bjc 3.1 3/21/11

C

HAPTER

3T

HE

R

AMJET

C

YCLE

, S

CRAMJETS

3.1 R

AMJET

FLOW

FIELD

Before we begin to analyze the ramjet cycle we will consider an example that can help us under-stand how the flow through a ramjet comes about. The key to understanding the flow field is theintelligent use of the relationship for mass flow conservation. In this connection there are twoequations that we will rely upon. The first is the expression for 1-D mass flow in terms of thestagnation pressure and temperature.

(3.1)

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

. (3.2)

This function is plotted below for three values of .

Figure 3.1 Area - Mach number relation.

m !UA "

" 1+2-------------# $

% &

" 1+2 " 1( )--------------------

-------------------------------------Pt A"RT t

----------------# $' (% & f M( )= =

f M( ) A*

A------" 1+

2-------------# $% &

" 1+2 " 1( )--------------------

M 1 " 12------------M2+# $

% &

" 1+2 " 1( )--------------------

# $' (' (% &

= =

"

1 2 3 4 5

0.2

0.4

0.6

0.8

1

M

f(M)

"=1.2

"=1.4"=1.66

Ramjet flow field

3/21/11 3.2 bjc

For adiabatic, isentropic flow of a calorically perfect gas along a channel this equation providesa 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 thatare also very helpful in guiding our understanding of the effect of heat addition on the flow inthe ramjet. These are the equations that describe the effect of heat addition on the Mach numberand stagnation pressure of the flow.

(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 ramjetflow. The first is that much more heat can be added to a subsonic flow than to a supersonic flowbefore thermal choking occurs; that is, before the flow is brought to Mach one. The second isthat stagnation pressure losses due to heat addition in subsonic flow are relatively small andcannot exceed about 20% of the stagnation pressure of the flow entering the region of heat addi-tion. In contrast stagnation pressure losses due to heat addition can be quite large in a supersonicflow.

With this background we will now construct a ramjet flow field beginning with supersonic flowthrough a straight, infinitely thin tube. For definiteness let the free stream Mach number be threeand the ambient temperature . Throughout this example we will assume that thefriction along the channel wall is negligible.

T t*

T t------ 1 "M

2+( )2

2 1 "+( )M2 1 " 12------------M2+# $

% &--------------------------------------------------------------------# $' (' (' (% &

=Pt

*

Pt------ 1 "M

2+1 "+----------------------# $

% &

" 1+2-------------

1 " 12------------# $% & M2+

-------------------------------------# $' (' (' (% &

"" 1( )-----------------

=

T 0 250K=

Ramjet flow field

bjc 3.3 3/21/11

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 shown.

Let the throat Mach number be 2.0 ( ). In Figure 3.4 the Mach number decreasesto the inlet throat ( increases), then increases again to the inlet value of three( ). The thrust of this system is clearly zero since the x-directed component ofthe pressure force on the inlet is exactly balanced on the upstream and downstream sides of theinlet.

M0 = 3

1 e

Me = 3

T0=250

Tt0=700

M0 = 3

1 e

Me = 3

1.5

M = 3

Tt0=700

T0=250

f M( )

M1.5 2.0=f M( )

f M( ) 0.236=

Ramjet flow field

3/21/11 3.4 bjc

Now add heat to the supersonic flow inside the engine. Neglect the mass flow of fuel addedcompared to the air mass flow.

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,

(3.4)

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

(3.5)

Conservation of mass (3.5) implies that must increase and the Mach number down-stream of the burner decreases. There is a limit to the amount of heat that can be added to thisflow and the limit occurs when attains its maximum value of one. At this point the flowlooks like the following.

M0 = 3

1 e

Me < 3

1.5

M = 3

3 4

3

1x

M = 2T0=250Tt0=700

M

m "

" 1+2-------------# $

% &

" 1+2 " 1( )--------------------

-------------------------------------Pt0 A3"RT t0

-------------------# $' (% & f 3( ) "

" 1+2-------------# $

% &

" 1+2 " 1( )--------------------

-------------------------------------Pt4 A4"RT t4

-------------------# $' (% & f M4( )= =

T t4 Pt4

Pt0T t0

------------ f 3( )Pt4T t4

------------ f M4( )=

f M4( )

f M4( )

Ramjet flow field

bjc 3.5 3/21/11

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 thisflow is

(3.6)

The corresponding stagnation pressure ratio across the system is

(3.7)

Now suppose the temperature at station 4 is increased very slightly. We have a problem; isup slightly, is down slightly but cannot increase. To preserve the mass flow rateimposed at the inlet the supersonic flow in the interior of the engine must undergo an unstartand the flow must switch to the configuration shown in Figure 3.7. The mass flow equation (3.5)

M0 = 3

1 e

Me = 1+)

1.5

M = 3

3 4

3

1x

M = 2T0=250Tt0=700

Tt4=1070

M

T t4T t3--------

M4 1=1.53=

Pt4Pt3--------

M4 1=# $' (' (% &

before unstart

0.292=

T t4Pt4 f M4( )

Ramjet flow field

3/21/11 3.6 bjc

can only be satisfied by a flow between the inlet throat and the burner that achieves the samestagnation pressure loss (3.7) since cannot exceed one and the stagnation temperatureratio is essentially the same.

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 thatthe engine internal pressure is still very large and the exit Mach number must remain one, thestagnation temperature has not changed and so, as was just pointed out, the mass balance tellsus that the stagnation pressure of the exit flow must be the same as before the unstart. Thus

(3.8)

The stagnation pressure loss is divided between two mechanisms, the loss across the shockwave and the loss due to heat addition across the burner. The stagnation pressure ratio across aMach three shock wave is

(3.9)

f M4( )

M0 = 3

1 e1.5

M = 3

3 4

3

1x

M = 2

shock

M = 0.475T0=250Tt0=700

Tt4=1070Me = 1

M

Pt4Pt0--------

M4 1=# $' (' (% &

after unstart

0.292=

Pt3Pt0--------

M 3=0.3285=

Ramjet flow field

bjc 3.7 3/21/11

The burner inlet Mach number is and the stagnation loss for thermal chokingacross the burner is

. (3.10)

The product of (3.9) and (3.10) is . Now lets look at the thrust generated by the flow depicted in Figure 3.7. The thrust definition,neglecting the fuel/air ratio, is

. (3.11)

The pressure ratio across the engine is

(3.12)

and the temperature ratio is

. (3.13)

This produces the velocity ratio

. (3.14)

Now substitute into (3.11).

(3.15)

The thrust is zero. We would expect this from the symmetry of the upstream and downstreamdistribution of pressure on the inlet. Now lets see if we can produce some thrust. First redesignthe inlet so that the throat area is reduced until the throat Mach number is just slightly largerthan one. This will only effect the flow in the inlet and all flow variables in the rest of the enginewill remain the same.

M3 0.475=

Pt4Pt3--------

M 0.475=0.889=

0.292

TP0 A0------------- "M0

2 UeU0------- 1# $' (% & Ae

A0------

PeP0------ 1# $' (% &+=

PeP0------

PtePt0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+------------------------------------

# $' (' (' (' (% &

"" 1------------

0.292 2.81.2-------# $% &

3.55.66= = =

T eT 0------

T teT t0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+------------------------------------

# $' (' (' (' (% &

1.53 2.81.2-------# $% & 3.5667= = =

UeU0-------

MeM0--------

T eT 0------ 0.6295= =

TP0 A0------------- 1.4 9( ) 0.6295 1( ) 1 5.66 1( )+ 4.66 4.66+ 0= = =

Ramjet flow field

3/21/11 3.8 bjc

With the flow in the engine subsonic and the shock positioned at the end of the diffuser we havea great deal of margin for further heat addition. If we increase the heat addition across the burnerthe mass balance (3.5) is still preserved and the exit Mach number remains one. Let the burneroutlet temperature be increased to . The flow now looks something like this.

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 hasmoved to the left to a lower upstream Mach number (higher ) while the mass flow (3.5)is preserved. Note that we now have some thrust arising from the x-component of the high pres-sure behind the shock which acts to the left on a portion of the inlet surface. This pressureexceeds the inlet pressure on the corresponding upstream portion of the inlet surface. The stag-nation pressure ratio across the engine is determined from the mass balance (3.5).

(3.16)

Lets check the thrust. The pressure ratio across the engine is

. (3.17)

The temperature ratio is

T t4 2100K=

M0 = 3

1 e

Me = 1

1.5

M < 3

3 4

3

1x

shock

M > 0.475T0=250Tt0=700

Tt4=2100M = 1+)

M

f M( )

PtePt0-------- f 3( )

T teT t0-------- 0.236 3 0.409= = =

Pe P0 0.4092.81.2-------# $% &

3.57.94= =

The role of the nozzle

bjc 3.9 3/21/11

(3.18)

and the velocity ratio is now

. (3.19)

The thrust is

. (3.20)

This is a pretty substantial amount of thrust. Note that the pressure term in the thrust definitionis the important thrust component in this design.

3.2 THE ROLE OF THE NOZZLE Lets see if we can improve the design. Add a convergent nozzle to the engine as shown below.

Figure 3.9 Step 6 - Add a convergent nozzle.

The mass balance is

T eT 0------

T teT t0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+------------------------------------

# $' (' (' (' (% &

3 2.81.2-------# $% & 7= = =

Ue U0 Me M0( ) T e T 013--- 7 0.882= = =

TP0 A0------------- 1.4 9( ) 0.882 1( ) 1 7.94 1( )+ 1.49 6.94+ 5.45= = =

M0 = 3

1 e

Me = 1

1.5 3 4

3

1x

M = 0.138T0=250Tt0=700

M = 1+) Tt4=2100

M

The role of the nozzle

3/21/11 3.10 bjc

(3.21)

How much can we decrease ? Begin with Figure 3.8. As the exit area is decreased the exitMach number remains one due to the high internal pressure in the engine. The shock movesupstream toward the inlet throat, the exit stagnation pressure increases and the product remains constant. The minimum exit area that can be reached without unstarting the inlet flowis when the inlet shock is very close to the throat and the shock becomes vanishingly weak. Atthis condition the only mechanism for stagnation pressure loss is the heat addition across theburner. The Mach number entering the burner is as shown in Figure 3.9. The stag-nation pressure loss across the burner is proportional to the square of the entering Mach number

(3.22)

To a reasonable approximation the stagnation loss across the burner can be neglected and wecan take . In this approximation, the area ratio that leads to the flow depicted in Figure3.9 is

(3.23)

This relatively large area ratio is expected considering the greatly increased temperature andlower 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

(3.24)

The temperatures and Mach numbers at the nozzle exit are the same so the velocity ratio doesnot change between Figure 3.8 and Figure 3.9. The dimensionless thrust is

(3.25)

Thats pretty good; just by adding a convergent nozzle and reducing the shock strength we haveincreased the thrust by about 20%. Where does the thrust come from in this ramjet design? Thefigure below schematically shows the pressure distribution through the engine. The pressureforces on the inlet and nozzle surfaces marked a roughly balance although the forward pres-sure is slightly larger compared to the rearward pressure on the nozzle due to the heat addition.

Pt0 A1.5700

-------------------Pte Ae

2100----------------=

Ae

Pte Ae

M3 0.138=

dPtPt

--------- "M2dT tT t

---------=

Pte Pt0*

AeA1.5----------

ideal

2100700------------ 1.732= =

Pe P02.81.2-------# $% &

3.519.41= =

TP0 A0------------- 1.4 9( ) 0.882 1( ) 1.732 0.236( ) 19.41 1( )+ = =

1.49 7.53+ 6.034=

The ideal ramjet cycle

bjc 3.11 3/21/11

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 netthrust is produced.

Figure 3.10 Imbalance of pressure forces leading to net thrust.

3.3 THE IDEAL RAMJET CYCLEBut we can do better still! The gas that exits the engine is at a very high pressure compared tothe ambient and it should be possible to gain thrust from this by adding a divergent section tothe nozzle as shown below.

Figure 3.11 Ideal ramjet with a fully expanded nozzle.

P

x

a a

aa

b

b

M0 = 3

1 e

M = 1

1.5 3 4

3

1x

M = 0.138T0=250Tt0=700

M = 1+) Tt4=2100

7

M

Pe=P0

The ideal ramjet cycle

3/21/11 3.12 bjc

The area ratio of the nozzle is chosen so that the flow is fully expanded, . The stag-nation pressure is constant through the engine and so we can conclude from

(3.26)

that . The temperature ratio is

(3.27)

Finally, the thrust equation is

(3.28)

Adding a divergent section to the nozzle at this relatively high Mach number increases the thrustby 50%. Now work out the other engine parameters. The fuel/air ratio is determined from

(3.29)

Assume the fuel added is JP-4 with . Equation (3.29) becomes

(3.30)

Pe P0=

PeP0------

PtePt0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

"" 1------------

=

1 11 " 12------------# $

% & M02+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

"" 1------------

=

Me M0=

T eT 0------

T teT t0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

T teT t0-------- 3= = =

TP0 A0------------- "M0

2 MeM0--------

T eT 0------ 1

# $' (% & 1.4 9( ) 3 1( ) 9.22= = =

m f h f ma m f+( )ht4 maht3=

h f 4.28 107J kg=

f

T t4T t3-------- 1

h fC pT t3----------------

T t4T t3--------

-------------------------------2100700------------ 1

60.8 2100700----------------------------------------- 0.0346= = =

The ideal ramjet cycle

bjc 3.13 3/21/11

The relatively small value of fuel/air ratio is the a posteriori justification of our earlier neglectof the fuel mass flow compared to the air mass flow. If we include the fuel/air ratio in the thrustcalculation (but still ignore the effect of mass addition on the stagnation pressure change acrossthe burner) the result is

. (3.31)

The error in the thrust is about 6% when the fuel contribution is neglected. The dimensionlessspecific impulse is

(3.32)

and the overall efficiency is ( )

. (3.33)

The propulsive efficiency is

. (3.34)

The thermal efficiency of the engine shown in Figure 3.11 can be expressed as follows

. (3.35)

The heat rejection is accomplished by mixing of the hot exhaust stream with surrounding air atconstant pressure. Noting (3.27) for the ideal ramjet the last term in brackets is one and thethermal efficiency becomes

TP0 A0------------- "M0

2 1 f+( )MeM0--------

T eT 0------ 1

# $' (% & 1.4 9( ) 1.0346( ) 3 1( ) 9.98= = =

Ispga0

---------- 1f---# $% & 1

"M0-----------# $% & T

P0 A0------------- 9.980.0346 1.4 3---------------------------------------- 68.7= = =

+ f h f C pT 0( ) 170.3= =

,ov" 1"

------------# $% & 1

f + f---------# $% & T

P0 A0-------------# $% & 0.4

1.4-------# $% & 9.22

0.0346 170.3------------------------------------# $% & 0.447= = =

,pr2U0

Ue U0+--------------------- 2

1 3+----------------- 0.732= = =

,thma m f+( )

Ue2

2--------- maU0

2

2----------m f h f

--------------------------------------------------------------ma m f+( ) hte he( ) ma ht0 h0( )

ma m f+( )hte maht0-------------------------------------------------------------------------------------------= =

,th 1Qrejected during the cycleQinput during the cycle

------------------------------------------------------ 1ma m f+( )he mah0

ma m f+( )hte maht0--------------------------------------------------------= =

,th 1T 0T t0--------

1 f+( )T eT 0------ 1

1 f+( )T teT t0-------- 1

-------------------------------------

- ./ // /0 1/ // /2 3

=

Optimization of the ideal ramjet cycle

3/21/11 3.14 bjc

. (3.36)

For the ramjet conditions of this example the thermal efficiency is . The Brayton cycle effi-ciency is

. (3.37)

In the ideal cycle approximation the Mach number at station 3 is very small thus andthe thermal and Brayton efficiencies are identical. Note that, characteristically for a Braytonprocess, 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. Infact the final design is what we would call the ideal ramjet. The ideal cycle will be the basis forcomparison with other engine cycles but it is not a practically useful design. The problem isthat the inlet is extremely sensitive to small disturbances in the engine. A slight increase inburner exit temperature or decrease in nozzle exit area or a slight decrease in the flight Machnumber will cause the inlet to unstart. This would produce a strong normal shock in front of theengine and a large decrease in air mass flow through the engine and a consequent decrease inthrust. A practical ramjet design for supersonic flight requires the presence of a finite amplitude inletshock for stable operation.

3.4 OPTIMIZATION OF THE IDEAL RAMJET CYCLEFor a fully expanded nozzle the thrust equation reduces to

. (3.38)

For the ideal cycle where , and the thrust equationusing is written

. (3.39)

This form of the thrust equation is useful because it expresses the thrust in terms of cycle param-eters that we can rationalize. The parameter is fixed by the flight Mach number. At a givenaltitude is determined by maximum temperature constraints on the hot section materials ofthe engine as well as fuel chemistry and gas dissociation. If the flight Mach number goes to zero

,thideal ramjet 1T 0T t0--------=

2 3

,B 1T 0T 3------=

T 3 T t0*

TP0 A0------------- "M0

2 1 f+( )MeM0--------

T eT 0------ 1

# $' (% &=

Pte Pto= Me M0= T te T t0 T e T 0=1 f+ + f +r( ) + f +4( )=

TP0 A0------------- 2"

" 1------------ +r 1( )+ f +r+ f +4------------------

+4+r----- 1

# $' (% &=

+r+4

Optimization of the ideal ramjet cycle

bjc 3.15 3/21/11

the thrust also goes to zero. As the flight Mach number increases for fixed the fuel flow mustdecrease until when fuel shut-off occurs and the thrust is again zero. A typical thrustplot is shown below.

Figure 3.12 Ramjet thrust

The optimization question is; at what Mach number should the ramjet operate for maximumthrust at a fixed ? Differentiate (3.39) with respect to and set the result to zero.

. (3.40)

The value of for maximum thrust is determined from

. (3.41)

+4

+4 +r=

TP0 A0-------------

+r

+4 8.4=+ f 170=" 1.4=

fuel shut-off

+4 +r

+r55 T

P0 A0-------------# $% & =

2"" 1------------

+4+r 1 3+r 2+4+r-----+ +r

# $' (% &

# $' (% &

+ f +4 +4+r 2+r2 +4

+r-----+

# $' (% &+

+ f +4( )+r2 +4

+r-----

----------------------------------------------------------------------------------------------------------------------------------------

# $' (' (' (' (' (% &

0=

+r

+4+r+ f

----------- 1 3+r 2+4+r-----+ +r

# $' (% &

+4 +4+r 2+r2 +4

+r-----+

# $' (% &+

Optimization of the ideal ramjet cycle

3/21/11 3.16 bjc

The quantity is quite large and so the second term in parentheses in (3.41) clearly dominatesthe first term. For the maximum thrust Mach number of a ramjet is found from

. (3.42)

For the case shown above with the optimum value of is corresponding to aMach number of 3.53. The ramjet is clearly best suited for high Mach number flight and theoptimum Mach number increases as the maximum engine temperature increases.The specific impulse of the ideal ramjet is

. (3.43)

The specific impulse also has an optimum but it is much more gentle than the thrust optimumas shown below.

Figure 3.13 Ramjet specific impulse

+ ff 1

+41 2 2 +rmax thrust( )

3 2

+rmax thrust 1+---------------------------------------=

+4 8.4= +r 3.5

Ispga0

----------2

" 1------------ +r 1( )# $% &

1 2

+4 +r+ f +4------------------

-----------------------------------------------+ f +r+ f +4------------------

+4+r----- 1

# $' (% &=

+r

Ispga0

----------

+4 8.4=+ f 170=" 1.4=

fuel shut-off

The non-ideal ramjet

bjc 3.17 3/21/11

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 shut-off.

3.5 THE NON-IDEAL RAMJETThe major non-ideal effects come from the stagnation pressure losses due to the inlet shock andthe burner heat addition. We have already studied those effects fairly thoroughly. In additionthere are stagnation pressure losses due to burner drag and skin friction losses in the inlet andnozzle where the Mach numbers tend to be quite high. A reasonable rule of thumb is that thestagnation pressure losses due to burner drag are comparable to the losses due to heat addition.

3.6 RAMJET CONTROLLets examine what happens when we apply some control to the ramjet. The two main controlmechanisms at our disposal are the fuel flow and the nozzle exit area. The engine we will usefor 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 sensitiveto unstart if the free stream conditions, burner temperature or nozzle area change. Changes areassumed to take place slowly so that unsteady changes in the mass, momentum and energy con-tained in the ramjet are negligible.

Figure 3.14 Ramjet control model.

The mass balance is

M0 = 3

1e

Me = 1

1.5 3 4

3

1x

shock

T0=250Tt0=700

M > 1 Tt4=2100

M

m f Ae

Ramjet control

3/21/11 3.18 bjc

. (3.44)

The pressure in the engine is likely to be very high at this free stream Mach number and so thenozzle is surely choked and we can write

. (3.45)

The thrust equation is

. (3.46)

Our main concern is to figure out what happens to the velocity ratio and pressure ratio as wecontrol the fuel flow and nozzle exit area. Nozzle exit area controlFirst, suppose is increased with constant. In order for (3.45) to be satisfied mustdrop keeping 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

remains constant. Note that the thrust decreases. This can be seen by writing the secondterm in (3.45) as

(3.47)

The left term in (3.47) is constant but the right term increases leading to a decrease in thrust. If 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 numberof the inlet throat.

me"

" 1+2-------------# $

% &

" 1+2 " 1( )--------------------

-------------------------------------Pte Ae"RT te

-------------------# $' (% & f Me( ) 1 f+( )ma= =

ma1

1 f+( )------------------"

" 1+2-------------# $

% &

" 1+2 " 1( )--------------------

-------------------------------------Pte Ae"RT te

-------------------# $' (% &=

TP0 A0------------- "M0

2 1 f+( )UeU0------- 1

# $' (% & Ae

A0------

PeP0------ 1# $' (% &+=

Ae T te PtePte Ae

Pe Ae

AeA0------

PeP0------

AeA0------

Ae

Example - Ramjet with unstarted inlet

bjc 3.19 3/21/11

Fuel flow controlNow, suppose is decreased with constant. In order for (3.45) to be satisfied, mustdrop keeping constant. Once again the shock moves downstream to a higher shockMach number. The velocity ratio goes down since the exit stagnation temperature is down andthe Mach numbers do not change.The pressure ratio also decreases since the exit stagnationpressure is down. The thrust clearly decreases in this case. If is increased, the reverse hap-pens, the inlet operates more efficiently and the thrust goes up. The amount by which thetemperature can be increased is again limited by the Mach number of the inlet throat.

3.7 EXAMPLE - RAMJET WITH UNSTARTED INLETFor simplicity, assume constant heat capacity with , M2/(sec2-K).The gas constant is M2/(sec2-K). The ambient temperature and pressure are

and . The fuel heating value is

. The sketch below shows a ramjet operating at a free stream Machnumber of 3.0. A normal shock stands in front of the inlet. Heat is added between 3 and 4

and the stagnation temperature at station 4 is . Relevant areas are , and . Determine the dimensionless thrust

. Do not assume f

Example - Ramjet with unstarted inlet

3/21/11 3.20 bjc

. (3.48)

For constant heat capacity

. (3.49)

Now we need to determine the flow batween stations 1 and 3. To get started we will neglectthe fuel addition for the moment. Knowing the Mach number at 4 and the stagnation tem-peratures at 3 and 4 we can use Rayleigh line results to estimate the Mach number at station3. The stagnation temperature ratio across the burner is

. (3.50)

The Rayleigh line tables give

. (3.51)

This is a reasonable approximation to the Mach number at station 3. The stagnation pres-sure ratio across the burner is

. (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 presureratio between the inlet throat and the exit can be determined from a mass balance betweenstations 1.5 and e.

. (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

m f h f ma m f+( )hte maht0=

fT t4 T t0 1

h f C pT t0 T t4 T t0------------------------------------------------------ 2000 604.8 1

4.28 107 1005 604.8( ) 2000 604.8-------------------------------------------------------------------------------------------------------- 0.0344= = =

T t4T t3--------

T t4T t

*--------M 0.2=

T t*

T t3--------

M ?=

2000604.8------------- 3.3069= = =

T t4T t3-------- 0.2066

T t*

T t3--------

M ?=

2000604.8------------- 3.3069= = =

T t*

T t3--------

M ?=

3.30690.2066---------------- 16.006 M36 0.103= = =

Pt4Pt3--------

Pt4Pt

*--------M 0.2=

Pt*

Pt3--------

M 0.103=

1.2351.258------------- 0.981= = =

Pt1.5 A1.5 1 f+( )

T t1.5-----------------------------------------

Pte AeT te

-------------- Pte

Pt1.5-----------6 1.0344( )38------------------------

2000604.8------------- 0.7054= = =

Example - Ramjet with unstarted inlet

bjc 3.21 3/21/11

. (3.54)

Thus far the ramjet flow looks as follows

Figure 3.15 State I

The stagnation pressure ratio across the external shock is(3.55)

and so the overall stagnation pressure ratio is

. (3.56)

The static pressure ratio is

. (3.57)

The temperature ratio is

7shock0.70540.981---------------- 0.719 Mshock6 2.004= = =

T0P0 1 3 4e

M0 = 3

shock

A1.5 P0

Me 1=

M4 0.1975=M3 0.103=M1.5 1=M1 0.0725=

M 1

Mbehind shock 0.475= Minlet shock 2.004=

T t0 604.8= T te 2000=

Pt1 Pt0 M 3=0.3283=

PtePt0--------

Pt1Pt0--------

PtePt1.5---------- 0.3283 0.7054 0.2316= = =

PeP0------

PtePt0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

"" 1------------

0.2316 2.81.2-------# $% &

3.54.494= = =

Example - Ramjet with unstarted inlet

3/21/11 3.22 bjc

. (3.58)

The velocity ratio is

(3.59)

Across the inlet the mass balance is(3.60)

and so(3.61)

Finally the thrust is

(3.62)

and

(3.63)

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 samesince it is determined by the choking at the nozzle exit and the fixed enthalpy rise acrossthe burner. The mass balance between the inlet throat and the nozzle exit is again

(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 is increased, the equality(3.64) is maintained and the inlet shock moves to the left increasing . At the point wherethe the inlet throat unchokes the shock is infinitely weak and the only stagnation pressureloss between station 1.5 and the nozzle exit is across the burner.

T eT 0------

T teT t0--------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

2000604.8-------------

2.81.2-------# $% & 7.72= = =

Ue U0 Me M0( ) T e T 013--- 7.72 0.926= = =

Pt1.5 A1.5 Pt0 A0 f M0( )=

A0 A1.5 Pt1.5 Pt0 f M0( )( ) 0.3283 1 4.235( ) 1.39= = =

TP0 A1------------- "M0

2 A0A1.5----------# $' (% & A1.5

A1----------# $' (% & 1 f+( )

UeU0------- 1

# $' (% & Ae

A1------

PeP0------ 1# $' (% &+=

TP0 A1-------------

State I1.4 9( ) 1.39( ) 1 8( ) 1.0344( )0.926 1( ) 1 3( ) 4.494 1( )+ = =

0.0923 1.165+ 1.0724=

Pt1.5 A1.5 1 f+( )

T t1.5-----------------------------------------

Pte AeT te

--------------=

A1.5Pte

Example - Ramjet with unstarted inlet

bjc 3.23 3/21/11

(3.65)

which corresponds to

(3.66)

The mass flow through the engine has increased by the ratio

. (3.67)

At this condition the stagnation pressure ratio across the system is

(3.68)

The static pressure ratio is

(3.69)

The Mach number at station 1 increases as increases and at the condition where theinlet is just about to unchoke reaches the same Mach number as station 3. At this conditionthe ramjet flow field looks like

A1.5 State IIAe

-------------------------- 11 f+( )-------------------Pte State II

Pt1.5------------------------

T t1.5T te

---------- 0.9821.0344( )---------------------604.82000------------- 0.522= = =

A1A1.5 State II--------------------------

A1Ae------# $' (% & Ae

A1.5 State II--------------------------# $' (% & 3

0.522------------- 5.747= = =

ma State IIma State I----------------------

A1.5 State IIA1.5 State I-------------------------

A1.5 State IIA1

-------------------------# $' (% & A1

A1.5 State I------------------------# $' (% & 8

5.747------------- 1.392A0 State IIA0 State I----------------------= = = = =

Pte State IIPt0

------------------------Pt1.5Pt0----------# $' (% & Pte State II

Pt1.5------------------------# $' (% & 0.3283 0.982 0.3224= = =

Pe State IIP0

----------------------Pte State II

Pt0------------------------

1 " 12------------# $% & M0

2+

1 " 12------------# $% & Me

2+-------------------------------------

# $' (' (' (' (% &

"" 1------------

0.3224 2.81.2-------# $% &

3.56.256= = =

A1.5

Example - Ramjet with unstarted inlet

3/21/11 3.24 bjc

Figure 3.16 State II

The inlet shock is gone, the inlet Mach number has increased to and the externalshock has moved somewhat closer to the inlet. Note that the capture area to throat area ratiois still

(3.70)

although both and have increased. Also

(3.71)

The thrust formula is

(3.72)

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

(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

T0P0 1 3 4e

M0 = 3

shock

A1.5 P0

Me 1=

M4 0.1975=M3 0.103=M1.5 1=M1 0.103=

M 1

Example - Ramjet with unstarted inlet

bjc 3.25 3/21/11

Now, suppose is increased until . The ramjet flow field looks likethe following.

Figure 3.17 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

(3.74)

Therefore the thrust is the same as the thrust for State II, equation (3.73). 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 fullyFirst increase the exit area to the point where . If we maintain

the Mach number at station 4 becomes one and the Mach number between1 and 3 is, from the Rayleigh solution, . The ramjet flow at this con-dition is sketched below

A1.5 A1 A1.5 A3= =

T0P0 1 3 4e

M0 = 3

shock

P0

Me 1=

M4 0.1975=M3 0.103=M1 0.103=

M 1

Example - Ramjet with unstarted inlet

3/21/11 3.26 bjc

Figure 3.18 State IV

The velocity ratio is still

(3.75)

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

(3.76)

Across the whole system

(3.77)

and the static pressure ratio is

(3.78)

The area ratio is

(3.79)

The thrust formula for state IV is

T0P0 1 3 4 e

M0 = 3

shock

P0

Me 1=

M4 1.0=M3 0.276=M1 0.276=

M 1=M 1