Introduction to High Speed Airbreathing Propulsion Systems

MAE 5540 - Propulsion Systems • Oxidizer … “liquid air” Have to take our own along So that the engine can breathe At high altitudes

• How Much Oxidizer? … depending op chosen propellants … 4-9 times as much As the fuel we carry!

MAE 5540 - Propulsion Systems Flight Without Oxidizer, I What if we could get enough oxygen from ambient air?

X-43 airbreathing SCRAMjet engine

What happens to our Isp?

OK .. Lets loose the Oxidizer … and our Isp goes up by a factor Of 7! … but where do we the the Oxidizer for combustion? ….

MAE 5540 - Propulsion Systems … from the atmosphere MAE 5540 - Propulsion Systems Flight Without Oxidizer, II

Do Not Need Oxidizer here

For a large portion of the Launch trajectory there is Plenty Enough air For combustion if you are going fast enough

Need Oxidizer here

MAE 5540 - Propulsion Systems Fuel Efﬁciencies of Various High Speed Propulsion Systems

Isp

Minimal turbo-machinery MAE 5540 - Propulsion Systems Operational Flight Envelope for Various Flight Vehicles

MAE 5540 - Propulsion Systems Operations Payoffs for Airbreathing Launch

• Decreased gross lift-off weight, resulting in smaller facilities and easier handling • Wider range of emergency landing sites for intact abort • Powered ﬂyback/go-around & more margin at reduced power • Self-ferry & taxi capabilities • Greatly expanded launch windows (double or triple) • Rapid orbital rendezvous (up to three times faster than rockets) • Wider array of landing sites from orbit, with 2,000-mile cross range and increased range • Reduced sensitivity to weight growth

MAE 5540 - Propulsion Systems Applications

MAE 5540 - Propulsion Systems Amount of Rocket propellant required to launch a 1000 kg payload into orbit

Titan IV Our total launch mass goes from space shuttle Rockets10,000 on kg/kg the Launch down Pad to are <1400 Mostly ALLkg/kg FUEL!! wow! Amount of Propellant required, kg "Venture-star"

MAE 5540 - Propulsion Systems Specific Impulse, sec. Airbreathing Propulsion Basics

Squeeze Bang Blow

Suck

MAE 5540 - Propulsion Systems Airbreathing (Jet) Propulsion Basics (cont’d)! • Ramjet Engine!

• Simplest of all Jet Engine= (Credit Narayanan Komerath, Georgia Tech)! Concepts!

• Compression and Power Extraction steps Performed Step 1-2! Step 3! Step 4! passively!

• Requires High Speed inlet ﬂow to operate! MAE 5540 - Propulsion Systems! Airbreathing (Jet) Propulsion Basics (cont’d)!

• Ramjet Engine!

Ramjets cannot produce thrust at zero airspeed; they cannot move an aircraft from a standstill. A ramjet powered vehicle, therefore, requires an assisted take-off like a rocket assist to accelerate it to a speed where it begins to produce thrust.! https://www.youtube.com/watch? v=1DG4f2umclk! MAE 5540 - Propulsion Systems! Airbreathing Propulsion Basics (cont’d)! Turbine • Turbojet/Turbo fan ! Combustor Nozzle engines Combustion! Compressor Cycle steps! Turbojet Compression and ! Power extraction! Inlet! steps use! Afterburner Turbo-machinery! Turbine To augment cycle! Compressor Afterburner

Turbofan

Inlet! Combustor Nozzle MAE 5540 - Propulsion Systems! Fan Airbreathing Propulsion Basics (cont’d)!

turbojet. Simple turbine engine that produces all of its thrust from the exhaust from the turbine section. However, because all of the air is passing through the whole turbine, all of it must burn fuel.!

MAE 5540 - Propulsion Systems! Airbreathing Propulsion Basics (cont’d)! Turbofan. Turbine primarily drives a fan at the front of the engine. Most engines drive the fan directly from the turbine. Part of the air enters the turbine section of the engine, and the rest is bypassed around the engine. In high-bypass engines, most of the air only goes through the fan and bypasses the rest of the engine and providing most of the thrust. ! A turbofan thus can be thought of as a turbojet being used to drive a ducted fan, with both of those contributing to the thrust. The ratio of the mass-ﬂow of air bypassing the engine core compared to the mass-ﬂow of air passing through the core is referred to as the bypass ratio. !

MAE 5540 - Propulsion Systems! Airbreathing Propulsion Basics (cont’d)!

Turbofans have a number of beneﬁts over simple turbojets. It is more energy efﬁcient to accelerate a large air mass by a small amount, than a small air mass by a lot. A turbofan effectively increases the volume of air propelled with a corresponding decrease in average exhaust gas velocity. !

Larger, slower fan, positioned upwind of the compressor pushes air (relatively) slowly through both the core and the bypass. The air that makes it to the core is compressed and undergoes combustion. The air that makes it to the bypass is ducted around the core, and mixes with the exhaust gases when it exits the back of the engine. !

MAE 5540 - Propulsion Systems! Ideal Ramjet Cycle Analysis Step 1-2 Step 3 Step 4 Region Process Ideal Behavior Real Behavior

A to 1(inlet) Isentropic ﬂow P0,T0 constant P0 drop

1-2 (diffuser) Adiabatic P,T increase P0 drop Compression P0 drop

2-3 (burner) Heat Addition P0 constant, T0 P0 drop s Increase " !q% !s = $ ' > 0 # T & rev

3-4 (nozzle) Isentropic T0,P0 constant s Increase expansion MAE 5540 - Propulsion Systems !s > !srev T0 drop Ideal Ramjet Cycle Analysis

T-s Diagram Step 1-2 Step 3 Step 4

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of Ideal Ramjet

• Net Work Available --> work perform by system in step 4 minus work required for step 1-2

• Net heat input --> heat input during step 3 (combustion) - heat lost in exhaust plume

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d) • Ideal Cycle Efﬁciency(η) = (Net work output)/(Net heat input}

Net Work • = (hA ! hB ) + (hC ! hD ) m Net Heat Input • = (hC ! hB ) m

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4 MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d) • Ideal Cycle Efﬁciency = (Net work output)/(Net heat input}

" Net Work % $ • ' # m & (hA ( hB ) + (hC ( hD ) (hD ( hA ) ! = = = 1( " Net Heat Input % (hC ( hB ) (hC ( hB ) $ • ' # m &

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4 MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d)

• Assume … Cpair ~ Cpproducts # C & p air %TD " TA ( C T " C T C p products D p air A $ p products ' ! = 1" ( ) = 1" C T " C T # C & ( p air C p products B ) p air % TC " TB ( C $ p products '

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4 MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d)

• For simplicity … let … Cpair ~ Cpproducts $ C ' p air &TD # TA ) C % p products ( (TD # TA ) ! " 1# = 1# $ C ' (TC # TB ) p air & TC # TB ) C % p products ( $ T T ' $ T T T ' D # A D # A B %& T T () %& T T T () = 1# C C = 1# C B C $ TB ' $ TB ' &1# ) &1# ) % TC ( % TC (

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d) • From C-->D ﬂow is isentropic …

( )1 (Credit Narayanan Komerath, Georgia Tech) ( TD " PD % ! = $ ' TC # PC &

Step 1-2 Step 3 Step 4 # ) "1 & # P & ) T T # T T T & % D " A B ( D " A B % $% P '( T T ( $% T T T '( $ C B C ' ! = 1" C B C = 1" # TB & # TB & %1" ( %1" ( $ TC ' $ TC '

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d) • Adiabatic compression across diffuser

(Credit Narayanan Komerath, Georgia Tech)

! P $ ! P P0 P0 $ A = A ' A ' B = Step 1-2 Step 3 Step 4 # P & # P P P & " B % " 0A 0B B %

( ( ( )1 ( )1 ! $ ( )1 T ( )1 P ( ! P $ ( T ! 0B $ 0A T ! P $ 0B A * T = T * A = A # T & # T & P 0A 0B T # P & # P & " 0A % " B % 0B B " B % " 0A %

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d)

• Sub into efﬁciency (Credit Narayanan Komerath, Georgia Tech) equation

Step 1-2 Step 3 Step 4

# ) "1 ) "1 ) "1 & # P & ) # P & ) # P & ) T % D " A 0B B ( % % ( % ( % ( ( $ PC ' $ PB ' $ P0 ' TC $% A '( ! = 1" # TB & %1" ( $ TC '

MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d)

(Credit Narayanan Komerath, Georgia Tech) ideal nozzle ! PA = PD ideal burner ! PB = PC

Step 1-2 Step 3 Step 4

# ) "1 & # ) "1 & # P & ) T # P & ) %1" 0B B ( %T " 0B T ( ) "1 % ( ) "1 C % ( B % P0 TC ( % P0 ( ) % $ A ' ( ) % $ A ' ( # PA & $ ' # PA & $ ' ! = 1" % ( = 1" % ( $ PB ' # TB & $ PB ' (TC " TB ) %1" ( $ TC ' MAE 5540 - Propulsion Systems Thermodynamic Efﬁciency of

Ideal Ramjet (cont’d)

# ) "1 & (Credit Narayanan Komerath, Georgia Tech) # P & ) %T " 0B T ( ) "1 C % ( B % P0 ( ) % $ A ' ( # PA & $ ' ! = 1" % ( $ PB ' (TC " TB ) Step 1-2 Step 3 Step 4

i) As engine pressure ratio, PB/PA, goes up … η goes up

ii) As combustor temperature difference TC-TB goes up … η goes up

iii) As inlet total pressure ratio (P0B/P0A) goes down … (stagnation pressure loss goes up) … η goes down

MAE 5540 - Propulsion Systems Why is total pressure recovery important?

• Choked • ! +1 (! (1) Nozzle Throat m ! " 2 % p0 massﬂow * = $ ' A Rg # ! + 1& T0

• Thrust is proportional to engine massﬂow

• • Thrust = me Ve ! mi Vi + ( peAe ! p" Ae ) MAE 5540 - Propulsion Systems Ideal Ramjet: Inlet and Diffuser

• Take a Rocket motor• and “lop the top off” m fuel

• normal • M > 1 m M < 1 m air shockwave air+fuel

• Works Ok for subsonic, but for supersonic ﬂow … can’t cram enough air down the tube

• Result is a normal shock wave at the inlet lip

MAE 5540 - Propulsion Systems Ideal Ramjet: Inlet and Diffuser (cont’d)

• m fuel

• normal • M > 1 m M < 1 m air shockwave air+fuel

M ∞ M2 • Mechanical Energy is Dissipated into Heat

• Huge Loss in Momentum

MAE 5540 - Propulsion Systems Ideal Ramjet: Inlet and Diffuser (cont’d) • So … we put a spike in front of the inlet • m fuel Oblique shockwave

• M > 1 m • air M < 1 m air+fuel

Much weaker normal shockwave

M! M2 • How does this spike Help?

• By forming an Oblique Shock wave ahead of the inlet

MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example

P0B M =4.0 1 B

β= 40° • Compare MB and P0 behind normal shockwaves

B P0B

Assume γ=1.4 MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d)

P0B M =4.0 1 B

• From Normal Shock wave solver

$ P0 ' B = 0.1388 & P ) 0 ! normal..shock & ) M " M = 0.434959 # & p ) ! B & B = 18.5 ) p & ! ) & ) % ( MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) β= 40°

• Across Oblique Shock wave

$ ! % M n=0.5064 • M1n= M1 sin β1 = 4 sin 40 =2.571 2 "180 #

$ 2 ! % 2 $4 sin2 $ 40% & 1% 2 2 180 ( " "180 # # ) 2{M1 sin (") # 1} atan ( ) tan(!) = ) ( ) 2 ! $ $ ! % % $ 2 $ $ ! % % % tan(")%2 + M1 &%$ + cos(2")('' ( tan 40 2 + 4 1.4 + cos 2'40 ) & ( " " "180 # # " " "180 # # # #

MAE 5540 - Propulsion Systems θ = 26.2° 2-D Ramjet Inlet Example (cont’d) β= 40°

• Across Oblique Shock wave 0.5064 M n M n = 0.5064 ! M = 2 = % ! & =2.123 2 2 sin # (40 " 26.2) $ sin("1-#) 180

P02/P01 = 0.4711

MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) β= 40°

• Across Normal Shock wave (behind oblique Shock) normal..shock M 2 = 2.123 ! M B = 0.557853 " P P P P 0 B = 0.663531 " 0 B = 0 B 0 2 =(0.663531)(0.4711)=0.3126 P P P P 0 2 0 ! 0 2 0 ! $ 1.4 % p p P P & $ 1.4 ! 1 2% (1.4 ! 1) ' B B 0 B 0 ! 0.3126 & "1 + 4 # ' = # # = " 2 # p P P p = 38.422 ! 0 B 0 ! ! $ 1.4 % & $ 1.4 ! 1 2% (1.4 ! 1) ' & "1 + 0.557853 # ' MAE 5540 - Propulsion Systems " 2 # 2-D Ramjet Inlet Example (cont’d)

M B = 0.4350 P 0 B = 0.1388 M =4.0 P 1 0 ! p B = 18.5 P0 • Compare ! • Spike aids in increasing Total Pressure recovery Reducing “ram drag”

M B = 0.557853 P 0 B = 0.3126 P 0 ! p B = 38.422 M =4.0 P 1 0 !

MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) • … Continuing example … Incoming Air to Ramjet

• Molecular weight = 28.96443 kg/kg-mole • γ = 1.40 • Rg = 287.056 J/°K-(kg) • T∞ = 216.65 °K • p∞ • =• 19.330 kPa • Combustor q = q/ m = 500 kJ/kg

• Assume that mass of added fuel is negligible, exhaust

and γ, Rg are the same

MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d)

• Compute free stream stagnation temperature 1.4 ! 1 $ " # 1 2 ' 216.65 $1 + 42% =909.93°K T0 = T! 1+ M ! = " # ! %& 2 () 2 • Compute BURNER stagnation temperature q + c T 3 T = p 0! = 500!10 + 1004.696 (909.93) 0C = 1407.6 °K cp 1004.696

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4 MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) • Compute efﬁciency …. Normal shock inlet

M B = 0.4350 P 0 B = 0.1388 P M1=4.0 0 ! p B = 18.5 P 0 !

• Compute TC, TB 909.93 TC = = 867.75°K 1.4 ! 1 $1 + 0.4352% " 2 # 1407.6 T = = 1356.3°K B 1.4 ! 1 $1 + 0.4352% " 2 # MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) • Compute efﬁciency …. Normal shock inlet

M B = 0.4350 P 0 B = 0.1388 M =4.0 P 1 0 ! p B = 18.5 P 0 !

# ) "1 & ) • Compute TC, TB # P0 & %T " B T ( ) "1 % C % ( B ( $ P0 ' T = 867.75 K # P & ) $% A '( C ° A = ! = 1" % ( $ PB ' (TC " TB ) TB = 1356.3°K ! (1.4 ! 1) & & (1.4 ! 1) ' ' 1.4 $ $ 1.4 % % 18.5 $1356.3 ! $0.1388 % 867.75% " " # # =0.2328 1 ! MAE 5540 - Propulsion Systems (1356.3 ! 867.75) 2-D Ramjet Inlet Example (cont’d) • Compute efﬁciency …. Oblique shock inlet

M B = 0.557853 P 0 B = 0.3126 P 0 ! p B = 38.422 M =4.0 P 1 0 !

• Compute TC, TB 909.93 T = = 856.61°K C $ 1.4 ! 1 2% "1 + 0.557853 # 2 1407.6 TB = 1.4 ! 1 $1 + 0.5578532% = 1325.1°K " 2 # MAE 5540 - Propulsion Systems 2-D Ramjet Inlet Example (cont’d) • Compute efﬁciency …. Oblique shock inlet

M B = 0.557853 P 0 B = 0.3126 P 0 ! p B = 38.422 M =4.0 P 1 0 ! # ) "1 & ) • Compute TC, TB # P0 & %T " B T ( ) "1 % C % ( B ( $ P0 ' T = 856.61 K # P & ) $% A '( C ° A = ! = 1" % ( $ PB ' (TC " TB ) TB = 1325.1°K ! (1.4 ! 1) & & (1.4 ! 1) ' ' 1.4 $ $ 1.4 % % 38.422 $1325.1 ! $0.3126 % 856.61% " " # # 1 ! = 0.4652 MAE 5540 - Propulsion Systems (1325.1 ! 856.61) 2-D Ramjet Inlet Example (cont’d) • Compute efﬁciency …. Oblique shock inlet

M1=4.0 η = 0.2328

η = 0.4652

M1=4.0 200% increase in efﬁciency! MAE 5540 - Propulsion Systems Supersonic Inlet: Condorde Inlet Design

• Mach 2 Cruise

Credit: http://www.concordesst.com/powerplant.html MAE 5540 - Propulsion Systems Conical versus 2-D Inlets

SR-71

Concorde

F-15 MAX-15E 5540 Ramjet - Propulsion Systems Physical Aspects of Cone Flow • Compare cone ﬂow to wedge • Cone ﬂow supports a much larger wedge angle Cone Flow before shock wave detaches

M∞=2.0 Wedge Flow

M∞=1.5

M∞=2.0

M∞=1.5

M∞=4.0 M∞=8.0

M∞=8.0 M∞=4.0

MAE 5540 - Propulsion Systems Physical Aspects of Cone Flow (cont’d) • Three-dimensional “relieving” effect

• Cone shock wave is Effectively weaker Than shock wave for Corresponding wedge angle

• Conical Inlet not at effective as 2-D inlet … but much less “draggy”

MAE 5540 - Propulsion Systems Air to Fuel Ratio Computations

(Credit Narayanan Komerath, Georgia Tech)

Step 1-2 Step 3 Step 4

• Stagnation enthalpy of the air and burned products leaving the combustor equals the enthalpy of the air entering the combustor + heat released by chemical reaction

• Sensible enthalpy of fuel assumed to be negligible when compared to heat of reaction

MAE 5540 - Propulsion Systems Fuel to Air Ratio Computations (cont’d)

0 (Credit Narayanan Komerath, Georgia Tech) T 2 -- stagnation temperature entering burner

T03 -- stagnation temperature exiting burner Step 1-2 Step 3 Step 4

• • • f -- fuel to air ratio mair 1 f C T mair C T m f q ( + ) p 3 0 3 = p 2 0 2 + r p • • C 2 -- speciﬁc heat of mair C T f mair q air entering burner = p 2 0 2 + r Solve for f Cp3 -- speciﬁc heat of C T combustion products p 3 0 3 ! 1 C T C T C T p 3 0 3 ! p 2 0 2 p 2 0 2 f = = q -- heat of reaction q ! C T " q C T % r ( r p 3 0 3 ) r ! p 3 0 3 $ C T C T ' MAE 5540 - Propulsion Sy#stemp 2s 0 2 p 2 0 2 & Equivalence Ratio versus Mixture Ratio • The equivalence ratio is used to characterize the mixture ratio Of airbreathing engines … analogous to mixture ratio

• The equivalence ratio,Φ , is deﬁned as the ratio of the actual fuel-air ratio to the stoichiometric fuel-air ratio.

… Φ > 1 ---> a rich mixture … Φ < 1 ---> lean mixture # • & m fuel • For Φ = 1, no oxygen is left in % ( exhaust products … combustion • % mair ( is called stoichiometric $ 'actual [ f ]actual ! " = # • & f m fuel [ ]stoich % • ( % mair ( $ 'stoich MAE 5540 - Propulsion Systems Equivalence Ratio versus Mixture Ratio (cont’d)

• Consider combustion of H2 fuel with air … what is Stoich fuel ratio (deﬁne only water as product of combustion)

!!" 2H 2 + O2 #!! H 2O • 2 2 • • • $ ' kg + kg* mol $ ' $ ' m H 2 kg* mol 1 m H 2 m H 2 0.21mO2 & • ) = = " ( f )stoich = & • ) = & • ) + • = 0.02625 & ) 32 kg + 1kg* mol 8 & ) & ) % mO2 ( % mair ( % mO2 ( mair stoich kg* mol stoich stoich Air/Fuel=38.09 • Assumed equilibrium chemistry is described by reaction above with only water vapor as combustion product, + left over atmospheric nitrogen

MAE 5540 - Propulsion Systems CEA Calculation

• Actual Equilibrium Calculations, Gaseous H2, air, φ=1, pc= 1000 kPa (145 psi) TB=856.6 °K (earlier ramjet problem) Assumed air composition (dry air) Nitrogen 79.00 Oxygen 21.00 Combustion Product MOLE FRACTIONS Burner M=1 H2O 0.31761 0.31761 N2 0.68239 0.68239 Combustion properties Burner M=1 T, K 2826.42 2521.42 H, KJ/KG 785.57 260.70 U, KJ/KG -160.58 -583.35 G, KJ/KG -28431.7 -25803.7 M, (1/n) 24.838 24.838 GAMMAs 1.2395 1.2437 O/F= 38.09

MAE 5540 - Propulsion Systems CEA Calculation, more realistic reaction

• Actual Equilibrium Calculations, Gaseous H2, air, φ=1, pc= 1000 kPa (145 psi) TB=856.6 °K (earlier ramjet problem) Gas % of Earth Atmosphere at sea level (dry air) Nitrogen 78.08 Oxygen 20.95 Equilibrium reactions Combustion Product Argon 0.94 Reduces ﬂame temperature MOLE FRACTIONS Carbon dioxide 0.03 Compared to previous Burner M=1 Ar 0.00564 0.00567 CO 0.00005 0.00003 Combustion properties CO2 0.00012 0.00014 Burner M=1 H 0.00205 0.00086 T, K 2662.82 2446.29 H2 0.01718 0.01062 H, KJ/KG 779.69 288.07 H2O 0.29080 0.30174 U, KJ/KG -120.74 -534.64 NO 0.00402 0.00229 G, KJ/KG -26714.1 -24970.0 N2 0.66569 0.67022 M, (1/n) 24.588 24.723 O 0.00065 0.00025 GAMMAs 1.1814 1.1951 OH 0.00936 0.00523 O/F= 37.88393 O2 0.00445 0.00296

MAE 5540 - Propulsion Systems CEA Calculations (cont’d)

• Actual Equilibrium Calculations, Gaseous H2, air, φ=2, pc= 1000 kPa (145 psi) TB=856.6 °K (earlier ramjet problem) (rich mixture) Gas % of Earth Atmosphere at sea level (dry air) Nitrogen 78.08 Reduced ﬂame temperature Oxygen 20.95 Combustion Product Compared to previous Argon 0.94 MOLE FRACTIONS But molecular weight is lower Carbon dioxide 0.03 Burner M=1 Ar 0.00433 0.00433 Combustion properties CO 0.00011 0.00010 Burner M=1 CO2 0.00002 0.00002 T, K 2422.36 2156.10 H 0.00274 0.00091 H, KJ/KG 963.90 382.97 H2 0.23958 0.24051 U, KJ/KG -76.701 -542.18 H2O 0.24021 0.24097 G, KJ/KG -28674.1 -25997.3 NH3 0.00001 0.00001 M, (1/n) 19.355 19.377 NO 0.00004 0.00001 GAMMAs 1.2399 1.2559 N2 0.51239 0.51300 O/F= 37.88393 OH 0.00057 0.00013

MAE 5540 - Propulsion Systems CEA Calculations (cont’d)

MAE 5540 - Propulsion Systems Speciﬁc Thrust of Air Breathing Engine • Analogous to speciﬁc impulse

• • Cruise design condition Thrust = me Ve ! mi Vi + ( peAe ! p" Ae ) When pe=p∞

• • • ! $ 'm + m *V - m V F ) f air , e air . ' 1 * 1 1 # thrust & ( + • = • = )1+ ,Ve - V. = Ve + (Ve - V. ) # m & m ( f + f f " f % opt f

MAE 5540 - Propulsion Systems Speciﬁc Thrust of Air Breathing Engine (cont’d) • Re writing in terms of mach number

! $ F ! f + 1$ 1 ! f + 1$ 1 # thrust & = V ' V = ) R T M ' ) R T M = • # & e ( # & e g e e e ( ( g ( ( ( ) # m & " f % f " f % f " f % opt ! $ ! f + 1$ Te 1 T( ) e R g To M e ' # ) ( R g T0 M ( & = "# f %& e e T f ( ( T oe " 0( % ! $ ! f + 1$ M 1 # M & ) R T e ' ) R T ( # & e g e oe # ( g ( 0( & " f % ) e ' 1 2 f # ) ( ' 1 2 & 1+ M e # 1+ M ( & 2 " 2 %

MAE 5540 - Propulsion Systems Speciﬁc Thrust of Air Breathing Engine (cont’d) • Re writing in terms of mach number

! $ ! $ F ! f + 1$ M 1 # M & # thrust & = ' R T e ( ' R T ) • # & e g e oe # ) g ) 0) & # & " f % ' e ( 1 2 f ' ) ( 1 2 m f # & " % opt 1+ M e # 1+ M ) & 2 " 2 %

Function of Nozzle geometry

f= Φ f Function of combustion stoich • Equivalence ratio, engine pressure ratio, nozzle drive process MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem

β= 40° M1=4.0

• Compute Speciﬁc Thrust for Φ=1, gaseous H2 fuel M B = 0.557853 P 0 B = 0.3126 • Nozzle Optimized for operation at ~10 km altitude P 0 ! (p∞ =pe= 26.43 kPa) …pburner ~ 1015.5 kPa p B = 38.422 P • Assume frozen chemistry at burner 0 ! MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) = 1253 kPa β= 40° M1=4.0

• Calculate exit stagnation pressure: M = 0.557853 $ 1.4 % B " # 1.4 1 1.4 ! 1 P0 $ ! 2% = 4008.5 kpa B P =26.4 "1 + 4 # = 0.3126 0 ! P0 2 $ 1.4 % ! $ " # % 1.4 ! 1 1.4 ! 1 p & $ 2% ' B P = 0.3126 &26.4 "1 + 4 # ' = 1253.0 kPa = 38.422 0 e " 2 # P 0 ! MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) From earlier CEA calculations

γe ~ 1.1814 ---> Φ=1, gaseous H2 fuel, pburner ~ 1000 kPa

) ! e "1 , 2 # P & ! e M = + 0 e " 1. e +% ( . ! e " 1 $ pe ' *+ -. • Calculate exit Me $ $ 1.1814 ! 1 % % 0.5 & 2 & $1253% 1.1814 ' ' & & " # ! 1' ' = 2.986 "1.1814 ! 1 " 26.4 # #

* Ae/A =7

MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d)

β= 40° M1=4.0

* • Assume Ae/A = 2.0

• Compute free stream stagnation temperature 1.4 ! 1 $ " # 1 2 ' 216.65 $1 + 42% =909.93°K T0 = T! 1+ M ! = " # ! %& 2 () 2

MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) • Exit stagnation temperature … From earlier CEA calculations

Tburner = 2662.82 °K • From inlet earlier problem

M B = 0.557853 P 0 B = 0.3126 P 0 ! p B = 38.422 T P0 0 burner = ! $ 1.1814 ! 1 2% 2662.82 "1 + 0.55785 # = 2738.2 °K 2 MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) • Calculate speciﬁc thrust ! $ ! $ F ! f + 1$ M 1 # M & # thrust & = ' R T e ( ' R T ) • # & e g e oe # ) g ) 0) & # & " f % ' e ( 1 2 f ' ) ( 1 2 m f # & " % opt 1+ M e # 1+ M ) & 2 " 2 %

1 f = Φ fstoich = (1) =0.0265 37.88393

MW = 24.6 --> Rg = 337.98 J/kg-degK

MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) • Calculate speciﬁc thrust ! $ ! $ F ! f + 1$ M 1 # M & # thrust & = ' R T e ( ' R T ) • # & e g e oe # ) g ) 0) & # & " f % ' e ( 1 2 f ' ) ( 1 2 m f # & " % opt 1+ M e # 1+ M ) & 2 " 2 %

#0.0265 + 1$ 0.5 (2.986) ! " (1.1814%337.98%2738.2) - 0.0265 1.1814 & 1 0.5 #1 + (2.986) 2$ ! 2 "

# 1 $ 0.5 4 ! " (1.4%287%909.93) = 45397 Nt/kg/sec 0.0265 1.4 & 1 0.5 #1 + (4) 2$ ! 2 " ! $ … convert to seconds 1 Fthrust 89928.2835 ! 44534.69 # • & = = 4628.4 sec g0 # m & 9.8077 " f % opt MAE 5540 - Propulsion Systems Revisit Earlier Ramjet problem (cont’d) • Calculate speciﬁc thrust • Idealized analysis … ! $ 1 F with a 2-D inlet # thrust & = 4628.4 sec • •If we consider losses of g0 # m & " f % opt 10% then we are right in the middle of the predicted operating range

MAE 5540 - Propulsion Systems Amount of Rocket propellant required to launch a 1000 kg payload into orbit

• Now one can see why air breathing options are so attractive

Isp > 4000! …. Off the chart

Titan IV

space shuttle Rockets on the Launch Pad are Mostly ALL FUEL!! Amount of Propellant required, kg "Venture-star"

MAE 5540 - Propulsion Systems Specific Impulse, sec. What is the thermodynamic efﬁciency

# ) "1 & # P & ) %T " 0B T ( ) "1 C % ( B % P0 ( ) % $ A ' ( # PA & $ ' ! = 1" % ( = $ PB ' (TC " TB )

! (1.4 ! 1) & & (1.4 ! 1) ' ' 1.4 $ $ 1.4 % % 38.422 $2662.82 ! $0.3126 % 856.61% " " # # 1 ! = 0.600 (2662.82 ! 856.61)

MAE 5540 - Propulsion Systems What is the thermodynamic efﬁciency (cont’d) • Unlike Rockets .. Ramjets … and air breathing propulsion systems in general tend to run more thermodynamically efﬁcient when the mixture ratio of leaner than stoich

• Combustion efﬁciency and stability limits are depending on several parameters : fuel, equivalence ratio, air stagnation pressure and temperature

• Airbreathers tend to run lean … that is why afterburners work … left over O2 MAE 5540 - Propulsion Systems What is the thermodynamic efﬁciency (cont’d)

• Additional fuel is introduced into the hot exhaust and burned using excess

O2 from main combustion

• The afterburner increases the temperature of the gas ahead of the nozzle Increases exit velocity

• The result of this increase in temperature is an increase of about 40 percent in thrust at takeoff and a much larger percentage at high speeds

.. But big loss in thrust to fuel consumption ratio

MAE 5540 - Propulsion Systems Thrust Speciﬁc Fuel Consumption

• m f 1 TSFC = ! Measure of fuel economy of Fthrust Ispg0 An airbreathing engine

• m f 1 TSFC = ! Fthrust Ispg0

Typical Turbojet ! TSFC = (2 " 4) lbm lbf "hr

MAE 5540 - Propulsion Systems Next: Supersonic Combustion RamJets (SCRAMjet)

MAE 5540 - Propulsion Systems Oblique Shock Waves:

Collected Algorithm (cont’d) • “Less Obvious” explicit solution

"1 2 ' 4$% + cos (&)* (M1 " 1) + 2# cos δ = 0 ---> Strong Shock () 3 +, tan(!) = . - " 1 2 1 δ = 1 ---> Weak Shock 3 1+ M1 tan(4) /0 2 23

2 2 $ # " 1 2 ' $ # + 1 2 ' 2 ! = (M1 " 1) " 3 1+ M1 1+ M1 tan (*) %& 2 () %& 2 ()

2 3 $ # " 1 2 ' $ # " 1 2 # + 1 4 ' 2 M1 " 1 " 9 1+ M1 1+ M1 + M1 tan (*) ( ) & 2 ) & 2 4 ) ! = % ( % ( + 3 MAE 5540 - Propulsion Systems Oblique Shock Waves:

Collected Algorithm (cont’d) • ... and the rest of the story …

2 !2 (" + 1)(M1 sin#) = 2 !1 2 + " $ 1 M sin# ( ( )( 1 ) ) p 2" 2 2 = 1+ M sin# $ 1 (( 1 ) ) p1 (" + 1) % 2 ( % 2 ( 2 + (" $ 1)(M1 sin#) T2 2" '( )* = '1+ (M1 sin#) $ 1 * 2 T 1 ( ) ' * 1 & (" + ) ) (" + 1) M1 sin# & ( ) )

MAE 5540 - Propulsion Systems Oblique Shock Waves:

Collected Algorithm (concluded) • ... and the rest of the story …

$ (! " 1) 2 ' 1+ (M1 sin#) %& 2 () Mn2 Mn2 = M 2 = $ 2 (! " 1)' sin ! " # ! (M1 sin#) " ( ) %& 2 ()

$ ! ' 2 & ) $ *(! + 1) - ' % ! #1( & (M1 sin") ) P 2 , 2 / 02 & + . ) = 1 & ! # 1 2 ) P01 ! #1 $ ' $ 2 (! # 1)' & &1+ (M1 sin") ) ) (! + 1) ! (M1 sin") # % 2 ( %& 2 () % (

MAE 5540 - Propulsion Systems