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

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

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

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 514 - Spring 2015 - Lecture 12

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

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

   "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 514 - Spring 2015 - Lecture 12

"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 514 - Spring 2015 - Lecture 12

"Conventional" ramjet example Specific thrust (ST) (assume FAR << 1) TSFC and overall efficiency AME 514 - Spring 2015 - Lecture 12

"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 514 - Spring 2015 - Lecture 12

"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

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 514 - Spring 2015 - Lecture 12

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

X43 AME 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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

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

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

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

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 = 0.0283, QR = 1.2 x 108 J/kg   = 35.6) AME 514 - Spring 2015 - Lecture 12

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) 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

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) 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

Hypersonic propulsion (const. T) - effect of M2 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) 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

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 = 0.0283, QR = 1.2 x 108 J/kg   = 35.6) AME 514 - Spring 2015 - Lecture 12

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 Const. P 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

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 = 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

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 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

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 514 - Spring 2015 - Lecture 12 2

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 2

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 514 - Spring 2015 - Lecture 12 2

Ideal DE (not P) performance AME 514 - Spring 2015 - Lecture 12 2

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 514 - Spring 2015 - Lecture 12

PDE analysis Wintenberger et al., 2001 AME 514 - Spring 2015 - Lecture 12

PDE analysis CJ pressure = P3 t1 = L/c1M1 Pressure at thrust surface Post-expansion pressure = P4 Time Wintenberger et al., 2001 AME 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

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 514 - Spring 2015 - Lecture 12

PDE analysis - hydrogen fuel AME 514 - Spring 2015 - Lecture 12

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

H2-air PDE – 1 tube @ 16 Hz Courtesy F. Schauer AME 514 - Spring 2015 - Lecture 12

H2-air PDE – 4 tubes @ 4 Hz each Do PDE's make thrust? Courtesy F. Schauer AME 514 - Spring 2015 - Lecture 12

H2-air PDE – 2 tubes @ high frequency Courtesy F. Schauer AME 514 - Spring 2015 - Lecture 12

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

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

References AME 514 - Spring 2015 - Lecture 12 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