AME 436. Energy and Propulsion. Lecture 11 Propulsion 1: Thrust and aircraft range

Transcription

1 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 2 1

2 Why gas turbines?! GE CT7-8 turboshaft (used in helicopters! Cummins QSK stroke 60.0 liter (3,672 in3 V-16 2-stage! turbocharged diesel (used in mining engines/ct7-engine trucks! Compressor/turbine stages: 6/4! Diameter 26, Length 48.8 = 426 liters = download.aspx?brochureid=14 5. hp/liter! 2.3 m long x 1.58 m wide x 2.31 m! Dry Weight 537 lb, max. power 2,634 hp high = 10,700 liters = 0.27 hp/liter (power/wt = 4.7 hp/lb! Dry weight 21,207 lb, 2850 hp at 100 RPM (power/wt = hp/lb = 35x! Pressure ratio at max. power: 21 (ratio per lower than gas turbine stage = 211/6 = 1.66! Specific fuel consumption at max. power:! BMEP = 22.1 atm! Volume compression ratio??? (not (units not given; if lb/hp-hr then given corresponds to 2.3 efficiency 3 Why gas turbines?! Pratt & Whitney R liter (2800 in3 18-cyl. 4-stroke supercharged gasoline engine! Total volume 53 dia x 81 length = 227 liters = 0.72 hp/liter! RPM! Dry weight 2360 lb. (power/wt = 0.8 hp/lb = 5.3x lower than gas turbine! BMEP = 14. atm! Volume compression ratio 6.8:1 (= pressure ratio 14.6 if isentropic! NuCellSys HY-80 Fuel cell engine (specs no longer on-line! Volume 220 liters = 0.41 hp/liter! 1 hp, 485 lb. (power/wt = 0.1 hp/lb! 41 efficiency (fuel to electricity at max. power; up to 58 at lower power! Uses hydrogen - NOT hydrocarbons! Does NOT include electric drive system ( 0.40 hp/lb at 0 electrical to mechanical efficiency! Fuel cell + motor overall 0.13 hp/lb at 37 efficiency, not including H2 storage 4 2

3 Why gas turbines?! Why do gas turbines have much higher power/weight & power/ volume than reciprocating engines? More air can be processed since steady flow, not start/stop! More air more fuel can be burned! More fuel more heat release! More heat more work (if thermal efficiency similar 5 Why gas turbines?! Disadvantages! Compressor is a dynamic device that pushes gas from low pressure (P to high P without positive sealing like 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 - turbine stays hot continuously and must rotate at high speeds (high stress» Severe materials and cooling engineering required (unlike reciprocating engine where components feel only average gas temperature during cycle» Turbine inlet temperature limit typically 1400 C - limits fuel input! As a result, turbines require more maintenance & are more expensive for same power 6 3

4 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?! Thrust depends primarily on the difference between inlet and exhaust velocity! Goal of propulsion analysis is to compute exhaust velocity for a given thermodynamic cycle 7 Thrust computation! Control volume for thrust computation - in frame of reference moving with the engine Incoming air Area A CV Thrust Velocity u e (Exit velocity Area A (Exit area Pressure P (Exit pressure Everywhere on this surface, located far upstream of the engine: Pressure P 1 (ambient pressure, velocity (flight velocity m a m f Weird control volume m a + m f Everywhere else on this surface: Pressure P 1 (ambient pressure, velocity (flight velocity 8 4

5 Thrust computation - steady flight! Newtons 2nd law: Force = rate of change of momentum d(mu Forces = Forces = Thrust + P 1 A CV P 1 (A CV A + P A = T + (P P A 1 d(mu = ( dm u + m du = (!mu + 0 (if steady, du = 0!mu = (!m f +!m a u (!m a (u fuel = 0 in moving reference frame Combine: Thrust = (!m a +!m f u!m a + ( P P 1 A Thrust =!m a ( 1+ FARu u 1 + ( P P 1 A ; FAR =!m f /!m a = Fuel to air mass ratio = f / (1 f (f = fuel mass fraction! At takeoff = 0; for rocket no inlet so = 0 always! For hydrocarbon-air combustion FAR << 1; typically 0.06 at stoichiometric, but maximum allowable FAR 0.03 due to turbine inlet temperature limitations (discussed later Thrust computation! But how to compute exit velocity (u and exit pressure (P as a function of ambient pressure (P 1, flight velocity (? Need compressible flow analysis, next lecture! Also - one can obtain a given thrust with large (P P 1 A and a small m a [(1+FARu - ] or vice versa - which is better, i.e. for given, P 1, m a and FAR, what P will give most thrust? Differentiate thrust equation and set = 0 d(thrust " = m a ( 1+ FAR d(u 0 $ + ( 1 0 A + (P P 1 d(a # & = 0! Momentum balance at exit (see next slide AdP + mdu = 0 A + m a (1+ FAR d(u d(p! Combine = 0 d(thrust = (P P 1 d(a = 0 P = P 1 Optimal performance occurs for exit pressure = ambient pressure Valid for any 1-D steady cycle (ideal or not, any material 10 5

6 1D momentum balance - constant-area duct Coefficient of friction (C f C f Wall drag force 1 2 ρu2 (Wall area Velocity u pressure P density ρ duct area A mass flux = ρua d(mu Forces = 1 Forces = PA (P + dpa C f 2 ρu2 (Cdx d(mu =!mu =!mu!m(u + du 1 Combine: AdP+!mdu +C f 2 ρu2 Cdx = 0 Velocity u + du pressure P + dp density ρ duct area A mass flux = ρua Thickness dx, circumference C, wall area Cdx 11 Thrust computation! Wait this says P = P 1 is an extremum but is it a minimum or maximum? d(thrust = (P P 1 d(a d 2 (Thrust 2 = (P P 1 d 2 (A 2 + d(a (1 but P = P 1 at the extremum cases so d 2 (Thrust = d(a 2! Maximum thrust if d 2 (Thrust/ 2 < 0 da /dp < 0 - we will show this is true for supersonic exit conditions! Minimum thrust if d 2 (Thrust/ 2 > 0 da /dp > 0 - we will show this is true for subsonic exit conditions, but for subsonic, P = P 1 always since acoustic (pressure waves can travel up the nozzle, equalizing the pressure to P, so its a moot point 12 6

7 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: Thrust =!m a ( 1+ FARu u 1 + P P ( 1 A +!m a u u 1! Note (1 + FAR term applies only to combusting stream! Note we assumed P = P 1 for fan stream; for any sane fan design u will be subsonic so P = P 1 will be true 13 Propulsive, thermal, overall efficiency! Thermal efficiency (η th η th Δ(Kinetic energy Heat input = ( m a + m f u 2 / 2 ( m a 2 / 2 m f If m f << m a (FAR <<1 then η th (u 2 u 2 1 / 2 FAR! Propulsive efficiency (η p η p Thrust power Δ(Kinetic energy = Thrust ( m a + m f u 2 / 2 ( m a 2 / 2 If m f << m a (FAR <<1 and P =P 1 then η p m (u u u a 1 1 m a (u 2 u 2 1 / 2 = 2u / 1+ / u! Overall efficiency (η o Thrust power Thrust power Δ(Kinetic energy η o = = η th η p Heat input Δ(Kinetic energy Heat input this is the most important efficiency in determining aircraft performance (see Breguet range equation, coming up 14 7

8 Propulsive, thermal, overall efficiency! Note on propulsive efficiency η p = m (u u u a 1 1 m a (u 2 u 2 1 / 2 = 2(u "u # 1 + (u $ " # 2 u1 $ = 1 ;Δu u 2 1+ Δu / 2u 1! η p 1 as Δu 0 u is only slightly larger than! But then you need large mass flow rate ( m a to get the required Thrust ~ m a Δu - but this is how turbofan engines work!! In other words, the best propulsion system accelerates an infinite mass of air by an infinitesimal Δu! Fundamentally this is because Thrust ~ Δu = u, but energy required to get that thrust ~ (u 2-2 /2! This issue will come up a lot in the next few weeks! 15 Ideal turbojet cycle - notes on thrust! Specific thrust thrust per unit mass flow rate, nondimensionalized by sound speed at ambient conditions ( Specific Thrust (ST Thrust!m a Thrust =!m a [(1+ FARu ]+ (P P 1 A ST Thrust!m a = (1+ FAR u + (P P 1 A!m a = (1+ FAR u c c M 1 + (P P 1 A ρ u A 1+ FAR P ( 1+ FAR For any 1D steady propulsion system γ RT = (1+ FARM M 1 + (P P ; M u (Mach number γ RT (P 1 / RT u c T = (1+ FARM M T P RT 1 ( 1+ FAR ; c = γ RT for ideal gas 1 γ RT M γ RT 1 T = (1+ FARM M T P T 1 1+ FAR For any 1D steady propulsion 1 P system if working fluid is an T 1 γm ideal gas with constant C P, γ 16 8

9 Other performance parameters! Specific thrust (ST continued if P = P 1 and FAR << 1 then ST Thrust T = M M 1 ( if FAR <<1, P 1 = P m a T 1! Thrust Specific Fuel Consumption (TSFC (PDRs definition m TSFC f " m = a $ FAR = FAR 2 2 Thrust # Thrust & ST m Note TSFC= f = M 1 Thrust η o! Usual definition of TSFC is just m f /Thrust, but this is not dimensionless; use to convert m f to heat input, one can use either or to convert the denominator to a quantity with units of power, but using would make TSFC blow up at = 0! Specific impulse (I sp = thrust per weight (on earth flow rate of fuel (plus oxidant if two reactants carried, e.g. rocket (units of seconds I sp = Thrust ;I sp = (Thrust = η Q o R Q = R m fuel m fuel M 1 M 1 (TSFC 17 Breguet range equation! Consider aircraft in level flight (Lift = Weight at constant flight velocity (Thrust = Drag L = m vehicle g; Thrust D = Thrust = η m o fuel Lift (L Weight (W = m vehicle g Drag (D = η o R / D g = dm vehicle D L η o m vehicle L D g L η o dm fuel = η o dm vehicle! Combine expressions for lift & drag and integrate from time t = 0 to t = R/ (R = range = distance traveled, i.e. travel time, to obtain Breguet Range Equation 0 g = η o initial dm vehicle m vehicle final R = ln m final R = L η o m initial D g ln m initial m final 18

10 Rocket equation! If acceleration (Δu rather than range in steady flight is desired [neglecting drag (D and gravitational pull (W], Force = mass x acceleration or Thrust = m vehicle du/! Since flight velocity is not constant, overall efficiency is not useful; instead use I sp, leading to Rocket Equation: dm Thrust = m fuel I sp = I fuel sp du Thrust = m vehicle g I earth sp dm du = I vehicle sp m vehicle Δu = I sp ln m initial & m final dm vehicle = I sp dm vehicle = m vehicle du Δu u final u initial = I sp ln m ( final * & ( * m initial! Gravity and aerodynamic drag will increase effective Δu requirement above that required by orbital mechanics alone 1 Brequet range equation - comments! Range (R for aircraft depends on! η o (propulsion system! (fuel! L/D (lift to drag ratio of airframe! g (gravity! Fuel consumption (m initial /m final ; m initial - m final = fuel mass burned (or fuel + oxidizer, if not airbreathing! R does not consider extra fuel mass required for taxi, takeoff, climb, decent, landing, fuel reserve, etc.! Note (irritating ln( or exp( term in both Breguet and Rocket $ m initial = exp R g m final & η o L (Breguet; m initial Δu = exp D ( m & final I (rocket sp ( that occurs because more thrust is required at the beginning of the flight to carry fuel that won t be used until the end of the flight - if not for ln( term it would be easy to fly around the world without refueling and the Chinese would have sent skyrockets into orbit thousands of years ago! $ 20 10

11 Examples! What initial to final mass ratio is needed to fly around the world without refueling? Assume distance traveled (R = 40,000 km, g =.8 m/s 2 ; hydrocarbon fuel ( = 4.3 x 10 7 J/kg; good propulsion system (η o = 0.25, good airframe (L/D = 25, m initial m final $ R g = exp & η o Q L R D ( ( (.81m /s 2 $ m = exp & ( 0.25 ( J /kg (25 = 4.31 ( So the aircraft takeoff mass has to be mostly fuel, i.e. m fuel /m initial = (m initial - m fina l/m initial = 1 - m final /m initial = 1-1/4.31 = 0.768! thats why no one flew around with world without refueling until 186 (solo flight 2005! What initial to final mass ratio is needed to get into orbit from the earths surface with a single-stage rocket propulsion system? For this mission Δu = 8000 m/s; using a good rocket propulsion system (e.g. Space Shuttle main engines, I SP 400 sec m # initial Δu & # = exp m final $ I sp ( = exp 8000m /s & ( (.81m /s 2 ( 400s ( = 7.68 $ Its practically impossible to obtain this large a mass ratio in a single stage, thus staging is needed where you jettison larger, heavier stages as fuel mass is consumed thats why no one put an object into earth orbit until 157, and no one has ever built a single stage to orbit vehicle. 21 Summary! Steady flow (e.g. gas turbine engines have much higher powerto-weight ratios than unsteady flow (e.g. reciprocating piston engines! A simple momentum balance on a steady-flow propulsion system shows that the best performance is obtained when! Exit pressure = ambient pressure! A large mass of gas is accelerated by a small Δu! 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 22 11