Flight Propulsion and Gas Turbines PDF

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Technische Universität Darmstadt

Dietmar K. Hennecke, Karl Wörrlein

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flight propulsion gas turbines turbomachinery engineering

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This document is a set of lecture notes on flight propulsion and gas turbines, covering topics such as turbojet engines and stationary gas turbines. The notes detail thermodynamic cycles, components, performance characteristics, and modifications of jet engines.

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Flight Propulsion and Gas Turbines Dietmar K. Hennecke, Karl Wörrlein GFA Fachgebiet Gasturbinen und Flugantriebe Technische Universität Darmstadt 2. Edition WS 2000 / 2001 Prof. Dietmar K. Hennecke, Ph.D. Dr.-Ing. Karl Wörrlein...

Flight Propulsion and Gas Turbines Dietmar K. Hennecke, Karl Wörrlein GFA Fachgebiet Gasturbinen und Flugantriebe Technische Universität Darmstadt 2. Edition WS 2000 / 2001 Prof. Dietmar K. Hennecke, Ph.D. Dr.-Ing. Karl Wörrlein Institute of Gas Turbines and Aerospace Propulsion Technical University of Darmstadt Overview i Overview The following reprint of the lecture “Flight Propulsion and Gas Turbines” will deal with turbojet engines and stationary gas turbines, with the first being the main focus of attention. However, since both aircraft engine and stationary and automotive gas turbines consist essentially of the same components, most of what is covered for turbojet engines can also be applied to stationary gas turbines. The lecture outline is divided into eleven chapters, where at first only flight propulsion is dis- cussed. In chapter one, the introduction, the subject of the lecture, the objective of this lecture, as well as the importance of the propulsion for the flight task are discussed.Then, based on sketches of principles and pictures of engines, the individual engine types are explained. This is followed by a short recap of the physical basics, whereby the focus is on defining the conservation laws as precisely as possible. The application of these conservation laws is then demonstrated on each jet engine and propeller engine. Chapter two is dealing with the simple jet engine. After the presentation of the basic structure, the individual changes of state of the engine are discussed and these are combined to form the thermodynamic cycle, which is then optimized for a design point to be defined. The next chapter deals with the components of the simple jet engine. Although some of these components specifically are handled with respect to jet engines (intake and nozzle), the expla- nations can also be applied to stationary gas turbines (compressor, combustor and turbine). In chapter four the performance of the engine is discussed. These remarks definitely also apply without exception to jet engines and stationary gas turbines. The points addressed in chapter five, such as engine control and engine start, are described in particular with respect to flight propulsion. The same applies to the engine installation described in chapter six, while the discussion about noise in the same chapter definitely also applies to stationary gas turbines. The seventh chapter is devoted to modifications of the simple jet engine. An analysis of the individual engine types, such as the jet engine with afterburner, the turbofan engine or the shaft power engine, shows that all these engines contain a “gas generator” at their core, as is already the case for the simple jet engine. By adding additional components, it is therefore possible to develop them from the simple jet engine. Chapter eight deals with stationary gas turbines and directly follows with the explanations on shaft power engines, since both are very similar in design. Here is discussed, under which conditions can gas turbines be used for power generation in peak load and base load operation (in combination with steam turbines). Chapters nine and ten deal with ramjet and rocket engines to complete the subject of flight propulsion systems. The ramjet engine is characterized by its suitability for very high Mach numbers. However, the absence of takeoff thrust must be stated as a disadvantage, so that rockets must always be equipped with a second engine for takeoff. The advantages of rockets ii Hennecke/Wörrlein: ,,Flight Propulsion and Gas Turbines” - in this lecture only chemical rocket propulsion systems will be discussed - are the absence of inlet momentum, so that thrust is still available even if the rocket flies faster than its exit jet, and that it is independent from the earth’s atmosphere. The latter, however, also implies that the burning time of the rocket is very limited, since besides the fuel also the oxygen has to be carried on board with the rocket. The final section is a bibliography, which lists all the books used in the making of this Script in alphabetical order. Darmstadt, Autumn 1999 CONTENTS iii Contents 1 Introduction 1 1.1 Subject and Objective............................... 1 1.2 Importance of the propulsion for the flight task................. 2 1.2.1 Hover flight................................ 3 1.2.2 Increase of the airspeed.......................... 6 1.3 Types of Flight Propulsion............................ 8 1.4 Theoretical Fundamentals............................. 10 1.4.1 General.................................. 10 1.4.2 Basic Laws of Physics.......................... 12 1.4.3 Description of The Properties of The Fluid............... 22 1.5 Thrust, Power and Efficiency........................... 25 1.5.1 Jet Engines................................ 25 1.5.2 Propeller engines............................. 33 2 The single-shaft Jet Engine 38 2.1 Schematic Structure................................ 38 2.2 Cycle and Process................................. 39 2.2.1 Process In The Compressor....................... 39 2.2.2 Process in the Turbine.......................... 42 2.2.3 The polytropic Efficiency......................... 42 2.2.4 Process in Intake and Nozzle....................... 45 2.2.5 Process in Combustor.......................... 48 2.2.6 The Cycle for the Design Point...................... 50 2.3 Optimization of the Cycles............................ 54 2.4 Determination of The Design Point........................ 56 iv CONTENTS 3 Components of a single-shaft Jet Engine 60 3.1 General...................................... 60 3.2 Intake and Nozzle................................. 60 3.2.1 Speed of Sound.............................. 60 3.2.2 Flow in ducts with different cross-section................ 64 3.2.3 Normal Shocks.............................. 67 3.2.4 Oblique Shocks.............................. 69 3.2.5 Design of the Intake........................... 71 3.2.6 Nozzle Design.............................. 74 3.3 Compressor and Turbine............................. 77 3.3.1 General.................................. 77 3.3.2 Operation of Turbomachines....................... 77 3.3.3 Blade Element Theory.......................... 83 3.3.4 Airfoil Theory of a Linear Blade Row.................. 86 3.3.5 Stage Design Parameters......................... 89 3.3.6 Stage Element.............................. 90 3.3.7 Operation of the Turbomachinery.................... 95 3.3.8 The Blade Row.............................. 97 3.3.9 The Radial Equilibrium......................... 97 3.3.10 Losses in Turbomachinery........................ 100 3.3.11 Special Characteristics of The Compressors............... 106 3.3.12 Special Characteristics of Turbines.................... 108 3.4 Combustor..................................... 112 3.4.1 General.................................. 112 3.4.2 Combustor Pressure Losses....................... 114 3.4.3 Fuels................................... 118 3.4.4 Combustion Chamber Design...................... 125 3.4.5 Total Temperature Distribution at The Combustion Chamber Outlet.. 128 Hennecke/Wörrlein: ,,Flight Propulsion and Gas Turbines” 1 1 Introduction 1.1 Subject and Objective The jet engines discussed in this lecture are thermal engines, in which chemical energy in the fuel is converted into kinetic energy or into mechanical work. They are used for the propulsion of aircraft, which cover an extremely wide range. Moving mass 0.1 t up to 1 000 t Range 0 km up to 15 000 km Airspeed 0 ≤ Ma∞ ≤ 25 Altitude range ground to space Very high requirements are specified for the propulsion system: Performance according to the flight task High reliability Small dimensions (volume, shape, weight) Low fuel consumption Low environmental impact (emissions, noise) Low life cycle costs (purchase price, maintenance, etc.) Energy and air supply of the aircraft The objectives of the lecture are as follows: Application of the physical basics and engineering approach on the example of a complex technical system with extreme requirements. Aircraft engines are very well suited for this purpose, since the various disciplines are very strongly interlinked. No specialist training The procedure shown here as an example can also be applied to other fields =⇒ Flexibil- ity of the engineer! Focus of the lecture: Thermodynamics - First and second law of thermodynamics - thermodynamic cycles - Combustion - Heat transfer Fluid mechanics including gas dynamics Construction 2 1 INTRODUCTION Further important fields: structural robustness Material science Production engineering Control engineering Fluid mechanic Thermodynamics Gas dynamic Fuel chemistry Flight propulsion Control engineering Machine dynamics Electronic Construction engineering Thermal turbomachines structural robustness Material science Figure 1.1: Interaction of the individual disciplines Figure 1.1 is intended to show that many individual disciplines interact in jet engines and that this subject is therefore ideal for demonstrating engineering thinking and work. 1.2 Importance of the propulsion for the flight task An aircraft must be equipped with a propulsion system for the following reasons: To overcome the drag imposed by the aircraft on the flight motion due to lift (induced drag) and friction. (In gliders, propulsion is provided by thermal lift and wind or by the towplane). Lift FT Drag gm Figure 1.2: Forces on aircraft, Lift, Drag, FT = Thrust, g·m = Weight Acceleration to airspeed and overcoming the height difference. (sometimes also deceler- ation). =⇒ Thrust reversal) Supply of air and energy for on-board supply and auxiliary units (air conditioning, gen- erators, etc.) Principle of propulsion =⇒ Impulse by fluid motion. In the CF6 engine, for example, approx. 500 kg/s are moved. 1.2 Importance of the propulsion for the flight task 3 Another kind of power transmission is not reasonable. Thus, a power transmission by friction, as in the case of land vehicles, would not be effective because of the low friction in the air. The energy required for propulsion can be generated from Muscle power (only for ultralight aircrafts, birds, insects) Solar energy (too low energy density) Chemical energy =⇒ Thermal engines =⇒ Gas turbines, Ramjets, Rocket engines Nuclear energy (so far still under study) Electrical energy =⇒ e.g. ion propulsion, so far still in experimental stage, suitable for long-term space flight, very low thrust In this lecture, only the chemical propulsion systems are dealt with, since they are the only relevant ones. An important parameter for the evaluation of a propulsion system is the ratio P ower (or T hrust) Engine weight This ratio is always driven upwards. Conclusion: build as light as possible, high flow velocities, high pressure stages in the compressor. 1.2.1 Hover flight The problem of flying starts at zero airspeed — Hovering. F G Figure 1.3: Helicopter hovering Figure 1.3 shows a helicopter in hovering flight. The generation of the force F is in principle possible in three different ways: Hydrostatic lift (Balloon) Impulse force F = ṁ · c Force due to the flow around a body, where the pressure distribution on the body is such that a force perpendicular to the direction of motion is generated. In order to generate such a lift force there must be a motion relative to the environment. (e.g. Rotation of an aerofoil =⇒ Helicopter rotor) 4 1 INTRODUCTION Basic physical principle of hovering: Figure 1.4 shows a flying apparatus. The force F which is directed in the opposite direction to the gravitational force is generated by a fluid flow ṁ that flows out with velocity c. The jet is completely homogeneous in its thermodynamic and flow properties. F c ; m; ρ ; T; p=p ; A Figure 1.4: Flying apparatus hovering Thrust =⇒ Momentum Equation; Newton’s second law FT = ṁc (1.1) Mass flow =⇒ Continuity Equation ṁ = Aρc (1.2) Kinetic energy in jet ṁ 2 Ė = c (1.3) 2 The heat energy stored in the outgoing jet can no longer be used. The “Thermodynamics” take place inside the aircraft. Substituting Eq. 1.2 in Eq. 1.1, we get : FT = ρAc2 (1.4) This means that the thrust changes quadratically with the velocity. Substituting Eq. 1.2 in Eq. 1.3 gives us: 1 Ė = ρAc3 (1.5) 2 Eq. 1.5 states that the kinetic energy changes with the third power of the jet velocity. Thus, with an increase of the velocity c, the kinetic energy increases faster than the thrust force FT. Since the kinetic energy has to be supplied to the jet in the machine (e.g., by the fuel), it can be regarded as a measurement for the consumption. In the above equations, the outflow velocity c and the outlet area A can be adjusted to change the thrust force and the kinetic energy. The density ρ is only slightly adjustable. From Eq. 1.4 the outflow velocity c can be determined as a function of the jet surface load FT /A. s FT /A c= (1.6) ρ 1.2 Importance of the propulsion for the flight task 5 4 10 °K 00 Influence of the jet surface load T= 30 c [m/s] on the jet velocity rockets 3 10 0°K FT/A 100 T= c= ρ lift engine 2 10 bypass lift engine fans °K propeller 288 T= 10 rotors valid for ISA sea level 1 2 3 4 5 6 7 10 10 10 10 10 10 10 FT [N/m ] A Figure 1.5: Outflow or jet velocity as a function of the jet surface load for different propulsion systems. As Eq. 1.6 shows, the outflow velocity changes with the square root of the jet surface load For the ratio of the thrust force to the power consumption, we obtain from Eq. 1.4 to 1.6: r FT ρ =2 (1.7) Ė FT /A Note that the thrust-to-power ratio increases with decreasing jet surface load. Since in heli- copters the jet surface load is small (large rotors), the thrust-to-power ratio is large. The opposite applies for rockets. The power Ė is taken from the chemical energy stored in the fuel and is given by Eq. 1.8: Ė = ṁf HL ηi (1.8) From the equation above is: HL lower calorific value of the fuel used ηi internal (thermal) efficiency - conversion of thermal energy into kinetic energy or mechanical work ṁf fuel massflow rate Especially the factor ṁf /FT is important, i.e. the quotient of fuel massflow rate and thrust. From Eq. 1.7 and 1.8 it follows: s ṁf Ė/FT 1 FT /A c = = ∼ ∼c (1.9) FT ηi HL 2ηi HL ρ HL The following table shows the fuel consumption in percentage of takeoff mass for a ten-minute hover. 6 1 INTRODUCTION 1 [N/W] rotors valid for T= ISA-sea level -1 28 FT 10 E 8°K propeller fans bypass lift engine -2 10 T= lift engine 10 00 °K -3 10 rockets Influence of the jet surface load T= on the Power consumption 3 00 0°K -4 10 6 7 10 102 103 104 105 10 10 FT [N/m 2] A Figure 1.6: Power consumption as a function of the jet surface load for different propulsion systems -3 10 Influence of the jet surface load on the fuel consumption 0°K [kg/Ns] 3 00 T= -4 rockets 10 mf FT/A FT mf = 0°K -5 FT H Lηi 2 ρ T= 100 lift engine 10 bypass lift engine fans propellers valid for -6 8°K ISA sea level 10 28 T= 7 H L =4.1868 10 J/kg rotors η i = 0.25 -7 10 2 3 4 5 6 7 10 10 10 10 10 10 10 FT [N/m ] A Figure 1.7: Fuel consumption as a function of jet surface load for different propulsion systems Helicopter, Rotors ∼1% Propeller ∼2% Lifting jet engine ∼ 10 to 20 % Rocket engine ∼ 200 to 300 % 1.2.2 Increase of the airspeed The starting point of the calculation is the unaccelerated horizontal flight with the speed c∞. The resulting drag force 1 Fd = Cd ρc2∞ Ad (1.10) 2 must be compensated by the thrust force of the engine. In Eq. 1.10 is: 1.2 Importance of the propulsion for the flight task 7 Cd Drag coefficient Ad Reference area for the drag c∞ Airspeed The required thrust performance is then calculated as: P = FT c∞ = Fd c∞ =⇒ P ∼ Cd c3∞ (1.11) The drag coefficient is a function of the Mach number and increases quite significantly with it. This has the consequence, as can be seen from Eq. 1.11, that a very high thrust performance is necessary for high-speed flight. 120 000 }F FT 1.00 Thrust with 0.90 FD afterburning 0.85 T 80 000 [N] FT 40000 afterburner switched off H = 0 km FD 0 0 0.4 0.8 1.2 1.6 Ma 2.0 120 000 FT FD }F FD 1.00 0.90 80 000 T 0.85 [N] 40000 FT H = 6 km FD 0 0 0.4 0.8 1.2 1.6 Ma 2.0 120 000 FT FD H = 11 km 80 000 }F 1.00 [N] 0.90 T 0.85 40000 FD FT 0 0 0.4 0.8 1.2 1.6 2.0 2.4 Ma Figure 1.8: Thrust FT and Drag FD of a supersonic aircraft 8 1 INTRODUCTION 1.3 Types of Flight Propulsion Basically, a distinction is made between two types of engine: Airbreathing engines =⇒ Only fuel is carried in the aircraft Rocket (exhaust) engines =⇒ Fuel and oxygen carriers are carried in the aircraft =⇒ Independent from the atmosphere Flight Propulsion Air breathing (Through-flow) engines Rocket (Exhaust) engines Energy on board, oxygen Energy and oxygen from the environment carrier on board Airspeed smaller Airspeed greater than jet velocity than jet velocity Static Thrust No Static Thrust Static Thrust Propeller Pulsating Liquid fuel rocket Turbo shaft engine Ram Jet Turbofan engine Ramjet engine Hybrid rocket Single-shaft Ramjet engine with Solid fuel rocket jet engine supersonic combustion Jet engine with after burner Figure 1.9: Classification of Flight Propulsion Types 1.3 Types of Flight Propulsion 9 Figures 1.10 and 1.11 show two examples of modern jet engines in operation. Figure 1.10: BMW - Rolls - Royce (D) BR 715; Turbofan engine FT,max ≈ 100 kN , ṁtotal = 286 kg/s, µ = 4.4, πt = 35, Ttmax = 1790 K Figure 1.11: Pratt&Whitney (USA) PW 4084; Turbofan engine FT,max ≈ 376 kN , ṁtotal = 1153 kg/s, µ = 6.4, πt = 34.2, Ttmax = 1634 K 10 1 INTRODUCTION 1.4 Theoretical Fundamentals 1.4.1 General The following physical laws are necessary for the analytical treatment of engine problems1 : Conservation of Mass theorem =⇒ Continuity equation Newton’s second law =⇒ Conservation of Linear Momentum Conservation of Angular Momentum theorem First law of thermodynamics =⇒ Energy law Second law of thermodynamics =⇒ Entropy theorem Furthermore, to describe the properties of the fluid, we need: Thermal and caloric equations of state The dependence of material properties (e.g. viscosity; thermal conductivity) on pressure and temperature A fluid (liquid or gas) is distinguished by the fact that: no form stability is present every smallest shear force leads to continuous deformation the resistance depends on the deformation speed (e.g. Newtonian fluids =⇒ shear stresses proportional to the velocity gradient) Liquids and gases are classified under the generic term Fluids. In general, the following classi- fication can be made between the two: Liquids: incompressible =⇒ No density changes occur with pressure changes Gases: compressible =⇒ Pressure and density changes are coupled with each other Since fluids are composed of countless molecules that move in an uncontrolled pattern, there are two ways to describe them mathematically: Kinetic theory =⇒ Tracking of the molecular motion from the initial state (statistical method). 1 Derivation of the conservation laws according to J.H. Spurk: ,,Strömungslehre” 1.4 Theoretical Fundamentals 11 Continuum Theory =⇒ Macroscopic approach. Introduction of meaningful continuous properties (Averaging). We further explain the idea of the continuum using the example of the density with the help of figure 1.12: Considering the fluid mass δm in a volume δV which contains the point P, thus the quotient δm/δV defines the average density of the fluid in the volume δV. If we now assume that the volume δV was initially relatively large and allow it to slowly decrease, this quotient changes, as Fig. 1.12 shows. This means that the density depends on the size of the volume Range of Range of Volume δ V molecular continuum P Mass δm effects δm δV δV ρ δV Figure 1.12: Definition of the density at a point of the fluid δV. If the volume δV becomes so small that it contains only a few molecules, the density also changes over time, since different numbers of molecules are contained in the volume δV at each time point. If we assume that the smallest volume at which the fluid can still be considered as a continuum is δV ′ , then the density at the point P can be defined according to Eq. 1.12. δm ρ ≡ lim (1.12) δV→δV ′ δV This definition explains the idea of the continuum and shows that it is an imaginary but very useful concept. The same applies to all other characteristics and properties of the continuum, such as veloc- ity, pressure, temperature, enthalpy, viscosity, thermal conductivity, etc. In the continuum, all properties can be given as continuous functions of place and time. Limitations of the applicability of this concept: The continuum theory fails if the smallest scale of the problem is in the order of magnitude of the free path length of the molecules (e.g. vacuum). According to the kinetic theory of gases, the dynamic viscosity is in the order of magnitude of µ ≈ ρcM lM where cM is the mean molecular velocity and lM is the mean free path of the molecules. Fur- thermore, since cM is proportional to the speed of sound a, the following expression for the Reynolds number is obtained when L denotes the characteristic length of a flowed body: ρcL ρcM lM c a L L Re ≡ = ∼ Ma µ µ a cM l M lM 12 1 INTRODUCTION The limit of validity thus results in L Re ∼ (1.13) lM Ma For flow problems, the smallest characteristic length can be considered to be the boundary layer thickness δ. This applies for a laminar flow along a flat plate: δ 1 ∼√ L Re Substituting this relation into Eq. 1.13, we obtain Tsien’s delimitation: √ δ Re ∼ > 100 (1.14) lM Ma This equation states that when applying the continuum theory, the boundary layers must be at least a hundred times thicker than the free path of the molecules. For the mathematical description of the continuum, there are two viewpoints, namely: Lagrange’s method: Considers particles of unchanged identity and describes location, forces, state and motion of these particles for each moment. Euler’s method: Considers a space-fixed region and describes forces, states and motions of those particles for each moment, which are at the observed place at the observed moment. For the calculation of propulsion problems, the Eulerian approach is more convenient, since fluids are not constant in shape, so it would be difficult to track particles of the same identity. 1.4.2 Basic Laws of Physics 1.4.2.1 Theorem of Conservation of Mass In the following, we always consider the same piece of the fluid, which is separated from the rest of the fluid by a smooth, closed surface. The enclosed part of the fluid always consists of the same fluid particles; its volume is therefore a material volume, and its surface is a material surface. In the motion, the shape of the material volume changes. The region occupied by the considered part of the fluid at time t is denoted by V (t). The mass m of the delimited part of the fluid is the sum of the mass elements dm over the set (M ) of material points in the body. Z m= dm (1.15) (M ) Since the density is supposed to be a continuous function of the place and the time, the mass can also be expressed as an integral of the density over the area occupied by the body (V (t)). Z Z Z Z m= dm = ρ(⃗x, t)dV (1.16) (M ) (V (t)) 1.4 Theoretical Fundamentals 13 According to the law of conservation of mass, the mass of the delimited piece of the fluid is constant in time, thus: Z Z Z Z Z Z Z   Dm D D ∂ρ ∂ = dm = ρdV = + (ρui ) dV = 0 (1.17) Dt Dt Dt ∂t ∂xi (M ) (V (t)) (V ) In Eq. 1.17 we use the fact that the result of the integration does not change, if instead of the time-varying range (V (t)) a fixed range (V ) is chosen, which at time t coincides with the different range. Eq. 1.17 is now valid for any shape of the volume occupied by the fluid under interest, i.e., for any choice of the integration domain (V ). This means that in Eq. 1.17 the integrand itself must vanish. Thus we get the differential form of the conservation law of mass. ∂ρ ∂ + (ρui ) = 0 (1.18) ∂t ∂xi This relation is also called the Continuity Equation. With the material derivative of the density, we can express Eq. 1.18 as Dρ ∂ui +ρ =0 (1.19) Dt ∂xi If now Dρ ∂ρ ∂ρ = + ui =0 (1.20) Dt ∂t ∂xi then the density of a material particle does not change in the course of its motion. But this is equivalent to ∂ui =0 (1.21) ∂xi i.e. the flow is volume-determined. The flowing fluid (gas or drippable fluid) can then be regarded as incompressible. Thus, if Eq. 1.20 is satisfied, the continuity equation takes the simple form according to Eq. 1.21, in which no derivatives with respect to time occur. but of course it is also valid for transient flows. Generally, the condition Dρ/Dt = 0 is fulfilled for drippable liquids. But it can also be fulfilled for gases, as shown in the following estimation for small Mach numbers M a ≪ 1. Compressibility is the ability of a fluid to be compressed by external pressure forces. A measure of compressibility is the so-called “volume elasticity modulus” E, which is defined by Eq. 1.22. ∆V ∆p = −E (1.22) V For gases, if the volume change remains relatively small and occurs at constant temperature, the E-modulus is equal to the pressure p0. Because of the conservation of mass is further valid (V0 + ∆V )(ρ0 + ∆ρ) = V0 ρ0 and thus ∆ρ ∆V =− ρ0 V0 14 1 INTRODUCTION so that Eq. 1.22 can also be written in the following form. ∆ρ ∆p = E (1.23) ρ0 Treating the flow as incompressible is acceptable as long as the relative density change remains very small, ∆ρ/ρ0 ≪ 1. Now the pressure change ∆p associated with the flow is in the order of the dynamic pressure, q = ρw2 /2, so that Eq. 1.23 can be expressed in the following form: ∆ρ q ≈ (1.24) ρ0 E According to Laplace’s formula for the speed of sound, a2 = E/ρ0. This gives the condition: ∆ρ ρ0 w 2 1  w 2 1 ≈ ≈ = M a2 ≪ 1 (1.25) ρ0 2 E 2 a 2 The compressibility can also be neglected in the case of gas flows if 1 M a2 ≪ 1 2 For example, M a = 0.3 gives ∆ρ/ρ0 ≈ 0.045. This value of M a should be considered as the uppermost limit of the fluid Mach number, up to which a gas flow can be treated as incompress- ible. The integral form of the continuity equation can be obtained by modifying Eq. 1.17 according to the Reynolds transport theorem Z Z Z Z Z Z Z Z D ∂φ φdV = dV + φui ni dS (1.26) Dt ∂t (V (t)) (V ) (S) In Eq. 1.26, S denotes the oriented bounding surface of V , φ denotes a tensor field of any order (ρ −→ tensor of zero order), and ni is the normal vector,which is set to be positive when it’s directed outward. Z Z Z Z Z Z Z Z Dm D ∂ρ = ρdV = dV + ρui ni dS = 0 (1.27) Dt Dt ∂t (V (t)) (V ) (S) or Z Z Z Z Z Z Z Z ∂ρ ∂ dV = ρdV = − ρui ni dS (1.28) ∂t ∂t (V ) (V ) (S) Here, we consider a fixed integration range, i.e. a so-called control volume. The statement of Eq. 1.28 can be interpreted as follows: The temporal change of the mass in the control volume is equal to the difference of the masses entering and leaving the surface of the control volume per time period. For stationary flows ∂ρ/∂t = 0. Thus the integral form of the continuity equation is Z Z ρui ni dS = 0 (1.29) (S) i.e., the same amount of mass flows into the control volume as it flows out per period of time. 1.4 Theoretical Fundamentals 15 1.4.2.2 Theorem of Conservation of Linear Momentum The balance of the linear momentum is a pure empirical theorem of classical mechanics and states: In an inertial system the temporal change of the linear momentum of a body is equal to the force acting on this body.” This theorem is represented in symbolic notation by Eq. 1.30. DI⃗ = F⃗ (1.30) Dt Since the body is a piece of the fluid, which always consists of the same material points, its momentum can be calculated. Z Z Z I⃗ = ρ⃗udV (1.31) (V (t)) The forces acting on the body are mass or volume forces and surface or contact forces. Mass forces are forces with a large range, they affect all material particles in the body and usually are caused by force fields. The most important example is the gravitational field of the earth. Other technically important mass or volume forces occur due to electromagnetic fields, or are so called pseudo forces (e.g. centrifugal force), if the motion is described in respect to an accelerated coordinate system. The contact or surface forces are applied from its surroundings to the considered fluid part. The total force acting on the considered part of the fluid is obtained by integration over the volume occupied by the fluid, or over its surface. With ⃗k as the mass force and ⃗t as the stress vector: Z Z Z Z Z F⃗ = ρ⃗kdV + ⃗tdS (1.32) (V (t)) (S(t)) With the Eqs. 1.30 to 1.32 the linear momentum theorem then takes the following form: Z Z Z Z Z Z Z Z D ⃗ ⃗tdS ρ⃗udV = ρkdV + (1.33) Dt (V (t)) (V ) (S) or, if also on the left side the time varying region (V (t)) is replaced by a fixed region (V ), which at time t coincides with the different range: Z Z Z Z Z Z Z Z D⃗u ρdV = ρ⃗kdV + ⃗tdS (1.34) Dt (V ) (V ) (S) Now the following applies for the stress vector ⃗t = ⃗n · T whereby ⃗n is the normal vector and T is the stress tensor, for which can be written in matrix form:   τ11 τ12 τ13 T = τ21 τ22  τ23  τ31 τ32 τ33 The elements of the main diagonals are the normal stresses, those of the diagonals are the shear stresses. 16 1 INTRODUCTION With the help of Gauss’s integral theorem, the surface integral in Eq. 1.34 can be transformed into a volume integral. Z Z Z   D⃗u ρ − ρ⃗k − ▽ · T dV = 0 (1.35) Dt (V ) Since the integrand is supposed to be continuous and furthermore the integration domain (V ) can be assumed arbitrarily, Eq. 1.35 is with the differential form of the linear momentum theorem in symbol notation equivalent to D⃗u ρ = ρ⃗k + ▽ · T (1.36) Dt or in index notation Dui ∂τji ρ = ρki + (1.37) Dt ∂xj The integral form of the linear momentum theorem plays an important role in the technical application especially if the integrals can be written as surface integrals. For this purpose Eq. 1.33 is transformed with the Reynolds transport theorem. Z Z Z Z Z Z Z Z Z Z ∂(ρ⃗u) ⃗ ⃗tdS dV + ρ⃗u(⃗u · ⃗n)dS = ρkdV + (1.38) ∂t (V ) (S) (V ) (S) The first integral of the left side cannot be converted into a surface integral, therefore this in- tegral must vanish, which is the case for a stationary flow. On the other hand, the first integral of the right side can be written as a surface integral if the volume force ρ⃗k can be calculated as a gradient of a scalar function, i.e., if it has a potential. If the potential of the volume force is denoted by Ω = −ρgi xi , where ρ⃗k = − ▽ Ω (1.39) shall apply, then the volume integral can be written as a surface integral. Z Z Z Z Z Z Z Z ⃗ ρkdV = − ▽ΩdV = − Ω⃗ndS (1.40) (V ) (V ) (S) Thus, Eq. 1.38 can be rewritten in the following form: Z Z Z Z Z Z ρ⃗u(⃗u · ⃗n)dS = − Ω⃗ndS + ⃗tdS (1.41) (S) (S) (S) The meaning of the linear momentum theorem in this form becomes clear if we consider that by knowing the linear momentum flux and the potential Ω , the force on the surface of the control volume is known. If we only want to know the force coming from the linear momentum flux alone, then Eq. 1.41 becomes: Z Z Z Z ρ⃗u(⃗u · ⃗n)dS = ⃗tdS (1.42) (S) (S) This represents the linear momentum theorem in the most commonly used form. The great advantage of Eq. 1.42 is that the often unknown − and possibly also non-calculable − flow 1.4 Theoretical Fundamentals 17 processes inside the control volume do not appear. Only the variables at the surface are relevant, and since the control volume is freely chosen, the surface in the specific case can be defined in such a way, that the integrals are easy to evaluate (see also section 1.5.1.2.2). 1.4.2.3 Theorem of Conservation of Angular Momentum The theorem of angular momentum represents the second empirical theorem of classical me- chanics. It states: In the inertial frame of reference the temporal change of the angular momen- tum is equal to the torque of the external forces acting on the body. In symbolic notation this is represented by Eq. 1.43. D ⃗ ⃗ (L) = M (1.43) Dt The angular momentum L ⃗ is calculated as an integral over the volume occupied by the fluid body. Z Z Z ⃗ L= ⃗x × (ρ⃗u)dV (1.44) (V (t)) The angular momentum according to Eq. 1.44 is calculated with the origin of the coordinate system as point of reference; therefore the moment of the outer forces must be also calculated with the same point of reference. Z Z Z Z Z M⃗ = ⃗x × (ρ⃗k)dV + ⃗x × ⃗tdS (1.45) (V (t)) (S(t)) Thus, the angular momentum theorem takes the form represented by Eq. 1.45. Z Z Z Z Z Z Z Z D ⃗ ⃗x × (ρ⃗u)dV = ⃗x × (ρk)dV + ⃗x × ⃗tdS (1.46) Dt (V (t)) (V ) (S) Like the integral form of the linear momentum theorem, the integral form of the angular mo- mentum theorem plays a significant role in technical applications. The important torque is only the one which can be attributed to the swirl flow through the control surface at stationary flow. It is written in symbolic notation Z Z Z Z ⃗x × ⃗uρ⃗u · ⃗ndS = ⃗x × ⃗tdS (1.47) (S) (S) and in index notation Z Z Z Z ϵijk xj uk ρul nl dS = ϵijk xj tk dS (1.48) (S) (S) A special form of the angular momentum theorem according to Eq. 1.48 is the Euler’s rotation equation, which is derived in section 3.3.2.1. 1.4.2.4 Theorem of Conservation of Energy Since mechanical energy can be transformed into heat and vice versa, the conservation laws of mechanics described so far are not sufficient for a complete description of the motion of a fluid. As the third fundamental empirical theorem, the conservation law of energy should be derived, 18 1 INTRODUCTION which can be formulated verbally as follows: ,,The temporal change of the total energy of a body is equal to the power of the external forces plus the energy supplied from outside per unit of time.” With e as internal energy per unit of mass, the internal energy of a material particle is given by edm. The internal energy E of a body, i.e. of a confined part of the fluid, is then given as an integral over the area occupied by the body. Z Z Z E= eρdV (1.49) (V (t)) In order to obtain the total energy of the particle under consideration, the kinetic energy must also be taken into account, which for a material particle is given by ui2ui dm. The kinetic energy of the body is then given by: Z Z Z ui ui K= ρdV (1.50) 2 (V (t)) The external forces are the surface and volume forces. The power of the surface forces is ⃗u ·⃗tdS, whereas the power of the volume forces ⃗u · ⃗kρdV. The power of the external forces on the body is then given by: Z Z Z Z Z P = ρui ki dV + ui ti dS (1.51) (V (t)) (S(t)) For the energy supplied from the outside, we need to introduce the heat flux through an element of the surface, for which −⃗q · ⃗ndS is valid, where ⃗q is called the heat flux vector. The negative sign is necessary to make the inflowing energy (⃗q and ⃗n form an obtuse angle) positive. This results in the energy supplied to the body per time unit: Z Z Q̇ = − ⃗q · ⃗ndS (1.52) (S(t)) The conservation law of energy can thus be written as follows: D (K + E) = P + Q̇ (1.53) Dt Substituting Eqs. 1.49 to 1.52 into Eq. 1.53, we get: Z Z Z  Z Z Z Z Z Z Z D ui ui  + e ρdV = ui ki ρdV + ui ti dS − qi ni dS (1.54) Dt 2 (V ) (V ) (S) (S) Here again we used the fact that for the time varying range (V (t) a fixed range (V ) can be chosen, which coincides with the time varying range at time t. If in the first surface integral the stress vector is expressed by the tensor, then both of the surface integrals can be converted into volume integrals using Gauss’s theorem. Z Z Z   D  ui ui  ∂ ∂qi ρ + e − ρki ui − (τji ui ) + dV = 0 (1.55) Dt 2 ∂xj ∂xi (V ) 1.4 Theoretical Fundamentals 19 If the integrand is continuous and the integration range is arbitrary, the integrand must vanish and the differential form of the energy theorem is given by: Dui De ∂τji ∂ui ∂qi ρui +ρ = ρki ui + ui + τji − (1.56) Dt Dt ∂xj ∂xj ∂xi If the enthalpy h = e + p/ρ is added to Eq. 1.55 and we consider, that the stress tensor can be split into τji = −pδji + Pji (1.57) where Pji is the tensor of the frictional stresses, then with respect to the continuity equation 1.19 and the definition of the total enthalpy we get ht = h + ui2ui : Z Z Z     D ui ui p ∂ ∂qi ρ +h− − ρki ui − (τji ui ) + dV = 0 Dt 2 ρ ∂xj ∂xi (V ) Z Z Z   Dht Dp p Dρ ∂(pui ) ∂ ∂qi ρ − + − ρki ui + − (Pji ui ) + dV = 0 Dt Dt ρ Dt ∂xi ∂xj ∂xi (V ) Z Z Z    Dht ∂p ∂p p Dρ ∂ui ∂p ρ − − ui + +ρ − ρki ui + ui Dt ∂t ∂xi ρ Dt ∂xi ∂xi (V )  ∂ ∂qi (Pji ui ) + dV = 0 ∂xj ∂xi Z Z Z   Dht ∂p ∂Pji ∂ui ∂qi ρ − − ρki ui − ui − Pji + dV = 0 (1.58) Dt ∂t ∂xj ∂xj ∂xi (V ) In Eq. 1.58 implies: Z Z Z Z Z Z Z Z Dht ∂(ρht ) ρ dV = dV + ρht ui ni dS (1.59) Dt ∂t (V ) (V ) (S) the change of total enthalpy in the control volume, Z Z Z Z Z Z Z Z ∂Ψ − ρki ui dV = ρui dV = ρui Ψni dS (1.60) dxi (V ) (V ) (S) the change of potential energy, where Ψ = −gi xi represents the potential of the mass force, Z Z Z   ∂p ∂Pji PS = + ui dV (1.61) ∂t ∂xj (V ) the shaft power, Z Z Z ∂ui PF = Pji dV (1.62) ∂xj (V ) the frictional power and Z Z Z Z Z ∂qi Q̇ = dV = − qi ni dS (1.63) ∂xi (V ) (S) 20 1 INTRODUCTION the heat supplied. For a stationary flow we get: Z Z Q̇ + PS + PF = (ht + Ψ)ρui ni dS (1.64) (S) and furthermore the total enthalpy is constant both at the inlet cross section as well as at the outlet cross-section, i.e. a one-dimensional flow is given, that is the first law for stationary flow processes known from thermodynamics. Q̇ + PS + PF = ṁ[(ht + g0 z)a − (ht + g0 z)e ] (1.65) The potential energy g0 z can be formally summarized with the static enthalpy h and the kinetic energy ui2ui into the total enthalpy ht. However, since the potential energy has no mass property like the kinetic and the internal energy, it is rather assigned to the mass only due to the external force field, the notation as in Eq. 1.65 is preferable. For a frictionless adiabatic system we get: PS = (ht + g0 z)a − (ht + g0 z)e (1.66) ṁ This means that in an adiabatic, frictionless system, the technical work can be calculated directly from the change of the sum of total enthalpy and potential energy. Here, frictionless means that there is no friction at the boundaries of the system. Friction inside the control volume is expressed by a change in the total enthalpy and is therefore taken into account in Eq. 1.66. 1.4.2.5 Theorem of Conservation of Entropy Assuming the Gibbs relation T ds = de + pdv (1.67) Assuming the Gibbs relation, which is supposed to be valid for reversible as well as for irre- versible processes, applying it to a material particle leads to the following relation Ds De Dv T = +p (1.68) Dt Dt Dt Applying the 1st main theorem for closed systems De Dt = δ ẇ + δ q̇, where δ ẇ = −p Dv Dt + ρ1 Φ ∂qi (with Φ as dissipation function) and δ q̇ = − ρ1 ∂xi can be set, it becomes: Ds Φ 1 ∂qi ρ = − (1.69) Dt T T ∂xi The last term of the right side can be transformed using the identity ∂  qi  1 ∂qi qi ∂T = − 2 (1.70) ∂xi T T ∂xi T ∂xi and thus we get the entropy balance equation Ds Φ qi ∂T ∂  qi  ρ = − 2 − (1.71) Dt T T ∂xi ∂xi T 1.4 Theoretical Fundamentals 21 In this equation the temporal change of the entropy of a material particle is divided into two parts. An entropy production with the rate Dsirr Φ qi ∂T ρ = − 2 ≥0 (1.72) Dt T T ∂xi qi which is always greater than or equal to zero and a divergence of an entropy current T which can be greater than, equal to or less than zero. Dsrev ∂  qi  ρ =− (1.73) Dt ∂xi T Eq. 1.72 represents the irreversible processes due to friction and heat conduction. The inequality ∂T 1.72 is sufficiently given by the conditions Φ ≥ 0 and qi ∂x i ≤ 0. The first inequality states that mechanical energy can be dissipated into heat by friction, but conversely no mechanical energy can be generated from heat. The second inequality states that heat can only flow in the direction of decreasing temperature. Eq. 1.73 represents the entropy change which a particle experiences through its environment, it can be positive or negative. The entropy change of a delimited part of the fluid is given by integration of Eq. 1.69 over the volume occupied by the fluid. Z Z Z Z Z Z   Z Z D DS Φ qi ∂T qi sρdV = = − 2 dV − ni dA (1.74) Dt Dt T T ∂xi T (V ) (V ) (A) Since, as shown, the volume integral on the right-hand side can never be negative, Eq. 1.74 directly gives the statement of the second law of thermodynamics: Z Z DS qi ≥− ni dA (1.75) Dt T (A) The equal sign applies to reversible processes and the greater sign to irreversible processes. If neither heat is added nor removed and the process is adiabatic, the surface integral on the right-hand side of Eq. 1.75 disappears and we obtain: DS ≥0 (1.76) Dt This correlation shows the well known fact that in an adiabatic process the entropy cannot decrease. (In Eqs. 1.74 and 1.75, to distinguish the entropy S and the surface of the control volume, we denoted the latter with A). The second law of thermodynamics is based, like the first law, on experience. According to M. Planck it can be expressed as follows ,,All processes in which friction occurs are irreversible” and according to C. Clausius ,,Heat can never be transfered by itself from a lower temperature to higher temperature body.” The second law thus provides hints in which direction a process can run. 22 1 INTRODUCTION 1.4.3 Description of The Properties of The Fluid Fluids can be understood as all liquids and gases. They can be distinguished by the following characteristics: Liquids =⇒ incompressible; this also applies for gases at low flow velocities Ma < 0.2 (0.3) ideal gases =⇒ compressible; Idealization of the real gases for pressure p > Tcr for aircraft engines, these requirements are fulfilled. 1.4.3.1 The Thermal Equation of State for The Ideal Gas The thermal state of the fluid is locally defined in the continuum as a boundary value (see derivation of the local density ρ) and is defined by the direct properties pressure p, volume V and temperature T. The variables p and T are so-called intensive properties. They are independent from the size of the system and therefore keep their values unchanged when the system is divided into subsystems. The functional correlation between the direct properties is given by thermal equations of state of the form F(p; V ; T ) = 0 From the direct properties, the derived properties internal energy U , enthalpy H and entropy S can be derived. As with the volume V , they are extensive properties which are proportional to the mass within the system. If we divide an extensive property by the mass within the system, we get a specific property, which is indicated by a lowercase letter. An important specific V property is the specific volume v = m , whose inverse is the density ρ = m V. Other specific quantities are the specific enthalpy h = Hm , the specific internal energy u = U m and the specific S entropy s = m. At low pressures all gases show a very simple behavior, which can be described by the ideal gas equation. pV = mRT (1.77) Where R represents the gas constant of a specific gas, which is given by Eq. 1.78 from the J universal gas constant R = 8314.7 kmolK and the molar mass M of the gas. R R= (1.78) M Dividing Eq. 1.77 by the gas mass m, we get the thermal equation of state with specific vari- ables. p pv = = RT (1.79) ρ In table 1.1 the molar masses of some gases are given, so that it is possible to calculate the special gas constants of these gases by using Eq. 1.78. 1.4.3.2 The Caloric Equation of State of Ideal Gases According to the first law, the specific internal energy is a variable of state, as are the specific volume, the pressure and the temperature. Since the equilibrium state of an ideal gas is already 1.4 Theoretical Fundamentals 23 Air N2 O2 H2 H2 O He CO CO2 CH4 M= 28.964 28.0134 32.000 2.0159 18.0159 4.026 28.0106 44.0106 16.043 Table 1.1: Molar masses of some gases in kg/kmol determined by two (independent) properties, the internal energy u must be a function of two properties. Thus there is the relation u = u(T, v) (1.80) which is called the caloric equation of state. Since the specific internal energy is a property (value depending only on the state, not on the path by which this state was reached), it has a full derivative.     ∂u ∂u du = dT + dv (1.81) ∂T v ∂v T From overflow experiments of Gay-Lussac (1807) and Joule (1845) we know, that for ideal gases (∂u/∂v)T = 0. The partial derivative   ∂u cv (T, v) = cv (T ) ≡ (1.82) ∂T v has a special name, it is called the specific heat at constant volume. Thus, for the internal energy follows: ZT u(T ) = cv (τ )dτ + u0 (1.83) T0 In Eq. 1.83 u0 represents the specific internal energy at T0. Likewise for the specific enthalpy: ZT h(T ) = cp (τ )dτ + h0 (1.84) T0 Here, the definition equation for the specific heat at constant pressure is given by Eq. 1.85, while h0 represents the value of the specific enthalpy at reference temperature T0.   ∂h cp (T ) ≡ (1.85) ∂T p From the definition equation of enthalpy h = u + pv = u + RT follows by differentiation: cp (T )dT = cv (T )dT + RdT cp (T ) − cv (T ) = R = constant (1.86) The ratio of the two specific heats is called the heat capacity ratio γ. cp (T ) γ(T ) ≡ (1.87) cv (T ) Figure 1.13 shows the specific heat referred to the gas constant and the heat capacity ratio for dry air as a function of temperature. At temperatures above 2000 K, the air can no longer be considered an ideal gas, since the material values become a function of pressure and temperature due to dissociation. 24 1 INTRODUCTION 5,0 1,40 4,5 cp/R 1,35 4,0 γ 1,30 3,5 1,25 3,0 250 500 750 1000 1250 1500 1750 1,20 250 500 750 1000 1250 1500 1750 T [K] T [K] Figure 1.13: Spez. W”arme und Isentropenexponent von Luft als Funktion der Temperatur For some applications, it is advantageous to calculate with a constant average specific heat. The averaging rule is then: RT2 cp (τ )dτ T1 cpm |TT21 = (1.88) T2 − T1 Such a suitably averaged specific heat often allows a closed solution of the differential equations and thus leads to simple and easily understandable correlations. This will be shown by the example of an isentropic process. The combination of the first and second law provides: dq du pdv ds = = + (1.89) T T T Since it has to be ds = 0 for an adiabatic and frictionless (=⇒ isentropic) process, with equa- tions 1.79, 1.81, 1.86 and 1.87 we get: dT dv + (γ − 1) = 0 (1.90) T v or with v = 1/ρ; dv/v = −dρ/ρ and dp/p = dρ/ρ + dT /T dT dρ − (γ − 1) = 0 (1.91) T ρ dT γ − 1 dp − =0 (1.92) T γ p By eliminating dT /T from Eqs. 1.91 and 1.92, we get finally: dp dρ =γ (1.93) p ρ The Eqs. 1.90 to 1.93 represent isentropic relations in a differential form. Eq. 1.93 is particularly significant because it shows that for γ → ∞ the density no longer changes with pressure. Thus, for isentropic changes of state, the behavior of incompressible media can always be derived from the compressible media by applying the boundary transition γ → ∞. An averaging of the heat capacity ratio between the temperatures T1 and T2 allows the integration of all isentropic relations. For example, Eq. 1.92. gives:   γmγ −1 T2 p2 m = T1 p1 1.5 Thrust, Power and Efficiency 25 In the case of an isentropic process, the effective work in a stationary system can be represented as the difference of the enthalpies. It is valid: Zh2 ZT2 "  γm −1 # p 2 γm ∆h12 = dh = cp (τ )dτ = cpm |TT21 (T2 − T1 ) = cpm |TT21 T1 −1 p1 h1 T1 For a flowed through control volume, the total enthalpy ht occurs instead of h, Zht2 ZTt2 "  γm −1 # Tt2 Tt2 pt2 γm ∆ht12 = dh = cp (τ )dτ = cpm |Tt (Tt2 − Tt1 ) = cpm |Tt Tt1 −1 1 1 pt1 ht1 Tt1 where Tt is the total temperature and pt is the total pressure. 1.4.3.3 Gas Mixtures For gas mixtures as the working medium (e.g. air and air + combustion products), the relation- ships derived in 1.4.3.1 and 1.4.3.2 also apply if the material values are appropriately averaged. According to Dalton’s law, the total pressure of a gas mixture is equal to the sum of the partial pressures of the individual components. Thus, however: X T X T p= pi = mi Ri = mtotal Rm (1.94) V V From Eq. 1.94 the mixing rule for the mean gas constant can be derived immediately. X mi X Rm = Ri = ξi Ri (1.95) mtotal P In Eq. 1.95 mi means the mass of the individual gases, mtotal = mi the total mass of the gases, ξi = mi /mtotal the mass fractions of the individual gases. All variables, which are related to “kg”, are calculated similarly to Eq. 1.95, i.e. over the mass fractions, all variables which are related to “kmol” are averaged over the volume fractions ri. Mass fraction and volume fraction can be ”merged” according to Eq. 1.96, r i Mi ξi = (1.96) Mm P where for the average molar mass Mm = r i Mi 1.5 Thrust, Power and Efficiency 1.5.1 Jet Engines 1.5.1.1 Choice of The Control Volume Figure 1.14 schematically shows an engine in the x-z- or in the x∞ -z∞ coordinate system. Here, the x∞ -z∞ system is an inertial system, i.e. a coordinate system which is fixed to the earth. The x-z system, on the other hand, is a coordinate system (relative system) which is fixed to the 26 1 INTRODUCTION aircraft and which moves in the x direction with respect to the inertial system at the flight speed c∞. The choice of the control volume is crucial for the calculation of the forces on the engine. For example, a fixed control volume in the inertial system would make the flow problem transient − and unnecessarily complicate the thrust calculation. By choosing the control volume as a relative system fixed to the aircraft a steady-state flow is obtained, which is easy to handle. In any case, the individual control surfaces must be selected in such a way that all properties are clearly defined and known. inertial system X relative system C ∞ control volume mf m1 FT m2 w1 w2 p1 p 2 A1 A2 mI p I e m II ; pII wI AI wII ; A II m III ; wIII Z Z ∞ 1 2 Figure 1.14: Engine in the inertial or relative system − Definition of the control volume The consequence of this is that the inlet control surface (plane 1) must be located as far in front of the engine as possible, so that there is no interference from the engine itself. The captured streamtube must therefore be free of any curvature, since otherwise both the pressure p1 and the velocity w1 can no longer be regarded as constant over the inlet surface and would therefore be unknown. This means in particular that the engine inlet plane should under no circumstances be chosen as the inlet surface into the control volume, since both the pressure and the velocity over this cross-section would have to be measured experimentally and then averaged. The exit control surface (plane 2) must be located directly at the nozzle exit, since this is the only place where pressure, velocity and mass flow can be accurately specified. If we were to choose a plane behind the engine outlet, the pressure, flow velocity, jet surface and mass flow would be completely unknown due to the admixture of ambient air. The side control surface, which for the purpose of simplification is assumed to be a cylinder surface axially parallel to the direction of airspeed (this does not imply any restriction of the generality), must be selected so far away from the engine that no more disturbances can be induced by it on the flow. The fixed walls of the engine must be excluded from the control volume so that the forces of these walls on the flow medium can be considered as external forces. If all these conditions are taken into account, it is possible to calculate the thrust of a jet engine with relatively simple mathematical tools. 1.5.1.2 Application of The Conservation Laws 1.5 Thrust, Power and Efficiency 27 1.5.1.2.1 Conservation of Mass The continuity equation is generally as follows: ∂ρ + div(ρw) ⃗ =0 (1.97) ∂t Since this case is a steady-state and one-dimensional flow problem, Eq. 1.97 simplifies to ṁ = ρwA = constant (1.98) For the flow through the engine, if ṁf denotes the mass flow of the fuel: ṁ1 + ṁf = ṁ2 (1.99) The continuity equation for the external flow is given by Eq. 1.100. ṁI = ṁII + ṁIII (1.100) The mass flow ṁIII exceeds the outer control surface, assuming that the angle between vector w ⃗ III and the axis of the control volume is so small that its cosine can be set equal to one. 1.5.1.2.2 Conservation Law of Linear Momentum According to Eq. 1.42 the linear momentum theorem is in vector notation: Z Z Z Z ρw( ⃗ w⃗ · ⃗n)dA = ⃗tdA (A) (A) The integration needs to be performed only over the inlet and outlet surfaces (Ai ; Ao ), as well as over the part of the control surface which encloses the engine. Z Z Z Z Z Z ρw( ⃗ w⃗ · ⃗n)dA + ρw( ⃗ · ⃗n)dA + ⃗ w ρw( ⃗ w⃗ · ⃗n)dA+ (Ai ) (Ao ) (AEi ) Z Z Z Z Z Z Z Z Z Z (1.101) ρw( ⃗ w⃗ · ⃗n)dA = ⃗tdA + ⃗tdA + ⃗tdA + ⃗tdA (AEo ) (Ai ) (Ao ) (AEi ) (AEo ) This equation is further simplified if we consider that w ⃗ · ⃗n on the inlet surface is replaced by −w1 or −wI and on the outlet surface by w2 ; wII or wIII. At the surface of the engine w ⃗ · ⃗n disappears, because the fixed parts are not flown through, thus the normal component of the velocity is zero. The flow at the inlet and outlet surfaces is presumably homogeneous. Thus the frictional stresses in Newtonian fluids disappear there. The stress tensor is then ⃗t = −p⃗n. The last two integrals represent the force exerted by the engine on the flow (or the negative of the force exerted by the flow on the engine). Z Z F⃗T = − ⃗tdA (1.102) (AEi ) and the external drag of the engine nacelle (force from the flow on the engine) Z Z ⃗ Fd = − ⃗tdA (1.103) (AEo ) 28 1 INTRODUCTION The external drag of an engine, which is only a few percent of the thrust, depends basically on its installation on the aircraft – a jet engine can be installed under the wing of an aircraft, on the fuselage, or even in the fuselage itself – which is why it is generally considered to contribute to the drag of the aircraft. This means, however, that for the calculation of the thrust of a jet engine Fd = 0 can be set. Assuming that A1 + AI = A2 + AII (1.104) which must always be true for a cylindrical control volume, and that p1 = pI = pII = p∞ (1.105) w1 = wI = wII = wIII = c∞ the thrust of a jet engine is obtained from Eqs. 1.99 to 1.105: F⃗T · ⃗e1 = FT = ṁ2 w2 − ṁ1 c∞ + A2 (p2 − p∞ ) (1.106) | {z } | {z } Momentum thrust Pressure thrust For the c

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