AME 436 Energy and Propulsion Lecture 5 Propulsion 5: Hypersonic propulsion Outline!!!!!! Why hypersonic propulsion? What's different about it? Conventional ramjet heat addition at M << Heat addition in compressible flows at M AirCycles4Hypersonics.xls spreadsheet Scramjet cycles and performance AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion
Hypersonic propulsion - motivation! Why use air even if you're going to space?! Carry only fuel, not fuel + O, while in atmosphere» 8x mass savings H -O, 4x hydrocarbons» Actually more than this when ln term in Breguet range equation is considered! Use aerodynamic lifting body rather than ballistic trajectory» Ballistic: need Thrust/weight >» Lifting body, steady flight: Lift L = weight mg; Thrust T = Drag D, Thrust/weight = L/D > for any decent airfoil, even at hypersonic conditions L D T mg AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 What's different about hypersonic propulsion?! Stagnation temperature T t - measure of total energy thermal + kinetic of flow - is really large even before heat addition - materials problems T t = T + M '! Stagnation pressure - measure of usefulness of flow ability to expand flow is really large even before heat addition - structural problems P t = P + M '! Large P t means no mechanical compressor needed at large M! Why are T t and P t so important? Recall isentropic expansion to P e = P a optimal exit pressure yielding maximum thrust yields u =! but it's difficult to add heat at high M without major loss of stagnation pressure RT P 9 t P t AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4
What's different about hypersonic propulsion?! High temperatures: not constant, also molecular weight not constant - dissociation - use GASEQ http://www.gaseq.co.uk to compute stagnation conditions! Example calculation: standard atmosphere at, ft! T = 7K, P =.8 atm, c = 3.7 m/s, h = 7.79 kj/kg atmospheric data from http://www.digitaldutch.com/atmoscalc/! Pick P > P, compress isentropically, note new T and h! st Law: h + u / = h + u /; since u =, h = h + M c / or M = [h -h /c ] /! Simple relations ok up to M 7! Dissociation not as bad as might otherwise be expected at ultra high T, since P increases faster than T! Limitations of these estimates! Ionization not considered! Stagnation temperature relation valid even if shocks, friction, etc. only depends on st law but stagnation pressure assumes isentropic flow! Calculation assumed adiabatic deceleration - radiative loss from surfaces and ions in gas may be important AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 5 What's different about hypersonic propulsion? WOW! HOT WARM COLD 5K 3K K K N+O+e - N +O N +O N +O 5 4 Stagnation pressure atm 4 T ideal T with dissoc. P ideal P with dissoc., ft. standard atmosphere. Ambient temp. = 7K Ambient press. =.8 atm. 5 5 Mach number AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion Stagnation temperature K 6 3
Conventional ramjet! Incoming air decelerated isentropically to M = - high T, P! No compressor needed, so only parameters are M! Heat addition at M = - no loss of P t - to max. allowable T 4 = τ λ T! Expand to P 9 = P! Doesn't work well at low M - P t /P T t /T low - Carnot efficiency low! As M increases, P t /P and T t /T increases, cycle efficiency increases, but if M too high, limited ability to add heat T t close to T max - high efficiency but less thrust m f m a m a + m f Diffuser Compressor Combustor Turbine Nozzle 3 4 5 9 AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 7 Conventional ramjet - effect of M! Banana shaped cycles for low M, tall skinny cycles for high M, fat cycles for intermediate M Basic ramjet τ λ = 7 Temperature K Diffuser Compressor Combustor Turbine comp Turbine fan Afterburner Nozzle Close the cycle 3 4 5 6 7 9 5 T-s diagram T-s diagram Intermediate T-s diagram M high M Low M 5 5-5 5 5 75 5 5 75 Entropy J/kg-K AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 8 4
Conventional ramjet example! Example: M = 5, τ λ =, =.4! Initial state : M = 5! State : decelerate to M = T = T + M.4 = T + 5 = 6T P = P + M.4.4 = P + 5.4 = 59.P! State 4: add at heat const. P; M 4 =, P 4 = P = 59.P, T 4 = T! State 9: expand to P 9 = P = P P 4 = 59.P = P 9 + M 9 T 4 =T = T 9 + M 9 = T 9 + +.4.4 5.4 M 9.4 M 9 = 5. T 9 = T AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 9 Conventional ramjet example! Specific thrust ST assume FAR << Thrust =!m a + FARu 9 u + P P 9! TSFC and overall efficiency A 9 ;FAR <<, P 9 = P ST = Thrust!m a c = u 9 c u c = u 9 c 9 c 9 c M = M 9 T 9 T M = 5 5 =.7 TSFC = ST Heat input =!m C T T a P 4t t =!m c C a P Thrust c Thrust c Thrust c [ / ]R T RT 6T = η o = M TSFC = 5 7.4 =.69.7.4 6 T 4t T t = 7.4; AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 5
Conventional ramjet - effect of M.9 9 Thrust fuel consumption.8 Basic ramjet τ λ = 7 Specific thrust TSFC 8 7 6 5 4 3 Specific_thrust TSFC Overall_eff.7.6.5.4.3. Overall efficiency..... 3. 4. 5. 6. Flight mach number M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion Scramjet Supersonic Combustion RAMjet! What if T t > T max allowed by materials or P t > P max allowed by structure? Can't decelerate to M =!! Need to mix fuel burn supersonically, never allowing air to decelerate to M =! Many laboratory studies, very few successful test flights e.g. X-43 below AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 6
Scramjet Supersonic Combustion RAMjet! Australian project HyShot:! http://hypersonics.mechmining.uq.edu.au/hyshot-about! US project X-43! https://www.nasa.gov/centers/dryden/history/pastprojects/hyperx/index.html! Steady flight thrust drag achieved at M 9.65 at, ft altitude u 934 m/s = 656 mi/hr! 3.8 lbs. H burned during - second test! Rich H -air mixtures φ. -.3, ignition with silane SiH 4, 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 33K my model, slide 36: 35K, surface temperatures up to 5K!! USAF X-5! http://www.af.mil/about-us/fact-sheets/display/article/4467/x-5a-waverider/! Acceleration to steady flight achieved at M 5 at 7, ft for 4 seconds using hydrocarbon fuel AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 D steady flow of ideal gases Lecture! Assumptions! Ideal gas, steady, quasi-d! Constant C P, C v, C P /C v! Unless otherwise noted: adiabatic, reversible, constant area! Note since nd Law states ds δq/t = for reversible, > for irreversible, reversible + adiabatic isentropic ds =! Governing equations! Equations of state h h = C P T T P = ρrt; S S = C p lnt T RlnP P /! Isentropic S = S where applicable: P /P = T /T! Mass conservation: m = ρ u A = ρ u A! Momentum conservation, constant area duct see lecture : AdP + m du +C f ρu /Cdx =» C f = friction coefficient; C = circumference of duct» No friction: P + ρ u = P + ρ u! Energy conservation: h + u / + q w = h + u / q = heat input per unit mass = fq R if due to combustion w = work output per unit mass AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4 7
D steady flow of ideal gases Lecture! Types of analyses: everything constant except! Area isentropic nozzle flow already covered! Entropy shock - skip! Momentum Fanno flow constant area with friction - skip! Diabatic q - several possible assumptions - covered here» Constant area A Rayleigh flow useful if limited by space» Constant T useful if limited by materials sounds weird, heat addition at constant T» Constant P useful if limited by structure! Products of analyses! Stagnation temperature! Stagnation pressure! Mach number = u/c = u/rt / recall c = sound speed at local conditions in the flow NOT at ambient condition!! From this, can get exit velocity u 9, exit pressure P 9 and thus thrust AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 5 Heat addition at const. Area Rayleigh flow! Mass, momentum, energy, equation of state all apply! Reference state : use M = not a throat in this case!! Energy equation not useful except to calculate heat input q = C p T t - T t P + ρ u = P + ρ u ; u = RT ; u = RT ; ρ = P / RT ; ρ = P / RT Combine: P = +M P +M ; use reference state where M = : P P = + +M Mass conservation with A = const.: ρ u = ρ u ; combine with above to obtain T T = + +M P t P = t + M ; T +M + t T = M t +M + +M + M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 6 8
Heat addition at constant area T/TRayleigh P/P Rayleigh Tstag/Tstag Rayleigh Pstag/PstagRayleigh Rayleigh flow P t /P t T t /T t. P/P T/T 3 4 5 Mach number AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 7 Heat addition at const. area! Implications! Stagnation temperature always increases towards M =! Stagnation pressure always decreases towards M = stagnation temperature increasing, more heat addition! Can't cross M = with constant area heat addition!! M = corresponds to the maximum possible heat addition! but there's no particular reason we have to keep area A constant when we add heat!! What if neither the initial state nor final state is the choked state? Use P /P = P /P /P /P etc. M P = +M P +M ;T = +M T +M M ; T t = M +M T t M +M + M + M ; P t = +M P t +M + M + M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 8 9
Heat addition at constant pressure! Relevant for hypersonic propulsion if maximum allowable pressure i.e. structural limitation is the reason we can't decelerate the ambient air to M =! Momentum equation: AdP + m du = u = constant! Reference state : use M = again but nothing special happens there! Again energy equation not useful except to calculate q T T = A A = P M P = T t T = + t +M M ' P t P = # + # t +' M '! Implications! Stagnation temperature increases as M decreases, i.e. heat addition corresponds to decreasing M! Stagnation pressure decreases as M decreases, i.e. heat addition decreases stagnation P! Area increases as M decreases, i.e. as heat is added AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 9 Heat addition at constant P P t /P t Constant P heat addition T/T, A/A ConstP P/P ConstP Tstag/Tstag ConstP Pstag/PstagConstP P/P T/T, A/A T t /T t. 3 4 5 Mach number AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion
Heat addition at constant pressure! What if neither the initial state nor final state is the reference state? Again use P /P = P /P /P /P etc. T = A = M P + t = M ' T t = M + M T A M P t + T t M M + M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion Heat addition at constant temperature! Probably most appropriate case for hypersonic propulsion since temperature materials limits is usually the reason we can't decelerate the ambient air to M =! T = constant c sound speed = constant! Momentum: AdP + m du = dp/p + MdM =! Reference state : use M = again T T = P P = exp M ' T t T = + t + M ' A P t P = # exp t +' M -# +,. / + M A = exp M M ' '! Implications! Stagnation temperature increases as M increases! Stagnation pressure decreases as M increases, i.e. heat addition decreases stagnation P! Minimum area i.e. throat at M = -/! Large area ratios needed due to exp[ ] term AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion
Heat addition at constant T Constant T heat addition A/A T t /T t T/T P/P P t /P t T/T ConstT P/P ConstT Tstag/Tstag ConstT Pstag/PstagConstT A/A ConstT. 3 4 5 Mach number AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 Heat addition at constant temperature! What if neither the initial state nor final state is the reference state? Again use P /P = P /P /P /P etc. P P = A A = M M P t P t = exp M ' exp = exp M ' M ' M exp M ' exp = M exp M ' M M ' M exp M ' + M -, / +. exp M ' + = exp M - M ' + M M -, /, + /, / M, / +. +. T t T t = + M + M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4
T-s diagram for diabatic flows T-s diagram for the various diabatic flows reference state Mach = T/TRayleigh T/T Const P T/T Const T Const P T/T Rayleigh Const A Const T Rayleigh Const A. -6-4 - S - S/R AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 5 T-s diagram for diabatic flows Pstag / Pstag ref Rayleigh Const A Comparison of stagnation pressure temperature for the various diabatic flows Rayleigh Const A Const P Rayleigh Const P Const T Const T. 3 4 Tstag / Tstag ref AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 6 3
Area ratios for diabatic flows Const T Const T Const P A / A ref Const A Rayleigh Const P Const T Comparison of area ratio stagnation temperature for the various diabatic flows. 3 4 Tstag / Tstag ref AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 7 What is the best way to add heat?! If maximum T or P is limitation, obviously use that case! What case gives least P t loss for given increase in T t?! Minimize dp t /dt t subject to mass, momentum, energy conservation, eqn. of state! Result lots of algebra - many trees died to bring you this result dp t = M dt t P t or dln P t T t dlnt t = M! Adding heat increasing T t always decreases P t! Least decrease in P t occurs at lowest possible M doesn t matter if it's at constant A, P, T, etc. AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 8 4
Summary of heat addition processes M Const. A Const. P Const. T Goes to M = Decreases Increases Area Constant Increases Min. at M = -/ P Decr. M < Incr. M > Constant Decreases P t Decreases Decreases Decreases T Incr. except for a small region at M < Increases Constant T t Increases Increases Increases AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 9 AirCycles4Hypersonics.xls! Diffuser same as AirCycles4Propulsion.xls except that M is specified rather than M =! Compressor NONE! 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 and included to show constant area / no friction on T-s! Turbine NONE! Nozzle same as AirCycles4Propulsion.xls AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 5
AirCycles4Hypersonics.xls! Combustion parameter τ λ = T 4t /T specifies stagnation temperature, not static temperature, after combustion! Caution on choosing τ λ! If τ λ T < τ r T τ r = + -/ M 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 τ λ T is too large, you can t add that much heat beyond thermal choking point spreadsheet chokes! For constant-t heat addition, if τ λ T 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 τ λ is calculated - make sure this doesn t exceed f stoichiometric! AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 Hypersonic propulsion const. T - T-s diagrams! With Const. T combustion, maximum temperature within sane limits, but as more heat is added, P decreases, eventually P 4 < P 9! Also, latter part of cycle has low Carnot-strip efficiency since constant T and P lines will converge Temperature K Diffuser Combustor Nozzle Diffuser Combustor Nozzle Area Close the cycle Rayleigh from Rayleigh from Area 4 Area 9 Area Diffuser 4 Combustor Nozzle Mach Mach 9 5 4 Mach.E+ 9 Mach 45 4 35 3 5 5 5 T-s diagram -5 5 5 5 3 Entropy J/kg-K Area m^.e-.e-.e-3.e-4 Area and Mach number.e-5 4 6 8 Step number Const. T combustion, M = ; M =.6 T = K Stoich. H -air f =.83, Q R =. x 8 J/kg τ λ = 35.6 8 6 4 Mach number AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 3 6
Hypersonic propulsion const. T - effect of M! Minimum M = 6.8 - below that T < even if M =! No maximum M! η overall improves slightly at high M due to higher η thermal lower T 9 M, Tau_lamda, ST OE 5 4.5 4 3.5 3.5.5.5 Tau_lamda/ Mach No after diffuser Specific Thrust Overall efficiency Specific Impulse 4 35 3 5 5 5 Specific impulse sec 6 7 8 9 3 Flight Mach No. M Const. T combustion, M = varies; M adjusted so that T = K; H -air Q R =. x 8 J/kg, τ λ adjusted so that f = f stoichiometric AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 33 Hypersonic propulsion const. T - effect of τ λ! M =, T = K specified M =.6! At τ λ =. no heat can be added! At τ λ = 35.6, f =.83 stoichiometric H -air! At τ λ = 4.3 assuming one had a fuel with higher heating value than H, P 4 = P 9! f Specific Thrust increase as more fuel is added τ λ increasing, η overall I SP decrease due to low η thermal at high heat addition see T-s diagram f, ST OE 4 3.5 3.5.5.5 Fuel mass fraction Stoich. fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 45 4 35 3 5 5 5 Specific impulse sec 5 3 35 4 45 Tau lamda Const. T combustion, M = ; M =.6 T = K H -air Q R =. x 8 J/kg, τ λ thus f varies AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 34 7
Hypersonic propulsion const. T - effect of M! Maximum M = 3. - above that P 4 < P 9 after combustion you could go have higher M but why would you want to - heat addition past P 4 = P 9 would reduce thrust!! No minimum M, but lower M means higher T - maybe beyond materials limits after all, high T t is the whole reason we re looking at alternative ways to burn at hypersonic Mach numbers! η overall decreases at higher M due to lower η thermal lower T 4 4 f, ST OE 3.5 3.5.5.5 Fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 35 3 5 5 5 Specific impulse sec.5.5.5 3 3.5 Mach No. after diffuser M Const. T combustion, M = ; M varies; H -air Q R =. x 8 J/kg, τ λ adjusted so that f = f stoichiometric AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 35 Hypersonic propulsion const. T - effect of η d! Obviously as η d decreases, all performance parameters decrease! If η d too low, pressure after stoichiometric heat addition < P, so need to decrease heat addition thus τ λ! Diffuser can be pretty bad η d.5 before no thrust f, ST OE 3.5.5.5 Fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 35 3 5 5 5 Specific impulse sec..4.6.8. Diffuser efficiency -5 Const. T combustion, M = ; M =.6; H -air Q R =. x 8 J/kg, τ λ adjusted so that f = f stoichiometric or P 5 = P, whichever is smaller AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 36 8
Hypersonic propulsion const. T - effect of η n! Obviously as η n decreases, all performance parameters decrease! Nozzle can be pretty bad η n.3 before no thrust, but not as bad as diffuser 3 35 f, ST OE.5.5.5 Fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 3 5 5 5 Specific impulse sec..4.6.8. Nozzle efficiency Const. T combustion, M = ; M =.6; H -air Q R =. x 8 J/kg, τ λ adjusted so that f = f stoichiometric AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 37 Hypersonic propulsion const. P - T-s diagrams! With Const. P combustion, no limitations on heat input, but maximum temperature becomes insane actually 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 Temperature K Diffuser Combustor Nozzle Diffuser Combustor Nozzle Area Close the cycle Rayleigh from Rayleigh from Area 4 Area 9 Area Diffuser 4 Combustor Nozzle Mach Mach 9 6 4 Mach.E+ 9 Mach 5 4 3 T-s diagram Area m^.e-.e-.e-3 Area and Mach number 8 6 4 Mach number.e-4-5 5 5 Entropy J/kg-K.E-5 4 6 8 Step number Const. P combustion, M = ; M =.6 T = K Stoich. H -air f =.83, Q R =. x 8 J/kg τ λ = 35.6 AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 38 9
Hypersonic propulsion const. P - performance! M =, T = K specified M =.6! Still, at τ λ =. no heat can be added! At τ λ = 35.6, f =.83 stoichiometric H -air! No upper limit on τ λ assuming one has a fuel with high enough Q R! f Specific Thrust increase as more fuel is added τ λ increasing, η overall I SP decrease only slightly at high heat addition due to lower η propulsive f, ST OE 4.5 4 3.5 3.5.5.5 Fuel mass fraction Stoich. fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 4 39 38 37 36 35 34 Specific impulse sec 33 5 3 35 4 45 Tau lamda Const. P combustion, M = ; M =.6 T = K H -air Q R =. x 8 J/kg, τ λ thus f varies AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 39 Hypersonic propulsion const. A - T-s diagrams! With Const. A combustion, heat input limited by thermal choking, maximum temperature even more insane than constant P! but Carnot-strip efficiency is awesome! Temperature K Diffuser Combustor Nozzle Diffuser Combustor Nozzle Area Close the cycle Rayleigh from Rayleigh from Area 4 Area 9 Area Diffuser 4 Combustor Nozzle Mach Mach 9 6 4 Mach.E+ 9 Mach 5 4 3 T-s diagram Area m^.e-.e-.e-3 Area and Mach number 8 6 4 Mach number.e-4-4 6 8 Entropy J/kg-K.E-5 4 6 8 Step number Const. A combustion, M = ; M =.6 T = K H -air f =.7, Q R =. x 8 J/kg τ λ = 3.; can t add stoichiometric amount of fuel at constant area for this M and M AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4
Hypersonic propulsion const. A - performance! M =, T = K specified M =.6! Still, at τ λ =. no heat can be added! At τ λ = 3.5, thermal choking at f =.93 <.83! f Specific Thrust increase as more fuel is added τ λ increasing, η overall I SP decrease slightly at high heat addition due to lower η propulsive f, ST OE 3.5.5.5 Fuel mass fraction Stoich. fuel mass fraction Specific Thrust Overall efficiency Specific Impulse 395 39 385 38 375 37 Specific impulse sec 365 4 6 8 3 3 Tau lamda Const. A combustion, M = ; M =.6 T = K H -air Q R =. x 8 J/kg, τ λ thus f varies AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4 Example Consider a very simple propulsion system in a standard atmosphere at, feet 7K and.7 atm, with =.4 in which Incoming air is decelerated isentropically from M = 5 until T = 3K Heat is added at constant T until ambient pressure is reached not a good way to operate, but this represents a sort of maximum heat addition a To what Mach number could the air be decelerated if the maximum allowable gas temperature is 3K? What is the corresponding pressure? " T + M " ' = T + # M ' # ".4 " 7K + 5.4 ' = 3K + M ' # # M 7K ".4 + = + 5 ' - =.43 M = 3.5.4 3K #, P " = T " ' = 3K.4 '.4 = 839;P = 839P = 839.7 = 89.79atm P # # 7K T AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 4
Example - continued b What is the exit Mach M 9 number? Note that heat is added until P 3 = P, so there is no separate nozzle 3 9 in this case. What is the area ratio? P 3 P = exp M M 3 ;ln P 3 P M 3 = M ln P P 3 = 3.5 c What is the specific thrust? = M M 3 ln / 839 = 5.3 M 3 = 5.3.4 A 3 A = M exp M 3 M M 3 = 3.5.4 exp 5.3 5.3 3.5 = 5869 ST = Thrust/ m c = u 9 u / c = M 9 c 9 M c /c = M e T e /T / a m a m a M, ST = M 9 T 9 /T / M = 5.33K/7K / 5 = 3.89 AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 43 Example - continued d What is the thrust specific fuel consumption? TSFC = Heat input/thrustc = [ m a C P T 3t T t c ]/[Thrustc ] = [ m a c /Thrust] [/ RT 3t T t /RT ] = [/ST] [/ ] [T 3t T t /T ] " T t = T + M " ';T 3t = T 3 + # M e ';T = T 3 = T e # T 3t T t = T " 3 + T M e ' T " + # M ' = T " + # M e + M ' # T T 3t T t = T T T T M e M = 3 7.4 Requirement is higher by a factor of.3, so H -air cannot provide this much heat release AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 44 T 5.3 3.5 = 34. TSFC = T 3t T t = 3.79 ST T.4 34. = 6. lousy! e Can stoichiometric hydrogen-air mixtures generate enough heat to accomplish this? Determine if the heat release per unit mass = f stoich Q R is equal to or greater than the heat input needed = C P T 3t T t. C P T 3t T t = R T 3t T t.4 "# 8.34J / molek.897kg / mole T = 34. 7K T.4 f stoich Q r =.83. 8 J / kg = 3.396 6 J / kg;c P T 3t T t = 7.777 6 J / kg
Summary! Propulsion at high Mach numbers is very different from conventional propulsion because! The optimal thermodynamic cycle decelerate to M = yields impracticably high T P! Deceleration from high M to low M without major P t losses is difficult! Propulsive efficiency u /u +u 9 is always high! 3 ways of adding heat discussed! Constant T» Probably most practical case» Low efficiency with large heat addition» Large area ratios! Constant P - best performance but very high T! Constant A - thermal choking limits heat input AME 436 - Spring 8 - Lecture 5 - Hypersonic Propulsion 45 3