# AME 436 Energy and Propulsion - Paul D. Ronney

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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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