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:
∂ρ
∫∫∫∂t dV + ∫∫ρU ⋅ndS = 0
V
Momentum:
∂
ρUdV +
∂t ∫V∫∫
(1.1)
S
∫∫ρ (U ⋅n )UdS = − ∫∫pdS + ∫∫∫ρ fdV +
S
S
Fviscous
(1.2)
V
Energy:
 U2 
∂
ρ
e +
dV +
∂t ∫V∫∫ 
2 
 U2 
ρ
∫S∫ e + 2 U ⋅ndS = −
(1.3)
∂q
∫∫pU ⋅ndS + ∫∫∫ρ ∂t dV + ∫∫∫ρ ( f ⋅U )dV
S
V
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, ρ
A+dA,u+du, ρ +d ρ
V
Mass:
− ρuA + ( u + du )( ρ + d ρ )( A + dA ) = 0
− ρuA + ρuA + ρudA + ρduA + d ρuA + higher order terms = 0
divide by through ρuA
dA du d ρ
+
+
=0
A
u
ρ
or
(1.4)
(1.5)
(1.6)
d ( ρuA) = 0
Momentum:
 PdA 
− ρu 2 A + ( ρ + d ρ )( u + du )( u + du )( A + dA) = PA − ( P + dP )( A + dA) + 2 
 (1.7)
 2 
− ρu 2 A + ρu 2 A + ρu 2 dA + u 2 Ad ρ + ρuAdu + ρuAdu = PA − PA − PdA − AdP + PdA (1.8)
(
)
u ρudA + uAd ρ + ρ Adu + ρuAdu = − AdP
(1.9)
dP = − ρudu
(1.10)
dP dP d ρ
dP 
=
= − udu a2 =

ρ dρ ρ
d ρ s
dρ
u
= − 2 du
ρ
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  u2  dA du
+
+
1 − M 2 )= 0
(
1 − 2  =
A
u  a  A
u
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 c P dT = − udu
(1.16)
Total-Static-Mach Relations
Isentropic Relations:
γ
γ
P2  ρ2  T2 γ− 1
=  = 
P1  ρ1   T1 
(1.17)
Static-Total
From Energy Equation with uo=0 (total)
c PTo = cPT1 +
u12
2
(1.18)
Rearranging
To
u2
= 1+
T
2cP T
(1.19)
To
γ− 1 u 2
= 1+
T
2 γRT
(1.20)
using cP = γR γ− 1 ,
Inserting a = γRT and M = u a gives
To
γ− 1 2
= 1+
M
T
2
(1.21)
γ
Po  γ− 1 2 γ− 1
= 1+
M 
P 
2


(1.22)
1
ρo  γ− 1 2 γ− 1
= 1+
M 
ρ 
2


Given Eq.’s (1.21),(1.22), and (1.23) we can now:
(1.23)
• Find any static property in an isentropic flow given Mach #, Po,To,ρo.
• Use/control known total conditions to find mach # through nozzle
Area-Mach Relations
From mass
ρ ∗u∗ A∗ = ρuA
(1.24)
at A∗ M =
u∗
= 1 → u ∗ = a∗ , giving
∗
a
2
2
∗
∗
 A   ρ  a 
=
 A∗   ρ   u 
    
or
2
2
2
(1.25)
2
∗
∗
 A   ρ   ρo   a 
=
 A∗   ρ   ρ   u 
   o    
using isentropic relations for the density terms
2
(1.26)
γ+ 1
1  2  γ− 1 2 γ− 1
A
=
M 
 ∗
1 +
2 
2
 A  M γ+ 1 

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
γ− 1


2  Po  γ 
Me =
  − 1
γ− 1  Pe 




Can now determine A* and entire Mach # distribution
P
A
If: o = 1 Then: M e = 0, ∗e → ∞ , A∗ = 0 NO FLOW
Pe
A
P
A
As o increases, Me increases, e∗ decreases, A* increases
Pe
A
(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)
γ
 γ− 1 2 1− γ
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 ,3 rd ≈60 psig
Po ,2 nd ≈20 psig
Po ,1st ≈8 psig