UNIFIED PROPULSION

8/7/2019 UNIFIED PROPULSION

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http://ocw.mit.edu/ans7870/16/16.unified/propulsionS04/UnifiedPropulsion7/UnifiedPropulsion7.htm

http://ocw.mit.edu/ans7870/16/16.unified/propulsionS04/learning_obj.html

UNIFIED PROPULSION: Learning Objectives:

Given the basic geometry and idealized component performance, to be able toestimate the thrust and specific impulse of a gas turbine and a rocket engine from fluidand thermodynamic principles

Measurable outcomes (assessment method):

A.To be able to explain at a level understandable by a high school senior or non-technical person what the various terms are in the integral momentum equationand how jet propulsion works. (quiz, self-assessment)

B.T

o be able to apply control volume analysis and the integral momentumequation to estimate the forces produced by aerospace propulsion systems(homework, quiz, self-assessment)

C.To be able to describe the principal figures of merit for aircraft engine androcket motor performance and explain how they are related to vehicle

performance. (quiz, self-assessment)D.Given weight, geometry, and aerodynamic and propulsion system performance

information, to be able to estimate the power required for flight, the range, theendurance, and the time-to-climb for an aircraft. (homework, quiz, self-assessment)

E. Given mass fractions, and propulsion system performance information, to beable to estimate the range and velocity of single-stage rockets. (homework andquiz, self-assessment)

F. To be able to describe the principal design parameters and constraints that setthe performance of gas turbine engines, and to apply ideal-cycle analysis to agas turbine engine to relate thrust and fuel burn to component-level

performance parameters and flight conditions. (homework, quiz, self-assessment)

G.To be able to explain at a level understandable by a high school senior or non-technical person the energy exchange processes that underlie the workings of

multistage compressor or turbine, and to be able to use velocity triangles andthe EulerTurbine Equation to estimate the performance of a compressor orturbine stage. (homework, quiz, self-assessment)


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I. Introduction to Propulsion

A. Goal: Create a Force to Propel a Vehicle

Two options:

1. Take mass stored in a vehicle and throw it backwards (rocket propulsion). Usethe reaction force to propel the vehicle.

Propellant--->

burn --->expand through

nozzle

(chem.energy)

(thermalenergy)

(kinetic energy& momentum)

Q1 (PDF)

2. Seize mass from the surroundings and set the mass in motion backwards. Usethe reaction force to propel vehicle (air-breathing propulsion).

Continuously: a) Draw in air.

b) Compress it.

c) Add fuel and burn (convert chemical energy to thermal energy).

d) Expand through a turbine to drive compressor (extract work).e.1) Then expand in a nozzle to convert thermal energy to kinetic energy& momentum (turbojet).

e.2) Or expand in a second turbine (extract work), use this to drive ashaft for a fan (turbofan), or a propeller (turboshaft). The fan orpropeller impartk.e. & mom. to the air.

*Remember:

Overall goal: take at Vo (flight speed), throw it out at Vo + DV

Q2 (PDF)


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Figure 1.1Schematics of typical military gas turbine engine: J57 turbojet withafterburning.

Figure 1.2 A typical high bypass-ratio turbofan (Adapted from Pratt & Whitney).

For more examples of real world powerplants, refer to Hill, P. and C. Peterson.Thermodynamics of Propulsion. 2nd Ed. Addison-Wesley, 1991.

B. Performance Parameters

The two performance parameters of greatest interest for a propulsion system are theforce it produces (thrust, T), and the overall efficiency with which it uses energy to

produce this force (hoverall). We will begin by looking at the production of thrust using


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the integral form of the momentum theorem. In the second lecture we will discuss theefficiency of propulsion systems.

C. Propulsion is a systems endeavor

There are a multitude of other factors which a propulsion engineer must take intoaccount when designing a device including weight, cost, manufacturability, safety,environmental effects, etc. Thus propulsion is truly a systems endeavor, requiringknowledge of a variety of disciplines:

Fluids + thermo + structures + dynamics + controls + chemistry +acoustics +

We will focus mostly on these two disciplines in the Unified propulsion lectures.


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II. Integral Momentum Theorem

We can learn a great deal about the overall behavior of propulsion systems using theintegral form of the momentum equation. The equation is the same as that used in

fluid mechanics.

A. An Expression of Newton's 2nd Law (e.g. = d/dt

(mv))

1. Consider two coordinate systems:

a) Inertial (labeled with subscript I in

diagram at right)

b) Fixed to vehicle (labeled with subscriptV in diagram at right). Moves with

velocity relative to the inertialcoordinate system. All velocities relative tothe vehicle-fixed coordinate frame aredenoted

2. Newtons second law for a control volume of fixed mass

or

All external forces oncontrol volume(pressure forces, shear

Force due to change ininertia for accelerating

Change inmomentum of massin c.v.


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forces, body forces) vehicle

To explain the above equation further, consider Figure 2.1

Figure 2.1 Falling blocks.

The falling block labeled (a) has a control volume fixed to it. In this case, the firstterm on the right hand side of the above equation is nonzero since the control volumeis accelerating relative to an inertial reference frame. The second term is zero becausethe block is not accelerating relative to a coordinate system fixed to the controlvolume. The opposite is true for the falling block labeled (b), which is falling within a

fixed control volume. The first term of the above equation is zero in this case becausethe control volume is not accelerating relative to an inertial reference frame. Thesecond term is nonzero because the block is moving relative to a coordinate systemfixed to the control volume. The mathematical result of both cases is as follows,

(a)


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(b)

As expected, the result is the same for both. The integral momentum equation

reduces to a familiar form, .

To continue, the equation can be rewritten as follows,

From conservation of mass,

so

NOTE: This is a vectorequation.

3. Considering only the components in the x-direction


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Then, by the divergence theorem,

where is outward unit normal vector.

so

where the forces acting on the control volume may be composed ofpressureforces, body forces, and skin friction

Q6 (PDF)

4. For steady flow, with no acceleration of the vehicle then

This is the form we will use most frequently in this class.

Q8 (PDF)


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B. Application of the Integral Momentum Equation to

Rockets

Figure 2.2 Control volume for application of momentum theorem to a rocket.

Static thrust for a rocket engine

C. Application of the Momentum Equation to an Aircraft

Engine

Figure 2.3 Control volume for application of the momentum theorem to a gas turbineengine.


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So we have:

Everything that relates to flow through the engine is conventionally called thrust.Everything that relates to the flow on the outside of the engine is conventionally

call drag. Therefore, gathering only those terms that relate to the fluid that passesthrough the engine, we have:

Q3 (PDF)Q9 (PDF)Q10 (PDF)

T

he thrust is largely composed of the net change in momentum of the air entering andleaving the engine, with a typically small adjustment for the differences in pressurebetween the inlet and the exit. We could have arrived at the same equation byconsidering only the streamtube that passes through the engine as shown below:


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Figure 2.4 Control volume for flow through engine.


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III. Efficiencies of A/C Engines

In the first lecture we arrived at general expressions that related the thrust of apropulsion system to the net changes in momentum, pressure forces, etc. Now we will

look at how efficiently the propulsion system converts one form of energy to anotheron its way to producing thrust.

A. Overall Efficiency

Thus

B. Thermal and Propulsive Efficiency

It is often convenient to break the overall efficiency into two parts: thermal efficiencyand propulsive efficiency where

such that


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hoverall = hthermal hprop

The thermal efficiency in this expression is the same as that which we usedextensively in Thermodynamics during Fall term. For an ideal Brayton cycle it is afunction of the temperature ratio across the compressor.

Note that we can use our expression for thrust to rewrite the equation for propulsiveefficiency in a more convenient form

Then

C. Implications of propulsive efficiency for engine design

If we consider our expressions for thrust and propulsive efficiency together

and

we see that

as but

and as and


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Also note that for

The balance between propulsive efficiency and specific thrust (~ thrust per unit mass

flow) is shown in Figure 3.1.

Figure 3.1 Propulsive efficiency and specific thrust as a function of exhaust velocity.

For fighter aircraft (e.g., the F-22 Raptor) that need high thrust/weight and fly at highspeed, it is typical to employ engines with smaller inlet areas and higher thrust perunit mass flow

and

However, transport aircraft (e.g., the Boeing 777-200) that require higher efficiencyand fly at lower speeds usually employ engines with relatively larger inlet areas andlower thrust per unit mass flow

and


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At low flight velocities, the highest propulsive efficiency is typically obtained with apropeller or an unducted fan. A propeller gives a relatively small impulse (Du) to arelatively large mass flow.

Figure 3.2 An advanced, contour-rotating, unducted fan concept. Copyright Rolls-Royce plc. Reproduced with the kind permission of Rolls-Royce plc.

Figure 3.3 Propulsive efficiency comparison for various gas turbine engineconfigurations. Copyright Rolls-Royce plc. Reproduced with the kind permission of

Rolls-Royce plc.

Q13 (PDF)

D. Other expressions for efficiency


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Sometimes the overall efficiency of aircraft engines is expressed in alternativeparameters: specific impulse, I, and thrust specific fuel consumption, TSFC or justSFC. Both of these parameters have dimensions.

Specific Impulse (I or Isp):

(units of seconds)

Specific Fuel Consumption (SFC orTSFC):

(lbm/hr/lbf or kg/s/N)

E. Trends in thermal and propulsive efficiency

Figure 3.4Trends in aircraft engine efficiency (after Pratt & Whitney).

Trends in thermal efficiency are driven by increasing compression ratios andcorresponding increases in turbine inlet temperature as shown in Figures 3.5 and 3.6.


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Whereas trends in propulsive efficiency are due to generally higher bypass ratioengines.

Figure 3.5 Pressure ratio trends for commercial transport engines (Epstein, 1998).

Figure 3.6Trends in turbine inlet temperature (Koff, 1991).


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Figure 3.7Trends in engine bypass ratio (Epstein, 1998).


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IV. Aircraft Performance

In this lecture we will make the connections between aircraft performance andpropulsion system performance.

For a vehicle in steady, level flight, the thrust force is equal to the drag force, and liftis equal to weight. Any thrust available in excess of that required to overcome thedrag can be applied to accelerate the vehicle (increasing kinetic energy) or to causethe vehicle to climb (increasing potential energy).

Figure 4.1 Force balance for aircraft in steady level flight.

Q14 (PDF)

A. Vehicle Drag

Recall from fluids that drag takes the form shown below, being composed of a parttermedparasitic dragthat increases with the square of the flight velocity, and a partcalled induced drag, or drag due to lift, that decreases in proportion to the inverse ofthe flight velocity.


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Figure 4.2 Components of vehicle drag.

where and

Thus

or

The minimum drag is a condition of interest. We can see that for a given weight, itoccurs at the condition of maximum lift-to-drag ratio

We can find a relationship for the maximum lift-to-drag ratio by setting

from which we find that

and

and


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B. Power Required

Now we can look at the propulsion system requirements to maintain steady levelflight since

Thus the power required (for steady level flight) takes the form

Figure 4.3Typical power required curve for an aircraft.

The velocity for minimum power is obtained by taking the derivative of the equationfor Preq with respect to V and setting it equal to zero.


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As we will see shortly, maximum endurance (time aloft) occurs when the minimumpower is used to maintain steady level flight. Maximum range(distance traveled) isobtained when the aircraft is flown at the most aerodynamically efficient condition(maximum CL/CD).

Homework P4 (PDF)

To see the implications

ofexcess power, visit NASA Glenn -

GO!

C. Aircraft Range, the Breguet Range Equation

Again, for steady, level flight,

The weight of the aircraft changes in response to the fuel burned

or

applying the initial conditions, at t = 0 W = Winitial \ const. = lnWinitial


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the time the aircraft has flown corresponds to the amount of fuel burned, therefore

then multiplying by the flight velocity we arrive at the Breguet Range Equation whichapplies for situations where Isp and flight velocity are constant over the flight.

This can be re-written in other forms:

where

or

NASA Glenn has a great interactive Java

program for aircraft range - GO!


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D. Aircraft Endurance

For a given amount of available fuel energy (Joules), the maximum endurance (timealoft) is obtained at a flight condition corresponding to the minimum rate of energy

expenditure (Joules/second), or Preqmin, as shown in Figure 4.3.We can determine the aerodynamic configuration which provides the minimumenergy expenditure:

so

where

Then

So the minimum power required (maximum endurance) occurs when is amaximum.

With a little algebra we can arrive at an expression for the maximum endurance.Setting

we find that

and


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and

Thus the minimum power (maximum endurance) condition occurs at a speed which is3-1/4 = 76% of the minimum drag (maximum range) condition. The correspondinglift-to-drag ratio is 86.6% of the maximum lift-to-drag ratio.

Figure 4.4 Relationship between condition for maximum endurance and maximumrange.

Continuing

which can be substituted into


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Such that, for maximum endurance

which can be integrated (assuming constant Isp) to yield

Have some interactive

endurance fun at NASA Glenn -

GO!

E. Climbing FlightAny excess in power beyond that required to overcome drag will cause the vehicleincrease kinetic or potential energy. We consider this case by resolving forces aboutthe direction of flight and equating these with accelerations.

Figure 4.5 Force balance for an aircraft in climbing flight.


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where is the accel. normal to theflight path

where is the accel. tangent to the flightpath

So the change in height of the vehicle (the rate of climb, R/C) is:

which is instructive to rewrite in the form

or

in words:

excess power = change in potential energy + change in kinetic energy

Q15 (PDF)

For steady climbing flight,

and the time-to-climb is


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where

for example, and

The power available is a function of the propulsion system, the flight velocity,altitude, etc. Typically it takes a form such as that shown in Figure 4.6. The shortesttime-to-climb occurs at the flight velocity where Pavail Preq is a maximum.

Figure 4.6Typical behavior of power available as a function of flight velocity.


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V. Rocket Performance

A. Thrust and Specific Impulse for Rockets

During the Fall semester thermodynamics lectures we used thesteady flow energyequation to relate the exhaust velocity of a rocket motor to the conditions in thecombustion chamber and the exit pressure

Figure 5.1Schematic of rocket nozzle and combustion chamber

The steady flow energy equation

then with no heat transfer or shaft work

which can be written as

and manipulated to obtain

Then considering the relationship we derived for thrust


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and assuming pe = po = 0

then

and

so

Thus the specific impulse can be directly related to the exhaust velocity leaving therocket

Q16 (PDF)Q17 (PDF)

B. The Rocket Equation

We can now look at the role of specific impulse in setting the performance of arocket. A large fraction (typically 90%) of the mass of a rocket is propellant, thus it is

important to consider the change in mass of the vehicle as it accelerates.Q6 (PDF)


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Figure 5.2Schematic for application of the momentum theorem.

There are several ways to do this through applying conservation of momentum. Here

we will apply the momentum theorem differentially by considering a small mass, dm,expelled from the rocket during time dt.

The initial momentum of the mass in the control volume (the vehicle) is

mvu

The final momentum of mass in the control volume (the vehicle and the massexpelled, dm) is

(mv - dm)(u + du) + dm(u - ue)

= mvu + mv du - udm dudm +udm -uedm

The change in momentum during the interval dt is

= momentumfinal momentuminitial

= mv du - uedm (since dudm is a higher order term)Now consider the forces acting on the system which is composed of the masses m (therocket), and dm (the small amount of propellant expelled from the rocket during timedt):

Applying conservation of momentum, the resulting impulse, SFdt, must balance thechange in momentum of the system.

then since where = propellant mass flow rate


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we have

or for pe = po

The Rocket Equation

The above can be integrated as a function of time to determine the velocity of therocket.

If we set ue = constant, and assume that at t=0, u=0, and neglect drag, and set q = 0,then we arrive at

which can be integrated to give

where is the initial mass of therocket

or

We can view this equation as being similar to the Breguet Range Equation foraircraft. It presents the overall dependence of the principal performance parameter fora rocket (velocity, u), on the efficiency of the propulsion system (Isp), and the

structural design (ratio of total mass to structural mass since the initial mass is thefuel mass plus the structural mass and the final mass. is only the structural mass).

Assuming the rate of fuel consumption is constant, the mass of the rocket varies overtime as


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wheretb is the time at which all of the propellant is used. This expression can be

substituted into the equation for velocity and then integrated to find the height at theend of burnout:

which for a single stage sounding rocket with no drag and constant gravity yields

the final height of the rocket can then be determined by equating the kinetic energy ofthe vehicle at burnout with its change in potential energy between that point and themaximum height. This is left as an exercise for the reader.

Q19 (PDF)

There is a great site on the web that describes

performance analysis for multistage model

rockets - GO!


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Figure 5.3The Saturn V rocket stood 365 feet tall and had 5 stages. It produced over

7.5 million pounds of thrust at liftoff (NAS

A, 1969).


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VI. Rocket Nozzles: Connection of Flow to

Geometry

We have considered the overall performance of a rocket and seen that is directlydependent on the exit velocity of the propellant. Further, we have used the steadyflow energy equation to determine the exhaust velocity using the combustion chamberconditions and the nozzle exit pressure. In this brief section, we will apply conceptsfrom thermodynamics and fluids to relate geometrical (design) parameters for a rocketnozzle to the exhaust velocity.

We will make the following assumptions:

1. The propellant gas obeys the perfect gas law2. The specific heat is constant3. The flow in the nozzle is one-dimensional4. There are no losses due to friction5. There is no heat transfer6. The flow velocity in the combustion chamber is negligible (zero)7. The flow is steady

A. Quasi-one-dimensional compressible flow in a variable

area duct

p = rRT (ideal gas)

(isentropic flow)

(energy equation)

imply that


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Then from conservation of mass

(cons. of mass)

The above equation relates the flow area, the mass flow, the Mach number and thestagnation conditions. It is frequently rewritten in a non-dimensional form bydividing through by the value at M=1 (where the area at M=1 is A*):

which takes a form something like that shown in Figure 6.1 below

Figure 6.1 General form of relationship between flow area and Mach number.

See this interactively at

NASA Glenn - GO!

B. Thrust in Terms of Nozzle Geometry

We can use these equations to rewrite our expression for rocket thrust in terms ofnozzle geometry (throat area = A*, and exit area Ae).


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evaluate at M = 1 (throat)

We can now specify geometry (A* and Ae) to determine Me. Then use Me with thecombustion chamber conditions to determine thrust and Isp.


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VII. Production of Thrust with a Propeller

A. Overview of propeller performance

Each propeller blade is a rotating airfoil which produces lift and drag, and because ofa (complex helical) trailing vortex system has an induced upwash and an induceddownwash.


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Figure 7.1 Adapted from McCormick, 1979.

The two quantities of interest are the thrust (T) and the torque (Q). We can write

expressions for these for a small radial element (dr) on one of the blades:

where

andIt is possible to integrate the relationships as a function of r with the appropriate liftand drag coefficients for the local airfoil shape, but determining the induced upwash(ai) is difficult because of the complex helical nature of the trailing vortex system. Inorder to learn about the details of propeller design, it is necessary to do this.However, for our purposes, we can learn a about the overall performance featuresusing the integral momentum theorem, some further approximations called actuatordisk theory, and dimensional analysis.

NASA Glenn has a nice

explanation of propeller

thrust- GO!

B. Application of the Integral Momentum Theorem to

Propellers


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Figure 7.2 Adapted from McCormick, 1979

The control volume shown in Figure 7.2 has been drawn far enough from the deviceso that the pressure is everywhere equal to a constant. This is not required, but itmakes it more convenient to apply the integral momentum theorem. We will alsoassume that the flow outside of the propeller streamtube does not have any change intotal pressure. Then since the flow is steady we apply:

Since the pressure forces everywhere are balanced, then the only force on the controlvolume is due to the change in momentum flux across its boundaries. Thus byinspection, we can say that

Or we can arrive at the same result in a step-by-step manner as we did for the jetengine example in Section II:


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Note that the last term is identically equal to zero by conservation of mass. If themass flow in and out of the propeller streamtube are the same (as we have defined),then the net mass flux into the rest of the control volume must also be zero.

So we have:

as we reasoned before.

The power expended is equal to the power imparted to the fluid which is the change inkinetic energy of the flow as it passes through the propeller

The propulsive power is the rate at which useful work is done which is the thrustmultiplied by the flight velocity

The propulsive efficiency is then the ratio of these two:

Which is the same expression as we arrived at before for the jet engine (as you mighthave expected).


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C. Actuator Disk Theory

To understand more about the performance of propellers, and to relate thisperformance to simple design parameters, we will apply actuator disk theory. Wemodel the flow through the propeller as shown in Figure 7.3 below and make thefollowing assumptions:

y Neglect rotation imparted to the flow.y Assume the Mach number is low so that the flow behaves as an

incompressible fluid.

y Assume the flow outside the propeller streamtube has constant stagnationpressure (no work is imparted to it).

y Assume that the flow is steady. Smear out the moving blades so they are onethin steady disk that has approximately the same effect on the flow as the

moving blades (the actuator disk).y Across the actuator disk, assume that the pressure changes discontinuously,

but the velocity varies in a continuous manner.


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Figure 7.3 Adapted from Kerrebrock.

We then take a control volume around the disk as shown in Figure 7.4

Figure 7.4 Control volume around actuator disk.

The force, T, on the disk is

So the power is

We also know that the power is

Thus we see that the velocity at the disk is

Half of the axial velocity change occurs upstream of the disk and half occursdownstream of the disk.


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We can now find the pressure upstream and downstream of the disk by applying theBernoulli equation in the regions of the flow where the pressure and velocity arevarying continuously.

and

From which we can determine

We generally dont measure or control udiskdirectly. Therefore, it is more useful towrite our expressions in terms of flight velocity uo, thrust, T, (which must equal dragfor steady level flight) and propeller disk area, A

disk.

So

From which we can obtain an expression for the exit velocity in terms of thrust andflight velocity which are vehicle parameters

The other parameters of interest become


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This is the ideal (minimum) power required to drive the propeller. In general , theactual power required would be about 15% greater than this.

There are several important trends that are apparent upon consideration of theseequations. We see that the propulsive efficiency is zero when the flight velocity iszero (no useful work, just a force), and tends towards one when the flight velocityincreases. In practice, the propulsive efficiency typically peaks at a level of around0.8 for a propeller before various aerodynamic effects act to decay its performance aswill be shown in the following section.

D. Dimensional AnalysisWe will now use dimensional analysis to arrive at a few important parameters for thedesign and choice of a propeller. Dimensional analysis leads to a number ofcoefficients which are useful for presenting performance data for propellers.

Parameter Symbol Units propeller diameter D m

propeller speed n rev/storque Q Nm

thrustT

Nfluid density r kg/m3fluid viscosity n m2/s

fluid bulk elasticity modulus K N/m2flight velocity uo m/s

1. Thrust Coefficient


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T = f(D; n; r; n; K; uo) = Const. Da nb rc nd Ke uo

f

Then putting this in dimensional form

[MLT-2] = [(L)a(T)-b(ML-3)c(L2T-1)d(ML-1T-2)e(LT-1)f]

Which implies

(M) 1= c+ e

(L) 1 = a - 3c + 2d e + f

(T) 2 = b + d + 2e + f

So

a = 4 2e 2d f

b = 2 d 2e f

c = 1 - e

T = Const. D4-2e-2d-fn2-d-2e-fr1-e nd Ke uof

We can now consider the three terms in the square brackets

: Dn is proportional to the tip speed, so this term is like

: K/r = a2 where a is the speed of sound, this is like

: uo/n is the distance advanced by the propeller in one revolution, here non-dimensionalized by the propeller diameter.

This last coefficient is typically called the advance ratio and given the symbol J.


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Thus we see that the thrust may be written as

which is often expressed as

wherekT is called the thrust coefficient andin general is a function of propeller design, Re, Mtip and J.

2. Torque Coefficient

We can follow the same steps to arrive at a relevant expression and functional

dependence for the torque or apply physical reasoning. Since torque is a forcemultiplied by a length, it follows that

wherekQ is called the thrust coefficientand in general is a function of propeller design, Re, Mtip and J.

3. Efficiency

The power supplied to the propeller is P in where

The useful power output is Pout where

Therefore the efficiency is given by

4. Power Coefficient

The power required to drive the propeller is


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which is often written using a power coefficient Cp = 2pkQ

then

E. Typical propeller performance

Typical propellor performance curves can be found in McCormick, B.W.Aerodynamics, Aeronautics, and Flight Mechanics. John Wiley & Sons Inc. 1979.


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VIII. Ideal Cycle Analysis of Aircraft Gas

Turbine Engines

A. Introduction

In thermodynamics we represented a gas turbine engine using a Brayton cycle, asshown in Figure 8.1, and derived expressions for efficiency and work as functions ofthe temperature at various points in the cycle. In this section we will perform idealcycle analysis, which is a method for expressing the thrust and thermal efficiency ofengines in terms ofuseful design variables. If you take a further course in propulsion,this ideal cycle analysis will be extended to take account of various inefficiencies inthe different components of the enginethat type of analysis is called non-ideal cycleanalysis.

Figure 8.1Schematic of a Brayton cycle.

1. Objective of ideal cycle analysis

Our objective is to express thrust, T, and thermal efficiency, h (or alternatively I)

as functions of 1) typical design limiters, 2) flight conditions, and 3) design

choices so that we can analyze the performance of various engines. The

expressions will allow us to define a particular mission and then determine theoptimum component characteristics (e.g. compressor, combustor, turbine) for anengine for a given mission. Note that ideal cycle analysis addresses only thethermodynamics of the airflow within the engine. It does not describe the details ofthe components (the blading, the rotational speed, etc.), but only the results thevarious components produce (e.g. pressure ratios, temperature ratios). In Section IX


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we will look in greater detail at how some of the components (the turbine and thecompressor) produce these effects.

Q21 (PDF)Q22 (PDF)

2. Notation and station numbering

Figure 8.2 Gas turbine engine station numbering.

Notation:

( )

Stagnation properties, TT & PT, are more easily measured quantities than staticproperties (T and p). Thus, it is standard convention to express the performance ofvarious components in terms of stagnation pressure and temperature ratios:


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p total or stagnation pressure ratio across component (d, c, b, t, a, n)

t total or stagnation temperature ratio across component (d, c, b, t, a, n)

where d= diffuser (or inlet), c= compressor, b= burner (or combustor), t= turbine, a=

afterburner, and n=nozzle.

3. Ideal Assumptions

1) Inlet/Diffuser: pd = 1, td = 1 (adiabatic, isentropic)

2) Compressor or fan: tc = pcg-1/g , tf= pf

g-1/g

3) Combustor/burner or afterburner: pb = 1, pa = 1

4) Turbine: tt = ptg-1/g

5) Nozzle: pn = 1, pn = 1

Check out the Thermo notes for a refresher

on cycle analysis done in the fall - GO!

B. Ideal Cycle Analysis Example: Turbojet Engine

Figure 8.3Schematic with appropriate component notations added.


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Methodology:

1) Find thrust by finding uexit/uo in terms of qo, temperature ratios, etc.

2) Use a power balance to relate turbine parameters to compressor parameters

3) Use an energy balance across the combustor to relate the combustor temperaturerise to the fuel flow rate and fuel energy content.

First write-out the expressions for thrust and I:

where f is the fuel/air mass flow ratio

(can neglect fuel)

That is the easy part! Now we have to do a little algebra to manipulate theseexpressions into more useful forms.

First we write an expression for the exit velocity:

Noting that

We can write


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Thus

(*)

Which expresses the exit temperature as a function of the inlet temperature, the Machnumber, and the temperature changes across each component. Since we will use thisexpression again later we will mark it with an asterisk (*).

We now write the pressure at the exit in a similar manner:

and then equate this to our expression for the temperature (*)

(**)

and again label it (**) for use later as we do the following expression:


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(***)

We now continue on the path to our expression for u7/uo.

Therefore

Now we have two steps left. First we write tc in terms of tt, by noting that they arerelated by the condition that the power used by the compressor is equal to the powerextracted by the turbine. Second, we put the burner temperature ratio in terms of the

exit temperature of the burner, (TT4 or more specifically qt=TT4/To) since this is thehottest point in the engine and is a frequent benchmark used for judging variousdesigns.

The steady flow energy equation tells us that

Assuming that the compressor and turbine are adiabatic, then

= - rate of shaft work done by the system = rate of shaft work done on thesystem

Since the turbine shaft is connected to the compressor shaft

assuming and Cp are the same


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This can be rewritten as

where

so

or

That was the first step relating the temperature rise across the turbine to that acrossthe compressor. The remaining step is to write the temperature rise across thecombustor in terms of qt=TT4/To.

and for an engine with an afterburner

Now substituting our expressions for tb, and tt into our expression for u7/uo, and finally

into the first expression we wrote for thrust, we get:

Specific thrust for a turbojet

which is what we were seeking, an expression for thrust in terms of important designparameters and flight parameters:

With algebra

We may write this write in another form which is


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often used

Our final step involves writing the specific impulse and other measures ofefficiency in terms of these same parameters. We begin by writing the First Lawacross the combustor to relate the fuel flow rate and heating value of the fuel to thetotal enthalpy rise.

and

where again, f is the fuel/air mass flow ratio

So the specific impulse becomes

Specific Impulse for an ideal turbojet

where I is expressed in terms of typical design parameters, flight conditions, andphysical constants

Similarly, we can write our overall efficiency, hoverall


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or

The ideal thermal efficiency is

and the propulsive efficiency can be found from hprop = hoverall / hthermal

We can now use these equations to better understand the performance of a simpleturbojet engine. We will use the following parameters (with g=1.4):

Mach number Altitude Ambient Temp. Speed of sound0 Sea level 288K 340m/s

0.85 12km 217K 295m/s2.0 18km 217K 295m/s

Stanford has a great standard atmosphere calculator

that can be used to find conditions at any altitude -

GO!

Note it is more typical to work with the compressor pressure ratio (pc) rather than thetemperature ratio (tc) so we will substitute the isentropic relationship:

tc = pcg-1/g

into the equations before plotting the results.


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Figure 8.4 Performance of an ideal turbojet engine as a function of flight Mach

number.


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Figure 8.5 Performance of an ideal turbojet engine as a function of compressorpressure ratio and flight Mach number.


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Figure 8.6 Performance of an ideal turbojet engine as a function of compressorpressure ratio and turbine inlet temperature.

Homework P7 (PDF)


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IX. Energy Exchange with Moving Blades

A. Introduction

So far we have only looked at the thermodynamic results of compressors and turbines(ps and ts). Now we will look in more detail at how the components of a gas turbineengine produce these effects. You will learn later (in 16.50 for example) that withoutheat transfer, it is only possible to change the total enthalpy of a fluid with anunsteady process (e.g. moving blades). Still we will use many of the steady flow toolsthat we have discussed in thermodynamics and propulsion by considering the steadyflow in and out of a component as shown in Figure 8.1.

Figure 9.1 Control volume around compressor or turbine.

B. The Euler Turbine Equation

The Euler turbine equation relates the power added to or removed from the flow, tocharacteristics of a rotating blade row. The equation is based on the conceptsofconservation of angular momentum and conservation of energy. We will workwith the following model of the blade row:

Figure 9.2 Control volume for EulerTurbine Equation.


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Applying conservation of angular momentum, we note that the torque, T, must beequal to the time rate of change of angular momentum in a streamtube that flowsthrough the device (In the text: For more information about angular momentum androtational energy, see pages 246 and 558 in Hibbler).

Q31 (PDF)

This is true whether the blade row is rotating or not. The sign matters (i.e. angularmomentum is a vector -- positive means it is spinning in one direction, negativemeans it is spinning in the other direction). So depending on how things are defined,there can be positive and negative torques, and positive and negative angularmomentum. In Figure 9.2, torque is positive when Vtangential out >Vtangential in ---- thesame sense as the angular velocity.

If the blade row is moving, then work is done on/by the fluid. The work per unit time,or power, P, is the torque multiplied by the angular velocity, w

If torque and angular velocity are of like sign, work is being done on the fluid (acompressor). If torque and angular velocity are of opposite sign work is beingextracted from the fluid (a turbine). Here is another approach to the same idea:

y If the tangential velocity increases across a blade row (where positivetangential velocity is defined in the same direction as the rotor motion) thenwork is added to the flow (a compressor).

y If the tangential velocity decreases across a blade row (where positivetangential velocity is defined in the same direction as the rotor motion) thenwork is removed from the flow (a turbine).

From the steady flow energy equation

with

Then equating this expression of conservation of energy with our expression fromconservation of angular momentum, we arrive at:


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or for a perfect gas with Cp = constant

Which is called the Euler Turbine Equation. It relates the temperature ratio (andhence the pressure ratio) across a turbine or compressor to the rotational speed and thechange in momentum per unit mass. Note that the velocities used in this equation arewhat we will later call absolute frame velocities (as opposed to relative framevelocities).

yIf angular momentum increases across a blade row, then

TTc >TT

b and workwas done on the fluid (a compressor).y If angular momentum decreases across a blade row, then TTc


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Q24 (PDF)

C. Multistage Axial Compressors

An axial compressor is typically made up of many alternating rows of rotating andstationary blades called rotors and stators, respectively, as shown in Figures 9.3 and9.4. The first stationary row (which comes in front of the rotor) is typically called theinlet guide vanes or IGV Each successive rotor-stator pair is called a compressorstage. Hence compressors with many blade rows are termed multistage compressors.

Figure 9.3 A typical multistage axial flow compressor. Copyright Rolls-Royce plc.Reproduced with the kind permission of Rolls-Royce plc.

Figure 9.4Schematic representation of an axial flow compressor.


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One way to understand the workings of a compressor is to consider energyexchanges. We can get an approximate picture of this using the Bernoulli Equation,where PT is the stagnation pressure, a measure of the total energy carried in the flow,

p is the static pressure a measure of the internal energy, and the velocity terms are ameasure of the kinetic energy associated with each component of velocity (u is radial,v is tangential, w is axial).

The rotor adds swirl to the flow, thus increasing the total energy carried in the flow byincreasing the angular momentum (adding to the kinetic energy associated with thetangential or swirl velocity, 1/2rv2).

The stator removes swirl from the flow, but it is not a moving blade row and thuscannot add any net energy to the flow. Rather, the stator rather converts the kineticenergy associated with swirl to internal energy (raising the static pressure of theflow). Thus typical velocity and pressure profiles through a multistage axialcompressor look like those shown in Figure 9.5.

Figure 9.5 Pressure and velocity profiles through a multi-stage axial compressor.Copyright Rolls-Royce plc. Reproduced with the kind permission of Rolls-Royce plc.


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Note that the IGV also adds no energy to the flow. It is designed to add swirl in thedirection of rotor motion to lower the Mach number of the flow relative to the rotor

blades, and thus improve the aerodynamic performance of the rotor.

D. Velocity Triangles for an Axial Compressor Stage

Velocity triangles are typically used to relate the flow properties and blade designparameters in the relative frame (rotating with the moving blades), to the propertiesin the stationary orabsolute frame.

We begin by unwrapping the compressor. That is, we take a cutting plane at aparticular radius (e.g. as shown in Figure 9.3) and unwrap it azimuthally to arrive atthe diagrams shown in Figure 9.6. Here we have assumed that the area of the annulusthrough which the flow passes is nearly constant and the density changes are small sothat the axial velocity is approximately constant.

Figure 9.6 Velocity triangles for an axial compressor stage. Primed quantities are inthe relative frame, unprimed quantities are in the absolute frame.


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In drawing these velocity diagrams it is important to note that the flow typically

leaves the trailing edges of the blades at approximately the trailing edge angle in

the coordinate frame attached to the blade (i.e. relative frame for the rotor,

absolute frame for the stator).

Q25 (PDF)Q30 (PDF)

Interactive program for calculating

velocity triangles (by Rodin Lyasoff)-

GO!

[There may be a delay while Javaloads.]

We will now write the EulerTurbine Equation in terms of stage design parameters: w,the rotational speed, and bb and bc the leaving angles of the blades.

From geometry,

vb = wb tan bb and vc = wc tan bc = wrc - wc tan

so

or


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So we see that the total or stagnation temperature rise across the stage increases withthe tip Mach number squared, and for fixed positive blade angles, decreases withincreasing mass flow. This behavior is represented schematically in Figure 9.7.

Figure 9.7 Compressor behavior

Homework P8 (PDF)

E. Velocity Triangles for an Axial Flow Turbine Stage

We can apply the same analysis techniques to a turbine. Again, the stator does no

work. It adds swirl to the flow, converting internal energy into kinetic energy.T

heturbine rotor then extracts work from the flow by removing the kinetic energyassociated with the swirl velocity.


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Figure 9.8Schematic of an axial flow turbine.

The appropriate velocity triangles are shown in Figure 9.9, where again the axialvelocity was assumed to be constant for purposes of illustration.

As we did for the compressor, we can write the EulerTurbine Equation in terms ofuseful design variables:


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Figure 9.9 Velocity triangles for an axial flow turbine stage.

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