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 2013 - Lecture 11 - Thrust and Aircraft Range
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Why gas turbines?
 GE CT7-8 turboshaft (used in
helicopters)
 http://www.geaviation.com/engines/co
mmercial/ct7/ct7-8.html
 Compressor/turbine stages: 6/4
 Diameter 26”, Length 48.8” = 426 liters
= 5.9 hp/liter
 Dry Weight 537 lb, max. power 2,520 hp
(power/wt = 4.7 hp/lb)
 Pressure ratio at max. power: 21 (ratio
per stage = 211/6 = 1.66)
 Specific fuel consumption at max.
power: 0.450 (units not given; if lb/hphr then corresponds to 29.3%
efficiency)
 Cummins QSK60-2850 4-stroke 60.0
liter (3,672 in3) V-16 2-stage
turbocharged diesel (used in mining
trucks)
 http://cumminsengines.com/assets/p
df/4087056.pdf
 2.93 m long x 1.58 m wide x 2.31 m
high = 10,700 liters = 0.27 hp/liter
 Dry weight 21,207 lb, 2850 hp at 1900
RPM (power/wt = 0.134 hp/lb = 35x
lower than gas turbine)
 BMEP = 22.1 atm
 Volume compression ratio ??? (not
given)
3
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
Why gas turbines?
 Lycoming IO-720 11.8 liter (720 cu in) 4stroke 8-cyl. gasoline engine
 http://www.lycoming.com/engines/series/
pdfs/Specialty%20insert.pdf
 Total volume 23” x 34” x 46” = 589 liters =
0.67 hp/liter
 400 hp @ 2650 RPM
 Dry weight 600 lb. (power/wt = 0.67 hp/lb
= 7x lower than gas turbine)
 BMEP = 11.3 atm (4 stroke)
 Volume compression ratio 8.7:1 (=
pressure ratio 20.7 if isentropic)
 NuCellSys HY-80 “Fuel cell engine”
 http://www.nucellsys.com/dyn/mediaout.
dhtml/234f73a2c579fb27158i/mime/PDF/H
Y-80-PDF/HY-80_2009.pdf
 Volume 220 liters = 0.41 hp/liter
 91 hp, 485 lb. (power/wt = 0.19 hp/lb)
 41% efficiency (fuel to electricity) at max.
power; up to 58% at lower power
 Uses hydrogen only - NOT hydrocarbons
 Does NOT include electric drive system
(≈ 0.40 hp/lb) at ≈ 90% electrical to
mechanical efficiency
(http://www.gm.com/company/gmability/a
dv_tech/images/fact_sheets/hywire.html)
(no longer valid)
 Fuel cell + motor overall 0.13 hp/lb at
37% efficiency, not including H2 storage
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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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 2013 - Lecture 11 - Thrust and Aircraft Range
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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 2013 - Lecture 11 - Thrust and Aircraft Range
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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 2013 - Lecture 11 - Thrust and Aircraft Range
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Thrust computation
 Control volume for thrust computation - in frame of
reference moving with the engine
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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Thrust computation - steady flight
 Newton’s 2nd law: Force = rate of change of momentum
d(mu)
Forces
=
å
å dt
å Forces = Thrust + P A
1
CV
- [ P1 (ACV - A9 ) + P9 A9 ] = T + (P1 - P9 )A9
d(mu)
dm
du
=
(
u
+
m
) = å ( m˙ u + 0) (if steady, du/dt = 0)
å dt å dt
dt
å m˙ u = (m˙ f + m˙ a )u9 - (m˙ a )u1 (u fuel = 0 in moving reference frame)
(
)
Combine : Thrust = m˙ a + m˙ f u9 - m˙ a u1 + ( P9 - P1 ) A9
Þ Thrust = m˙ a [(1+ FAR) u9 - u1 ] + ( P9 - P1) A9;
FAR = m˙ f / m˙ a = 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 2013 - Lecture 11 - Thrust and Aircraft Range
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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
˙ a[(1+FAR)u9 - u1] or vice versa - which is better, i.e. for given
small m
˙ aand FAR, what P9 will give most thrust? Differentiate
u1, P1, m
thrust equation and set = 0
é
d(Thrust)
d(u9 ) ù
d(A9 )
= ma ê(1+ FAR)
- 0ú + (1- 0) A9 + (P9 - P1 )
=0
d(P9 )
d(P9 ) û
d(P9 )
ë
 Momentum balance at exit (see next slide)
AdP + mdu = 0 Þ A9 + ma (1+ FAR)
d(u9 )
=0
d(P9 )
 Combine
d(Thrust)
d(A9 )
= (P9 - P1 )
= 0 Þ P9 = P1
d(P9 )
d(P9 )
 Optimal performance occurs for exit pressure = ambient pressure
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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1D momentum balance - constant-area duct
Coefficient of friction (Cf)
Cf º
Wall drag force
1 ru 2 × (Wall area)
2
d(mu)
dt
å Forces = PA - (P + dP)A - C f (1/2ru2 )(Cdx)
å Forces = å
d(mu)
å dt = å m˙ u = m˙ u - m˙ (u + du)
Combine : AdP + m˙ du + C f (1/2 ru 2 )Cdx = 0
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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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 - P1 )
Þ
= (P9 - P1 )
+
(1)
2
2
d(P9 )
d(P9 )
d(P9 )
d(P9 ) d(P9 )
but P9 = P1 at the extremum cases so
d 2 (Thrust) d(A9 )
=
2
d(P9 )
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 2013 - Lecture 11 - Thrust and Aircraft Range
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Thrust computation
 Turbofan: same as turbojet except that there are two
streams, one hot (combusting) and one cold (noncombusting, fan only, use prime (‘) superscript):
(
)
(
)
Thrust = ma éë 1+ FAR u9 - u1 ùû + P9 - P1 A9 + ma' éëu9' - u1 ùû
 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 2013 - Lecture 11 - Thrust and Aircraft Range
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Propulsive, thermal, overall efficiency
 Thermal efficiency (th)
2
2
D(Kinetic energy) (ma + m f )u9 / 2 - (ma )u1 / 2
hth º
=
Heat input
m f QR
(u92 - u12 ) / 2
If m f << ma (FAR << 1) then hth »
FAR × QR
 Propulsive efficiency (p)
hp º
Thrust × u1
Thrust power
=
D(Kinetic energy) (ma + m f )u92 / 2 - (ma )u12 / 2
ma (u9 - u1 ) × u1 2u1 / u9
If m f << ma (FAR << 1) and P9 =P1 then h p »
=
2
2
ma (u9 - u1 ) / 2 1+ u1 / u9
 Overall efficiency (o)
this is the most important efficiency in determining aircraft
performance (see Breguet range equation, coming up…)
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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Propulsive, thermal, overall efficiency
 Note on propulsive efficiency
ma (u9 - u1 )u1
2(u9 - u1 )u1
1
hp =
=
=
;Du º u9 - u1
2
2
2
2
ma (u9 - u1 ) / 2 éu + (u - u )ù - éu ù 1+ Du / 2u1
ë 1
ë 1û
9
1 û
 p  1 as Du  0  u9 is only slightly larger than u1
˙ a) to get the
 But then you need large mass flow rate ( m
˙ aDu - but this is how commercial
required Thrust ~ m
turbofan engines work!
 In other words, the best propulsion system accelerates an
infinite mass of air by an infinitesimal Du
 Fundamentally this is because Thrust ~ Du = 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 2013 - Lecture 11 - Thrust and Aircraft Range
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Ideal turbojet cycle - notes on thrust
 Specific thrust – thrust per unit mass flow rate, nondimensionalized 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
= (1+ FAR)M 9
g RT9
(P9 - P9 )
- M1 +
(P9 / RT9 )u9 c1
g RT1
= (1+ FAR)M 9
æ P1 ö RT9
T9
- M1 + ç1- ÷
T1
è P9 ø c9 M 9 c1
= (1+ FAR)M 9
æ P1 ö
T9
RT9
- M1 + ç1- ÷
T1
è P9 ø g RT9 M 9 g RT1
= (1+ FAR)M 9
æ P1 ö T9 1
T9
- M1 + ç1- ÷
T1
è P9 ø T1 g M 9
For any 1D steady propulsion
system if working fluid is an
ideal gas with constant CP, 
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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Other performance parameters
 Specific thrust (ST) continued… if P9 = P1 and FAR << 1 then
ST º
Thrust
T
= M 9 9 - M1 (if FAR <<1, P1 = P9 )
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
=
÷
2
Thrust c1 è Thrust ø c1
ST × c12
m f QR u1
M1
Note TSFC=
=
Thrust u1 c1
ho
˙ f /Thrust, but this is not
 Usual definition of TSFC is just m
˙ f to heat input, one can use
dimensionless; use QR to convert m
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)
I sp =
Thrust
(Thrust)u1 QR
hoQR
QR
; I sp =
=
=
m fuel gearth
m fuelQR gearth c1M1 M1c1gearth (TSFC)c1gearth
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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Breguet range equation
 Consider aircraft in level flight
(Lift = weight) at constant flight
velocity u1 (thrust = drag)
L = mvehicle g;
D = Thrust =
=
Thrust
Drag (D)
ho m˙ fuel QR
hoQR dm fuel
u1
Lift (L)
dt
u1
=
Weight (W = mvehicleg)
hoQR -dmvehicle
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
D u1g
dmvehicle
dt = Þ
L hoQR
mvehicle
R / u1
ò
0
final
D u1g
dmvehicle
dt = - ò
L hoQR
initial m vehicle
m final
D u1g R
L hoQR minitial
Þ
= -ln
Þ R=
ln
L hoQR u1
minitial
D g
m final
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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Rocket equation
 If acceleration (Du) 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
æ m final ö
dmvehicle
Þ du = -gearth I sp
Þ Du º u final - uinitial = -gearth I sp ln ç
÷
mvehicle
m
è initial ø
æm
ö
initial
÷÷
Þ Du = gearth I sp ln çç
è m final ø
 Of course gravity and atmospheric drag will increase effective Du
requirement beyond that required strictly by orbital mechanics
AME 436 - Spring 2013 - Lecture 11 - Thrust and Aircraft Range
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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 2013 - Lecture 11 - Thrust and Aircraft Range
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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
çh Q L ÷
m final
(
) ø
è
è o R Dø
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 Du = 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 2013 - Lecture 11 - Thrust and Aircraft Range
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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 Du
 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 2013 - Lecture 11 - Thrust and Aircraft Range
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