t he r amjet c ycle , s cramjets

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 " 1–2

------------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 "M2+( )

2

2 1 "+( )M2 1 " 1–2

------------M2+# $% &

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

# $' (' (' (% &

=Pt

*

Pt------ 1 "M2+

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

" 1+2

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

1 " 1–2

------------# $% &M2+

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

# $' (' (' (% &

"" 1–( )

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

=

T 0 250°K=

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 decreases

to 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

1

x

M = 2T0=250

Tt0=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

Pt0

T t0------------ f 3( )

Pt4

T 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; is

up 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

1

x

M = 2T0=250

Tt0=700

Tt4=1070

M

T t4T t3--------

M4 1=

1.53=

Pt4Pt3--------

M4 1=# $' (' (% &

before unstart

0.292=

T t4

Pt4 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

1

x

M = 2

shock

M = 0.475T0=250

Tt0=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 let’s 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 let’s 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 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

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

3.55.66= = =

T eT 0------

T teT t0--------

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &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)

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

. (3.17)

The temperature ratio is

T t4 2100°K=

M0 = 3

1 e

Me = 1

1.5

M < 3

3 4

3

1

x

shock

M > 0.475T0=250

Tt0=700

Tt4=2100M = 1+)

M

f M( )

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

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

Pe P0 0.409 2.81.2-------# $% &

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 Let’s 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 " 1–2

------------# $% &M0

2+

1 " 1–2

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

1

x

M = 0.138T0=250

Tt0=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)

That’s 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 A1.5

700-------------------

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

1

x

M = 0.138T0=250

Tt0=700

M = 1+) Tt4=2100

7

M

Pe=P0


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 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

=

1 11 " 1–

2------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

=

Me M0=

T eT 0------

T teT t0--------

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &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= = =


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------------- 9.98

0.0346 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= = =

,th

ma m f+( )Ue

2

2--------- ma

U02

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 equation

using 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 given

altitude 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 ramjet1

T 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


bjc 3.15 3/21/11

the thrust also goes to zero. As the flight Mach number increases for fixed the fuel flow must

decrease 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-----–+

# $' (% &

+


3/21/11 3.16 bjc

The quantity is quite large and so the second term in parentheses in (3.41) clearly dominates

the 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

+ f

f 1«

+41 2

2 +rmax thrust( )

3 2

+rmax thrust1+

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

+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 CONTROLLet’s 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

1

x

shock

T0=250

Tt0=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 control

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

drop 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 Pte

Pte Ae

Pe Ae

AeA0------

PeP0------

AeA0------–

Ae

Example - Ramjet with unstarted inlet

bjc 3.19 3/21/11

Fuel flow control

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

drop 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 INLET

For 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<<1. Neglect stagnation pressure losses due to wall frictionand burner drag. Assume that the static pressure outside the nozzle has recovered to theambient value. Suppose can be increased until . By what propor-tion would the air mass flow change?

Solution - The first point to recognize is that the stagnation pressure at station 4 exceedsthe ambient by more than a factor of two - note the pressure outside the nozzle is assumedto have recovered to the ambient value. Thus the exit Mach number is one and the Machnumber at station 4 is . The stagnation temperature at station 3 is

. The fuel-air ratio is determined from the enthalpy balance

T te Ae Pte

Pte T te

T te

" 1.4= C p 1005 =

R 287=

T 0 216K= P0 2 104× N M2

=

h f 4.28 107 J/kg×=

T0P0 1 3 4e

M0 = 3

shock

A1.5 P0

T t4 2000K=

A3 A1.5 8= A1 A3 A4= = A4 Ae 3=

T P0 A1( )

A1.5 A1 A1.5 A3= =

M4 0.1975=

T t3 604.8°K=


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 t4

T 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--------

Pt4

Pt*

--------

M 0.2=

Pt*

Pt3--------

M 0.103=

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

Pt1.5A1.5 1 f+( )

T t1.5

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

Ae

T te

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

Pt1.5

-----------61.0344( )3

8------------------------ 2000

604.8------------- 0.7054= = =


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<M 1<M 1< M 1>

Mbehind shock 0.475= Minlet shock 2.004=

T t0 604.8= T te 2000=

Pt1 Pt0M 3=

0.3283=

PtePt0--------

Pt1Pt0--------

Pte

Pt1.5

---------- 0.3283 0.7054× 0.2316= = =

PeP0------

PtePt0--------

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

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

3.54.494= = =


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 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

2000604.8------------- 2.8

1.2-------# $% & 7.72= = =

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

Pt1.5A1.5 Pt0

A0 f M0( )=

A0 A1.5 Pt1.5Pt0

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 I

1.4 9( ) 1.39( ) 1 8( ) 1.0344( )0.926 1–( ) 1 3( ) 4.494 1–( )+ = =

0.0923– 1.165+ 1.0724=

Pt1.5A1.5 1 f+( )

T t1.5

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

Ae

T te

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

A1.5

Pte


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)


(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 t1.5

T 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------------- 1.392

A0 State IIA0 State I----------------------= = = = =

Pte State IIPt0

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

Pt0----------# $' (% & Pte State II

Pt1.5

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

0.3283 0.982× 0.3224= = =

Pe State IIP0

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

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

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

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

3.56.256= = =

A1.5


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<M 1<M 1<

Mbehind shock 0.475=

T t0 604.8= T te 2000=

M1 M3=

A0 State II A1.5 State II Pt1.5Pt0

f M0( )( ) 0.3283 1 4.235( ) 1.39= = =

A1.5 A0

A0 State II A1A0 State II A1.5 State II

A1 A1.5 State II------------------------------------------------------ 1.39

5.747------------- 0.242= = =

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

StateII

"M02 A0 State II

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

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

1 f+( )UeU0------- 1–

# $' (% & Ae

A1------

Pe State IIP0

--------------------- 1–# $' (% &

+=

TP0 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=


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 fully

First increase the exit area to the point where . If we maintain

the Mach number at station 4 becomes one and the Mach number between

1 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<M 1<


T t0 604.8= T te 2000=

A0 State III A1 A0 State II A1A0 State II A1.5 State II

A1 A1.5 State II------------------------------------------------------ 1.39

5.747------------- 0.242= = = =

A1 A3 A4 Ae= = =

T t4 2000K=

M1 M3 0.276= =


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<


T t0 604.8= T te 2000=

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

PtePt1--------

State IV

0.8278=

Pte State IVPt0

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

Pt0----------# $' (% & Pte State IV

Pt1.5

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

0.3283 0.8278× 0.2718= = =

Pe State IVP0

-----------------------Pte State IV

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

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

0.2718 2.81.2-------# $% &

3.55.274= = =

A0 State IV A1Pt1

f M1( )( )

Pt0f M0( )( )

------------------------------ 0.3283 0.45580.2362----------------× 0.634= = =


bjc 3.27 3/21/11

(3.80)

which evaluates to

(3.81)

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

(3.82)

which accounts for much of the increased thrust in spite of the increase in stagnation pres-sure loss across the burner.

State V - Reduce the burner outlet temperature until the shock is very close to station 1

Now reduce until the Mach number at 3 matches the Mach number behind the shock.

From the Rayleigh solution, this occurs when . At this condition the ramjetflow field looks like

Figure 3.19 State V

The static temperature ratio is

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

StateIV

"M02 A0 State IV

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

1 f+( )UeU0------- 1–

# $' (% & Ae

A1------

Pe State IVP0

----------------------- 1–# $' (% &

+=

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

StateIV

1.4 9( ) 0.634( ) 1.0344( )0.926 1–( ) 1( ) 5.274 1–( )+=

0.3367– 4.274+ 3.937=

ma State IVma State III------------------------

A0 State IVA1

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

A0 State III------------------------# $' (% & 0.634

0.242------------- 2.62= = =

T te

T te 924.8=

T0P0 1 3 4 e

M0 = 3

shock

P0

Me 1=

M4 1.0=M3 0.475=

M 0.475= M 1=


T t0 604.8= T te 924.8=


3/21/11 3.28 bjc

(3.83)

The velocity ratio at this temperature is

(3.84)

The stagnation pressure ratio across the burner at this condition is

(3.85)

and across the system

(3.86)


(3.87)

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

(3.88)

State VI - Reduce the burner temperature slightly to establish supersonic flow up to theburner.

If the temperature at station 4 is decreased by an infinitesimal amount then supersonic flowwill be established through the engine. Finally the flow is

T eT 0------

T teT t0--------

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

924.8604.8------------- 2.8

1.2-------# $% & 3.57= = =

Ue U0 Me M0( ) T e T 013--- 3.57 0.630= = =

PtePt3-------- 0.889=

PtePt0-------- 0.889 0.3283× 0.292= =

PeP0------

PtePt0--------

1 " 1–2

------------# $% &M0

2+

1 " 1–2

------------# $% &Me

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

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

"" 1–------------

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

3.55.67= = =

A0 A1 Ae= =

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

2 UeU0------- 1–# $' (% & Pe

P0------ 1–# $' (% &

+ = =

1.4 9( ) 0.629 1–( ) 5.67 1–( )+ 4.67– 4.67+ 0= =

Very High Speed Flight - Scramjets

bjc 3.29 3/21/11

Figure 3.20 State VI

The mass flow, velocity ratio, pressure ratio and thrust all remain the same as in state V. Ifwe 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 - SCRAMJETSAs 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 relatedto the very high stagnation temperatures of high Mach number flow. To get started let’srecall the thrust equation for the ramjet with a fully expanded nozzle.

(3.89)

Define the thrust coefficient as

(3.90)

If we assume ideal behavior ( ) the thrust coefficient becomes

. (3.91)

When we carried out the energy balance across the burner,

(3.92)

T0P0 1 3 4 e

M0 = 3

shock

P0

Me 1=M 3= M 1=

T t0 604.8= T te 924.8=

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

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

T eT 0------ 1–

# $' (% &

=

CthrustThrust

12---!0U0

2 A0

------------------------ T"2---M0

2 P0 A0( )------------------------------= =

Me M0=

Cthrust 2 1 f+( )T teT t0-------- 1–

# $' (% &

=

m f h f ma m f+( )ht4 maht3–=


3/21/11 3.30 bjc

we assumed that the fuel enthalpy was simply added to the flow without regard to the chem-istry of the process. In fact the chemistry highly limits the range of fuel-air ratios that arepossible. The stoichiometric reaction of JP-4 with air where the fuel and oxygen are com-pletely consumed is

. (3.93)

corresponding to a fuel-air ratio

. (3.94)

This is roughly the value of the fuel-air ratio that produces the maximum outlet temperaturefrom the burner. If we choose the much more energetic hydrogen fuel the reaction is

(3.95)

corresponding to a fuel-air ratio

. (3.96)

The fuel enthalpies are generally taken to be

(3.97)

In the earlier discussion we took the perspective that the ramjet cycle was limited by a red linetemperature in the hot part of the engine. Let’s relax this assumption and allow the maximumtemperature to be free while keeping the fuel-air ratio constant. Furthermore let’s continue toretain the assumption of constant heat capacities even at high Mach numbers. We will correctthis eventually in Chapter 9, but for now we just want to see what happens to the ideal thrustcoefficient (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 canexpress the stagnation temperature ratio across the engine as

. (3.98)

where we recall that . Now the thrust coefficient becomes

CH1.94 1.485O2 5.536N2+ CO2 0.97H2O 5.536N2+ +8+

f 12.01 1.94 1.008×+( )1.485 32.00 5.536 28.02×+×( )

------------------------------------------------------------------------------- 0.0689= =

H2 0.5O2 1.864N2+ H2O 1.864N2+8+

f 2 1.008×16.00 1.864 28.02×+( )

----------------------------------------------------------- 0.0295= =

h f JP 4–4.28 107J kg×=

h f H212.1 107J kg×=

T teT t0--------

f + f1 f+--------------# $% & 1

1 " 1–2

------------# $% &M0

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

# $' (' (' (% &

11 f+( )

-------------------+=

+ f h f C pT 0( )=


bjc 3.31 3/21/11

. (3.99)

The temperature ratio and thrust coefficient are plotted below.

Figure 3.21 Temperature ratio of an ideal ramjet at constant fuel-air ratio.

Figure 3.22 Thrust coefficient of an ideal ramjet at constant fuel-air ratio.

Cthrust 2f 1 f+( )+ f

1 " 1–2

------------# $% &M0

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

# $' (' (' (% & 1 2

1–

# $' (' (' (% &

=

2.5 5 7.5 10 12.5 15 17.5 20

2

4

6

8

10

12

14

M0

T 0 216K=

" 1.4=H2

JP 4–

1 1

T te T t0

JP 4 real chemistry–

5 10 15 20

1

2

3

4

5

M0

T 0 216K=

" 1.4=

Cthrust

H2

JP 4–


3/21/11 3.32 bjc

The drag coefficient is

(3.100)

At high Mach numbers the drag coefficient of a body tends toward a constant value. For a spherethe drag coefficient tends toward a constant somewhat less than one, about 0.75. More stream-lined bodies have lower drag and coefficients as low as 0.2 may be feasible with adequatedevelopment. This observation together with the above figures indicates that as the Mach num-ber increases it becomes harder and harder to produce thrust that exceeds drag. The thrustcoefficient drops below one at a Mach number of about 7 to 8 for both fuels. As the Mach num-ber increases a conventional ramjet (even an ideal one) simply cannot produce enough thrust toovercome drag. The limiting thrust coefficient at infinite Mach number is

. (3.101)

This last result suggests that a scramjet can benefit from the choice of a fuel-rich mixture ratioas long as it does not exceed the flammability limit of the fuel. This would also provide addi-tional fuel for cooling the vehicle. An advantage of a hydrogen system is that it can operatequite fuel rich. In addition hydrogen has a higher heat capacity than any other fuel enabling itto be used to provide cooling for the vehicle that, in contrast to a re-entry body, has to operatein a very high temperature environment for long periods of time. The down side of hydrogenis that liquid hydrogen has to be stored at very low temperatures and the liquid density is onlyabout 1/10 of that of JP-4. It is clear that small effects can be important. For example when wedeveloped the thrust formula we neglected the momentum of the injected fuel. In a realisticscramjet 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 thecombustion of JP-4 with air. Notice that the temperatures reached by the combustion gases issubstantially less than the ideal and for high Mach numbers the temperature rise from the reac-tion 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 theburner operating with a supersonic Mach number on the order of two or so thereby cutting thestatic temperature of the air into the combustor by almost a factor of two. This is the conceptof a supersonically burning ramjet or scramjet. A generic sketch of such a system is shown inFigure 3.23. Two Homework problems are used to illustrate basic concepts.

CdragDrag

12---!0U0

2 A0

------------------------ D"2---M0

2 P0 A0( )------------------------------= =

CthrustM0 98lim f=


bjc 3.33 3/21/11

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 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 operating envelope of a scramjet.

Mach number


3/21/11 3.34 bjc

This figure somewhat accurately illustrates the challenges of scramjet flight. These can be listedas follows.

1) At high Mach numbers the vehicle is enveloped in an extremely high temperature gas for along period of time perhaps more than an hour.

2)At high altittude, combustion is hard to sustain even at high Mach numbers because of thelow 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 thepoint 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 pre-sented as a line of constant dynamic pressure coinciding with increasing Mach number andaltitude. But in supersonic flow the free stream dynamic pressure is really not a sufficient mea-sure of the actual loads that are likely to act on the vehicle. Again let’s make a constant heatcapacity assumption and compare the free stream stagnation pressure to the free streamdynamic pressure as follows. Form the pressure coefficient

. (3.102)

This function is plotted below.

Figure 3.25 Comparison of stagnation pressure and dynamic pressure in super-sonic flow.

If one repeats this calculation with real gas effects the stagnation pressures one calculates areeven larger because of reduced values of .

Pt9 P9–q9

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

P9 1 " 1–2

------------# $% &M9

2+# $% &

"" 1–------------

1–# $' (' (% &

"2---P9M9

2------------------------------------------------------------------------------

1 " 1–2

------------# $% &M9

2+# $% &

"" 1–------------

1–# $' (' (% &

"2---M9

2----------------------------------------------------------------------= =

Pt9 P9–q9

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

M

"

Problems

bjc 3.35 3/21/11

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 num-ber at the entrance to the combustor the combuster will experience a pressure 218 times the freestream dynamic pressure. For example, at an altitude of about 26 kilometers and a low Mach number combuster would operate at about 3200 psia (10,000 times the ambientatmospheric pressure at that altitude). This would require a very heavy structure making thewhole idea impractical.

If instead, the flow Mach number entering the combuster can be maintained at about Mach 2then 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 impor-tant as the thermochemistry issue in forcing the designer to consider operating the combusterat supersonic Mach numbers in order to attain hypersonic flight.

3.9 PROBLEMSProblem 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.

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 and . The Mach number at station1, is subsonic and the pressure on the annular step is approximately equal to . Theheat of reaction is denoted by Q. (i) Show that the exit Mach number is given by

. (3.103)

M9 8.0=

1 2

Pt1 T t1M1 P1

M22 1 " 1–

2------------M2

2+# $% &

1 "M22+( )

2--------------------------------------------- 1 Q

C pT t1----------------+# $

% &M1

2 1 " 1–2

------------M12+# $

% &

A2A1------ "M1

2+# $' (% & 2

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

Problems

3/21/11 3.36 bjc

(ii) Show that when and goes to infinity

. (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).

. (3.105)

Problem 3 - The sketch below shows a ramjet operating at a free stream Mach numberof 0.7. Heat is added between stations 3 and 4 and the stagnation temperature at sta-tion 4 is . The Mach number at station 3 is very low. The ambient

temperature and pressure are and . Assume that f<<1.

Using appropriate assumptions estimate the dimensionless thrust and the

area ratio .

Problem 4 - The sketch below shows a ramjet operating at a free stream Mach number ( ) 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. The Mach number at station 3 is very low. There is no shock in the

inlet. The ambient temperature and pressure are and .

Q 0= A2 A1

Pt1Pt2-------- 1 " 1–

2------------M1

2+# $% &

"" 1–------------

=

,th" 1–

2------------M0

2# $% & 1 " 1–

2------------M0

2+# $% &=

T t4 1000K=

T 0 216K= P0 2 104× N M2

=

13 4

eP0 T0

M0 = 0.7

T P0 A0( )

A0 Ae

M0 1.5= f 1.5( ) 0.8502=

T t4 1400K=

T 0 216K= P0 2 104× N M2

=

Problems

bjc 3.37 3/21/11

Using appropriate assumptions estimate the dimensionless thrust and the

area ratio .

Problem 5 - A ramjet operates at a freestream Mach number of 3. The inlet is a straightduct and the Mach number of the flow entering the burner at station 3 is three. Heat is addedacross the burner such that the Mach number of the flow exiting the burner at station 4 is

two. The ambient temperature and pressure are and .

Assume that .

The exit flow is expanded to . Determine the dimensionless thrust, .

Problem 6 - A ramjet operates at a freestream Mach number of 2.5. The ambient temper-

ature and pressure are and . The engine operates with a

straight duct after the burner, . Supersonic flow is established at the entranceof the inlet and a normal shock is stabilized somewhere in the diverging part of the inlet.The stagnation temperature exiting the burner is . Neglect wall friction.

Determine the dimensionless thrust, .

13 4

eP0 T0

M0 = 1.5

shock

T P0 A0( )

A0 Ae

T 0 216K= P0 2 104× N M2

=

f 1«1 3 4 e

M3 = 3 M4 = 2

Pe P0= T P0 A0

T 0 216K= P0 2 104× N M2

=

A1 Ae 1=

T t4 1800K=

T P0 A0

Problems

3/21/11 3.38 bjc

Problem 7 - The figure below shows a ramjet test facility. A very large plenum containsAir at constant stagnation pressure and temperature, . Jet fuel

( ) is added between stations 3 and 4 ( ) where combustiontakes place. The flow exhausts to a large tank which is maintained at pressure . Let

. The upstream nozzle area ratio is . The exit area, canbe varied in order to change the flow conditions in the engine. The gas temperature in theplenum is . To simplify the analysis, assume adiabatic flow, neglect wallfriction and assume constant specific heat throughout with .

Initially, the valve controlling is closed, and the fuel mass flow is shut off.Consider a test procedure where the nozzle area is opened, then closed. In the process

is slowly increased from zero causing Air to start flowing. The nozzle is openeduntil . Then the nozzle area is slowly reduced until once again.Plot the thrust normalized by the plenum pressure and upstream throat area,

T0

P01 3 4 e

M0 = 2.5

shock

A1t

Pt0 , Tt0

h f 4.28 107 J/kg×= A3 A4=P0

Pt0 P0 100= A3 A1.5 8= Ae

T t0 805.2K=" 1.4=

Tt0

Pt0

P0

1.5 3e

4

m f

ma

Ae Ae 0=

Ae A1.5

Ae A3= Ae A1.5 0=T Pt0 A1.5( )

Problems

bjc 3.39 3/21/11

and the fuel-Air ratio as a function of . Distinguish points corresponding toincreasing and decreasing . The fuel flow is adjusted to maintain the stagnation temper-ature at 4 at a constant value. Plot the results for three cases.

(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 . What flight Mach number andaltitude are being simulated at this condition?

Problem 8 - A ramjet operates at a freestream Mach number of 3. The area ratio across theengine is . Supersonic flow is established at the entrance of the inlet and anormal shock is stabilized in the diverging part of the inlet. The inlet throat Mach numberis 1.01. The stagnation temperature exiting the burner is . Assume ,

M2/(sec2-°K), M2/(sec2-°K). The ambient temperature and pres-

sure are and . Assume throughout that .

i) Suppose the fuel flow is increased with the geometry of the engine held fixed. Estimatethe value of that would cause the inlet to unstart. Plot the dimensionlessthrust, , specific impulse and overall efficiency of the engine as a function of .Plot your result beyond the point where the inlet unstarts and assume the engine does notflame out. Assume the fuel flow is throttled so as to keep the fuel/air ratio constant. Notethat the thrust is normalized by a fixed geometric area rather than the capture area thatchanges when the engine unstarts.

ii) With suppose is reduced keeping the fuel flow the same. Estimate

the value of that would cause the inlet to unstart? Plot the dimensionless

thrust, specific impulse and overall efficiency of the engine as a function of

. As in part (i) plot your result beyond the point where the inlet unstarts and assumethe engine does not flame out.

f Ae A1.5

Ae

T t4 805.2K (zero fuel flow), T t4 1200K T t4, 2700K= = =

f 1«

A1 Ae 2=

T t4 1944K= " 1.4=

R 287= C p 1005 =

T 0 216K= P0 2 104× N M2

= f 1«

T0

P01 3 4

eM0 = 3

shock

A1.5

fT P0 A1.5 f

A0

T t4 1944K= Ae

A1 Ae

T P0 A1.5

A1 Ae

Problems

3/21/11 3.40 bjc

Problem 9 - A ramjet operates at a freestream Mach number of 3. The area ratio acrossthe engine is . Supersonic flow is established at the entrance of the inlet anda normal shock is stabilized in the diverging part of the inlet. The stagnation temperatureexiting the burner is .The ambient temperature and pressure are

and . Do not assume that . Do not neglect theeffects of wall friction.

i) Determine the fuel-air ratio, .

ii) Determine the dimensionless thrust, .

Problem 10 -Assume , M2/(sec2-°K), M2/(sec2-°K). The

ambient temperature and pressure are and . The sketchbelow 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.

A1 Ae 2=

T t4 1944K=

T 0 216K= P0 2 104× N M2

= f 1«

T0

P01 3 4

eM0 = 3

shock

A1.5

f

T P0 A0

" 1.4= R 287= C p 1005 =

T 0 216K= P0 2 104× N M2

=

0 1 3 4 e

M0 = 3.0M3 = 2.0 M4 = 1.0

Problems

bjc 3.41 3/21/11

Heat is added between 3 and 4 bringing the Mach number at station 4 to one. The flow isthen ideally expanded to . This type of engine with the combustion of fuel occur-ring in a supersonic stream is called a SCRAMJET (supersonic combustion ramjet).Determine the dimensionless thrust . Assume f<<1 if you wish and neglect stag-nation pressure losses due to friction.

Problem 11 - A ramjet operates in the upper atmosphere at a high supersonic Mach numberas shown in the sketch below. Supersonic flow is established at the entrance of the inlet anda normal shock is stabilized in the diverging part of the inlet. The stagnation temperatureexiting the burner is sufficient to produce substantial positive thrust. A small flat plate isplaced downstream of the burner as shown causing a small drop in stagnation pressurebetween station “a” and station “b”. The plate can be positioned so as to produce high dragas shown or it can be rotated 90° so that the long dimension is aligned with the flow pro-ducing lower drag. The engine operates with a convergent-divergent nozzle. The nozzlethroat is choked and the nozzle exit is fully expanded.

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) 5)

2) 6)

3) 7)

4) 8) Engine thrust

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

Pe P0=

T P0 A0( )

1 3 4 e

M = 3

shock

A1.5 8a b

M3 T e

Ma Pe

Mb Ue

Me

Problems

3/21/11 3.42 bjc

Problem 12 - A ramjet operates at a freestream Mach number of 3. The area ratio acrossthe engine is . Supersonic flow is established at the entrance of the inlet anda normal shock is stabilized in the diverging part of the inlet. The stagnation temperatureexiting the burner is . The ambient temperature and pressure are and

.

Do not neglect wall friction.

i) Determine the dimensionless thrust, .

ii) Suppose the wall friction is increased slightly. Determine if each of the followingincreases, 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

Problem 13 - A ramjet operates at a freestream Mach number of 3. The area ratio acrossthe engine is . Supersonic flow is established at the entrance of the inletand a normal shock is stabilized in the diverging part of the inlet. The stagnation temper-ature exiting the burner is . The nozzle is fully expanded . The fuel

enthalpy is . The ambient temperature and pressure are and

.

i) Determine the dimensionless thrust, .

A1 Ae 2=

T t4 1512K= T 0 216K=

P0 2 104× N M2

=

T0

P01 3 4

eM0 = 3

shockA1.5

Ae

T P0 A0

T P0 A0

A1 A8 1.75=

T t4 1512K= Pe P0=

h f 4.28 107 J/kg×= T 0 216K=

P0 2 104× N M2

=

1 3 4 e

M0 = 3

shockA1.5 8

T P0 A0

Problems

bjc 3.43 3/21/11

ii) Suppose 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

T t4

Pe

Problems

3/21/11 3.44 bjc

t he r amjet c ycle , s cramjets

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