Bar-Meir G. Fundamentals of Compressible Fluid

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10.3. NON–DIMENSIONALIZATION OF THE EQUATIONS 181

The term

dρ

can be eliminated by utilizing equation (10.18) and substituting it into

equation (10.22) and rearrangement yields

= −

1 + (k − 1)M

(10.23)

The term dU

can be eliminated by using (10.23)

= −

1 + (k − 1)M

2(1 −M

)

4fdx

(10.24)

The second equation for Mach number, M variable is obtained by combining equation

(10.20) and (10.21) by eliminating dT /T . Then dρ/ρ and U are eliminated by utilizing

equation (10.18) and equation (10.22). The only variable that is left is P (or dP/P )

which can be eliminated by utilizing equation (10.24) and results in

4fdx

1 −M

(1 +

k−1

)

(10.25)

Rearranging equation (10.25) results in

1 +

k−1

1 −M

4fdx

(10.26)

After similar mathematical manipulation one can get the relationship for the

velocity to read

2 (1 − M

)

4fdx

(10.27)

and the relationship for the temperature is

= −

k(k − 1)M

2(1 −M

)

4fdx

(10.28)

density is obtained by utilizing equations (10.27) and (10.18) to obtain

dρ

= −

2 (1 − M

)

4fdx

(10.29)

The stagnation pressure is similarly obtained as

= −

4fdx

(10.30)

The second law reads

ds = C

− R ln

(10.31)

182 CHAPTER 10. FANNO FLOW

The stagnation temperature expresses as T

= T (1 + (1 −k)/2M

). Taking derivative

of this expression when M remains constant yields dT

= dT (1 + (1 − k)/2M

) and

thus when these equations are divided they yield

dT/T = dT

(10.32)

In similar fashion the relationship between the stagnation pressure and the pressure can

be substituted into the entropy equation and result in

ds = C

− R ln

(10.33)

The ﬁrst law requires that the stagnation temperature remains constant, (dT

= 0).

Therefore the entropy change is

= −

(k − 1)

(10.34)

Using the equation for stagnation pressure the entropy equation yields

(k − 1)M

4fdx

(10.35)

10.4 The Mechanics and Why the Flow is Choked?

The trends of the properties can be examined by looking in equations (10.24) through

(10.34). For example, from equation (10.24) it can be observed that the critical point

is when M = 1. When M < 1 the pressure decreases downstream as can be seen

from equation (10.24) because fdx and M are positive. For the same reasons, in

the supersonic branch, M > 1, the pressure increases downstream. This pressure

increase is what makes compressible ﬂow so diﬀerent from “conventional” ﬂow. Thus

the discussion will b e divided into two cases: One, ﬂow above speed of sound. Two,

ﬂow with speed below the speed of sound.

Why the ﬂow is choked?

Here, the explanation is based on the equations developed earlier and there is no known

explanation that is based on the physics. First, it has to be recognized that the critical

point is when M = 1. It will be shown that a change in location relative to this point

change the trend and it is singular point by itself. For example, dP (@M = 1) = ∞ and

mathematically it is a singular point (see equation (10.24)). Observing from equation

(10.24) that increase or decrease from subsonic just below one M = (1 − ²) to ab ove

just above one M = (1 + ²) requires a change in a sign pressure direction. However,

the pressure has to be a monotonic function which means that ﬂow cannot crosses over

the point of M = 1. This constrain means that because the ﬂow cannot “crossover”

M = 1 the gas has to reach to this speed, M = 1 at the last point. This situation is

called choked ﬂow.

10.5. THE WORKING EQUATIONS 183

The Trends

The trends or whether the variables are increasing or decreasing can be observed from

looking at the equation developed. For example, the pressure can be examined by look-

ing at equation (10.26). It demonstrates that the Mach number increases downstream

when the ﬂow is subsonic. On the other hand, when the ﬂow is supersonic, the pressure

decreases.

The summary of the properties changes on the sides of the branch

Subsonic Supersonic

Pressure, P decrease increase

Mach number, M increase decrease

Velocity, U increase decrease

Temperature, T decrease increase

Density, ρ decrease increase

10.5 The Working Equations

Integration of equation (10.25) yields

max

fdx =

1 −M

k + 1

k+1

1 +

k−1

(10.36)

A representative friction factor is deﬁned as

f =

max

fdx (10.37)

In the isothermal ﬂow model it was shown that friction factor is constant through the

process if the ﬂuid is ideal gas. Here, the Reynolds number deﬁned in equation (9.28) is

not constant because the temperature is not constant. The viscosity even for ideal gas

is complex function of the temperature (further reading in “Basic of Fluid Mechanics”

chapter one, Potto Project). However, the temperature variation is very limit. Simple

improvement can be done by assuming constant constant viscosity (constant friction

factor) and ﬁnd the temperature on the two sides of the tube to improve the friction

factor for the next iteration. The maximum error can be estimated by looking at the

maximum change of the temperature. The temperature can be reduced by less than

20% for most range of the spesiﬁc heats ratio. The viscosity change for this change is

for many gases about 10%. For these gases the maximum increase of average Reynolds

number is only 5%. What this change in Reynolds number does to friction factor? That

depend in the range of Reynolds number. For Reynolds number larger than 10,000 the

change in friction factor can be considered negligible. For the other extreme, laminar

ﬂow it can estimated that change of 5% in Reynolds number change about the same

184 CHAPTER 10. FANNO FLOW

amount in friction factor. With the exception the jump from a laminar ﬂow to a

turbulent ﬂow, the change is noticeable but very small. In the light of the about

discussion the friction factor is assumed to constant. By utilizing the mean average

theorem equation (10.36) yields

max

1 −M

k + 1

k+1

1 +

k−1

(10.38)

It is common to replace the

f with f which is adopted in this book.

Equations (10.24), (10.27), (10.28), (10.29), (10.29), and (10.30) can be

solved. For example, the pressure as written in equation (10.23) is represented by

4fL

, and Mach number. Now equation (10.24) can eliminate term

4fL

and describe

the pressure on the Mach number. Dividing equation (10.24) in equation (10.26) yields

= −

1 + (k − 1M

1 +

k−1

(10.39)

The symbol “*” denotes the state when the ﬂow is choked and Mach number is equal

to 1. Thus, M = 1 when P = P

∗

equation (10.39) can be integrated to yield:

∗

k+1

1 +

k−1

(10.40)

In the same fashion the variables ratio can be obtained

∗

k+1

1 +

k−1

(10.41)

∗

1 +

k−1

k+1

(10.42)

∗

−1

= M

k+1

1 +

k−1

(10.43)

10.5. THE WORKING EQUATIONS 185

The stagnation pressure decreases and can be expressed by

∗

(

1−k

)

k−1

z}|{

∗

|{z}

(

k+1

)

k−1

∗

(10.44)

Using the pressure ratio in equation (10.40) and substituting it into equation (10.44)

yields

∗

1 +

k−1

k+1

k−1

1 +

k−1

k+1

(10.45)

And further rearranging equation (10.45) provides

∗

1 +

k−1

k+1

2(k−1)

(10.46)

The integration of equation (10.34) yields

s −s

∗

= ln M

k + 1

1 +

k−1

k+1

(10.47)

The results of these equations are plotted in Figure (10.2) The Fanno ﬂow is in many

cases shockless and therefore a relationship between two points should be derived. In

most times, the “star” values are imaginary values that represent the value at choking.

The real ratio can be obtained by two star ratios as an example

∗

(10.48)

A special interest is the equation for the dimensionless friction as following

4fL

dx =

max

4fL

dx −

max

4fL

dx (10.49)

Hence,

4fL

max

4fL

max

−

4fL

(10.50)

186 CHAPTER 10. FANNO FLOW

0.1 1 10

Mach number

0.01

0.1

1e+01

1e+02



4fL



T/T

U/U*

Fanno Flow

P/P

, ρ/ρ

and T/T

as a function of M

Tue Sep 25 10:57:55 2007

Fig. -10.2. Various parameters in Fanno ﬂow as a function of Mach number

10.6 Examples of Fanno Flow

Example 10.1:

Fig. -10.3. Schematic of Example (10.1)

Air ﬂows from a reservoir and enters a uni-

form pipe with a diameter of 0.05 [m] and

length of 10 [m]. The air exits to the at-

mosphere. The following conditions prevail

at the exit: P

= 1[bar] temperature T

◦

C M

= 0.9

. Assume that the average

friction factor to be f = 0.004 and that the

ﬂow from the reservoir up to the pipe inlet

is essentially isentropic. Estimate the total

temperature and total pressure in the reservoir under the Fanno ﬂow model.

Solution

For isentropic, the ﬂow to the pipe inlet, the temperature and the total pressure at the

This property is given only for academic purposes. There is no Mach meter.

10.6. EXAMPLES OF FANNO FLOW 187

pipe inlet are the same as those in the reservoir. Thus, ﬁnding the total pressure and

temperature at the pipe inlet is the solution. With the Mach number and temperature

known at the exit, the total temperature at the entrance can be obtained by knowing

the

4fL

. For given Mach number (M = 0.9) the following is obtained.

4fL

∗

0.90000 0.01451 1.1291 1.0089 1.0934 0.9146 1.0327

So, the total temperature at the exit is

∗

300

1.0327

= 290.5[K]

To “move” to the other side of the tube the

4fL

is added as

4fL

4 ×0.004 × 10

0.05

+ 0.01451 ' 3.21

The rest of the parameters can be obtained with the new

4fL

either from Table (10.1)

by interpolations or by utilizing the attached program.

4fL

∗

0.35886 3.2100 3.0140 1.7405 2.5764 0.38814 1.1699

Note that the subsonic branch is chosen. The stagnation ratios has to be added

for M = 0.35886

A×P

∗

×P

∗

0.35886 0.97489 0.93840 1.7405 0.91484 1.5922 0.78305

The total pressure P

can be found from the combination of the ratios as follows:

z }| {

∗

z }| {

∗

=1 ×

1.12913

× 3.014 ×

0.915

= 2.91[Bar]

188 CHAPTER 10. FANNO FLOW

z }| {

∗

z }| {

∗

=300 ×

1.0327

× 1.17 ×

0.975

' 348K = 75

◦

End solution

Another academic question/example:

Example 10.2:

shock

d-c nozzle

atmosphere

conditions

Fig. -10.4. The schematic of Example

(10.2).

A system is composed of a convergent-

divergent nozzle followed by a tube with

length of 2.5 [cm] in diameter and 1.0 [m]

long. The system is supplied by a vessel.

The vessel conditions are at 29.65 [Bar], 400

K. With these conditions a pipe inlet Mach

number is 3.0. A normal shock wave occurs

in the tube and the ﬂow discharges to the

atmosphere, determine:

(a) the mass ﬂow rate through the system;

(b) the temperature at the pipe exit; and

Take k = 1.4, R = 287 [J/kgK] and f = 0.005.

Solution

(a) Assuming that the pressure vessel is very much larger than the pipe, therefore the

velocity in the vessel can be assumed to be small enough so it can be neglected.

Thus, the stagnation conditions can be approximated for the condition in the

tank. It is further assumed that the ﬂow through the nozzle can be approximated

as isentropic. Hence, T

= 400K and P

= 29.65[P ar]

10.6. EXAMPLES OF FANNO FLOW 189

The mass ﬂow rate through the system is constant and for simplicity point 1 is

chosen in which,

˙m = ρAMc

The density and speed of sound are unknowns and need to be computed. With

the isentropic relationship the Mach number at point one (1) is known, then the

following can be found either from Table (10.1) or the Potto–GDC

A×P

∗

×P

∗

3.0000 0.35714 0.07623 4.2346 0.02722 0.11528 0.65326

The temperature is

= 0.357 × 400 = 142.8K

Using the temperature, the speed of sound can be calculated as

√

kRT =

√

1.4 ×287 × 142.8 ' 239.54[m/sec]

The pressure at point 1 can be calculated as

= 0.027 × 30 ' 0.81[Bar]

The density as a function of other properties at point 1 is

8.1 ×10

287 ×142.8

' 1.97

The mass ﬂow rate can be evaluated from equation (10.2)

˙m = 1.97 ×

π × 0.025

× 3 × 239.54 = 0.69

sec

(b) First, check whether the ﬂow is shockless by comparing the ﬂow resistance and

the maximum p ossible resistance. From the Table (10.1) or by using the Potto–

GDC, to obtain the following

4fL

∗

3.0000 0.52216 0.21822 4.2346 0.50918 1.9640 0.42857

190 CHAPTER 10. FANNO FLOW

and the conditions of the tube are

4fL

4 ×0.005 × 1.0

0.025

= 0.8

Since 0.8 > 0.52216 the ﬂow is choked and with a shock wave.

The exit pressure determines the location of the shock, if a shock exists, by

comparing “possible” P

exit

to P

. Two possibilities are needed to be checked;

one, the shock at the entrance of the tube, and two, shock at the exit and

comparing the pressure ratios. First, the possibility that the shock wave occurs

immediately at the entrance for which the ratio for M

are (shock wave Table

(6.1))

3.0000 0.47519 2.6790 3.8571 10.3333 0.32834

After the shock wave the ﬂow is subsonic with “M

”= 0.47519. (Fanno ﬂow

Table (10.1))

4fL

∗

0.47519 1.2919 2.2549 1.3904 1.9640 0.50917 1.1481

The stagnation values for M = 0.47519 are

A×P

∗

×P

∗

0.47519 0.95679 0.89545 1.3904 0.85676 1.1912 0.65326

The ratio of exit pressure to the chamber total pressure is

z }| {

∗

¶µ

∗

¶µ

z }| {

= 1 ×

2.2549

× 0.8568 × 0.32834 × 1

= 0.12476