Gas Dynamics 1 Fanno Flow
Gas Dynamics 1
Fanno Flow
Gas Dynamics
* The Flow of a compressible fluid in a duct is Always accompanied by :-
Simple frictional flow ( Fanno Flow ) Adiabatic frictional flow in a constant-area duct
Friction
Heat transfer
Variation in the cross –sectional area of the duct
*
* Although, it is difficult in many cases to separate the effects of each of these parameters , yet in order to provide an insight into the effect of friction , adiabatic flow thermally insulated in a constant area duct is analyzed in this chapter.
The above parameters contribute to changes the flow properties.
Gas Dynamics
* The entropy of the system still increase because of friction ….
- Friction is associated with the turbulence and viscous shear of molecules of the gas . - Irreversibility associated with friction causes a decrease in the stagnation pressure , steady 1-D flow as a perfect gas .
Assumptions: - Adiabatic frictional - flow in a constant cross- sectional area duct .
V dVV +
21 wτ
X dXX +ρ ρρ d+V dVV +
Gas Dynamics
Momentum equation
)()( VdVVAVBdxdPPAAP w −+=−+− ρτ
AVdVBdxAdP w ρτ =−− 1
= wetted perimeter HDAB 4=
)21( 2VFw ρτ = Dynamic pressure 2
21 Vρ
F is the conventional Fanning friction factor ( circular duct ) F for flow over as the drag coefficient
gV
DLf
gV
DLFh WDFpipeinloss 22
42
.
2
==
Fanning coefficient of friction
Darcy-weisbech coefficient of friction
diameterhydraulicDH =
V dVV +21
wτX dXX +ρρρ d+
VdVV +
Gas Dynamics
From eq. 1 ( eq.1 / A)
0=++ VdVdXABdP w ρτ
022
42
222
=++VdVVdXV
DFdPH
ρρ 2
2VFwρ
τ =
BADH4
=(fF =The friction factor for turbulent in smooth ducts is the Von-Karman Nikurdse formula
)4(Relog28.041
10 fF
+−=
2
Reynold’s number , roughness of boundaries )
2
22
2 VdVVVdV ρ
ρ =AVdVBdxAdP w ρτ =−− 1
VdV= dV2/2
Gas Dynamics
Perfect gas law RTP ρ=
TdTd
PdP
+=ρρ
Continuity Eq. == VAm ρ
0=+VdVd
ρρ
Constant
Energy Eq. =+=2
2Vhho Constant
3
4
0=+VdVdh 5
Mach number RTVMγ
22 =
TdT
VdV
MdM
−= 2
2
2
2
6
2nd law of thermodynamics 0≥ds 7
Gas Dynamics
Mach number TdT
VdV
MdM
−= 2
2
2
2
6
2nd law of thermodynamics 0≥ds 7
τρ (4&,,,,,HDFdXSTPVM Variable ) 6 equation
Momentum equation 022
42
222
=++VdVVdXV
DFdPH
ρρ 2
Perfect gas law TdTd
PdP
+=ρρ
Continuity Eq. 0=+VdVd
ρρ
Energy Eq.
3
4
0=+VdVdh 5
Gas Dynamics
022
42
222
=++VdV
PVdX
PV
DF
PdP
H
ρρ
From eq. 2 Dividing by P 022
42
222
=++VdVVdXV
DFdPH
ρρ
2
22
2
PMRTVPV γ
γγ
ρ ==
022
42
222
=++VdVMdXM
DF
PdPtherefore
H
γγ8
5._ Eqfrom
2
2dVdTCP −=
2
22
22
21
2)1(
2 VdVM
RTdV
TCdV
TdT
P
−−=
−−=
−=
γγ
γ 9
0=+VdVdh 5
Gas Dynamics
Combining eq.(9) with eq. (6)
2
22
2
2
2
2
21
VdVM
MdM
VdV −
−=−γ
2
2
22
2
*
211
1MdM
MVdV
−+
=γ
10
From eq. 3 & 4 (3), TdT
VdV
PdP
+−=From eq. 9
2
22
21
VdVM
VdV
PdP −
−−=γ
2
2
2VdV
VdV
=
(9), 2
12
)1(2 2
22
22
VdVM
RTdV
TCdV
TdT
P
−−=
−−=
−=
γγ
γ(6) 2
2
2
2
TdT
VdV
MdM
−=
(4) 0=+VdVd
ρρ
Gas Dynamics
02
42
)21
21(
2
2
222
2
2
=++−
−− dXMDF
VdVMM
VdV
H
γγγ
)21
21( 2
2
2
MVdV
PdP −
−−=γ
11
The PdP term and the
2
2
VdV term in eq. 8 can
be replaced to give
02
4))1(1](
211
1[21 2
222
2
2=++−−−
−+
dXMDFMM
MdM
M H
γγγ
γ
(8) 022
42
222
=++VdVMdXM
DF
PdP
H
γγ
From Eq. 10 (10) *
211
12
2
22
2
MdM
MVdV
−+
=γ
Gas Dynamics
2
2
22
2
*)
211(
)1(4MdM
MM
MDFdX
H−
+
−=
γγ
12
02
4))1(1](
211
1[21 2
222
2
2=++−−−
−+
dXMDFMM
MdM
M H
γγγ
γ
HDFdX
M
MM
MdM 4.
)1(
)211(2
22
2
2
−
−+
=
γγ
M<1
dM +ve dM -ve
M>1
Gas Dynamics
From continuity eq. ( eq. 4 ) & from eqs. 10 & 12 (10) 2
2*
22
11
12
2
M
dM
MV
dV−
+=
γ (11) 22
22
2
VdV
V
VdV
V
dV==
HDFdX
M
MM
MdM 4.
)1(
)211(2
22
2
2
−
−+
=
γγ
)211(
4.)1(2 22
2
MM
dMDFdX
MMd
VdV
H−
+=
−=−=
γγ
ρρ 13
M<1
dV +ve ρ -ve dV -ve
ρ +ve
M>1
12
Gas Dynamics
From eqs. 11 & 13
dMMM
MDFdX
MMM
PdP
H
])
2)1(1(
))1(1([4.)1(2
))1(1(2
2
2
22
−+
−+−=
−
−+−=
γγγγ
14 From eqs. 3,13& 14
)211(
)()(1(4.)1(2
)1(2
2
4
M
MdMDFdX
MM
TdT
H−
+−−=
−
−−=
γγ
γγ
15
Gas Dynamics
0≥ds F∴
HP DFdXM
Cds 4.
2)1( 2−
=γ
MdM
MM
MRDFdXRMdsH )
211(
)1(4.2 2
22
−+
−==
γγ
Then
16
Is positive
Entropy changes are determined from
PdPR
TdTCds P −= From eqs. 14 & 15 &
1−=γγRCP
(14) ])
2)1(1(
))1(1([4.)1(2
))1(1(2
2
2
22
dMMM
MDFdX
MMM
PdP
H−
+
−+−=
−
−+−=
γγγγ
(15) )
211(
)()(1(4.)1(2
)1(2
2
4
M
MdMDFdX
MM
TdT
H−
+−−=
−
−−=
γγγγ
Gas Dynamics
Also as a result of frictional flow
- For subsonic flow the Mach number increases
1<M )1( 2M− ve+
veds += vedM +=∴
- For supersonic flow the Mach number decreases
1>M )1( 2M− ve−veds += vedM −=∴
(16) )
211(
)1(2
2
MdM
MM
MRds−
+
−=
γ
Gas Dynamics
The stagnation pressure can be calculated from
120 )
211( −−
+= γγ
γ MPP
2
2
2
2
0
211
2/ MdM
M
MPdP
PdPO
−+
+=∴γγ
From eqs. 12 & 14 then
MdM
MM
MDFdXM
PdP
H )2
11(
)1(4.2
2
22
0
0
−+
−−=
−=
γγ
17
(12), 4.)1(
)2
11(2
22
2
2
HDFdX
M
MM
MdM
−
−+
=
γγ
(14) ])
2)1(1(
))1(1([4.)1(2
))1(1(2
2
2
22
dMMM
MDFdX
MMM
PdP
H−
+
−+−=
−
−+−=
γγγγ
Gas Dynamics
- For subsonic flow
1<M )1( 2M− ve+
veds +=- For supersonic flow
1>M )1( 2M− ve−
veds +=
vedM += vedPO −=∴
vedM −= vedPO −=∴
(17) )
211(
)1(4.2
2
22
0
0
MdM
MM
MDFdXM
PdP
H−
+
−−=
−=
γγ
Gas Dynamics
dT
M>1 M<1
dP
dV
dM
dρ
+ve -ve
+ve -ve
-ve +ve
-ve +ve
-ve +ve
dS
dPo
ho
+ve +ve
-ve -ve
Constant
HDFdX
M
MM
MdM 4.
)1(2
)211(2
22
−
−+
=
γγ
HD
FdXMMd
VdV 4.
)1(2 2
2
−=−=
γρρ
HDFdX
MMM
PdP 4.
)1(2))1(1(
2
22
−
−+−=
γγ
HDFdX
MM
TdT 4.
)1(2)1(2
4
−
−−=
γγ
MdM
MM
MRDFdXRMdsH )
211(
)1(4.2 2
22
−+
−==
γγ
MdM
MM
MDFdXM
PdP
H
O
)2
11(
)1(4.2
2
22
0−
+
−−=
−=
γγ
Gas Dynamics
Fanno table relations. M, 4FL*/D, V/V*, P/P*, T/T*, ρ/ρ*, Po/Po
*, (S-S*)/Cp
HDFdX
M
MM
MdM 4.
)1(2
)211(2
22
−
−+
=
γγ
HDFdX
MMd
VdV 4.
)1(2 2
2
−=−=
γρρ
HDFdX
MMM
PdP 4.
)1(2))1(1(
2
22
−
−+−=
γγ
HDFdX
MM
TdT 4.
)1(2)1(2
4
−
−−=
γγ
MdM
MM
MR
HDFdXRMds
)2211(
)21(4.2
2
−+
−==
γγ
MdM
MM
MDFdXM
PdP
H
O
)2
11(
)1(4.2
2
22
0−
+
−−=
−=
γγ
By separating the variables and by establishing the limits M=M1 at the duct entrance. M=M2 at distance L. If M2=1, then L2 =L*, V2 =V*, P2 =P*, T2 =T*, ρ2 =ρ*, Po2 =Po
*, and S2 =S*
Gas Dynamics
The Fanno Line
* It is defined by the Continuity eq.
Energy eq.
Eq. of state From Continuity eq. (4)
VdVd
−=ρρ
From Energy eq. (5) )(2 0 TTCV P −= )(2)(TTTTd
VdV
O
O
−
−=∴
The change of entropy for the perfect gas ρρdR
TdTCds v −=
)(2)(TTTTdR
TdTCds
O
Ov −
−−= 1−= γ
vCR
)(2)(
21
TTTTd
TdT
Cds
O
O
v −
−−−=∴γ
19
Gas Dynamics
Eq. 19 is integrated , the change of entropy may be described in terms Of temperature for a given value of constant ) and m./ A=constant .
Th,
S
OO Th ,
MaxS =
.consthO =
(19) )(2)(
21
TTTTd
TdT
Cds
O
O
v −
−−−=∴γ
=oTT (0
Th,
S
1=M
OO Th ,
MaxS =
.consthO =
2/2V1<M
1>M
Fanno Line
h
The Fanno Line equation is explained as a function of Mach Prove that, this
curve is right.
Gas Dynamics
From the 1st & 2nd law of thermodynamics PdPR
Tdhds −=
dhdP
PR
Tdsdh
−= 1
120
Continuity ,0=+VdVd
ρρ Energy 0=+VdVdh VdVdh −=
ρρdVdh 2= Eq. of state
TdT
PdPd
−=ρρ
)(2TdT
PdPVdh −=∴
dTdh∝∴
)(2TCdTC
PdPVdh
P
P−=PdPVdh
TCV
P
22
)1( =+
TCP
VP
dhdP
P
+= 2 21
From 20 & 21 TCR
VR
TdSdh
P
−−=
21
1
1−=γγRCP
The Fanno Line equation is a function of Mach
Gas Dynamics
])[1( 22
2
22
2
RVTRCCVRTV
RVTRCCVTCV
dSdh
PPPP
P
−−−=
−−=
γγ
RTMV γ22 =
12
2
−=MTM
dSdh γ
22
TCR
VR
TdSdh
P
−−=
21
11−
=γγRCP
Th,
S
1=M
OO Th ,
MaxS =
.consthO =
2/2V1<M
1>M
This equation describe fanno line as a function on Mach
At M<1 dh/dS -ve dS ≥ 0 ∴ dh -ve dT –ve ∴ dV +ve At M>1 dh/dS +ve dS ≥ 0 ∴ dh +ve dT +ve ∴ dV -ve
This means that, A Fanno line moving only in the direction from left to right M1 >1, or M1 <1 then Po -ve.
At M=1 dh/dS h which corresponding to the state of maximum entropy h
If m./A increase Smax decease.
Gas Dynamics
It was noted that L*,the maximum length of duct which doesn’t cause choking.
L1-2 (L*) M1 (L*) M2 = -
(4FL1-2)/DH ((4F L*)/ DH) M1 = - ((4F L*)/ DH) M2
L1-2
(L*) M1
(L*) M2
M2 M1 M=1
2 1 * Flow
Multiply the equation by (4F / DH)
Gas Dynamics
Fanno flow in a constant cross-sectional area duct proceeded by an isentropic nozzle
Fanno flow ( Ld) isentropic nozzle
1
2 3 e
X
P/Po
Back pressure Pb
Flow
The flow in nozzle& duct are affected by Ld & Pb
M2 <1
Gas Dynamics
M2<1 the gas will accelerate in the duct owing to friction and pressure decrease
Me <1 or 1 depend on Pb & Ld If Me =1 (m./A)max choked condition
To increase mass, by decreasing To and/or increasing Po at the nozzle inlet. In this case Me=1, but Pe would be higher.
ToPo
RγG*=(m. /A)max = [2/(γ+1)] (γ+1)/(2(γ-1))
Gas Dynamics
M2>1, the gas will decelerate in the duct due to friction, if no shock occur, the flow will approach Me=1 directly.
In this case (Me=1) , m. max is determined by the throat area of the nozzle and also by the area of the duct
Changes in duct length do not affect the mass flow rate
Gas Dynamics
Effect of increasing duct length isentropic nozzle
1
2 3 4
( L2d)
L1d Fanno flow
h ho ho=const.
2
3 4
P4*
P3* P2
*isen
Po
S
(m./A)2> (m./A)3 >(m./A)4
M=1
P/P o
1 (M duct entr <1)
Gas Dynamics
(M duct entr >1) At certain length, the M=1 at exit duct cross-sectional area Subsonic conditions can also be attained at exit area, if discontinuity occurs.
If [ Ld > Ld*
at Me =1 ] , a shock appears in the duct
If the duct length is increased further, the shock will position itself further upstream.
If the duct is very long, the shock will be at the throat of the nozzle. Beyond that length, there will be no shock at all. Then, the flow is subsonic at all point.
Gas Dynamics
Effect of reducing back pressure (Mach at duct inlet ≤1)
P/P o
X
Pback/Po1
1
1
m,
Pback<Pexit a
b c
a
Pb Pa
Pc
h
S
ho=constant a
b c
1
M=1
2
Gas Dynamics
Fanno flow critical length L1* Supersonic
nozzle
1
P/Po Pb
Flow
The flow in nozzle& duct are affected by Ld & Pb
M1 >1
L12 < L1*
P*/Po= Pb/Po
Pe=Pb Pb<Pe
Pb>Pe M=1
Effect of reducing back pressure (Mach at duct inlet >1)
X
Oblique shock wave
In the duct L12 <L*, there are 5 cases:-
Design case Pe=Pb Expansion wave Shock wave
stand somewhere inside the duct N.S.W stand
at duct exit 2