1 / 49

AME 436 Energy and Propulsion

AME 436 Energy and Propulsion. Lecture 12 Propulsion 2: 1D compressible flow. Outline. Governing equations Analysis of 1D flows Isentropic, variable area Shock Constant area with friction (Fanno flow) Heat addition Constant area (Rayleigh) Constant P Constant T.

fifi
Download Presentation

AME 436 Energy and Propulsion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. AME 436Energy and Propulsion Lecture 12 Propulsion 2: 1D compressible flow

  2. Outline • Governing equations • Analysis of 1D flows • Isentropic, variable area • Shock • Constant area with friction (Fanno flow) • Heat addition • Constant area (Rayleigh) • Constant P • Constant T AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  3. 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 11): • 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  4. 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(defined later) • Stagnation pressure (defined later) • 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  5. Isentropic nozzle flow • Reversible, adiabatic  S = constant, A ≠ constant, w = 0 • Momentum equation not needed - turns out to be redundant with energy equation • Define stagnation temperature Tt = 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  6. Isentropic nozzle flow • Pressure is related to temperature through isentropic compression law: • Define 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  7. Stagnation temperature and pressure • Stagnation temperature Tt - measure of total energy (thermal + kinetic) of flow • T = static temperature - T measured by thermometer moving with flow • Tt = temperature of the gas if it is decelerated adiabatically to M = 0 • Stagnation pressure Pt - measure of usefulness of flow (ability to expand flow) • P = static pressure - P measured by pressure gauge moving with flow • Pt = pressure of the gas if it is decelerated reversibly and adiabatically to M = 0 • These relations are basically definitions of Tt & Pt at a particular state and can be used even if Tt &Ptchange during the process • These relations assumed constant g & R, i.e. constant CP and M (molecular weight); what if this assumption is invalid? To be discussed in Lecture 15 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  8. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  9. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  10. Isentropic nozzle flow • Mass flow and velocity can be determined similarly: AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  11. 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 - 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  12. Isentropic nozzle flow A/A* T/Tt P/Pt AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  13. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  14. Stagnation temperature and pressure • Why are Tt and Pt so important? Recall 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  15. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  16. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  17. Stagnation temperature and pressure • Note also (Tt) ~ heat or work transfer • Recall (lecture 6) for control volume, steady flow q12 - w12 = CP(T2 - T1) • Why T not Tt in that case? KE not included in lecture 6 since KE almost always small (more specifically, M << 1) in reciprocating engines! AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  18. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  19. Constant everything except S (shock) • Momentum conservation (constant area, dx = 0) AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  20. Constant everything except S (shock) • Complete results • Implications • M2 = M1 = 1 is a solution (acoustic wave, P2 ≈ P1) • 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  21. Constant everything except S (shock) T2/T1 P2/P1 T2t/T1t M2 P2t/P1t AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  22. Everything const. but momentum (Fanno flow) • Since friction loss is path dependent, we need to use differential form of momentum equation (constant A by assumption) AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  23. Everything const. but momentum (Fanno flow) AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  24. Everything const. but 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  25. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  26. Everything constant but momentum Pt/Pt* Tt/Tt* T/T* Length P/P* Length AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  27. Everything constant but momentum Pt/Pt* P/P* Tt/Tt* T/T* Length Length AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  28. Heat addition at const. 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)) • 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  29. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  30. Heat addition at constant area Pt/Pt* Tt/Tt* T/T* P/P* AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  31. T-s diagram - reference state M = 1 Fanno M < 1 Shock Rayleigh M < 1 M > 1 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  32. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  33. 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 + 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 P • Area increases as M decreases, i.e. as heat is added AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  34. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  35. Heat addition at constant P Pt/Pt* P/P* Tt/Tt* T/T*, A/A* AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  36. 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  a (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 P • Minimum area (i.e. throat) at M = -1/2 • Large area ratios needed due to exp[ ] term AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  37. 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  38. Heat addition at constant T Tt/Tt* A/A* T/T* Pt/Pt* P/P* AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  39. T-s diagram for diabatic flows Const P Const T Rayleigh (Const A) Rayleigh (Const A) AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  40. T-s diagram for diabatic flows Rayleigh (Const A) Const P Rayleigh (Const A) Const T AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  41. Area ratios for diabatic flows Const T Const T Const P Const A AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  42. Summary - 1D compressible flow • 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 - doesn’t really matter if it’s at constant A, P, T, etc. AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  43. Summary - 1D compressible flow AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  44. Summary of heat addition processes AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  45. Example Consider a very simple propulsion system operating at a flight Mach number of 5 that consists of 2 processes: Process 1: Shock at entrance to duct Process 2: Heat addition in a constant-area duct until thermal choking occurs a) Compute all of the following properties of this system: (i) Static (not stagnation) temperature relative to T1after the shock (ii) Static (not stagnation) pressure relative to P1 after the shock AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  46. Example (iii) Static (not stagnation) temperature and pressure relative to T1 at the exit (iv) Dimensionless heat addition {qin/RT1= CP(T3t-T2t)/RT1 = [/(‑1)](T3t-T2t)/T1} AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  47. Example (v) Specific thrust (assume FAR << 1 in the thrust calculation) (vi) Overall efficiency (vii) Draw this cycle on a T - s diagram. Include appropriate Rayleigh and Fanno curves. AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  48. Example b) Repeat (a) if a nozzle is added after station 3 to expand the flow isentropically back to P = P1. Everything is the same up to state 3, but now we have a state 4, i.e. isentropic expansion (P4t = P3t, T4t = T3t until P4 = P1. Heat addition is the same as before, so AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

  49. Summary - compressible flow • The 1D conservation equations for energy, mass and momentum along with the ideal gas equations of state yield a number of unusual phenomenon • Choking - isentropic, diabatic (Rayleigh), friction (Fanno) • Garden hose in reverse (rule of thumb: for supersonic flow, all of your intuitions about flow should be reversed) • Shocks • Stagnation conditions • Temperature - a measure of the total energy (thermal + kinetic) contained by a flow • Pressure - a measure of the “usefulness” (ability to expand) of a flow AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow

More Related