#### Transcript Document

```Chap 5
Quasi-OneDimensional Flow
5.1 Introduction
Good approximation for practicing gas dynamicists
eq. nozzle flow、flow through wind tunnel & rocket engines
5.2 Governing Equations
The continuity equation :
  
  v  ds 
t
s
  d


1u1 A1  2u2 A2
The momentum equation :


( v )
  

d   f  d   p  ds
s (v  ds )v  
t


s
X-dir
Y-dir
p1 A1   u A  (A
2
1 1 1
A2
1

pdA) x  p2 A2  2u22 A2
Automatically balainced
The energy equation
 
 
 q  d   pv  ds    ( f  v )d


s

V2
V2 
  [  (e  )]d    (e  )v  ds
2
2
 t
s
h  e p

2
1
2
2
u
u
h1 
 h2 
 h0  const
2
2
total enthalpy is
constant along the flow
Actually, the total enthalpy is constant along a streamline in any
P
A
u
ρ
In differential forms
uA  const
 d ( uA)  0
pA  u 2 A  pdA 
P +dP
A +dA
u +du
ρ+dρ
dx
( p  dp)( A  dA)  (   d )(u  du) ( A  dA)
2
Dropping 2nd order terms
2
d ( uA)  0
2
u dA uAdu Au d  0
2
2
(1)
(2)
(1) - (2) =
dp   udu
u2
 h   const
2
u2
d (h  )  0
2
dh  udu  0
Euler’s equation
dp (  )  du ( )
dp ( )  du (  )
5.3 Area-Velocity Relation
d ( uA)  0
dP d

 udu
 d 
dP
udA Adu  Aud  0
uA
∴ no dissipation mechanism
→ isentropic
d
du dA


0

u
A
dA
du
2
 ( M  1)
A
u
d
2
udu
u du
2 du
  2   2  M

a
au
u
Important information
1.
2.
3.
4.
M→0 incompressible flow
Au=const consistent with the familiar continuity eq for
incompressible flow
0≦M＜1 subsonic flow
an increase in velocity (du，+) is associated with a decrease
in area (dA，－) and vice versa.
M>1 supersonic flow
an increase in velocity is associated with an increase in area ,
and vice versa
M=1 sonic flow →dA/A=0
a minimum or maximum in the area
A subsonic flow is to be
accelerated isentropically
from
subsonic
to
supersonic
Supersonic flow is to be
decelercted isentropically
from
supersonic
to
subsonic
Application of area-velocity
relation
1.Rocket engines
2.Ideal supersonic wind tunnel
Diffuser is to slow down the flow in the convergent duct to sonic
flow at the second throat, and then futher slowed to low subsonic
speeds in the divergent duct.
(finally being exhausted to the atmosphere for a blow-down wind
tunnel)
“chocking”
Handout – Film Note by Donald Coles
(When both nozzle
“blocking” with M=1)
5.4 Isentropic Flow of a Calorically Perfect
Gas through Variable-Area Duct
 u A  uA  u  a
* *
*
*
A  0 a

*
A
0  u
*
*
Stagnation density
(constant throughout an isentropic flow)
*
0
r 1
 (1 
M )

2
1
2 r 1
0
r 1
 (1 
)
*

2
1
r 1
(1)
r 1
(
)
2
1
r 1
(2)
r 1 2
M
u 2
2
( *) 
 M * (ch.3)
r 1 2
a
1
M
2
2
(3)
r 1
r 1
A 2
1
2
r 1 2
( *)  2 [
(1 
M )]
A
M r 1
2
Area – Mach Number Relation
M  f (A
*
A
)
There are two values of M
which correspond to a given
A/A* >1 , a subsonic & a
supersonic value
Boundary conditions will
determine the solution is
subsonic or supersonic
1.
For a complete shock-free
isentropic supersonic flow, the
exit pressure ratio Pe /P0 must be
precisely equal to Pb /P0
2.
Pe /P0、Te /T0 & Pe /P0 = f(Ae
/A*) and are continuously
decreasing.
3.
To start the nozzle flow, Pb must
be lower than P0
4.
For a supersonic wind tunnel, the
test section conditions are
determined by (Ae /A*)、P0、T0
gas property & Pb
Pb=P0 at the beginning ∴there is no
flow exists in the nozzle
Minutely reduce Pb，this small
pressure difference will cause a small
wind to blow through the duct at low
subsonic speeds
Futher reduce Pb ，sonic
conditions are reached (Pb=Pe3)
Pe /P0 & A/At are the controlling
factors for the local flow properties
at any given section
p*
r  1 rr1
 (1 
)  0.528
p0
2
for r=1.4
  tAtUt
m
What happens when Pb is further reduced below Pe3 ?
Should use dash-line
to indicate irreversible
process
Note: quasi-1D consideration does not tell us much about how to
design the contour of a nozzle – essentially for ensuring a
shockfree supersonic nozzle
Method of characteristics
Wave reflection from a free boundary
Waves incident on a solid (free) boundary reflect in like
(opposite) manner , i.e, a compression wave as a compression
(expansion wave ) and an expansion wave reflects as an
expansion ( compression ) wave
5.5 diffusers
 Assume that we want to
design a supersonic wind
tunnel with a test section
M=3
Ae/A*=4.23
P0/Pe=36.7
 3 alternatives
(a) Exhaust the nozzle directly to
the atmosphere
(b) Exhaust the nozzle into a constent area duct which
serves as the test section
P0 
P0 Pe
1
P  (36.7)(
)P  3.55atm
Pe P02
10.33
∴the resvervair pressure required to drive the wind tunnel
has markedly dropped from 36.7 to 3.55 atm
(c) Add a divergent duct behind the normal the normal shock
to even slow down the already subsonic flow to a lower
velocity
P0 
P0 Pe P2
1  1 
P  (36.7)(
)
1  3.04atm
F
Pe P2 P02
10.33  1.17 
o
r
Note:
M 3
P01
1


P02 0.328
P02
 0.328
P01
 3.04
∴ the reservoir pressure required to drive a supersonic wind
tunnel (and hence the power required form the compressors) is
considerably reduced by the creation of a normal shock and
subsequent isentropic diffusion to M～0 at the tunnel exit
Diffuser - the mechanism to slow the flow with as small a
loss total pressure as possible
Consider the ideal supersonic wind funnel again
If shock-free →P02/P01=1 no lose in total pressure
→a perpetual motion machine!!← something is wrong
(1) in real life , it hard to prevent oblique shock wave from occuring inside
the duct
(2) even without shocks , friction will cause a lose of P0
∴the design of a perfect isentropic diffuser is physically impossible
Replace the normal shock diffuser with an oblique shock
diffuser provide greater pressure recovery
Diffuser efficiency
(
Pd0
)
P0 actual
 D
(mostl common one)
P02
(
)
P01 Normal shock at Me
If ηD=1→normal shock diffuser
for low supersonic test section Me, ηD>1
for hypersonic conditions ηD<1
(normal shock recovery is
about the best to be expected)
At2>At1(due to the entropy increase in the diffuser)
proof: assume the sonic flow exists at both throats
1* At1a1*  2* At 2a2*
P1*
At 2
1* a1*
1*
RT1* P1* P01
 * ( *)  *  *
 * 
At1
 2 a2
 2 P2
P2
P02
RT2*
At 2 P01

At1 P02
 P02  P01 always  At 2  At1
 D Is very sensitive to At 2
At2ηD=max is slightly larger than (P01/P02)At1
∴the fix- geometry diffuser will operate at an efficiency less than ηD,m to start properly
ηD is low it is because At2 is too large
∴ the flow pass though a series of
oblique shock waves id still “very”
supersonic
∴ a strong normal shock form before
exit of the diffuser
∴ defeats the purpose of are oblique
shock diffuser
```