1. AME 514 Applications of Combustion
Lecture 12: Hypersonic Propulsion Applications

2. 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 Spring Lecture 12

3. 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 Spring Lecture 12

4. 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 Spring Lecture 12

5. T-s for control volume: work, heat & KE
1st Law for a control volume, steady flow, with PE = 0
For ideal gas with constant CP, dh = CPdT  h2 - h1 = CP(T2 - T1)
If 2 = outlet, 1 = inlet, and noting
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 Spring Lecture 12

6. T-s & P-v diagrams: work, heat & KE for CV3
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 Spring Lecture 12

7. T-s diagrams: work, heat, KE
Going back to the 1st Law again
Around a closed path, since E = U + KE + PE; since U is a property of the system, , 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 Spring Lecture 12

8. T-s diagrams: work, heat & KE
Animation: using T-s diagram to determine heat & work
Heat transfer in
Net heat transfer = net work + KE
Heat transfer out
AME Spring Lecture 12

9. Constant P curves Recall for ideal gas with constant specific heats
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 Spring Lecture 12

10. Constant P and V curves AME 514 - Spring 2015 - Lecture 12 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 Spring Lecture 12

11. Constant P curves Net work + KE decrease ~ (T3 - T4) - (T2 - T1)
“Payback” for compression work and KE decrease
(T3 - T4) - (T2 - T1) ~ net work + KE increase 4 2
T2 - T1 ~ Work input + KE decrease during compression 1
Net work + KE decrease ~ (T3 - T4) - (T2 - T1)
= (T2 - T1)[exp(s/CP) - 1] > 0
AME Spring Lecture 12

12. Inferring efficiencies
TH,i
Carnot cycle “strip”
th,i = 1 - TL,i/TH,i
TL,i
Double-click chart to open Excel spreadsheet
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 Spring Lecture 12

13. Compression & expansion efficiency
Animation: comparison of ideal Brayton cycle with non-ideal compression & expansion
Same parameters as before but with comp = exp =0.9
2 charts on top of each other; double-click each to open Excel spreadsheets
AME Spring Lecture 12

14.    "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   
AME Spring Lecture 12

15. "Conventional" ramjet example
Example: M1 = 5, T/T1 = 12,  = 1.4
Initial state (1): M1 = 5
State 2: decelerate to M2 = 0
State 4: add at heat const. P; M4 = 0, P4 = P3 = P2 = 529.1P1, T4 = T = 12T1
State 9: expand to P1
AME Spring Lecture 12

16. "Conventional" ramjet example
Specific thrust (ST) (assume FAR << 1)
TSFC and overall efficiency
AME Spring Lecture 12

17. "Conventional" ramjet - effect of M1
"Banana" shaped cycles for low M1, tall skinny cycles for high M1, "fat" cycles for intermediate M1
Basic ramjet
T/T1 = 7
AME Spring Lecture 12

18. "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 Spring Lecture 12

19. 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
AME Spring Lecture 12

20. 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:
NASA X-43:
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 second test
Rich H2-air mixtures ( ≈ ), 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:
Acceleration to steady flight achieved at M1 ≈ 5 at 60,000 ft for 240 seconds using hydrocarbon fuel
AME Spring Lecture 12

21. X43
AME Spring Lecture 12

22. 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 n(-1)/ < 1 even though d = 1, n = 1
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 Spring Lecture 12

23. 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
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 Spring Lecture 12

24. 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 Spring Lecture 12

25. 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 Spring Lecture 12

26. 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 Spring Lecture 12

27. 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 Spring Lecture 12

28. 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
Const. T combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
Stoich. H2-air (f = , QR = 1.2 x 108 J/kg   = 35.6)
AME Spring Lecture 12

29. Hypersonic propulsion (const. T) - effect of M1
Minimum M1 = below that T2 < 2000 even if M2 = 0
No maximum M2
overall improves slightly at high M1 due to higher thermal (lower T9)
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 Spring Lecture 12

30. 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 = (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)
Const. T combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
AME Spring Lecture 12

31. Hypersonic propulsion (const. T) - effect of M2
Maximum M2 = 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)
Const. T combustion, M1 = 10; M2 varies;
H2-air (QR = 1.2 x 108 J/kg),  adjusted so that f = fstoichiometric
AME Spring Lecture 12

32. 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
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 Spring Lecture 12

33. 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
Const. T combustion, M1 = 10; M2 = 2.611; H2-air (QR = 1.2 x 108 J/kg),
 adjusted so that f = fstoichiometric
AME Spring Lecture 12

34. 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)
Const. P combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
Stoich. H2-air (f = , QR = 1.2 x 108 J/kg   = 35.6)
AME Spring Lecture 12

35. 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 = (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
Const. P combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
AME Spring Lecture 12

36. 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!
Const. A combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (f = , 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 Spring Lecture 12

37. 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 = <
f & Specific Thrust increase as more fuel is added ( increasing), overall & ISP decrease slightly at high heat addition due to lower propulsive
Const. A combustion, M1 = 10; M2 = 2.61 (T2 = 2000K)
H2-air (QR = 1.2 x 108 J/kg),  (thus f) varies
AME Spring Lecture 12

38. 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
Courtesy Fred Schauer
AME Spring Lecture 12 2

39. 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 Spring Lecture 12 2

40. 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
AME Spring Lecture 12 2

41. Ideal DE (not P) performance
AME Spring Lecture 12 2

42. 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:
"Uninformed" gas (u1 = 0)
ushock = c1M1,CJ
Shock
Reaction zone (2)
Post - reaction zone (u3 = c1M1 - c3M3 = c1M1 - c3)
Thrust surface (Area A)
Outside pressure = P1
"Stretched" or expanded gas (u4 = 0)
Tube length L
AME Spring Lecture 12

43. PDE analysis Wintenberger et al., 2001
AME Spring Lecture 12

44. PDE analysis CJ pressure = P3 t1 = L/c1M1 Pressure at thrust surface
Post-expansion pressure = P4
Time
Wintenberger et al., 2001
AME Spring Lecture 12

45. 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
AME Spring Lecture 12

46. 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
AME Spring Lecture 12

47. PDE analysis - hydrogen fuel
AME Spring Lecture 12

48. 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 Spring Lecture 12

49. H2-air PDE – 1 tube @ 16 Hz Courtesy F. Schauer
AME Spring Lecture 12

50. H2-air PDE – 4 tubes @ 4 Hz each
Do PDE's make thrust?
Courtesy F. Schauer
AME Spring Lecture 12

51. H2-air PDE – 2 tubes @ high frequency
Courtesy F. Schauer
AME Spring Lecture 12

52. 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 Spring Lecture 12

53. 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 Spring Lecture 12

54. References AME 514 - Spring 2015 - Lecture 12
K. Kailasanath, "Review of Propulsion Applications of Detonation Waves," AIAA J. 38, (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,
F. Schauer, J. Stutrud, R. Bradley (2001). "Detonation Initiation Studies and Performance Results for Pulsed Detonation Engine Applications," AIAA paper No (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).
AME Spring Lecture 12