1. AME 514 Applications of Combustion
Lecture 11: 1D compressible flow

2. 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 Spring Lecture D 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 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 Spring Lecture D 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
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 Spring Lecture D compressible flow

5. Isentropic nozzle flow
Reversible, adiabatic  S = constant, A ≠ constant, w = 0
Momentum equation not used - constant-area form doesn’t apply
Recall 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 Spring Lecture D compressible flow

6. 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 Spring Lecture D compressible flow

7. 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 Spring Lecture D compressible flow

8. 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 Spring Lecture D compressible flow

9. Isentropic nozzle flow
Mass flow and velocity can be determined similarly:
AME Spring Lecture D compressible flow

10. 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 Spring Lecture D compressible flow

11. Isentropic nozzle flow
A/A*
T/Tt
P/Pt
AME Spring Lecture D compressible flow

12. 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 Spring Lecture D compressible flow

13. 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 Spring Lecture D compressible flow

14. 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 Spring Lecture D compressible flow

15. 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 Spring Lecture D compressible flow

16. Stagnation temperature and pressure
Note also (Tt) ~ heat or work transfer
AME Spring Lecture D compressible flow

17. 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 Spring Lecture D compressible flow

18. Constant everything except S (shock)
Momentum conservation (constant area, dx = 0)
AME Spring Lecture D compressible flow

19. 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 Spring Lecture D compressible flow

20. Constant everything except S (shock)
T2/T1
P2/P1
T2t/T1t M2
P2t/P1t
AME Spring Lecture D compressible flow

21. 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 Spring Lecture D compressible flow

22. 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 Spring Lecture D compressible flow

23. Everything constant except momentum
Pt/Pt*
Tt/Tt*
T/T*
Length
P/P*
Length
AME Spring Lecture D compressible flow

24. Everything constant except momentum
Pt/Pt*
P/P*
Length
Tt/Tt*
T/T*
Length
AME Spring Lecture D compressible flow

25. 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 Spring Lecture D compressible flow

26. 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 Spring Lecture D compressible flow

27. Heat addition at constant area
Pt/Pt*
Tt/Tt*
T/T*
P/P*
AME Spring Lecture D compressible flow

28. T-s diagram - reference state M = 1
Fanno
M < 1
Shock
Rayleigh
M < 1
M > 1
AME Spring Lecture D compressible flow

29. T-s diagram - Fanno, 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 Spring Lecture D compressible flow

30. 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 Spring Lecture D compressible flow

31. 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 Spring Lecture D compressible flow

32. Heat addition at constant P
Pt/Pt*
P/P*
Tt/Tt*
T/T*, A/A*
AME Spring Lecture D compressible flow

33. 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 Spring Lecture D compressible flow

34. 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 Spring Lecture D compressible flow

35. Heat addition at constant T
A/A*
Tt/Tt*
T/T*
Pt/Pt*
P/P*
AME Spring Lecture D compressible flow

36. T-s diagram for diabatic flows
Const P
Const T
Rayleigh
(Const A)
Rayleigh
(Const A)
AME Spring Lecture D compressible flow

37. Pt vs. Tt for diabatic flows
Rayleigh
(Const A)
Const P
Rayleigh
(Const A)
Const T
AME Spring Lecture D compressible flow

38. Area ratios for diabatic flows
Const T
Const T
Const P
Const A
AME Spring Lecture D compressible flow

39. AME 514 - Spring 2015 - Lecture 11 - 1D compressible flow
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 Spring Lecture D compressible flow

40. AME 514 - Spring 2015 - Lecture 11 - 1D compressible flow
Summary
Const. A?
Adia-batic?
Friction-less?
Tt const.?
Pt const.?
Isentropic No
Yes
Fanno
Shock
Rayleigh
Const. T heat addition
Const. P heat addition
AME Spring Lecture D compressible flow

41. Summary of heat addition processes
Const. A
Const. P
Const. T M
Goes to
M = 1
Decreases
Increases
Area
Constant
Min. at M = -1/2 P
Decr. M < 1
Incr. M > 1 Pt T
Incr. except for a small region at M < 1 Tt
AME Spring Lecture D compressible flow