ME 131B Fluid Mechanics

Solutions to Week Six Problem Session:

Fanno Flow (2/17/98)

1. From which conservation principle(s) do we derive the Fanno curve?

� The Fanno curve is derived from the conservation of mass and the conservation ofenergy principles.

� All states on the same Fanno curve have the same

{ mass ux, _m=A

{ stagnation enthalpy, h0

� E�ects of changing the mass ux and stagnation enthalpy on the Fanno curve aredisplayed in the following �gures:

Changing _m=A Changing h0T

s

T0

T*

.increasing

m

A

T

s

increasing h0

2. Trace out a Fanno curve on a T � s plane and locate the corresponding stagnation andsonic states on the same diagram.

� Since the stagnation temperature is constant for a Fanno ow (adiabatic), all thestagnation states lies on the horizontal line (T = T0).

� The ratio of sonic to stagnation temperature is a constant for any local staticstate

T0T �

= 1 +k � 1

2

Hence, the sonic temperature is also a constant. All the sonic states lies on thehorizontal line (T = T �).

1

T

s

T01 (static)

T*

Locus of all stagnation states

1*

10

Locus of all sonic states

Subsonic regime

Supersonic regime

3. Show that the Mach number corresponds to the maximum entropy point on a Fannocurve is unity.

� The road map to solve this problem is as follows:

(a) Derive the equation of a Fanno curve on the T � s plane.

(b) Di�erentiate the equation with respect to temperature.

(c) Solve for the maximum entropy point:

ds

dT= 0

� We �rst start with the Gibbs equation:

T ds = du �P

�2d�

ds = Cv

dT

T� R

d�

�(ideal gas)

= Cv

dT

T+ R

dV

V(� V = constant)

� Integrate the above equation (assume constant Cv), we obtain

s = Cv logT + R logV + constant (1)

� From energy conservation, we have

V 2

2= h0 � h = Cp (T0 � T ) (perfect gas)

) V =q2Cp (T0 � T ) (2)

2

� Combine Equation (1) and Equation (2), we obtain the equation of a Fanno curveon the T � s plane:

s = Cv logT +R

2log[ 2Cp (T0 � T ) ] + constant

� At the maximum entropy point,

ds

dT= 0

Cv

T�

R

2

1

T0 � T= 0

Cv

T�

R

2

2Cp

V 2= 0 From Equation (2)

) V 2 =Cp

Cv

RT

= k R T

= c2

We conclude that the ow speed at the maximum entropy point equals the speedof sound. Hence, it corresponds to a Mach 1.0 point.

4. Complete the following table with increases, decreases, remains constant for a owthrough a constant-area, frictional duct:

M < 1 M > 1P decreases increases� decreases increasesT decreases increasesV increases decreasesc decreases increasesM increases decreasesP0 decreases decreases�0 decreases decreasesT0 remains constant remains constants increases increasesA� increases increasesP � decreases decreases�� decreases decreasesT � remains constant remains constant

P + � V 2 decreases decreases

3

5. A ow is supplied by a converging nozzle (unchoked).

(a) Will the addition of a diverging section increase or decrease the mass ow rate?

� The addition of a diverging section will increase the mass ow rate. Thereasons are outlined as follows:

{ Pressure rises in the diverging section for a subsonic ow which demandsthe pressure at the nozzle throat to be lower than the back pressure.

{ This leads to a higher Mach number at the nozzle throat, hence, a higher

mass ow rate in the system.

(b) What about adding a constant-area pipe? Will it increase or decrease the mass ow rate?

� The addition of a constant-area pipe will decrease the mass ow rate. Thereasons are outlined as follows:

{ Pressure drops along the pipe for a subsonic ow which demands thepressure at the nozzle exit to be higher than the back pressure.

{ This leads to a lower Mach number at the nozzle exit, hence, a lower

mass ow rate in the system.

Pressure distribution of the above three cases can be compared in the following�gure:

P / P0

x

Pb / P0

Diverging Section(Pressure rises)

Pipe Section (Pressure drops)

Converging Nozzle Only

Throat

(c) Sketch the variation of the mass ow rate, _m, with back-to-stagnation pressureratio, Pb=P0, for the above two cases on the same plot and highlight the di�erences.

� The C-D nozzle is choked at a much higher pressure ratio (determined by theexit-to-throat area ratio) than the frictional pipe (determined by fLmax=D).

� The mass ow rate out of a C-D nozzle is higher than that out of a frictionalpipe.

Mass ow rate of the above three cases can be compared in the following �gure:

4

.

Pb / P0

Converging Nozzle

C−D Nozzle

0.528

m

Constant−Area Pipe

Pchoke, CD > 0.528Pchoke, PIPE < 0.528

mpipe < mnozzle

. .mnozzle

.

1.0

6. Consider the following system:

Frictional PipeConverging Nozzle

P0 , T0Pamb < P0

Control Volume

In what direction will the cart move? Explain your answer.

� As a �rst step of our analysis, let us choose a control volume as indicated aboveby the dotted line.

� We need to turn to the momentum equation

~FS + ~FB| {z }external forces

=@

@t

ZCV

~V (� dV )| {z }storage (acceleration)

+ZCS

~V�� ~V � d ~A

�| {z }

net momentum out ow

and examine the di�erences in

{ momentum ux across the control surfaces

{ pressure forces on the control volume

before we conclude the motion of the cart.

� As we start lowering the back-to-stagnation pressure ratio, the exit plane pressureat the converging nozzle and the pipe are equal to the back pressure (subsonic ow). Hence, the net pressure force on the cart is zero.

5

� Let us examine the momentum ux

�e V2

e Ae =Pe

RTeV 2

e Ae

for both cases. The two exit states can be represented on the T � s diagram asfollows:

T

s

T

T

Nozzle Exit

Pipe Exit

Pb = Pe

From the above diagram, we observe that

{ Pe;nozzle = Pe;pipe

{ Ve;nozzle > Ve;pipe{ Te;nozzle < Te;pipe

Hence, we can conclude that the momentum ux out of the converging nozzle isgreater than that out of the frictional pipe.

� As a conclusion, The cart will move to the right.

7. Fanno Flow in Subsonic Regime

Frictional Pipe:

Converging NozzleP0,1 , T0,1

Pb

f = 0.02

L = 5 m

D = 0.2 m

(1) (e)

Consider the above setup.

(a) Where can Mach 1.0 be realized?

6

� Mach 1.0 can only be realized at the pipe exit.

(b) Unchoked Case

In this part, we operate the above setup at a pressure ratio of Pb=P0;1 = 0.60.What is the ow conditions at the exit (P; T;M)?

� We have learned from Question 5 that the pipe is choked at a lower pressureratio than the converging nozzle.

� In our case here, Pb=P0;1 = 0:60 > 0:528 (choking condition for convergingnozzle). Hence, we can conclude

{ the pipe is not choked

{ ow is subsonic at the exit

{ exit plane pressure, Pe, equals the back pressure, Pb

� We can write the pressure ratio as follows:

Pb

P0;1

=Pb

Pe

Pe

P �

e;f

P �

e;f

P �

1;f

P �

1;f

P1

P1

P0;1

� Let us examine the individual terms in the above equation:

{ Pb=P0;1 = 0:60 as given.

{ Pb=Pe = 1 because exit pressure equals back pressure for subsonic ow.

{ Pe=P�

e;f depends on the exit Mach number, Me (from Fanno ow table).

{ P �

e;f=P�

1;f = 1 because Station (1) and Station (e) are on the same Fannocurve and are driven to the same reference �

f state.

{ P �

1;f=P1 depends on the inlet Mach number, M1 (from Fanno ow table).

{ P1=P0;1 also depends on the inlet Mach number, M1 (from isentropic owtable).

We then simplify to the following relationship:

0:60 =Pe

P �

e;f

P �

1;f

P1

P1

P0;1

(3)

� The pressure ratios in Equation (3) depend on two unknown Mach numbers(M1;Me). We can relate them by using the pipe geometry:

f L

D=

f Lmax

D

!1

�

f Lmax

D

!e

where

{ fL=D = 0:5 as given by system speci�cation.

{ (fLmax=D)1 depends on the inlet Mach number, M1 (from Fanno owtable).

7

{ (fLmax=D)e depends on the exit Mach number, Me (from Fanno owtable).

We have

0:5 =

f Lmax

D

!1

�

f Lmax

D

!e

(4)

� The solution procedure is an iterative one. It is outlined as follows:

i. Guess M1.

ii. Look up the Fanno ow table to �nd out (fLmax=D)1.

iii. Calculate (fLmax=D)e from Equation (4).

iv. Look up the Fanno ow table to �nd the corresponding Me.

v. With these two Mach numbers (M1;Me),

{ look up the Fanno ow table to �nd Pe=P�

e;f

{ look up the Fanno ow table to �nd P �

1;f=P1

{ look up the isentropic ow table to �nd P1=P0;1

vi. If the product of the above pressure ratios equals 0.60 (Equation (3)), theguess is correct in Step (i). Otherwise, keep guessing di�erent values forM1 and repeat the above procedure until the product of pressure ratiosconverges to 0.60.

(c) Choked Case

How much do we need to lower the back-to-stagnation pressure ratio to reach thechoking condition? What does this ratio depend on? (Recall that it depends onthe exit-to-throat area ratio for a C-D nozzle.)

� The solution procedure for the choked case is simpler than the unchoked casebecause we know

{ Me = 1

{ Pb = P �

1;f

� This also implies that

f L

D=

f Lmax

D

!1

= 0:5

� From the Fanno ow table, we obtain the inlet Mach number to be

M1 = 0:5978

� With M1 = 0:5978,

{ the Fanno ow table gives

P1

P �

1;f

= 1:7705

8

{ the isentropic ow table gives

P1

P0;1

= 0:78538

� The choking pressure ratio can then be computed as

Pb

P0;1

=P �

1;f

P1

P1

P0;1

=�

1

1:77051

�(0:78538) = 0:4436

� This choking pressure ratio depends on the value of fL=D of the pipe.

(d) L > Lmax

After we lower the pressure ratio to the value we computed in Part (c), the ow ischoked. What do you expect to happen if an extra two meters of pipe section isadded to the existing system? Do you expect the ow is still choked at the exit?Explain your answer by showing the corresponding states on a T � s diagram.

� After the ow is choked, addition of extra pipe section will reduce the mass ow rate inside the pipe. This corresponds to switching to another Fannocurve with a smaller mass ux value ( _m=A) on the T � s diagram.

� Since the exit pressure cannot be greater than the back pressure in subsonic ow (no shock mechanism), the ow leaves the exit subsonically with Pe = Pb.In other words, the addition of extra pipe section unchokes the system.

� This adjustment of ow conditions within the system is possible because sub-

sonic ow can communicate. The addition of extra pipe section downstreamcan a�ect the pipe inlet condition upstream. In this case, it reduces the localMach number at the pipe inlet. As we shall see in the next question, super-sonic ow does not have this communication means. It can only adjust toextra pipe section by shock/expansion mechanism.

� The above conclusion can be summarized graphically in the following �gure:

Flow changes from Fanno Curve (1) to Fanno Curve (2)

T

s

T0

choked

T*

Pb

unchoked (higher entropy due to longer pipe)

Asmaller

m

(1)

(2)

.

9

8. Fanno Flow in Supersonic Regime

Frictional Pipe:

C−D Nozzle (A2 / A1 = 2.0)P0,1 , T0,1

Pb

f = 0.02

D = 0.2 m

(1) (2) (e)

L

(a) Slightly di�erent from the last problem, there are two possible locations at whichMach 1.0 is attainable in the above setup. Where are they?

� Mach 1.0 can be realized at

{ the throat of the C-D nozzle

{ exit of the frictional pipe.

(b) L = Lmax

In the supersonic operation mode, determine the pipe length Lmax which gives asonic ow right at the pipe exit.

� Under supersonic operation mode, the inlet Mach number, M2, is governedby the area ratio of the C-D nozzle.

� For an area ratio of A2=A1 = 2:0, we obtain from the isentropic ow table

M2 = 2:197

� From the Fanno ow table, this corresponds to f Lmax

D

!2

= 0:36017

which gives a critical pipe length of

Lmax = 3:602 m

(c) L < Lmax

i. For L = 2:0 m, determine the range of the back-to-stagnation pressure ratio,Pb=P0;1, over which

10

A. a normal shock appears in the diverging section of the C-D nozzle

� There are two limiting cases to consider here:

Case I Case II

(1) (2) (e) (1) (2) (e)

2a 2b

� Case I: upper limit

{ With the normal shock right at the nozzle throat, the ow goes sub-sonic in the diverging section of C-D nozzle.

{ Pipe inlet Mach number, M2, is governed by the area ratio of theC-D nozzle (subsonic solution from isentropic ow table):

M2 = 0:3060;P2

P0;2

= 0:93712

{ With M2 = 0:3060, the Fanno ow table gives

P2

P �

2;f

= 3:5479

f Lmax

D

!2

= 5:031

{ This further gives the value of fLmax=D at Station (e) f Lmax

D

!e

=

f Lmax

D

!2

�f L

D= 4:831

{ From the Fanno ow table, this corresponds to an exit Mach numberof 0.3105 and a pressure ratio

Pe

P �

e;f

= 3:4947

{ This gives the back-to-stagnation pressure ratio for Case I to be

Pb

P0;1

=Pb

Pe

Pe

P �

e;f

P �

e;f

P �

2;f

P �

2;f

P2

P2

P0;2

P0;2

P0;1

= (1) (3:4947) (1)�

1

3:5479

�(0:93712) (1)

= 0:9231

11

� Case II: lower limit

{ With the normal shock right at the pipe inlet, the ow goes super-

sonic in the diverging section of C-D nozzle until it hits a shock at the

nozzle exit, then goes subsonic right before entering the pipe section.

{ The Mach number upstream of the shock is governed by the arearatio of C-D nozzle (supersonic solution from isentropic ow table):

M2a = 2:197P2a

P0;2a

= 0:093936

{ With M2a = 2:197, the normal shock table gives

M2b = 0:54744P2b

P2a

= 5:4656

{ With M2b = 0:54744, the Fanno ow table gives

P2b

P �

2b;f

= 1:9438

f Lmax

D

!2b

= 0:74305

{ This gives the value of fLmax=D at Station (e) f Lmax

D

!e

=

f Lmax

D

!2b

�f L

D= 0:54305

{ From the Fanno ow table, this corresponds to an exit Mach numberof 0.5874 and a pressure ratio

Pe

P �

e;f

= 1:8037

{ This gives the back-to-stagnation pressure ratio for Case II to be

Pb

P0;1

=Pb

Pe

Pe

P �

e;f

P �

e;f

P �

2b;f

P �

2b;f

P2b

P2b

P2a

P2a

P0;2a

P0;2a

P0;1

= (1) (1:8037) (1)�

1

1:9438

�(5:4656) (0:093936) (1)

= 0:4764

� Hence, a normal shock appears in the diverging section of the C-Dnozzle when

0:4764 <Pb

P0;1

< 0:9231

12

B. a normal shock appears in the pipe

� There are two limiting cases to consider here:

Case II Case III

(1) (2) (e)

2a 2b

(1) (2) (e)

ea eb

� Case II: upper limit It has already been studied in the previous part.

� Case III: lower limit

{ With the normal shock right at the pipe exit, the ow remains su-

personic from the nozzle throat all the way up to the pipe exit just

before the shock, then exits subsonically after the shock.

{ The pipe inlet Mach number, M2, is again governed by the area ratioof C-D nozzle (supersonic solution from isentropic ow table). It hasbeen found from the previous part:

M2 = 2:197P2

P0;2

= 0:093936

{ With M2 = 2:197, the Fanno ow table gives

P2

P �

2;f

= 0:35567

f Lmax

D

!2

= 0:36012

{ This gives the value of fLmax=D at Station (ea) f Lmax

D

!ea

=

f Lmax

D

!2

�f L

D= 0:16012

{ From the Fanno ow table, this corresponds to a Mach number ofMea = 1:566 and a pressure ratio

Pea

P �

ea;f

= 0:57292

{ With Mea = 1:566, the normal shock table gives

Meb = 0:6790Peb

Pea

= 2:6944

13

{ This gives the back-to-stagnation pressure ratio for Case III to be

Pb

P0;1

=Pb

Peb

Peb

Pea

Pea

P �

ea;f

P �

ea;f

P �

2;f

P �

2;f

P2

P2

P0;2

P0;2

P0;1

= (1) (2:6944) (0:57292) (1)�

1

0:35567

�(0:093936) (1)

= 0:4077

� Hence, a normal shock appears in the pipe section when

0:4077 <Pb

P0;1

< 0:4764

For any back-to-stagnation pressure ratio which is lower than the criticalvalue corresponds to Case III, the ow within the C-D nozzle and pipesection will be una�ected. All the pressure adjustment will take placeoutside the pipe. We will expect

� oblique shocks if the back pressure is higher than the design condition

� oblique expansion waves if the back pressure is lower than the designcondition

The back-to-stagnation pressure ratio corresponds to the design condition(free of shock/expansion) is

Pb

P0;1

=Pb

Pe

Pe

P �

e;f

P �

e;f

P �

2;f

P �

2;f

P2

P2

P0;2

P0;2

P0;1

= (1) (0:57292) (1)�

1

0:35567

�(0:093936) (1)

= 0:1513

C. oblique shocks appear outside the pipe

� Oblique shocks appear outside the pipe when the back-to-stagnation

pressure ratio is between the design condition and the critical value

corresponds to Case III:

0:1513 <Pb

P0;1

< 0:4077

D. oblique expansion waves appear outside the pipe

� Oblique expansion waves appear outside the pipe when the back-to-

stagnation pressure ratio is below the design condition:

Pb

P0;1

< 0:1513

14

ii. For each of the above cases,

A. sketch the process path from the nozzle inlet to the pipe exit on a T � sdiagram.

Normal Shock in Nozzle Normal Shock in PipeT

s

T0

T* (1)

(a)

(b)

(2)

(e)

Pe = Pb

T

s

T0

T* (1)

(a)

(b)

(2)

(e)

Pe = Pb

Oblique Shock Outside Pipe Oblique Expansion Outside PipeT

s

T0

T* (1)

(2)

(e)

Pb

Pe < Pb

T

s

T0

T* (1)

(2)

(e)

Pb

Pe > Pb

Remarks:State (a) and State (b) are the upstream and downstream states of thenormal shock respectively.

15

B. sketch the pressure distribution along the streamwise location from thenozzle inlet to the pipe exit.

Normal Shock in Nozzle Normal Shock in Pipe

(1) (2) (e)

P

P0

x (1) (2) (e)

a b

ba

0.528

(1) (2) (e)

P

P0

x (1) (2) (e)

a b

ba

0.528

Oblique Shock Outside Pipe Oblique Expansion Outside Pipe

(1) (2) (e)

P

P0

x (1) (2) (e)

0.528

(1) (2) (e)

P

P0

x (1) (2) (e)

0.528

C. outline the solution procedure to locate the shock position in cases wherenormal shock appears.

� Similar to our previous procedure in locating a normal shock in the di-verging section of a C-D nozzle, we need to solve this problem iteratively.

� The solution procedure is outlined below for the case with a normal shockstanding in the diverging section of C-D nozzle (more di�cult case):

16

(1) (2) (e)

ba

A. Guess the Mach number upstream of the shock, Ma.

B. Obtain the pressure ratio, Pa=P0;a, from the isentropic ow table (func-tion of Ma).

C. Obtain the Mach number downstream of the shock, Mb, and the pres-sure ratios (Pb=Pa; P0;b=P0;a) from the normal shock table (function ofMa).

D. Relate Mb to the Mach number at the nozzle exit, M2, as follows:

A2

A1

=A2

A�

2

A�

2

A�

b

A�

b

A�

a

A�

a

A1

{ A2=A1 = 2:0 as given by system geometry.

{ A�

2=A�

b = 1 because the ow is isentropic from Station (b) to Station(2).

{ A�

b=A�

a = P0;a=P0;b because the shock process is adiabatic. This ratiodepends on Ma.

{ A�

a=A1 = 1 because the isentropic ow from Station (1) to Station(a) is choked at Station (1).

E. We then obtain the simpli�ed equation:

2:0 =A2

A�

2

P0;a

P0;b

F. With the pressure ratio, P0;a=P0;b, determined in Step (C), the aboveequation gives the value of A2=A

�

2which de�nes the Mach number at

the nozzle exit, M2 (from the isentropic ow table).

G. Obtain P2=P0;2 from the isentropic ow table (function of M2).

H. Obtain P2=P�

2;f ; (fLmax=D)2 from the Fanno ow table (function ofM2).

I. Relate the Mach number at pipe exit, Me, with the Mach number atpipe inlet, M2, as follows:

f Lmax

D

!e

=

f Lmax

D

!2

�

f L

D

!

17

J. With fL=D = 0:2 (given by system speci�cation), the above equationgives the value of (fLmax=D)e which de�nes the Mach number at thepipe exit, Me (subsonic solution from the Fanno ow table).

K. Obtain Pe=P�

e;f from the Fanno ow table (function of Me).

L. The overall back-to-stagnation pressure ratio can then be computed asfollows:

Pb

P0;1

=Pb

Pe

Pe

P �

e;f

P �

e;f

P �

2;f

P �

2;f

P2

P2

P0;2

P0;2

P0;b

P0;b

P0;a

P0;a

P0;1

{ Pb=Pe = 1 because exit pressure equals back pressure for subsonicexit.

{ P �

e;f=P�

2;f = 1 because Station (2) and Station (e) are on the sameFanno curve and are driven to the same reference �

f state.

{ P0;2=P0;b = 1 because the ow is isentropic from Station (b) toStation (2).

{ P0;a=P0;1 = 1 because the ow is isentropic from Station (1) toStation (a).

M. We then simplify to the following relationship:

Pb

P0;1

=Pe

P �

e;f| {z }Step K

P �

2;f

P2| {z }Step H

P2

P0;2| {z }Step G

P0;b

P0;a| {z }Step C

All the pressure ratios have been determined in the previous steps. Wejust need to multiply all these pressure ratios together and check if theirproduct equals the given back-to-stagnation pressure ratio. If it is, theguess is correct in Step (A). Otherwise, keep guessing di�erent valuesof Ma and repeat the above procedure until the product of pressureratios converges to the required value.

� The case with a normal shock standing in the pipe section can be analyzedin a similar manner and is easier!

(d) L > Lmax (Common case)

� Since the fLmax=D values for supersonic ow is much smaller than those ofsubsonic ow, for most applications, L > Lmax.

� Since L > Lmax and supersonic ow cannot communicate with downstream,a shock is unavoidable within the pipe section.

� The ow goes subsonic after the shock. It can communicate with the down-stream condition. It either exits the pipe subsonically, matching the back

pressure or sonically with an exit pressure higher than the back pressure.

� To decide between these two possible situations, we need to compare the backpressure, Pb, with the sonic pressure on the Fanno curve, P �

f .

18

{ If Pb > P �

f , the ow exits subsonically and Pe = Pb.

{ If Pb < P �

f , the ow exits sonically and Pe > Pb. Expansion waves areexpected to occur outside the pipe to adjust to the lower back pressure.

� In our present case, the sonic pressure on the Fanno curve, P �

f , can be foundas follows:

i. From Part (b), we know that M2 = 2:197 in supersonic operation mode.

ii. From the isentropic ow table, we obtain

P2

P0;2

= 0:093936

iii. From the Fanno ow table, we obtain

P2

P �

2;f

= 0:35567

iv. Combine these two pressure ratios, we obtain

P �

2;f

P0;1

=P �

2;f

P2

P2

P0;2

P0;2

P0;1

=�

1

0:35567

�(0:093936) (1)

= 0:2641

i. For L = 5:0 m, qualitatively describe the ow in the system for the followingpressure ratios:

� In both cases, there is a normal shock within the system because L >Lmax.

A. Pb=P0;1 = 0:50

� Since Pb > P �

f , the ow exits subsonically with exit pressure equalsthe back pressure.

� No pressure adjustment is necessary outside the pipe.

B. Pb=P0;1 = 0:10

� Since Pb < P �

f , the ow exits sonically with exit pressure higher thanthe back pressure.

� Pressure adjustment in the form of oblique expansion is expected to

occur outside the pipe.

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ii. For each of the above cases,

A. sketch the process path from the nozzle inlet to the pipe exit on a T � sdiagram.

Pb > P �

f Pb < P �

fT

s

T0

T* (1)

(a)

(b)

(2)

(e)

Pe = Pb

Pf*

T

s

T0

T* (1)

(a)

(b)

(2)

(e)

Pb

Pe = Pf*

B. sketch the pressure distribution along the streamwise location from thenozzle inlet to the pipe exit.

Pb > P �

f Pb < P �

f

(1) (2) (e)

P

P0

x (1) (2) (e)

a b

ba

0.528

(1) (2) (e)

P

P0

x (1) (2) (e)

a b0.528

ba

20