Unified Propulsion Professor Ian Waitz Spring 2001 UNIFIED PROPULSION SECTION I Spring 2001 UNIFIED PROPULSION: Learning Objectives: Given the basic geometry and idealized component performance, to be able to estimate the thrust and specific impulse of a gas turbine and a rocket engine from fluid and thermodynamic principles Measurable outcomes (assessment method): A. To be able to explain at a level understandable by a high school senior or non-technical person what the various terms are in the integral momentum equation and how jet propulsion works. (quiz, self-assessment) B. To be able to apply control volume analysis and the integral momentum equation to estimate the forces produced by aerospace propulsion systems (homework, quiz, self-assessment) C. To be able to describe the principal figures of merit for aircraft engine and rocket motor performance and explain how they are related to vehicle performance. (quiz, self-assessment) D. Given weight, geometry, and aerodynamic and propulsion system performance information, to be able to estimate the power required for flight, the range, the endurance, and the time-to- climb for an aircraft. (homework, quiz, self-assessment) E. To be able to explain at a level understandable by a high school senior or non-technical person the arrangement and layout of the major components of gas turbine and rocket engines. (quiz, self-assessment) F. Given mass fractions, and propulsion system performance information, to be able to estimate the range and velocity of single-stage rockets. (homework and quiz, self-assessment) G. To be able to describe the principal design parameters and constraints that set the performance of gas turbine engines. (quiz, self-assessment) H. To be able to apply ideal-cycle analysis to a gas turbine engine to relate thrust and fuel burn to component-level performance parameters and flight conditions. (homework, self- assessment) I. To be able to explain at a level understandable by a high school senior or non-technical person the energy exchange processes that underlie the workings of multistage compressor or turbine. (quiz, self-assessment) J. To be able to use velocity triangles and the Euler Turbine Equation to estimate the performance of a compressor or turbine stage. (homework, quiz, self-assessment) - 1 - UNIFIED PROPULSION SECTION I Spring 2001 I. Introduction to Propulsion A. Goal: Create a Force to Propel a Vehicle Two options: 1. Take mass stored in a vehicle and throw it backwards (rocket propulsion). Use the reaction force to propel the vehicle. Propellant –––→ burn ––––→ expand through nozzle (chem. energy) (thermal energy) (kinetic energy & momentum) Figure 1.1 Typical liquid propellant rocket motor (Hill and Peterson, 1992). 2. Seize mass from the surroundings and set the mass in motion backwards. Use the reaction force to propel vehicle (air-breathing propulsion). Continuously: a) Draw in air. b) Compress it. c) Add fuel and burn (convert chemical to thermal energy). d) Expand through a turbine to drive the compressor (extract work). e.1) Then expand in a nozzle to convert thermal energy to kinetic energy & momentum (turbojet). e.2) Or expand in a second turbine (extract work), use this to drive a shaft for a fan (turbofan), or a propeller (turboshaft). The fan or propeller impart k.e. & mom. to the air. *Remember: ∆ Overall goal: take m at Vo (flight speed), throw it out at Vo + V - 2 - UNIFIED PROPULSION SECTION I Spring 2001 Figure 1.2 Schematics of typical military gas turbine engines. Top: turbojet with afterburning, bottom: GE F404 low bypass ratio turbofan with afterburning (Hill and Peterson, 1992). Figure 1.3 Schematics of a PW PT6A-65, a typical turboprop (Hill and Peterson, 1992). - 3 - UNIFIED PROPULSION SECTION I Spring 2001 Figure 1.4 The RB211-535E4, a typical high bypass-ratio turbofan (Hill and Peterson, 1992). B. Performance parameters The two performance parameters of greatest interest for a propulsion system are the force it produces (thrust, T), and the overall efficiency with which it uses energy to produce η this force ( overall). We will begin by looking at the production of thrust using the integral form of the momentum theorem. In the second lecture we will discuss the efficiency of propulsion systems. C. Propulsion is a systems endeavor There are a multitude of other factors which a propulsion engineer must take into account when designing a device including weight, cost, manufacturability, safety, environmental effects, etc. Thus propulsion is truly a systems endeavor, requiring knowledge of a variety of disciplines: Fluids + thermo + structures + dynamics + controls + chemistry + acoustics + … We will focus mostly on these two disciplines in the Unified propulsion lectures. - 4 - UNIFIED PROPULSION SECTION II Spring 2001 II. Integral Momentum Theorem We can learn a great deal about the overall behavior of propulsion systems using the integral form of the momentum equation. The equation is the same as that used in fluid mechanics. nd A. An Expression of Newton’s 2 Law (e.g. ΣF = d/dt (mv)) 1. Consider two coordinate systems: a) Inertial b) Fixed to vehicle • Moves with velocity u* relative to the inertial o coordinate system • All velocities relative to the vehicle-fixed coordinate frame are denoted u* 2. Newton’s second law for a control volume of fixed mass ⌠ (* + *) ∑=* ρD u o u F ⌡ dV Dt C.V. V or * ⌠ ⌠ Du* ∑ F = a& ρdV + ρ dV o ⌡ ⌡V V Dt All external Force due to Change in momentum forces on change in inertia of mass in c.v. c.v. for accel. vehicle * ≡ F o • Pressure forces • Shear forces • Body forces To explain the above equation further, consider Figure 2.1 - 1 - UNIFIED PROPULSION SECTION II Spring 2001 Figure 2.1 Falling blocks. The falling block labeled (a) has a control volume fixed to it. In this case, the first term of the above equation is nonzero since the control volume is accelerating relative to an inertial reference frame. The second term is zero because the block is not accelerating relative to a coordinate system fixed to the control volume. The opposite is true for the falling block labeled (b), which is falling within a fixed control volume. The first term of the above equation is zero in this case because the control volume is not accelerating relative to an inertial reference frame. The second term is nonzero because the block is moving to a coordinate system fixed to the control volume. The mathematical result of both cases is as follows, * * = & ⌠ +⌠ Du ∑ F a o ρdV ρ dV ⌡V ⌡V Dt = & + (a) ma o 0 = ma* * * = & ⌠ +⌠ Du ∑ F a o ρdV ρ dV ⌡V ⌡V Dt Du* (b) = 0 + m Dt = ma* As expected, the result is the same for both. The integral momentum equation reduces to a * familiar form,F = m a* . - 2 - UNIFIED PROPULSION SECTION II Spring 2001 To continue, the integral momentum equation can be rewritten as follows, * * ⌠ ∂u* * ∑F − F = ρ + u*∇u* dV o ∂ ⌡V t ⌠ ∂u* ∂ρ ∂ρ * = ρ + u* − u* + ρu* ⋅ ∇u* dV ∂ ∂ ∂ ⌡V t t t ⌠ ∂ ()ρu* ∂ρ * = − u* + ρu* ⋅ ∇u* dV ∂ ∂ ⌡V t t From conservation of mass, ∂ρ * ∂ρ * + ∇() ρu* = 0oru* + u* ∇ ⋅() ρu* = 0 ∂ ∂ t t so ⌠ ∂()ρ* ∑−* * = u + *∇* ⋅() ρ* +ρ*⋅∇* * F Fo ⌡ u u u u dV V ∂t NOTE: This is a vector equation. 3. Considering only the components in the x-direction ⌠ ∂ ()ρu * * ∑ F − F = x + u ∇ ⋅ ()ρu* + ρu* ⋅ ∇u dV x ox x x ⌡V dt ⌠ ∂ ()ρu ∂()u ∂()u ∂()u ∂()u ∂()u ∂()u x + x + y + z + x + x + x ρu x i ρu x j ρu x k ρu x i ρu y j ρu z kdV dt ∂x ∂y ∂z ∂x ∂y ∂z ⌡V ⌠ ∂ ()ρu ∂()u ∂()u ∂()u ∂()u ∂()u ∂()u x + x + x + y + x + z + x ρu x ρu x i ρu x ρu y j ρu x ρu z k dV dt ∂x ∂x ∂y ∂y ∂z ∂z ⌡V ⌠ ∂ ()ρu = x + ∇* ⋅ ()* ρu x u dV ⌡V dt Then, by the divergence theorem, - 3 - UNIFIED PROPULSION SECTION II Spring 2001 * * * ∫∇ ⋅ AdV = ∫ n* ⋅ Ads where n* is outward unit normal vector. V S so ⌠ ∂ρ ()ux ⌠ ∑ F − F = dV + u ()ρu* ⋅ n* ds x ox ⌡ ∂t ⌡ x V s sum of forces on C.V. change in momentum of change in momentum mass contained in C.V. flux across surface of C.V. over time where the forces acting on the control volume may be composed of pressure forces, body forces, and skin friction 4. For steady flow, with no acceleration of the vehicle then ⌠ ∑= ρ*⋅ * Fx ⌡ ux u n ds s This is the form we will use most frequently in this class. - 4 - UNIFIED PROPULSION SECTION II Spring 2001 B. Application of the Integral Momentum Equation to Rockets Figure 2.2 Control volume for application of momentum theorem to a rocket. ∑ F = T + A ()p − p , u* ⋅ n* = 0 everywhere except A x e o e e + ()− = = T A e po pe ρu e u e A e m u e = + ()− T m u e A e pe po Static thrust for a rocket engine C. Application of the Momentum Equation to an Aircraft Engine Figure 2.3 Control volume for application of the momentum theorem to a gas turbine engine. - 5 - UNIFIED PROPULSION SECTION II Spring 2001 ⌠ ()& ⋅ & = − + ρu x u n dA T D ∑ PressureForces ⌡s ⌠ ⌠ ρu ()u& ⋅ n& dA = T − D + − P A + P A − (P − P )dA ⌡ x e e o e ⌡ R o R s AR ⌠ ⌠ ()& ⋅ & = − + & ⋅ & ρu x u n dA ρ e u e A e u e ρo u o A o u o ρu x u ndA ⌡ ⌡ − − s Cs Ao Ae m e m o ⌠ = − + & ⋅ & m e u e m o u o ρu x u ndA ⌡ − − Cs Ao Ae So we have: ⌠ ⌠ − − ()− − (− )= − + & ⋅ & T D Pe Po A e PR Po dA R m e u e m o u o ρu x u ndA ⌡ ⌡ − − AR Cs Ae Ao Everything that relates to flow through the engine is conventionally called thrust.