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AME 436
Energy and Propulsion
Lecture 12Propulsion 2: 1D compressible flow
2AME 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
3AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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
combustionw = work output per unit mass
1D steady flow of ideal gases
4AME 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 u9, exit pressure P9 and thus
thrust
5AME 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 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
6AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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
7AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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 & Pt change 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
8AME 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 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
9AME 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)?
10AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Isentropic nozzle flow
Mass flow and velocity can be determined similarly:
11AME 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
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)
12AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Isentropic nozzle flow
P/Pt
T/Tt
A/A*
13AME 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 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)
14AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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
15AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Stagnation temperature and pressure From previous page
No thrust if P1t = P9t,P9 = P1 & T1t = T9t; to get thrust we need either
A. 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)
P1 = P1t
(M1 = 0)
P9t = P1t
P9 < P1
T9t = T1tCase A
P9t = P1t
P9 = P1 T9t > T1t
P1t > P1
(M1 > 0)
Case B
16AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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
17AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Stagnation temperature and pressure
Note also (Tt) ~ heat or work transfer
Recall (lecture 6) for control volume, steady flowq12 - w12 = 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!
18AME 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
19AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Constant everything except S (shock) Momentum conservation (constant area, dx = 0)
20AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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
21AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Constant everything except S (shock)
M2
T2t/T1t
T2/T1
P2t/P1t
P2/P1
22AME 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)
23AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Everything const. but momentum (Fanno flow)
24AME 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 (Pt, Tt) and duct properties (C/A, Cf) Note C/A = Circumference/Area = 4/diameter for round duct
25AME 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 P2/P1 = (P2/P*)/(P1/P*) etc., except for L, where we subtract to get net length L
26AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Everything constant but momentum
Length
Tt/Tt*
T/T*
Pt/Pt*
P/P*
Length
27AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Everything constant but momentum
Length
Tt/Tt*
T/T*
Pt/Pt*
P/P*
Length
28AME 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 = 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!
29AME 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 P2/P1 = (P2/P*)/(P1/P*) etc.
30AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Heat addition at constant area
Tt/Tt*
T/T*
Pt/Pt*
P/P*
31AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
T-s diagram - reference state M = 1
Shock
Rayleigh
Fanno
M < 1
M > 1
M < 1
32AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
T-s diagram - Fanno, Rayleigh, shock
Shock
Rayleigh
Fanno
M < 1 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!
33AME 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
34AME 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 P2/P1 = (P2/P*)/(P1/P*) etc.
35AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Heat addition at constant P
Tt/Tt*
T/T*, A/A*
Pt/Pt*
P/P*
36AME 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
37AME 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 P2/P1 = (P2/P*)/(P1/P*) etc.
38AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Heat addition at constant T
Tt/Tt*
T/T*
Pt/Pt*
P/P*
A/A*
39AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
T-s diagram for diabatic flows
Const TRayleigh(Const A)
Const P
Rayleigh(Const A)
40AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
T-s diagram for diabatic flows
Const T
Rayleigh(Const A)
Const P
Rayleigh(Const A)
41AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Area ratios for diabatic flows
Const T
Const A
Const PConst T
42AME 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 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.
43AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Summary - 1D compressible flow
Const. A?
Adia-batic?
Friction-less?
Tt const.?
Pt const.?
Isentropic No Yes Yes Yes Yes
Fanno Yes Yes No Yes No
Shock Yes Yes Yes Yes No
Rayleigh Yes No Yes No No
Const. T heat addition
No No Yes No No
Const. P heat addition
No No Yes No No
44AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
Summary of heat addition processes
Const. A Const. P Const. T
M Goes to M = 1
Decreases Increases
Area Constant Increases Min. at M = -1/2
P Decr. M < 1Incr. M > 1
Constant Decreases
Pt Decreases Decreases Decreases
T Incr. except for a small region
at M < 1
Increases Constant
Tt Increases Increases Increases
45AME 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 T1 after the shock
(ii) Static (not stagnation) pressure relative to P1 after the shock
46AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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}
47AME 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.
48AME 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 = 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
49AME 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