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Advanced propulsion systems 3 lectures

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mdot/A is maximum at M = 1 - flow is 'choked' at throat - if flow is choked then ... Momentum equation: AdP mdot*du = 0 u = constant ... – PowerPoint PPT presentation

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Title: Advanced propulsion systems 3 lectures


1
Advanced propulsion systems (3 lectures)
  • Hypersonic propulsion background (Lecture 1)
  • Why hypersonic propulsion?
  • Whats 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

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, gt for irreversible), reversible
    adiabatic ? isentropic (dS 0)
  • Governing equations
  • Equations of state
  • Isentropic (S2 S1) (where applicable)
  • Mass conservation
  • Momentum conservation, constant area duct
  • 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

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, lets 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 ue, exit
    pressure Pe and thus thrust

4
Isentropic nozzle flow
  • Reversible, adiabatic ? S constant, A ?
    constant, w 0
  • Momentum equation not used - simple constant-area
    form (slide 15) doesnt 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

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

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

7
Isentropic nozzle flow
  • Mass flow and velocity can be determined
    similarly

8
Isentropic nozzle flow
  • Summary
  • A area at M 1
  • Implications
  • P and T decrease monotonically as M increases
  • Area is minimum at M 1 - need a throat to
    transition from M lt 1 to M gt 1 or vice versa
  • mdot/A is maximum at M 1 - flow is choked at
    throat - if flow is choked then any change in
    downstream conditions cannot affect mdot
  • 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)

9
Isentropic nozzle flow
A/A
T/Tt
P/Pt
10
Isentropic nozzle flow
  • When can choking occur? If M 1 or
  • so need pressure ratio gt 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)

11
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 lt P1
  • (e.g. tank of high-P, ambient-T gas,
  • reversible adiabatic expansion)
  • B. T9t gt T1t, P9t P1t gt P1 P9
  • (e.g. high-M ramjet/scramjet,
  • no Pt losses)
  • C. T9t gt T1t, P9t gt P1t P1 P9
  • (e.g. low-M turbojet or fan)
  • Fan T9t/T1t (P9t/P1t)(?-1)/?
  • due to adiabatic compression
  • Turbojet T9t/T1t gt (P9t/P1t)(?-1)/?
  • due to adiabatic compression
  • plus heat addition
  • Could get thrust even with

P9t P1t P9 lt P1 T9t T1t
P1 P1t (M1 0)
Case A
P9t P1t P9 P1 T9t gt T1t
P1t gt P1 (M1 gt 0)
Case B
P9t gt P1t P9 P1 T9t gt T1t
P1t P1 (M1 0)
Case C
12
Stagnation temperature and pressure
  • Note also ?(Tt) heat or work transfer

13
Constant everything except S (shock)
  • Q what if A constant but S ? constant? Can
    anything happen while still satisfying mass,
    momentum, energy, eqn. of state?
  • A YES! (shock)
  • Implications
  • One possibility is no change in state ( )1 ( )2
  • If M1 gt 1 then M2 lt 1 and vice versa - equations
    dont show a preferred direction, but only M1 gt
    1, M2 lt 1 results in dS gt 0, thus M1 lt 1, M2 gt 1
    is impossible
  • Tt constant (no change in total enthalpy) but Pt
    decreases across shock (a lot if M gtgt 1!), P, T
    increase
  • Note there are only 2 states, ( )1 and ( )2 - no
    continuum of states

14
Constant everything except S (shock)
T2/T1
P2/P1
T2t/T1t
M2
P2t/P1t
15
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
  • Cant 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)

16
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

17
Everything constant except momentum
Pt/Pt
Tt/Tt
T/T
Length
P/P
Length
18
Everything constant except momentum
Pt/Pt
P/P
Length
Tt/Tt
T/T
Length
19
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 pressure always decreases towards M
    1
  • Stagnation temperature always increases towards M
    1
  • Cant cross M 1 with constant area heat
    addition!
  • M 1 corresponds to the maximum possible heat
    addition
  • but theres no particular reason we have to keep
    area (A) constant when we add heat!

20
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.

21
Heat addition at constant area
Pt/Pt
Tt/Tt
T/T
P/P
22
T-s diagram - reference state M 1
Fanno
M lt 1
Shock
Rayleigh
M lt 1
M gt 1
23
T-s diagram - Fanno, Rayleigh, shock
Rayleigh
Shock
Constant area, with friction, no heat addition
M lt 1
Constant area, no friction, with heat addition
Fanno
M lt 1
M gt 1
This jump constant area, no friction, no heat
addition ? SHOCK!
M gt 1
24
Heat addition at constant pressure
  • Relevant for hypersonic propulsion if maximum
    allowable pressure (i.e. structural limitation)
    is the reason we cant decelerate the ambient air
    to M 0)
  • Momentum equation AdP mdotdu 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

25
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.

26
Heat addition at constant P
Pt/Pt
P/P
Tt/Tt
T/T, A/A
27
Heat addition at constant temperature
  • Probably most appropriate case for hypersonic
    propulsion since temperature (materials) limits
    is usually the reason we cant decelerate the
    ambient air to M 0
  • T constant ? c (sound speed) constant
  • Momentum AdP mdotdu 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

28
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.

29
Heat addition at constant T
Tt/Tt
A/A
T/T
Pt/Pt
P/P
30
T-s diagram for diabatic flows
Const P
Const T
Rayleigh (Const A)
Rayleigh (Const A)
31
Pt vs. Tt for diabatic flows
Rayleigh (Const A)
Const P
Rayleigh (Const A)
Const T
32
Area ratios for diabatic flows
Const T
Const T
Const P
Const A
33
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
    - its 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

34
Summary
35
Summary of heat addition processes
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