AME 436 Energy and Propulsion Lecture 10 Propulsion 1: Thrust and aircraft range

AME 436AME 436

Energy and PropulsionEnergy and Propulsion

Lecture 10Lecture 10Propulsion 1: Thrust and aircraft rangePropulsion 1: Thrust and aircraft range

AME 436 - Lecture 10 - Thrust and Aircraft Range - Spring 2009AME 436 - Lecture 10 - Thrust and Aircraft Range - Spring 2009 22

OutlineOutline Why gas turbines?Why gas turbines? Computation of thrustComputation of thrust Propulsive, thermal and overall efficiencyPropulsive, thermal and overall efficiency Breguet range equationBreguet range equation


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Why gas turbines?Why gas turbines?

GE CT7-8 turboshaft (used in GE CT7-8 turboshaft (used in helicopters)helicopters)

http://www.geae.com/engines/http://www.geae.com/engines/commercial/ct7/ct7-8.htmlcommercial/ct7/ct7-8.html

Compressor/turbine stages: 6/4Compressor/turbine stages: 6/4 Diameter 26”, Length 48.8” = 426 liters Diameter 26”, Length 48.8” = 426 liters

= = 5.9 hp/liter5.9 hp/liter Dry Weight 537 lb, max. power 2,520 hp Dry Weight 537 lb, max. power 2,520 hp

((power/wt = 4.7 hp/lbpower/wt = 4.7 hp/lb)) Pressure ratio at max. power: 21 (ratio Pressure ratio at max. power: 21 (ratio

per stage = 21per stage = 211/61/6 = 1.66) = 1.66) Specific fuel consumption at max. Specific fuel consumption at max.

power: 0.450 (units not given; if lb/hp-hr power: 0.450 (units not given; if lb/hp-hr then corresponds to 29.3% efficiency)then corresponds to 29.3% efficiency)

Cummins QSK60-2850 4-stroke Cummins QSK60-2850 4-stroke 60.0 60.0 liter (3,672 inliter (3,672 in33)) V-16 2-stage V-16 2-stage turbocharged diesel (used in mining turbocharged diesel (used in mining trucks)trucks)

http://http://www.everytime.cummins.com/www.everytime.cummins.com/assets/pdf/4087056.pdfassets/pdf/4087056.pdf

2.93 m long x 1.58 m wide x 2.31 m 2.93 m long x 1.58 m wide x 2.31 m high = 10,700 liters = high = 10,700 liters = 0.27 hp/liter0.27 hp/liter

Dry weight 21,207 lb, 2850 hp at 1900 Dry weight 21,207 lb, 2850 hp at 1900 RPM (RPM (power/wt = 0.134 hp/lb = 35x power/wt = 0.134 hp/lb = 35x lower than gas turbinelower than gas turbine))

BMEP = 22.1 atm BMEP = 22.1 atm Volume compression ratio ??? (not Volume compression ratio ??? (not

given)given)


Ballard HY-80 “Fuel cell engine”Ballard HY-80 “Fuel cell engine” http://www.ballard.com/resources/transportation/http://www.ballard.com/resources/transportation/

XCS-HY-80_Trans.pdf (no longer valid link!)XCS-HY-80_Trans.pdf (no longer valid link!) Volume 220 liters = 0.41 hp/literVolume 220 liters = 0.41 hp/liter 91 hp, 485 lb. (91 hp, 485 lb. (power/wt = 0.19 hp/lbpower/wt = 0.19 hp/lb)) 48% efficiency (fuel to electricity)48% efficiency (fuel to electricity) Uses hydrogen only - NOT hydrocarbonsUses hydrogen only - NOT hydrocarbons Does NOT include electric drive systemDoes NOT include electric drive system (≈ 0.40 (≈ 0.40

hp/lb) at ≈ 90% electrical to mechanical efficiency hp/lb) at ≈ 90% electrical to mechanical efficiency (http://www.gm.com/company/gmability/adv_tech/i(http://www.gm.com/company/gmability/adv_tech/images/fact_sheets/hywire.html) (no longer valid)mages/fact_sheets/hywire.html) (no longer valid)

Fuel cell + motor overall 0.13 hp/lb at 43% Fuel cell + motor overall 0.13 hp/lb at 43% efficiency, not including Hefficiency, not including H22 storage storage

NiMH battery - 26.4 kW-hours, 1147 pounds = 1.83 NiMH battery - 26.4 kW-hours, 1147 pounds = 1.83 x 10x 1055 J/kg = 246x lower than HC fuel Q J/kg = 246x lower than HC fuel QRR (http://www.gmev.com/power/power.htm) (no (http://www.gmev.com/power/power.htm) (no longer valid)longer valid)

Structure: 290 pounds, < 10% of total vehicle; Structure: 290 pounds, < 10% of total vehicle; Aerodynamics: CAerodynamics: CDD = 0.19, “world's most energy- = 0.19, “world's most energy-efficient vehicle platform”efficient vehicle platform”


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Why gas turbines?Why gas turbines?


are needed to see this picture.

Lycoming IO-720 11.8 liter (720 Lycoming IO-720 11.8 liter (720 cu in) 4-stroke 8-cyl. gasoline cu in) 4-stroke 8-cyl. gasoline engine engine ((http://www.lycoming.com/engines/series/pdfs/Specialty%20insert.pdf))

Total volume 23” x 34” x 46” = Total volume 23” x 34” x 46” = 589 liters = 0.67 hp/liter589 liters = 0.67 hp/liter

400 hp @ 2650 RPM400 hp @ 2650 RPM Dry weight 600 lb. (Dry weight 600 lb. (power/wt = power/wt = 0.67 hp/lb = 7x lower than gas 0.67 hp/lb = 7x lower than gas turbineturbine))

BMEP = 11.3 atm (4 stroke)BMEP = 11.3 atm (4 stroke) Volume compression ratio 8.7:1 (= Volume compression ratio 8.7:1 (= pressure ratio 20.7 if pressure ratio 20.7 if isentropic)isentropic)


Why gas turbines?Why gas turbines? Why does gas turbine have much higher power/weight & Why does gas turbine have much higher power/weight &

power/volume than recips? power/volume than recips? More air can be processedMore air can be processed since steady since steady flow, not start/stop of reciprocating-piston enginesflow, not start/stop of reciprocating-piston engines More air More air more fuel can be burned more fuel can be burned More fuel More fuel more heat release more heat release More heat More heat more work (if thermal efficiency similar) more work (if thermal efficiency similar)

What are the disadvantages?What are the disadvantages? CompressorCompressor is a is a dynamicdynamic device that makes gas move from low pressure device that makes gas move from low pressure

to high pressure without a positive seal like a piston/cylinderto high pressure without a positive seal like a piston/cylinder» Requires very precise aerodynamicsRequires very precise aerodynamics» Requires blade speeds ≈ sound speed, otherwise gas flows back to low P Requires blade speeds ≈ sound speed, otherwise gas flows back to low P

faster than compressor can push it to high Pfaster than compressor can push it to high P» Each stage can provide only 2:1 or 3:1 pressure ratio - need many stages for Each stage can provide only 2:1 or 3:1 pressure ratio - need many stages for

large pressure ratiolarge pressure ratio Since steady flow, each component sees a constant temperature - at end Since steady flow, each component sees a constant temperature - at end

of combustor - of combustor - turbineturbine stays hot continuously and must rotate at high stays hot continuously and must rotate at high speeds (high stress)speeds (high stress)» Severe materials and cooling engineering required (unlike recip, where engine Severe materials and cooling engineering required (unlike recip, where engine

components only feel components only feel average gas temperature during cycleaverage gas temperature during cycle))» Turbine inlet temperature limit typically 1400K - limits fuel inputTurbine inlet temperature limit typically 1400K - limits fuel input

As a result, turbines require more maintenance & are more expensive for As a result, turbines require more maintenance & are more expensive for same powersame power

Simple intro to gas turbines: Simple intro to gas turbines: http://geae.com/education/engines101/http://geae.com/education/engines101/


Thrust computationThrust computation In gas turbine and rocket propulsion we need THRUST (force In gas turbine and rocket propulsion we need THRUST (force

acting on vehicle)acting on vehicle) How much push can we get from a given amount of fuel?How much push can we get from a given amount of fuel? We’ll start by showing that thrust depends primarily on the We’ll start by showing that thrust depends primarily on the

difference between the engine inlet and exhaust gas velocity, difference between the engine inlet and exhaust gas velocity, then compute exhaust velocity for various types of flows then compute exhaust velocity for various types of flows (isentropic, with heat addition, with friction, etc.)(isentropic, with heat addition, with friction, etc.)


Control volume for thrust computation - Control volume for thrust computation - in frame of in frame of reference moving with the enginereference moving with the engine

Thrust computationThrust computation

Everywhere on this surface, located far upstream of the engine: Pressure Pa (ambient pressure), velocity u1 (flight velocity)

Area ACV

Weird control volume

Velocity ue (Exit velocity) Area Ae (Exit area) Pressure Pe (Exit pressure)

Thrust

Everywhere else on this surface: Pressure Pa (ambient pressure), velocity u1 (flight velocity)

Ý m f

Ý m a

Ý m a Ý m f

Incoming air


Forces d(mu)

dt

Forces Thrust Pa ACV Pa (ACV Ae ) Pe Ae T (Pa Pe )Ae

d(mu)

dt (

dm

dtu m

du

dt) ( Ý m u 0) (if steady, du/dt = 0)

Ý m u ( Ý m f Ý m a )ue ( Ý m a )u1 (ufuel 0 in moving reference frame)

Combine : Thrust Ý m a Ý m f ue Ý m au1 Pe Pa Ae

Thrust Ý m a 1 FAR ue u1 Pe Pa Ae;

FAR Ý m f / Ý m a Fuel to air mass ratio f /(1 f ) ( f = fuel mass fraction)

Newton’s 2nd law: Force = rate of change of momentumNewton’s 2nd law: Force = rate of change of momentum

At takeoff uAt takeoff u11 = 0; for rocket no inlet so u = 0; for rocket no inlet so u11 = 0 always = 0 always For hydrocarbon-air usually FAR << 1; typically 0.06 at For hydrocarbon-air usually FAR << 1; typically 0.06 at

stoichiometric, but in practice maximum allowable FAR ≈ 0.03 due stoichiometric, but in practice maximum allowable FAR ≈ 0.03 due to turbine inlet temperature limitations (discussed later…)to turbine inlet temperature limitations (discussed later…)

Thrust computation - steady flightThrust computation - steady flight


But how to compute exit velocity (uBut how to compute exit velocity (uee) and exit pressure (P) and exit pressure (Pee) as a ) as a

function of ambient pressure (Pfunction of ambient pressure (Paa), flight velocity (u), flight velocity (u11)? Need )? Need

compressible flow analysis, coming next…compressible flow analysis, coming next… Also - one can obtain a given thrust with large (PAlso - one can obtain a given thrust with large (Pe e - P- Paa)A)Aee and and

small mdotsmall mdotaa[(1+FAR)u[(1+FAR)ue e - u- u11] or vice versa - which is better, i.e. ] or vice versa - which is better, i.e.

for given ufor given u11, P, Paa, mdot, mdotaa and FAR, what P and FAR, what Pee will give most thrust? will give most thrust?

Differentiate thrust equation and set = 0Differentiate thrust equation and set = 0

Momentum balance on exit (see next slide)Momentum balance on exit (see next slide)

CombineCombine

Optimal performance occurs for exit pressure = ambient Optimal performance occurs for exit pressure = ambient pressurepressure


d(Thrust)

d(Pe ) Ý m a 1 FAR d(ue )

d(Pe ) 0

1 0 Ae (Pe Pa )

d(Ae )

d(Pe )0

AdP + Ý m du 0 Ae Ý m a (1 FAR )d(ue )

d(Pe )0

d(Thrust)

d(Pe )(Pe Pa )

d(Ae )

d(Pe )0 Pe Pa


1D momentum balance - constant-area duct

Coefficient of friction (CCoefficient of friction (Cff))

Forces d(mu)

dt

Forces PA (P dP)A C f (1/2u2)(Cdx)

d(mu)

dt Ý m u Ý m u Ý m (u du)

Combine : AdP + Ý m du C f (1/2u2)Cdx 0

Thickness dx,circumference C,wall area Cdx

Velocity upressure Pdensity duct area Amass flux = uA

Velocity u + dupressure P + dPdensity duct area Amass flux = uA

C f Wall drag force

12 u2 (Wall area)


But wait - this just says PBut wait - this just says Pee = P = Paa is an extremum - is it a is an extremum - is it a

minimum or maximum? minimum or maximum?

but Pbut Pee = P = Paa at the extremum cases so at the extremum cases so

Maximum thrust if dMaximum thrust if d22(Thrust)/d(P(Thrust)/d(Pee))22 < 0 < 0 dA dAee/dP/dPee < 0 - we will < 0 - we will

show this is true for supersonic exit conditionsshow this is true for supersonic exit conditions Minimum thrust if dMinimum thrust if d22(Thrust)/d(P(Thrust)/d(Pee))22 > 0 > 0 dA dAee/dP/dPee > 0 - we will > 0 - we will

show this is would be true for subsonic exit conditions, but show this is would be true for subsonic exit conditions, but for subsonic, Pfor subsonic, Pee = P = Paa always since acoustic (pressure) waves always since acoustic (pressure) waves

can travel up the nozzle, equalizing the pressure to Pcan travel up the nozzle, equalizing the pressure to Paa, so it’s , so it’s

a moot point for subsonic exit velocitiesa moot point for subsonic exit velocities


d(Thrust)

d(Pe )(Pe Pa )

d(Ae )

d(Pe )

d2(Thrust)

d(Pe )2 (Pe Pa )d2(Ae )

d(Pe )2 d(Ae )

d(Pe )(1)

d2(Thrust)

d(Pe )2 d(Ae )

d(Pe )


Turbofan: same as turbojet except that there are two Turbofan: same as turbojet except that there are two streams, one hot (combusting) and one cold (non-streams, one hot (combusting) and one cold (non-combusting, fan only, use prime (‘) superscript):combusting, fan only, use prime (‘) superscript):

Note (1 + FAR) term applies only to combusting streamNote (1 + FAR) term applies only to combusting stream Note we assumed PNote we assumed Pee = P = Paa for fan stream; for any reasonable for fan stream; for any reasonable

fan design, ufan design, uee’’ will be subsonic so this will be true will be subsonic so this will be true


Thrust Ý m a 1 FAR ue u1 Pe Pa Ae Ý m a' ue

' u1


Propulsive, thermal, overall Propulsive, thermal, overall efficiencyefficiency Thermal efficiencyThermal efficiency ( (thth) )

Propulsive efficiencyPropulsive efficiency ( (pp) )

Overall efficiencyOverall efficiency ( (oo) )

this is the most important efficiency in determining aircraft this is the most important efficiency in determining aircraft performance (see Breguet range equation, coming up…)performance (see Breguet range equation, coming up…)

th (Kinetic energy)

Heat input

( Ý m a Ý m f )ue2 /2 ( Ý m a )u1

2 /2

Ý m f QR

If Ý m f Ý m a (FAR 1) then th (ue2 u1

2) /2

FAR QR

p Thrust power

(Kinetic energy)

Thrust u1

( Ý m a Ý m f )ue2 /2 ( Ý m a )u1

2 /2

If Ý m f Ý m a (FAR 1) and Pe = Pa then p Ý m a (ue u1)u1

Ý m a (ue2 u1

2) /2 2u1 /ue

1 u1 /ue

o Thrust power

Heat input

Thrust power

(Kinetic energy) (Kinetic energy)

Heat inputthp


Propulsive, thermal, overall Propulsive, thermal, overall efficiencyefficiency Note on propulsive efficiency Note on propulsive efficiency

pp 1 as u 1 as u11/u/uee 1 1 u ue e is only slightly larger than uis only slightly larger than u11

But then you need large mdotBut then you need large mdotaa to get required Thrust ~ to get required Thrust ~

mdotmdotaa(u(uee - u - u11) - but this is how commercial turbofan engines ) - but this is how commercial turbofan engines

work!work! In other words, the best propulsion system accelerates an In other words, the best propulsion system accelerates an

infinite mass of air by an infinitesimal infinite mass of air by an infinitesimal uu Fundamentally this is because Thrust ~ (uFundamentally this is because Thrust ~ (uee - u - u11), but energy ), but energy

required to get that thrust ~ (urequired to get that thrust ~ (uee22 - u - u11

22)/2)/2 This issue will come up a lot in the next few weeks!This issue will come up a lot in the next few weeks!

p 2u1 /ue

1 u1 /ue


D

L

u1g

oQR

dt dmvehicle

mvehicle

D

L

u1g

oQR

d0

R / u1

t dmvehicle

mvehicleinitial

final

D

L

u1g

oQR

R

u1

lnm final

minitial

R L

D

oQR

gln

minitial

m final

Breguet range equation

Consider aircraft in Consider aircraft in level flightlevel flight (Lift = weight) at (Lift = weight) at constant flightconstant flightvelocityvelocity u u11 (thrust = drag) (thrust = drag)

Combine expressions for lift & drag and integrate from time t = 0 to Combine expressions for lift & drag and integrate from time t = 0 to t = R/ut = R/u11 (R = range = distance traveled), i.e. time required to reach (R = range = distance traveled), i.e. time required to reach

destination, to obtain destination, to obtain Breguet Range EquationBreguet Range Equation

Thrust (T)

Lift (L)

Weight (W = mvehicle g)

Drag (D)

Flight velocity = u1

L mvehicleg;

D Thrust o Ý m fuelQR

u1

oQR

u1

dm fuel

dt

oQR

u1

dmvehicle

dt


Rocket equation

If acceleration (If acceleration (u) rather than range in steady flight is desired u) rather than range in steady flight is desired [neglecting drag (D) and gravitational pull (W)], Force = mass x [neglecting drag (D) and gravitational pull (W)], Force = mass x acceleration or Thrust = macceleration or Thrust = mvehiclevehicledu/dtdu/dt

Since flight velocity uSince flight velocity u11 is not constant, overall efficiency is not an is not constant, overall efficiency is not an

appropriate performance parameter; instead use appropriate performance parameter; instead use specific impulsespecific impulse (I(Ispsp) = thrust per unit weight (on earth) flow rate of fuel (+ oxidant if ) = thrust per unit weight (on earth) flow rate of fuel (+ oxidant if

two reactants carried), i.e. Thrust = mdottwo reactants carried), i.e. Thrust = mdotfuelfuel*g*gearthearth*I*Ispsp

Integrate to obtain Integrate to obtain Rocket EquationRocket Equation

Of course gravity and atmospheric drag will increase effective Of course gravity and atmospheric drag will increase effective u u requirement beyond that required strictly by orbital mechanicsrequirement beyond that required strictly by orbital mechanics

Thrust Ý m fuel gearthIsp gearthIsp

dm fuel

dt gearthIsp

dmvehicle

dt

Thrust mvehicle

du

dt gearthIsp

dmvehicle

dtmvehicle

du

dt

u gearthIsp lnminitial

m final


Brequet range equation - comments Range (R) for aircraft depends on Range (R) for aircraft depends on

oo (propulsion system) - dependd on u (propulsion system) - dependd on u11 for airbreathing propulsion for airbreathing propulsion QQRR (fuel) (fuel) L/D (lift to drag ratio of airframe)L/D (lift to drag ratio of airframe) g (gravity)g (gravity) Fuel consumption (mFuel consumption (minitialinitial/m/mfinalfinal); m); minitialinitial - m - mfinalfinal = fuel mass used (or fuel + = fuel mass used (or fuel +

oxidizer, if not airbreathing)oxidizer, if not airbreathing) This range does not consider fuel needed for taxi, takeoff, climb, This range does not consider fuel needed for taxi, takeoff, climb,

decent, landing, fuel reserve, etc.decent, landing, fuel reserve, etc. Note (irritating) ln( ) or exp( ) term in both Breguet and Rocket:Note (irritating) ln( ) or exp( ) term in both Breguet and Rocket:

because you have to use more fuel at the beginning of the flight, 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 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 - if not for this it would be easy to fly around the world without refueling and the Chinese would have sent skyrockets into orbit refueling and the Chinese would have sent skyrockets into orbit thousands of years ago! thousands of years ago!

minitial

m final

= expRg

oQRL

D

(Breguet);

minitial

m final

= expu

gearthIsp

(rocket)


Brequet range equation - examples Fly around the world (g = 9.8 m/sFly around the world (g = 9.8 m/s22) without refueling ) without refueling

R = 40,000 kmR = 40,000 km Use hydrocarbon fuel (QUse hydrocarbon fuel (QRR = 4.5 x 10 = 4.5 x 1077 J/kg), J/kg), Good propulsion system (Good propulsion system (oo = 0.25) = 0.25) Good airframe (L/D = 20), Good airframe (L/D = 20),

you need you need mminitialinitial/m/mfinalfinal ≈ 5.7 ≈ 5.7 - aircraft has to be mostly fuel - - aircraft has to be mostly fuel -

mmfuelfuel/m/minitial initial = (m= (minitialinitial - m - mfinalfinal)/m)/minitialinitial = 1 - m = 1 - mfinalfinal/m/minitialinitial = 1 - 1/5.7 = = 1 - 1/5.7 =

0.825! - that’s why no one flew around with world without 0.825! - that’s why no one flew around with world without refueling until 1986 (solo flight 2005)refueling until 1986 (solo flight 2005)

To get into orbit from the earth’s surfaceTo get into orbit from the earth’s surface u = 8000 m/su = 8000 m/s Use a good rocket propulsion system (e.g. Space Shuttle main Use a good rocket propulsion system (e.g. Space Shuttle main

engines, Iengines, ISPSP ≈ 400 sec) ≈ 400 sec)

need need mminitialinitial/m/mfinalfinal ≈ 7.7 ≈ 7.7 - can’t get this good a mass ratio in a - can’t get this good a mass ratio in a

single vehicle - need single vehicle - need staging staging - that’s why no one put an - that’s why no one put an object into earth orbit until 1957object into earth orbit until 1957


Summary Steady flow (e.g. gas turbine) engines have much higher Steady flow (e.g. gas turbine) engines have much higher

power/weight ratio than unsteady flow (e.g. reciprocating piston) power/weight ratio than unsteady flow (e.g. reciprocating piston) enginesengines

When used for thrust, a simple momentum balance on a steady-When used for thrust, a simple momentum balance on a steady-flow engine shows that the best performance is obtained whenflow engine shows that the best performance is obtained when Exit pressure = ambient pressureExit pressure = ambient pressure A large mass of gas is accelerated by a small A large mass of gas is accelerated by a small uu

Two types of efficiencies for propulsion systems - thermal Two types of efficiencies for propulsion systems - thermal efficiency and propulsive efficiency (product of the two = overall efficiency and propulsive efficiency (product of the two = overall efficiency)efficiency)

Range of an aircraft depends critically on overall efficiency - effect Range of an aircraft depends critically on overall efficiency - effect more severe than in ground vehicles, because aircraft must more severe than in ground vehicles, because aircraft must generate enough lift (thus thrust, thus required fuel flow) to carry generate enough lift (thus thrust, thus required fuel flow) to carry entire fuel load at first part of flight)entire fuel load at first part of flight)

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AME 436 Energy and Propulsion Lecture 10 Propulsion 1: Thrust and aircraft range