1. dse +; thus dp0 −, or p0 decreases.
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
dse +; thus dp0 −, or p0 decreases.
For cooling,
dse −; thus dp0 +, or p0 increases.
In practice, the latter condition is difficult to achieve because the friction
that is inevitably present introduces a greater drop in stagnation pressure
than the rise created by the cooling process, unless the cooling is done by
vaporization of an injected liquid.
6.4 WORKING EQUATIONS FOR PERFECT GASES
By this time you should have a good idea of the property changes that
are occurring in both subsonic and supersonic Rayleigh flow. Remember
that we can progress along a Rayleigh line in either direction, depending on
whether the heat is being added to or removed from the system. We now
proceed to develop relations between properties at arbitrary sections.
Recall that we want these working equations to be expressed in terms of
Mach numbers and the specific heat ratio. To obtain explicit relations, we
assume the fluid to be a perfect gas.
 Momentum
We start with the momentum equation developed since this will lead
directly to a pressure ratio:
p + GV = canst.
or this can be written as
p + pV2 = canst.
Substitute for density from the equation of state: p RT
and for the velocity
V2 = M2a2 = M2 yRT
p =

2. Show that equation becomes
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Show that equation becomes
If we apply this between two arbitrary points, we have
which can be solved for
 Continuity
From previous section we have
ρV = G = constant
Again, if we introduce the perfect gas equation of state together with the
definition of Mach number and sonic velocity,
Written between two points, this gives us
which can be solved for the temperature ratio:
The introduction of the pressure ratio results in the following working
equation for static temperatures:

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A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
The density relation can easily be obtained from equations and the perfect
gas equation of state:
 Stagnation Conditions
This is the first flow that we have examined in which the stagnation
enthalpy does not remain constant. Thus we must seek a stagnation
temperature ratio for use with perfect gases. We know that y
If we write this for each location and then divide one equation by the other,
we will have
1 + fy - 1l M2
T T fy - 1l M2
Since we already have solved for the static temperature ratio, this can
immediately be written as y
T yM M fy - 1l M2
Similarly, we can obtain an expression for the stagnation pressure ratio,
since we know that y
p0 = p ( M )
which means that
T0 = T (1 +
M ) 2
T02
T2 ( = 2 1
T02
1 + yM M2
1 + f l M2 ( = 2 2 2 1
y - 1
(y-1) 2

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A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi y
p02 = p2 (1 + r 2 l M2
Substitution for the pressure ratio yields
p02 = 1 + yM r 2 l M2 1
 Energy
h01 + q = h02
For perfect gases we express enthalpy as
h = cpT
which can also be applied to the stagnation conditions
h0 = cpT0
Thus the energy equation can be written as
cpT01 + q = cpT02 or
q = cp(T02 - T01)
Note carefully that
q = &'∆)* ≠ &'∆)
6.5 REFERENCE STATE ANDTHE RAYLEIGH TABLE
The equations developed in Section 6.4 provide the means of
predicting properties at one location if sufficient information is known
concerning a Rayleigh flow system. Although the relations are
y - 1 2
(y-1)
p01 p1
1 + ry - 1l M2
y - 1 2
(y-1) 2
p01
1 + yM2 2
1 + ry - 1l M2 2

5. We introduce still another (∗) reference state defined as before, in
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
straightforward, their use is frequently cumbersome and thus we turn to
techniques used previously that greatly simplify problem solution.
We introduce still another (∗) reference state defined as before, in
that the Mach number of unity must be reached by some particular process.
In this case we imagine that the Rayleigh flow is continued (i.e., more heat
is added) until the velocity reaches sonic. Figure 6.6 shows a T –s diagram
for subsonic Rayleigh flow with heat addition. A sketch of the physical
system is also shown. If we imagine that more heat is added, the entropy
continues to increase and we will eventually reach the limiting point where
sonic velocity exists. The dashed lines show a hypothetical duct in which
the additional heat transfer takes place. At the end we reach the ∗ reference
point for Rayleigh flow.

6. Figure 6.6 The ∗ reference for Rayleigh flow.
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Figure 6.6 The ∗ reference for Rayleigh flow.
We now rewrite the working equations in terms of the Rayleigh flow ∗
reference condition. Consider first:
Let point 2 be any arbitrary point in the flow system and let its Rayleigh ∗
condition be point 1. Then
and equation becomes

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A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
6.6 APPLICATIONS
The procedure for solving Rayleigh flow problems is quite similar to
the approach used for Fanno flow except that the tie between the two
locations in Rayleigh flow is determined by heat transfer considerations
rather than by duct friction. The recommended steps are, therefore, as
follows:
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. Determine the unknown Mach number.
5. Calculate the additional properties desired.

8. Table 6.1Fluid Property Variation for Rayleigh Flow
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Table 6.1Fluid Property Variation for Rayleigh Flow
6.7 CORRELATION WITH SHOCKS
We can now picture a supersonic Rayleigh flow followed by a normal
shock, with additional heat transfer taking place subsonically. Such a
situation is shown in Figure 6.7. Note that the shock merely jumps the flow
from the supersonic branch to the subsonic branch of the same Rayleigh
line. This also brings to light another reason why the supersonic stagnation
curve must lie above the subsonic stagnation curve. If this were not so, a
shock would exhibit a decrease in entropy, which is not correct.

9. Figure 6.7 Combination of Rayleigh flow and normal shock.
A course in Gas Dynamics…………………………………..….…Lecturer: Dr.Naseer Al-Janabi
Figure 6.7 Combination of Rayleigh flow and normal shock.
Example Air enters a constant-area duct with a Mach number of 1.6,
a temperature of 200 K, and a pressure of 0.56 bar (as shown in figure).
After some heat transfer a normal shock occurs, whereupon the area is
reduced as shown. At the exit the Mach number is found to be 1.0 and the
pressure is 1.20 bar. Compute the amount and direction of heat transfer.

10. q = cp (T02 - T01) = (1005)(226 - 302) = -76.38 k]/kg
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It is not known whether a heating or cooling process is involved. The flow
from 3 to 4 is isentropic; thus p 4
Note that point 3 is on the same Rayleigh line as point 1 and this permits us
to compute M2 through the use of the Rayleigh table
p p03 p1 p
From the Rayleigh table we find M3 = and from the shock table, M2 =
2.906.
Now we can compute the stagnation temperatures: T 1
T02 T* 01
and the heat transfer:
q = cp (T02 - T01) = (1005)( ) = k]/kg
The minus sign indicates a cooling process that is consistent with the Mach
number’s increase from 1.60 to
p p04 = =
p4 = (0.5283) (1.2) = bar p =
= (
) (0.2353)(1.1756) = p* p p p*
0.56
T01 =
T1 = (0.6614) (200) = 302 K T
T02 =
T01
= (0.6629) ( ) (302) = 226 K
0.8842 T* T