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

2. 2 AME 436 - Spring 2013 - Lecture 12 - 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

3. 3 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow  Assumptions  Ideal gas, steady, quasi-1D  Constant C P, C v,   C P /C v  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 (S 2 = S 1 ) (where applicable):  Mass conservation:  Momentum conservation, constant area duct (see lecture 11): »C f = friction coefficient; C = circumference of duct »No friction:  Energy conservation: q = heat input per unit mass = fQ R if due to combustion w = work output per unit mass 1D steady flow of ideal gases

4. 4 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 u 9, exit pressure P 9 and thus thrust

5. 5 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 T t = temperature of gas stream when decelerated adiabatically to M = 0  Thus energy equation becomes simply T 1t = T 2t, which simply says that the sum of thermal energy (the 1 term) and kinetic energy (the (  -1)M 2 /2 term) is a constant

6. 6 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Isentropic nozzle flow  Pressure is related to temperature through isentropic compression law:  Define stagnation pressure P t = pressure of gas stream when decelerated adiabatically and reversibly to M = 0  Thus the pressure / Mach number relation is simply P 1t = P 2t

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

8. 8 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Isentropic nozzle flow  Relation of P & T to duct area A determined through mass conservation  But for adiabatic reversible flow T 1t = T 2t and P 1t = P 2t; also define throat area A * = area at M = 1 then  A/A * shows a minimum at M = 1, thus it is indeed a throat

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

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

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

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

13. 13 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 P t come from? Mechanical compressor (turbojet) or vehicle speed (high flight Mach number M 1 )  Where did T t come from? Combustion! (Even if at high M thus high T t, no thrust unless T t increased!) (Otherwise just reversible compression & expansion)

14. 14 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Stagnation temperature and pressure  Why are T t and P t so important? Recall isentropic expansion of a gas with stagnation conditions T t and P t to exit pressure P yields  For exit pressure = ambient pressure and FAR << 1,  Thrust increases as T t and P t increase, but everything is inside square root, plus P t is raised to small exponent - hard to make big improvements with better designs having larger T t or P t

15. 15 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Stagnation temperature and pressure  From previous page  No thrust if P 1t = P 9t,P 9 = P 1 & T 1t = T 9t ; to get thrust we need either A.T 9t = T 1t, P 9t = P 1t = P 1 ; P 9 < P 1 (e.g. tank of high-P, ambient-T gas, reversible adiabatic expansion) B. T 9t > T 1t, P 9t = P 1t > P 1 = P 9 (e.g. high-M ramjet/scramjet, no P t losses) P 1 = P 1t (M 1 = 0) P 9t = P 1t P 9 < P 1 T 9t = T 1t Case A P 9t = P 1t P 9 = P 1 T 9t > T 1t P 1t > P 1 (M 1 > 0) Case B

16. 16 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Stagnation temperature and pressure C. T 9t > T 1t, P 9t > P 1t = P 1 = P 9 (e.g. low-M turbojet or fan) »Fan: T 9t /T 1t = (P 9t /P 1t ) (  -1)/  due to adiabatic compression »Turbojet: T 9t /T 1t > (P 9t /P 1t ) (  -1)/  due to adiabatic compression plus heat addition »Could get thrust even with T 9t = T 1t, but how to pay for fan or compressor work without heat addition??? P 9t > P 1t P 9 = P 1 T 9t > T 1t P 1t = P 1 (M 1 = 0) Case C

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

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

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

20. 20 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Constant everything except S (shock)  Complete results  Implications  M 2 = M 1 = 1 is a solution (acoustic wave, P 2 ≈ P 1 )  If M 1 > 1 then M 2 1 then M 2 < 1 and vice versa - equations don’t show a preferred direction  Only M 1 > 1, M 2 0, thus M 1 1 is impossible  P, T increase across shock which sounds good BUT…  T t constant (no change in total enthalpy) but P t decreases across shock (a lot if M 1 >> 1!);  Note there are only 2 states, ( ) 1 and ( ) 2 - no continuum of states

21. 21 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Constant everything except S (shock) M2M2M2M2 T 2t /T 1t T 2 /T 1 P 2t /P 1t P 2 /P 1

22. 22 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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)

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

24. 24 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 (P t, T t ) and duct properties (C/A, C f )  Note C/A = Circumference/Area = 4/diameter for round duct

25. 25 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Everything const. but momentum (Fanno flow)  What if neither the initial state (1) nor final state (2) is the choked (*) state? Again use P 2 /P 1 = (P 2 /P * )/(P 1 /P * ) etc., except for L, where we subtract to get net length  L

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

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

28. 28 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 = C p (T 2t - T 1t ))  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!

29. 29 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Heat addition at const. Area (Rayleigh flow)  What if neither the initial state (1) nor final state (2) is the choked (*) state? Again use P 2 /P 1 = (P 2 /P * )/(P 1 /P * ) etc.

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

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

32. 32 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow T-s diagram - Fanno, Rayleigh, shock Shock Rayleigh Fanno M < 1 M > 1 M < 1 Constant area, no friction, with heat addition Constant area, with friction, no heat addition This jump: constant area, no friction, no heat addition  SHOCK!

33. 33 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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

34. 34 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Heat addition at constant pressure  What if neither the initial state (1) nor final state (2) is the reference (*) state? Again use P 2 /P 1 = (P 2 /P * )/(P 1 /P * ) etc.

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

36. 36 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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

37. 37 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Heat addition at constant temperature  What if neither the initial state (1) nor final state (2) is the reference (*) state? Again use P 2 /P 1 = (P 2 /P * )/(P 1 /P * ) etc.

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

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

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

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

42. 42 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 P t loss for given increase in T t ? »Minimize d(P t )/d(T t ) subject to mass, momentum, energy conservation, eqn. of state »Result (lots of algebra - many trees died to bring you this result)  Adding heat (increasing T t ) always decreases P t  Least decrease in P t occurs at lowest possible M - doesn’t really matter if it’s at constant A, P, T, etc.

43. 43 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Summary - 1D compressible flow Const. A? Adia- batic? Friction- less? T t const.? P t const.? IsentropicNoYesYesYesYes FannoYesYesNoYesNo ShockYesYesYesYesNo RayleighYesNoYesNoNo Const. T heat addition NoNoYesNoNo Const. P heat addition NoNoYesNoNo

44. 44 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Summary of heat addition processes Const. A Const. P Const. T M Goes to M = 1 DecreasesIncreases AreaConstantIncreases Min. at M =  -1/2 P Decr. M < 1 Incr. M > 1 ConstantDecreases PtPtPtPtDecreasesDecreasesDecreases T Incr. except for a small region at M < 1 IncreasesConstant TtTtTtTtIncreasesIncreasesIncreases

45. 45 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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 T 1 after the shock (ii) Static (not stagnation) pressure relative to P 1 after the shock

46. 46 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow Example (iii) Static (not stagnation) temperature and pressure relative to T 1 at the exit (iv) Dimensionless heat addition {q in /RT 1 = C P (T 3t -T 2t )/RT 1 = [  /(  ‑ 1)](T 3t -T 2t )/T 1 }

47. 47 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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.

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

49. 49 AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow 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