Lecture - Supersonic

8/2/2019 Lecture - Supersonic

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September 2009

FLUIDS II

(305-430)

SUPERSONIC JETS AND DIFFUSERS

1. Supersonic JetsThe discharge of a jet into the atmosphere is an interesting phenomenon, particularly for the case of asupersonic jet. For a subsonic jet, the exit pressure of the jet pe must necessarily be adjusted to match the

atmospheric pressure pa. Ifpe > pa, the jet is underexpanded and hence the jet continues to expand. In asubsonic flow, the pressure increases as the flow expands as the streamline diverges (i.e. a diffuser). Thus the

jet pressure increases further and will not be able to adjust to the environment. If the jet is overexpanded and

pe < pa, then upon exit, the jet streamlines will converge. In a subsonic flow, the pressure decreases in aconverging flow (i.e. a nozzle). Thus upon exit, the jet pressure continues to decrease and again cannot match

the environment. Hence, in a subsonic jet, the exit pressure must necessarily be the same as the environment

i.e.pe = pa. This condition is achieved by internal adjustments of the flow during the transient starting process

Compression or expansion waves can propagate upstream in a subsonic flow and adjust the flow field prior tothe jet exit in order to satisfy the condition ofpe = pa.

For a supersonic jet, the phenomenon is more complex. Consider a deLaval nozzle connected to a reservoir offixed stagnation conditions. Let us connect the nozzle exit to a large vacuum tank in which the pressure can be

varied. If the tank pressure is above the value required to cause choking at the throat, then the flow in the

nozzle will be subsonic throughout. The pressure variation in the nozzle would correspond to the area variationi.e. the pressure decreases first in the converging section (i.e. nozzle) and then increases in the diverging section

(i.e. diffuser). The pressure variations for subsonic flow in the nozzle are shown by curves c in the figure

below.


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When the tank pressure is reduced to some critical value, the flow will be choked at the throat. The pressure a

the throat is p* and the ratio 528.0*

0

p

p(for 4.1 ) when choking occurs. Downstream of the throat, the

flow decelerates toM< 1 and the pressure then increases in the subsonic diffuser to match the tank pressure atthe nozzle exit (curve b).

If the tank pressure is further reduced after choking had occurred at the throat, no adjustment of the flow can be

made upstream of the throat because M = 1 and the expansion waves are trapped there. The flow thencontinues to expand to supersonic speeds downstream of the throat. However, in order to match the jet exi

pressure to the tank pressure, a shock is now formed inside the diverging section. The location of the shock issuch that the subsonic flow downstream of the shock can decelerate in the remaining diverging section of the

nozzle so that the jet exit pressure can match the tank pressure (curve a). With further decrease in the outside

tank pressure, the normal shock wave moves downstream. It is interesting to note that the strength of the shock

wave increases at first as it moves downstream from the throat, reaches a maximum and then decreases as itnears the nozzle exit. When the tank pressure is reduced to the value corresponding to the jet exit pressure for

supersonic flow throughout the nozzle, we have the perfect operating condition ofpe = paand the supersonic jet

discharges parallel from the nozzle into the tank.

It is of interest to outline the steps in determining the shock location to match a given back pressure. Assuming

the upstream stagnation conditions are given and the area ratio of the nozzle is known, the first step is to assumea certain shock location in the diverging section, hence the area A. The Mach number at the assumed shocklocation is then given by the equation

121

2

1

12

11

1

21

*

M

MA

A

With the upstream shock Mach number M1 determined, the downstream flow Mach number M2 behind the

normal shock can be obtained from the equation

12

12

2

1

2

1

2

M

MM

The stagnation pressure downstream of the shock is given by

1

1

2

1

1

2

1

2

1

0

0

12

1

1

12

1

2

MM

M

p

p

Knowing the area ratioeA

Awhere A is the area where the shock is located and Ae is the nozzle exit area, the

Mach number (downstream of the shock atA), the exit Mach no. can now be found from


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3

121

2

2

2

2

2

11

2

11

e

e

e M

M

M

M

A

A

With the exit Mach numberMe and the stagnation pressure downstream of the shock2

0p determined, the static

pressure of the jet at the nozzle exit can be found from the equation

12

0

2

11

2

e

e

M

pp

We now check if the computed pressure pe is equal to the tank pressure. If not, we change the shock location

and go through the computations again and we iterate for the correct shock location. In practice, it is more

convenient to use the normal shock and isentropic flow tables to carry out the iteration.

Consider now the adjustment of the flow outside the nozzle exit if the tank pressure is further reduced past the

ideal operating condition when the jet exit pressure matches the outside tank pressure, i.e. ae pp . When the

tank pressure is reduced below the jet exit pressure, the jet is said to be underexpanded and further expansionmust occur downstream of the nozzle exit. For an underexpanded subsonic jet, expansion waves can propagate

upstream into the nozzle and readjust the flow so that the exit pressure equal to the outside pressure. For an

underexpanded supersonic jet, the further expansion of the jet to match the outside pressure must take placeoutside the jet exit because the expansion waves cannot propagate upstream. Thus expansion waves originate

from the jet boundary and penetrate into the flow causing the jet to diverge downstream of the nozzle exit with a

reduction of the jet pressure as it expands further. Due to inertia, the jet expands past its equilibrium position

and the jet pressure drops below the outside pressure resulting in the jet to converge again. The jet boundary

then becomes periodic and undergoes a few cycles before turbulent mixing in the shear layer of the jet boundarydiffuses the jet momentum as outside air is entrained into the jet. Schlieren photographs illustrating the

supersonic jet downstream of the nozzle exit for various ratios of the jet exit pressure p1 to the environment

presure Ep are illustrated below.


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Now reverse the condition and start to increase the tank pressure after the supersonic jet has been establishedFor the outside tankpressure, slightly higher than the jet exit pressure (the jet is now said to be overexpanded)readjustment of the pressures must take place outside the exit plane of the nozzle. For overexpanded supersonic

jets, oblique shocks are formed to raise the jet pressure to match the tank pressure. The oblique shock turns the

jet inwards to form a converging boundary (i.e. diffuser) and the jet pressure rises. The oblique shocks reflectfrom the jet axis as oblique shocks which, when reflected off the jet boundary, becomes expansion waves and

result in a periodic phenomenon until the growth of the turbulent jet boundary shear layer and an entrainment

diffuses the jet momentum sufficiently to form a subsonic stream. If the back pressure is increased further, theexternal oblique shock then moves into the nozzle. Under ideal consideration, we assume a normal shock to

form inside the nozzle with the location of the shock governed by the matching of the pressure at the exit plane.


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In practice, the phenomenon is quite complex as the oblique shocks start to move into the nozzle. There isalways a subsonic boundary layer at the nozzle wall, thus pressure disturbances can propagate upstream of a

supersonic flow through the subsonic boundary layer. The sharp increase in pressure due to the oblique shock

interacting with the boundary layer causes the boundary layer to separate. The jet boundary then detaches fromthe wall and the complex flow pattern is illustrated below.

If the shock angle is large, the intersection of the oblique shocks can result in a Mach disc formed near the jetaxis. As the back pressure increases, this oblique shock-Mach disc complex moves upstream. The flow

downstream is highly complex and three dimensional.

2. SUPERSONIC WIND TUNNELFor a channel with two throats (e.g. a supersonic wind tunnel) where a nozzle first expands the flow to

supersonic speeds followed by a diffuser to decelerate the flow to recover the pressure, the stagnation pressure

losses associated with the shock waves that are present during the transient starting process must be accountedfor by the appropriate choice of the throat areas. A typical continuous flow supersonic wind tunnel is illustrated

in the schematic sketch below


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The mass flow rate depends on the stagnation pressure. If there is a stagnation pressure loss upstream of the

diffuser, then the throat area of the diffuser, i.e.*

DA must be larger than the throat area*

NA of the nozzle to

accommodate the same mass flow rate. Thus

sN

sD

D

N

p

p

A

A

*

*

where sNp and sDp the stagnation pressure for the nozzle and the diffuser respectively. Since shock waves in

the nozzle and test section result in a stagnation pressure loss, i.e. sNsD pp , we see that**

DN AA . In other

words the throat area for the diffuser must be larger than that of the nozzle otherwise the tunnel cannot be

started. On the other hand, if*

DA is too large, the diffuser flow will be highly supersonic and a shock wave will

be formed downstream of the throat of the diffuser. This shock wave will cause a large stagnation pressure loss

which would have to be compensated for by the compressor resulting in a higher power requirement to run the

compressor.

Let us consider the starting process of the wind tunnel. When the compressor is turned on to increase theupstream stagnation pressure of the nozzle and decrease the pressure downstream of the diffuser, the flow will

progressively accelerate from subsonic to supersonic flow in the nozzle, subsequently to choking at the nozzlethroat, a shock will be formed which then moves progressively downstream of the diverging section of the

nozzle towards the exit into the test section. This is illustrated by curves a, b, c, d, e of the figure below.


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As a result of this shock wave, there is a loss in stagnation pressure and hence the throat area of the diffusermust be greater than that of the nozzle in order to accommodate the same mass flow rate. The worst case is

when the shock reaches the test section (curve e) when the shock Mach no. is the highest. Thus, the minimum

throat area of the diffuser must be such that it can account for the stagnation pressure loss of a shock wavecorresponding to the Mach number at the test section. Note that the flow in the test section and diffuser is

subsonic for curves a to e. Further decrease in the back pressure downstream of the diffuser now causes the

shock wave to move rapidly through the test section and get swallowed by the diffuser and stabilizing

somewhere downstream of the diffuser throat (curve f). Note that the flow in the nozzle, the test section, and in

the converging part of the diffuser is now shock free. Since ** ND AA , the diffuser throat is not choked and the

flow is supersonic there. Since the Mach no. increases further downstream of the diffuser throat, it would

minimize the stagnation pressure loss if we could stabilize the shock wave just downstream of the diffuserthroat where the Mach number is smallest. Thus, after the tunnel has been started and the shock wave has been

swallowed, we can now adjust the compressor pressure ratio to move the shock back upstream to the diffuser

throat where the Mach no. is a minimum (curves a, b, c of figure below).


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This will reduce the stagnation pressure loss across the shock and save power in running the compressor.

Alternately, we can vary the diffuser throat area after the shock has been swallowed to drive the shock closer tothe throat to minimize the stagnation pressure loss. Note that the shock is not stable in the converging part of

the diffuser. If the shock is at the converging part of the diffuser and some perturbation causes it to move

slightly upstream, then the Mach no. will be higher and the stagnation pressure loss across the shock will be

greater. The mass flow through the diffuser throat will be decreased as a result of the stagnation pressure loss.

This causes a transient mass accumulation between the shock and throat which will drive the shock further

upstream to an even higher Mach no. Thus the shock will eventually be driven out to the test section and into

the nozzle causing an unstart of the wind tunnel. On the other hand, a small displacement of the shocktowards the throat results in a smaller stagnation pressure loss and a higher mass flow rate which then sucks

the shock in further until it stabilizes in the downstream diverging section of the diffuser. The optimumoperating conditions will have be to have the shock stabilized slightly downstream of the diffuser throat where

the Mach number is near the minimum hence where the stagnation pressure loss is near the minimum also.

To start the wind tunnel, the diffuser throat area must be sufficiently large to compensate for the stagnationpressure loss corresponding to a shock wave of the Mach number of the test section. For example, a M= 3

shock wave will have a stagnation pressure ratio of 328.0

1

2 s

s

p

p. Thus the ratio of the diffuser to nozzle throat

area 04.3328.0

1

*

*

N

D

A

A

In other words, the diffuser throat area is three times larger than the nozzle throat area in order to swallow the

starting shock wave. When the shock wave has been swallowed and stabilizes downstream of the diffuser

throat, the Mach number at the diffuser throat can be found from the area ratio, i.e. 3*

*

N

D

A

Aand the Mach

number. at the diffuser throat is 2.635. For a normal shock where

M1 = 2.635, the stagnation pressure ratio is about 0.44. Thus we have a loss of about 56% which the

compressor has to compensate for. To overcome this large stagnation pressure loss, a variable geometry tunnewhere the throat area of the diffuser (or nozzle) could be adjusted to should be used. For example, we could

open up the diffuser throat to swallow the shock during starting, then reduce the throat area afterwards to reduce

the Mach no. at the throat toward sonic condition. We could also simultaneously raise the back pressure of thediffuser. This is illustrated in the series of curves a, b and c of the figure below.


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In the ideal limiting case, the Mach number at the diffuser throat should be unity and the entire flow in the wind

tunnel is then isentropic. However, because of friction, there will always be some stagnation pressure loss inthe nozzle and test section so that the diffuser throat must always be larger than the nozzle throat even in the

absence of shock waves.

3. SUPERSONIC INLETS (DIFFUSERS)For air-breathing engines, the air flow must be decelerated from the free stream Mach no. (i.e. flight Mach no.)

to less than about M= 0.4 prior to entering the turbocompressor which then brings the Mach number further

down to about 2.0M before the flow enters the combustor. Whether the free stream is subsonic orsupersonic, it must be decelerated in the intake diffuser. The design of the intake diffuser must be such that the

flow is uniform and the stagnation pressure losses are kept to a minimum. Furthermore, the inlet should no

produce a large drag due to the air spillage flow over the external surface of the inlet. Consider first subsonicdiffusers, i.e. the free stream is subsonic. There are two types of subsonic diffusers, the external compression

inlet and the internal compression inlet. For the external diffusion inlet, the diffuser is just a straight duct and

the flow diffusion (i.e. compression) occurs ahead of the inlet plane as seen in the figure below.


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The flow entering the diffuser inlet A1 corresponds to the flow through the so-called capture area A0upstream. The flow area divergence from A0 to A1 occurs prior to the entrance to the inlet and for subsonic

flow, the flow divergence results in a pressure rise. Since there is no solid surface to generate shear, theexternal diffusion is isentropic. The flow between A0 and A1 will spill over the lip of the diffuser inlet

creating a large drag. Thus the external diffusion inlet is not satisfactory for high subsonic flight Mach

numbers. For the internal compression subsonic inlet, the capture area is the same as the inlet area anddiffusion is achieved internally by area increase in the duct itself.


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For isentropic flow, the pressure increase due to area increase is given by

p

dp

M

M

A

dA

2

21

For internal diffusion, the most severe problem is due to boundary layer separation. Because of the adverse

pressure gradient in the diffuser, the boundary layer tends to separate. Thus, one must maintain a small pressure

gradient, hence a gentle area divergence (of the order of 5 to 8 degrees). A gentle area divergence results in along diffuser length, thus increasing the friction losses. Boundary layer separation will result in a loss of the

internal diffusion due to area change.

For supersonic inlets, we have the simple normal shock diffuser, the convergent-divergent diffuser, and the

spike diffuser. In the normal shock diffuser, a normal shock is stabilized either in front of the inlet, at the

inlet or gets swallowed inside the inlet and thus becomes a series of oblique shock waves. The operating

characteristic of a normal shock diffuser for design condition, for increasing and decreasing back pressure areillustrated in the sketch below.


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When the normal shock stabilizes ahead of the inlet, the capture area10AA , external diffusion of the

subsonic flow downstream of the shock occurs prior to entering the inlet followed by further internal diffusion

if there is an area divergence in the duct itself. Since10AA , flow spillage occurs and results in large external

drag. When the normal shock stabilizes at the inlet plane, further internal diffusion can occur downstream of

the normal shock if the duct area diverges. The capture area is now the same as the inlet area and no flow

spillage occurs. In a normal shock diffuser, there is a large stagnation pressure loss associated with the normal

shock. Thus the diffuser is inefficient at high free stream or flight Mach numbers (e.g. 8.1M ). For highersupersonic flows, the normal shock may be swallowed inside the diffuser and turned into a series of obliqueshock waves and Mach stems. The stagnation pressure loss in oblique shocks is less than that of a normashock. However, the reflection of the oblique shocks off the wall causes complex shock boundary layer

interactions and flow separations. Normal shock diffusers are simple and give acceptable performance for fligh

Mach numbers less than aboutM= 1.8.

For higher supersonic speeds, the more efficient converging-diverging diffuser should be used. The idea

operating condition would be for a normal shock to be stabilized slightly downstream of the diffuser throat as in

the case of the diffuser of a supersonic wind tunnel. The Mach number at the diffuser throat should be slightlysupersonic. If the free stream Mach number is decreased while the back pressure is increased, one can achieve

isentropic flow throughout the throat choked and subsonic flow downstream of it (If we assume no friction inthe duct).

The ideal operating condition must be approached from some transient starting conditions where the flow Mach

number accelerates to the operating Mach number. As the free stream Mach number increases, the detachednormal shock (ahead of the inlet) will eventually reach the inlet and the capture area equals the inlet area A1. If

the throat is choked, then mass flow rate is a maximum for the value of the stagnation pressure downstream of

the normal shock. For the shock to be swallowed and move downstream of the throat, the throat area must be

increased. If the throat area is slightly decreased, the throat cannot accommodate the mass flow rate and masswill accumulate and the shock will be expelled and detached from the inlet. The detachment distance will be

such to allow for the accumulated mass to be spilled out of the inlet and flow externally over the lip of the inlet.

So for each free stream Mach number M, we can look up the normal shock table for the downstream Machnumber and then the isentropic flow tables for

1

*

A

Aas a function of this downstream Mach number behind the

shock and plot a curve1

*

A

A(M) as shown in figure below.


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For any Mach number, we see that for an area ratio11

*

A

A

A

At where the shock will be swallowed and stabilizes

downstream of the throat. For11

*

A

A

A

At , the shock will be detached. The1

*

A

A(M) curve marks the boundary

between condition of the shock being swallowed or expelled and detached from the inlet.

Consider next the case when the diffuser has already been started and supersonic flow now occurs in the

converging section. Again there corresponds a certain area ratio1

*

A

Afor a given free stream Mach number M

when the throat becomes choked. ForA

A

A

At *

1

, the throat becomes unchoked the flow becomes supersonic


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and the shock is swept downstream. However, for11

*

A

A

A

At the throat can no longer handle the captured mass

flow for the given supersonic Mach number. Thus the normal shock will be expelled and detached from the

inlet. Looking up the isentropic flow tables for1

*

A

Afor various Mach numbers, we can plot M

A

A

1

*as shown

in the figure below.

The difference between this curve and the curve in the previous figure is that in the previous case the isentropicflow is based on the downstream Mach number of the normal shock at the free stream Mach number. The curve

in the figure above is just based on the free stream Mach number since no shock wave is involved.


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The starting process for a supersonic diffuser can now be clearly illustrated by plotting these two curvestogether. These two curves define the boundaries where the shock is swallowed or expelled with and without a

normal shock at the inlet entrance.

Consider a fixed area ratio1

A

At inlet. As the free stream Mach M number increases, we move along the

horizontal line a, c, b. The shock will be detached until we reach point b on the boundary curve. At b, thethroat is just choked, and further increase in Mach number will cause the shock to be swallowed. Once the

shock is swallowed, we have isentropic flow throughout the converging section including the throat. The shock

is now stabilized somewhere downstream of the throat. If we now reduce the throat area keeping the Machnumber constant, we move down the vertical line from b to dand the shock remains swallowed and the flow


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supersonic throughout (including the throat) up to the shock location downstream of the throat. However, if wedecrease the throat area past point c (which is the limit boundary where the throat is now choked) for the free

stream Mach number, the normal shock will be expelled out of the diffuser and we have a so-called unstart

phenomenon. Hence, there is a sort of hysterisis where the shock, once swallowed, will remain downstream ofthe throat until the area ratio goes way down past the value when the shock was first swallowed before it gets

expelled when the second boundary curve is reached. Thus, for a fixed geometry diffuser, we can overspeed

to first swallow the shock to start the diffuser, then slow down and operate at any free stream Mach number

between point b and c. The shock will not be expelled until the Mach number has decreased past point c

Because of the high stagnation pressure loss across a normal shock with increasing Mach number, we see that

we must overspeed considerably to be able to swallow the shock as the area ratio1

A

At decreases. In fact, we

reach a limit when 6.01

A

At . For this area ratio, the operating free stream Mach numberM= 2 before the

shock gets expelled if the diffuser can be started. However, to start the diffuser for 6.01

A

At , we have to

overspeed to M . Thus, overspeeding is not an efficient way to start the diffuser when the area ratio issmall.

The alternative is to use a variable throat area diffuser. For example, if we want to operate at a flight Mach

numberM= 3.2, we first open up the throat area to 7.01

A

At , i.e. point b. The shock will then be swallowed

and once swallowed, we can close up the throat area and move vertically downwards from point b to point d

We can operate atM= 3.2 for 2.01

A

At without causing the diffuser to unstart.

We can also open up the throat area larger, say to 8.0

1

A

At . In that case, the shock will be swallowed when

2.2M (point e in the above figure). Once the shock has been swallowed, we can close the throat area andsimultaneously increase the free stream Mach number. We will follow the dotted curve to the operating point din this case.

The variable throat diffuser is a much more efficient way to start the diffuser. However, for a three-dimensional

axisymmetrical inlet, it is difficult to design the mechanism to vary the throat area. The spike inlet offers a nice

solution to the variable area design as well as reducing the stagnation pressure loss by replacing the normal

shock by a series of oblique shocks. The spike inlet was suggested by the famous German aerodynamicistKlans Oswatitsch. A sketch is shown below.


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As can be observed, a conical oblique shock is now attached to the spike instead of a normal shock and thatreduces the free stream Mach number somewhat (depends on cone angle), but the flow downstream of the

oblique shock is still supersonic. A second normal shock now reduces the flow to subsonic. However, the

normal shock is now weaker and the loss in stagnation pressure is reduced. Further subsonic diffusion can also

be achieved by area divergence between the central spike body and the external housing. With the spikecentral body configuration, the mechanical design to vary the throat area is also simpler. The spike inlet is used

in the Lockheed SR-71.


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MECH 430 FLUIDS 2

Flow of a compressible fluid in a constant area duct with friction

For an incompressible fluid where the density remains constant, friction results in

a pressure drop in the flow direction. For a compressible flow, viscous dissipation heats

up the gas resulting in a density change. Thus, the flow velocity and pressure can

increase or decrease depending on the rate of expansion due to density decrease and theconvective mass flux through the cross section of the pipe. As in the case of area

change, friction has opposite effect if the flow is subsonic or supersonic.

Let us first review the steady flow of an incompressible fluid in constant area pipe

with friction (which results in a wall shear stress w

). Referring to the sketch below:

p + dp

w

L

p

The conservation of mass for steady flow gives u=constant and since = constant, the

flow velocity is constant throughout the length of the duct. The momentum equation

gives:

( )pA p dp A Aw w + = 0

where A is the area of the duct ( AD

= 2

4for a circular duct) andAw is the wetted area

( A Dw = L for a circular pipe). Rearranging the momentum equation gives:

= =dpA

A

L

Dww w

4

where we have assumed a circular pipe for convenience. The wall shear stress is givenby:


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w

r R

u

r=

=

where is the coefficient of viscosity and

u

rr R

=

is the velocity gradient at the wall (if

we assume a laminar flow). The velocity profile for steady laminar incompressible flowin a circular pipe is given by:

u

u

r

R=

2 1

2

where u is the mean flow velocity (i.e. the classical Hagen-Poiseuille Flow.) Thisparabolic velocity profile gives the wall shear stress as:

w

r R

u

r

u

R

u

D=

= =

=

4 8

The negative sign denotes the direction ofw that gives the parabolic profile and we have

already considered the appropriate sign (i.e. direction) ofw when we write the

momentum equation. Thus, we write:

=

= = =

dp

L

p p

L D

u

D

u

D

w

D

1 2

2

24 32 64

2

1

Re

where ReDu D

=

is the Reynolds number based on the pipe diameter. In general, we

define a friction factor f as:

fu

w=4

2

2

where is determined from experiments. (Moody diagram where fis given for a

wide range of Reynolds number and wall roughness). For the present case of laminar

Poiseuille Flow:

(f DRe )

2


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fD

=64

Re

We may write the pressure drop as:

2

1

2Re

64 2221 u

D

f

D

u

L

pp

D

=

=

For incompressible flow, we need not worry about the energy equation since = constant

and the problem is a dynamic problem with pressure forces balancing the friction forces.

Compressible Flow (Fanno Flow)

For steady compressible flow in a constant area, adiabatic duct with friction as illustrated

below, i.e.:

dx

P+dp

+ d

T+dT

u+du

w

u

p

T

The conservation of mass gives:

m uA= = constant

d du

u + =0 1.

since dA=0 for a constant area duct.

The conservation of momentum can be written as:

3


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DsxdpAmdu w= 2.

where S andA are the perimeter and the area of the cross-section of the tube, respectively.Defining a friction factor:

fu

w=

2

2

and a hydraulic diameter DH by:

Darea A

SH= =

4 4*

wetted perimeter

Equation 2 becomes:

udu dpu fdx

DH=

2

2

43.

If the flow is adiabatic, the energy equation is given by:

hu

ho+ =2

2= constant

Note that there is no work done by the viscous stress w since the velocity at the wallvanishes. The no slip (i.e. u=0) condition at the wall generates a velocity profile resulting

in viscous dissipation. The viscous heating is at the expense of the kinetic energydecrease in the flow but since there is no heat transfer, the viscous heating remains in the

flow. In the present assumption of a quasi one dimensional flow, the flow velocity is

uniform across the cross section of the duct. We should consider the flow velocity as anaveraged value to permit the presence of a velocity gradient for the viscous stress to

occur. In an inviscid flow, there will be no shear stress at the wall. For adiabatic (not

isentropic) flow where

tconsHu

h tan2 0

2

==+

we get

dh+udu=0

4


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Since h=cpT, and cR

p =

1and c , the energy equation can be written as:R2 = T

( )dT

TM

du

u+ = 1 2 0 4.

The equation of state for a perfect gas is:

p RT=

and hence

dp

p

d dT

T= +

5.

Rewriting Equation 3 as:

udu

p

dp

p

u

p

fdx

DH=

2

2

4

and noting that

pc =2 , we obtain an expression for

dp

pas

dp

pM

du

u

M fdx

DH=

2

2

2

46.

From Equations 1, 4, 5 and 6, we can get an expression fordu

uas

( )du

u

M

M

fdx

DH=

2

22 1

47.

From Equation 1, we see that:

( )

d du

u

M

M

fdx

DH

= =

2

2

2 1

48.

From Equation 4, we get

( )dT

TM

du

u= 1 2

5


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and substituting Equation 7 into the above yields

( )

( )

4

2

- -1 4

2 1-

=

H

MdT fdx

T DM9.

Substituting Equations 8 and 9 into Equation 5, gives:

( )( )( )

dp

p

M M

M

fdx

DH=

+

2 2

2

1 1

2 1

410.

To get an expression for the Mach number, we note Mu

c= , thus

dM

M

du

u

dc

c=

and since c RT dcc

dTT

2 12

= = ,

Hence,

dM

M

du

u

dT

T=

1

2

And substituting Equations 7 and 9 into the above, gives:

( )dM

M

M M

M

fdx

DH=

+

2 2

2

11

2

2 1

411.

For the variation of the stagnation pressure along the pipe, we note that the definition ofthe stagnation pressure is the pressure one would obtain if we decelerate a flow

isentropically to zero velocity. Thus,

p

pM

o= +

11

2

21

and differentiating, we get:

dp

p

dp

p

M

M

dM

M

o

o

= +

+

2

211

2

6


24/50

The variation ofp andMalong the pipe is given by Equations 10 and 11, thus, we get:

dp

p

M fd

D

o

o h

= 2

2

4 x12.

The variation of entropy along the pipe can be found from:

Tds = dh-vdp

with h c TRT

p= =

1,pv=RT, the above becomes:

dsc

dTT

dppp

=

1

Equations 9 and 10 give the variation ofdT

Tand

dp

p, thus we obtain for the change in

entropy along the pipe the following expression:

ds

c

M fd

Dp H=

1

2

42 x

ords

R

M f

Ddx

dp

pH

o

o

= = 2

2

413.

since( )c

Rp

=

1. As in the case of the normal shock, the entropy increase can be

correlated to the stagnation pressure loss due to viscous dissipation.

Equations 7-13 give the variation of u,, T, P, M, po and s along the pipe

respectively. Except fordp

p

o

o

andds

R, all the equations contain the term (1-M2) in the

denominator. Thus, depending on whether the flow is subsonic M0 or

supersonic, M>1 and (1-M2)


25/50

decrease due to friction. The table below summarizes the effect of friction on the fluidand thermodynamic states. The sign + or denotes whether the variable increases or

decreases along the pipe (i.e. increased value ofx or positive dx).

Variable Subsonic Supersonic

Velocity u + -

Mach Number M + -

Pressure p - +

Temperature T - +

Density - +

Stagnation pressure po - -

Entropy s + +

The difference between subsonic and supersonic flow is a result of thecompetition between the rate of expansion due to viscous heating of the flow and the rate

of mass convected through the cross section of the pipe. Thus, for subsonic flow, the rate

of expansion (characterized by the sound speed) dominates, and the density decreases

resulting in an increase in the flow velocity, a decrease in temperature and pressure due tothe expansion. The reverse happens for supersonic flow when the rate of convection

(characterized by the flow velocity) dominates over the rate of expansion. However, for

the stagnation pressure and entropy, both subsonic and supersonic flows indicate thesame behavior of loss in stagnation pressure and increase in entropy from irreversible

viscous dissipation.

Integration of the equations

To get the variation of the various state variables along the duct, the differentialequations have to be integrated. The important one to start is the variation of the Mach

number with distance given by Equation 11. Rewriting it in the following form:

( )2 2 M dM1

we can integrate the above using the method of partial fraction. The above can be

expanded as:

4fdx=

1

1

2

4 2D

M MH +

8


26/50

1 2dM2

dM24fdx + +dM1 1 2

Since friction always drives the flow towards M=1, we integrate the above from

0


27/50

4 4 41 2

fL

D

fL

D

fL

DH

M

H

M

H

= * *

We know L and M1, we can look up the table for such that*1L4

1

fL

D

M

H

*

corresponds

to 1=M . Then, we can find the value for:

4 42 2

fL

D

fL

D

fL

D

M

H

M

H H

* *

= 4

We can then find the value of M2 corresponding to the value of4

2fL

D

M

H

*

.

Equation 14 gives M as a function of L*. If we want to know the other flow variables

like p, T, , etc. we can integrate the other equations fordp

p

dT

T, , etc. However, it is

more convenient to express them in terms of the Mach number so that once M at asection is known, we can find the other variables. From Equations 10 and 11, we write:

( )( )( )

( )

dp

p

dM

M

M M

M

fdx

D

MM

Mfdx

D

H

H

=

+

+

1 1

2 1

4

2

11

21

4

2 2

2

2

2

2

and simplifying the above yields:

( )( )dpp

M

M

dM

M=

+

+

1 1

2 11

2

2

2

2

2

Using partial fraction, the above can be written as:

22

22

1

212

2 12

dMdp dM

p MM

= +

10


28/50

which integrates to yield:

p

p MM

* =+

+

1 1

2 11

2

2

15.

The above equation gives the pressure p at M and p = p* when M=1. Similarly, we

can find expressions for the other variables in terms of the Mach number as follows:

T

T

c

cM

* *=

=

+

+

2

2

1

2 11

2

16.

( )uu

M

M* =

+

+

1

2 11

2

2

17.

*

*

= =

+

+uu M

M

1

2 11

21

2

18.

( )

( )

p

p M

Mo

o

* =

+

+

+

12 1

1

2

1

2

1

2 1

19.

( )s sc

M

M Mp

=

+

+

+

*

ln2

2 2

1

2

1

2 11

2

20.

11


29/50

Equations 15 to 20 are also computed for various Mach numbers and tabulated together

with4fL

DH

*

in the Fanno tables for convenience in solving problems of compressible flow

in ducts with friction.

The Fanno Line

The adiabatic flow of a compressible fluid in a constant area duct with friction

satisfies the conservation of mass and energy i.e.

constant=u

0=+u

dud

2

constant2

o

uh h+ = =

0=+ ududh

From these two equations, we can obtain an expression for the variation of the enthalpy

(or equivalently the temperature) and the entropy along the duct. From the Tds equation,

i.e.:

vdpdhTds =

we get:

p

dp

h

dh

c

ds

p

1=

and from the equation of state RTp = , we get:

h

dhd

T

dTd

p

dp+=+=

Hence,

12


30/50


31/50

+

=

2

1

*

2

11

*

*

2

1

1

2ln

h

h

h

h

c

ss

p

21.

The Fanno line gives the variation of the enthalpy as a function of the entropy for

a given mass flow rate in and stagnation enthalpy ho (or equivalently h*). The lower

branch is for supersonic flow and the upper branch is for subsonic flow. For any givenvalue of M, friction will drive the flow towards sonic conditions M=1 where the entropy

is a maximum. To prove that entropy is a maximum when the flow becomes sonic, one

could differentiate Equation 20 and equate 0=dM

dsand solve for M. Alternately, we do

the following: from the Tds equation, we write:

1Tds dh dp=

and from the energy equation ,ududh = we get:

dpuduTds

1=

From the conservation of mass constant=u , we get :

0=+ uddu

If we express ( ps, ) , we can write:

( ) dpc

dss

dpp

dss

psdpsp

2

1, +

=

+

=

since the sound speed c is defined by:

2

s

pc

=

Replacing d in the continuity equation by the above expression yields :

02

=

++=+ dss

udpc

uduuddu

p

22.

14


32/50

From the Tds equation, we get:

Tdsududp =

and replacing dp in Equation 22 by the above and rearranging gives:

( ) 012

2 =

psc

TudsMdu

23.

Since 2cT is always positive and

ps

is always negative (since density decreases

when heat is added at constant pressure and heat added means entropy increases, i.e.

00 at constant p). The bracketed term is always positive, i.e.:

02 >

psc

pT

Thus, as , ds must vanish, thus s corresponds to an extremum (maximum) value

when M=1.

1M

Another alternate expression for the Fanno Line that involves the Mach number can beobtained as follows: From the Tds equation, we write:

p

Rdp

T

dh

ds =

and rearranging, we write:

dh

dp

p

R

T

ds

dh

=

1

1

From the energy equation, we get:

02

2 =+ dudh

And the conservation of mass gives:

2

2

12

d du duu u

= =

15


33/50

we can express dh as:

dududh

22

2==

And using the equation of state ,RTp = we can express:

T

dT

p

dpd=

Hence,

==

T

dT

p

dpu

dudh

22

And solving for the slopedh

dp, we get:

( )( )22

11 Mu

p

dh

dp+=

We can now writedh

dsas:

( )

+

=

2

2

211

1

upM

up

pR

Tds

dh

or12

2

=

M

TM

ds

dh

The above is an alternate form of the equation for the Fanno Line and it is clear that for

subsonic flow where M1, 0>dsdh , the locus of states is represented by the lower branch of the

Fanno Line. When M=1,dh

ds and .maxs

To plot a Fanno Line, we first note that the Fanno Line involves the conservation of mass

and energy, i.e.

16


34/50

constant== mu

constant2

2

==+ ohu

h

Thus, a Fanno Line corresponds to a value for and . Combining conservation of

energy and mass, we get:

m oh

constant2

22

==+ ohvm

h

24.

Where

1=v is the specific value. So, given and , we pick different values form oh

and compute h from Equation 24. And Equation 21 gives s as a function of h. Note

that *2

1oh

=

+h , thus the value ofh*can be found for a given stagnation enthalpy ho.

Effect of duct length and back pressure

The flow inside a duct with friction is influenced by the duct length and the back

pressure of the exit of the duct. Consider first the case of subsonic flow in the duct. Weassume the inlet to the duct is from a converging nozzle connected to a large plenum

chamber where the stagnation pressure and temperature is po and To. Consider firstjust the nozzle. For a subsonic jet, the exit pressure must be the same as the backpressure. Thus, for a given exit pressure, the Mach number at the jet exit is given by the

isentropic relationship

12

2

11

+=

M

p

po

If we now connect a pipe to the exit of the converging nozzle, then the exit of the nozzlecorresponds to the entrance to the duct. Let the end of the duct be connected to a large

plenum chamber where the pressure can be controlled. The system is shown

schematically in the sketch below.

17


35/50


36/50

duct length corresponds to the critical valueL* for the inlet Mach number (curve c). If

the back pressure is further decreased, i.e. pB


37/50

downstream of the duct exit. Forpe=p*, there is only one location where the shock inside

the duct is located.

For supersonic inlet flow and a very high pB so that a shock is driven inside the

duct, the process is illustrated in the h-s diagram below.

ho

normal shock

transition

Me , Pe = PB

M> 1

M< 1

isentropic flowin nozzle

20


38/50

Mech 430A Fluid Mechanics II:

Effect of heat addition on the compressible

flow in a constant area duct

For incompressible flows, the density stays constant and heat addition hasno effect on the dynamics of the fluid motion (except if one considers the tem-perature dependence of the viscosity). For a compressible fluid, heat additionchanges the density and depending on whether the flow is subsonic or super-sonic, competition between expansion (due to density change) and convectionresults in different behaviors of the flow. We shall consider frictionless flow in

a constant area duct and we are interested to know the variation of the fluidproperties along the duct as a result of heat addition (or extraction). Again, westart from the conservation laws and refer to the simple sketch below, i.e.

The conservation of mass gives:

m = uA = constant

dm = 0 dA = 0

henced

=du

u(1)

The conservation of momentum gives:

mdu = pA (p + dp) A

Fig. 1: Sketch of the control volume

1


39/50

2

Adp + mdu = 0, dp + udu = 0 (2)

The conservation of energy gives:

mdh +u2

2 = dqdh + d

u2

2

= dq (3)

where dq is the heat per unit mass added to the control volume Adx. We canwrite dq = dho where ho is the stagnation enthalpy and heat addition essentiallyincreases the stagnation enthalpy of the flow. Hence, we can also write Equation3 as:

d

h + u2

2

= dho

The equation relates p, , and T, i.e.:

dp

p=

d

+

dT

T(4)

From Equations 1 to 4, we can solve for the variation of the flow properties as afunction of the heat addition. For example, consider the velocity variation first.Equation 2 can be written as:

dp = udu

dp

p=u

pdu = M2

du

u(5)

Equation 3 can be written as:

dh + udu = dho

dT

T+ ( 1) M2

du

u=

dTo

T(6)

since

h = cpT =c2

1

Combining Equations 4, 5 and 6, i.e.:

dp

p= M2

du

u=

d

+

dT

T=

du

u ( 1) M2

du

u+

dTo

T

or

M2duu

= duu ( 1) M2 du

u+ dT

o

T


40/50

3

we obtain:du

u=

1

(1M2)

dTo

T

Since

To

T= 1 + 1

2M2

we get:du

u=

1 + 12 M

2

(1M2)

dTo

To(7)

Hence,

d

=

1 + 12 M

2

(1M2)

dTo

To(8)

We can get dpp

and dTT

from the equation of state and the energy equationas:

dTT =

1 +

1

2 M

2 1

M

2(1M2)

dToTo

(9)

dp

p=M2

1 + 12 M

2

(1M2)

dTo

To(10)

Since the Mach number is defined as:

M =u

c

hence

dM

M=

du

u

dc

c=

du

u

1

2

dT

T

substitution of Equations 7 and 9 into the above, yields:

dM

M=

1 + M2

1 + 12 M

2

2 (1M2)

dTo

To(11)

and since the stagnation pressure is defined as:

po

p=

1 +

1

2M2

1

hence

dpo

po=

dp

p+

1

12

2M dM

1 + 12 M2

and rearranging, we get:


41/50

4

Fig. 2: Table showing the change in flow properties with heat addition

dpo

po=M2

2

dTo

To(12)

For the entropy change, we use the T ds equation, i.e.:

T ds = dh vdp

or

ds

cp=

dT

T

1

dp

p

and substituting Equations 9 and 10 into the above, yields:

ds

cp=

1 +

1

2M2

dTo

To(13)

Thus, Equations 7 to 13 give the variation of duu

, d

, dTT

, dpp

, dMM

, dpopo

anddscp

respectively. Note that, except for Equations 12 and 13, for the stagnation

pressure and entropy, all the other expressions have the term 1M2. Thus, the

behavior differs for subsonic and supersonic flows. Table 1 indicates whetherthe flow properties increases (+) or decreases (-) with heat addition +dTo.

For heat extraction dTo < 0 and hence the behavior is reversed for subsonicand supersonic flows. We can also integrate these equations with reference tosome reference state (choose the critical state where M = 1). We first integrateEquation 11 for the Mach number variation with heat addition and we thenexpress the other flow quantities in terms of the Mach number. Equation 11can be rewritten as:

dTo

To=

1M2

dM2

(1 + M2)

1 + 12 M2

M2


42/50

5

and using the method of partial fraction, we can split up the right hand side ofthe above expression and obtain:

dTo

To=

2d

M2

1 + M2

dM2

M2+

12

dM2

1 + 12 M2

The above can readily be integrated and we obtain:

To

To=

2 (+ 1) M2

1 + 12 M2

(1 + M2)2 (14)

The other flow properties can readily be obtained as a function of the Machnumber M as:

u

u=

(+ 1) M2

1 + M2(15)

p

p=

+ 1

1 + M2(16)

= 1 + M2(+ 1) M2

(17)

T

T=

(+ 1)2

M2

(1 + M2)2 (18)

po

po=

+ 1

(1 + M2)

2

1 + 12 M2

+ 1

1

(19)

s s

cp= ln M2

+ 1

1 + M2

+1

(20)

Equations 14 to 20 have been computed for a chosen value of and a range

of values of the Mach number for both subsonic and supersonic flows. Theseare tabulated in the Rayleigh flow tables. A graph showing the variations of theflow parameters with the Mach number is illustrated in Figure 3.

The results of our analysis indicate that heat addition always drives a flowtowards sonic condition. Thus, for a given initial value of M (subsonic or super-sonic), there corresponds a maximum value of the heat addition (i.e. stagnationtemperature increases) when the flow becomes sonic. If more heat is addedbeyond this maximum value that chokes the flow, upstream conditions have tobe modified to accommodate for the additional heat addition.


43/50

6

Fig. 3: Variations of flow parameters with Mach number for Rayleigh flow


44/50

7

The Rayleigh LineHeat addition to a compressible flow can be represented graphically on the

Rayleigh Line on the h s diagram. The Rayleigh Line is based on the conser-vation of mass and momentum, i.e.

m = u = constant

p+u2 = F = constant

Using the conservation of mass, the momentum equation can be written as:

p +m2

= F = p

1 + M2

(21)

Thus, given initial conditions where m and F are found, corresponding values ofp for chosen values of can be obtained. Since h = p

(1) and the entropy

is given by:

s s1

cv= ln

p

p1

1

values for h and s corresponding to given values of p and can beobtained and the locus of states along the Rayleigh Line can be plotted onthe h s diagram. The shape of the Rayleigh Line is illustrated in Figure 4.An analytical expression for the Rayleigh Line can also be obtained. The T dsequation gives:

T ds = dh vdp

ords

cp=

dh

h

1

dp

p

which intergrates to yield:

s

s

cp= ln h

h

1

ln p

p(22)

where we have used the critical condition M = 1 as the reference state. Thenext task is to eliminate p

pin terms of h

h. From the equation of state, we

write:

p

p=

T

T

and using the conservation of mass, we get:

p

p=

u

u

T

T

In terms of the Mach number, the above can be written as:


45/50

8

Fig. 4: Sketch showing the shape of the Rayleigh line


46/50

9

p

p=

u

cc

u

c

c

T

T=

M

M

T

T

12

=1

M

h

h

12

where we have used c2 = RT and M = 1, h = cpT. Squaring the aboveequation gives:

p

p

2=

1

M2

h

h

(23)

From the Equation 16, i.e.:

p

p=

+ 1

1 + M2

we can solve for M2 and obtain:

M2 =( + 1) p

p 1

Replacing M2 in Equation 23 by the above, we get the following quadratic

equation for pp , i.e.: p

p

2 ( + 1)

p

p+

h

h

= 0

and the solution can be written as:

p

p=

(+ 1)

(+ 1)2 4

hh

2

the negative root recovers the solution that when hh

= 1, pp

= 1. Substituting

the above expression for pp

into the entropy equation (Equation 22) gives thedesired equation for the Rayleigh Line:

s s

cp= ln

h

h

2(+ 1)

(+ 1)

2 4

hh

1

(24)

The above equation is a bit difficult to illustrate the qualitative features of theRayleigh Line. An alternate expression can be derived. From the T ds equation,we write:

T ds = dh vdp = dh1

dp

From the momentum equation, we write:

1

dp = udu


47/50

10

thusT ds = dh + udu

from which we obtain:

hds

cp= dh + udu

and re-arranging, gives:

dh

ds=

h

cp

1 + ududh

(25)From the momentum equation and the equation of state, i.e.

dpp

= M2 duu

dp

p= d

+ dT

T= du

u+ dh

h

where we have used the continuity equation and h = cpT in the above. Com-bining the two expressions, gives:

udu

dh=

( 1) M2

(1 M2)

Substituting the above expression for ududh

in Equation 25 leads to the followingresult:

dh

ds=

h

1 M2

cp (1M2)(26)

Note that for subsonic flow where M < 1, 1M2 > 0, then dhds

> 0 ifM2 < 1

or M 1, 1M2 < 0 but 1M2 is also negative. Thus dhds > 0. Thusheat addition tends to always drive the flow towards M = 1. This is referredto as thermal choking. For reversible heat transfer, the heat addition can beequated to T ds. Thus, the area under the Rayleigh Line represents the heatadded. If we write:

dq = T ds =hds

cp= dho

Then, the maximum heat that can be added to a flow with initial Mach numberM is:

qmax =

s

s1

h (s) ds

cp =

ho

ho1

dho = h

o

ho1 = cpTo1 To

To1

1


48/50

11

orqmax

cpT1=

To1T1

ToT1 1

Since:

To

T= 1 +

1

2M2

andTo

Tois given by Equation 14 as:

To

To=

2 (+ 1) M2

1 + 12 M2

(1 + M2)2

the maximum heat that can be added for an initial flow Mach number M isthus:

qmax

cpT=

1 +

1

2M2

1 + M2

22 (+ 1) M2

1 + 12 M

2 1

which simplifies to:

qmax

cpT=

M2 1

22 (+ 1) M2

(27)

A plot of qmaxcpT

as a function of the Mach number M is shown in figure 4.

For a given initial Mach number M, if the heat added is greater thanthe maximum value (corresponding to the initial value of M), then the initialconditions must be modified and the mass flow rate reduced to permit the addi-tional heat addition. For subsonic flow, information can propagate upstream toeffect this change of mass flow rate. For supersonic flow, a shock wave is formedand propagates upstream. However, since there is no change in the stagnationenthalpy across a normal shock, having a shock upstream in the constant areaduct would not permit more heat to be added. Thus, the shock has to advance

all the way back to the nozzle which supplies the supersonic inlet flow to theduct in order to lower the inlet Mach number. Note that the mass flow rateremains the same since the flow is chocked at the throat of the nozzle and themass flow rate remains invariant irrespective of the downstream flow conditions.However, the inlet to the duct is subsonic and permits a larger amount of heataddition before choking occurs. If the amount of heat addition in the duct isstill too large, then the shock propagates past the nozzle throat upstream andthe whole flow is subsonic. The mass flow can now be reduced to accommodatethe heat addition to give choking condition at the exit of the pipe.

The analysis we have carried out can be applied to the study of combustionwaves. Flames and detonations are, in general, very thin compared to thedimension of the duct and hence can be treated as a discontinuity like a normalshock. Heat addition occurs inside this thin front in the form of energy release

by chemical reactions. Applying the conservation laws across the combustion


49/50

12

Fig. 5: A plot of qmaxcpT

as a function of the Mach number M


50/50

Documents

Lecture - Supersonic