric or P5 = P1, whichever is smaller
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. T) - effect of n
 Obviously as n decreases, all performance parameters decrease
 Nozzle can be pretty bad (n ≈ 0.32) before no thrust, but not as
bad as diffuser
3500
3
Fuel mass fraction (%)
3000
Specific Thrust
f, ST & OE
Overall efficiency
2
2500
Specific Impulse
2000
1.5
1500
1
1000
0.5
500
Specific impulse (sec)
2.5
0
0
0
0.2
0.4
0.6
0.8
1
1.2
Nozzle efficiency
Const. T combustion, M1 = 10; M2 = 2.611; H2-air (QR = 1.2 x 108 J/kg),
 adjusted so that f = fstoichiometric
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Hypersonic propulsion (const. P) - T-s diagrams
 With Const. P combustion, no limitations on heat input, but maximum T
becomes insane (though dissociation & heat losses would decrease
this T substantially)
 Carnot-strip (thermal) efficiency independent of heat input; same as
conventional Brayton cycle (s-P-s-P cycle)
Diffuser
Close the cycle
1
9
6000
Nozzle
Rayleigh from 2
4
T-s diagram
Combustor
4 Area
Nozzle
9 Mach
Nozzle
9 Area
1 Mach
1 Area
Diffuser
2 Mach
12
Area and Mach number
10
1.E-01
Area (m^2)
4000
3000
2000
8
1.E-02
6
1.E-03
4
1.E-04
1000
0
-500
Diffuser
2 Area
Combustor
4 Mach
1.E+00
0
500
Entropy (J/kg-K)
1000
1500
Mach number
Temperature (K)
5000
Combustor
Rayleigh from 1
2
2
1.E-05
0
0
20
40
60
80
Step number
Const. P combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
Stoich. H2-air (f = 0.0283, QR = 1.2 x 108 J/kg   = 35.6)
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. P) - performance
M1 = 10, T2 = 2000K specified  M2 = 2.61
Still, at  = 21.1 no heat can be added
At  = 35.6, f = 0.0283 (stoichiometric H2-air)
No upper limit on  (assuming one has a fuel with high enough QR)
f & Specific Thrust increase as more fuel is added ( increasing), overall & ISP
decrease only slightly at high heat addition due to lower propulsive
4.5
4000
Fuel mass fraction (%)
4
Stoich. fuel mass fraction
Overall efficiency
3800
Specific Impulse
3
2.5
3700
2
3600
1.5
3500
1
3400
0.5
Specific impulse (sec)
3900
Specific Thrust
3.5
f, ST & OE





3300
0
20
25
30
35
40
45
Tau lamda
Const. P combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
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Hypersonic propulsion (const. A) - T-s diagrams
 With Const. A combustion, heat input limited by thermal choking,
maximum T becomes even more insane than constant P
 … but Carnot-strip efficiency is awesome!
Diffuser
Close the cycle
1
9
6000
Nozzle
Rayleigh from 2
4
T-s diagram
Combustor
4 Area
Nozzle
9 Mach
Nozzle
9 Area
1 Mach
1 Area
Diffuser
2 Mach
12
Area and Mach number
10
1.E-01
Area (m^2)
4000
3000
2000
8
1.E-02
6
1.E-03
4
1.E-04
1000
0
-200
Diffuser
2 Area
Combustor
4 Mach
1.E+00
0
200
400
600
Entropy (J/kg-K)
800
1000
Mach number
Temperature (K)
5000
Combustor
Rayleigh from 1
2
2
1.E-05
0
0
20
40
60
80
Step number
Const. A combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (f = 0.0171, QR = 1.2 x 108 J/kg   = 30.1; (can't add stoichiometric
amount of fuel at constant area for this M1 and M2))
AME 514 - Spring 2015 - Lecture 12
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Hypersonic propulsion (const. A) - performance
M1 = 10, T2 = 2000K specified  M2 = 2.61
Still, at  = 21.1 no heat can be added
At  = 30.5, thermal choking at f = 0.0193 < 0.0283
f & Specific Thrust increase as more fuel is added ( increasing), overall &
ISP decrease slightly at high heat addition due to lower propulsive
3950
3
Fuel mass fraction (%)
3900
Stoich. fuel mass fraction
Specific Thrust
Overall efficiency
2
3850
Specific Impulse
1.5
3800
1
3750
0.5
3700
0
3650
20
22
24
26
28
30
Specific impulse (sec)
2.5
f, ST & OE




32
Tau lamda
Const. A combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
AME 514 - Spring 2015 - Lecture 12
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Pulse detonation engine concept
 References: Kailasanath (2000), Roy et al. (2004)
 Simple system - fill tube with detonable mixture, ignite, expand
exhaust
 Something like German WWII "buzz bombs" that were "Pulse
Deflagration Engines"
 Advantages over conventional propulsion systems
 Nearly constant-volume cycle vs. constant pressure - higher ideal
thermodynamic efficiency
 No mechanical compressor needed
 (In principle) can operate from zero to hypersonic Mach numbers
Fill Tube
Fuel
Air
(or other
oxidizer)
Detonate
Mixture
Exhaust
Refill Tube, Repeat
Courtesy Fred Schauer
AME 514 - Spring 2015 - Lecture 12
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Pulse detonation engine concept
Kailasanath (2000)
 Challenges (i.e. problems…)




Detonation initiation in small tube lengths
Deceleration of gas to low M at high flight M
Fuel-air mixing
Noise
AME 514 - Spring 2015 - Lecture 12
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Steady Chapman-Jouget detonation engine
 Recall (lecture 1) detonation is shock followed by heat release; whole spectrum
of M3, but only stable case is M3 = 1
 Exploit pressure rise across shock to generate thrust after expansion
 Expand post-detonation gas to Pexit = Pambient to evaluate performance
 … but have to ignore low-H region - not realistic since temperature rise due to
weak shock is insufficient to ignite mixture
 Not practical since
 Need to have exactly correct heat release (H), i.e. fuel-air ratio, to match
inlet (flight) Mach number M1
 Not self-starting at M1 = 0
é H(g 2 -1) ù1/ 2 é H(g 2 -1) ù1/ 2
M1 = ê1+
ú +ê
ú
2g û
ë
ë 2g û
q
H=
(heat release parameter)
RT1
where q = CP (T3t - T2t ) = fQR
H » 33 (stoichiometric hydrocarbon - air)
H » 40 (stoichiometric hydrogen - air)
AME 514 - Spring 2015 - Lecture 12
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Ideal DE (not P) performance
100
10
Steady detonation wave engine
80
Steady detonation wave engine
60
M1
M2
Me
1
40
20
P2/P1
P3/P1
0
0.1
0
10
20
30
40
Dimensionless heat input
50
0
60
10
20
30
40
50
60
Dimensionless heat input
100
30
Steady detonation wave engine
Steady detonation wave engine
25
Specific thrust
TSFC
20
15
10
10
T2/T1
T3/T1
Te/T1
5
0
1
0
10
20
30
40
Dimensionless heat input
50
60
0
10
20
30
40
50
60
Dimensionless heat input
AME 514 - Spring 2015 - Lecture 12
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PDE analysis




Gas in tube initially at rest
Propagate shock through gas (M = M1 CJ detn), compute T2, P2, M2
Add heat to sonic condition (M3 = 1) for CJ detonation, compute T3, P3
… (again) have to ignore low-H region - not realistic since temperature rise due
to weak shock is insufficient to ignite mixture
 Gas behind detonation is moving toward open end of tube, so "stretches" gas
near thrust surface (closed end of tube),which must have u4 = 0
 Detonation structure is like piston with velocity u = c1M1 - c3, causes expansion
wave:
Tube length L
"Stretched" or
expanded gas
(u4 = 0)
Post - reaction zone
(u3 = c1M1 - c3M3 =
c1M1 - c3)
Thrust surface
(Area A)
Outside pressure = P1
Reaction
zone (2)
AME 514 - Spring 2015 - Lecture 12
"Uninformed
" gas (u1 = 0)
ushock = c1M1,CJ
Shock
42
PDE analysis
Wintenberger et al., 2001
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PDE analysis
Pressure at thrust surface
CJ pressure = P3
t1 = L/c1M1
t2 = L/c4
Post-expansion
pressure = P4
Time
Wintenberger et al., 2001
AME 514 - Spring 2015 - Lecture 12
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Simplified PDE analysis
 P4 acts on thrust surface for the time it takes the detonation
front to reach the end of the tube and for the reflected wave to
propagate back to the thrust surface ≈ L/(c1M1) + L/c4
 Not exactly correct because
 Reflected wave speed > c4 (finite-amplitude, not sound wave)
 Even after reflected wave hits thrust surface, still some additional
expansion, but hard to compute because "non-simple region"
(interacting waves)
 More accurate analysis: Shepherd & collaborators (Wintenberger
et al., 2001)
 Impulse = Thrust x time = (P4 - P1)A[L/(c1M1) + L/c4]
 Mass of fuel = mf = fuel mass fraction x density x volume =
f1LV
 Specific impulse = Impulse/(weight of fuel) = Impulse/mfgearth
 Specific thrust & TSFC calculated as usual
ISP
QR æ c1 c1 ö æ P4 ö QR æ 1
T1 ö
+
ç
÷
÷
ç + ÷ = ç -1÷
T4 ø
ø Hgc1 è u1 c 4 ø è P1 ø Hgc1 è M1
P4 - P1 ) AL(1/u1 + 1/c 4 ) æ P4 ö
(
=
=
-1
P1 ALHg /QR
ç
è P1
AME 514 - Spring 2015 - Lecture 12
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PDE analysis - hydrocarbon fuel
 Performance unimpressive compared to turbofan engines (Isp >
10,000 sec) but simple, far fewer moving parts, can work at M = 0
up to hypersonic speeds
8
Pulse detonation engine
Flight Mach number = 0
7
Specific thrust
6
TSFC
TSFC
5
Isp/1000 (sec) (PDR)
Isp/1000 (sec) (Shepherd)
4
3
ST
Isp (Shep)
2
Isp (PDR)
1
Stoichiometric HC-air
0
0
10
20
30
40
50
60
Dimensionless heat input
AME 514 - Spring 2015 - Lecture 12
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PDE analysis - hydrogen fuel
14
Specific thrust
12
TSFC
Pulse detonation engine
Flight Mach number = 0
Isp/1000 (sec) (PDR)
Isp/1000 (sec) (Shepherd)
10
8
6
Isp (Shep)
TSFC
4
Isp (PDR)
ST
2
Stoichiometric H2-air
0
0
10
20
30
40
50
60
Dimensionless heat input
AME 514 - Spring 2015 - Lecture 12
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PDE Research Engine - Wright-Patterson Air Force Base
 Pontiac Grand Am engine driven by
electric motor used as air pump to
supply PDE
 Allows study of high frequency
operation, multi-tube effects
 Stock Intake Manifold with
Ball Valve Selection of 1-4
Detonation Tubes
Photos courtesy F. Schauer
AME 514 - Spring 2015 - Lecture 12
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H2-air PDE – 1 tube @ 16 Hz
Courtesy F. Schauer
AME 514 - Spring 2015 - Lecture 12
49
H2-air PDE – 4 tubes @ 4 Hz each
Do PDE's
make thrust?
Courtesy F. Schauer
AME 514 - Spring 2015 - Lecture 12
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H2-air PDE – 2 tubes @ high frequency
Courtesy F. Schauer
AME 514 - Spring 2015 - Lecture 12
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Performance of "laboratory" PDE
 Performance (Schauer et al., 2001) using H2-air similar to
predictions unless too lean (finite-rate chemistry, not included in
calculations)
AME 514 - Spring 2015 - Lecture 12
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Effect of tube fill fraction
 Better performance with lower tube fill fraction - better propulsive
efficiency (accelerate large mass by small u), but this is of little
importance at high M1 where u << u1
AME 514 - Spring 2015 - Lecture 12
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References
 K. Kailasanath, "Review of Propulsion Applications of Detonation Waves," AIAA J. 38, 1698-1708
(2000).
 G. D. Roy, S. M. Frolov, A. A. Borisov, D. W. Netzer (2004). "Pulse Detonation Engines:
challenges, current status, and future perspective," Progress in Energy and Combustion Science 30,
545-672.
 F. Schauer, J. Stutrud, R. Bradley (2001). "Detonation Initiation Studies and Performance Results
for Pulsed Detonation Engine Applications," AIAA paper No. 2001-1129 (2001).
 E. Wintenberger, J. M. Austin, M. Cooper, S. Jackson, J. E. Shepherd, "An analytical model for the
impulse of a single-cycle pulse detonation engine," AIAA paper no. 2001-3811 (2001).
AME 514 - Spring 2015 - Lecture 12
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