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AME 514 Applications of Combustion Lecture 10: Hypersonic Propulsion I: Motivation, performance parameters.

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Presentation on theme: "AME 514 Applications of Combustion Lecture 10: Hypersonic Propulsion I: Motivation, performance parameters."— Presentation transcript:

1 AME 514 Applications of Combustion Lecture 10: Hypersonic Propulsion I: Motivation, performance parameters

2 2 AME 514 - Spring 2015 - Lecture 10 Advanced propulsion systems (3 lectures)  Hypersonic propulsion background (Lecture 1)  Why hypersonic propulsion?  What's different at hypersonic conditions?  Real gas effects (non-constant C P, 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)  Ramjet/scramjets  Pulse detonation engines

3 3 AME 514 - Spring 2015 - Lecture 10Lmg T D Why use air even if you're going to space?  Carry only fuel, not fuel + O 2, while in atmosphere  8x mass savings (H 2 -O 2 ), 4x (hydrocarbons)  Actually even more than this when the ln( ) term in the Breguet range equation is considered  Use aerodynamic lifting body rather than ballistic trajectory  Ballistic: need Thrust/weight > 1  Lifting body, steady flight: Lift (L) = weight (mg); Thrust (T) = Drag (D), Thrust/weight = L/D > 1 for any decent airfoil, even at hypersonic conditions

4 4 AME 514 - Spring 2015 - Lecture 10 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 = static temperature - T measured by thermometer moving with flow  T t = temperature of the gas if it is decelerated adiabatically to M = 0   = gas specific heat ratio = C p /C v ; M = Mach number = u/(  RT) 1/2  Stagnation pressure - measure of usefulness (ability to expand flow) is large even before heat addition - structural problems  P = static pressure - P measured by pressure gauge moving with flow  P t = pressure of the gas if it is decelerated reversibly and adiabatically to M = 0  Large P t means no mechanical compressor needed at large M

5 5 AME 514 - Spring 2015 - Lecture 10 What's different about hypersonic propulsion?  Why are T t and P t so important? Isentropic expansion until exit pressure (P 9 ) = ambient pressure (P 1 ) (optimal exit pressure yielding maximum thrust) yields  … but it's difficult to add heat at high M without major loss of stagnation pressure

6 6 AME 514 - Spring 2015 - Lecture 10 What's different about hypersonic propulsion?  High temperatures:  and molecular mass not constant - dissociation - use GASEQ (http://www.gaseq.co.uk) to compute stagnation conditionshttp://www.gaseq.co.uk  Example calculation: standard atmosphere at 100,000 ft  T 1 = 227K, P 1 = 0.0108 atm, c 1 = 302.7 m/s, h 1 = 70.79 kJ/kg (atmospheric data from http://www.digitaldutch.com/atmoscalc/)http://www.digitaldutch.com/atmoscalc/  Pick P 2 > P 1, compress isentropically, note new T 2 and h 2  1st Law: h 1 + u 1 2 /2 = h 2 + u 2 2 /2; since u 2 = 0, h 2 = h 1 + (M 1 c 1 ) 2 /2 or M 1 = [2(h 2 -h 1 )/c 1 2 ] 1/2  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  Problems  Ionization not considered  Stagnation temperature relation valid even if shocks, friction, etc. (only depends on 1st law) but stagnation pressure assumes isentropic flow  Calculation assumed adiabatic deceleration - radiative loss (from surfaces and ions in gas) may be important

7 7 AME 514 - Spring 2015 - Lecture 10 What's different about hypersonic propulsion? WOW! HOT WARM COLD 5000K 3000K 1000K 200K N+O+e - N 2 +O N 2 +O 2 N 2 +O 2

8 8 AME 514 - Spring 2015 - Lecture 10 Thrust computation  In airbreathing and rocket propulsion we need THRUST (force acting on vehicle)  How much push can we get from a given amount of fuel?  We'll start by showing that thrust depends primarily on the difference between the engine inlet and exhaust gas velocity, then compute exhaust velocity for various types of flows (isentropic, with heat addition, with friction, etc.)

9 9 AME 514 - Spring 2015 - Lecture 10  Control volume for thrust computation - in frame of reference moving with the engine Thrust computation

10 10 AME 514 - Spring 2015 - Lecture 10  Newton's 2nd law: Force = rate of change of momentum  At takeoff u 1 = 0; for rocket no inlet so u 1 = 0 always  For hydrogen or hydrocarbon-air FAR << 1; typically 0.06 at stoichiometric Thrust computation - steady flight

11 11 AME 514 - Spring 2015 - Lecture 10  But how to compute exit velocity (u 9 ) and exit pressure (P 9 ) as a function of ambient pressure (P 1 ), flight velocity (u 1 )? Need compressible flow analysis, next lecture …  And you can obtain a given thrust with small [(1+FAR)u 9 - u 1 ] and large large (P 9 – P 1 )A 9 or vice versa - which is better, i.e. for given, u 1, P 1 and FAR, what P 9 will give most thrust? Differentiate thrust equation and set = 0  Momentum balance on exit (see next slide)  Combine  Optimal performance occurs for exit pressure = ambient pressure Thrust computation

12 12 AME 514 - Spring 2015 - Lecture 10 1D momentum balance - constant-area duct Coefficient of friction (C f )

13 13 AME 514 - Spring 2015 - Lecture 10  But wait - this just says P 9 = P 1 is an extremum - is it a minimum or a maximum? but P e = P a at the extreme cases so  Maximum thrust if d 2 (Thrust)/d(P 9 ) 2 < 0  dA 9 /dP 9 < 0 - we will show this is true for supersonic exit conditions  Minimum thrust if d 2 (Thrust)/d(P 9 ) 2 > 0  dA 9 /dP 9 > 0 - we will show this is would be true for subsonic exit conditions, but for subsonic, P 9 = P 1 always since acoustic (pressure) waves can travel up the nozzle, equalizing the pressure to P 9, so it's a moot point for subsonic exit velocities Thrust computation

14 14 AME 514 - Spring 2015 - Lecture 10 Propulsive, thermal, overall efficiency  Thermal efficiency (  th )  Propulsive efficiency (  p )  Overall efficiency (  o ) this is the most important efficiency in determining aircraft performance (see Breguet range equation, coming up…)

15 15 AME 514 - Spring 2015 - Lecture 10 Propulsive, thermal, overall efficiency  Note on propulsive efficiency for FAR << 1   p  1 as u 1 /u 9  1  u 9 is only slightly larger than u 1  But then you need large to get required Thrust ~ (u 9 - u 1 ); but this is how commercial turbofan engines work!  In other words, the best propulsion system accelerates an infinite mass of air by an infinitesimal  u  Fundamentally this is because Thrust ~ (u 9 - u 1 ), but energy required to get that thrust ~ (u 9 2 - u 1 2 )/2  For hypersonic propulsion systems, u 1 is large, u 9 - u 1 << u 1, so propulsive efficiency usually high (i.e. close to 1)

16 16  Specific thrust – thrust per unit mass flow rate, non- dimensionalized by sound speed at ambient conditions (c 1 ) AME 514 - Spring 2015 - Lecture 10 Specific Thrust For any 1D steady propulsion system if working fluid is an ideal gas with constant C P,  For any 1D steady propulsion system

17 17 AME 514 - Spring 2015 - Lecture 10 Specific Thrust  Specific thrust (ST) continued… if P 9 = P 1 and FAR << 1 then  Thrust Specific Fuel Consumption (TSFC) (PDR's definition)  Usual definition of TSFC is just, but this is not dimensionless; use Q R to convert to heat input, use c 1 to convert the denominator to a quantity with units of power  Specific impulse (I sp ) = thrust per weight (on earth) flow rate of fuel (+ oxidant if two reactants, e.g. rocket) (units of seconds)

18 18 AME 514 - Spring 2015 - Lecture 10 Breguet range equation  Consider aircraft in level flight (Lift = weight) at constant flight velocity u 1 (thrust = drag)  Combine expressions for lift & drag and integrate from time t = 0 to t = R/u 1 (R = range = distance traveled), i.e. time required to reach destination, to obtain Breguet Range Equation Lift (L) Thrust Drag (D) Weight (W = m vehicle g)

19 19 AME 514 - Spring 2015 - Lecture 10 Rocket equation  If acceleration (  u) rather than range in steady flight is desired [neglecting drag (D) and gravitational pull (W)], Force = mass x acceleration or Thrust = m vehicle du/dt  Since flight velocity u 1 is not constant, overall efficiency is not an appropriate performance parameter; instead use specific impulse (I sp ) = thrust per unit weight (on earth) flow rate of fuel (+ oxidant if two reactants carried), i.e. Thrust = mdot fuel *g earth *I sp  Integrate to obtain Rocket Equation  Of course gravity and atmospheric drag will increase effective  u requirement beyond that required strictly by orbital mechanics

20 20 AME 514 - Spring 2015 - Lecture 10 Breguet & rocket equations - comments  Range (R) for aircraft depends on   o (propulsion system) - depends on u 1 for airbreathing propulsion  Q R (fuel)  L/D (lift to drag ratio of airframe)  g (gravity)  Fuel consumption (m initial /m final ); m initial - m final = fuel mass used (or fuel + oxidizer, if not airbreathing)  This range does not consider fuel needed for taxi, takeoff, climb, decent, landing, fuel reserve, etc.  Note (irritating) ln( ) or exp( ) term in both Breguet and Rocket: because you have to use more fuel at the beginning of the flight, since you're carrying fuel you won't use until the end of the flight - if not for this it would be easy to fly around the world without refueling and the Chinese would have sent skyrockets into orbit thousands of years ago!

21 21 AME 514 - Spring 2015 - Lecture 10 Breguet & rocket equations - examples  Fly around the world (g = 9.8 m/s 2 ) without refueling  R = 40,000 km  Use hydrocarbon fuel (Q R = 4.5 x 10 7 J/kg),  Good propulsion system (  o = 0.25)  Good airframe (L/D = 20),  Need m initial /m final ≈ 5.7 - aircraft has to be mostly fuel - m fuel /m initial = (m initial - m final )/m initial = 1 - m final /m initial = 1 - 1/5.7 = 0.825! - that's why no one flew around with world without refueling until 1986  To get into orbit from the earth's surface   u = 8000 m/s  Use a good rocket propulsion system (e.g. Space Shuttle main engines, I SP ≈ 400 sec)  Need m initial /m final ≈ 7.7 can't get this good a mass ratio in a single vehicle - need staging – that's why no one put an object into earth orbit until 1957


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