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
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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, > for
irreversible), reversible + adiabatic  isentropic (dS = 0)
 Governing equations
 Equations of state h2 - h1 = CP (T2 - T1 )
P = rRT; S2 - S1 = C p ln(T2 T1) - Rln(P2 P1)
g /(g -1)
 Isentropic (S2 = S1) (where applicable): P2 /P1 = (T2 /T1 )
˙ = r1u1A1 = r2u2 A2
 Mass conservation: m
 Momentum conservation, constant area duct (see lecture 11):
AdP + m˙ du +Cf (ru2/2)Cdx = 0
» Cf = friction coefficient; C = circumference of duct
2
2
» No friction: P1 + r1u1 = P2 + r2u2
 Energy conservation: h1 + u1 /2 + q - w = h2 + u2 /2
q = heat input per unit mass = fQR if due to combustion
w = work output per unit mass
2
2
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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, 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
4
Isentropic nozzle flow
 Reversible, adiabatic  S = constant, A ≠ constant, w = 0
 Momentum equation not needed - turns out to be redundant
with energy equation
Energy equation, adiabatic, no work : h1 + u12 /2 = h2 + u22 /2
h2 - h1 = CP (T2 - T1 );M1 = u1 /c1 = u1 / gRT1 ;M 2 = u2 / gRT2
Þ CP T1 + M12gRT1 /2 = CP T2 + M 22gRT2 /2
æ
æ
gR 2 ö
gR 2 ö
R g -1
Þ T1ç1+
M1 ÷ = T2 ç1+
M 2 ÷, but
=
CP
g
è 2CP
ø
è 2CP
ø
æ g -1 2 ö
æ g -1 2 ö
Þ T1ç1+
M1 ÷ = T2 ç1+
M2 ÷
è
ø
è
ø
2
2
 Define stagnation temperature Tt = temperature of gas stream
when decelerated adiabatically to M = 0
æ g -1 2 ö
æ g -1 2 ö
Þ T1t = T1ç1+
M1 ÷;T2t = T2ç1+
M 2 ÷;
è
ø
è
ø
2
2
 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
5
Isentropic nozzle flow
 Pressure is related to temperature through isentropic
g
compression law:
æ
gR 2 ö g -1
g
g
M1 ÷
g -1
g -1ç 1+
æ
ö
æ
ö
P2
T
P
T
2CP
÷
=ç 2÷
Þ 2 = ç 2t ÷ ç
g
R
P1 è T1 ø
P1 è T1t ø ç1+
2÷
M
ç
2÷
è 2CP
ø
æ
gR 2 ö
but T2t = T1t Þ P2 ç1+
M2 ÷
è 2CP
ø
g
g -1
æ
gR 2 ö
= P1ç1+
M1 ÷
è 2CP
ø
g
g -1
 Define stagnation pressure Pt = pressure of gas stream
when decelerated adiabatically and reversibly to M = 0
g
g
g
-1
æ g -1 2 ö
æ g -1 2 ö g -1
Þ P1t = P1ç1+
M1 ÷ ;P2t = P2 ç1+
M2 ÷ ;
è
ø
è
ø
2
2
 Thus the pressure / Mach number relation is simply P1t = P2t
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
6
Stagnation temperature and pressure
 Stagnation temperature Tt - measure of total energy (thermal +
kinetic) of flow
æ g -1 2 ö
Tt = Tç1+
M ÷
è
ø
2
 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)
æ g -1 2 ö
Pt = Pç1+
M ÷
è
ø
2
g
g -1
 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  & 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
7
Isentropic nozzle flow
 Relation of P & T to duct area A determined through mass
conservation
P
P
P
P
m˙ = r1u1 A1 = r2 u2 A2 Þ 1 u1 A1 = 2 u2 A2 Þ 1 M1 gRT1 A1 = 2 M 2 gRT2 A2
RT1
RT2
T1
T2
g
g
æ g -1 2 ö
æ g -1 2 ö g -1
P1t ç1+
M1 ÷
P2t ç1+
M2 ÷
P1
P2
è
ø
è
ø
2
2
Þ
M1 A1 =
M 2 A2 Þ
M1 A1 =
M 2 A2
RT1
RT2
g -1 2
g -1 2
RT1t
1+
M1
RT2t
1+
M2
2
2
Þ
P1t
RT1t
æ g -1 2 ö
M1 ÷
ç1+
è
ø
2
-(g +1)
2(g -1)
M1 A1 =
P2t
RT2t
g -1
æ g -1 2 ö
M2 ÷
ç1+
è
ø
2
-(g +1)
2(g -1)
M 2 A2
 But for adiabatic reversible flow T1t = T2t and P1t = P2t; also
define throat area A* = area at M = 1 then
æ g -1 2 ö
(1) ÷
ç1+
è
ø
2
-(g +1)
2(g -1)
æ g -1 2 ö
(1)A = ç1+
M ÷
è
ø
2
*
-(g +1)
2(g -1)
(g +1)
A(M ) 1 é 2 æ g -1 2 öù
MA Þ
= ê
M ÷ú
ç1+
*
øû
A
M ëg +1 è
2
2(g -1)
 A/A* shows a minimum at M = 1, thus it is indeed a throat
AME 436 - Spring 2013 - Lecture 12 - 1D 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)?
(g +1)
A1
1 é 2 æ g -1 2 öù
=
M1 ÷ú
ç1+
ê
*
è
øû
A
M1 ëg + 1
2
2(g -1)
(g +1)
A2
1 é 2 æ g -1 2 öù
; *=
M 2 ÷ú
ç1+
ê
è
øû
A
M 2 ëg + 1
2
(g +1)
éæ g -1 2 ö ù
M 2 ÷ú
ç1+
ê
*
A
A /A
M è
øú
2
Þ 2 = 2 * = 1ê
A1 A1 / A
M 2 êæ g -1 2 ö ú
M1 ÷
ç1+
êëè
ø úû
2
2( g -1)
2(g -1)
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
9
Isentropic nozzle flow
 Mass flow and velocity can be determined similarly:
-(g +1)
æ g -1 2 ö
m
P
m
P
= ru =
M g RT Þ = t
g M ç1+
M ÷
è
ø
A
RT
A
2
RTt
-(g +1)
æ g -1 2 ö
m
Pt
Þ *=
g (1)ç1+
(1) ÷
è
ø
A
2
RTt
2(g -1)
2(g -1)
(g +1)
Þm=A
*
æ 2 ö
Pt
gç
÷
RTt
è g +1 ø
2(g -1)
æ g -1 2 ö
u
2 æ Tt ö
Tt = T ç1+
M ÷Þ M =
=
ç -1÷
2
g -1 è T ø
è
ø
gRT
g -1
æ
æ Tö
æPö g
2g
2g
ç
Þu=
RTt ç1- ÷ Þ u =
RTt 1- ç ÷
g -1 è Tt ø
g -1 ç è Pt ø
è
ö
÷
÷
ø
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
10
Isentropic nozzle flow
 Summary
æ g -1 2 ö
Tt = Tç1+
M ÷ = constant
è
ø
2
m˙
P
= t
A
RTt
æ g -1 2 ö
Pt = Pç1+
M ÷
è
ø
2
(g +1)
æ g -1 2 ö
g Mç1+
M ÷
è
ø
2
(g +1)
A
1 é 2 æ g -1 2 öù
= ê
M ÷ú
ç1+
*
øû
A
M ëg + 1 è
2
2(g -1)
g
g -1
= constant
(g +1)
m=A
*
æ 2 ö
Pt
gç
÷
RTt
è g +1 ø
2(g -1)
2(g -1)
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
˙ / A is maximum at M = 1 - flow is “choked” at throat - any change
 m
˙
in downstream conditions cannot affect m
 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
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Isentropic nozzle flow
A/A*
T/Tt
P/Pt
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
12
Isentropic nozzle flow
 When can choking occur? If M ≥ 1 or
Pt æ g -1 2 ö
³ ç1+
(1) ÷
*
ø
P è
2
g
g -1
æ g + 1ö
=ç
÷
è 2 ø
g
g -1
= 1.89 for g = 1.4
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
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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
g -1 ö
æ
2g
ç æPö g ÷
u=
RT 1- ç ÷
g -1 t çç è Pt ø ÷÷
è
ø
 For exit pressure = ambient pressure and FAR << 1,
Thrust = m˙ a (u9 - u1 ) = m˙ a
é
g -1 ö
æ
2gRT1t ê T9t ç æ P9 ö g ÷
1- ç ÷ ÷ ê
ç
g -1 ê T1t ç è P9t ø ÷
è
ø
ë
g -1 öù
æ
ç æ P1 ö g ÷ú
çç1- çè P ÷ø ÷÷ú
1t
è
øúû
 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
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Stagnation temperature and pressure
 From previous page
Thrust = m˙ a (u9 - u1 ) = m˙ a
é
g -1 ö
æ
2gRT1t ê T9t ç æ P9 ö g ÷
1- ç ÷ ÷ ê
ç
g -1 ê T1t ç è P9t ø ÷
è
ø
ë
g -1 öù
æ
ç æ P1 ö g ÷ú
çç1- çè P ÷ø ÷÷ú
1t
è
øúû
 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)
P9t = P1t
P9 < P1
T9t = T1t
P1 = P1t
(M1 = 0)
Case A
B. T9t > T1t, P9t = P1t > P1 = P9
(e.g. high-M ramjet/scramjet,
no Pt losses)
P9t = P1t
P9 = P1
T9t > T1t
P1t > P1
(M1 > 0)
Case B
AME 436 - Spring 2013 - Lecture 12 - 1D 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 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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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
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
æ g -1 2 ö
æ g -1 2 ö
T1t = T1ç1+
M1 ÷ = T2t = T2 ç1+
M 2 ÷ = constant
è
ø
è
ø
2
2
 Mass conservation
m˙ = ruA; m˙ = constant, A = constant Þ ru = constant Þ r1u1 = r2 u2
Þ
Þ
P1
P
PM
PM
M1 gRT1 = 2 M 2 gRT2 Þ 1 1 = 2 2
RT1
RT2
T1
T2
P1 M1
=
T1
1+
g -1
M12
P2 M 2
M
2
Þ P2 = P1 1
g -1 2
M2
g -1 2
1+
M2
1+
M1
2
2
T1
g -1 2
1+
M2
2
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
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Constant everything except S (shock)
 Momentum conservation (constant area, dx = 0)
P1 +r1u12 =P2 +r 2u22 =P2 +
r1u1
u2
u22 = P2 +r1u1u2
æ
æ P ö 2
æ P ö
PM
Þ P1 + çç 1 ÷÷ M 1 g RT1 = P2 + çç 1 ÷÷ M 1 M 2 g R T1T2 = P2 + ç 1 1
ç T
è RT1 ø
è RT1 ø
è
1
(
(
Þ P1 1+g M
(
Þ P1 1+g M
(
2
1
2
1
)
)
(
)
)
ö
÷ M g T
2
÷ 2
ø
æ
ö
P
M
= P2 + ç 2 2 ÷ M 2 g T2 Þ P1 1+g M 12 = P2 1+g M 22
ç T ÷
è
2 ø
(
M
= P1 1
M2
1+g M 12
)
g -1
)
M 12
(
)
(
(
)
2
(
)
)
(
)
2
1+g M
1+g M
2
2
1+g M 2 Þ
=
æ
ö
æ g -1 2 ö
g -1 2
g
-1
2
2
2
1+
M2
M 1 ç1+
M 1 ÷ M 2 ç1+
M2 ÷
2
2
2
è
ø
è
ø
1+
(
)
2
1
2
2
2
æ
1+ 2g x + g 2 x 2
g -1 ö 2
Let
= C, M = x Þ C =
Þ çg 2 C ÷ x + 2g - C x +1= 0
æ
ö
æ
ö
2 ø
g -1 2
g -1
è
M 12 ç1+
M1 ÷
x ç1+
x÷
2
2 ø
è
ø
è
2
M 12 +
g -1
Quadratic equation for x, 2 solutions: M 2 = M 1 (nothing happens) or M 22 =
2g
M 12 -1
g -1
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
2
2
(
)
19
Constant everything except S (shock)
 Complete results
2
M12 +
g -1
M 22 =
2g
M12 -1
g -1
P2t
=
P1t
P2
2g
g -1
=
M12 P1 g + 1
g +1
éæ g + 1 2 ö æ g -1 2 öù
M1 ÷ ç1+
M1 ÷ú
êç
è
ø
è
øû
2
ë 2
æ 2g
g -1ö
2
M1 ç
÷
g + 1ø
èg +1
 Implications
1
g -1
g
g -1
T2
=
T1
T2t
=1
T1t
ö
æ g -1 2 öæ 2g
2
M1 ֍
M1 -1÷
ç1+
è
øè g -1
2
ø
(g + 1) M 2
1
2(g -1)
2
 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
20
Constant everything except S (shock)
P2/P1
T2/T1
T2t/T1t
M2
P2t/P1t
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
21
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)
AdP + mdu + C f ( ru 2 / 2)Cdx = 0,m = ruA
mC f C
m
Þ dP + du +
udx = 0 (1)
2
A
2A
C
u2
g
C PT + = constant, P = r RT ,m = ruA, P =
2
R g -1
g -1 m æ K u ö
g A * * u*2
ÞP=
Pu +
; ( )* = any reference state
ç - ÷; K =
g Aè u 2ø
g -1 m
2
g -1 m æ K 1 ö
Þ dP = ç 2 + ÷ du (2)
g Aèu 2ø
Combine (1) and (2) and integrate:
g +1 du
g -1 du C f C
2A ég +1
g -1 1 ù
-K
+
dx = 0 Þ x =
ln(u) - K
+ constant
ê
3
2ú
2g u
g u
2A
C f C ë 2g
2g u û
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
22
Everything const. but momentum (Fanno flow)
Evaluate at reference state x = L*
æ g A * * u*2 ö g -1 1 ù
2A ég +1
*
*
L =
ê
ln(u ) - ç
Pu +
ú + constant
÷
*2
C f C êë 2g
2 ø 2g u úû
è g -1 m
*
ég +1
æ
2A
1
A
P
g -1öù
*
*
constant = L ln(u ) - ç
+
ê
÷ú
*
C f C êë 2g
4g øúû
è2 m u
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
23
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!)
æ
ö
2
ç (g + 1) M 2 ÷ T
g +1
1M
g
+
1
*
=
÷
(C / A)C f L =
+
lnç
*
æ g -1 2 ö
2
T
æ
ö
g -1 2 ÷
gM
2g ç
2ç1+
M ÷
M ÷÷
ç 2ç1+
è
ø
2
øø
2
è è
æ
ö1/ 2
(g +1)
ç
÷
P
Tt
1
g +1
Pt
1 æ 2 æ g - 1 2 ö ö 2 (g -1)
=
1
ç
÷
=
=
M ÷÷
ç1+
ç
Tt*
*
P * M ç æ g -1 2 ö ÷
øø
Pt
M è g +1 è
2
M ÷÷
ç 2ç1+
øø
2
è è
 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
24
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
æ g -1 2 ö
P2 M1 ç1+ 2 M1 ÷
=
ç
÷
P1
M 2 ç1+ g -1 M 2 ÷
2
è
ø
2
1/ 2
( g +1)
æ g -1 2 ö
P2t M1 ç 1+ 2 M 2 ÷
=
ç
÷
P1t M 2 ç 1+ g -1 M 2 ÷
1
è
ø
2
2( g -1)
1+
g -1
M12
T2
2
=
T1 1+ g -1 M 2
2
2
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
25
Everything constant but momentum
10
Fanno flow
Pt/Pt*
Tt/Tt*
1
T/T*
Length
0.1
T/T*(Fanno)
P/P* (Fanno)
Tstag/Tstag* (Fanno)
Pstag/Pstag*(Fanno)
Dimensionless length (4CfL*/D)
Length
P/P*
0.01
0
1
2
3
4
5
Mach number
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
26
Everything constant but momentum
3
Fanno flow
Pt/Pt*
2.5
T/T*(Fanno)
P/P* (Fanno)
Tstag/Tstag* (Fanno)
Pstag/Pstag*(Fanno)
Dimensionless length (4CfL*/D)
2
P/P*
1.5
Tt/Tt*
1
T/T*
0.5
Length
Length
0
0
1
2
3
4
5
Mach number
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
27
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))
g -1 2 ö
2
2æ
2(g + 1)M ç1+
M ÷
1+ g
T æ 1+ g ö 2 P
Tt
è
ø
2
÷ M
= çç
=
=
*
2÷
*
2
T è1+ gM ø
P 1+ gM
2 2
Tt *
1+ gM )
(
g
g
g
-1
æ
ö
Pt æ 2 ö
1+ g æ g -1 2 ö g -1
÷ 1+
=ç
M ÷
÷ çç
*
2 ֍
ø
Pt è g + 1ø è1+ gM øè
2
 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
28
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.
g -1 2
2
2
1+
M2
2
2
1+ gM1
2ö
2
2
T
M
æ
2t
2
2
P2 1+ gM1
T2 1+ gM1 M 2
=
=
=ç
2
2
2
2÷
2
2
T
M
1t
1
P1 1+ gM 2
T1 è1+ gM 2 ø M1
1+ gM 2 1+ g -1 M12
2
(
(
æ g -1 2 ö
1+
M2 ÷
P2t æ1+ gM12 öç
2
=ç
÷
2 ֍
P1t è1+ gM 2 øç1+ g -1 M 2 ÷
1
è
ø
2
g
)
)
g -1
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
29
Heat addition at constant area
10
T/T*(Rayleigh)
P/P* (Rayleigh)
Tstag/Tstag* (Rayleigh)
Pstag/Pstag*(Rayleigh)
Rayleigh flow
Pt/Pt*
1
Tt/Tt*
T/T*
P/P*
0.1
0
1
2
3
4
5
Mach number
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
30
T-s diagram - reference state M = 1
10
T/T*(Rayleigh)
T/T* (Fanno)
T/T* (shock)
T-s diagram (reference state Mach = 1)
T/T*
Fanno M < 1
Shock
1
Rayleigh
M<1
M>1
0.1
-6
-4
-2
0
2
(S - S*)/R
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
31
T-s diagram - Fanno, Rayleigh, shock
T-s diagram (reference state Mach = 4)
7
T/Tref (Rayleigh)
T/Tref (Fanno)
T/Tref (shock)
6
Constant area,
with friction, no
heat addition
T/Tref
5
4
Rayleigh
Shock
M<1
Constant area,
no friction, with
heat addition
Fanno
M<1
3
M>1
This jump: constant
area, no friction, no heat
addition  SHOCK!
2
M>1
1
0
0.5
1
1.5
2
2.5
3
3.5
4
(S - Sref)/R
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
32
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)
˙ du = 0  u = constant
 Momentum equation: AdP + m
 Reference state ( )*: use M = 1 again but nothing special happens
there
 Again energy equation not useful except to calculate q
T
A
1
=
=
T * A* M 2
Pt æ 2 ö
=ç
÷
*
Pt è g + 1ø
g
P
=1
*
P
æ g -1 2 ö
M ÷
ç1+
è
ø
2
g -1
g
æ g -1 2 ö
Tt
2
=
1+
M ÷
*
2ç
ø
(g + 1)M è
2
Tt
g -1
 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
33
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.
T2 A2 M12
=
= 2
T1 A1 M 2
g -1
T2t M12 1+ 2 M
= 2
T1t M 2 1+ g -1 M 2
1
2
2
2
æ g -1 2 ö
1+
M2 ÷
P2t ç
2
=ç
÷
P1t ç1+ g -1 M 2 ÷
1
è
ø
2
g
g -1
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
34
Heat addition at constant P
10
Constant P heat addition
Pt/Pt*
T/T*, A/A* (ConstP)
P/P* (ConstP)
Tstag/Tstag* (ConstP)
Pstag/Pstag*(ConstP)
1
P/P*
Tt/Tt*
T/T*, A/A*
0.1
0
1
2
3
4
5
Mach number
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
35
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
˙ du = 0  dP/P + MdM = 0
 Momentum: AdP + m
 Reference state ( )*: use M = 1 again
T
=1
*
T
ég
P
2 ù
= expê (1- M )ú
*
ë2
û
P
é-g
A
1
2 ù
= expê (1- M )ú
*
ë2
û
A
M
 Implications
Tt
2 æ g -1 2 ö
=
M ÷
ç1+
*
ø
g + 1è
2
Tt
Pt æ 2 ö
=ç
÷
*
Pt è g + 1ø
g
g -1
ég
g -1 2 ö
2 ùæ
expê (1- M )úç1+
M ÷
ë2
ûè
ø
2
g
g -1
 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
36
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.
ég
2 ù
expê (1- M 2 )ú
ég
P2
ë2
û
2
2 ù
=
= expê ( M1 - M 2 )ú
é
ù
g
ë2
û
P1
expê (1- M12 )ú
ë2
û
1+
g -1
T2t
2
=
T1t 1+ g -1 M 2
1
2
é-g
2 ù
expê (1- M 2 )ú
ég
A2 M1
ë2
û M1
2
2 ù
=
=
expê ( M 2 - M1 )ú
é
ù
-g
ë2
û
A1 M 2
M2
expê (1- M12 )ú
ë2
û
g
ég
g -1 2 ö g -1
2 ùæ
expê 1- M 2 ú ç1+
M2 ÷
ég
P2t
ë2
ûè
ø
2
2
2
=
=
exp
M
M
2
g
êë2 1
ég
P1t
2 ùæ
g
-1
expê 1- M1 ú 1+ g -1 M 2 ö
1÷
ë2
û çè
ø
2
(
)
(
)
M 22
(
æ g -1 2 ö
ùç 1+ 2 M 2 ÷
)úûç g -1 2 ÷
ç 1+
M1 ÷
è
ø
2
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
g
g -1
37
Heat addition at constant T
10
Constant T heat addition
Tt/Tt*
A/A*
T/T*
1
T/T* (ConstT)
P/P* (ConstT)
Tstag/Tstag* (ConstT)
Pstag/Pstag*(ConstT)
A/A* (ConstT)
Pt/Pt*
P/P*
0.1
0
1
2
3
4
5
Mach number
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
38
T-s diagram for diabatic flows
10
T-s diagram for the various diabatic flows
(reference state Mach = 1)
T/T*
T/T*(Rayleigh)
T/T* (Const P)
T/T* (Const T)
Const P
1
Const T
Rayleigh
(Const A)
Rayleigh
(Const A)
0.1
-6
-4
-2
0
2
(S - S*)/R
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
39
T-s diagram for diabatic flows
10
Comparison of stagnation pressure & temperature
for the various diabatic flows
Pstag / Pstag (ref)
Rayleigh
(Const A)
Rayleigh
Const P
Const T
1
Const P
Rayleigh
(Const A)
Const T
0.1
0
1
2
3
4
Tstag / Tstag (ref)
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
40
Area ratios for diabatic flows
10
Const T
A / A (ref)
Const T
Const P
1
Const A
Rayleigh
Const P
Const T
Comparison of area ratio & stagnation temperature
for the various diabatic flows
0.1
0
1
2
3
4
Tstag / Tstag (ref)
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
41
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)
dPt
gM 2 Pt d(ln Pt )
gM 2
=or
=dTt
2 Tt d(lnTt )
2
 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
42
Summary - 1D compressible flow
Const.
A?
Adia- FrictionTt
Pt
batic?
less?
const.? 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
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
43
Summary of heat addition processes
Const. A
Const. P
Const. T
Goes to
M=1
Decreases
Increases
Constant
Increases
Min. at M = -1/2
P
Decr. M < 1
Incr. M > 1
Constant
Decreases
Pt
Decreases
Decreases
Decreases
T
Incr. except for
a small region
at M < 1
Increases
Constant
Tt
Increases
Increases
Increases
M
Area
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
44
Example
Consider a very simple propulsion system operating at a flight Mach number of 5 that consists of
Shock
q
2 processes:
in
Process 1: Shock at entrance to duct
Process 2: Heat addition in a constant-area
duct until thermal choking occurs
Air, g = 1.4
a) Compute all of the following properties of this system:
M3 = ???
P3 /P1 = ???
T3/T1 = ???
M2 = ???
P2 /P1 = ???
T2/T1 = ???
M1 = 5
P = P1
T = T1
1 2
3
(i) Static (not stagnation) temperature relative to T1 after the shock
T2
=
T1
ö
æ g - 1 2 öæ 2g
2
M1 ֍
M1 - 1÷
ç 1+
2
è
øè g - 1
ø
(g + 1) M
2(g - 1)
2
2
1
=
æ 1.4 - 1 2 öæ 2(1.4) 2 ö
5 ֍
5 - 1÷
ç1+
2
1.4
1
è
øè
ø
(1.4 + 1) 5
2(1.4 - 1)
2
= 5.8
2
(ii) Static (not stagnation) pressure relative to P1 after the shock
P2
2g
g -1 2(1.4) 2 1.4 -1
=
M12 =
5 = 29
g + 1 1.4 + 1
1.4 + 1
P1 g + 1
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
45
Example
(iii) Static (not stagnation) temperature and pressure relative to T1 at the exit
2
2
T æ 1+ g ö 2
T2 æ 1+ g ö 2
T3 T3 T2 æ1+ gM 22 ö 1 T2
÷ M Þ =ç
=ç
=ç
÷
÷ M2 Þ =
T3 è 1+ gM 22 ø
T1 T2 T1 è 1+ g ø M 22 T1
T * çè1+ gM 2 ÷ø
2
2
g -1
2
1.4 -1 = 0.1724 Þ M = 0.4152
M 22 =
=
2
2g
2(1.4) 2
æ1+ 1.4(0.4152)2 ö2
2
T
1
M -1
5 -1
÷÷
Þ 3 = çç
(5.8) = 9
g -1 1
1.4 -1
2
T1 è
1+ 1.4
0.4152
ø
M12 +
52 +
P
1+ g
P2
1+ g
P3 P3 P2 1+ gM 22 P2 1+ 1.4(0.4152)
=
Þ =
Þ =
=
=
(29) = 15
P3 1+ gM 22
P1 P2 P1
1+ g P1
1+ 1.4
P * 1+ gM 2
2
(iv) Dimensionless heat addition {qin/RT1 = CP(T3t-T2t)/RT1 = [/(-1)](T3t-T2t)/T1}
qin
g T3t - T2t
g T2t (T3t / T2t -1)
g T1t (T3t / T2t -1)
g T1t æ T3t ö g æ g -1 2 öæ T3t ö
=
=
=
=
M1 ÷ç -1÷
ç -1÷ =
ç1+
øè T2t ø
RT1 g -1 T1
g -1
T1
g -1
T1
g -1 T1 è T2t ø g -1 è
2
æ g -1 2 ö
æ 1.4 -1
ö
2(g +1)M 22 ç1+
M 2 ÷ 2(1.4 +1)0.4152 2 ç1+
0.4152 2 ÷
T2t
è
ø
è
ø
2
2
=
=
= 0.5555
2
2
2
2
T3t
1+
g
M
1+1.4(0.4152)
(
(
)
2)
T1t
g -1 2
1.4 -1 2
= 1+
M1 = 1+
5 = 6;T2t = T1t = 6T1;T3t = 1.8T2t = 10.8T1
T1
2
2
ö
q
1.4 æ 1.4 -1 2 öæ 1
Þ in =
5 ֍
-1÷ = 16.8
ç1+
øè 0.5555 ø
RT1 1.4 -1 è
2
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
46
Example
(v) Specific thrust (assume FAR << 1 in the thrust calculation)
Specific Thrust (ST) º Thrust / ma c1
FAR << 1: Thrust = ma [u9 - u1 ]+ (P9 - P1 )A9
ST º
Thrust u9 u1 (P9 - P1 )A9 u9 c9
(P - P )A
= - +
=
- M1 + 9 1 9
ma c1
c1 c1
ma c1
c9 c1
( r9u9 A9 )c1
= M9
æ P ö RT9
g RT9
(P9 - P9 )
T
- M1 +
= M 9 9 - M1 + ç1- 1 ÷
(P9 / RT9 )u9 c1
T1
g RT1
è P9 ø c9 M 9 c1
= M9
æ P1 ö
æ P1 ö T9 1
T9
RT9
T9
- M1 + ç1- ÷
= M9
- M1 + ç1- ÷
T1
T1
è P9 ø g RT9 M 9 g RT1
è P9 ø T1 g M 9
æ 1ö
1
= 1 9 - 5 + ç1- ÷ 9
= 0(???)
è 15 ø 1.4(1)
(vi) Overall efficiency
æ Thrust ö (g -1) T1M1
(1.4 -1) (5) = 0
ho = ç
=
(0)
÷
(10.8 - 6)
è ma c1 ø (T3t - T2t )
(vii) Draw this cycle on a T - s diagram.
Include appropriate Rayleigh and Fanno curves.
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
47
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.
æ g -1 2 ö
P4t = P3t Þ P4ç1+
M4 ÷
2
è
ø
g
g -1
æ g -1 2 ö
= P3 ç1+
M3 ÷
2
è
ø
g
g -1
g -1
é
ù
æ
ö
2 ê P3 g æ g -1 2 ö ú
Þ M4 =
M 3 ÷ -1
ç ÷ ç1+
ê
g -1 è P4 ø è
2
ø ú
ë
û
P3
2 é g -1g æ 1.4 -1 2 ö ù
M 3 = 1, P4 = P1 , = 15 Þ M 4 =
1 ÷ -1ú = 2.829
ê(15) ç1+
P1
g -1 ë
2
è
ø û
g -1 2
1.4 -1 2
1+
1
æ g -1 2 ö
æ g -1 2 ö T 4 T 3 1+ 2 M 3
2
T4t = T3t Þ T4ç1+
M 4 ÷ = T3 ç1+
M3 ÷ Þ
=
= (9)
= 4.152
g
-1
1.4
-1
2
2
T
T
è
ø
è
ø
1
1
1+
M 42
1+
2.8292
2
2
T9
P9 = P1 : ST = M 9
- M1 = 2.829 4.152 - 5 = 0.764
T1
Heat addition is the same as before, so
æ Thrust ö (g -1) T1M1
(1.4 -1) (5) = 0.3183
ho = ç
= (0.764)
÷
(10.8 - 6)
è ma c1 ø (T3t - T2t )
AME 436 - Spring 2013 - Lecture 12 - 1D Compressible Flow
48
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
49