CHAPTER 3
THE RAMJET CYCLE, SCRAMJETS
3.1
RAMJET FLOW FIELD
Before we begin to analyze the ramjet cycle we will consider an example that can help us understand how the flow through a ramjet comes about. The key to understanding the flow field is the
intelligent use of the relationship for mass flow conservation. In this connection there are two
equations that we will rely upon. The first is the expression for 1-D mass flow in terms of the
stagnation pressure and temperature.
% Pt A &
"
ṁ = ! UA = ------------------------------------- ' ----------------( f ( M )
"+1
-------------------- # " RT t$
2
(
" + 1& " – 1 )
% -----------# 2 $
(3.1)
The second is the all-important area-Mach number function.
*
"+1
A
f ( M ) = ------ = % -------------&
# 2 $
A
"+1
-------------------- %
2(" – 1)
"+1
--------------------&
2
(
2& " – 1 )(
–1
' M  % 1 + "-----------M
'
#
$
2
#
(
$
.
(3.2)
This function is plotted below for three values of " .
1
0.8
f(M)
0.6
"=1.4
"=1.66
0.4
0.2
"=1.2
1
2
3
M
4
5
Figure 3.1 Area - Mach number relation.
bjc
3.1
3/21/11
Ramjet flow field
For adiabatic, isentropic flow of a calorically perfect gas along a channel this equation provides
a direct connection between the local channel cross sectional area and Mach number.
In addition to the mass flow relations there are two relationships from Rayleigh line theory that
are also very helpful in guiding our understanding of the effect of heat addition on the flow in
the ramjet. These are the equations that describe the effect of heat addition on the Mach number
and stagnation pressure of the flow.
%
&
2 2
'
(
(1 + " M )
------ = ' --------------------------------------------------------------------(
Tt
– 1 2& (
' 2 ( 1 + " )M 2 % 1 + "-----------M $
#
#
$
2
*
Tt
"
"+1
%
& ---------------(" – 1)
------------'
(
1
+
"
M
2
------ = % ----------------------& ' -------------------------------------(
# 1 + " $'
Pt
% " – 1-& M 2(
# 1 + # ----------$
2 $
*
Pt
2
(3.3)
These equations are plotted below.
Figure 3.2 Effects of heat exchange on Mach number and stagnation pressure.
There are several features shown in these plots that have important implications for the ramjet
flow. The first is that much more heat can be added to a subsonic flow than to a supersonic flow
before thermal choking occurs; that is, before the flow is brought to Mach one. The second is
that stagnation pressure losses due to heat addition in subsonic flow are relatively small and
cannot exceed about 20% of the stagnation pressure of the flow entering the region of heat addition. In contrast stagnation pressure losses due to heat addition can be quite large in a supersonic
flow.
With this background we will now construct a ramjet flow field beginning with supersonic flow
through a straight, infinitely thin tube. For definiteness let the free stream Mach number be three
and the ambient temperature T 0 = 250°K . Throughout this example we will assume that the
friction along the channel wall is negligible.
3/21/11
3.2
bjc
Ramjet flow field
e
1
M0 = 3
Me = 3
T0=250
Tt0=700
Figure 3.3 Step 1 - Initially uniform Mach three flow.
Add an inlet convergence and divergence.
1
e
1.5
M0 = 3
M=3
Me = 3
T0=250
Tt0=700
Figure 3.4 Step 2 - Inlet convergence and divergence with f ( M ) shown.
Let the throat Mach number be 2.0 ( M 1.5 = 2.0 ). In Figure 3.4 the Mach number decreases
to the inlet throat ( f ( M ) increases), then increases again to the inlet value of three
( f ( M ) = 0.236 ). The thrust of this system is clearly zero since the x-directed component of
the pressure force on the inlet is exactly balanced on the upstream and downstream sides of the
inlet.
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3.3
3/21/11
Ramjet flow field
Now add heat to the supersonic flow inside the engine. Neglect the mass flow of fuel added
compared to the air mass flow.
1
M0 = 3
3
1.5
M=2
e
4
Me < 3
M=3
T0=250
Tt0=700
3
M
1
x
Figure 3.5 Step 2 - Introduce a burner and add heat to the flow.
As the heat is added the mass flow is conserved. Thus, neglecting the fuel added,
% P t0 A 3 &
% P t4 A 4 &
"
"
ṁ = ------------------------------------- ' -------------------( f ( 3 ) = ------------------------------------- ' -------------------( f ( M 4 )
"+1
"+1
-------------------- # " RT t0$
-------------------- # " RT t4$
2
(
"
–
1
)
2
(
" + 1&
" + 1& " – 1 )
% -----------% -----------# 2 $
# 2 $
(3.4)
As the heat is added T t4 goes up and P t4 goes down while the following equality must be
maintained
P t0
P t4
------------ f ( 3 ) = ------------ f ( M 4 )
T t0
T t4
(3.5)
Conservation of mass (3.5) implies that f ( M 4 ) must increase and the Mach number downstream of the burner decreases. There is a limit to the amount of heat that can be added to this
flow and the limit occurs when f ( M 4 ) attains its maximum value of one. At this point the flow
looks like the following.
3/21/11
3.4
bjc
Ramjet flow field
1
M0 = 3
3
1.5
M=2
4
M=3
Tt4=1070
T0=250
e
Me = 1+)
Tt0=700
3
M
1
x
Figure 3.6 Step 3 - Introduce sufficient heat to bring the exit Mach number to a
value slightly greater than one.
The Rayleigh line relations tell us that the temperature rise across the burner that produces this
flow is
T t4
-------T t3
= 1.53
(3.6)
M4 = 1
The corresponding stagnation pressure ratio across the system is
%P
t4
' ------'P
t3
#
M4 =
&
(
(
1$
= 0.292
(3.7)
before unstart
Now suppose the temperature at station 4 is increased very slightly. We have a problem; T t4 is
up slightly, P t4 is down slightly but f ( M 4 ) cannot increase. To preserve the mass flow rate
imposed at the inlet the supersonic flow in the interior of the engine must undergo an unstart
and the flow must switch to the configuration shown in Figure 3.7. The mass flow equation (3.5)
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3.5
3/21/11
Ramjet flow field
can only be satisfied by a flow between the inlet throat and the burner that achieves the same
stagnation pressure loss (3.7) since f ( M 4 ) cannot exceed one and the stagnation temperature
ratio is essentially the same.
1
M0 = 3
1.5
4
M = 0.475
M=2
T0=250
3
shock
Tt4=1070
e
Me = 1
M=3
Tt0=700
3
M
1
x
Figure 3.7 Step 4 - Increase the heat added very slightly to unstart the flow.
As a result of the unstart, a shock wave now sits at the end of the diffuser section. Notice that
the engine internal pressure is still very large and the exit Mach number must remain one, the
stagnation temperature has not changed and so, as was just pointed out, the mass balance tells
us that the stagnation pressure of the exit flow must be the same as before the unstart. Thus
%P
t4
' ------'P
# t0 M 4 =
&
(
(
1$
= 0.292
(3.8)
after unstart
The stagnation pressure loss is divided between two mechanisms, the loss across the shock
wave and the loss due to heat addition across the burner. The stagnation pressure ratio across a
Mach three shock wave is
P t3
-------P t0
3/21/11
= 0.3285
(3.9)
M =3
3.6
bjc
Ramjet flow field
The burner inlet Mach number is M 3 = 0.475 and the stagnation loss for thermal choking
across the burner is
P t4
-------P t3
= 0.889 .
(3.10)
M = 0.475
The product of (3.9) and (3.10) is 0.292 .
Now let’s look at the thrust generated by the flow depicted in Figure 3.7. The thrust definition,
neglecting the fuel/air ratio, is
& Ae % Pe
&
2% Ue
T
------------- = " M 0 ' ------- – 1( + ------ ' ------ – 1( .
P0 A0
#U0
$ A0 # P0
$
(3.11)
The pressure ratio across the engine is
"
%
– 1& 2& ----------% "----------- M 0( " – 1
+
1
'
P te
Pe
# 2 $
2.8 3.5
------ = -------- ' ------------------------------------(
= 0.292 % -------&
= 5.66
# 1.2$
P t0 '
P0
" – 1& 2(
%
' 1 + ------------ M e (
# 2 $
#
$
(3.12)
and the temperature ratio is
%
– 1& 2&
% "----------- M
+
1
'
T te
Te
# 2 $ 0(
2.8
'
------ = -------- ------------------------------------( = 1.53 % -------& = 3.5667 .
(
#
T t0 '
1.2$
T0
"–1
2
' 1 + % ------------& M e (
# 2 $
#
$
(3.13)
This produces the velocity ratio
Me T e
Ue
------- = -------- ------ = 0.6295 .
M0 T 0
U0
(3.14)
T
------------- = 1.4 ( 9 ) ( 0.6295 – 1 ) + 1 ( 5.66 – 1 ) = – 4.66 + 4.66 = 0
P0 A0
(3.15)
Now substitute into (3.11).
The thrust is zero. We would expect this from the symmetry of the upstream and downstream
distribution of pressure on the inlet. Now let’s see if we can produce some thrust. First redesign
the inlet so that the throat area is reduced until the throat Mach number is just slightly larger
than one. This will only effect the flow in the inlet and all flow variables in the rest of the engine
will remain the same.
bjc
3.7
3/21/11
Ramjet flow field
With the flow in the engine subsonic and the shock positioned at the end of the diffuser we have
a great deal of margin for further heat addition. If we increase the heat addition across the burner
the mass balance (3.5) is still preserved and the exit Mach number remains one. Let the burner
outlet temperature be increased to T t4 = 2100°K . The flow now looks something like this.
1
M0 = 3
1.5
3
shock
M = 1+)
M > 0.475
M<3
T0=250
e
4
Tt4=2100
Me = 1
Tt0=700
3
M
1
x
Figure 3.8 Step 5 - Increase the heat addition to produce some thrust.
The stagnation temperature at the exit is up, the stagnation pressure is up and the shock has
moved to the left to a lower upstream Mach number (higher f ( M ) ) while the mass flow (3.5)
is preserved. Note that we now have some thrust arising from the x-component of the high pressure behind the shock which acts to the left on a portion of the inlet surface. This pressure
exceeds the inlet pressure on the corresponding upstream portion of the inlet surface. The stagnation pressure ratio across the engine is determined from the mass balance (3.5).
P te
T te
-------- = f ( 3 ) -------- = 0.236 3 = 0.409
P t0
T t0
(3.16)
Let’s check the thrust. The pressure ratio across the engine is
2.8 3.5
P e  P 0 = 0.409 % -------&
= 7.94 .
# 1.2$
(3.17)
The temperature ratio is
3/21/11
3.8
bjc
The role of the nozzle
%
– 1& 2&
% "----------T te ' 1 + # 2 $ M 0(
Te
2.8
'
------ = -------- ------------------------------------( = 3 % -------& = 7
#
'
(
1.2$
T t0
T0
"–1
2
' 1 + % ------------& M e (
# 2 $
#
$
(3.18)
and the velocity ratio is now
1
U e  U 0 = ( M e  M 0 ) T e  T 0 = --- 7 = 0.882 .
3
(3.19)
T
------------- = 1.4 ( 9 ) ( 0.882 – 1 ) + 1 ( 7.94 – 1 ) = – 1.49 + 6.94 = 5.45 .
P0 A0
(3.20)
The thrust is
This is a pretty substantial amount of thrust. Note that the pressure term in the thrust definition
is the important thrust component in this design.
3.2 THE ROLE OF THE NOZZLE
Let’s see if we can improve the design. Add a convergent nozzle to the engine as shown below.
1
3
1.5
M0 = 3
M = 1+)
T0=250
M = 0.138
4
e
Tt4=2100
Me = 1
Tt0=700
3
M
1
x
Figure 3.9 Step 6 - Add a convergent nozzle.
The mass balance is
bjc
3.9
3/21/11
The role of the nozzle
P te A e
P t0 A 1.5
------------------- = ---------------700
2100
(3.21)
How much can we decrease A e ? Begin with Figure 3.8. As the exit area is decreased the exit
Mach number remains one due to the high internal pressure in the engine. The shock moves
upstream toward the inlet throat, the exit stagnation pressure increases and the product P te A e
remains constant. The minimum exit area that can be reached without unstarting the inlet flow
is when the inlet shock is very close to the throat and the shock becomes vanishingly weak. At
this condition the only mechanism for stagnation pressure loss is the heat addition across the
burner. The Mach number entering the burner is M 3 = 0.138 as shown in Figure 3.9. The stagnation pressure loss across the burner is proportional to the square of the entering Mach number
dP t
2 dT t
--------- = – " M --------Tt
Pt
(3.22)
To a reasonable approximation the stagnation loss across the burner can be neglected and we
can take P te * P t0 . In this approximation, the area ratio that leads to the flow depicted in Figure
3.9 is
Ae
---------A 1.5
=
ideal
2100
------------ = 1.732
700
(3.23)
This relatively large area ratio is expected considering the greatly increased temperature and
lower density of the exhaust gases compared to the gas that passes through the upstream throat.
What about the thrust? Now the static pressure ratio across the engine is
2.8 3.5
P e  P 0 = % -------&
= 19.41
# 1.2$
(3.24)
The temperatures and Mach numbers at the nozzle exit are the same so the velocity ratio does
not change between Figure 3.8 and Figure 3.9. The dimensionless thrust is
T
------------- = 1.4 ( 9 ) ( 0.882 – 1 ) + 1.732 ( 0.236 ) ( 19.41 – 1 ) =
P0 A0
(3.25)
– 1.49 + 7.53 = 6.034
That’s pretty good; just by adding a convergent nozzle and reducing the shock strength we have
increased the thrust by about 20%. Where does the thrust come from in this ramjet design? The
figure below schematically shows the pressure distribution through the engine. The pressure
forces on the inlet and nozzle surfaces marked “a” roughly balance although the forward pressure is slightly larger compared to the rearward pressure on the nozzle due to the heat addition.
3/21/11
3.10
bjc
The ideal ramjet cycle
But the pressure on the inlet surfaces marked “b” are not balanced by any force on the nozzle.
These pressures substantially exceed the pressure on the upstream face of the inlet and so net
thrust is produced.
a
a
a
a
b
b
P
x
Figure 3.10 Imbalance of pressure forces leading to net thrust.
3.3 THE IDEAL RAMJET CYCLE
But we can do better still! The gas that exits the engine is at a very high pressure compared to
the ambient and it should be possible to gain thrust from this by adding a divergent section to
the nozzle as shown below.
1
3
1.5
M0 = 3
M = 1+)
T0=250
M = 0.138
4
Tt4=2100
7
e
M=1
Pe=P0
Tt0=700
3
M
1
x
Figure 3.11 Ideal ramjet with a fully expanded nozzle.
bjc
3.11
3/21/11
The ideal ramjet cycle
The area ratio of the nozzle is chosen so that the flow is fully expanded, P e = P 0 . The stagnation pressure is constant through the engine and so we can conclude from
"
%
– 1& 2& ----------% "----------- M 0( " – 1
+
1
'
P te
Pe
# 2 $
------ = -------- ' -------------------------------------(
P t0 '
P0
" – 1 2(
' 1 + % ------------& M e (
# 2 $
#
$
"
– 1& 2& "----------% "----------–
1
- M 0(
# 2 $
(3.26)
%
'1 +
1 = 1 ' -------------------------------------(
'
" – 1 2(
' 1 + % ------------& M e (
# 2 $
#
$
that M e = M 0 . The temperature ratio is
%
– 1& 2&
% "----------T te ' 1 + # 2 $ M 0(
T te
Te
------ = -------- ' -------------------------------------( = -------- = 3
T t0 '
T0
T t0
" – 1 2(
' 1 + % ------------& M e (
#
$
2
#
$
(3.27)
Finally, the thrust equation is
&
2% Me T e
T
------------- = " M 0 ' -------- ------ – 1( = 1.4 ( 9 ) ( 3 – 1 ) = 9.22
P0 A0
# M0 T 0
$
(3.28)
Adding a divergent section to the nozzle at this relatively high Mach number increases the thrust
by 50%.
Now work out the other engine parameters. The fuel/air ratio is determined from
ṁ f h f = ( ṁ a + ṁ f )h t4 – ṁ a h t3
(3.29)
7
Assume the fuel added is JP-4 with h f = 4.28 × 10 J  kg . Equation (3.29) becomes
T t4
2100
-------- – 1
------------ – 1
T t3
700
f = ------------------------------- = ----------------------------- = 0.0346
2100
hf
T t4
60.8 – --------------------------- – -------700
C p T t3 T t3
3/21/11
3.12
(3.30)
bjc
The ideal ramjet cycle
The relatively small value of fuel/air ratio is the a posteriori justification of our earlier neglect
of the fuel mass flow compared to the air mass flow. If we include the fuel/air ratio in the thrust
calculation (but still ignore the effect of mass addition on the stagnation pressure change across
the burner) the result is
Me T e
&
2%
T
------------- = " M 0 ' ( 1 + f ) -------- ------ – 1( = 1.4 ( 9 ) ( ( 1.0346 ) 3 – 1 ) = 9.98 .
M0 T 0
P0 A0
#
$
(3.31)
The error in the thrust is about 6% when the fuel contribution is neglected. The dimensionless
specific impulse is
I sp g
1
1
T
9.98
---------- = % ---& % -----------& ------------- = ---------------------------------------- = 68.7
#
$
#
$
f " M 0 P0 A0
0.0346 × 1.4 × 3
a0
(3.32)
and the overall efficiency is ( + f = h f  ( C p T 0 ) = 170.3 )
9.22
1
T
0.4
"–1
, ov = % ------------& % ---------& % -------------& = % -------& % ------------------------------------& = 0.447
# 1.4$ # 0.0346 × 170.3$
# " $ # f + $ #P A $
f
0 0
.
(3.33)
The propulsive efficiency is
2U 0
2
, pr = --------------------- = ----------------- = 0.732 .
Ue + U0
1+ 3
(3.34)
The thermal efficiency of the engine shown in Figure 3.11 can be expressed as follows
2
2
Ue
U0
( ṁ a + ṁ f ) --------- – ṁ a ---------( ṁ a + ṁ f ) ( h te – h e ) – ṁ a ( h t0 – h 0 )
2
2
, th = -------------------------------------------------------------- = ------------------------------------------------------------------------------------------( ṁ a + ṁ f )h te – ṁ a h t0
ṁ f h f
( ṁ a + ṁ f )h e – ṁ a h 0
Q rejected during the cycle
, th = 1 – ------------------------------------------------------ = 1 – -------------------------------------------------------Q input during the cycle
( ṁ a + ṁ f )h te – ṁ a h t0
.
(3.35)
Te
2
3
-----–1/
(
1
+
f
)
/
T0 /
T0
/
, th = 1 – -------- 0 ------------------------------------- 1
T t0 /
T te
/
------- – 1/
(
1
+
f
)
/
T t0
.
The heat rejection is accomplished by mixing of the hot exhaust stream with surrounding air at
constant pressure. Noting (3.27) for the ideal ramjet the last term in brackets is one and the
thermal efficiency becomes
bjc
3.13
3/21/11
Optimization of the ideal ramjet cycle
, th
ideal ramjet
T0
= 1 – -------- .
T t0
(3.36)
For the ramjet conditions of this example the thermal efficiency is 2  3 . The Brayton cycle efficiency is
T0
, B = 1 – ------ .
T3
(3.37)
In the ideal cycle approximation the Mach number at station 3 is very small thus T 3 * T t0 and
the thermal and Brayton efficiencies are identical. Note that, characteristically for a Brayton
process, the thermal efficiency is determined entirely by the inlet compression process.
The ramjet design shown in Figure 3.11 represents the best we can do at this Mach number. In
fact the final design is what we would call the ideal ramjet. The ideal cycle will be the basis for
comparison with other engine cycles but it is not a practically useful design. The problem is
that the inlet is extremely sensitive to small disturbances in the engine. A slight increase in
burner exit temperature or decrease in nozzle exit area or a slight decrease in the flight Mach
number will cause the inlet to unstart. This would produce a strong normal shock in front of the
engine and a large decrease in air mass flow through the engine and a consequent decrease in
thrust.
A practical ramjet design for supersonic flight requires the presence of a finite amplitude inlet
shock for stable operation.
3.4 OPTIMIZATION OF THE IDEAL RAMJET CYCLE
For a fully expanded nozzle the thrust equation reduces to
Me T e
&
2%
T
------------- = " M 0 ' ( 1 + f ) -------- ------ – 1( .
M0 T 0
P0 A0
#
$
(3.38)
For the ideal cycle where P te = P to , M e = M 0 and T te  T t0 = T e  T 0 the thrust equation
using 1 + f = ( + f – + r )  ( + f – + 4 ) is written
% + f – +r +4
&
2"
T
------------- = ------------ ( + r – 1 ) ' ------------------ ----- – 1( .
"–1
P0 A0
# + f – +4 +r
$
(3.39)
This form of the thrust equation is useful because it expresses the thrust in terms of cycle parameters that we can rationalize. The parameter + r is fixed by the flight Mach number. At a given
altitude + 4 is determined by maximum temperature constraints on the hot section materials of
the engine as well as fuel chemistry and gas dissociation. If the flight Mach number goes to zero
3/21/11
3.14
bjc
Optimization of the ideal ramjet cycle
the thrust also goes to zero. As the flight Mach number increases for fixed + 4 the fuel flow must
decrease until + 4 = + r when fuel shut-off occurs and the thrust is again zero. A typical thrust
plot is shown below.
+ 4 = 8.4
+ f = 170
T
------------P0 A0
" = 1.4
fuel shut-off
+r
Figure 3.12 Ramjet thrust
The optimization question is; at what Mach number should the ramjet operate for maximum
thrust at a fixed + 4 ? Differentiate (3.39) with respect to + r and set the result to zero.
5 % T &
------------- =
5 + r # P 0 A 0$
+4 & &
%
%
%%
2 + 4& &
' ' + 4 + r ' 1 – 3 + r + 2 ----- + r( ( + + f ' + 4 + + 4 + r – 2 + r -----( (
+r $ $
+r $ (
#
#
2" ' #
------------ ' ----------------------------------------------------------------------------------------------------------------------------------------( = 0
" – 1'
(
2 +4
'
(
( + f – + 4 ) + r ----+r
#
$
.
(3.40)
The value of + r for maximum thrust is determined from
+4 +r %
+4 & %
2 + 4&
----------- ' 1 – 3 + r + 2 ----- + r( + ' + 4 + + 4 + r – 2 + r -----( .
+f #
+r $ #
+r $
bjc
3.15
(3.41)
3/21/11
Optimization of the ideal ramjet cycle
The quantity + f is quite large and so the second term in parentheses in (3.41) clearly dominates
the first term. For f « 1 the maximum thrust Mach number of a ramjet is found from
32
2 ( +r
)
12
max thrust
= --------------------------------------.
+4
+r
+1
(3.42)
max thrust
For the case shown above with + 4 = 8.4 the optimum value of + r is 3.5 corresponding to a
Mach number of 3.53. The ramjet is clearly best suited for high Mach number flight and the
optimum Mach number increases as the maximum engine temperature increases.
The specific impulse of the ideal ramjet is
12
2
% ----------- ( + r – 1 )&
I sp g
#" – 1
$
% + f – +r +4
&
---------- = ----------------------------------------------- ' ------------------ ----- – 1(
+4 – +r
a0
# + f – +4 +r
$
-----------------+ f – +4
.
(3.43)
The specific impulse also has an optimum but it is much more gentle than the thrust optimum
as shown below.
fuel shut-off
I sp g
---------a0
+ 4 = 8.4
+ f = 170
" = 1.4
+r
Figure 3.13 Ramjet specific impulse
3/21/11
3.16
bjc
The non-ideal ramjet
Optimizing the cycle with respect to thrust essentially gives close to optimal specific impulse.
Notice that the specific impulse of the ideal cycle has a finite limit as the fuel flow reaches shutoff.
3.5 THE NON-IDEAL RAMJET
The major non-ideal effects come from the stagnation pressure losses due to the inlet shock and
the burner heat addition. We have already studied those effects fairly thoroughly. In addition
there are stagnation pressure losses due to burner drag and skin friction losses in the inlet and
nozzle where the Mach numbers tend to be quite high. A reasonable rule of thumb is that the
stagnation pressure losses due to burner drag are comparable to the losses due to heat addition.
3.6 RAMJET CONTROL
Let’s examine what happens when we apply some control to the ramjet. The two main control
mechanisms at our disposal are the fuel flow and the nozzle exit area. The engine we will use
for illustration is a stable ramjet with an inlet shock and simple convergent nozzle shown below.
The inlet throat is designed to have a Mach number well above one so that it is not so sensitive
to unstart if the free stream conditions, burner temperature or nozzle area change. Changes are
assumed to take place slowly so that unsteady changes in the mass, momentum and energy contained in the ramjet are negligible.
ṁ f
1
1.5
3
shock
Ae
e
M0 = 3
Tt4=2100
M>1
T0=250
4
Me = 1
Tt0=700
3
M
1
x
Figure 3.14 Ramjet control model.
The mass balance is
bjc
3.17
3/21/11
Ramjet control
% P te A e &
"
ṁ e = ------------------------------------- ' -------------------( f ( M e ) = ( 1 + f )ṁ a
"+1
-------------------- # " RT te$
2
(
" + 1& " – 1 )
% -----------# 2 $
.
(3.44)
The pressure in the engine is likely to be very high at this free stream Mach number and so the
nozzle is surely choked and we can write
% P te A e &
1
"
ṁ a = ------------------ ------------------------------------- ' -------------------(
"+1
(1 + f )
-------------------- # " RT te$
2
(
" + 1& " – 1 )
% -----------# 2 $
.
(3.45)
The thrust equation is
Ue
&
& Ae % Pe
2%
T
------------- = " M 0 ' ( 1 + f ) ------- – 1( + ------ ' ------ – 1(
U0
P0 A0
$
#
$ A0 # P0
.
(3.46)
Our main concern is to figure out what happens to the velocity ratio and pressure ratio as we
control the fuel flow and nozzle exit area.
Nozzle exit area control
First, suppose A e is increased with T te constant. In order for (3.45) to be satisfied P te must
drop keeping P te A e constant. The shock moves downstream to a higher shock Mach number.
The velocity ratio remains the same and since the Mach numbers do not change the product
P e A e remains constant. Note that the thrust decreases. This can be seen by writing the second
term in (3.45) as
Ae Pe Ae
------ ------ – -----A0 P0 A0
(3.47)
The left term in (3.47) is constant but the right term increases leading to a decrease in thrust. If
A e is decreased, the reverse happens, the inlet operates more efficiently and the thrust goes up.
But remember, the amount by which the area can be decreased is limited by the Mach number
of the inlet throat.
3/21/11
3.18
bjc
Example - Ramjet with unstarted inlet
Fuel flow control
Now, suppose T te is decreased with A e constant. In order for (3.45) to be satisfied, P te must
drop keeping P te  T te constant. Once again the shock moves downstream to a higher shock
Mach number. The velocity ratio goes down since the exit stagnation temperature is down and
the Mach numbers do not change.The pressure ratio also decreases since the exit stagnation
pressure is down. The thrust clearly decreases in this case. If T te is increased, the reverse happens, the inlet operates more efficiently and the thrust goes up. The amount by which the
temperature can be increased is again limited by the Mach number of the inlet throat.
3.7 EXAMPLE - RAMJET WITH UNSTARTED INLET
For simplicity, assume constant heat capacity with " = 1.4 , C p = 1005 M2/(sec2-°K).
The gas constant is R = 287 M2/(sec2-°K). The ambient temperature and pressure are
T 0 = 216K
4
2
P 0 = 2 × 10 N  M .
and
The
fuel
heating
value
is
7
h f = 4.28 × 10 J/kg . The sketch below shows a ramjet operating at a free stream Mach
number of 3.0. A normal shock stands in front of the inlet. Heat is added between 3 and 4
P0
T0
1
A1.5
3
P0
4
e
M0 = 3
shock
and the stagnation temperature at station 4 is T t4 = 2000K . Relevant areas are
A 3  A 1.5 = 8 , A 1 = A 3 = A 4 and A 4  A e = 3 . Determine the dimensionless thrust
T  ( P 0 A 1 ) . Do not assume f<<1. Neglect stagnation pressure losses due to wall friction
and burner drag. Assume that the static pressure outside the nozzle has recovered to the
ambient value. Suppose A 1.5 can be increased until A 1 = A 1.5 = A 3 . By what proportion would the air mass flow change?
Solution - The first point to recognize is that the stagnation pressure at station 4 exceeds
the ambient by more than a factor of two - note the pressure outside the nozzle is assumed
to have recovered to the ambient value. Thus the exit Mach number is one and the Mach
number at station 4 is M 4 = 0.1975 . The stagnation temperature at station 3 is
T t3 = 604.8 ° K . The fuel-air ratio is determined from the enthalpy balance
bjc
3.19
3/21/11
Example - Ramjet with unstarted inlet
ṁ f h f = ( ṁ a + ṁ f )h te – ṁ a h t0 .
(3.48)
For constant heat capacity
T t4  T t0 – 1
2000  604.8 – 1
f = ------------------------------------------------------ = -------------------------------------------------------------------------------------------------------- = 0.0344 .
7
h f  C pT t0 – T t4  T t0
4.28 × 10  ( 1005 × 604.8 ) – 2000  604.8
(3.49)
Now we need to determine the flow batween stations 1 and 3. To get started we will neglect
the fuel addition for the moment. Knowing the Mach number at 4 and the stagnation temperatures at 3 and 4 we can use Rayleigh line results to estimate the Mach number at station
3. The stagnation temperature ratio across the burner is
*
T t4
Tt
T t4
2000
-------= ------------- = 3.3069
-------- = -------*
T t3
604.8
T t3
Tt
M =?
M = 0.2
.
(3.50)
*
Tt
T t4
2000
= ------------- = 3.3069
-------- = 0.2066 -------T
604.8
T t3
t3
M =?
The Rayleigh line tables give
*
Tt
-------T t3
M =?
3.3069
= ---------------- = 16.006 6 M 3 = 0.103 .
0.2066
(3.51)
This is a reasonable approximation to the Mach number at station 3. The stagnation pressure ratio across the burner is
*
P t4
Pt
P t4
1.235
-------= ------------- = 0.981 .
-------- = -------*
P t3
1.258
P t3
Pt
M = 0.103
M = 0.2
(3.52)
The subsonic critical Mach number for an area ratio of 8 is 0.0725. The fact that the Mach
number at station 3 is higher than this value implies that there is a shock in the diverging
part of the inlet and the inlet throat Mach number is equal to one. The stagnation presure
ratio between the inlet throat and the exit can be determined from a mass balance between
stations 1.5 and e.
Pt Ae
P t A 1.5 ( 1 + f )
e
1.5
----------------------------------------- = -------------Tt
Tt
1.5
e
Pt
( 1.0344 )3 2000
e
6 ----------- = ------------------------ ------------- = 0.7054 .
Pt
8
604.8
(3.53)
1.5
The results (3.52) and (3.53) determine the stagnation pressure ratio across the inlet shock
and this determines the Mach number of the inlet shock
3/21/11
3.20
bjc
Example - Ramjet with unstarted inlet
0.7054
7 shock = ---------------- = 0.719
0.981
6 M shock = 2.004 .
(3.54)
Thus far the ramjet flow looks as follows
M behind shock = 0.475
P0
T0
1
M inlet shock = 2.004
A1.5
3
P0
4
e
M0 = 3
M<1
T t 0 = 604.8
M>1
T te = 2000
M<1
M<1
M 3 = 0.103
M 4 = 0.1975
Me = 1
shock
M 1 = 0.0725
Figure 3.15
M 1.5 = 1
State I
The stagnation pressure ratio across the external shock is
P t1  P t0
M =3
= 0.3283
(3.55)
and so the overall stagnation pressure ratio is
P t1 P t e
P te
-------- = -------- ---------- = 0.3283 × 0.7054 = 0.2316 .
P t0 P t
P t0
1.5
(3.56)
The static pressure ratio is
"
%
" – 1& 2& "----------% ----------- M 0( – 1
+
1
'
P te
Pe
# 2 $
2.8 3.5
------ = -------- ' -------------------------------------(
= 0.2316 % -------&
= 4.494 .
# 1.2$
P t0 '
P0
" – 1& 2(
%
' 1 + ------------ M e (
# 2 $
#
$
(3.57)
The temperature ratio is
bjc
3.21
3/21/11
Example - Ramjet with unstarted inlet
%
– 1& 2&
% "----------- M
+
1
'
T te
Te
# 2 $ 0(
2000 2.8
'
------ = -------- -------------------------------------( = ------------- % -------& = 7.72 .
# 1.2$
(
T t0 '
604.8
T0
"–1 2
' 1 + % ------------& M e (
# 2 $
#
$
(3.58)
The velocity ratio is
1
U e  U 0 = ( M e  M 0 ) T e  T 0 = --- 7.72 = 0.926
3
(3.59)
Across the inlet the mass balance is
Pt
1.5
A 1.5 = P t A 0 f ( M 0 )
0
(3.60)
and so
A 0  A 1.5 = P t  ( P t f ( M 0 ) ) = 0.3283  ( 1  4.235 ) = 1.39
1.5
0
(3.61)
Finally the thrust is
Ue
&
& Ae % Pe
2 % A 0 & % A 1.5& %
T
------------- = " M 0 ' ----------( ' ----------( ' ( 1 + f ) ------- – 1( + ------ ' ------ – 1(
U0
P0 A1
# A 1.5$ # A 1 $ #
$
$ A1 # P0
(3.62)
and
T
------------P0 A1
= 1.4 ( 9 ) ( 1.39 ) ( 1  8 ) ( ( 1.0344 )0.926 – 1 ) + ( 1  3 ) ( 4.494 – 1 ) =
State I
(3.63)
– 0.0923 + 1.165 = 1.0724
State II - Now increase the inlet throat area to the point where the inlet unchokes.
As the inlet throat area is increased the Mach number at station 3 will remain the same
since it is determined by the choking at the nozzle exit and the fixed enthalpy rise across
the burner. The mass balance between the inlet throat and the nozzle exit is again
Pt Ae
P t A 1.5 ( 1 + f )
1.5
e
----------------------------------------- = ------------Tt
Tt
1.5
e
(3.64)
The stagnation pressure at station 1.5 is fixed by the loss across the external shock. The
fuel-air ratio is fixed as are the temperatures in (3.64). As A 1.5 is increased, the equality
(3.64) is maintained and the inlet shock moves to the left increasing P te . At the point where
the the inlet throat unchokes the shock is infinitely weak and the only stagnation pressure
loss between station 1.5 and the nozzle exit is across the burner.
3/21/11
3.22
bjc
Example - Ramjet with unstarted inlet
Tt
A 1.5 State II
1 P te State II
0.982
604.8
1.5
- = --------------------- ------------- = 0.522
-------------------------- = ------------------- ------------------------ --------( 1.0344 ) 2000
( 1 + f ) Pt
Tt
Ae
1.5
(3.65)
e
which corresponds to
Ae
A1
% A 1& %
&
3
-------------------------- = ' ------( ' --------------------------( = ------------- = 5.747
0.522
A 1.5 State II
# A e$ # A 1.5 State II$
(3.66)
The mass flow through the engine has increased by the ratio
A1 &
ṁ a State II
A 1.5 State II
A 0 State II
% A 1.5 State II& %
8
---------------------- = ------------------------- = ' -------------------------( ' ------------------------( = ------------- = 1.392 = ---------------------A1
ṁ a State I
A 1.5 State I
5.747
A 0 State I
#
$ # A 1.5 State I$
.
(3.67)
At this condition the stagnation pressure ratio across the system is
P te State II
% P t 1.5& % P te State II&
------------------------ = ' ----------( ' ------------------------( = 0.3283 × 0.982 = 0.3224
P t0
# P t0 $ # P t 1.5 $
(3.68)
The static pressure ratio is
"
%
– 1& 2& ----------% "----------- M 0( " – 1
+
1
'
P te State II
P e State II
# 2 $
2.8 3.5
---------------------- = ------------------------ ' -------------------------------------(
= 0.3224 % -------&
= 6.256
# 1.2$
P t0 '
P0
" – 1& 2(
%
' 1 + ------------ M e (
# 2 $
#
$
(3.69)
The Mach number at station 1 increases as A 1.5 increases and at the condition where the
inlet is just about to unchoke reaches the same Mach number as station 3. At this condition
the ramjet flow field looks like
bjc
3.23
3/21/11
Example - Ramjet with unstarted inlet
M behind shock = 0.475
P0
T0
1
A1.5
3
P0
4
e
M0 = 3
T t 0 = 604.8
M<1
M<1
T te = 2000
M<1
Me = 1
shock
M 1 = 0.103
Figure 3.16
M 1.5 = 1
M 3 = 0.103
M 4 = 0.1975
State II
The inlet shock is gone, the inlet Mach number has increased to M 1 = M 3 and the external
shock has moved somewhat closer to the inlet. Note that the capture area to throat area ratio
is still
A 0 State II  A 1.5 State II = P t  ( P t f ( M 0 ) ) = 0.3283  ( 1  4.235 ) = 1.39
1.5
0
(3.70)
although both A 1.5 and A 0 have increased. Also
A 0 State II  A 1.5 State II
1.39
A 0 State II  A 1 = ------------------------------------------------------ = ------------- = 0.242
5.747
A 1  A 1.5 State II
(3.71)
The thrust formula is
T
------------P0 A1
StateII
Ue
& A e % P e State II
&
2 % A 0 State II & % A 1.5 State II& %
= " M 0 ' -------------------------( ' -------------------------( ' ( 1 + f ) ------- – 1( + ------ ' --------------------- – 1(
A1
U0
# A 1.5 State II$ #
$#
$ A1 # P0
$
(3.72)
The velocity ratio across the engine is unchanged by the increase in inlet throat area. The
thrust of state II is
T
------------P0 A1
State II
1
= 1.4 ( 9 ) ( 1.39 ) % -------------& ( ( 1.0344 )0.926 – 1 ) + ( 1  3 ) ( 6.256 – 1 ) =
# 5.747$
(3.73)
– 0.1284 + 1.752 = 1.624
The reduced loss of stagnation pressure leads to almost a 60% increase in thrust at this
condition.
State III - Now remove the inlet throat altogether
3/21/11
3.24
bjc
Example - Ramjet with unstarted inlet
Now, suppose A 1.5 is increased until A 1 = A 1.5 = A 3 . The ramjet flow field looks like
the following.
M behind shock = 0.475
P0
T0
1
3
P0
4
e
M0 = 3
T t 0 = 604.8
T te = 2000
M<1
M<1
Me = 1
shock
M 1 = 0.103
Figure 3.17
M 3 = 0.103
M 4 = 0.1975
State III
With the inlet throat absent, the Mach number is constant between 1 and 3. There is no
change in mass flow, fuel-air ratio, stagnation pressure, or the position of the upstream
shock. The capture area remains
A 0 State II  A 1.5 State II
1.39
A 0 State III  A 1 = A 0 State II  A 1 = ------------------------------------------------------ = ------------- = 0.242
5.747
A 1  A 1.5 State II
(3.74)
Therefore the thrust is the same as the thrust for State II, equation (3.73). If we want to
position the upstream shock very near the entrance to the engine we have to increase the
nozzle exit area and reduce the heat addition.
State IV - Open the nozzle exit fully
First increase the exit area to the point where A 1 = A 3 = A 4 = A e . If we maintain
T t4 = 2000K the Mach number at station 4 becomes one and the Mach number between
1 and 3 is, from the Rayleigh solution, M 1 = M 3 = 0.276 . The ramjet flow at this condition is sketched below
bjc
3.25
3/21/11
Example - Ramjet with unstarted inlet
M behind shock = 0.475
P0
T0
1
3
M0 = 3
T t 0 = 604.8
4
P0
e
T te = 2000
Me = 1
M = 1
M<1
shock
M 1 = 0.276
Figure 3.18
M 3 = 0.276
M 4 = 1.0
State IV
The velocity ratio is still
1
U e  U 0 = ( M e  M 0 ) T e  T 0 = --- 7.72 = 0.926
3
(3.75)
The stagnation pressure ratio across the burner is, from the Rayleigh solution,
P te
-------P t1
= 0.8278
(3.76)
State IV
Across the whole system
P te State IV
% P t 1.5& % P te State IV&
------------------------- = ' ----------( ' -------------------------( = 0.3283 × 0.8278 = 0.2718
P t0
# P t0 $ # P t 1.5 $
(3.77)
and the static pressure ratio is
"
%
– 1& 2& "----------% "----------–1
M
+
1
P te State IV '
P e State IV
# 2 $ 0(
2.8 3.5
----------------------- = ------------------------- ' -------------------------------------(
= 0.2718 % -------&
= 5.274
# 1.2$
'
P t0
P0
" – 1& 2(
%
' 1 + ------------ M e (
# 2 $
#
$
(3.78)
The area ratio is
( Pt f ( M 1 ) )
0.4558
1
- = 0.3283 × ---------------- = 0.634
A 0 State IV  A 1 = ----------------------------0.2362
( Pt f ( M 0 ) )
0
(3.79)
The thrust formula for state IV is
3/21/11
3.26
bjc
Example - Ramjet with unstarted inlet
T
------------P0 A1
Ue
&
& A e % P e State IV
2 % A 0 State IV& %
= " M 0 ' ------------------------( ' ( 1 + f ) ------- – 1( + ------ ' ----------------------- – 1(
A1 $ #
U0
P0
#
$
$ A1 #
StateIV
(3.80)
which evaluates to
T
------------P0 A1
= 1.4 ( 9 ) ( 0.634 ) ( ( 1.0344 )0.926 – 1 ) + ( 1 ) ( 5.274 – 1 )
(3.81)
StateIV
– 0.3367 + 4.274 = 3.937
Note the considerable increase in mass flow for state IV compared to state III. From (3.74)
A1 &
ṁ a State IV
% A 0 State IV& %
0.634
------------------------ = ' ------------------------( ' ------------------------( = ------------- = 2.62
A
A
0.242
ṁ a State III
#
$ # 0 State III$
1
(3.82)
which accounts for much of the increased thrust in spite of the increase in stagnation pressure loss across the burner.
State V - Reduce the burner outlet temperature until the shock is very close to station 1
Now reduce T te until the Mach number at 3 matches the Mach number behind the shock.
From the Rayleigh solution, this occurs when T te = 924.8 . At this condition the ramjet
flow field looks like
M behind shock = 0.475
P0
T0
1
M0 = 3
3
T t 0 = 604.8
M = 0.475
4
e
T te = 924.8
P0
Me = 1
M = 1
shock
M 3 = 0.475
Figure 3.19
M 4 = 1.0
State V
The static temperature ratio is
bjc
3.27
3/21/11
Example - Ramjet with unstarted inlet
%
– 1& 2&
% "----------- M
+
1
'
T te
Te
# 2 $ 0(
924.8 2.8
'
------ = -------- -------------------------------------( = ------------- % -------& = 3.57
(
T t0 '
604.8 # 1.2$
T0
"–1 2
' 1 + % ------------& M e (
# 2 $
#
$
(3.83)
The velocity ratio at this temperature is
1
U e  U 0 = ( M e  M 0 ) T e  T 0 = --- 3.57 = 0.630
3
(3.84)
The stagnation pressure ratio across the burner at this condition is
P te
-------- = 0.889
P t3
(3.85)
P te
-------- = 0.889 × 0.3283 = 0.292
P t0
(3.86)
and across the system
The static pressure ratio is
"
%
" – 1& 2& ----------% ----------- M 0( " – 1
+
1
'
P te
Pe
# 2 $
2.8 3.5
------ = -------- ' -------------------------------------(
= 0.292 % -------&
= 5.67
# 1.2$
P t0 '
P0
" – 1& 2(
%
' 1 + ------------ M e (
# 2 $
#
$
(3.87)
Neglecting the fuel flow, the thrust is (Note that at this condition A 0 = A 1 = A e ).
& % Pe
&
2% Ue
T
------------- = " M 0 ' ------- – 1( + ' ------ – 1( =
P0 A1
#U0
$ # P0
$
(3.88)
1.4 ( 9 ) ( 0.629 – 1 ) + ( 5.67 – 1 ) = – 4.67 + 4.67 = 0
State VI - Reduce the burner temperature slightly to establish supersonic flow up to the
burner.
If the temperature at station 4 is decreased by an infinitesimal amount then supersonic flow
will be established through the engine. Finally the flow is
3/21/11
3.28
bjc
Very High Speed Flight - Scramjets
P0
T0
1
M0 = 3
3
4
T t 0 = 604.8
e
T te = 924.8
M = 3
P0
Me = 1
M = 1
shock
Figure 3.20
State VI
The mass flow, velocity ratio, pressure ratio and thrust all remain the same as in state V. If
we were to reduce the fuel flow to the burner to zero we would be back to the state of
undisturbed Mach three flow through a straight tube.
3.8 VERY HIGH SPEED FLIGHT - SCRAMJETS
As the Mach number reaches values above 5 or so the ramjet cycle begins to become
unusable and a new design has to be considered where the heat addition across the burner
is carried out at supersonic Mach numbers. There are several reasons why this is so related
to the very high stagnation temperatures of high Mach number flow. To get started let’s
recall the thrust equation for the ramjet with a fully expanded nozzle.
Me T e
&
2%
T
------------- = " M 0 ' ( 1 + f ) -------- ------ – 1(
M0 T 0
P0 A0
#
$
(3.89)
Define the thrust coefficient as
T
Thrust
C thrust = ------------------------ = -----------------------------1
" 2
2
--- ! 0 U 0 A 0
--- M 0 ( P 0 A 0 )
2
2
(3.90)
If we assume ideal behavior ( M e = M 0 ) the thrust coefficient becomes
T te
%
&
C thrust = 2 ' ( 1 + f ) -------- – 1( .
T t0
#
$
(3.91)
When we carried out the energy balance across the burner,
ṁ f h f = ( ṁ a + ṁ f )h t4 – ṁ a h t3
bjc
3.29
(3.92)
3/21/11
Very High Speed Flight - Scramjets
we assumed that the fuel enthalpy was simply added to the flow without regard to the chemistry of the process. In fact the chemistry highly limits the range of fuel-air ratios that are
possible. The stoichiometric reaction of JP-4 with air where the fuel and oxygen are completely consumed is
C H 1.94 + 1.485O 2 + 5.536N 2 8 CO 2 + 0.97H 2 O + 5.536N 2 .
(3.93)
corresponding to a fuel-air ratio
( 12.01 + 1.94 × 1.008 )
f = ------------------------------------------------------------------------------- = 0.0689 .
( 1.485 × 32.00 + 5.536 × 28.02 )
(3.94)
This is roughly the value of the fuel-air ratio that produces the maximum outlet temperature
from the burner. If we choose the much more energetic hydrogen fuel the reaction is
H 2 + 0.5O 2 + 1.864N 2 8 H 2 O + 1.864N 2
(3.95)
corresponding to a fuel-air ratio
2 × 1.008
f = ----------------------------------------------------------- = 0.0295 .
( 16.00 + 1.864 × 28.02 )
(3.96)
The fuel enthalpies are generally taken to be
hf
7
JP – 4
hf
= 4.28 × 10 J  kg
7
H
(3.97)
= 12.1 × 10 J  kg
2
In the earlier discussion we took the perspective that the ramjet cycle was limited by a red line
temperature in the hot part of the engine. Let’s relax this assumption and allow the maximum
temperature to be free while keeping the fuel-air ratio constant. Furthermore let’s continue to
retain the assumption of constant heat capacities even at high Mach numbers. We will correct
this eventually in Chapter 9, but for now we just want to see what happens to the ideal thrust
coefficient (3.91) as we increase the free stream Mach number at constant fuel-air ratio. Using
(3.92), constant heat capacities and assuming adiabatic flow in the inlet and nozzle we can
express the stagnation temperature ratio across the engine as
%
&
f+f '
T te
(
1
1
-------- = % --------------& ' -------------------------------------( + ------------------- .
#
$
1+ f '
T t0
% " – 1-& M 2( ( 1 + f )
# 1 + # ----------2 $ 0$
(3.98)
where we recall that + f = h f  ( C p T 0 ) . Now the thrust coefficient becomes
3/21/11
3.30
bjc
Very High Speed Flight - Scramjets
%%
&
&1  2
'' f (1 + f )+ f
(
(
C thrust = 2 ' ' ------------------------------------- + ( 1 + f )(
– 1( .
– 1& 2
' ' 1 + % "----------(
(
- M
##
$
$
# 2 $ 0
(3.99)
The temperature ratio and thrust coefficient are plotted below.
14
T 0 = 216K
H2
12
" = 1.4
10
T te  T t0
JP – 4
8
JP – 4 real chemistry
6
4
2
1
1
5
2.5
Figure 3.21
5
7.5
10
12.5
15
17.5
20
M0
Temperature ratio of an ideal ramjet at constant fuel-air ratio.
T 0 = 216K
H2
" = 1.4
4
C thrust
JP – 4
3
2
1
5
10
15
20
M0
Figure 3.22 Thrust coefficient of an ideal ramjet at constant fuel-air ratio.
bjc
3.31
3/21/11
Very High Speed Flight - Scramjets
The drag coefficient is
D
Drag
C drag = ------------------------ = -----------------------------1
" 2
2
--- ! 0 U 0 A 0
--- M 0 ( P 0 A 0 )
2
2
(3.100)
At high Mach numbers the drag coefficient of a body tends toward a constant value. For a sphere
the drag coefficient tends toward a constant somewhat less than one, about 0.75. More streamlined bodies have lower drag and coefficients as low as 0.2 may be feasible with adequate
development. This observation together with the above figures indicates that as the Mach number increases it becomes harder and harder to produce thrust that exceeds drag. The thrust
coefficient drops below one at a Mach number of about 7 to 8 for both fuels. As the Mach number increases a conventional ramjet (even an ideal one) simply cannot produce enough thrust to
overcome drag. The limiting thrust coefficient at infinite Mach number is
lim C thrust = f .
M0 8 9
(3.101)
This last result suggests that a scramjet can benefit from the choice of a fuel-rich mixture ratio
as long as it does not exceed the flammability limit of the fuel. This would also provide additional fuel for cooling the vehicle. An advantage of a hydrogen system is that it can operate
quite fuel rich. In addition hydrogen has a higher heat capacity than any other fuel enabling it
to be used to provide cooling for the vehicle that, in contrast to a re-entry body, has to operate
in a very high temperature environment for long periods of time. The down side of hydrogen
is that liquid hydrogen has to be stored at very low temperatures and the liquid density is only
about 1/10 of that of JP-4. It is clear that small effects can be important. For example when we
developed the thrust formula we neglected the momentum of the injected fuel. In a realistic
scramjet analysis that would have to be taken into acount.
3.8.1
REAL CHEMISTRY EFFECTS
The real chemistry of combustion shows that the problem is even worse than just discussed.
The plotpoints in Figure 3.21 are derived from an equilibrium chemistry computation of the
combustion of JP-4 with air. Notice that the temperatures reached by the combustion gases is
substantially less than the ideal and for high Mach numbers the temperature rise from the reaction is actually less than one due to the cooling effect of the added fuel.
The solution to this problem that has been pursued since the 1960s is to try to add heat with the
burner operating with a supersonic Mach number on the order of two or so thereby cutting the
static temperature of the air into the combustor by almost a factor of two. This is the concept
of a supersonically burning ramjet or scramjet. A generic sketch of such a system is shown in
Figure 3.23. Two Homework problems are used to illustrate basic concepts.
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Very High Speed Flight - Scramjets
Figure 3.23
Conceptual figure of a ramjet.
3.8.2
SCRAMJET OPERATING ENVELOPE
Figure 3.24 shows a widely circulated plot showing the altitude and Mach number regime
where a scramjet might be expected to operate. Contours of constant free stream stagnation
temperature and flow dynamic pressure are indicated on the plot.
Mach number
Figure 3.24
bjc
Conceptual operating envelope of a scramjet.
3.33
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Very High Speed Flight - Scramjets
This figure somewhat accurately illustrates the challenges of scramjet flight. These can be listed
as follows.
1) At high Mach numbers the vehicle is enveloped in an extremely high temperature gas for a
long period of time perhaps more than an hour.
2)At high altittude, combustion is hard to sustain even at high Mach numbers because of the
low atmospheric density and long chemical times. This defines the combustor blowout limit.
3) At lower altitude and high Mach number the free stream dynamic pressure increases to the
point where the structural loads on the vehicle become untenable.
Let’s take a look at the vehicle structural limit (item 3) in a little more detail. This limit is presented as a line of constant dynamic pressure coinciding with increasing Mach number and
altitude. But in supersonic flow the free stream dynamic pressure is really not a sufficient measure of the actual loads that are likely to act on the vehicle. Again let’s make a constant heat
capacity assumption and compare the free stream stagnation pressure to the free stream
dynamic pressure as follows. Form the pressure coefficient
"
-----------%
&
"–1
"
–
1
2
'
%
&
%
&
P 9 1 + ------------ M 9
– 1(
# 2 $
'#
$
(
Pt 9 – P9
#
$
----------------------- = ------------------------------------------------------------------------------ =
"
q9
2
--- P 9 M 9
2
"
-----------%
&
"–1
"
–
1
2
' % 1 + % ------------& M &
– 1(
# 2 $ 9$
'#
(
#
$
---------------------------------------------------------------------- . (3.102)
" 2
--- M 9
2
This function is plotted below.
Pt9 – P9
----------------------q9
M
Figure 3.25 Comparison of stagnation pressure and dynamic pressure in supersonic flow.
If one repeats this calculation with real gas effects the stagnation pressures one calculates are
even larger because of reduced values of " .
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Problems
Consider the downstream end of the inlet entering the combuster. According to Figure 3.25 if,
at a free stream Mach number of 8, the internal flow is brought isentropically to low Mach number at the entrance to the combustor the combuster will experience a pressure 218 times the free
stream dynamic pressure. For example, at an altitude of about 26 kilometers and M 9 = 8.0
a low Mach number combuster would operate at about 3200 psia (10,000 times the ambient
atmospheric pressure at that altitude). This would require a very heavy structure making the
whole idea impractical.
If instead, the flow Mach number entering the combuster can be maintained at about Mach 2
then the combuster will operate at a much more feasible value of 28 bar or about 410 psia
(somewhat higher with real gas effects accounted for). This structural issue is at least as important as the thermochemistry issue in forcing the designer to consider operating the combuster
at supersonic Mach numbers in order to attain hypersonic flight.
3.9 PROBLEMS
Problem 1 - Review 1-D gas dynamics with heat addition and area change. Consider the
flow of a combustible gas mixture through a sudden expansion in a pipe.
2
1
Combustion occurs between section 1 at the exit of the small pipe and section 2 in the large
pipe where the flow is uniform. Wall friction may be assumed to be negligible throughout.
The flow at station 1 has stagnation properties P t1 and T t1 . The Mach number at station
1, M 1 is subsonic and the pressure on the annular step is approximately equal to P 1 . The
heat of reaction is denoted by Q.
(i) Show that the exit Mach number is given by
"–1 2
2
"–1 2
2
M 1 % 1 + ------------ M 1&
M 2 % 1 + ------------ M 2&
#
$
#
$
2
2
Q
--------------------------------------------- = % 1 + ----------------& --------------------------------------------#
2
C p T t1$ % A
2 2
2&
2
(1 + " M2)
----+
"
M
'
1(
# A1
$
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3.35
.
(3.103)
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Problems
(ii) Show that when Q = 0 and A 2  A 1 goes to infinity
"
------------
P t1
" –1 2 "–1
.
-------- = % 1 + ------------ M 1&
#
$
2
P t2
(3.104)
This limit is the case of a simple jet coming from an orifice in an infinite plane?
Problem 2 - Show that for an ideal ramjet (do not assume f<<1).
"–1 2
"–1 2
, th = % ------------ M 0&  % 1 + ------------ M 0&
$
# 2
$ #
2
.
(3.105)
Problem 3 - The sketch below shows a ramjet operating at a free stream Mach number
of 0.7. Heat is added between stations 3 and 4 and the stagnation temperature at station 4 is T t4 = 1000K . The Mach number at station 3 is very low. The ambient
4
2
temperature and pressure are T 0 = 216K and P 0 = 2 × 10 N  M . Assume that f<<1.
3
P0 T0
1
4
e
M0 = 0.7
Using appropriate assumptions estimate the dimensionless thrust T  ( P 0 A 0 ) and the
area ratio A 0  A e .
Problem 4 - The sketch below shows a ramjet operating at a free stream Mach number
M 0 = 1.5 ( f ( 1.5 ) = 0.8502 ) with a normal shock in front of the engine. Heat is
added between stations 3 and 4 and the stagnation temperature at station 4 is
T t4 = 1400K . The Mach number at station 3 is very low. There is no shock in the
4
2
inlet. The ambient temperature and pressure are T 0 = 216K and P 0 = 2 × 10 N  M .
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Problems
3
4
e
1
P0 T0
M0 = 1.5
shock
Using appropriate assumptions estimate the dimensionless thrust T  ( P 0 A 0 ) and the
area ratio A 0  A e .
Problem 5 - A ramjet operates at a freestream Mach number of 3. The inlet is a straight
duct and the Mach number of the flow entering the burner at station 3 is three. Heat is added
across the burner such that the Mach number of the flow exiting the burner at station 4 is
4
2
two. The ambient temperature and pressure are T 0 = 216K and P 0 = 2 × 10 N  M .
Assume that f « 1 .
1
3
M3 = 3
e
4
M4 = 2
The exit flow is expanded to P e = P 0 . Determine the dimensionless thrust, T  P 0 A 0 .
Problem 6 - A ramjet operates at a freestream Mach number of 2.5. The ambient temper4
2
ature and pressure are T 0 = 216K and P 0 = 2 × 10 N  M . The engine operates with a
straight duct after the burner, A 1  A e = 1 . Supersonic flow is established at the entrance
of the inlet and a normal shock is stabilized somewhere in the diverging part of the inlet.
The stagnation temperature exiting the burner is T t4 = 1800K . Neglect wall friction.
Determine the dimensionless thrust, T  P 0 A 0 .
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Problems
1
P0
T0
A1t
3
e
4
M0 = 2.5
shock
Problem 7 - The figure below shows a ramjet test facility. A very large plenum contains
Air at constant stagnation pressure and temperature, P t0 , T t0 . Jet fuel
7
( h f = 4.28 × 10 J/kg ) is added between stations 3 and 4 ( A 3 = A 4 ) where combustion
takes place. The flow exhausts to a large tank which is maintained at pressure P 0 . Let
P t0  P 0 = 100 . The upstream nozzle area ratio is A 3  A 1.5 = 8 . The exit area, A e can
be varied in order to change the flow conditions in the engine. The gas temperature in the
plenum is T t0 = 805.2K . To simplify the analysis, assume adiabatic flow, neglect wall
friction and assume constant specific heat throughout with " = 1.4 .
P0
ṁ f
1.5
3
4
e
Pt0
Tt0
ṁ a
Initially, the valve controlling A e is closed, A e = 0 and the fuel mass flow is shut off.
Consider a test procedure where the nozzle area is opened, then closed. In the process
A e  A 1.5 is slowly increased from zero causing Air to start flowing. The nozzle is opened
until A e = A 3 . Then the nozzle area is slowly reduced until A e  A 1.5 = 0 once again.
Plot the thrust normalized by the plenum pressure and upstream throat area, T  ( P t0 A 1.5 )
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Problems
and the fuel-Air ratio f as a function of A e  A 1.5 . Distinguish points corresponding to
increasing and decreasing A e . The fuel flow is adjusted to maintain the stagnation temperature at 4 at a constant value. Plot the results for three cases.
(3.106)
T t4 = 805.2K (zero fuel flow), T t4 = 1200K , T t4 = 2700K
Neglect stagnation pressure loss across the burner due to aerodynamic drag of the burner,
retain the loss due to heat addition. Do not assume f « 1 . What flight Mach number and
altitude are being simulated at this condition?
Problem 8 - A ramjet operates at a freestream Mach number of 3. The area ratio across the
engine is A 1  A e = 2 . Supersonic flow is established at the entrance of the inlet and a
normal shock is stabilized in the diverging part of the inlet. The inlet throat Mach number
is 1.01. The stagnation temperature exiting the burner is T t4 = 1944K . Assume " = 1.4 ,
R = 287 M2/(sec2-°K), C p = 1005 M2/(sec2-°K). The ambient temperature and pres4
2
sure are T 0 = 216K and P 0 = 2 × 10 N  M . Assume throughout that f « 1 .
1
P0
T0
A1.5
3
4
e
M0 = 3
shock
i) Suppose the fuel flow is increased with the geometry of the engine held fixed. Estimate
the value of f that would cause the inlet to unstart. Plot the dimensionless
thrust, T  P 0 A 1.5 , specific impulse and overall efficiency of the engine as a function of f .
Plot your result beyond the point where the inlet unstarts and assume the engine does not
flame out. Assume the fuel flow is throttled so as to keep the fuel/air ratio constant. Note
that the thrust is normalized by a fixed geometric area rather than the capture area A 0 that
changes when the engine unstarts.
ii) With T t4 = 1944K suppose A e is reduced keeping the fuel flow the same. Estimate
the value of A 1  A e that would cause the inlet to unstart? Plot the dimensionless
thrust, T  P 0 A 1.5 specific impulse and overall efficiency of the engine as a function of
A 1  A e . As in part (i) plot your result beyond the point where the inlet unstarts and assume
the engine does not flame out.
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Problems
Problem 9 - A ramjet operates at a freestream Mach number of 3. The area ratio across
the engine is A 1  A e = 2 . Supersonic flow is established at the entrance of the inlet and
a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature
exiting the burner is T t4 = 1944K .The ambient temperature and pressure are
4
2
T 0 = 216K and P 0 = 2 × 10 N  M . Do not assume that f « 1 . Do not neglect the
effects of wall friction.
1
P0
A1.5
3
4
e
T0
M0 = 3
shock
i) Determine the fuel-air ratio, f .
ii) Determine the dimensionless thrust, T  P 0 A 0 .
Problem 10 -Assume " = 1.4 , R = 287 M2/(sec2-°K), C p = 1005 M2/(sec2-°K). The
4
2
ambient temperature and pressure are T 0 = 216K and P 0 = 2 × 10 N  M . The sketch
below shows a ramjet operating at a free stream Mach number of 3.0. The incoming air is
decelerated to a Mach number of 2.0 at station 3.
0
M0 = 3.0
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1
3
M3 = 2.0
3.40
4
e
M4 = 1.0
bjc
Problems
Heat is added between 3 and 4 bringing the Mach number at station 4 to one. The flow is
then ideally expanded to P e = P 0 . This type of engine with the combustion of fuel occurring in a supersonic stream is called a SCRAMJET (supersonic combustion ramjet).
Determine the dimensionless thrust T  ( P 0 A 0 ) . Assume f<<1 if you wish and neglect stagnation pressure losses due to friction.
Problem 11 - A ramjet operates in the upper atmosphere at a high supersonic Mach number
as shown in the sketch below. Supersonic flow is established at the entrance of the inlet and
a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature
exiting the burner is sufficient to produce substantial positive thrust. A small flat plate is
placed downstream of the burner as shown causing a small drop in stagnation pressure
between station “a” and station “b”. The plate can be positioned so as to produce high drag
as shown or it can be rotated 90° so that the long dimension is aligned with the flow producing lower drag. The engine operates with a convergent-divergent nozzle. The nozzle
throat is choked and the nozzle exit is fully expanded.
1
A1.5
3 4
a b
8
e
M=3
shock
Suppose the plate is rotated from the high drag position to the low drag position. State
whether each of the following area-averaged quantities increases, decreases or remains the
same.
1) M 3
5) T e
2) M a
6) P e
3) M b
7) U e
4) M e
8) Engine thrust
Explain the answer to part 8) in terms of the drag force on the plate and the pressure forces
that act on the engine inlet and nozzle.
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Problems
Problem 12 - A ramjet operates at a freestream Mach number of 3. The area ratio across
the engine is A 1  A e = 2 . Supersonic flow is established at the entrance of the inlet and
a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature
exiting the burner is T t4 = 1512K . The ambient temperature and pressure are T 0 = 216K and
4
2
P 0 = 2 × 10 N  M .
1
P0
T0
A1.5 shock
3
4
e
M0 = 3
Ae
Do not neglect wall friction.
i) Determine the dimensionless thrust, T  P 0 A 0 .
ii) Suppose the wall friction is increased slightly. Determine if each of the following
increases, decreases or remains the same. Assume the inlet does not unstart.
a) The Mach number at station 4
c) The shock Mach number
b) The Mach number at station 1.5
d) The dimensionless thrust T  P 0 A 0
Problem 13 - A ramjet operates at a freestream Mach number of 3. The area ratio across
the engine is A 1  A 8 = 1.75 . Supersonic flow is established at the entrance of the inlet
and a normal shock is stabilized in the diverging part of the inlet. The stagnation temperature exiting the burner is T t4 = 1512K . The nozzle is fully expanded P e = P 0 . The fuel
enthalpy is
7
h f = 4.28 × 10 J/kg . The ambient temperature and pressure are T 0 = 216K and
4
2
P 0 = 2 × 10 N  M .
1
A1.5 shock
3
4
8
e
M0 = 3
i) Determine the dimensionless thrust, T  P 0 A 0 .
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Problems
ii) Suppose T t4 is decreased slightly. Determine if each of the following increases,
decreases or remains the same.
a) The Mach number at station 4
c) The shock Mach number
b) The Mach number at station 3
d) The nozzle exit pressure P e
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3.43
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Problems
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3.44
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