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AME 436 Energy and Propulsion Lecture 11 Propulsion 1: Thrust and aircraft range Outline Why gas turbines? Computation of thrust Propulsive, thermal and overall efficiency Specific thrust, thrust specific fuel consumption, specific impulse Breguet range equation AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 2 Why gas turbines? GE CT7-8 turboshaft (used in helicopters) Cummins QSK60-2850 4-stroke 60.0 http://www.geaviation.com/engines/co liter (3,672 in3) V-16 2-stage mmercial/ct7/ct7-8.html turbocharged diesel (used in mining Compressor/turbine stages: 6/4 trucks) http://cumminsengines.com/assets/p Diameter 26”, Length 48.8” = 426 liters = 5.9 hp/liter df/4087056.pdf Dry Weight 537 lb, max. power 2,520 hp 2.93 m long x 1.58 m wide x 2.31 m (power/wt = 4.7 hp/lb) high = 10,700 liters = 0.27 hp/liter Pressure ratio at max. power: 21 (ratio Dry weight 21,207 lb, 2850 hp at 1900 per stage = 211/6 = 1.66) RPM (power/wt = 0.134 hp/lb = 35x Specific fuel consumption at max. lower than gas turbine) BMEP = 22.1 atm power: 0.450 (units not given; if lb/hp- hr then corresponds to 29.3% Volume compression ratio ??? (not efficiency) given) AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 3 Why gas turbines? NuCellSys HY-80 “Fuel cell engine” http://www.nucellsys.com/dyn/mediaout. Lycoming IO-720 11.8 liter (720 cu in) 4- dhtml/234f73a2c579fb27158i/mime/PDF/H stroke 8-cyl. gasoline engine Y-80-PDF/HY-80_2009.pdf http://www.lycoming.com/engines/series/ Volume 220 liters = 0.41 hp/liter pdfs/Specialty%20insert.pdf 91 hp, 485 lb. (power/wt = 0.19 hp/lb) Total volume 23” x 34” x 46” = 589 liters = 41% efficiency (fuel to electricity) at max. power; up to 58% at lower power 0.67 hp/liter Uses hydrogen only - NOT hydrocarbons 400 hp @ 2650 RPM Does NOT include electric drive system Dry weight 600 lb. (power/wt = 0.67 hp/lb (≈ 0.40 hp/lb) at ≈ 90% electrical to = 7x lower than gas turbine) mechanical efficiency BMEP = 11.3 atm (4 stroke) (http://www.gm.com/company/gmability/a Volume compression ratio 8.7:1 (= dv_tech/images/fact_sheets/hywire.html) pressure ratio 20.7 if isentropic) (no longer valid) Fuel cell + motor overall 0.13 hp/lb at 37% efficiency, not including H2 storage AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 4 Why gas turbines? Simple intro to gas turbines: http://www.geaviation.com/education/engines101/ Why does gas turbine have much higher power/weight & power/volume than recips? More air can be processed since steady flow, not start/stop of reciprocating-piston engines More air more fuel can be burned More fuel more heat release More heat more work (if thermal efficiency similar) AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 5 Why gas turbines? What are the disadvantages? Compressor is a dynamic device that makes gas move from low pressure to high pressure without a positive seal like a piston/cylinder » Requires very precise aerodynamics » Requires blade speeds ≈ sound speed, otherwise gas flows back to low P faster than compressor can push it to high P » Each stage can provide only 2:1 or 3:1 pressure ratio - need many stages for large pressure ratio Since steady flow, each component sees a constant temperature - at end of combustor - turbine stays hot continuously and must rotate at high speeds (high stress) » Severe materials and cooling engineering required (unlike reciprocating engine where components only feel average gas temperature during cycle) » Turbine inlet temperature limit typically 1300˚C - limits fuel input As a result, turbines require more maintenance & are more expensive for same power AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 6 Thrust computation In gas turbine 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.) AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 7 Thrust computation Control volume for thrust computation - in frame of reference moving with the engine AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 8 Thrust computation - steady flight Newton’s 2nd law: Force = rate of change of momentum d(mu) å Forces = å dt å Forces = Thrust + P A 1 CV - [ P1 (ACV - A9 ) + P9 A9 ] = T + (P1 - P9 )A9 d(mu) dm du å dt = å( dt u + m ) = å ( mu + 0) (if steady, du/dt = 0) dt ˙ å mu = (m f + ma )u9 - (ma )u1 (u fuel = 0 in moving reference frame) ˙ ˙ ˙ ˙ ( ) Combine : Thrust = ma + m f u9 - ma u1 + ( P9 - P1 ) A9 ˙ ˙ ˙ Þ Thrust = ma [(1+ FAR) u9 - u1 ] + ( P9 - P1) A9; ˙ FAR = m f / ma = Fuel to air mass ratio = f /(1 - f ) ( f = fuel mass fraction) ˙ ˙ At takeoff u1 = 0; for rocket no inlet so u1 = 0 always For hydrocarbon-air usually FAR << 1; typically 0.06 at stoichiometric, but in practice maximum allowable FAR ≈ 0.03 due to turbine inlet temperature limitations (discussed later…) AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 9 Thrust computation But how to compute exit velocity (u9) and exit pressure (P9) as a function of ambient pressure (P1), flight velocity (u1)? Need compressible flow analysis, coming next… Also - one can obtain a given thrust with large (P9 – P1)A9 and a ˙ small ma[(1+FAR)u9 - u1] or vice versa - which is better, i.e. for given ˙ u1, P1, maand FAR, what P9 will give most thrust? Differentiate thrust equation and set = 0 d(Thrust) é d(u9 ) ù d(A9 ) = ma ê(1+ FAR) - 0ú + (1- 0) A9 + (P9 - P ) 1 =0 d(P9 ) ë d(P9 ) û d(P9 ) Momentum balance on exit (see next slide) d(u9 ) AdP+mdu = 0 Þ A9 + ma (1+ FAR) =0 d(P9 ) Combine d(Thrust) d(A9 ) = (P9 - P ) 1 = 0 Þ P9 = P1 d(P9 ) d(P9 ) Optimal performance occurs for exit pressure = ambient pressure AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 10 1D momentum balance - constant-area duct Coefficient of friction (Cf) Wall drag force Cf º 1 ru 2 × (Wall area) 2 d(mu) å Forces = å dt å Forces = PA - (P + dP)A - C f (1/2ru2 )(Cdx) d(mu) å dt = å mu = mu - m(u + du) ˙ ˙ ˙ Combine : AdP + mdu + C f (1/2 ru 2 )Cdx = 0 ˙ AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 11 Thrust computation But wait - this just says P9 = P1 is an extremum - is it a minimum or maximum? d(Thrust) d(A9 ) d 2 (Thrust) d 2 (A9 ) d(A9 ) = (P9 - P ) 1 Þ = (P9 - P ) 1 + (1) d(P9 ) d(P9 ) d(P9 ) 2 d(P9 ) 2 d(P9 ) but P9 = P1 at the extremum cases so d 2 (Thrust) d(A9 ) = d(P9 ) 2 d(P9 ) Maximum thrust if d2(Thrust)/d(P9)2 < 0 dA9/dP9 < 0 - we will show this is true for supersonic exit conditions Minimum thrust if d2(Thrust)/d(P9)2 > 0 dA9/dP9 > 0 - we will show this is would be true for subsonic exit conditions, but for subsonic, P9 = P1 always since acoustic (pressure) waves can travel up the nozzle, equalizing the pressure to P9, so it’s a moot point for subsonic exit velocities AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 12 Thrust computation Turbofan: same as turbojet except that there are two streams, one hot (combusting) and one cold (non- combusting, fan only, use prime (‘) superscript): ë( ) Thrust = ma é 1+ FAR u9 - u1 ù + P9 - P1 A9 + ma éu9 - u1 ù û ( ) ' ë ' û Note (1 + FAR) term applies only to combusting stream Note we assumed P9 = P1 for fan stream; for any reasonable fan design, u9’ will be subsonic so this will be true AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 13 Propulsive, thermal, overall efficiency Thermal efficiency (th) D(Kinetic energy) (ma + m f )u9 / 2 - (ma )u1 / 2 2 2 hth º = Heat input m f QR (u9 - u12 ) / 2 2 If m f << ma (FAR << 1) then hth » FAR × QR Propulsive efficiency (p) Thrust power Thrust × u1 hp º = D(Kinetic energy) (ma + m f )u9 / 2 - (ma )u12 / 2 2 ma (u9 - u1 )× u1 2u1 / u9 If m f << ma (FAR << 1) and P9 =P then h p » = ma (u9 - u12 ) / 2 1+ u1 / u9 1 2 Overall efficiency (o) this is the most important efficiency in determining aircraft performance (see Breguet range equation, coming up…) AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 14 Propulsive, thermal, overall efficiency Note on propulsive efficiency 2u1 / u9 hp » 1+ u1 / u9 p 1 as u1/u9 1 ue is only slightly larger than u1 ˙ But then you need large mass flow rate ( ma) to get the ˙ required Thrust ~ ma(u9 - u1) - 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 ~ (u9 - u1), but energy required to get that thrust ~ (u92 - u12)/2 This issue will come up a lot in the next few weeks! AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 15 Ideal turbojet cycle - notes on thrust Specific thrust – thrust per unit mass flow rate, non- dimensionalized by sound speed at ambient conditions (c1) Specific Thrust (ST) º Thrust / ma c1 Thrust = ma [(1+ FAR)u9 - u1 ]+ (P9 - P1 )A9 For any 1D steady propulsion system Thrust u u (P - P )A u c (P - P )A ST º = (1+ FAR) 9 - 1 + 9 1 9 = (1+ FAR) 9 9 - M1 + 9 1 9 ma c1 c1 c1 ma c1 c9 c1 ( r 9 u9 A9 )c1 g RT9 (P9 - P9 ) = (1+ FAR)M 9 - M1 + g RT1 (P9 / RT9 )u9 c1 T9 æ P1 ö RT9 = (1+ FAR)M 9 - M1 + ç1- ÷ T1 è P9 ø c9 M 9 c1 T9 æ P1 ö RT9 = (1+ FAR)M 9 - M1 + ç1- ÷ T1 è P9 ø g RT9 M 9 g RT1 T9 æ P1 ö T9 1 For any 1D steady propulsion = (1+ FAR)M 9 - M1 + ç1- ÷ system if working fluid is an T1 è P9 ø T1 g M 9 ideal gas with constant CP, AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 16 Other performance parameters Specific thrust (ST) continued… if Pe = Pa and FAR << 1 then Thrust T ST º = M 9 9 - M1 (if FAR <<1, P = P9 ) 1 ma c1 T1 Thrust Specific Fuel Consumption (TSFC) (PDR’s definition) m f QR æ ma c1 ö FAR ×QR FAR × QR TSFC º =ç ÷ = Thrust c1 è Thrust ø c1 2 ST × c12 mf QR u1 M1 Note TSFC= = Thrust u1 c1 ho ˙ Usual definition of TSFC is just m f /Thrust, but this is not ˙ dimensionless; use QR to convert m f to heat input, one can use either u1 or c1 to convert the denominator to a quantity with units of power, but using u1 will make TSFC blow up at u1 = 0 Specific impulse (Isp) = thrust per weight (on earth) flow rate of fuel (+ oxidant if two reactants carried, e.g. rocket) (units of seconds) Thrust (Thrust)u1 QR hoQR QR I sp = ; I sp = = = m fuel gearth m fuelQR gearth c1M1 M1c1gearth (TSFC)c1gearth AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 17 Breguet range equation Consider aircraft in level flight (Lift = weight) at constant flight Lift (L) velocity u1 (thrust = drag) L = mvehicle g; Thrust Drag (D) ho m fuel QR ˙ D = Thrust = u1 Weight (W = mvehicleg) hoQR dm fuel hoQR -dmvehicle = = u1 dt u1 dt Combine expressions for lift & drag and integrate from time t = 0 to t = R/u1 (R = range = distance traveled), i.e. time required to reach destination, to obtain Breguet Range Equation R / u1 final D u1g dmvehicle D u1g dmvehicle L hoQR dt = - mvehicle Þ ò L hoQR dt = - ò 0 initial m vehicle D u1g R m final L hoQR minitial Þ = -ln Þ R= ln L hoQR u1 minitial D g m final AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 18 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 = mvehicledu/dt Since flight velocity u1 is not constant, overall efficiency is not an appropriate performance parameter; instead use Isp, leading to Rocket equation: dm fuel dm Thrust = m fuel gearth I sp = gearth I sp = -gearth I sp vehicle dt dt du dm du Thrust = mvehicle Þ -gearth I sp vehicle = mvehicle dt dt dt dmvehicle æ m final ö Þ du = -gearth I sp Þ Du º u final - uinitial = -gearth I sp ln ç ÷ mvehicle è minitial ø æm ö çm ÷ Þ Du = gearth I sp ln ç initial ÷ è final ø Of course gravity and atmospheric drag will increase effective u requirement beyond that required strictly by orbital mechanics AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 19 Brequet range equation - comments Range (R) for aircraft depends on o (propulsion system) - dependd on u1 for airbreathing propulsion QR (fuel) L/D (lift to drag ratio of airframe) g (gravity) Fuel consumption (minitial/mfinal); minitial - mfinal = 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! AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 20 Examples What initial to final mass ratio is needed to fly around the world without refueling? Assume distance traveled (R) = 40,000 km, g = 9.8 m/s2; hydrocarbon fuel (QR = 4.3 x 107 J/kg); good propulsion system (o = 0.25), good airframe (L/D = 25), æ ö æ ( 40 ´ 10 6 m)(9.81m /s2 ) ö minitial R× g ÷ = expç = expç ÷ ç (0.25) 4.3 ´ 10 7 J /kg (25) ÷ = 4.31 m final çh Q L ÷ è o R Dø è ( ) ø So the aircraft has to be mostly fuel, i.e. mfuel/minitial = (minitial - mfinal)/minitial = 1 - mfinal/minitial = 1 - 1/4.31 = 0.768! – that’s why no one flew around with world without refueling until 1986 (solo flight 2005) What initial to final mass ratio is needed to get into orbit from the earth’s surface with a single stage rocket propulsion system? For this mission u = 8000 m/s; using a good rocket propulsion system (e.g. Space Shuttle main engines, ISP ≈ 400 sec It’s practically impossible to obtain this large a mass ratio in a single vehicle, thus staging is needed – that’s why no one put an object into earth orbit until 1957 AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 21 Summary Steady flow (e.g. gas turbine) engines have much higher power/weight ratio than unsteady flow (e.g. reciprocating piston) engines When used for thrust, a simple momentum balance on a steady-flow engine shows that the best performance is obtained when Exit pressure = ambient pressure A large mass of gas is accelerated by a small u Two types of efficiencies for propulsion systems - thermal efficiency and propulsive efficiency (product of the two = overall efficiency) Definitions - specific thrust, thrust specific fuel consumption, specific impulse Range of an aircraft depends critically on overall efficiency - effect more severe than in ground vehicles, because aircraft must generate enough lift (thus thrust, thus required fuel flow) to carry entire fuel load at first part of flight AME 436 - Spring 2012 - Lecture 11 - Thrust and Aircraft Range 22

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