February 11, 2017 | Author: Juan García | Category: N/A
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Lecture Notes on Aerospace Propulsion Bioengineering and Aerospace Engineering Department Universidad Carlos III de Madrid 2015-2016
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Aerospace Propulsion
Lecture notes
These lecture notes were originally based on a course that has been taught several years at MIT by Manuel Mart´ınez-S´ anchez. Many of the figures have been taken or adapted from the lecture notes from that course and from the book Aircraft Engines and Gas Turbines by J.L. Kerrebrock published by MIT press. The lecture notes have been gradually improved upon by Manuel Mart´ınez-S´anchez, Manuel Garc´ıa Villalba, Pablo Fajardo, Mario Merino, and Andrea Ianiro. If you discover any errata, or have any suggestions, please contact us at: •
[email protected]
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Bioengineering and Aerospace Engineering Dept.
v. 2015-2016
Aerospace Propulsion
Lecture notes
Contents 1 Introduction to aerospace propulsion 1.1 Thrust generation and jet propulsion . 1.2 Effect of external expansion on thrust 1.3 Global performance parameters . . . . 1.3.1 Range of aircraft . . . . . . . . 1.3.2 Efficiencies . . . . . . . . . . .
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6 . 6 . 7 . 8 . 9 . 11
2 Aircraft Engine Modeling: the Turbojet 2.1 Thrust equation . . . . . . . . . . . . . . . . . . 2.2 Shaft balance for the turbojet . . . . . . . . . . . 2.3 Fuel consumption . . . . . . . . . . . . . . . . . . 2.4 Design parameters. Effect of mass flow on thrust. 2.4.1 Note on Ramjets . . . . . . . . . . . . . . 2.5 Propulsive efficiency . . . . . . . . . . . . . . . . 2.6 Thermal and overall efficiencies . . . . . . . . . .
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13 13 17 17 17 19 20 20
3 Introduction to Component Matching and Off-Design 3.1 Discussion on nozzle choking . . . . . . . . . . . . . . . 3.2 Component matching . . . . . . . . . . . . . . . . . . . . 3.3 Effects of Mach number . . . . . . . . . . . . . . . . . . 3.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Compressor-turbine matching. Gas generators. . . . . .
Operation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4 Turbofan Engines 4.1 Ideal turbofan model . . . . 4.2 Shaft balance . . . . . . . . 4.3 Velocity matching condition 4.4 Optimal compression ratio .
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31 31 33 33 34
5 Inlets and Nozzles 5.1 Inlets or Diffusers 5.2 Subsonic Inlets . 5.3 Supersonic Inlets 5.4 Exhaust nozzles .
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36 36 36 37 42
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6 Principles of Compressors and Fans 47 6.1 Euler’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 6.2 Velocity triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 6.3 Isentropic efficiency and compressor map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 7 Compressor Blading, design and multi-staging 7.1 Diffusion factor. Stall and surge . . . . . . . . . 7.2 Compressor blading and radial variations. . . . 7.3 Multi-staging and flow area variation . . . . . . 7.4 Mach Number Effects . . . . . . . . . . . . . . 7.5 The Polytropic Efficiency . . . . . . . . . . . . 7.6 Starting and Low-Speed Operation . . . . . . .
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Aerospace Propulsion
Lecture notes
8 Turbines. Stage characteristics. Degree of reaction 8.1 Euler’s Equation . . . . . . . . . . . . . . . . . . . . 8.2 Degree of Reaction . . . . . . . . . . . . . . . . . . . 8.3 Radial variations . . . . . . . . . . . . . . . . . . . . 8.4 Rotating blade temperature . . . . . . . . . . . . . .
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60 60 60 64 64
9 Turbine solidity. Mass flow limits. Internal cooling. 66 9.1 Solidity and aerodynamic loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 9.2 Mass flow per unit of annulus area and blade stress . . . . . . . . . . . . . . . . . . . . . . . . . . 67 9.3 Turbine cooling. General trends and systems. Internal cooling. . . . . . . . . . . . . . . . . . . . 68 10 Film cooling. Thermal stresses. Impingement. 10.1 Film cooling . . . . . . . . . . . . . . . . . . . . 10.2 Impingement cooling . . . . . . . . . . . . . . . 10.3 Thermal stresses . . . . . . . . . . . . . . . . . 10.4 How to design cooled blades . . . . . . . . . . .
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72 72 74 74 75
11 Combustion: Combustors and Pollutants 11.1 Combustion process . . . . . . . . . . . . 11.2 Combustor chambers . . . . . . . . . . . . 11.3 Combustor sizing . . . . . . . . . . . . . . 11.4 Afterburners . . . . . . . . . . . . . . . . 11.5 Pollutants: regulations . . . . . . . . . . . 11.6 Mechanisms for pollutant formation . . . 11.7 Upper-Atmospheric Emissions . . . . . . .
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77 77 79 80 81 82 84 86
12 Introduction to engine noise and aeroacoustics 12.1 Noise propagation . . . . . . . . . . . . . . . . . 12.2 Acoustic energy density and power flux . . . . . 12.3 Noise sources and noise modeling . . . . . . . . . 12.4 Jet Noise . . . . . . . . . . . . . . . . . . . . . . 12.5 Turbomachinery noise . . . . . . . . . . . . . . .
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89 89 90 91 94 95
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13 Engine rotating structures 97 13.1 Blade loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 13.2 Centrifugal stresses and disc design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 14 Fundamentals of rotordynamics 14.1 Bearings and engine arrangements 14.2 Lumped mass model . . . . . . . . 14.3 Critical speed . . . . . . . . . . . . 14.4 Forces on bearings . . . . . . . . . 14.5 Comments on blade vibrations . .
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Aerospace Propulsion
Lecture notes
Main nomenclature
Γ(γ)
General engine and flight magnitudes Specific impulse [m/s] (NB: also used in [s] in older texts) Thrust [N] Drag [N] Lift [N] Weight [N] Fuel mass flow [kg/s] Mass flow (of air) [kg/s] Nondimensional mass-flow (with respect to the critical mass flow rate) [-] Specific fuel comsumption [s/m] Fuel mass ratio to air mass flow, f = m ˙ f /m ˙ 0 [-] Specific heat value of the fuel [J/kg] Overall, propulsive and thermal efficiency; η = ηt ηp [-] Flight Mach number [-] Flow Mach number at stage i [-] Flight velocity [m/s] Exhaust velocity [m/s] Aircraft range [m] Component subindices Subindex for ‘compressor’ Subindex for ‘turbine’ Subindex for ‘burner’ (combustor) Subindex for ‘afterburner’ Subindex for ‘diffuser’ Subindex for ‘nozzle’ Subindex for ‘fan’ Thermodynamic properties: stagnation states and ratios Stagnation (total) pressure and temperature at station i [Pa, K] Static pressure and temperature at station i [Pa, K] Stagnation pressure ratios of compressor, turbine, etc. [-] Stagnation temperature ratios of compressor, turbine, etc. [-] Non-dimensional stagnation temperature at station i (normalized with T0 ) [-] Non-dimensional stagnation pressure at station i (normalized with p0 ) [-] Non-dimensional stagnation temperature at turbine inlet, θt = Tt4 /T0 [-] turbine inlet to free-stream Stagnation temperature ratio, pΘ = Tt4 /Tt0 [-] Sound velocity at station i, [m/s]. For an ideal gas, a = γRg T Mach number at station i, [-] γ+1 √ 2 2(γ−1) Choked nondimensional mass flow parameter, Γ(γ) = γ γ+1 [-]
u, v, w V D ψz R φ, ψ
Turbomachinery Radial, Azimuthal (tangential), and axial velocity components [m/s] Velocity modulus [m/s] Diffusion factor [-] Zweifel coefficient [-] Degree of reaction [-] Flow and power coefficients [-]
Isp F D L W m ˙f m ˙ 0; m ˙ m ¯i SFC f h η, ηp , ηt M0 Mi u0 u9 R c t b a d n f pti , Tti pi , Ti πc , πt , etc τc , τt , etc θi δi θt Θ ai Mi
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1 1.1
Lecture notes
Introduction to aerospace propulsion Thrust generation and jet propulsion
Propulsion is the1.-Introductory action of exerting a forward force F (termed thrust) on a vehicle. According to Newton’s Session L ecture second law, d (mv) = F + D (v) , (1.1) Propulsion O verview. dt this thrust is necessary to accelerate the vehicle (i.e., change its momentum) or to compensate any resistive ofmovement exerting in a forward force onaauniform vehicle.velocity This force on forcesPropulsion that oppose is tothe the action vehicle’s order to maintain (in themust case react of aircraft, this force is the aerodynamic drag). something else, like the ground for a ground vehicle. For an aircraft in flight, the reaction is Newton’s third law establishes that the existance of this force is necessarily linked to the existance of a exerted on air that has been admitted for the purpose, and that is therefore accelerated reaction force equal in magitude but opposite in direction, acting somewhere else, e.g. the ground for a ground backwards, together with fuelis products. On air a rocket space, wherebyno can be vehicle. For an aircraft in flight, thesome reaction exerted on the that hasinbeen admitted theair engine for the purpose, and that is therefore accelerated downstream, together with some fuel products. On a rocket in space, admitted, the reaction can only be exerted on some material that is carried on board and where there is no air to be admitted, the reaction force can only be exerted on some material that is carried ex!"##"$%&'%()*(%+!""$,%-+%."%.)##%+""/%)'%)+%012213%'1%"4&'"%'("%25'5&#%6170"%89'(75+':;% on board and expelled from the vehicle at high speed. As we will see, it is common to evaluate thrust in terms of theinmomentum gained by the medium (i.e., force), it must keptininmind mind that that this this is terms of the momentum gained bythe thereaction medium, butbut it must bebekept isan an indirect procedure, and the thrust force is, by definition, the force exerted by the medium on the vehicle (in indirect procedure, and the thrust force is truly exerted by the medium on the vehicle (in complicated ways, to be sure). complicatedtheways, to be sure). propulsion device is the Rocket. In a rocket combustion chamber, Conceptually, simplest aerospace some propellants carried on board are reacted together to generate heated gas at high pressure, and this gas is allowed to flow through ansimplest open ductpropulsion (a nozzle), where it pushes the walls and is pushed backwards Conceptually, the device is the forward Rocket.onSome propellants carried on by the walls. The speed u9 of the gas as it leaves the nozzle is a measure of the momentum per unit mass it boardand are so, reacted together to generate has gained, if the mass flow rate is m˙ 9 , theheated thrust gas is at high pressure, and this gas is allowed to
flow through an open duct 8&%931 p0 , one could add positive thrust by adding some extra diverging area to the nozzle. This fact will become clear in the study of nozzle flows. These comments can be extended easily to other engine types. As a final point, you may be wondering how it is possible to have p9 different from p0 at the exit from the nozzle. Indeed, this would not be possible if the flow were subsonic there, because there cannot be a significant pressure difference across a low-speed nearly parallel flow. But if the flow is supersonic, as it almost always is in propulsive nozzles, the pressure equalization takes place through more complex flow features, such as oblique shocks anchored at the nozzle lip, and the internal pressure at the exit plane can be different from the outside pressure.
1.3
Global performance parameters
For all the devices discussed, thrust is proportional to fuel flow rate, and it is advantageous to minimize this flow rate, or to maximize the ratio known as specific impulse Is =
8
Thrust F = Fuel mass flow rate m ˙f
(in m/s).
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Aerospace Propulsion
Lecture notes
This is the rational definition, but historically, specific impulse has been defined as thrust per unit weight flow rate of fuel F I˜s = (in s). (1.7) gm ˙f In this notes we are going to try to use most of the times the first definition, although students should always check units carefully to be sure about which definition is being used. For a rocket, the only flow rate is that of the fuel, but for an airbreathing engine, care must be taken to use the fuel flow rate in the calculation of Is , not the air or the total flow rate. The inverse of Is is also in common use, especially for airbreathing engines, and it is called the Specific Fuel Consumption (or Thrust Specific Fuel Consumption), often expressed in picturesque units such as “pounds of fuel mass per hour-per pound of thrust force”, or the metric equivalent, SFC or TSFC =
m ˙f 1 = Is F
(in s/m).
(1.8)
EF7$0)G)2GH'1 p0 so F (u9 − u0 ) p9 m(u ˙ 9 − u0 ) + (p9 − p0 )A9 p0 A9 ϕ2 = + = = m2 Γ √ − , (3.31) pt0 A2 pt0 A2 pt0 pt0 A2 RTt0 and this time M9 = 1, leading to u9 =
p
r γRT9 = γR
2 Tt5 = γ+1
r
2γ RTt0 Θτt γ+1
(3.32)
u9 depends on Θ alone. Since we also know that m2 and p9 /pt0 are functions of Θ alone, it RTt0 makes sense to separate out equation (3.31) in the form u9 p 9 A9 u0 p 0 A9 ϕ2 = m2 Γ √ + − m2 Γ √ + , (3.33) pt0 A2 pt0 A2 RTt0 RTt0 " # " # γ r γ−1 r 2γ A9 γ 2 1 A9 ϕ2 = m2 Γ − m2 ΓM0 . (3.34) Θτt + τt τc + γ γ+1 γ+1 A2 θ0 θ γ−1 A2 so that √
0
In equation (3.34) the first bracket is a function of Θ only, ϕ∗2 (Θ). Once again, the normalized thrust depends on both, Θ and M0 , but the structure is fairly simple, and in particular, the portion ϕ∗2 of ϕ2 (neglecting the incoming momentum and the external pressure) is a function of Θ alone. This portion can be very easily scaled between conditions, and the rest can be subtracted separately. 8. The Operating Line in the compressor map. Compressor performance is typically presented as a map of πc vs. m2 , with lines of constant normalized rotational speed ω and ηc , the isentropic efficiency of the compressor (to be defined in future lectures), superimposed. The details are the subject of later Lectures, but the general shape is as shown in Fig. 3.2, where the flow and speed variables are renormalized by the “Design” values. Actually the “nominal operating line” shown in the figure is not a property of the compressor, but rather of the rest of the engine. We can calculate this line with the information we have now, before deciding what particular compressor to use. From eq. (3.17), γ
A4 τcγ−1 √ , m2 = A2 Θ
(3.35)
and from the shaft power balance (Eq. 3.12), τc − 1 , 1 − τt where we recall that τt is fixed for a fixed geometry. Eliminating Θ, r γ A4 γ−1 1 − τt m2 = τc , A2 τc − 1 Θ=
(3.36)
(3.37)
or, in terms of πc
A4 m2 = πc A2
s
1 − τt γ−1
(3.38)
πc γ − 1 which is the equation for the operating line (written in reverse). If the compressor is already available, we see from eq. (3.38) that we can adjust the nozzle area A4 to place this line in a ”good” place on the map, i.e., below the stall line and through the best efficiency points. Since m2 depends on Θ = Tt4 /Tt0 , varying Tt4 moves the operating point along the operating line, and this is what the pilot does with the throttle stick to power the engine up or down. At each selected Θ, the engine settles to a πc , a M2 , a (normalized) rotation rate, etc. 26
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Aerospace Propulsion
Lecture notes
Figure The corrected airflow p 3.2: Performance map for a typical high-pressure-ratio p compressor. m ˙ 0 T2 /298.15K/(p2 /101325Pa) and corrected rotational velocity N T2 /298.15K are customarily used in these diagrams to account for ambient condition variations at the entrance of the compressor.
3.3
Effects of Mach number
If we look at operation of a given engine at different flight Mach numbers, we may try to maintain the same non-dimensional conditions throughout, which, as we have seen, can be done by maintaining for example a constant compressor inlet Mach number M2 . This, in turn guarantees a constant Θ = Tt4 /Tt0 , but since now we have a varying Mach number, so that Tt0 increases with M0 , we may find that the turbine inlet temperature Tt4 needs to become too high at the higher Mach numbers. For example, Tt4 would have to be 1.8 times higher at M0 = 2 than at static conditions, and 2.25 times at M0 = 2.5. A more reasonable assumption is that the ratio θt = Tt4 /T0 can be maintained the same at all Mach numbers, since at least in the tropopause (between 11 km and 20 km above mean sea level), T0 is almost θt (1 − τt ) invariant. The compressor temperature ratio now follows from τc = 1 + , where the numerator θ0 is a constant; thus, τc will be lowered as the Mach number increases, less strongly than as would √ but be required in order to maintain maximum thrust per unit flow θt /θ0 . The flow parameter is now determined by Eq. (3.17), i.e. compressor-turbine flow matching, and then the compressor-inlet Mach number from Eq. (3.15). Once these parameters are known, we can use Eq. (3.27) to calculate the normalized thrust; since we are interested in the effect of Mach number, it makes sense to re-normalize
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Aerospace Propulsion
thrust by p0 A2 , or
3.4
Lecture notes
γ F = ϕ2 θ0γ−1 . p0 A2
(3.39)
Examples
A note on Θ: the near-constancy of the engine operating point In this section different examples of aircraft operation are given in order to show the possibility of maintaining a near-constant value of the parameter Θ. Two important points in the flight envelope of an aircraft engine are: (a) Take-off conditions (M0 ' 0.25, T0 ' 290K) , and (b) End-of-climb conditions (M0 ' 0.85, T0 ' 220K).
Using γ = 1.4, the total temperatures are Tt0 = 290 1 + 0.2 × 0.252 = 294K (take-off) and Tt0 = 220 1 + 0.2 × 0.852 = 252K (end of climb). Suppose the engine is dimensioned for end-of-climb, which is common, and that the peak temperature Tt4 , which will have to be maintained for many hours of cruise, is selected at a conservative Tt4 = 1600K. We then have Θ = Tt4 /Tt0 = 6.35 at this condition. If we now decided to maintain Θ = 6.35 also for take-off, we would need then Tt4 = 6.35 × 294 = 1868K. While this is too high for long-term operation (creep, corrosion), it may be acceptable for the few minutes per cycle that the engine will be at take-off maximum power. If this actually done, the engine operates at a fixed nondimensional point all the way from take-off through start of cruise.
Consider now a commercial jet in a long cruise. As the fuel is consumed and the weight decreases, 1 the lift must decrease L = ρ0 u20 Aw cL . Now, the lift coefficient will be kept close to that for optimum 2 aerodynamic efficiency, L/D|max , and the Mach number M0 is unlikely to change much, as it will stay just below the transonic drag peak, and so u20 will be proportional to T0 due to the speed of sound variation. Together with the density part of lift, we can see that the ambient pressure p0 must be decreasing in proportion to the airplane’s weight, i.e., the plane must be climbing gradually. Turning now to the forward force balance, given a constant L/D, the drag, and hence the engine thrust, must also be decreasing in time in the same proportion as the ambient pressure. Therefore, from Eq. (3.27), the nondimensional thrust ϕ2 (Θ, M0 ) will remain constant, and since M0 does too, the peak temperature ratio Θ will also remain constant, and with it all the important ratios like τc , M2 , etc. In other words, Θ may not vary much among (important) flight conditions, and the engine will be operating at a fixed nondimensional condition (constant compression ratio, nondimensional flow, compressor inlet Mach number, etc.). But of course, the dimensional quantities (flow rate, peak pressure, etc.) will be different, depending on p0 , etc. A numerical example We want to analyze the parameters during the engine operation at different regimes (different values of M0 ), but keeping a constant θt = 7, or Tt4 = 1540K (T0 = 220K) in the stratosphere. The geometry of the engine must have been specified in advance. This means that the turbine temperature ratio (Eq. 3.10) is a known fixed number. For the example, we select τt such as to obtain maximum thrust at M0 = 1. From the shaft balance equation, θ0 τt = 1 − (τc − 1) (3.40) θt √ and we put now θ0 = 1.2 and τc = 7/1.2 = 2.2048 (at M0 = 1). This fixes τt = 0.7935. Similarly, the area ratio A4 /A2 must have been fixed, and we select it here so as to obtain at M0 = 1 a compressor-face Mach number M2 = 0.5, which, from Eq. (3.15) implies m2 = 0.7464.
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Lecture notes
In order to analyze how m2 is scaled at different conditions, using eq. (3.17) we can write, τ 3.5 r θ c 0 m2 = 0.7464 2.2048 1.2
(3.41)
and the rest of the steps are as described above. The table 3.2 summarizes the results. M0 θ0 τc m2 M2 ϕ2 F/(p0 A2 )
0 1 2.4458 0.9796 0.8486 2.9117 2.9117
1 1.2 2.2048 0.7464 0.5 1.5531 2.9399
2 1.8 1.8032 0.4523 0.2737 0.5795 4.534
2.5 2.25 1.6426 0.3648 0.2172 0.3503 5.985
Table 3.2: Summary of results of the numerical example We find that at a fixed altitude the thrust is nearly constant up to Mach 1, then it increases rapidly. Actually the increase is less rapid than this simple model predicts, because of losses in the supersonic flow in the engine inlet. The previous model has been computed assuming constant altitude. However, we should note that an aircraft normally flies at increasing altitude as the Mach Number increases, so that dynamic pressure γp0 M02 is roughly constant. In this case the change in F between Mach 1 and Mach 2 is actually a thrust reduction.
3.5
Compressor-turbine matching. Gas generators.
Assembling the compressor and the turbine with a combustor between gives us a gas generator, which is the heart of any gas turbine engine. Once we understand its behavior we can appreciate most of the real characteristics of aircraft engines, and graduate from thinking of them as a lot of abstract equations. Actually, we have already done in the previous section most of the work needed to understand the performance of an ideal Gas Generator, including the important concept of the “Compressor operating line”, which is set by the flow passing characteristics of the rest of the engine. We summarize here the main findings 1. If the nozzle and the turbine stators are both choked, the turbine temperature ratio is fixed once the flow area ratio A4 /A8 is set. 2. One additional single parameter is sufficient to specify all the other gas generator parameters. This can be the temperature ratio Θ = Tt4 /Tt2 , or the compressor temperature ratio τc , or the engine-face Mach number M2 , or the normalized air mass flow, or the fuel flow ratio. 3. A “Compressor Operating Line” in the plane of compressor pressure ratio vs. normalized flow can be calculated once the ratios of all the flow areas are known. For a given choice of one of the parameters listed above, a point is selected along this operating line. 4. All the above is independent of the compressor specifics. After the compressor has been selected, its performance map (pressure ratio vs. normalized flow) contains normalized rotational speed lines as additional information. This parameter is therefore to be added to our list of possible parameters (as is done in the figure at the end of this lecture), each of which uniquely specifies the state of the gas generator. It follows from the above that a single degree of freedom is left to the pilot (or to the engine controller), unless geometry can be varied. It is probably most intuitive to think of this unique freedom as the normalized fuel factor, or the peak temperature ratio Tt4 /Tt2 , since these closely relate to the engine throttle control. 29
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It is to be noted, however, that many idealizations have been made to obtain these simple results. If the turbine or the nozzle un-choke, or if the engine inefficiencies are rigorously accounted for, the overall detailed behavior is more complex, but its main qualitative features are not too different. Using these ideas, one can generate and plot a set of Gas Generator Characteristics, such as those below. With this Gas Generator, we can analyze several engines 1. Turbojet 2. Turbofan 3. Turboprop 4. Unducted fan 5. Helicopter-Turboshaft Notice that the single free variable chosen for this particular plot is the normalized rotational speed (as a fraction of its design value). The quantity in the denominator is the non-dimensional compressor-face temperature θ2 = T2 /T0 , because the blade speed ωr is made non-dimensional with the speed of sound at station 2.
Figure 3.3: Pumping characteristics for a gas generator with Tt4 /Tt2 = 6
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Lecture notes
Turbofan Engines
In §2.4 we saw that for low M0 , ηp is not very high for turbojets. In essence, there is too much kinetic energy in the exhaust jet (per unit mass). This is the main reason for using the Turbofan engine. The turbofan separates the inflow into a core airstream and a secondary or bypass airstream. The core stream goes into a gas generator which is essentially a turbojet. The bypass stream is simply moved by a fan, which is essentially a big ducted propeller that increases the total pressure and temperature of the air. This fan is driven by the turbine inside the core part of the engine, which usually has two shafts: the inner shaft rigidly connects the low-pressure turbine with the fan, and the outer shaft (or spool) connects the high-pressure turbine with the high-pressure compressor. The spool and the shaft rotate freely using a bearing system. The jets that result from the core and bypass streams can be merged at the exit or before a possible afterburner stage, or be released independently into the ambient. If it is designed for subsonic cruise flight it looks like the sketch of Fig. 4.1.
13
19
9
Figure 4.1: Schematic diagram of turbofan engine If designed for both subsonic cruise and for supersonic flight with afterburning it looks more like Fig. 4.2. In both of these diagrams the inlet has been greatly simplified, of course. The numbering of the stations follows the standard SAE AS755: numbers in the bypass stream are prepended with a “1”, to express that this is a different air stream from the core one (could be regarded as prepended by index “0”).
4.1
Ideal turbofan model
Let us now see how we can model these engines thermodynamically, following the same process as with the turbojet. The total thrust of the device is given by Eq. 1.4. Notice first that we now have two air streams: the core stream through the gas generator, with a mass flow m, ˙ and the bypass stream, with a mass flow that can be written as a fraction of the former, αm. ˙ We call α the bypass ratio (BPR). To obtain a simplified model of the device sketched in Fig. 4.1, we will take the following assumptions, which will be familiar from the turbojet model: • isentropic processes in the fan, compressor, turbine 31
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13
7
9
Figure 4.2: Schematic diagram of turbofan engine with afterburner • constant total pressure through combustor • fuel mass flow neglected with respect to air flow • the nozzle of the core jet and the bypass stream are both adapted to the ambient pressure (p9 = p19 = p0 ). Of course, this may be not the case in many circumstances: as in the turbojet, a fixed convergent nozzle is more likely to be sonic and under-expanded, namely, to have M8 = M18 = 1 and p9 , p19 > p0 . The derivation of the model with non-matched conditions is left to the student (like in the turbojet case). Let us start with the core jet. With these hypotheses, the expression for the thrust contribution of the core jet is the same as for the turbojet before applying the shaft balance. Thus from equation (2.24) we get for the core jet5 s Fc 2 θt (θ0 τc τt − 1) = − M0 . (4.1) ma ˙ 0 γ−1 θ0 τc For the bypass stream, there is no turbine, so let us repeat the argument γ−1 2 M19 = T0 θ0 τf , Tt19 = T19 1 + 2 γ γ γ − 1 2 γ−1 pt19 = p19 1 + M19 = p0 δ0 πf = p0 (θ0 τf ) γ−1 . 2 So, for p19 = p0
γ−1 2 M19 = θ0 τf , 2 r 2 = (θ0 τf − 1). γ−1
1+ M19
Using (4.4), from eq. (4.2) we also have that T19 = T0 so r u19 2 FBP = M0 −1 = (θ0 τf − 1) − M0 . αma ˙ 0 u0 γ−1
(4.2)
(4.3)
(4.4) (4.5)
(4.6)
5 Quick observation: τ = T t13 /Tt2 . For congruency with the definitions of previous chapters, we keep τc = Tt3 /Tt2 , including f in it the possible compression of the core stream due to the central part of the fan.
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Adding this thrust to the thrust of the core jet, we find the total thrust "s # r F 2 2 θt = (θ0 τc τt − 1) − M0 + α (θ0 τf − 1) − M0 ma ˙ 0 γ−1 θ0 τc γ−1
4.2
(4.7)
Shaft balance
Now we need τt , which has to be calculated with the applicable shaft balance in this case. We will see later that most engines have two shafts, so each will set a separate balance equation. For now, however, we will lump their power together; at off-design conditions, this would have to be modified. Applying the energy equation to the compressor, fan and turbine, and forcing the work extracted by the latter to be consumed by the former, mC ˙ p (Tt4 − Tt5 )
= mC ˙ p (Tt3 − Tt2 ) + αmC ˙ p (Tt13 − Tt2 ),
θt (1 − τt )
(4.8)
= θ0 (τc − 1) + αθ0 (τf − 1),
(4.9)
θ0 [(τc − 1) + α(τf − 1)]. θt
(4.10)
leading to τt = 1 −
Substituting this gives us our result for the thrust of the turbofan. It doesn’t really help to carry out the substitution at this point. Instead, let us think about how to simplify the expressions to make them more easily understandable.
4.3
Velocity matching condition
For this engine there are more parameters than for the turbojet • θt , τc - as before • α, τf - characterizing the fan flow. We can relate some of the parameters to others by noting that the highest propulsive efficiency is realized momentum when u19 = u9 : this maximizes the ratio of in the jets6 . Setting the two velocities to be the energy same is referred to as velocity matching, and is usually imposed as a design constrain for the nominal operation √ point. For this situation, the two ’s in the thrust equation are equal, and this requires that: (θ0 τc τt − 1) 6A
θt = θ0 τf − 1 θ 0 τc
(4.11)
simple derivation: for two streams 1 and 2, momentum
=
kinetic energy
=
m ˙ 1 u1 + m ˙ 2 u2 , 1 1 2 m ˙ 1 u1 + m ˙ 2 u22 . 2 2
Now we maximize the momentum at a constrained energy. Using a Lagrange multiplier λ, we form the auxiliary function 1 1 φ=m ˙ 1 u1 + m ˙ 2 u2 + λ m ˙ 1 u21 + m ˙ 2 u22 , 2 2 and equate to zero the derivatives w.r.t. u1 and u2 , leading to m ˙ 1 + λm ˙ 1 u1
=
0,
m ˙ 2 + λm ˙ 2 u2
=
0.
−1 So that u1 = u2 = . λ
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Solving for τf as a function of α, using eq. (4.10), after some algebra we obtain
τf =
1 + θt + θ0 (1 + α − τc ) − θ0 (1 + α)
θt θ0 τc
.
(4.12)
When this equation is satisfied, the thrust equation becomes simply r 2 F = (1 + α) (θ0 τf − 1) − M0 ma ˙ 0 γ−1
4.4
(4.13)
Optimal compression ratio
Now, as for the turbojet, there is still the choice of the compression ratio. Generally we want to choose it for maximum power, which in this case for given turbine inlet temperature θt and bypass ratio α means maximum thrust. To maximize F in this velocity-matching condition, we choose τc to maximize τf , since F increases monotonically with τf . From the expression for τf , eq. (4.12), √ θt θt ∂τf = 0 → τc |max(F ) = = −θ0 + . (4.14) 2 ∂τc θ0 τc θ0 Notice that this is precisely the same result as for the Turbojet. Substituting in the expression for τf τf |max(F ) F m ˙ 0 a0 max(F )
√
2 θt − 1 +1 = θ0 (1 + α) v u u 2 = (1 + α) t γ−1
(4.15) √
! 2 θt − 1 + θ0 − 1 − M 0 1+α
(4.16)
Now we have 3 parameters, • θt - which we set at the maximum feasible value • M0 - flight speed • α - the prime variable distinguishing the turbofan As before, for u9 = u19 we still have: ηpropulsive =
2 u9 +1 u0
=
2 F ma ˙ 0 +2 M0 (1 + α)
(4.17)
The variation of F/(ma ˙ 0 ) and ηp with M0 and α is shown in Figures 4.3 and 4.4 for θt = 6.25. Of course for the higher bypass ratios we are really only interested in the range of M0 < 1, but the lower bypass and higher M0 range may be of interest for a supersonic transport.
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Lecture notes
10 9 8
α 20
7 6 5
F 5 ma ˙ 0
1
4 3 2 1 0 0
0.5
1
1.5
2
2.5
3
M0 Figure 4.3: Variation of F/(ma ˙ 0 ) with M0 and α for θt = 6.25 and γ = 1.4.
1 0.9
α 0.8 20 0.7 5 0.6 1
ηp
0.5 0.4 0.3 0.2 0.1 0 0
0.5
1
1.5
2
2.5
3
M0 Figure 4.4: Variation of ηp with M0 and α for θt = 6.25 and γ = 1.4.
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Lecture notes
Inlets and Nozzles
5.1
Inlets or Diffusers
While the Gas Generator, composed of the compressor, combustor and turbine, is the heart of any gas turbine engine, the overall performance of the propulsion system is strongly influenced by the inlet and the nozzle. This is especially true for high M0 flight, when a major portion of the overall temperature and pressure rise of the cycle are in the inlet, and a correspondingly large part of the expansion in the nozzle. So it is important to understand how these components function and how they limit the performance of the propulsion system. Inlets or Diffusers These two titles are used interchangeably for the component that captures the oncoming propulsive streamtube and conditions it for entrance to the engine. The function of the inlet is to adjust the flow from the ambient flight condition, to that required for entry into the fan or compressor of the engine. It must do this over the full flight speed range, from static (takeoff) to the highest M0 the vehicle can attain. Comparing the simple diagrams of subsonic and supersonic inlets, we can appreciate that the subsonic inlet has the simpler task:
Figure 5.1: Schematic diagram of diffusers For any inlet the requirements are two • To bring the inlet flow to the engine with the highest possible stagnation pressure. This is measured by t2 the Inlet Pressure Recovery, πd = ppt0 . • To provide the required engine mass flow. As we shall see the mass flow can be limited by choking of the inlet.
5.2
Subsonic Inlets
The subsonic inlet must satisfy two basic requirements: • Diffusion of the free-stream flow to the compressor inlet condition at cruise. • Acceleration of static air to the compressor inlet condition at takeoff. There is a compromise to be made because a relatively thin ”lip” aligned with the entering flow is best for the cruise condition. This is to avoid accelerating the flow, already at a Mach number of the order of 0.8, to supersonic speeds that will lead to shock losses. But a more rounded lip will better avoid separation for the takeoff condition because the air must be captured from a wide range of angles: Usually subsonic diffusers consist of a divergent duct. Therefore, the minimum area is located at the inlet. The inlet area, A1 , is set to just avoid choking (M1 < 1) when M2 (compressor inlet) has its largest value (this 36
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Figure 5.2: Schematic diagram of a subsonic diffuser is set by the maximum blade Mach number and the blade shape). Since the total temperature and pressure are common, we must have m1 (M1 )A1 = m2 (M2 )A2 . (5.1) A well designed subsonic inlet will produce a stagnation pressure recovery πd in the order of 0.97 at its design condition.
5.3
Supersonic Inlets
At supersonic flight speeds the pressure and temperature rise in the inlet can be quite large. For the best thermodynamic efficiency it is important that this compression is as nearly reversible (isentropic) as possible. At a flight Mach number of 3 the ideal pressure ratio is pt0 δ0 = = p0 while the temperature rise is θ0 =
γ−1 2 1+ M0 2
γ γ−1
= 36.7
Tt0 γ−1 2 =1+ M0 = 2.8 T0 2
(5.2)
(5.3)
For the turbojet cycle the compression is partly in the inlet and partly in the compressor: If the diffusion from point 0 to point 2 is not reversible, the entropy increases, and this results in a lower value of pt2 . This has two effects: • The expansion ratio of the nozzle is decreased, so the jet velocity and thrust are lower. • The lower pressure at point 2 limits the mass flow through the compressor to a lower value than for isentropic diffusion.
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There is, thus, a double penalty for losses in the inlet. Unfortunately some losses are inevitable and are higher as M0 is larger. To see why, we will discuss the flow in supersonic diffusers, beginning with the simplest. 1. Normal-Shock diffuser All existing compressors and fans require subsonic flow at their inlet with 0.5 < M2 < 0.8 at high power conditions. So the inlet must reduce the flow Mach number from M0 > 1 to M2 < 1. The simplest way to do this is with a Normal Shock Wave.
Figure 5.3: Normal shock wave Here M1 < 1 is entirely determined by M0 , according to the normal shock relation
M12
γ−1 2 M0 2 = γ−1 γM02 − 2 1+
(5.4)
The stagnation and static pressures are also determined by M0 pt0 pt1 p1 p0
1 γ γ − 1 γ−1 (γ − 1)M02 + 2 γ−1 2γM02 − γ+1 γ+1 (γ + 1)M02 2γ M02 − 1 1+ γ+1
= =
(5.5) (5.6) (5.7)
For low supersonic speeds, such diffusers are adequate because the stagnation pressure loss is small, but at M0 = 2, pt2 /pt0 ≈ .71, a serious penalty, and at M0 = 3, pt2 /pt0 ≈ .32. For example the F-16 fighter has a simple normal shock diffuser, while the F-15 has an oblique shock diffuser such as will be discussed next. 2. Oblique - Shock diffusers The losses can be greatly reduced by decelerating the flow through one or more oblique shock waves, the deflection and the pressure rise of each being small enough to be in the range where the stagnation pressure ratio is close to unity. It is very important to understand that an Oblique Shock Wave is in fact just a normal shock wave standing at an angle with respect to the flow. For an oblique shock wave the relevant Mach number is that corresponding to the normal component of the velocity. 38
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1 0.9 0.8 0.7 0.6 0.5
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igure 5.4: Stagnation pressure ratio pt1 /pt0 (dashed line) and static pressure ratio p0 /p1 (solid line). Red, γ = 1.66. Green,,7)#$42+4!4#!(+.*$'42+.!4024!2+!D5%,3(*!E0#-B!,'!,+!12-4!K('4!2!+#$72%!'0#-B! γ = 1.4. Blue, γ = 1.3. '42+.,+6!24!2+!2+6%*!4#!40*!1%#&=!!! !
! ! L%%!#1!40*!-02+6*!2-$#''!40*!'0#-B!42B*'!)%2-*!5*-2('*!40*!82-0!+(75*$!+#$72%!4#! 40*!'0#-B!,'!%2$6*$!402+!(+,4H=!!G0*!J*%#-,4H!+#$72%!4#!40*!'0#-B!.*-$*2'*'!&,40! Figure 5.5: Oblique-shock diffuser -#+'*3(*+4!,+-$*2'*'!,+!4*7)*$24($*!2+.!)$*''($*/!5(4!40*!J*%#-,4H!)2$2%%*%!4#!40*! '0#-B!,'!(+-02+6*.F!
All of the change across the shock wave takes place because the Mach number corresponding to the velocity component normal to the shock is larger than unity. The velocity normal to the shock decreases with consequent increase in temperature and pressure; at the same time the velocity parallel to the shock is unchanged. M1n is given in terms of M0n by the same relation given for M1 as a function of M0 . The relations obtained for the pressure ratios through a shock wave as a function of M0 are here function of M0n . The advantage is that even for M0 very large, M0n can be made close to 1. Of course the condition for having a weak wave is to have at least M0n = 1, or sin(θ) →
1 M0
(5.8)
By choosing the wedge angle (or deflection angle) δ we can set the shock angle. A series of weak oblique shocks, for each of which the Mn is near unity, hence all lying in the range of small stagnation pressure loss, can yield an efficient diffuser. 3. Diffusers with internal contraction One might ask why we do not just use a convergent-divergent nozzle in reverse as an inlet 39
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igure 5.7: Is it possible to use a convergent-divergent nozzle in reverse as an inlet? ! F$)!,%'1*!2&;!:10!:)!8-!$-*! At A∗ and considering that the flow is chocked at the throat, At = A∗ , then A1 > A(M0 )
(5.10)
(5.11)
This means that the streamtube that can be captured by the diffuser (A(M0 )) is smaller than the diffuser inlet (A1 ); the rest of the flow into the frontal area A1 has to be somehow ”spilled”. This ”spillage” cannot occur in supersonic regime since information is not able to travel upstream. Therefore a detached shock forms ahead of the lip and the spill of excess mass flow occurs in the subsonic flow behind it. This situation is shown in Fig. 5.8 (b). It has to be remarked that after the normal shock wave the stagnation pressure decreases and the mass flow that has to be ”spilled” is thus even higher. We have now to consider the subsonic Mach number after the normal shock wave M0,1 When the flight Mach number, M0 , is increased A(M0,1 ) 1 , the normal shock will stand just at the lip, as in figure to the critical value M0 c such that A At = A∗ 5.8 (c). But in this position it is unstable and will move downstream if perturbed. Unfortunately, the shock at the full flight Mach number is very lossy, and it is not practical to simply force the aircraft to 41
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3
2.5 A1/A*
A/A*
2
A /A*
1.5
0
1
M=1
0.5
0 0
0.5
1
M0 0, βa > 0 we get a functional dependence as shown in Fig. 6.8.
Figure 6.8: Temperature ratio vs. ratio of the axial flow velocity to blade speed for various tangential Mach numbers.
6.3
Isentropic efficiency and compressor map
In reality, compressors provide a lower stagnation pressure raise than the ideal one for a given work input, γ/(γ−1) πc = τc , due to entropy increase by viscosity and other factors. Conversely, this means that we need to provide more work than the ideal (i.e., we need to increase more the total temperature) to achieve the same pressure ratio. If we consider a given stagnation pressure raise, the ratio of the ideal temperature increase to
51
Bioengineering and Aerospace Engineering Dept.
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Aerospace Propulsion
Lecture notes
the actual, necessary one is called the (isentropic) compressor efficiency,
(∆Tt )ideal ηc = = (∆Tt )actual γ
Tta
ptb pta
! γ−1 γ
Ttb − Tta
− Tta
=
ptb pta
! γ−1 γ
Ttb −1 Tta
−1
.
(6.17)
γ
Since ηc < 1, we have πc = [1 + ηc (τc − 1)] γ−1 < τcγ−1 . We usually plot Euler’s equation in terms of the pressure ratio rather than the temperature ratio, and include superimposed lines of efficiency, so that the map looks like in Fig. 6.9.
Figure 6.9: Performance map for a typical high-pressure-ratio compressor (refer to the caption of Fig. 3.2.
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7 7.1
Lecture notes
Compressor Blading, design and multi-staging Diffusion factor. Stall and surge
By decreasing βa and βb0 , or even making them negative, we can increase Ttb /Tta . What limits this increase? The answer is best given in terms of a Diffusion Factor, which describes the tendencies for the boundary layer to separate under the influence of the pressure rise in the blade passage, as sketched in Fig. 7.1.
Figure 7.1: Schematic diagram of velocity distributions on suction and pressure surfaces of a blade, showing the diffusion from a maximum velocity Vmax to the final velocity Vb and the resultant thickening of the boundary layer on the suction surface. The critical region is the suction surface (extrados) of the blade. This region feels a pressure rise due to the decrease of V 0 from Va0 to Vb0 , (the prime is a reminder that these velocities in the relative frame) and also due to the acceleration, followed by deceleration on the suction side. For compressors and fans, we define the Diffusion Factor as V0 |vb − va | D = 1 − b0 + , (7.1) Va 2σVa0 which is a crude but effective way to account for these two flow deceleration components. Here, σ = c/s is the solidity of the rotor, where c is the chord of the blades and s is their spacing (in the peripheral direction). A value of about 0.5 for D is the upper limit for good efficiency. As the angles of atack increase in the compressor a point is reached when the flow begins to separate from the blades, with serious consequences. Large incidence angles occur when the ratio (axial velocity)/(blade speed) is low, i.e., for a fixed rotational velocity, at reduced flow, and hence at increased pressure ratio. All compressors have a fairly well defined “stall line” that runs diagonally from low flow and low πc , to high values of both. Operation must be restricted to the region below and to the right of this line. The operational line for the compressor runs roughly parallel to this stall line. As we saw in §3, the reason for the compressor to be restricted to its operation line has to do with flow continuity between it and the turbine and nozzle. It is a fortunate fact that the line thus defined tends to correspond to a constant blade incidence, which, when chosen properly, is close to that which optimizes blade performance, and so the operating line is more or less the peak compressor efficiency line. 53
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Lecture notes
The phenomena that occur at and beyond stall are complex, dynamical and highly nonlinear. A detailed understanding of these effects has only emerged in the last decades, and remedial measures based on this understanding are still under evaluation. In most axial compressors, incipient separation first leads to the Rotating Stall phenomenon: sections of the stalling rotor then operate in deep stall, with almost zero flow, while the rest carries normal or high flow per unit frontal area. These regions move backwards in the rotating frame at ∼ 0.4 − 0.6ωr, so that, when observed from rest they rotate forward, but only at a fraction of the rotor speed. The reason for the bimodal flow distribution is that adjacent streamtubes become unstable with respect to flow exchange: if the rotor as a whole is near stall conditions, and one streamtube loses some flow and diverts it to its neighbors, the streamtube with less flow goes into stall and loses even more flow, while the neighbors gain flow and remain stable. The reason the “stall cells” move backwards relative to the rotor is an elaboration of the same argument: if one flow passage stalls, the swirling incoming flow is re-routed such that the passage ahead of the one stalled sees a more axial flow, while the one behind sees a larger incidence angle. The stall moves to this trailing passage, and the stalled passage clears. Rotating stall, since it moves rapidly about, tends to average out and, aside from high-frequency excitation, may not be dynamically significant. On the other hand, the net compressor performance drops strongly, and in addition, it is not easy to reverse once started, except by stopping and re-starting the engine. Detecting rotating stall and instituting the appropriate control reaction is very important. If the engine controls simply detect a loss of pumping performance (pressure loss) they may react by increasing fuel flow, which combined with the reduced airflow, may lead to overheating and burnout. Under some conditions having to do mainly with the ratio of flow inertia to flow passage capacitance, the complete “pumping system” (compressor plus choked turbine nozzles) can enter a global oscillation, called Surge. Unlike rotating stall, surge involves deep oscillations or even reversals of the whole flow through the compressor, and can be mechanically destructive (certainly quite noticeable, in the form of loud, repeated bangs, accompanied by flame ejection from both ends of the engine). When stall is reached, the engine may go into either Rotating Stall or Surge. The detailed mechanisms that determine which of the phenomena will prevail are encapsulated in Greitzer’s “B parameter” r ωr Vp B= (7.2) 2a Vc where a is an average speed of sound in the burner, Vp is the “plenum” volume (mainly the burner volume) and Vc is the volume in the compressor flow passages. In a single-stage compressor, values of B below ∼ 0.8 lead to rotating stall, while higher values lead to surge. For multistage (N ) compressors, Bcrit is lower by somewhere √ between N and N . One favorable aspect of surge (as opposed to rotating stall) is that it can usually be cleared by simply reducing fuel flow (or, in a test stand, opening the downstream throttle).
7.2
Compressor blading and radial variations.
In all of the discussion so far the blade speed has been taken as a parameter. In fact the blade linear speed varies with the radial location in the compressor, so the “designs” that we have discussed are applicable only at one radius in any real compressor. In practice it is usual to begin a compressor design with such a treatment, called a “mean line design” carried out at some mean radius. But the effects of the blade speed variation with radius are important, so we must be aware of the constraints they impose on the design. Normally it is desirable for the total pressure (or temperature) ratio of a rotor blade row to be approximately constant over the radial length of the blade, so that the outlet airflow has uniform pressure. From the Euler equation, ω(rb vb − ra va ) Ttb −1= . (7.3) Tta Cp Tta
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Bioengineering and Aerospace Engineering Dept.
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Subjects: Compressor Blading; Design; M ulti-Staging Compressor blading; Radial variations In all of the discussion so far the blade speed has been taken as a parameter. In fact the Lecture notes blade speed varies with the radial location in the comp!"##$!%&#$&'("&)*"#+,-#.&'(/'&0"& have discussed are applicable only at one radius in any real compressor. In practice it is 1#1/2&'$&3",+-&/&4$56!"##$!&*"#+,-&0+'((&/&'!"/'5"-'%&4/22"*&/&)5"/-&2+-"&*"#+,-.& carried out at some mean radius. But the effects of the blade speed variation with radius are important, so we must be aware of the constraints they impose on the design.
Aerospace Propulsion
Figure 7.2: Cross-section of an axial compressor showing the variation of the radius with axial distance. As a simple example, suppose that va = 0 (no inlet guide vanes). We see the condition that the total temperature Normally it is desirable for the pressure (or temperature) ratio of a rotor blade row to be ratio remains constant with radius requires
approximately constant over the radial length of the blade, so that the outlet airflow has const . (7.4) b vb = const. ; vb = uniform pressure. From the Eulerrequation rb r1 1 )Vortex flow (Fig. 7.3). Now let us draw the Tt 2should generate ( r2 2 a Free This implies that the rotor blade row 1 = velocity triangles for tip and hub (seeTFig. 7.4), further cpTt1assuming rH /rT = 1/2, vb (rH ) = ωrH (that is that t1 βb0 = 0), and ωrT Supposing for simplicity that 1 = 0wa(no guide vane), = winlet . we see this condition requires(7.5) b = const. = 2 const Hence, for this example, r2 2 = const.; ωr 2 = ωr H T vb (rT ) = = r2 . (7.6) 2 4 This implies that the rotor blade row should generate a Free Vortex in the flow
Now let us draw the velocity triangles and flow hub,(vassuming rH/rT = !, and w1 = w2 = Figure 7.3: for Freetip vortex ∝ 1/r) const = rT /2.
The approximate blade shapes as sketched are determined by the condition that the leading and trailing edges are aligned with the flow. We see that the blades are strongly “twisted” from hub to tip. Now let us see what these variations imply for the Diffusion Factor, D (Remember that V 0 are velocities relative to the rotor blade). V0 |vb − va | D = 1 − b0 + (7.7) Va 2σVa0 55
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Lecture notes
Figure 7.4: Sketch of the hub and the tip profiles and corresponding velocity triangles. From the geometry, for the TIP and taking va = 0 as in the previous example, |vb − va | = Va0 Vb0 DT
ωrT /4 r
(7.8)
ωrT
(7.9)
1 = 1.12ωrT 4 s 2 2 1 3 = ωrT + = 0.901ωrT 2 4 0.116 = 0.195 + σT =
1+
(7.10) (7.11)
and for the HUB |vb − va | = Va0 Vb0
=
DH
=
=
ωrH = ωrT /2 p p ωrH (2) = ωrT (2)/2 ωrH = ωrT /2 0.353 0.293 + σH
(7.12) (7.13) (7.14) (7.15)
The solidities at hub and tip, σH = cH /sH and σT = cT /sT are of course design choices, but since the spacings sH and sT are related by sH rH = (7.16) sT rT σH cH sT cH rT = = (7.17) σT cT sH cT rH Let us choose σT = 1, CH /CT = 1, then σH = 2, leading to DT
=
DH
=
0.195 + 0.116 = 0.311 0.353 0.293 + = 0.470 2
(7.18) (7.19)
Both of these are acceptable from the viewpoint of losses, but DH barely so. It is generally true that the hub section of the blade is limiting from the viewpoint of diffusion.
7.3
Multi-staging and flow area variation
The reason behind multi-stage compressors is the limited pressure ratio that can be achieved with a single stage. Typically, several rotor-stator stages are stacked up to produce the required compression. 56
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Lecture notes
Because the density ρ increases, the axial velocity will decrease unless we reduce the flow area accordingly. Axial deceleration in a compressor is severely limited, as it can stall the airfoils of the blades. Thus, the flow area must decrease as we go through the compressor to maintain near-constant axial velocity. We can choose to effect this area variation by decreasing the tip radius, by increasing the hub radius, or something in between, like keeping a constant mean radius. All these solutions have both advantages and disadvantages. The constant outer diameter design maintains the blade velocity at the maximum for the entire length of the compressor, hence gives the highest pressure ratio for a given number of stages. But because the length of the blades becomes small at the final stages of the compressor, leakage through the tip clearance is most serious for this design choice. The constant inner diameter design minimizes the tip clearance problem and also yields lower stresses in the discs of the last stages, but requires more stages for a given pressure ratio (since the lower-radius blades have a lower speed and exert less work than higher-radius ones).
7.4
Mach Number Effects
The performance of compressor blades deteriorates once the relative inlet Mach number exceeds 0.7–0.8, since on the extrados of the airfoil the accelerated flow typically reaches supersonic conditions, creating shock waves and thicker boundary layers which decrease the total pressure of the flow, incurring in efficiency losses. Since the linear velocity is maximal at the blade tip, it is also there where the Mach number is higher. There are however reasons that encourage us to go beyond this limit: a higher Mach number implies a higher mass flow per unit area, so that the compressor is more compact. Second, a higher Mach number is caused by a higher blade velocity, which enables more work to be deposited into the flow, and hence higher pressure ratios at the cost of lower compressor efficiency (there is of course a trade-off here, as stronger shock waves will ultimately cancel the increased pressure gain). Note that, while on-ground compressors aim to maximize efficiency, aircraft compressors also have to consider the work exerted by each stage to reduce the number of stages and result in a lighter device (even at the cost of efficiency, which is usually reduced a bit). Hence, in many circumstances, compressors are designed to operate in a transonic regime in their first stages (later stages have lower Mach numbers since temperature — and sound speed — increase downstream). Typical values of first-stage blade tip relative Mach numbers can reach as high as 1.5 in modern engines. As an example, for an axial Mach number ≈ 0.7 and blade tangential Mach number MT ≈ 1.3, we have at the tip q 0 (7.20) Ma,T = Ma2 + MT2 = 1.48 so the tips of the blades see supersonic flow. For rH /rT = 0.5, at the hub √ 0 Ma,H = 0.49 + 0.42 = 0.95
(7.21)
so the hub is at a very high subsonic speed. Just as for inlets, hub and tip therefore require quite different diffusion techniques. To alleviate the effects of the high relative Mach number, very thin blades are used to reduce their blockage, with relative thicknesses of only a few percent (see Fig. 7.5). At the hub, instead, a blading that is the equivalent of a subsonic inlet is required. The shockwaves that form lead to a rise in entropy of the flow, but it does not necessarily lead to excessive losses if the blades are designed carefully.
7.5
The Polytropic Efficiency
What is almost nearly constant among stages of a multi-stage compressor is the limiting isentropic efficiency for a small, differential-like compression, which is called the Polytropic efficiency:
ηpoly 57
dpt (dht )s RTt d ln pt γ − 1 d ln pt ρt = = = = dht Cp dTt Cp dTt γ d ln Tt
Bioengineering and Aerospace Engineering Dept.
(7.22) v. 2015-2016
Aerospace Propulsion
Lecture notes
Figure 7.5: Sketch of the supersonic flow on the first rotor row and the system of shockwaves and expansion waves. (remember that dh = T ds + (1/ρ)dp). If we assume ηpoly remains constant along the compression path, this definition can be integrated to γ−1 γ−1 Tt,out pt,out γηpoly γη = → τc = πc poly . (7.23) Tt,in pt,in Inserting this into the definition of the global Isentropic Efficiency allows one to compute it in terms of ηpoly and the finite pressure or temperature ratio: γ−1
γ−1
πc γ − 1 πc γ − 1 ηc ≡ = γ−1 τc − 1 γη πc poly − 1
(7.24)
As an example, take a multi-stage compressor with an overall pressure ratio πc = 16 and assume the polytropic efficiency is constant and equal to 0.9 in all stages. Its isentropic efficiency (ideal work required divided by actual work) is then calculated to be ηc = 0.856. This is less than ηpoly , the small-increment efficiency, because the last stages of the compressor receive gas that is hotter than it should, due to the inefficiencies in the previous stages, and hence more work is needed to compress it. It could be mentioned here that a similar argument can be made for turbines, and in that case one finds ηpoly < ηt , because the extra thermal energy due to inefficiencies of the early stages is now available for conversion to work by the latter stages. Likewise: the global isentropic efficiency of a compressor with N stages, each with an isentropic stage efficiency ηs (close to ηpoly ) is not ηsN (actually, ηsN > ηc ).
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Aerospace Propulsion Starting and Low-Speed Operation
7.6
Lecture notes
Because the Low-Speed density variationOperation through the compressor is much less at low speed Starting and
conditions, the compressor develops adverse blade loading situations in both the inlet and Since the density variation through the compressor is much less at low speed conditions, but it is designed for the outlet stages at low operating speeds. As shown in the sketch, the axial velocity in the near-cruise conditions, the compressor develops adverse blade loading situations in both the inlet and the outlet inlet stages tends to be lower, and that in the outlet stages higher, relative to the blade stages at low operating speeds. As shown in Fig. 7.6, the axial velocity in the inlet stages tends to be lower, speed, than at design, so the front stages tend to be stalled and the rear ones to and that in the outlet stages higher, relative to the blade speed, than at design, so the front stages tend to !"#$%&#''()*+,#-*&./0-*#1*%#22#34'1*15*%0-#6$*.*35&780--58*1,.1*"#''*5708.10*-.1#-2.3158#'9* be stalled and the rear ones to “windmill”. This makes it difficult to design a compressor that will operate over a wide range of speeds. satisfactorily over a wide range of speeds.
Figure 7.6: Sketch of the velocities at design and low-speed operation. The solution to this problem has taken two forms: one is to divide the compressor into The solution to this problem has taken two forms: one is to divide the compressor into two or even three 1"5*58*0:0$*1,800*!-755'-(;*5708.1#$6*.1*%#22080$1*-700%-;*1,0*51,08*#-*15*4-0*-1.158*
