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Advanced propulsion systems (3 lectures)

Advanced propulsion systems (3 lectures). Hypersonic propulsion background (Lecture 1) Why hypersonic propulsion? What ’ s different at hypersonic conditions? Real gas effects (non-constant C P , dissociation) Aircraft range How to compute thrust? Idealized compressible flow (Lecture 2)

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Advanced propulsion systems (3 lectures)

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  1. Advanced propulsion systems (3 lectures) • Hypersonic propulsion background (Lecture 1) • Why hypersonic propulsion? • What’s different at hypersonic conditions? • Real gas effects (non-constant CP, dissociation) • Aircraft range • How to compute thrust? • Idealized compressible flow (Lecture 2) • Isentropic, shock, friction (Fanno) • Heat addition at constant area (Rayleigh), T, P • Hypersonic propulsion applications (Lecture 3) • Ramjet/scramjets • Pulse detonation engines • Combustion instability (Lecture 3) AME 514 - Spring 2013 - Lecture 11

  2. 1D steady flow of ideal gases • Assumptions • Ideal gas, steady, quasi-1D • Constant CP, Cv,   CP/Cv • Unless otherwise noted: adiabatic, reversible, constant area • Note since 2nd Law states dS ≥ Q/T (= for reversible, > for irreversible), reversible + adiabatic  isentropic (dS = 0) • Governing equations • Equations of state • Isentropic (S2 = S1) (where applicable): • Mass conservation: • Momentum conservation, constant area duct (see lecture 10): • Cf = friction coefficient; C = circumference of duct • No friction: • Energy conservation: q = heat input per unit mass = fQR if due to combustion w = work output per unit mass AME 514 - Spring 2013 - Lecture 11

  3. 1D steady flow of ideal gases • Types of analyses: everything constant except… • Area (isentropic nozzle flow) • Entropy (shock) • Momentum (Fanno flow) (constant area with friction) • Diabatic (q ≠ 0) - several possible assumptions • Constant area (Rayleigh flow) (useful if limited by space) • Constant T (useful if limited by materials) (sounds weird, heat addition at constant T…) • Constant P (useful if limited by structure) • Constant M (covered in some texts but really contrived, let’s skip it) • Products of analyses • Stagnation temperature • Stagnation pressure • Mach number = u/c = u/(RT)1/2(c = sound speed at local conditions in the flow(NOT at ambient condition!)) • From this, can get exit velocity u9, exit pressure P9 and thus thrust AME 514 - Spring 2013 - Lecture 11

  4. Isentropic nozzle flow • Reversible, adiabatic  S = constant, A ≠ constant, w = 0 • Momentum equation not used - constant-area form doesn’t apply • Recall stagnation temperatureTt = temperature of gas stream when decelerated adiabatically to M = 0 • Thus energy equation becomes simply T1t = T2t, which simply says that the sum of thermal energy (the 1 term) and kinetic energy (the (-1)M2/2 term) is a constant AME 514 - Spring 2013 - Lecture 11

  5. Isentropic nozzle flow • Pressure is related to temperature through isentropic compression law: • Recall stagnation pressure Pt = pressure of gas stream when decelerated adiabatically and reversibly to M = 0 • Thus the pressure / Mach number relation is simply P1t = P2t AME 514 - Spring 2013 - Lecture 11

  6. Isentropic nozzle flow • Relation of P & T to duct area A determined through mass conservation • But for adiabatic reversible flow T1t = T2t and P1t = P2t; also define throat area A* = area at M = 1 then • A/A* shows a minimum at M = 1, thus it is indeed a throat AME 514 - Spring 2013 - Lecture 11

  7. Isentropic nozzle flow • How to use A/A* relations if neither initial state (call it 1) nor final state (call it 2) are at the throat (* condition)? AME 514 - Spring 2013 - Lecture 11

  8. Isentropic nozzle flow • Mass flow and velocity can be determined similarly: AME 514 - Spring 2013 - Lecture 11

  9. Isentropic nozzle flow • Summary A* = area at M = 1 • Recall assumptions: 1D, reversible, adiabatic, ideal gas, const.  • Implications • P and T decrease monotonically as M increases • Area is minimum at M = 1 - need a “throat” to transition from M < 1 to M > 1 or vice versa • is maximum at M = 1 - flow is “choked” at throat - if flow is choked then any change in downstream conditions cannot affect • Note for supersonic flow, M (and u) INCREASE as area increases - this is exactly opposite subsonic flow as well as intuition (e.g. garden hose - velocity increases as area decreases) AME 514 - Spring 2013 - Lecture 11

  10. Isentropic nozzle flow A/A* T/Tt P/Pt AME 514 - Spring 2013 - Lecture 11

  11. Isentropic nozzle flow • When can choking occur? If M ≥ 1 or so need pressure ratio > 1.89 for choking (if all assumptions satisfied…) • Where did Pt come from? Mechanical compressor (turbojet) or vehicle speed (high flight Mach number M1) • Where did Tt come from? Combustion! (Even if at high M thus high Tt, no thrust unless Tt increased!) (Otherwise just reversible compression & expansion) AME 514 - Spring 2013 - Lecture 11

  12. Stagnation temperature and pressure • Why are Tt and Pt so important? Isentropic expansion of a gas with stagnation conditions Tt and Pt to exit pressure P yields • For exit pressure = ambient pressure and FAR << 1, • Thrust increases as Tt and Pt increase, but everything is inside square root, plus Pt is raised to small exponent - hard to make big improvements with better designs having larger Tt or Pt AME 514 - Spring 2013 - Lecture 11

  13. Stagnation temperature and pressure • From previous page • No thrust if P1t = P9t,P9 = P1 & T1t = T9t; to get thrust we need either • T9t = T1t,P9t = P1t = P1; P9 < P1 (e.g. tank of high-P, ambient-T gas, reversible adiabatic expansion) B. T9t > T1t, P9t = P1t > P1= P9 (e.g. high-M ramjet/scramjet, no Pt losses) P9t = P1t P9 < P1 T9t = T1t P1 = P1t (M1 = 0) Case A P9t = P1t P9 = P1 T9t > T1t P1t > P1 (M1 > 0) Case B AME 514 - Spring 2013 - Lecture 11

  14. Stagnation temperature and pressure C. T9t > T1t, P9t > P1t = P1 = P9 (e.g. low-M turbojet or fan) • Fan: T9t/T1t = (P9t/P1t)(-1)/ due to adiabatic compression • Turbojet: T9t/T1t > (P9t/P1t)(-1)/ due to adiabatic compression plus heat addition • Could get thrust even with T9t = T1t, but how to pay for fan or compressor work without heat addition??? P9t > P1t P9 = P1 T9t > T1t P1t = P1 (M1 = 0) Case C AME 514 - Spring 2013 - Lecture 11

  15. Stagnation temperature and pressure • Note also (Tt) ~ heat or work transfer AME 514 - Spring 2013 - Lecture 11

  16. Constant everything except S (shock) • Q: what if A = constant but S ≠ constant? Can anything happen while still satisfying mass, momentum, energy & equation of state? • A: YES! (shock) • Energy equation: no heat or work transfer thus • Mass conservation AME 514 - Spring 2013 - Lecture 11

  17. Constant everything except S (shock) • Momentum conservation (constant area, dx = 0) AME 514 - Spring 2013 - Lecture 11

  18. Constant everything except S (shock) • Complete results • Implications • One possibility is no change in state ( )1 = ( )2 • If M1 > 1 then M2 < 1 and vice versa - equations don’t show a preferred direction • Only M1 > 1, M2 < 1 results in dS > 0, thus M1 < 1, M2 > 1 is impossible • P, T increase across shock which sounds good BUT… • Tt constant (no change in total enthalpy) but Pt decreases across shock (a lot if M1 >> 1!); • Note there are only 2 states, ( )1 and ( )2 - no continuum of states AME 514 - Spring 2013 - Lecture 11

  19. Constant everything except S (shock) T2/T1 P2/P1 T2t/T1t M2 P2t/P1t AME 514 - Spring 2013 - Lecture 11

  20. Everything constant except momentum (Fanno flow) • Since friction loss is path dependent, need to use differential form of momentum equation (constant A by assumption) • Combine and integrate with differential forms of mass, energy, eqn. of state from Mach = M to reference state ( )* at M = 1 (not a throat in this case since constant area!) • Implications • Stagnation pressure always decreases towards M = 1 • Can’t cross M = 1 with constant area with friction! • M = 1 corresponds to the maximum length (L*) of duct that can transmit the flow for the given inlet conditions (Pt, Tt) and duct properties (C/A, Cf) • Note C/A = Circumference/Area = 4/diameter for round duct AME 514 - Spring 2013 - Lecture 11

  21. Everything const. but momentum (Fanno flow) • What if neither the initial state (1) nor final state (2) is the choked (*) state? Again use P2/P1 = (P2/P*)/(P1/P*) etc., except for L, where we subtract to get net length L AME 514 - Spring 2013 - Lecture 11

  22. Everything constant except momentum Pt/Pt* Tt/Tt* T/T* Length P/P* Length AME 514 - Spring 2013 - Lecture 11

  23. Everything constant except momentum Pt/Pt* P/P* Length Tt/Tt* T/T* Length AME 514 - Spring 2013 - Lecture 11

  24. Heat addition at constant area (Rayleigh flow) • Mass, momentum, energy, equation of state all apply • Reference state ( )*: use M = 1 (not a throat in this case!) • Energy equation not useful except to calculate heat input (q = Cp(T2t - T1t)) or dimensionless q/CPTt* = 1 - Tt/Tt*) • Implications • Stagnation temperature always increases towards M = 1 • Stagnation pressure always decreases towards M = 1 (stagnation temperature increasing, more heat addition) • Can’t cross M = 1 with constant area heat addition! • M = 1 corresponds to the maximum possible heat addition • …but there’s no particular reason we have to keep area (A) constant when we add heat! AME 514 - Spring 2013 - Lecture 11

  25. Heat addition at const. Area (Rayleigh flow) • What if neither the initial state (1) nor final state (2) is the choked (*) state? Again use P2/P1 = (P2/P*)/(P1/P*) etc. AME 514 - Spring 2013 - Lecture 11

  26. Heat addition at constant area Pt/Pt* Tt/Tt* T/T* P/P* AME 514 - Spring 2013 - Lecture 11

  27. T-s diagram - reference state M = 1 Fanno M < 1 Shock Rayleigh M < 1 M > 1 AME 514 - Spring 2013 - Lecture 11

  28. T-s diagram - Fanno, Rayleigh, shock Rayleigh Shock Constant area, with friction, no heat addition M < 1 Constant area, no friction, with heat addition Fanno M < 1 M > 1 This jump: constant area, no friction, no heat addition  SHOCK! M > 1 AME 514 - Spring 2013 - Lecture 11

  29. Heat addition at constant pressure • Relevant for hypersonic propulsion if maximum allowable pressure (i.e. structural limitation) is the reason we can’t decelerate the ambient air to M = 0) • Momentum equation: AdP + mdot*du = 0  u = constant • Reference state ( )*: use M = 1 again but nothing special happens there • Again energy equation not useful except to calculate q • Implications • Stagnation temperature increases as M decreases, i.e. heat addition corresponds to decreasing M • Stagnation pressure decreases as M decreases, i.e. heat addition decreases stagnation T • Area increases as M decreases, i.e. as heat is added AME 514 - Spring 2013 - Lecture 11

  30. Heat addition at constant pressure • What if neither the initial state (1) nor final state (2) is the reference (*) state? Again use P2/P1 = (P2/P*)/(P1/P*) etc. AME 514 - Spring 2013 - Lecture 11

  31. Heat addition at constant P Pt/Pt* P/P* Tt/Tt* T/T*, A/A* AME 514 - Spring 2013 - Lecture 11

  32. Heat addition at constant temperature • Probably most appropriate case for hypersonic propulsion since temperature (materials) limits is usually the reason we can’t decelerate the ambient air to M = 0 • T = constant  c (sound speed) = constant • Momentum: AdP + du = 0  dP/P + MdM = 0 • Reference state ( )*: use M = 1 again • Implications • Stagnation temperature increases as M increases • Stagnation pressure decreases as M increases, i.e. heat addition decreases stagnation T • Minimum area (i.e. throat) at M = -1/2 • Large area ratios needed due to exp[ ] term AME 514 - Spring 2013 - Lecture 11

  33. Heat addition at constant temperature • What if neither the initial state (1) nor final state (2) is the reference (*) state? Again use P2/P1 = (P2/P*)/(P1/P*) etc. AME 514 - Spring 2013 - Lecture 11

  34. Heat addition at constant T Tt/Tt* A/A* T/T* Pt/Pt* P/P* AME 514 - Spring 2013 - Lecture 11

  35. T-s diagram for diabatic flows Const P Const T Rayleigh (Const A) Rayleigh (Const A) AME 514 - Spring 2013 - Lecture 11

  36. Pt vs. Tt for diabatic flows Rayleigh (Const A) Const P Rayleigh (Const A) Const T AME 514 - Spring 2013 - Lecture 11

  37. Area ratios for diabatic flows Const T Const T Const P Const A AME 514 - Spring 2013 - Lecture 11

  38. Summary • Choking - mass, heat addition at constant area, friction with constant area - at M = 1 • Supersonic results usually counter-intuitive • If no friction, no heat addition, no area change - it’s a shock! • Which is best way to add heat? • If maximum T or P is limitation, obviously use that case • What case gives least Pt loss for given increase in Tt? • Minimize d(Pt)/d(Tt) subject to mass, momentum, energy conservation, eqn. of state • Result (Lots of algebra - many trees died to bring you this result) • Adding heat (increasing Tt) always decreases Pt • Least decrease in Pt occurs at lowest possible M AME 514 - Spring 2013 - Lecture 11

  39. Summary AME 514 - Spring 2013 - Lecture 11

  40. Summary of heat addition processes AME 514 - Spring 2013 - Lecture 11

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