AME 514
Applications of
Combustion
Lecture 10: Hypersonic Propulsion I:
Motivation, performance parameters
Advanced propulsion systems (3 lectures)
 Hypersonic propulsion background (Lecture 1)
 Why hypersonic propulsion?
 What's different at hypersonic conditions?
 Real gas effects (non-constant CP, 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
AME 514 - Spring 2015 - Lecture 10
2
Why use air even if you're going to space?
 Carry only fuel, not fuel + O2, while in atmosphere
 8x mass savings (H2-O2), 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
L
T
D
mg
AME 514 - Spring 2015 - Lecture 10
3
What's different about hypersonic propulsion?
 Stagnation temperature Tt - measure of total energy (thermal +
kinetic) of flow - is really large even before heat addition - materials
problems
æ g -1 2 ö
Tt = Tç1+
M ÷
è
ø
2
 T = static temperature - T measured by thermometer moving with flow
 Tt = temperature of the gas if it is decelerated adiabatically to M = 0
  = gas specific heat ratio = Cp/Cv; M = Mach number = u/(RT)1/2
 Stagnation pressure - measure of usefulness (ability to expand flow)
is large even before heat addition - structural
problems
g
g
æ g -1 2 ö
Pt = Pç1+
M ÷
è
ø
2
-1
 P = static pressure - P measured by pressure gauge moving with flow
 Pt = pressure of the gas if it is decelerated reversibly and adiabatically
to M = 0
 Large Pt means no mechanical compressor needed at large M
AME 514 - Spring 2015 - Lecture 10
4
What's different about hypersonic propulsion?
 Why are Tt and Pt so important? Isentropic expansion until exit
pressure (P9) = ambient pressure (P1) (optimal exit pressure
yielding maximum thrust) yields
g -1 ö
æ
2g
ç æ Pa ö g ÷
u=
RTt ç1- ç ÷ ÷
g -1 ç è Pt ø ÷
è
ø
 … but it's difficult to add heat at high M without major loss of
stagnation pressure
AME 514 - Spring 2015 - Lecture 10
5
What's different about hypersonic propulsion?
 High temperatures:  and molecular mass not constant dissociation - use GASEQ (http://www.gaseq.co.uk) to compute
stagnation conditions
 Example calculation: standard atmosphere at 100,000 ft
 T1 = 227K, P1 = 0.0108 atm, c1 = 302.7 m/s, h1 = 70.79 kJ/kg
(atmospheric data from http://www.digitaldutch.com/atmoscalc/)
 Pick P2 > P1, compress isentropically, note new T2 and h2
 1st Law: h1 + u12/2 = h2 + u22/2; since u2 = 0, h2 = h1 + (M1c1)2/2 or M1 =
[2(h2-h1)/c12]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
AME 514 - Spring 2015 - Lecture 10
6
What's different about hypersonic propulsion?
WOW!
HOT
WARM
COLD
5000K
N+O+e-
3000K
N2+O
1000K
N2+O2
200K
N2+O2
AME 514 - Spring 2015 - Lecture 10
7
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.)
AME 514 - Spring 2015 - Lecture 10
8
Thrust computation
 Control volume for thrust computation - in frame of reference
moving with the engine
AME 514 - Spring 2015 - Lecture 10
9
Thrust computation - steady flight
 Newton's 2nd law: Force = rate of change of momentum
d(mu)
Forces
=
å
å dt
å Forces = Thrust + P1ACV - [ P1 (ACV - A9 ) + P9 A9 ] = T + (P1 - P9 )A9
d(mu)
dm
du
å dt = å( dt u + m dt ) = å(mu + 0) (if steady, du/dt = 0)
å mu = (mf + ma )u9 - (ma )u1 (u fuel = 0 in moving reference frame)
Combine: Thrust = ( ma + mf ) u9 - mau1 + ( P9 - P1 ) A9
Þ Thrust = ma éë(1+ FAR) u9 - u1 ùû + ( P9 - P1 ) A9 ;
FAR= mf / 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 hydrogen or hydrocarbon-air FAR << 1; typically 0.06 at
stoichiometric
AME 514 - Spring 2015 - Lecture 10
10
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, next lecture …
 And you can obtain a given thrust with small ma[(1+FAR)u9 - u1]
and large large (P9 – P1)A9 or vice versa - which is better, i.e. for
given ma, u1, P1 and FAR, what P9 will give most thrust?
Differentiate 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 on exit (see next slide)
d(u9 )
AdP+ mdu = 0 Þ A9 + ma (1+ FAR)
=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 514 - Spring 2015 - Lecture 10
11
1D momentum balance - constant-area duct
Coefficient of friction (Cf)
Wall drag force
Cf º
1 ru2 × (Wall area)
2
å Forces = å
d(mu)
dt
2
Forces
=
PA
(P
+
dP)A
C
(1/2
r
u
)(Cdx)
å
f
d(mu)
å dt = å m˙ u = m˙ u - m˙ (u + du)
˙ du + C f (1/2 ru2 )Cdx = 0
Combine : AdP + m
AME 514 - Spring 2015 - Lecture 10
12
Thrust computation
 But wait - this just says P9 = P1 is an extremum - is it a minimum
or a maximum?
d(Thrust)
d(A9 ) d2 (Thrust)
d2 (A9 ) d(A9 )
= (P9 - P1 )
Þ
= (P9 - P1 )
+
(1)
2
2
d(P9 )
d(P9 )
d(P9 )
d(P9 ) d(P9 )
but Pe = Pa at the extreme cases so
d2 (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 514 - Spring 2015 - Lecture 10
13
Propulsive, thermal, overall efficiency
 Thermal efficiency (th)
2
2
D(Kinetic energy) (ma + mf )u9 / 2 - (ma )u1 / 2
hth º
=
Heat input
mf QR
(u92 - u12 ) / 2
If mf << ma (FAR<< 1) then hth »
FARºQR
 Propulsive efficiency (p)
Thrust power
Thrust ºu1
hp º
=
D(Kinetic energy) (ma + mf )u92 / 2 - (ma )u12 / 2
If mf << ma (FAR<< 1) and P9 =P1 then h p »
 Overall efficiency (o)
ma (u9 - u1 )ºu1
2u1 / u9
=
ma (u92 - u12 ) / 2 1+ u1 / u9
this is the most important efficiency in determining aircraft
performance (see Breguet range equation, coming up…)
AME 514 - Spring 2015 - Lecture 10
14
Propulsive, thermal, overall efficiency
 Note on propulsive efficiency for FAR << 1
hp »
2u1 / u9
1+ u1 / u9
 p  1 as u1/u9  1  u9 is only slightly larger than u1
 But then you need large
tomget
required Thrust ~
a
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
 For hypersonic propulsion systems, u1 is large, u9 - u1 << u1, so
propulsive efficiency usually high (i.e. close to 1)
AME 514 - Spring 2015 - Lecture 10
15
Specific Thrust
 Specific thrust – thrust per unit mass flow rate, nondimensionalized by sound speed at ambient conditions (c1)
Specific Thrust (ST) º Thrust / mac1
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
r9u9 A9
mac1
c1 c1
mac1
c9 c1
c1
1+ FAR
(P - P ) (1+ FAR)
g RT9
= (1+ FAR)M 9
- M1 + 9 9
(P9 / RT9 )u9 c1
g RT1
= (1+ FAR)M 9
æ P1 ö RT9 (1+ FAR)
T9
- M1 + ç1- ÷
T1
è P9 ø c9 M 9 c1
= (1+ FAR)M 9
æ P1 ö RT9 (1+ FAR)
T9
- M1 + ç1- ÷
T1
è P9 ø g RT9 M 9 g RT1
= (1+ FAR)M 9
æ P1 ö T9 1+ FAR
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 514 - Spring 2015 - Lecture 10
16
Specific Thrust
 Specific thrust (ST) continued… if P9 = P1 and FAR << 1 then
ST º
Thrust
T
= M 9 9 - M1 (if FAR<<1, P1 = P9 )
mac1
T1
 Thrust Specific Fuel Consumption (TSFC) (PDR's definition)
mf QR æ mac1 ö FAR×QR FAR× QR
TSFC º
=ç
=
÷
2
Thrust c1 è Thrust ø c1
ST × c12
mf 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, use c1 to
dimensionless; use QR to convert m
convert the denominator to a quantity with units of power
 Specific impulse (Isp) = thrust per weight (on earth) flow rate of
fuel (+ oxidant if two reactants, e.g. rocket) (units of seconds)
Thrust
(Thrust)u1 QR
hoQR
QR
I sp =
; I sp =
=
=
mfuel gearth
mfuel QR gearthc1M1 M1c1gearth (TSFC)c1gearth
AME 514 - Spring 2015 - Lecture 10
17
Breguet range equation
 Consider aircraft in level flight
(Lift = weight) at constant flight
velocity u1 (thrust = drag)
L = mvehicleg;
D = Thrust =
=
Lift (L)
Thrust
˙ fuel QR
h om
hoQR dmfuel
Drag (D)
u1
=
hoQR -dmvehicle
Weight (W = mvehicleg)
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
D u1g
dm
dt = - vehicle Þ
L hoQR
mvehicle
R / u1
ò
0
final
D u1g
dmvehicle
dt = - ò
L hoQR
initial mvehicle
mfinal
D u1g R
L hoQR minitial
Þ
= -ln
Þ R=
ln
L hoQR u1
minitial
D g
mfinal
AME 514 - Spring 2015 - Lecture 10
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 specific impulse
(Isp) = thrust per unit weight (on earth) flow rate of fuel (+ oxidant
if two reactants carried), i.e. Thrust = mdotfuel*gearth*Isp
dmfuel
dmvehicle
˙ fuel gearth I sp = gearth I sp
Thrust = m
= -gearth I sp
dt
dt
du
dmvehicle
du
Thrust = mvehicle
Þ -gearth I sp
= mvehicle
dt
dt
dt
 Integrate to obtain Rocket Equation
 Of course gravity and atmospheric drag will increase effective u
requirement beyond that required strictly by orbital mechanics
19
AME 514 - Spring 2015 - Lecture 10
Breguet & rocket equations - comments
 Range (R) for aircraft depends on
 o (propulsion system) - depends 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!
20
AME 514 - Spring 2015 - Lecture 10
Breguet & rocket equations - examples
 Fly around the world (g = 9.8 m/s2) without refueling





R = 40,000 km
Use hydrocarbon fuel (QR = 4.5 x 107 J/kg),
Good propulsion system (o = 0.25)
Good airframe (L/D = 20),
Need minitial/mfinal ≈ 5.7 - aircraft has to be mostly fuel - mfuel/minitial =
(minitial - mfinal)/minitial = 1 - mfinal/minitial = 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, ISP ≈ 400 sec)
 Need minitial/mfinal ≈ 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
AME 514 - Spring 2015 - Lecture 10
21