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INTRODUCTORY ENGINEERING NOTESY. Dai, J. Waranyuwat, A. Burkit

1. Aerospace Engineering

1.1. Introduction. Aerospace Engineering is a branch of engineering that deals with design, buildingand testing of aircraft and spacecraft. It is divided into several subdisciplines, from which the mainones are listed below:

(1) Aerodynamics(2) Propulsion(3) Structural technology(4) Materials technology(5) Stability and Control

At UIUC, all of these areas of specialization are covered during the course of an AE Baccalaureateprogram, which gives students an opportunity to get introduced to each of these disciplines and tochoose the area of specialization they want to pursue.

1.2. Aerodynamics.

1.2.1. Introduction. Aerodynamics is a subset of Fluid Dynamics and is a study of the flow of airaround ob jects. Aerodynamics is applied in a variety of industries, ranging from aircraft and automo-bile design to wind turbine and bicycle equipment design. Since we will study aerospace engineering,we will be primarily concerned with aerodynamics of aircraft and spacecraft.

Aerodynamics explains the physics behind the flight, defines the forces acting on an aircraft andenables us to find out performance parameters of the aircraft and flow parameters of the flow aroundit. Aerodynamics relies on governing laws of physics, such as principle of mass conservation, Newton’ssecond law and principle of energy conservation.

We will cover the following topics in this section:

• Wing terminology• Principle of continuity• Bernoulli’s equation• Lift and Drag, Thrust and Weight• Performance parameters• Flow similarity parameters

1.2.2. Wing terminology. When deriving equations later in this section, we need to be familiar withwing terminology.

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In the figure above, a top view of the wing is on the left, and a side view of the wing is on the right.A side view of the wing shows a side cut of the wing, which is called an airfoil.

Aerospace engineers utilize the following conventions:b - wingspanS - wing area

LE - leading edge, the edge of the wing going in the flow of air firstT E - trailing edge, the edge of the wing that the air reaches lastc - chordα - angle of attack

Question 1.2.1 Aspect ratio is a ratio of the square of the wing span to the wing area, or, AR = b2

S .

It is a useful wing parameter that allows to understand the purpose of the aircraft. F-16 fighter jet has wingspan of 32.8 ft and wing area of 300 sq.ft. Boeing Phantom Eye HALE aircraft has 150 ft wingspan and approximately 4 ft chord length. You can assume a rectangular wing. Calculate the aspect ratios of both these aircraft and think about the reasons for why they are different .

1.2.3. Principle of continuity. In aerodynamics, the flow of air is considered a continuum flow sincethe scale at which aerodynamics studies the forces and stresses around the aircraft is much largerthan the scale of a separate air molecule. That is, aerodynamics does not study forces between themolecules, but rather the forces between the air as a whole and the aircraft, particularly its wings.Thus, the air can be considered a continuous mass.

Another assumption that is made in aerodynamics to simplify the study of flowing air is a notion of acontrol volume . The control volume, or V , is an imaginary volume that encloses a finite volume of air and in which the laws of aerodynamics can be applied. Generally, it is assumed that the controlvolume is fixed in space and the air flows through the control volume. It makes it easier to derive andapply the equations of aerodynamics. The surface of the control volume is called a control surface,and is denoted as S . The principle of continuity is derived from the principle of mass conservation,that states that mass is conserved. That is, the mass that enters the control volume should be thesame mass that exits it. In equation form, the principle of continuity looks as follows:

(1) ∂ ∂t

V

ρdV + S

ρv · d S = 0

where V is the control volume of interest, S is the surface of this control volume, ρ is the density of air, and v is velocity of the flow through the control surface. The first integral in the equation denotesthe rate of change of mass inside the control volume and the second integral is the net amount of massthat flows in and out of the control volume through its surfaces. Equation (1) is called the continuity

equation.

Note that density, ρ, of the flow can change with time, that is why it is inside the time derivative. If density is not changing with time, the flow is said to be incompressible, otherwise it is compress-

ible. Generally, compressible flow occurs at high speeds of air flow. In this section we will assumeincompressible flow in order to derive a simplified version of the continuity equation.

So, assuming that density is constant, and that the control volume does not change with time, thefirst integral becomes zero, and that leaves

S

ρv · d S = 0.

Assuming a control volume like in Figure 1, where the air flow cannot exit through upper, lower orside surfaces, we can derive by calculating the dot product between the velocity and the surface area:

(2) ρ1A1v1 = ρ2A2v2

where A1 and A2 are respective surfaces of the control volume. Since we assumed that the flow isincompressible, ρ1 = ρ2, and we know that A1 = A2.

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Figure 1. Control volume.

Hence, v1 = v2 from equation (2). This means that the speed with which the air enters the controlvolume as shown in Figure 1, will remain the same when the air exits this control volume.

Now define mass flow rate as:

(3) m = ρAv

where m is the mass flow rate, that is, the change of mass per unit time. Thus, restating equation (2):

(4) m1 = m2

Now, if we assume an airfoil in a wind tunnel, as shown in Figure 2, and define a control volume asthe volume inside the wind tunnel from inlet 1 to outlet 3, the principle of continuity will still hold,yielding:

(5) m1 = m3

(6) ρ1A1v1 = ρ3A3v3

Figure 2. Airfoil in a wind tunnel (schematic).

And assuming incompressible flow:(7) A1v1 = A3v3

Since we know that A1 > A3, we conclude that v1 < v3. That is, the speed of the air flow in a narrowsection of the wind tunnel is greater than that in a wide section.

Question 1.2.2 Consider a convergent duct with an inlet area A1 = 5 m 2. Air enters this duct with velocity v 1 = 10 m/s and leaves the duct exit with a velocity v 2 = 30 m/s. Assuming incompressible flow, what is the area of the duct exit?

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1.2.4. Bernoulli’s equation. Bernoulli’s equation is derived from Newton’s second law, that states thatforce = mass × acceleration, or more generally, force is equal to time rate of change of momentum.That is,

(8) F = ∂

∂t(mv)

Applying it to a control volume, we obtain:

(9) ∂

∂t

V

ρvdV +

S

(ρv · d S )v = −

S

pd S +

V

ρ f dV + F viscous

where p is the pressure exerted on the surfaces of the control volume, f is the body force on the airinside the control volume, and F viscous are the forces resulting from friction on the surfaces of controlvolume. Equation (9) is called the momentum equation in integral form. When expanded to x, yand z coordinates, the previous equation becomes a set of differential equations. This set of equationsis called the momentum equation in differential form. We present the equation for only one of the dimensions: x-direction, since the other two are analogous.

(10) ∂ (ρu)

∂t + ∇ · (ρu)v = −

∂p

∂x + ρf x + (F x)viscous

where u is the velocity component in x-direction, and∇ is gradient, or ∂u

∂x + ∂u

∂y + ∂u

∂z, and only ∂u

∂x is non-zero.

In our case, we assume an inviscid flow, which makes (F x)viscous = 0. Moreover, we can assumethat the body force is negligibly small compared to pressure forces acting on the surfaces, so we can

ignore body forces, that is f x = 0. Also, as before, the flow is incompressible, so ∂ (ρu)∂t becomes

zero. Integrating u ∂u∂x

and ∂ p, and getting a constant from integrating 0 on the right hand side of theequation, we obtain:

(11) 1

2ρu2 + p = const = p0

or more generally,

(12) 1

2ρv2 + pstatic = ptotal

Equation (12) is called Bernoulli’s equation, and states that the total pressure at a point in thecontrol volume equals to the sum of static and dynamic pressures, where

(13) q = 1/2ρv2

is dynamic pressure.

For our control volume,

(14) pt1 = ps1 + q 1 = ps3 + q 3 = pt3

Now, using the previous conclusion that v1 < v3, and knowing from Bernoulli’s equation that pt1 = pt3,we can conclude that when velocity increases at point 3 (i.e. q 3 increases), the static pressure ps3decreases to conserve the total pressure. Thus, the static pressure above the airfoil is less than thestatic pressure below it, which will induce an upward force, called lift.

Question 1.2.3 Consider an airfoil in a flow of air, where far ahead of the airfoil, the pressure,density and velocity are 1.01×10 5 N/m 2, 1.225 kg/m 3, and 150 km/h respectively. At a given point Aon the airfoil, the pressure is 9.95 ×10 4 N/m 2. What is the velocity at point A?

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1.2.5. Lift and Drag, Thrust and Weight. Lift is an aerodynamic force that enables the aircraft tofly. Lift results from pressure and stress distribution on the airfoil. The other aerodynamic forcethat results from pressure distribution on the airfoil is drag. Drag opposes the motion of the aircraftforward. For a well designed airfoil, drag is much less than lift. The force that enables the motionof the aircraft forward is called thrust. It is produced by engines of the aircraft. And the last main

force acting on the aircraft is weight. The forces acting on the are summarized in Figure 3 below.

Figure 3. Main aerodynamic forces acting on aircraft.

1.2.6. Performance parameters. Lift can be calculated using formula:

(15) L = qScL

where L - lift force,q - dynamic pressure,S - wing area,and cL - lift coefficient . It is a dimensionless number, and is an important performance parameterin aircraft design.

Drag is calculated using a similar formula:

(16) D = qScD

where D - drag force,cD - drag coefficient, and is also an important dimensionless performance parameter.

As mentioned earlier, for a well designed airfoil cL >> cD, that is drag coefficient is usually an orderof magnitude less than lift coefficient. Lift and drag coefficients allow us to evaluate the performanceof different airfoils without knowing their shapes or sizes. These coefficients are one of the maindeliverables in aircraft design reports.

Question 1.2.4 For a Boeing 747-200 in cruise flight at an altitude of 40,000 ft with the wing area of 5,500 ft 2, the following flight conditions have been reported:air density = 0.00058735 slug/ft 3,true airspeed = 871 ft/s,lift coefficient = 0.52,drag coefficient = 0.022.Calculate the lift force on the aircraft. What is the maximum weight at takeoff for Boeing 747-200? What is the required thrust power?

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1.2.7. Flow similarity parameters. When tests on designing new airfoils are conducted, they take placein wind tunnels, in which the sizes of wing models are much smaller than the full size aircraft wings.What enables aerospace engineers to use the results from wind tunnel tests to predict full size wingperformance? The flow similarity parameters. There are many parameters that are reported fromwind tunnel tests, but we will focus on two main ones in this section. They are Mach and Reynolds

numbers. These numbers are also dimensionless, but can deliver plenty of information about the flowof air around the wing. These numbers are used to mimic the conditions under which the wind tunneltests were performed. Moreover, they can predict the performance of a full size model based on thetests on the scaled model.

Mach number is defined as the ratio of the speed of the air flow to the speed of sound.

(17) M = v

a

where M - Mach number,v - speed of air flow,a - speed of sound at that attitude.

The speed of sound varies with altitude, according to a formula:(18) a =

γRT

where γ - specific heat, (usually equal to 1.4),R - universal gas constant, equal to 287 J/kg/K (1716 ft-lb/slug/oR),T - temperature at the altitude of interest.

Depending on Mach number, the flow can be:

• Subsonic M < 1• Transonic 0.8 < M < 1.2• Supersonic M > 1• Hypersonic M > 5

Reynolds number is a measure of the viscosity of the flow and is defined as:

(19) Re = ρvc

µ

where Re - Reynolds number,ρ - density of the flow,v - speed of the flow,c - chord length,µ - viscosity of the flow. Reynolds number is usually of the magnitude of 106 for an inviscid flow, andabout 104 and lower for a viscous flow.

Question 1.2.5 An aircraft flies at speed of 800 km/h at an altitude of 35,000 ft. The temperature of the surroundings is -51oC. What is the Mach number at which the aircraft is flying?

Lift and drag coefficients depend on both Mach and Reynolds numbers and also on an angle of attack,i.e. cL = f (M,Re,α) and cD = g(M,Re,α), and the exact expresisons for cL and cD are difficultto obtain. However, during wind tunnel tests, the values of lift and drag forces can be obtainedexperimentally, and the lift and drag coefficients can be calculated from Equations (15) and (16)respectively. A visual representation of how lift coefficient varies with angle of attack as shown inFigure 4 below:

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Figure 4. Angle of attack vs. lift coefficient

It can be seen that lift coefficient increases as the angle of attack increases, however only to somepoint. That point is called a critical angle of attack, or αstall. After this point, stall occurs, that isno more lift can be generated at that angle of attack. This is a dangerous condition for aircraft flight,that is why there are plenty of instruments onboard the aircraft that warn the pilot of the approachingof the angle of attack that will cause stalling.

Question 1.2.6 If the section lift acting on a two-dimensional wing of chord 2 m, flying at 250 km/h in sea level altitude is 3000 N/m, when the angle of attack is 4o and section lift curve slope is 0.11,determine the zero lift angle of attack of the wing.

*Note that ρ=1.225 kg/m 3 at sea level altitude and lift coefficient varies with angle of attack ac-cording to cL = k ∗ (α − α0), where k - slope of lift curve, α - angle of attack (in degrees), α0 - zerolift angle of attack (in degrees).

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1.3. Propulsion.

1.3.1. Introduction. Propulsion is what enables the aircraft and spacecraft propel forward. Most of thepropulsion systems work on the same principle that will be discussed later in this section. Propulsionsystems are divided into two categories, the ones typically used in aircraft and the ones used in rockets.The engines that use air as their working fluid are called air breathing engines. Rocket engines, onthe other hand, work on chemical reaction principles.

The basic parts of a typical airbreathing engine are shown in Figure 5 below.

Figure 5. Main parts of an air breathing engine.

The air enters the engine through the inlet and goes into compressor, where the pressure and thetemperature of the air is increased. Next, this air gets into a combustion chamber, where the fuelis added and starts burning under high-pressure high-temperature conditions. Then, this mixtureproceeds to a turbine section of the engine, where it cools and enables the turning of the turbine

blades, that enable the turning of the shaft that is connected to the compressor blades. After that theair exits through the nozzle and creates momentum on the aircraft.

Newton’s third law that states that for every action there is an equal and opposite reaction. Thatmeans, the force generated by the air exiting the engine will act on the aircraft in the opposite direction,that is forward. Thus, propulsive force is generated on an aircraft.

1.3.2. Basic Equations on Thrust. Thrust equations will be derived using a generic propulsive deviceas described below:

Figure 6. Generic schematic of a propulsive device.8



The reference frame is placed on the propulsion device moving at a constant speed of V 1 traveling inthe X direction. The purpose of this propulsion device is to produce thrust T , acting to the left of thedevice. The thrust produced is a result of the net pressure and shear forces distributions acting onthe surface areas at each point where air contacts the propulsive device. Using Newton’s third law,the propulsion device will exert on the air an equal and opposite force T acting to the right.

The propulsion device will be bound by a control volume. Air will enter the control volume at inlet 1and leave at outlet 2, as seen in the figure. The force T that the propulsion device exerts on the aircan be found using Newton’s second law, which can be expressed as follows:

(20)

F = ∂

∂t

V

vρdV +

S

(vρv) · d A

The above equation can be simplified using the following assumptions:

(1) The flow is steady inside the propulsion device.(2) The effects of pressure and shear stress acting on the side ”walls” of the stream tube are

ignored.

Using the above assumptions and taking into account that the flow travels in the X-direction, theprevious equation becomes:

(21) T + p1A1 − p2A2 = m2v2 − m1v1

where m = mass flow rate p = pressurev= air velocityA = areas of the inlet and outlet sections of the control volume

Remembering the conservation of mass principle, m1 = m2 + mfuel, where mfuel is the mass flow rateof fuel that is added in the combustion chamber. However, the fuel rate can be neglected, since it

is relatively low compared to the air flow rate. Thus, we can conclude that m = m1 = m2 and thethrust equation becomes:

(22) T = m(v2 − v1) + p2A2 − p1A1

In rocket engine propulsion systems, air will never enter the system. Hence, the air mass flow rateat the inlet m1 = 0. That makes the thrust produced by a rocket equal to:

(23) T = m2v2 + p2A2 − p1A1

Question 1.3.1 A jet engine on an airplane moving at 950 km/h has an exit area of 0.5 m 2. If the gas exiting the engine has a speed of 1950 km/h (with respect to the engine) and a density of 0.5 kg/m 3, how much thrust does the engine produce?

If the airplane has two identical engines producing the same thrust, what is the total drag on the airplane? You may assume equal cross sections for the inlet and the outlet of the engine.

1.3.3. Turbofans, Turboprops, and Turbojets. Airbreathing engines can be grouped into three cate-gories:

• Turbofan• Turboprop• Turbojet

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All of these categories of engines have a common component called core. The core is composed of thecompressor, the combustion chamber, and the turbine.

In a turbofan, thrust is produced by a fan that is located inside the casing, called duct. This fan isdriven by the shaft connected to the turbine, as explained in the beginning of the section. A basic

schematic of the turbofan is shown in below:

Figure 7. Turbofan.

In a turboprop, thrust is produced by a propeller that is located outside of the engine casing, in frontof the core. The propeller is also driven by the shaft connected to the turbine. A basic schematic of the turboprop is shown below:

Figure 8. Turboprop.

In a turbojet, the thrust is produced by a high-speed jet resulting from the expansion of high-pressurehot gas, coming from the turbine to the nozzle. A basic schematic of a turbojet engine is shown below:

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Figure 9. Turbojet.

In both turbofan and turboprop engines, the large flow rate of air at the exit of the nozzle has speedsonly slightly faster than the aircraft speed. In turbojet, the rate of air flow is faster, but the amountof flow is smaller. So, in turbofans and turboprops the smaller velocity increment of the flow leads toa higher propulsion efficiency. Thus, it can be concluded that: Turbofan and turboprop engines have higher propulsion efficiency and, in turn, a better fuel efficiency than turbojets.

1.3.4. Afterburners. During short periods of time during take-off, climb, and high thrust demandingmaneuvers of military aircraft, additional thrust may be required. This demand in thrust is providedby afterburners. The figure below shows the location of the afterburners in the engine.

Figure 10. Afterburners schematic.

Afterburners work by providing an additional increase to the temperature and pressure of air thatcomes out of the turbine. The fuel is injected into this air by use of additional nozzles and is burnedusing an igniter. Thus, the needed thrust is increased.

1.3.5. Ramjet. A special type of jet engine is a ramjet. It does not have any moving parts, as can beseen from its schematic:

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Figure 11. Ramjet.

In a ramjet, the air from the inlet is compressed by the forward motion of the engine ramming the airthrough the engine. Hence the term, ramjet.

Ramjets are capable of operating at subsonic speeds, but are inefficient. The best efficiency is achieved

at speeds ranging from Mach 3 to Mach 5. Above Mach 5, the ramjet efficiency decreases due to highlosses in the compression cycle. In order to have an efficient propulsion system at these speeds,supersonic combustion needs to take place. Ramjets that operate under supersonic combustion arereferred to as scramjets.

1.3.6. Rocket Propulsion. As seen in the figure below, a chemical rocket consists of a combustionchamber connected to a converging-diverging nozzle.

Figure 12. Rocket engine.

The rocket engines are subdivided into two main categories: solid propellant and liquid propellant

rockets. The thrust is produced as the result of burning of the propellant under a chemical reaction.Energy produced from the combustion of propellant heats up the reaction gases to very high temper-atures ranging from 2800 K to 4000 K. The pressure of those gases increases significantly. So, whenthese gases exit through the nozzle, they are accelerated to high velocities. Thus, the large amount of thrust is produced.

1.3.7. Rocket Thrust Equation. As derived in one of the previous sections,

(24) T = m2v2 + p2A2 − p1A1

Substituting T = F , m2 = m, v2 = ve, p2 = pe and p1 = pa we get:

(25) F = mv e + ( pe − pa)Ae

where ve = velocity at the nozzle exit pe =pressure at the nozzle exit

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Ae = area of the nozzle exit pa = ambient pressure (pressure outside the nozzle exit)

At the exit of the nozzle, the exhaust velocity is not uniform over the entire cross-sectional area.For simplicity, a uniform velocity will be assumed to allow a one-dimensional analysis. An averagesimplified velocity at the nozzle exit is denoted as the effective exhaust velocity, c . It is given by:

(26) c = F

m = v e + ( pe − pa)

Ae

m

1.3.8. Rocket Propulsion Parameters. In rocket propulsion, a very important parameter is the specificimpulse, I sp. It is of the same importance in rocket engine design as lift coefficient in airfoil design. Itis measured in seconds and it is the thrust generated per unit weight-flow-rate of the propellant. I spis given by:

(27) I sp =

t0 F dt

go t0 mdt

where go is the gravitational acceleration at sea-level (9.81 ms−2). Assuming constant propellant massflow rate and constant thrust, the equation can be simplified to:

(28) I sp = F

mg0

This equation also assumes pa = 0. Rockets operating at very high altitudes frequently encounternear-vacuum conditions, or close to zero ambient pressure, and thus Equation (28) for specific impulsewould hold.

Two additional parameters are used in rocket propulsion. The first is the characteristic exhaustvelocity c∗. The characteristic exhaust velocity is easily determined experimentally and is used oftento compare the performance of various different chemical rockets. The characteristic exhaust velocityis defined as:

(29) c∗ = poAt

mwhere At is the throat area of the nozzle, that is, the most narrow part of the nozzle,and p0 is the pressure outside of the nozzle

The second additional parameter is the thrust coefficient, C f . This parameter can be used to studyhow pressure and altitude changes can affect the nozzle performance. Thrust coefficient is given as:

(30) C f = F

poAt=

mc

poAt

The specific impulse can be written in terms of characteristic exhaust velocity and thrust coefficientas:

(31) I sp = C f c

∗

go

In order to achieve space travel and exploration, rockets are required to launch space vehicles toorbit. The use of rockets provide the necessary velocity, called escape velocity, to overcome thegravitational force present on earth. The formula for escape velocity can be found by equating thekinetic energy of a moving body to the work necessary to overcome gravity, and is given by:

(32) ue =

2go(Ro + h)

where R0 is the radius of the Earth, and h is the altitude of the orbit.13



1.3.9. Multi-Staging in Rockets. In a rocket vehicle, the total mass is composed of structural mass,payload mass, and propellant mass. Because an increase in mass demands higher thrust requiredduring the mission, it is not efficient to carry all the initial structural mass of the rocket throughoutthe length of the entire mission. A single stage rocket is limited by the payload it can carry duringthe mission. In multistage rockets, separate fuel tanks are commonly grouped with their own engines,

called a stage.After the propellant for a particular stage is burned out, the remaining empty structural mass of thatstage is dispensed from the vehicle, and the next stage is ignited. The last stage of the rocket willcarry the payload for the mission. Decreasing structural mass of the vehicle with each stage allowsfor a more efficient flight. In general, the benefits of having multistage rockets are higher velocities,increase in payload, and improved performance for long range missions.

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