1. For this type of flow, the stagnation temperature is constant, then
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
For this type of flow, the stagnation temperature is constant, then
Taking logarithmic of this expression and then differentiating gives
Substitute for dT⁄T
The last term can be manipulated to be:
Substitute this expression into last equation and rearrange gives:

2.  For supersonic the value of fLmax lies between O and M=1 and
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Integration of this equation gives
For Fanno flow, the integration limits are
This equation relates friction factor f, , to M directly. For air γ=1.4, then;
 For supersonic the value of fLmax lies between O and M=1 and
O.8215 at M=∞.
 For subsonic the value of fLmax becomes very large as M becomes
very small.
5.3 Reference state and Fanno Flow Table
The equations developed in this chapter are the means of computing the
properties at one location in terms of those given at some other location.
The key to problem solution is predicting the Mach number at the new
location through the use of last equation. The solution of this equation for
the unknown M2 presents a messy task, as no explicit relation is possible
between M2 and M1.
In reference case we imagine that we continue by Fanno flow (i.e., more
duct is added) until the velocity reaches M=1. Figure (5.2) shows a physical

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A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
system together with its T –s diagram for a subsonic Fanno flow. We know
that if we continue along the Fanno line (remember that we always move to
the right), we will eventually reach the limiting point where sonic velocity
exists. The dashed lines show assumed elongation duct of sufficient length
to enable the flow to traverse the remaining portion of the upper branch
and reach the limit point. This is the (*) reference point for Fanno flow.
Figure 5.2 The * reference for Fanno flow
We now rewrite the working equations in terms of the Fanno flow ∗ reference
condition
Where T* is the static temperature at M=1, and

4. Substitute from continuity equation for constant area duct
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Figure 5.3 Fanno line
Substitute from continuity equation for constant area duct

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Figure 5.4

6. Figure 5.6 mass flow reduction.
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Suppose now that the duct is long enough for a flow initially subsonic to
reach Mach 1, and an additional length is added, as shown in Figure (5.5).
The flow Mach number for the given mass flow cannot go past 1 without
decreasing entropy. This is impossible from the second law. Hence the
additional length brings about a reduction in mass flow. The flow jumps to
another Fanno Line (see Figure 5.6). Essentially, the duct is choked due to
friction. Corresponding to a given inlet subsonic Mach number, there is a
certain maximum duct length Lmax beyond which a flow reduction occurs.
Figure 5.5
Figure 5.6 mass flow reduction.

7. Table 1 Absolute Roughness of common materials
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Now suppose the inlet flow is supersonic and the duct length is made
greater than Lmax to produce Mach =1. With the supersonic flow unable to
sense changes in duct length occurring ahead of it, the flow adjusts to the
additional length by means of a normal shock rather than a flow reduction.
The location of the shock in the duct is determined by the back pressure
imposed on the duct.
5.4 Friction factor
Dimensional analysis of the fluid flow in fluid mechanics shows that the
friction factor can be expressed as = f(Re, cfDe ) . Where cfDe is the
relative roughness. The relationship among, Re, and cfDe is determined
experimentally and plotted on a chart called a Moody chart or a Moody
diagram. Typical values of , the absolute roughness are shown in the
following table.
Table 1 Absolute Roughness of common materials
If the flow rate is known together with the duct size and material, the
Reynolds number and relative roughness can easily be calculated and the
value of the friction factor is taken from the diagram. The curve in the
laminar flow region can be represented by
f = 64fRe

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A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
For noncircular cross sections the equivalent diameter can be used. 4A
This equivalent diameter may be used in the determination of relative
roughness and Reynolds number, and hence the friction factor. However,
care must be taken to work with the actual average velocity in all
computations. Experience has shown that the use of an equivalent diameter
works quite well in the turbulent zone. In the laminar flow region this
concept is not sufficient and consideration must also be given to the aspect
ratio of the duct. In some problems the flow rate is not known and thus a
trial-and-error solution results. As long as the duct size is given, the
problem is not too difficult; an excellent approximation to the friction
factor can be made by taking the value corresponding to where the ε/D
curve begins to level off. This converges rapidly to the final answer, as most
engineering problems are well into the turbulent range.
Figure 5.7 Moody diagram for friction factor in circular ducts
De =

9. Table 5.1 Fluid Property Variation for Fanno Flow
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5.5 APPLICATIONS
The following steps are recommended to develop good problem-solving
technique:
1. Sketch the physical situation (including the hypothetical ∗
reference point).
2. Label sections where conditions are known or desired.
3. List all given information with units.
4. Compute the equivalent diameter, relative roughness, and Reynolds
number.
5. Find the friction factor from the Moody diagram.
6. Determine the unknown Mach number.
7. Calculate the additional properties desired.
Table 5.1 Fluid Property Variation for Fanno Flow
5.6 CORRELATION WITH SHOCKS
Now we can imagine a supersonic Fanno flow leading into a normal
shock. If this is followed by additional duct, subsonic Fanno flow would

10. of the same Fanno line. [See Figure 5.8b].
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occur. Such a situation is shown in Figure 5.8a. Note that the shock merely
causes the flow to jump from the supersonic branch to the subsonic branch
of the same Fanno line. [See Figure 5.8b].
Figure 5.8a Combination of Fanno flow and normal shock (physical system).
Figure 5.8b Combination of Fanno flow and normal shock.
Example A large chamber contains air at a temperature of 300 K and a
pressure of 8 bar abs (Figure E5.6). The air enters a converging–diverging
nozzle with an area ratio of 2.4. A constant-area duct is attached to the
nozzle and a normal shock stands at the exit plane. Receiver pressure is 3
bar abs. Assume the entire system to be adiabatic and neglect friction in the
nozzle. Compute the f Δx/D for the duct.

11. A3 p3 A2 p03 We proceed to compute p5 pS pS p01 p03 p3 3 1
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Figure E5.6
A p3
A p03
We proceed to compute p5
pS pS p01 p03 p
For
= 2.4, M3 = 2.4, and = p* = p
= ( ) (1) ( ) (0.3111) = p* p* p p