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