Quasi-1D Nozzle Review
Example: Want to find P,T,M, etc. given Po, Pe, and nozzle shape.
Quasi – Area is allowed to vary along x coordinate, but flow variables are functions of x
only.
Start out with governing conservation equations:
Mass:

 t dV   U  ndS  0
(1.1)

UdV    U  n UdS   pdS    fdV  Fviscous
t 
V
S
S
V
(1.2)
V
Momentum:
S
Energy:
 U2 
 U2 

q

e

dV




e 
U  ndS    pU  ndS    dV     f  U  dV


t V 
2 
2 
t

S
S
V
V
(1.3)
Assuming:
1. steady
2. inviscid
3. no body forces
4. 2D flow
Quasi-1D: Area is allowed to vary but flow variables are a function of x only
A, u, 
A+dA,u+du,  +d
Mass:
 uA   u  du    d   A  dA  0
 uA  uA  udA   duA  d uA  higher order terms  0
(1.4)
(1.5)
divide by through  uA
dA du d 


0
A
u

or
(1.6)
d   uA   0
Momentum:
 PdA 
  u 2 A     d   u  du  u  du  A  dA   PA   P  dP  A  dA   2 
 (1.7)
 2 
  u 2 A   u 2 A   u 2 dA  u 2 Ad    uAdu   uAdu  PA  PA  PdA  AdP  PdA (1.8)


u udA  uAd    Adu  uAdu   AdP
(1.9)
dP    udu
(1.10)
dP d 
dP 
 udu a 2 

 d 
d  s
d
u
  2 du

a
Substituting equation (1.12) into (1.6) to get:
dA du u

 du  0
A
u a2
which can be rearranged to get:
dA du  u 2  dA du
 1  2  

1 M 2  0
A
u  a  A
u
dP



dA du

 M 2  1
A
u
(1.11)
(1.12)
(1.13)
(1.14)
(1.15)
which can be used to determine general flow behavior in a converging-diverging nozzle,
as below:
M
<1
subsonic
<1
>1
supersonic
>1
dA
<0
>0
<0
>0
du
>0
<0
<0
>0
dA <0 --> converging
dA >0 --> diverging
du <0 --> decelerating
du >0 --> accelerating
Energy:
We will not go through the derivation for the energy equation but,
applying analysis as before will give:
dh  udu  0 or cP dT  udu
(1.16)
Total-Static-Mach Relations
Isentropic Relations:


 T   1
P2   2 
   2
P1  1 
 T1 
(1.17)
Static-Total
From Energy Equation with uo=0 (total)
cPTo  cPT1 
u12
2
(1.18)
Rearranging
To
u2
1
T
2cPT
(1.19)
To
  1 u2
 1
T
2  RT
(1.20)
using cP   R  1 ,
Inserting a   RT and M  u a gives
To
 1 2
1
M
T
2
(1.21)

Po    1 2   1
 1 
M 
P 
2

(1.22)
1
o    1 2  1
 1
M 
 
2

(1.23)
Given Eq.’s (1.21),(1.22), and (1.23) we can now:


Find any static property in an isentropic flow given Mach #, Po,To,o.
Use/control known total conditions to find mach # through nozzle
Area-Mach Relations
From mass
 u A  uA
(1.24)
at A M 
u
 1  u   a , giving

a
2


 A    a 





 
A      u 
or
2
2
2
2


 A      o   a 

       
 A   o      u 
using isentropic relations for the density terms
2
(1.25)
2
(1.26)
 1
1  2    1 2    1
 A
M 
   2 
1 
M   1 
2
A 

Mach # is a function of this area ratio only. Must find A*.
2
(1.27)
Completely Subsonic Flow
Pe  Patm
(1.28)
From isentropic relation
 




2  Po 

Me 
   1

   Pe 


Can now determine A* and entire Mach # distribution
A
P
If: o  1 Then: M e  0, e  , A  0 NO FLOW
A
Pe
A
P
As o increases, Me increases, e decreases, A* increases
A
Pe
(1.29)
Note that A/A* < 1 is not physically possible. That is, after 1st critical is reached, must
have Amin = A*
Supersonic Flow
Subsonic Flow ahead of throat. Follow supersonic A/A* branch after throat.
A  At
A 
M e  f  e 
A 
(1.30)
(1.31)

   2 1
Pe  Po 1 
Me 
2


(1.32)
Suppose we pick Po so that Pe = Patm


If we decrease Po, then Pe < Patm because Me is unchanged. Need weak oblique
shocks to get a small pressure jump.
As Po decreases, need stronger oblique shocks until normal shock at exit, 2nd
critical.


As Po decreases, shock moves up the nozzle. Eventually get to 1st critical.
Increasing Po from 3rd critical, Pe > Patm. Get Prandtl-Meyer expansion fan to get
pressure decrease
Summary:
For our nozzle:
Po ,3rd  60 psig
Po ,2 nd  20 psig
Po ,1st  8 psig