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Chapter 3

The ramjet cycle

3.1 Ramjet flow field

Before we begin to analyze the ramjet cycle we will consider an example that can help usunderstand how the flow through a ramjet comes about. The key to understanding the flowfield is the intelligent use of the relationship for mass flow conservation. In this connectionthere are two equations that we will rely upon. The first is the expression for 1-D massflow in terms of the stagnation pressure and temperature.

m = ⇢UA =�

⇣

�+1

2

⌘

�+1

2(��1)

✓

PtAp�RTt

◆

f (M) (3.1)

The second is the all-important area-Mach number function.

f (M) =A⇤

A=

✓

� + 1

2

◆

�+1

2(��1) M⇣

1 + ��1

2

M2

⌘

�+1

2(��1)

(3.2)

This function is plotted in Figure 3.1 for three values of �.

For adiabatic, isentropic flow of a calorically perfect gas along a channel Equation (3.1)provides a direct connection between the local channel cross sectional area and Machnumber.

In addition to the mass flow relations there are two relationships from Rayleigh line theorythat are also very helpful in guiding our understanding of the e↵ect of heat addition on

3-1

CHAPTER 3. THE RAMJET CYCLE 3-2

Figure 3.1: Area - Mach number relation.

the flow in the ramjet. These are the equations that describe the e↵ect of heat addition onthe Mach number and stagnation pressure of the flow.

Tt⇤

Tt=

�

1 + �M2

�

2

2 (1 + �)M2

⇣

1 + ��1

2

M2

⌘

Pt⇤

Pt=

✓

1 + �M2

1 + �

◆

�+1

2

1 + ��1

2

M2

!

��1

(3.3)

These equations are plotted in Figure 3.2.

Figure 3.2: E↵ects of heat exchange on Mach number and stagnation pressure.

There are several features shown in these plots that have important implications for theramjet flow. The first is that much more heat can be added to a subsonic flow than to asupersonic flow before thermal choking occurs; that is, before the flow is brought to Machone. The second is that stagnation pressure losses due to heat addition in subsonic floware 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 heataddition can be quite large in a supersonic flow.

With this background we will now construct a ramjet flow field beginning with supersonicflow through a straight, infinitely thin tube. For definiteness let the free stream Machnumber be three and the ambient temperature T

0

= 216K. Throughout this example wewill assume that the friction along the channel wall is negligible.

Figure 3.3: Step 1 - Initially uniform Mach three flow.

Add an inlet convergence and divergence.

Figure 3.4: Step 2 - Inlet convergence and divergence with f(M) shown.

Let the throat Mach number be two (M1.5 = 2.0). In Figure 3.4 the Mach number de-

creases 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 componentof the pressure force on the inlet is exactly balanced on the upstream and downstreamsides of the inlet.

In Figure 3.5 heat is added to the supersonic flow inside the engine. Neglect the massflow of fuel added compared to the air mass flow. As the heat is added the mass flow isconserved. Thus, neglecting the fuel added, the mass balance is


Figure 3.5: Step 2 - Introduce a burner and add heat to the flow.

m =�

⇣

�+1

2

⌘

�+1

2(��1)

Pt0A3p�RTt0

f (3) =�

⇣

�+1

2

⌘

�+1

2(��1)

Pt4A4p�RTt4

f (M4

) . (3.4)

As the heat is added, Tt4 goes up and Pt4 goes down while the following equality must bemaintained

Pt0pTt0

f (3) =Pt4pTt4

f (M4

) . (3.5)

Conservation of mass (3.5) implies that f(M4

) must increase and the Mach number down-stream of the burner decreases. There is a limit to the amount of heat that can be addedto this flow and the limit occurs when f(M

4

) attains its maximum value of one. At thispoint the flow looks like Figure 3.6.

The Rayleigh line relations tell us that the temperature rise across the burner that producesthis flow is

Tt4

Tt3

�

�

�

�

M4

=1

= 1.53. (3.6)

The corresponding stagnation pressure ratio across the system is


Figure 3.6: Step 3 - Introduce su�cient heat to bring the exit Mach number to a valueslightly greater than one.

Pt4

Pt3

�

�

�

�

M4

=1

!

before unstart

= 0.292. (3.7)

Now, suppose the temperature at station 4 is increased very slightly. We have a problem;Tt4 is up slightly, Pt4 is down slightly but f(M

4

) cannot increase. To preserve the mass flowrate imposed at the inlet, the supersonic flow in the interior of the engine must undergo anun-start. The flow must switch to the configuration shown in Figure 3.7. The mass flowequation (3.5) can only be satisfied by a flow between the inlet throat and the burner thatachieves the same stagnation pressure loss (3.7), since f(M

4

) cannot exceed one and thestagnation temperature ratio is essentially the same.

As a result of the un-start, a shock wave now sits at the end of the di↵user section of theinlet. Notice that the engine internal pressure is still very large and the exit Mach numbermust remain one. The stagnation temperature has not changed and so, as was just pointedout, the mass balance tells us that the stagnation pressure of the exit flow must be thesame as before the un-start. Thus

Pt4

Pt0

�

�

�

�

M4

=1

!

after unstart

= 0.292. (3.8)

Now, the stagnation pressure loss is divided between two mechanisms, the loss across the


Figure 3.7: Step 4 - Increase the heat added very slightly to unstart the flow.

shock wave, and the loss due to heat addition across the burner. The stagnation pressureratio across a Mach three shock wave is

Pt3

Pt0

�

�

�

�

M=3

= 0.3285. (3.9)

The burner inlet Mach number is M3

= 0.475, and the stagnation loss due to thermalchoking across the burner is

Pt4

Pt3

�

�

�

�

M=0.475

= 0.889. (3.10)

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 thrustdefinition, neglecting the fuel/air ratio, is

T

P0

A0

= �M0

2

✓

Ue

U0

� 1

◆

+Ae

A0

✓

Pe

P0

� 1

◆

. (3.11)

The pressure ratio across the engine is


Pe

P0

=Pte

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

= 0.292

✓

2.8

1.2

◆

3.5

= 5.66 (3.12)

and the temperature ratio is

Te

T0

=Tte

Tt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

= 1.53

✓

2.8

1.2

◆

= 3.5667. (3.13)

This produces the velocity ratio

Ue

U0

=Me

M0

r

Te

T0

= 0.6295. (3.14)

Now substitute into (3.11).

T

P0

A0

= 1.4⇥ 9⇥ (0.6295� 1) + 1⇥ (5.66� 1) = �4.66 + 4.66 = 0 (3.15)

The thrust is zero. We would expect this from the symmetry of the upstream and down-stream distribution of pressure on the inlet. Now let’s see if we can produce some thrust.First adjust the inlet so that the throat area is reduced until the throat Mach number isjust slightly larger than one. This will only e↵ect the flow in the inlet and all flow variablesin the rest of the engine will remain the same.

With the flow in the engine subsonic, and the shock positioned at the end of the di↵user, wehave a great deal of margin for further heat addition. If we increase the heat addition acrossthe burner the mass balance (3.5) is still preserved and the exit Mach number remains one.Let the burner outlet temperature be increased to Tt4 = 1814.4K. The flow now lookssomething like Figure 3.8.

The stagnation temperature at the exit is up, the stagnation pressure is up, and the shockhas moved to the left to a lower upstream Mach number (higher f(M) ), while the massflow (3.5) is preserved. Note that we now have some thrust arising from the x-componentof the high pressure force behind the shock that acts to the left on an outer portion ofthe inlet surface. This pressure exceeds the inlet pressure on the corresponding upstreamportion of the inlet surface. The stagnation pressure ratio across the engine is determinedfrom the mass balance (3.5).

Pte

Pt0= f (3)

r

Tte

Tt0= 0.263

p3 = 0.409 (3.16)


Figure 3.8: Step 5 - Increase the heat addition to produce some thrust.

Let’s check the thrust. The pressure ratio across the engine is

Pe

P0

= 0.409

✓

2.8

1.2

◆

3.5

= 7.94. (3.17)

The temperature ratio is

Te

T0

=Tte

Tt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

= 3

✓

2.8

1.2

◆

= 7 (3.18)

and the velocity ratio is now

Ue

U0

=Me

M0

r

Te

T0

=

p7

3= 0.882. (3.19)

The thrust is

T

P0

A0

= 1.4⇥ 9⇥ (0.882� 1) + 1⇥ (7.94� 1) = �1.49 + 6.94 = 5.45. (3.20)

This is a pretty substantial amount of thrust. Note that, so far, the pressure term in thethrust 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 inFigure 3.9. The mass balance is

Figure 3.9: Step 6 - Add a convergent nozzle.

Pt0A1.5p604.8

=PteAep1814.4

. (3.21)

How much can we decrease Ae? Begin with Figure 3.8. As the exit area is decreasedthe exit Mach number remains one due to the high internal pressure in the engine. Theshock moves upstream toward the inlet throat, the exit stagnation pressure increases andthe product PteAe remains constant. The minimum exit area that can be reached withoutun-starting the inlet flow is when the inlet shock is very close to the throat, becomingvanishingly weak. At this condition, the only mechanism for stagnation pressure loss is theheat addition across the burner. The Mach number entering the burner is M

3

= 0.138 asshown in Figure 3.9. The stagnation pressure loss across the burner is proportional to thesquare of the entering Mach number.

dPt

Pt= ��M2

dTt

Tt(3.22)

To a reasonable approximation the stagnation loss across the burner can be neglected andwe can take Pte

⇠= Pt0. In this approximation, the area ratio that leads to the flow depicted


in Figure 3.9 is

Ae

A1.5

�

�

�

�

ideal

=

r

1814.4

604.8= 1.732. (3.23)

This relatively large area ratio is expected considering the greatly increased temperatureand lower density of the exhaust gases compared to the gas that passes through the up-stream throat. What about the thrust? Now the static pressure ratio across the engineis

Pe

P0

=

✓

2.8

1.2

◆

3.5

= 19.41. (3.24)

The temperatures and Mach numbers at the nozzle exit are the same as before so thevelocity ratio does not change between Figure 3.8 and Figure 3.9. The dimensionlessthrust is

T

P0

A0

= 1.4⇥ 9⇥ (0.882� 1)+1.732⇥ 0.236⇥ (19.41� 1) = �1.49+7.53 = 6.034. (3.25)

That’s pretty good; just by adding a convergent nozzle and reducing the shock strengthwe have increased the thrust by about 20%. Where does the thrust come from in thisramjet design? The figure below schematically shows the pressure distribution through theengine. 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 thenozzle due to the heat addition. But the pressure on the inlet surfaces marked ”b” are notbalanced by any force on the nozzle. These pressures substantially exceed the pressure onthe upstream face of the inlet and so net thrust is produced.

3.3 The ideal ramjet cycle

But we can do better still! The gas that exits the engine is at a very high pressure comparedto the ambient and it should be possible to gain thrust from this by adding a divergentsection to the nozzle as shown below.

The area ratio of the nozzle is chosen so that the flow is fully expanded, Pe = P0

. Thestagnation pressure is constant through the engine and so we can conclude from


Figure 3.10: Imbalance of pressure forces leading to net thrust.

Figure 3.11: Ideal ramjet with a fully expanded nozzle.


Pe

P0

=Pte

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

1 = 1⇥

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

(3.26)

that Me = M0

. The temperature ratio is

Te

T0

=Tte

Tt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

=Tte

Tt0= 3. (3.27)

Finally, the dimensionless thrust is

T

P0

A0

= �M0

2

Me

M0

r

Te

T0

� 1

!

= 1.4⇥ 9⇥⇣p

3� 1⌘

= 9.22. (3.28)

Adding a divergent section to the nozzle at this relatively high Mach number increases thethrust by 50%.

Now work out the other engine parameters. The fuel/air ratio is determined from

mfhf = (ma + mf )ht4 � maht3. (3.29)

Assume the fuel added is JP-4 with hf = 4.28⇥ 107J/kg. Equation (3.29) becomes

f =Tt4Tt3

� 1hf

CpTt3� Tt4

Tt3

=1814.4604.8 � 1

70.41� 1814.4604.8

= 0.0297. (3.30)

The relatively small value of fuel/air ratio is the a posteriori justification of our earlierneglect of the fuel mass flow compared to the air mass flow. If we include the fuel/airratio in the thrust calculation (but still ignore the e↵ect of mass addition on the stagnationpressure change across the burner) the result is

T

P0

A0

= �M0

2

(1 + f)Me

M0

r

Te

T0

� 1

!

= 1.4⇥ 9⇥⇣

1.0297⇥p3� 1

⌘

= 9.872. (3.31)


The error in the thrust is about 7% when the fuel contribution is neglected. The dimen-sionless specific impulse is

Ispg

a0

=

✓

1

f

◆✓

1

�M0

◆

T

P0

A0

=9.872

0.0297⇥ 1.4⇥ 3= 79.14 (3.32)

and the overall e�ciency is, (⌧f = hf/CpT0

= 197.2),

⌘ov =

✓

� � 1

�

◆✓

1

f⌧f

◆

T

P0

A0

=

✓

0.4

1.4

◆✓

9.872

0.0297⇥ 197.2

◆

= 0.482. (3.33)

The propulsive e�ciency is

⌘pr =2U

0

Ue + U0

=2

1 +p3= 0.732. (3.34)

The thermal e�ciency of the engine shown in Figure 3.11 can be expressed as follows

⌘th =(ma + mf )

Ue2

2

� maU0

2

2

mfhf=

(ma + mf ) (hte � he)� ma (ht0 � h0

)

(ma + mf )hte � maht0

⌘th = 1�Qrejected during the cycle

Qinput during the cycle= 1�

(ma + mf )he � mah0(ma + mf )hte � maht0

⌘th = 1� T0

Tt0

(1 + f) TeT0

� 1

(1 + f) TteTt0

� 1

!

.

(3.35)

The heat rejection is accomplished by mixing of the hot exhaust stream with surroundingair at constant pressure. Noting (3.27) for the ideal ramjet, the last term in brackets isone and the thermal e�ciency becomes

⌘th ideal ramjet= 1� T

0

Tt0. (3.36)

For the ramjet conditions of this example the thermal e�ciency is 2/3 . The Brayton cyclee�ciency is

⌘B = 1� T0

T3

. (3.37)


In the ideal cycle approximation, the Mach number at station 3 is very small thus T3

⇠= Tt0

and the thermal and Brayton e�ciencies are identical. Note that, characteristically fora Brayton process, the thermal e�ciency is determined entirely by the inlet compressionprocess. The ramjet design shown in Figure 3.11 represents the best we can do at this Machnumber. In fact the final design is what we would call the ideal ramjet. The ideal cycle willbe 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 slightdecrease in the flight Mach number will cause the inlet to un-start. This would produce astrong normal shock in front of the engine, a large decrease in air mass flow through theengine and a consequent decrease in thrust. A practical ramjet design for supersonic flightrequires 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

T

P0

A0

= �M0

2

(1 + f)Me

M0

r

Te

T0

� 1

!

. (3.38)

For the ideal cycle, where Pte = Pt0, Me = M0

and Tte/Tt0 = Te/T0

, the thrust equationusing (1 + f) = (⌧f � ⌧r) / (⌧f � ⌧�) becomes

T

P0

A0

=2�

� � 1(⌧r � 1)

✓

⌧f � ⌧r⌧f � ⌧�

r

⌧�⌧r

� 1

◆

. (3.39)

This form of the thrust equation is useful because it expresses the thrust in terms of cycleparameters that we can rationalize. The parameter ⌧r is fixed by the flight Mach number.At a given altitude ⌧� is determined by maximum temperature constraints on the hotsection materials of the engine, as well as fuel chemistry, and gas dissociation. If the flightMach number goes to zero the thrust also goes to zero. As the flight Mach number increasesfor fixed ⌧� the fuel flow must decrease until ⌧� = ⌧r when fuel shut-o↵ occurs and thethrust is again zero. A typical thrust plot is shown below.

The optimization question is; at what Mach number should the ramjet operate for max-imum thrust at a fixed ⌧�? Di↵erentiate (3.39) with respect to ⌧r and set the result tozero.


Figure 3.12: Ramjet thrust

@

@⌧r

✓

T

P0

A0

◆

=2�

� � 1

0

@

⌧�⌧r⇣

1� 3⌧r + 2⌧rq

⌧�⌧r

⌘

+ ⌧f⇣

⌧� + ⌧�⌧r � 2⌧r2q

⌧�⌧r

⌘

(⌧f � ⌧�) ⌧r2q

⌧�⌧r

1

A = 0

(3.40)

The value of ⌧r for maximum thrust is determined from

⌧�⌧r

✓

1� 3⌧r + 2⌧r

r

⌧�⌧r

◆

+ ⌧f

✓

⌧� + ⌧�⌧r � 2⌧r2

r

⌧�⌧r

◆

= 0. (3.41)

The quantity ⌧f is quite large and so the second term in parentheses in (3.41) clearlydominates the first term. For f ⌧ 1 the maximum thrust Mach number of a ramjet isfound from

⌧�1/2 =

2(⌧rmax thrust)

3/2

⌧rmax thrust + 1

. (3.42)

For the case shown above, with ⌧� = 8.4, the optimum value of ⌧r is 3.5 corresponding toa Mach number of 3.53. The ramjet is clearly best suited for high Mach number flight andthe optimum Mach number increases as the maximum engine temperature increases.

Ispg

a0

=

⇣

2

��1

(⌧r � 1)⌘

1/2

⌧��⌧r⌧f�⌧�

✓

⌧f � ⌧r⌧f � ⌧�

r

⌧�⌧r

� 1

◆

(3.43)


The specific impulse also has an optimum but it is much more gentle than the thrustoptimum, as shown in Figure 3.13.

Figure 3.13: Ramjet specific impulse.

Optimizing the cycle with respect to thrust essentially gives close to optimal specific im-pulse. Notice that the specific impulse of the ideal cycle has a finite limit as the fuel flowreaches shut-o↵.

3.5 The non-ideal ramjet

The major non-ideal e↵ects come from the stagnation pressure losses due to the inlet shockand the burner heat addition. We have already studied those e↵ects fairly thoroughly. Inaddition there are stagnation pressure losses due to burner drag and skin friction losses inthe inlet and nozzle where the Mach numbers tend to be quite high. A reasonable ruleof thumb is that the stagnation pressure losses due to burner drag are comparable to thelosses due to heat addition.

3.6 Ramjet control

Let’s examine what happens when we apply some control to the ramjet. The two maincontrol mechanisms at our disposal are the fuel flow and the nozzle exit area. The enginewe will use for illustration is a stable ramjet with an inlet shock and simple convergentnozzle shown below. The inlet throat is designed to have a Mach number well above oneso that it is not so sensitive to un-start if the free stream conditions, burner temperature


or nozzle area change. Changes are assumed to take place slowly so that unsteady changesin the mass, momentum, and energy contained in the ramjet are negligible.

Figure 3.14: Ramjet control model.

The mass balance is

me =�

⇣

�+1

2

⌘

�+1

2(��1)

PteAep�RTte

f (Me) = (1 + f) ma. (3.44)

The pressure in the engine is virtually certain to be very high at this free stream Machnumber, and so the nozzle is surely choked, and we can write

ma =1

(1 + f)

�⇣

�+1

2

⌘

�+1

2(��1)

PteAep�RTte

. (3.45)

The thrust equation is

T

P0

A0

= �M0

2

✓

(1 + f)Ue

U0

� 1

◆

+Ae

A0

✓

Pe

P0

� 1

◆

. (3.46)

Our main concern is to figure out what happens to the velocity ratio and pressure ratio aswe control the fuel flow and nozzle exit area.

Nozzle exit area control

First, suppose Ae is increased with Tte constant. In order for (3.45) to be satisfied Pte

must drop keeping PteAe constant. The shock moves downstream to a higher shock Mach


number. The velocity ratio remains the same and, since the exit Mach number does notchange, the product PeAe remains constant. Note that the thrust decreases. This can beseen by writing the second term in (3.46) as

Ae

A0

Pe

P0

� Ae

A0

. (3.47)

The left term in (3.47) is constant but the right term increases leading to a decrease inthrust. If Ae is decreased, the reverse happens, the inlet operates more e�ciently and thethrust goes up. But remember, the amount by which the area can be decreased is limitedby the Mach number of the inlet throat.

Fuel flow control

Now, suppose Tte is decreased with Ae constant. In order for (3.45) to be satisfied, Pte

must drop keeping Pte/p

Tte constant. Once again the shock moves downstream to a highershock Mach number. The velocity ratio goes down since the exit stagnation temperature isdown and the Mach numbers do not change.The pressure ratio also decreases since the exitstagnation pressure is down. The thrust clearly decreases in this case. If Tte is increased,the reverse happens, the inlet operates more e�ciently and the thrust goes up. The amountby which the burner exit temperature can be increased is again limited by the Mach numberof the inlet throat.

3.7 Example - Ramjet with un-started inlet

For simplicity, assume constant heat capacity with � = 1.4 , Cp = 1005M2/�

sec2 �K�

.The gas constant is R = 287M2/

�

sec2 �K�

. The ambient temperature and pressure areT0

= 216K and P0

= 2⇥ 104N/M2. The fuel heating value is hf = 4.28⇥ 107 J/kg. Thesketch below shows a ramjet operating at a free stream Mach number of 3.0. A normalshock stands in front of the inlet. Heat is added between stations 3 and 4.

Figure 3.15: Ramjet with normal shock ahead of the inlet.


The stagnation temperature at station 4 is Tt4 = 2000K. Relevant areas are A3

/A1.5 = 8,

A1

= A3

= A4

and A4

/Ae = 3. Determine the dimensionless thrust T/(P0

A1

). Do notassume 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 A1

= A1.5 = A

3

. By what proportion would the airmass flow change?

Solution

The first point to recognize is that the stagnation pressure at station 4 exceeds the ambientby more than a factor of two - note the pressure outside the nozzle is assumed to haverecovered to the ambient value. Thus the exit Mach number is one and the Mach numberat station 4 is M

4

= 0.1975. The stagnation temperature at station 3 is Tt3 = 604.8K .The fuel-air ratio is determined from the enthalpy balance

mfhf = (ma + mf )hte � maht0. (3.48)

For constant heat capacity

f =Tt4Tt0

� 1hf

CpTt0� Tt4

Tt0

=2000

604.8 � 14.28⇥10

7

1005⇥604.8 � 2000

604.8

= 0.0344. (3.49)

Now we need to determine the flow between stations 1 and 3. To get started we will neglectthe fuel addition for the moment. Knowing the Mach number at 4 and the stagnationtemperatures at 3 and 4 we can use Rayleigh line results to estimate the Mach number atstation 3. The stagnation temperature ratio across the burner is

Tt4

Tt3=

Tt4

Tt⇤

�

�

�

�

M=0.1975

Tt⇤

Tt3

�

�

�

�

M=?

=2000

604.8= 3.3069

Tt4

Tt3= 0.2066

Tt⇤

Tt3

�

�

�

�

M=?

=2000

604.8= 3.3069.

(3.50)

The Rayleigh line tables give

Tt⇤

Tt3

�

�

�

�

M=?

=3.3069

0.2066= 16.006 ) M

3

= 0.103. (3.51)

This is a reasonable approximation to the Mach number at station 3. The stagnationpressure ratio across the burner is


Pt4

Pt3=

Pt4

Pt⇤

�

�

�

�

M=0.1975

Pt⇤

Pt3

�

�

�

�

M=0.103

=1.235

1.258= 0.981. (3.52)

The subsonic critical Mach number for an area ratio of 8 is 0.0725. The fact that the Machnumber at station 3 is higher than this value implies that there is a shock in the divergingpart of the inlet and the inlet throat Mach number is equal to one. The stagnation pressureratio between the inlet throat and the exit can be determined from a mass balance betweenstations 1.5 and e.

Pt1.5A1.5 (1 + f)p

Tt1.5

=PteAep

Tte=

Pte

Pt1.5

=1.0344⇥ 3

8

r

2000

604.8= 0.7054

(3.53)

The results (3.52) and (3.53) determine the stagnation pressure ratio across the inlet shockand this determines the Mach number of the inlet shock.

⇡shock =0.7054

0.981= 0.719 ) Mshock = 2.004 (3.54)

Thus far the ramjet flow looks as shown in Figure 3.16.

Figure 3.16: State I.

The stagnation pressure ratio across the external shock is


Pt1

Pt0

�

�

�

�

M=3

= 0.3283 (3.55)

and so the overall stagnation pressure ratio is

Pte

Pt0=

Pt1

Pt0

Pte

Pt1.5= 0.3283⇥ 0.7054 = 0.2316. (3.56)

The static pressure ratio is

Pe

P0

=Pte

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

= 0.2316⇥✓

2.8

1.2

◆

3.5

= 4.494. (3.57)

The temperature ratio is

Te

T0

=Tte

Tt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

=2000

604.8⇥✓

2.8

1.2

◆

= 7.72. (3.58)

The velocity ratio is

Ue

U0

=Me

M0

r

Te

T0

=

p7.72

3= 0.926. (3.59)

Across the inlet, the mass balance is

Pt1.5A1.5 = Pt0A0

f (M0

) (3.60)

and so

A0

A1.5

=Pt1.5

Pt0f (M0

)= 0.3283⇥ 4.235 = 1.39. (3.61)

Finally the thrust is

T

P0

A1

= �M0

2

A0

A1.5

A1.5

A1

✓

(1 + f)Ue

U0

� 1

◆

+Ae

A1

✓

Pe

P0

� 1

◆

(3.62)

and


T

P0

A1

�

�

�

�

StateI

= 1.4⇥ 9⇥ 1.39⇥ (1/8)⇥ (1.0344⇥ 0.926� 1) + (1/3)⇥ (4.494� 1) =

�0.0923 + 1.165 = 1.0724.(3.63)

State II

Now increase the inlet throat area to the point where the inlet un-chokes. As the inletthroat area is increased the Mach number at station 3 will remain the same since it isdetermined 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

Pt1.5A1.5 (1 + f)p

Tt1.5

=PteAep

Tte. (3.64)

The stagnation pressure at station 1.5 is fixed by the loss across the external shock. Thefuel-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 Pte . At the pointwhere the the inlet throat un-chokes the shock is infinitely weak and the only stagnationpressure loss between station 1.5 and the nozzle exit is across the burner.

A1.5State II

Ae=

1

1 + f

Pte State II

Pt1.5

r

Tt1.5

Tte=

0.982

1.0344

r

604.8

2000= 0.522 (3.65)

This corresponds to

A1

A1.5State II

=

✓

A1

Ae

◆✓

Ae

A1.5State II

◆

=3

0.522= 5.747. (3.66)

The mass flow through the engine has increased by the ratio

maState II

maState I=

A1.5State II

A1.5State I

=

✓

A1.5State II

A1

◆✓

A1

A1.5State I

◆

=8

5.747= 1.392 =

A0State II

A0State I

.

(3.67)

At this condition the stagnation pressure ratio across the system is

Pte State II

Pt0=

✓

Pt1.5

Pt0

◆✓

Pte State II

Pt1.5

◆

= 0.3283⇥ 0.982 = 0.3224. (3.68)



Pe State II

P0

=Pte State II

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

= 0.3224

✓

2.8

1.2

◆

3.5

= 6.256. (3.69)

The Mach number at station 1 increases as A1.5 increases and at the condition where

the inlet is just about to un-choke reaches the same Mach number as station 3. At thiscondition the ramjet flow field looks like that shown in Figure 3.17.

Figure 3.17: State II.

The inlet shock is gone, the inlet Mach number has increased to M1

= M3

and the externalshock has moved somewhat closer to the inlet. Note that the capture area to throat arearatio is still

A0State II

A1.5State II

=Pt1.5

Pt0f (M0

)= 0.3283⇥ 4.235 = 1.39 (3.70)

although both A1.5 and A

0

have increased. Also

A0State II

A1

=A

0State II

A1.5State II

A1.5State II

A1

=1.39

5.747= 0.242. (3.71)

The thrust formula is


T

P0

A1

�

�

�

�

State II

= �M0

2

✓

A0State II

A1.5State II

◆✓

A1.5State II

A1

◆✓

(1 + f)Ue

U0

� 1

◆

+Ae

A1

✓

Pe State II

P0

� 1

◆

.

(3.72)

The velocity ratio across the engine is unchanged by the increase in inlet throat area. Thethrust of state II is

T

P0

A1

�

�

�

�

State II

= 1.4⇥ 9⇥ 1.39⇥✓

1

5.747

◆

⇥ (1.0344⇥ 0.926� 1) +

✓

1

3

◆

⇥ (6.256� 1) =

�0.1284 + 1.752 = 1.624.(3.73)

The reduced loss of stagnation pressure leads to almost a 60% increase in thrust at thiscondition.

State III

Now remove the inlet throat altogether.

Now, suppose A1.5 is increased until A

1

= A1.5 = A

3

. The ramjet flow field looks likeFigure 3.18.

Figure 3.18: State III.

With the inlet throat absent, the Mach number is constant between 1 and 3. There is nochange in mass flow, fuel-air ratio, stagnation pressure, or the position of the upstreamshock. The capture area remains


A0State III

A1

=A

0State II

A1

=A

0State II

A1.5State II

A1.5State II

A1

=1.39

5.747= 0.242. (3.74)

Therefore the thrust is the same as the thrust for State II, equation (3.74). If we want toposition the upstream shock very near the entrance to the engine we have to increase thenozzle exit area and reduce the heat addition.

State IV

Open the nozzle exit fully.

First increase the exit area to the point where A1

= A3

= A4

= Ae . If we maintainTt4 = 2000K the Mach number at station 4 becomes one and the Mach number between 1and 3 is, from the Rayleigh solution, M

1

= M3

= 0.276. The ramjet flow at this conditionis sketched in Figure 3.19.

Figure 3.19: State IV.

The velocity ratio is still

Ue

U0

=Me

M0

r

Te

T0

=

p7.72

3= 0.926. (3.75)

The stagnation pressure ratio across the burner is, from the Rayleigh solution,

Pte

Pt1

�

�

�

�

State IV

= 0.8278. (3.76)

Across the whole system


Pte

Pt0

�

�

�

�

State IV

=Pt1.5

Pt0

�

�

�

�

State IV

Pte

Pt1.5

�

�

�

�

State IV

= 0.3283⇥ 0.8278 = 0.2718 (3.77)

and the static pressure ratio is

Pe State IV

P0

=Pte State IV

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

= 0.2718

✓

2.8

1.2

◆

3.5

= 5.274. (3.78)

The area ratio is

A0State IV

A1

=Pt1f (M

1

)

Pt0f (M0

)= 0.3283⇥ 0.4558

0.2362= 0.634. (3.79)

The thrust formula for state IV is

T

P0

A1

�

�

�

�

State IV

= �M0

2

✓

A0State IV

A1

◆✓

(1 + f)Ue

U0

� 1

◆

+Ae

A1

✓

Pe State IV

P0

� 1

◆

(3.80)

which evaluates to

T

P0

A1

�

�

�

�

State IV

= 1.4⇥ 9⇥ 0.634⇥ (1.0344⇥ 0.926� 1) + 1⇥ (5.274� 1) =

�0.3367 + 4.274 = 3.937.

(3.81)

Note the considerable increase in mass flow for state IV compared to state III. From(3.74)

maState IV

maState III=

✓

A0State IV

A1

◆✓

A1

A0State III

◆

=0.634

0.242= 2.62 (3.82)

which accounts for much of the increased thrust in spite of the increase in stagnationpressure loss across the burner.

State V

Reduce the burner outlet temperature until the shock is very close to station 1.


Now reduce Tte until the Mach number at station 3 matches the Mach number behind theshock. From the Rayleigh solution, this occurs when Tte = 924.8. At this condition theramjet flow field looks like Figure 3.20.

Figure 3.20: State V.

The static temperature ratio is

Te

T0

=Tte

Tt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

=924.8

604.8⇥✓

2.8

1.2

◆

= 3.57. (3.83)

The velocity ratio at this temperature is

Ue

U0

=Me

M0

r

Te

T0

=

p3.57

3= 0.630. (3.84)

The stagnation pressure ratio across the burner at this condition is

Pte

Pt3= 0.889 (3.85)

and across the system

Pte

Pt0= 0.889⇥ 0.3283 = 0.292. (3.86)



Pe

P0

=Pte

Pt0

1 + ��1

2

M0

2

1 + ��1

2

Me2

!

��1

= 0.292⇥✓

2.8

1.2

◆

3.5

= 5.67. (3.87)

Neglecting the fuel flow, the thrust is (Note that at this condition A0

= A1

= Ae ).

T

P0

A1

= �M0

2

✓

Ue

U0

� 1

◆

+

✓

Pe

P0

� 1

◆

= 1.4⇥9⇥(0.629� 1)+(5.67� 1) = �4.67+4.67 = 0

(3.88)

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 flowwill be established through the engine. Finally the flow is as shown in Figure 3.21.

Figure 3.21: 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 ofundisturbed 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 becomeunusable and a new design has to be considered where the heat addition across the burneris carried out at supersonic Mach numbers. There are several reasons why this is so, allrelated to the very high stagnation temperature and stagnation pressure of high Machnumber flight. To get started let’s recall the thrust equation for the ramjet with a fullyexpanded nozzle.


T

P0

A1

= �M0

2

(1 + f)Me

M0

r

Te

T0

� 1

!

(3.89)

Define the thrust coe�cient as

Cthrust =Thrust

1

2

⇢0

U0

2A0

=T

�2

M0

2P0

A0

. (3.90)

If we assume ideal behavior (Me = M0

) the thrust coe�cient becomes

Cthrust = 2

(1 + f)

r

Tte

Tt0� 1

!

. (3.91)

When we carried out the energy balance across the burner

mfhf = (ma + mf )ht4 � maht3 (3.92)

we assumed that the fuel enthalpy was simply added to the flow without regard to thechemistry of the process. In fact the chemistry highly limits the range of fuel-air ratiosthat are possible. The stoichiometric reaction of JP-4 with air where the fuel and oxygenare completely consumed is

CH1.94 + 1.485O

2

+ 5.536N2

! CO2

+ 0.97H2

O + 5.536N2

. (3.93)

corresponding to a fuel-air ratio

f =12.01 + 1.94⇥ 1.008

1.485⇥ 32.00 + 5.536⇥ 28.02= 0.0689. (3.94)

This is roughly the value of the fuel-air ratio that produces the maximum outlet temper-ature from the burner. If we choose the much more energetic hydrogen fuel, the reactionis

H2

+ 0.5O2

+ 1.864N2

! H2

O + 1.864N2

(3.95)

corresponding to a fuel-air ratio


f =2⇥ 1.008

16.00 + 1.864⇥ 28.02= 0.0295. (3.96)

The fuel enthalpies are generally taken to be

hfJP�4

= 4.28⇥ 107 J/kg

hfH2

= 12.1⇥ 107 J/kg.(3.97)

In the earlier discussion, we took the perspective that the ramjet cycle was limited by a redline temperature in the hot part of the engine. Let’s relax this assumption and allow themaximum temperature to be free while keeping the fuel-air ratio constant. Furthermorelet’s continue to retain the assumption of constant heat capacities even at high Machnumbers. We will correct this eventually in Chapter 9, but for now we just want to seewhat happens to the ideal thrust coe�cient (3.91) as we increase the free stream Machnumber at constant fuel-air ratio. Using (3.92)), constant heat capacities, and assumingadiabatic flow in the inlet and nozzle we can express the stagnation temperature ratioacross the engine as

Tte

Tt0=

✓

f⌧f1 + f

◆

1

1 + ��1

2

M0

2

!

+1

1 + f(3.98)

where we recall that ⌧f = hf/ (CpT0

). Now the thrust coe�cient becomes

Cthrust = 2

0

@

f (1 + f) ⌧f

1 + ��1

2

M0

2

+ (1 + f)

!

1/2

� 1

1

A . (3.99)

The temperature ratio and thrust coe�cient are plotted in Figures 3.22 and 3.23. Thedrag coe�cient is defined as

Cdrag =Drag

1

2

⇢0

U0

2A0

=D

�2

M0

2P0

A0

. (3.100)

At high Mach numbers the drag coe�cient of a body tends toward a constant value. Fora sphere the drag coe�cient tends toward a constant slightly less than one, about 0.95.More streamlined bodies have lower drag and coe�cients as low as 0.2 can be achieved.This observation together with Figure 3.23 indicates that as the Mach number increasesit becomes harder and harder to produce thrust that exceeds drag. The thrust coe�cient


Figure 3.22: Temperature ratio of an ideal ramjet at constant fuel-air ratio.

Figure 3.23: Thrust coe�cient of an ideal ramjet at constant fuel-air ratio.


drops below one at a Mach number of about 7 to 8 for both fuels. As the Mach numberincreases a conventional ramjet (even an ideal one) simply cannot produce enough thrustto overcome drag. The limiting thrust coe�cient at infinite Mach number is

limM

0

!1Cthrust = f. (3.101)

This last result suggests that a scramjet can benefit from the choice of a fuel-rich mixtureratio as long as it does not exceed the flammability limit of the fuel. This would alsoprovide additional fuel for cooling the vehicle. An advantage of a hydrogen system is thatit can operate quite fuel rich. In addition, hydrogen has a higher heat capacity than anyother fuel enabling it to be used to provide cooling for the vehicle that, in contrast to are-entry body, has to operate in a very high temperature environment for long periods oftime. The down side of hydrogen is that liquid hydrogen has to be stored at very lowtemperatures and the liquid density is only about 1/10 of that of JP-4. It is clear thatsmall e↵ects can be important. For example when we developed the thrust formula weneglected the momentum of the injected fuel. In a realistic scramjet analysis that wouldhave to be taken into account.

3.8.1 Real chemistry e↵ects

The real chemistry of combustion shows that the problem is even worse than just discussed.The plot points in Figure 3.21 are derived from an equilibrium chemistry computation ofthe combustion of JP-4 with air. Notice that the temperature reached by the combustiongases is substantially less than the ideal, and for high Mach numbers the temperature risefrom the reaction is actually less than one due to the cooling e↵ect of the added fuel.

The solution to this problem, which has been pursued since the 1960s, is to try to addheat with the burner operating with a supersonic Mach number on the order of two orso thereby cutting the static temperature of the air flowing into the combustor by almosta factor of two. This is the concept of a supersonically burning ramjet or scramjet. Ageneric sketch of such a system is shown in Figure 3.24. Two Homework problems are usedto illustrate basic concepts.

3.8.2 Scramjet operating envelope

Figure 3.25 shows a widely circulated plot showing the altitude and Mach number regimewhere a scramjet might be expected to operate. Contours of constant free stream stagnationtemperature and flow dynamic pressure are indicated on the plot.


Figure 3.24: Conceptual figure of a ramjet.

Figure 3.25: Conceptual operating envelope of a scramjet.


This figure somewhat accurately illustrates the challenges of scramjet flight. These can belisted as follows.

1) At high Mach numbers the vehicle is enveloped in an extremely high temperature gasfor 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 ofthe low atmospheric density and long chemical times. This defines the combustor blowoutlimit.

3) At lower altitude and high Mach number the free stream dynamic pressure increases tothe 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 ispresented as a line of constant dynamic pressure coinciding with increasing Mach numberand altitude. But in supersonic flow, the free stream dynamic pressure is really not asu�cient measure of the actual loads that are likely to act on the vehicle. Again let’s makea constant heat capacity assumption and compare the free stream stagnation pressure tothe free stream dynamic pressure as follows. Form the pressure coe�cient

Pt1 � P1q1

=P1

�2

P1M12

✓

1 +� � 1

2M1

2

◆

��1

� 1

!

=

⇣

1 + ��1

2

M12

⌘

��1 � 1

�2

M12

.

(3.102)

This function is plotted in Figure 3.26.

Figure 3.26: Comparison of stagnation pressure and dynamic pressure in supersonic flow.

If one repeats this calculation with real gas e↵ects the stagnation pressures one calculatesare even larger because of reduced values of �.

Consider the downstream end of the inlet where the flow enters the combuster. According toFigure 3.26 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 apressure 218 times the free stream dynamic pressure. For example, at an altitude of about26 kilometers and M

0

= 8.0 a low Mach number combuster would operate at about 3200psia (10,000 times the ambient atmospheric pressure at that altitude). This would requirea very heavy structure making the whole idea impractical.

If instead, the flow Mach number entering the combuster can be maintained at about Mach2 then the combuster will operate at a much more feasible value of 28 bar or about 410psia (somewhat higher with real gas e↵ects accounted for). This structural issue is at leastas important as the thermochemistry issue in forcing the designer to consider operatingthe 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 theflow of a combustible gas mixture through a sudden expansion in a pipe shown in Figure3.27.

Figure 3.27: Dump combustor.

Combustion occurs between section 1 at the exit of the small pipe and section 2 in the largepipe where the flow is uniform. Wall friction may be assumed to be negligible throughout.The flow at station 1 has stagnation properties Pt1 and Tt1 . The Mach number at station1, M

1

is subsonic and the pressure on the annular step is approximately equal to P1

. Theheat of reaction is denoted by Q.

(i) Show that the exit Mach number is given by


M2

2

⇣

1 + ��1

2

M2

2

⌘

�

1 + �M2

2

�

2

=

✓

1 +Q

CpTt1

◆M1

2

⇣

1 + ��1

2

M1

2

⌘

⇣

A2

A1

+ �M1

2

⌘

2

. (3.103)

(ii) Show that when Q = 0 and A2

/A1

goes to infinity

Pt1

Pt2=

✓

1 +� � 1

2M

1

2

◆

��1

. (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

⌘th =��1

2

M0

2

1 + ��1

2

M0

2

. (3.105)

Do not assume f ⌧ 1.

Problem 3 - Figure 3.28 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 isTt4 = 1000K . The Mach number at station 3 is very low. The ambient temperature andpressure are T

0

= 216K and P0

= 2⇥ 104N/M2. Assume that f ⌧ 1.

Figure 3.28: Ramjet in subsonic flow.

Using appropriate assumptions, estimate the dimensionless thrust T/ (P0

A0

) and the arearatio A

0

/Ae.

Problem 4 - Figure 3.29 shows a ramjet operating at a free streamMach numberM0

= 1.5,with a normal shock in front of the engine. Heat is added between stations 3 and 4 andthe stagnation temperature at station 4 is Tt4 = 1400K. The Mach number at station 3


is very low. There is no shock in the inlet. The ambient temperature and pressure areT0

= 216K and P0


Figure 3.29: Ramjet in supersonic flow with a shock ahead of the inlet.

Using appropriate assumptions, estimate the dimensionless thrust T/ (P0

A0

) and the arearatio A

0

/Ae.

Problem 5 - In Figure 3.30 a ramjet operates at a freestream Mach number of 3. Theinlet is a straight duct and the Mach number of the flow entering the burner at station 3is three. Heat is added across the burner such that the Mach number of the flow exitingthe burner at station 4 is two. The ambient temperature and pressure are T

0

= 216K andP0


Figure 3.30: Supersonically burning ramjet.

The exit flow is expanded to Pe = P0

. Determine the dimensionless thrust, T/ (P0

A0

).

Problem 6 - In Figure 3.31 a ramjet operates at a freestream Mach number of 2.5. Theambient temperature and pressure are T

0

= 216K and P0

= 2 ⇥ 104N/M2. The engineoperates with a straight duct after the burner, A

1

/Ae = 1. Supersonic flow is establishedat the entrance of the inlet and a normal shock is stabilized somewhere in the divergingpart of the inlet. The stagnation temperature exiting the burner is Tt4 = 1800K. Neglect


wall friction. Determine the dimensionless thrust, T/(P0

A0

).

Figure 3.31: Ramjet with a constant area nozzle.

Problem 7 - Figure 3.32 shows a ramjet test facility. A very large plenum contains Air atconstant stagnation pressure and temperature, Pt0 , Tt0. Jet fuel (hf = 4.28 ⇥ 107 J/kg)is added between stations 3 and 4 ( A

3

= A4

) where combustion takes place. The flowexhausts to a large tank which is maintained at pressure P

0

. Let Pt0/P0

= 100. Theupstream nozzle area ratio is A

3

/A1.5 = 8. The exit area, Ae can be varied in order to

change the flow conditions in the engine. The gas temperature in the plenum is Tt0 =805.2K. To simplify the analysis, assume adiabatic flow, neglect wall friction and assumeconstant specific heat throughout with � = 1.4.

Figure 3.32: Ramjet test facility.

Initially, the valve controlling Ae is closed, Ae = 0 and the fuel mass flow is shut o↵.Consider a test procedure where the nozzle area is opened, then closed. In the processAe/A1.5 is slowly increased from zero causing Air to start flowing. The nozzle is openeduntil Ae = A

3

. Then the nozzle area is slowly reduced until Ae/A1.5 = 0 once again. Plotthe thrust normalized by the plenum pressure and upstream throat area, T/(Pt0A1.5) andthe fuel-Air ratio f as a function of Ae/A1.5. Distinguish points corresponding to increasingand decreasing Ae. The fuel flow is adjusted to maintain the stagnation temperature at 4


at a constant value. Plot the results for three cases.

Tt4 = 805.2K(zero fuel flow)

Tt4 = 1200K

Tt4 = 2700K

(3.106)

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 andaltitude are being simulated at this condition?

Problem 8 - In Figure 3.33 a ramjet operates at a free-stream Mach number of 3. The arearatio across the engine is A

1

/Ae = 2. Supersonic flow is established at the entrance of theinlet and a normal shock is stabilized in the diverging part of the inlet. The inlet throatMach number is 1.01. The stagnation temperature exiting the burner is Tt4 = 1944K.Assume � = 1.4 , R = 287M2/

�

sec2 �K�

, Cp = 1005M2/�

sec2 �K�

. The ambienttemperature and pressure are T

0

= 216K and P0

= 2 ⇥ 104N/M2. Assume throughoutthat f ⌧ 1.

Figure 3.33: Ramjet with a simple convergent nozzle.

i) Suppose the fuel flow is increased with the geometry of the engine held fixed. Estimate thevalue of f that would cause the inlet to unstart. Plot the dimensionless thrust, T/(P

0

A1.5),

specific impulse, and overall e�ciency of the engine as a function of f . Plot your resultbeyond the point where the inlet un-starts 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 thethrust is normalized by a fixed geometric area rather than the capture area A

0

that changeswhen the engine un-starts.

ii) With Tt4 = 1944K, suppose Ae is reduced keeping the fuel flow the same. Estimatethe value of A

1

/Ae that would cause the inlet to un-start? Plot the dimensionless thrust,T/(P

0

A1.5) specific impulse and overall e�ciency of the engine as a function of A

1

/Ae. Asin part (i), plot your result beyond the point where the inlet un-starts and assume theengine does not flame out.


Problem 9 - In Figure 3.34 a ramjet operates at a free-stream Mach number of 3. Thearea ratio across the engine is A

1

/Ae = 2. Supersonic flow is established at the entrance ofthe inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnationtemperature exiting the burner is Tt4 = 1944K. The ambient temperature and pressureare T

0

= 216K and P0

= 2 ⇥ 104N/M2. Do not assume that f ⌧ 1. Do not neglect thee↵ects of wall friction.

Figure 3.34: Ramjet with a convergent nozzle and inlet shock.

i) Determine the fuel-air ratio, f .

ii) Determine the dimensionless thrust, T/(P0

A0

).

Problem 10 - Assume � = 1.4 , R = 287M2/�

sec2 �K�

, Cp = 1005M2/�

sec2 �K�

.The ambient temperature and pressure are T

0

= 216K and P0

= 2 ⇥ 104N/M2. Figure3.35 shows a ramjet operating at a free stream Mach number of 3.0. The incoming air isdecelerated to a Mach number of 2.0 at station 3.

Figure 3.35: A scramjet concept.

Heat is added between 3 and 4 bringing the Mach number at station 4 to one. The flowis then ideally expanded to Pe = P

0

. This type of engine with the combustion of fueloccurring in a supersonic stream is called a SCRAMJET (supersonic combustion ramjet).Determine the dimensionless thrust T/(P

0

A0

). Assume f ⌧ 1 if you wish and neglectstagnation pressure losses due to friction.

Problem 11 - A ramjet operates in the upper atmosphere at a high supersonic Machnumber as shown in Figure 3.36. Supersonic flow is established at the entrance of the


inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnationtemperature exiting the burner is su�cient to produce substantial positive thrust. A smallflat plate is placed downstream of the burner as shown, causing a small drop in stagnationpressure between station ”a” and station ”b”. The plate can be positioned so as to producehigh drag as shown, or it can be rotated 90� so that the long dimension is aligned with theflow, producing lower drag. The engine operates with a convergent-divergent nozzle. Thenozzle throat is choked and the nozzle exit is fully expanded.

Figure 3.36: Ramjet with variable drag loss.

Suppose the plate is rotated from the high drag position to the low drag position. Statewhether each of the following area-averaged quantities increases, decreases or remains thesame.

1)M3

5)Te

2)Ma 6)Pe

3)Mb 7)Ue

4)Me 8)Engine thrust

Explain the answer to part 8) in terms of the drag force on the plate and the pressureforces that act on the engine inlet and nozzle.

Problem 12 - In Figure 3.37 a ramjet operates at a freestream Mach number of 3. Thearea ratio across the engine is A

1

/Ae = 2. Supersonic flow is established at the entrance ofthe inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnationtemperature exiting the burner is Tt4 = 1512K. The ambient temperature and pressureare T

0

= 216K and P0

= 2⇥ 104N/M2.

Do not neglect wall friction.

i) Determine the dimensionless thrust, T/(P0

A0

).

ii) Suppose the wall friction is increased slightly. Determine if each of the following in-creases, decreases or remains the same.


Figure 3.37: Ramjet in supersonic flow.

a) The Mach number at station 4

b) The Mach number at station 1.5

c) The shock Mach number

d) The dimensionless thrust T/(P0

A0

)

Assume the inlet does not un-start.

Problem 13 - In Figure 3.38 a ramjet operates at a freestream Mach number of 3. The arearatio across the engine is A

1

/A8

= 1.75. Supersonic flow is established at the entrance ofthe inlet and a normal shock is stabilized in the diverging part of the inlet. The stagnationtemperature exiting the burner is Tt4 = 1512K. The nozzle is fully expanded Pe = P

0

.The fuel enthalpy is hf = 4.28 ⇥ 107 J/kg. The ambient temperature and pressure areT0

= 216K and P0

= 2⇥ 104N/M2.

Figure 3.38: Ramjet with a converging-diverging nozzle.

i) Determine the dimensionless thrust, T/(P0

A0

).

ii) Suppose Tt4 is decreased slightly. Determine if each of the following increases, decreasesor remains the same.

a) The Mach number at station 4

b) The Mach number at station 3

c) The shock Mach number

d) The nozzle exit pressure Pe

the ramjet cycle - stanford universitycantwell/aa283_course_material/...chapter 3. the ramjet cycle...

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