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Unformatted text preview: IX. Compressible Flow
Compressible flow is the study of fluids flowing at speeds comparable to the local
speed of sound. This occurs when fluid speeds are about 30% or more of the local
acoustic velocity. Then, the fluid density no longer remains constant throughout
the flow field. This typically does not occur with fluids but can easily occur in
flowing gases.
Two important and distinctive effects that occur in compressible flows are (1)
choking where the flow is limited by the sonic condition that occurs when the flow
velocity becomes equal to the local acoustic velocity and (2) shock waves that
introduce discontinuities in the fluid properties and are highly irreversible.
Since the density of the fluid is no longer constant in compressible flows, there are
now four dependent variables to be determined throughout the flow field. These
are pressure, temperature, density, and flow velocity. Two new variables,
temperature and density, have been introduced and two additional equations are
required for a complete solution. These are the energy equation and the fluid
equation of state. These must be solved simultaneously with the continuity and
momentum equations to determine all the flow field variables.
Equations of State and Ideal Gas Properties:
Two equations of state are used to analyze compressible flows: the ideal gas
equation of state and the isentropic flow equation of state. The first of these
describe gases at low pressure (relative to the gas critical pressure) and high
temperature (relative to the gas critical temperature). The second applies to ideal
gases experiencing isentropic (adiabatic and frictionless) flow.
The ideal gas equation of state is = P
RT In this equation, R is the gas constant, and P and T are the absolute pressure and
absolute temperature respectively.
Air is the most commonly incurred
compressible flow gas and its gas constant is Rair = 1717 ft2/(s2oR) = 287 m2/(s2K).  v v IX1  eText Main Menu  Textbook Table of Contents  Study Guide Two additional useful ideal gas properties are the constant volume and constant
pressure specific heats defined as Cv = du
dT and C p = dh
dT where u is the specific internal energy and h is the specific enthalpy. These two
properties are treated as constants when analyzing elemental compressible flows.
Commonly used values of the specific heats of air are: cv = 4293 ft2/(s2oR) = 718
m2/(s2K) and cp = 6010 ft2/(s2oR) = 1005 m2/(s2K). Additional specific heat
relationships are R = C p − Cv and k= Cp
Cv The specific heat ratio k for air is 1.4.
When undergoing an isentropic process (constant entropy process), ideal
gases obey the isentropic process equation of state: P
k = constant Combining this equation of state with the ideal gas equation of state and
applying the result to two different locations in a compressible flow field
yields P2 T2 = P1 T1 k / ( k −1) = 2 1 k Note: The above equations may be applied to any ideal gas as it undergoes
an isentropic process.
Acoustic Velocity and Mach Number
The acoustic velocity (speed of sound) is the speed at which an infinitesimally
small pressure wave (sound wave) propagates through a fluid. In general, the
acoustic velocity is given by  v v IX2  eText Main Menu  Textbook Table of Contents  Study Guide a2 = P The process experienced by the fluid as a sound wave passes through it is an
isentropic process. The speed of sound in an ideal gas is then given by a = k RT
The Mach number is the ratio of the fluid velocity and speed of sound, Ma = V
a This number is the single most important parameter in understanding and
analyzing compressible flows.
Mach Number Example:
An aircraft flies at a speed of 400 m/s. What is this aircraft’s Mach number when
flying at standard sealevel conditions (T = 289 K) and at standard 15,200 m (T =
217 K) atmosphere conditions? k RT = (1.4)(287)( 289) = 341 m / s
and at 15,200 m, a = (1.4)(287)( 217) = 295 m / s . The aircraft Mach
At standard sealevel conditions, a =
numbers are then V 400
=
= 1.17
a 341
V 400
15,200 m : Ma = =
= 1.36
a 295 sea − level : Ma = Note: Although the aircraft speed did not change, the Mach number did change
because of the change in the local speed of sound.  v v IX3  eText Main Menu  Textbook Table of Contents  Study Guide Ideal Gas Steady Isentropic Flow
When the flow of an ideal gas is such that there is no heat transfer (i.e., adiabatic)
or irreversible effects (e.g., friction, etc.), the flow is isentropic. The steadyflow
energy equation applied between two points in the flow field becomes V12
V22
h1 +
= h2 +
= ho = constant
2
2
where h0 is called the stagnation enthalpy that remains constant throughout the
flow field. Observe that the stagnation enthalpy is the enthalpy at any point in an
isentropic flow field where the fluid velocity is zero or very nearly so.
The enthalpy of an ideal gas is given by h = Cp T over reasonable ranges of
temperature. When this is substituted into the adiabatic, steadyflow energy
equation, we see that h o = C p To = constant and To
k −1
=1 +
Ma 2
T
2
Thus, the stagnation temperature To remains constant throughout an isentropic or
adiabatic flow field and the relationship of the local temperature to the field
stagnation temperature only depends upon the local Mach number.
Incorporation of the acoustic velocity equation and the ideal gas equations of
state into the energy equation yields the following useful results for steady
isentropic flow of ideal gases. To
k −1
=1 +
Ma 2
T
2
1/2
ao To 1 / 2 k− 1
2
=
= 1+
Ma a T
2
k / (k −1 )
Po To k / ( k −1) k −1
2
=
= 1+
Ma P T
2
o T
= o T 1/ ( k −1) 1/ ( k −1) 1 + k − 1 Ma 2 = 2  v v IX4  eText Main Menu  Textbook Table of Contents  Study Guide The values of the ideal gas properties when the Mach number is 1 (i.e., sonic
flow) are known as the critical or sonic properties and are given by To
k −1
= 1+
*
T
2
ao To 1 / 2 k −1 1 / 2
=
= 1+ a * T* 2
Po To k /( k −1) k − 1 k /( k −1 )
=
= 1+ P* T * 2
o
* T
= o T* 1/ ( k −1) 1/ ( k −1)
1 + k − 1 =
2 given by
Isentropic Flow Example:
Air flowing through an adiabatic, frictionless duct is supplied from a large supply
tank in which P = 500 kPa and T = 400 K. What are the Mach number Ma. the
temperature T, density _, and fluid V at a location in this duct where the pressure is
430 kPa?
The pressure and temperature in the supply tank are the stagnation pressure and
temperature since the velocity in this tank is practically zero. Then, the Mach
number at this location is 2 Po ( k −1) / k Ma =
− 1 k − 1 P 2 500 Ma = 0.4 430 0.4/1.4 − 1 Ma = 0.469
and the temperature is given by  v v IX5  eText Main Menu  Textbook Table of Contents  Study Guide To
k −1
1+
Ma 2
2
400
T=
1 + 0.2 ( 0.469) 2
T = 383 K
T= The ideal gas equation of state is used to determine the density, = P
430,000
=
= 3.91 kg / m3
R T ( 287) (383) Using the definition of the Mach number and the acoustic velocity, V = Ma k RT = 0.469 (1.4)(287)( 383) = 184 m / s
Solving Compressible Flow Problems
Compressible flow problems come in a variety of forms, but the majority of
them can be solved by
1. Use the appropriate equations and reference states (i.e., stagnation and sonic
states) to determine the Mach number at all the flow field locations
involved in the problem.
2. Determine which conditions are the same throughout the flow field (e.g.,
the stagnation properties are the same throughout an isentropic flow field).
3. Apply the appropriate equations and constant conditions to determine the
necessary remaining properties in the flow field.
4. Apply additional relations (i.e., equation of state, acoustic velocity, etc.) to
complete the solution of the problem.
Most compressible flow equations are expressed in terms of the Mach number.
You can solve these equations explicitly by rearranging the equation, by using
tables, or by programming them with spreadsheet or EES software.  v v IX6  eText Main Menu  Textbook Table of Contents  Study Guide Isentropic Flow with Area Changes
All flows must satisfy the continuity and momentum relations as well as the
energy and state equations. Application of the continuity and momentum
equations to a differential flow (see textbook for derivation) yields: dV
1
dA
=
V
Ma 2 − 1 A
This result reveals that when Ma < 1 (subsonic flow) velocity changes are the
opposite of area changes. That is, increases in the fluid velocity require that the
area decrease in the direction of the flow. For supersonic flow (Ma > 1), the area
must increase in the direction of the flow to cause an increase in the velocity.
Changes in the fluid velocity dV can only be finite in sonic flows (Ma = 1) when
dA = 0. The effect of the geometry upon velocity, Mach number, and pressure is
illustrated in Figure 1 below. Figure 1 ˙
Combining the mass flow rate equation m =
preceding isentropic flow equations yields
* A V = constant with the 1 /( k −1) 2
k −1 =
1+
Ma 2 2
k + 1  v v IX7  eText Main Menu  Textbook Table of Contents  Study Guide V*
12
k −1 =
1+
Ma 2 V Ma k + 1 2 1/2 ( k + 1) / [ 2 (k −1 )] A
1 1 + 0.5( k − 1) Ma2 = A* Ma 0.5(k + 1) where the sonic state (denoted with *) may or may not occur in the duct. If the
sonic condition does occur in the duct, it will occur at the duct minimum or
maximum area. If the sonic condition occurs, the flow is said to be choked since
*
˙
the mass flow rate m = A V =
A* V * is the maximum mass flow rate the
duct can accommodate without a modification of the duct geometry.
Review Example 9.4 of the textbook.
Normal Shock Waves
Under the appropriate conditions, very thin, highly irreversible discontinuities can
occur in otherwise isentropic compressible flows. These discontinuities are known
as shock waves which when they are perpendicular to the flow velocity vector are
called normal shock waves. A normal shock wave in a onedimensional flow
channel is illustrated in Figure 2. Figure 2  v v IX8  eText Main Menu  Textbook Table of Contents  Study Guide Application of the second law of thermodynamics to the thin, adiabatic normal
shock wave reveals that normal shock waves can only cause a sharp rise in the gas
pressure and must be supersonic upstream and subsonic downstream of the normal
shock. Rarefaction waves that result in a decrease in pressure and increase in
Mach number are impossible according to the second law.
Application of the conservation of mass, momentum, and energy equations along
with the ideal gas equation of state to a thin, adiabatic control volume surrounding
a normal shock wave yields the following results. ( k − 1)Ma12 + 2 , Ma > 1
Ma =
1
2 k Ma12 − ( k − 1)
2
2 P2 1 + k Ma12
=
2
P1 1 + k Ma 2
2
1 V1
( k + 1) Ma12
=
=
V2 ( k − 1) Ma12 + 2 To1 = To 2
2
T2
2 2 k Ma − ( k − 1)
1
= [2 + (k − 1) Ma1 ]
2
T1
( k + 1) 2 Ma1
k / (k −1 ) Po 2
=
Po1 o2
o1 ( k + 1) Ma12 =
2 2 + ( k − 1) Ma1 A2* Ma 2 2 + (k − 1) Ma12 = 2
A1* Ma1 2 + (k − 1) Ma2 1/ ( k − 1) k +1 2 2 k Ma1 − (k − 1) ( k + 1) / [ 2 ( k −1) ] When using these equations to relate conditions upstream and downstream of a
normal shock wave, keep the following points in mind:  v v IX9  eText Main Menu  Textbook Table of Contents  Study Guide 1. Upstream Mach numbers are always supersonic while downstream Mach
numbers are subsonic.
2. Stagnation pressures and densities decrease as one moves downstream
across a normal shock wave while the stagnation temperature remains
constant.
3. Pressures increase greatly while temperature and density increase
moderately across a shock wave in the downstream direction.
4. The effective throat area increases across a normal shock wave in the
downstream direction.
5. Shock waves are very irreversible causing the specific entropy downstream
of the shock wave to be greater than the specific entropy upstream of the
shock wave.
Moving normal shock waves such as those caused by explosions, spacecraft
reentering the atmosphere, and others can be analyzed as stationary normal
shock waves by using a frame of reference that moves at the speed of the
shock wave in the direction of the shock wave.
ConvergingDiverging Nozzle Example: Also see Example 9.6 of textbook
Air is supplied to the convergingdiverging
nozzle shown here from a large tank where
P = 2 Mpa and T = 400 K. A normal shock
wave in the diverging section of this nozzle
forms at a point P 1 = Po2 = 2 MPa where
o
the upstream Mach number is 1.4. The
ratio of the nozzle exit area to the throat
area is 1.6. Determine (a) the Mach
number downstream of the shock wave, (b) the Mach number at the nozzle
exit, (c) the pressure at the nozzle exit, and (d) the temperature at the nozzle
exit.
This flow is isentropic from the supply tank (1) to just upstream of the
normal shock (2) and also from just downstream of the shock (3) to the exit
(4). Stagnation temperatures do not change in isentropic flows or across
shock waves, To1 = To 2 = To3 = To 4 = 400 K . Stagnation pressures do not
change in isentropic flows, P 1 = Po2 = 2 MPa and P 3 = Po 4 , but
o
o
stagnation pressures change across shocks, P 2 > Po3 .
o  v v IX10  eText Main Menu  Textbook Table of Contents  Study Guide Based upon the Mach number at 2 and the isentropic relations,
2
A2 A3 A2
1 (1 + 0.2 Ma2 )
=
=
=
= 1.115
At At At* Ma2
1.728
3 The normal shock relations can be used to work across the shock itself. The
answer to (a) is then:
2 (k − 1) Ma2 + 2 Ma3 = 2 2 k Ma2 − ( k − 1) 1/2 1/2 ( 0.4)(1.4 )2 + 2 =
2 2 (1.4) (1.4 ) − 0.4 = 0.740 Continuing to work across the shock,
k / ( k −1) ( k + 1) Ma22 Po 4 = Po3 = Po 2 2 2 + (k − 1) Ma2 1/ ( k −1) k+ 1 2 k Ma 2 − (k − 1) 2 3.5 2.5 (2.4 )( 0.74) 2 2.4 Po 4 = Po3 = 2 2
2 2 + ( 0.4) ( 0.74) 2 (1.4) ( 0.74) − 0.4 A3* Ma3 2 + (k − 1) Ma22 = 2
A2* Ma 2 2 + (k − 1) Ma3 = 1.92 MPa ( k + 1) / [ 2 ( k −1) ] = 1.044 Now, we know A4/At, and the flow is again isentropic between states 3 and 4.
Writing an expression for the area ratio between the exit and the throat, we have A4
A4 A4* A3* A2* A4
= 1.6 = * * *
= * (1)(1.044)(1.115)
At
A4 A3 A2 At
A4
Solving for A4
* we obtain
A4 A4
* = 1.374
A4 Using a previously developed equation for choked, isentropic flow, we can write  v v IX11  eText Main Menu  Textbook Table of Contents  Study Guide A4
1 1 + 0.5( k − 1) Ma2 * = 1.374 = A4
Ma 0.5( k + 1) or (k + 1) / [ 2 ( k −1 )] 2
1 (1 + 0.2 Ma4 )
1.374 =
Ma4
1.728 3 The solution of this equation gives answer (b) Ma4 = 0.483.
Now that the Mach number at 4 is known, we can proceed to apply the
isentropic relations to obtain answers (c) and (d). P4 = Po 4
1.92 MPa
k / ( k −1 ) =
1 + 0.2(0.483) 2
[1 + 0.5 (k − 1) Ma42 ] [ T4 = 3.5 = 1.637 MPa To4
400 K
= 382 K
2=
1 + 0.5( k − 1) Ma4 1 + 0.2( 0.483)2 Note: Observe how the sonic area downstream from the shock is not the
same as upstream of the shock. Also, observe the use of the area ratios
to determine the Mach number at the nozzle exit.
The following steps can be used to solve most onedimensional compressible flow
problems.
1. Clearly identify the flow conditions:e.g., isentropic flow, constant stagnation
temperature, constant stagnation pressure, etc.
2. Use the flow condition relationships, tables, or software to determine the
Mach number at major locations in the flow field.
3. Once the Mach number is known at the principal flow locations, one can
proceed to use the flow relations, tables, or software to determine other flow
properties such as fluid velocity, pressure, and temperature. This may
require the reduction of property ratios to the product of several ratios, as
was done with the area ratio in the above example to obtain the answer.  v v IX12  eText Main Menu  Textbook Table of Contents  Study Guide Operation of ConvergingDiverging Nozzles
A convergingdiverging nozzle like that shown in Figure 3 can operate in several
different modes depending upon the ratio of the discharge and supply pressure
Pd/Ps. These modes of operation are illustrated on the pressure ratio – axial
position diagram of Figure 3. Figure 3 Mode (a) The flow is subsonic throughout the nozzle, supply, and discharge
chambers. Without friction, this flow is also isentropic and the
isentropic flow equations may be used throughout the nozzle.
Mode (b) The flow is still subsonic and isentropic throughout the nozzle and
chambers. The Mach number at the nozzle throat is now unity.
At the throat, the flow is sonic, the throat is choked, and the mass
flow rate through the nozzle has reached its upper limit. Further
reductions in the discharge tank pressure will not increase the
mass flow rate any further.
Mode (c) A shock wave has now formed in the diverging section of the
nozzle. The flow is subsonic before the throat, same as mode (b),
the throat is choked, same as mode (b), and the flow is supersonic  v v IX13  eText Main Menu  Textbook Table of Contents  Study Guide and accelerating between the throat and just upstream of the shock.
The flow is isentropic between the supply tank and just upstream of
the shock. The flow downstream of the shock is subsonic and
decelerating. The flow is also isentropic downstream of the shock
to the discharge tank. The flow is not isentropic across the shock.
Isentropic flow methods can be applied upstream and downstream
of the shock while normal shock methods are used to relate
conditions upstream to those downstream of the shock.
Mode (d) The normal shock is now located at the nozzle exit. Isentropic flow
now exists throughout the nozzle. The flow at the nozzle exit is
subsonic and adjusts to flow conditions in the discharge tank, not
the nozzle. Isentropic flow methods can be applied throughout the
nozzle.
Mode (e) A series of twodimensional shocks are established in the discharge
tank downstream of the nozzle. These shocks serve to decelerate
the flow. The flow is isentropic throughout the nozzle, same as
mode (d).
Mode (f) The pressure in the discharge tank equals the pressure predicted by
the supersonic solution of the nozzle isentropic flow equations. The
pressure ratio is known as the supersonic design pressure ratio.
Flow is isentropic everywhere in the nozzle, same as mode (d) and
(e), and in the discharge tank.
Mode (g) A series of twodimensional shocks are established in the discharge
tank downstream of the nozzle. These shocks serve to decelerate
the flow. The flow is isentropic throughout the nozzle, same as
modes (d), (e), and (f). Review Example 9.9 of the textbook.  v v IX14  eText Main Menu  Textbook Table of Contents  Study Guide Adiabatic, Constant Duct Area Compressible Flow with Friction
When compressible fluids flow through insulated, constantarea ducts, they are
subject to Moodylike pipefriction which can be described by an average DarcyWeisbach friction factor f . Application of the conservation of mass, momentum,
and energy principles as well as the ideal gas equation of state yields the following
set of working equations.
*
f L 1 − Ma 2 k + 1
( k + 1) Ma 2
=
+
ln
D
k Ma2
2k
2 + (k − 1) Ma 2 P
1
( k + 1) = P* Ma 2 + ( k − 1) Ma 2 1/2 V*
1 2 + ( k − 1) Ma 2 =
*= V Ma k +1 T
a
( k + 1)
= *2 =
T* a
2 + ( k − 1) Ma 2 1/2 P
=
Po* o
*
o 1 2 + (k − 1) Ma2 = Ma k +1 ( k +1) / [ 2 ( k −1) ] where the asterisk state is the sonic state at which the flow Mach number is one.
This state is the same throughout the duct and may be used to relate conditions at
one location in the duct to those at another location. The length of the duct enters
these calculations by
*
*
f ∆L f L f L = − D 1 D 2
D Thus, given the length ∆L of the duct and the Mach number at the duct entrance or
exit, the Mach number at the other end (or location) of the duct can be determined.  v v IX15  eText Main Menu  Textbook Table of Contents  Study Guide Compressible Flow with Friction Example:
Air enters a 0.01mdiameter duct ( f = 0.05) with Ma = 0.05. The pressure and
temperature at the duct inlet are 1.5 MPa and 400 K. What are the (a) Mach
number, (b) pressure, and (c) temperature in the duct 50 m from the entrance?
At the duct entrance, with f = 0.05, D = 0.01 m, and Ma = 0.05, we obtain f L* 1 − Ma 2 k + 1
( k + 1) Ma 2 +
ln = D 1 k Ma 2
2k
2 + (k − 1) Ma 2 1 f L* 1 − 0.052
2.4
(2.4) 0.052 +
ln = = 280 D 1 1.4 (0.05)2 2.8 2 + ( 0.4) 0.052 1
Then, at the duct exit we obtain f L* f L* f ∆L
( 0.05) 50
= 280 −
= 30 =
− D 2 D 1
D
0.01
We can not write for the duct exit that 1 − Ma2 k + 1 f L* ( k + 1) Ma 2 ln = 30 = 2+ D 2
2k
2 + (k − 1) Ma 2 2 k Ma
or 1 − Ma22 2.4
2.4 Ma22
30 =
ln
2+
1.4 Ma2 2.8 2 + 0.4 Ma22 The solution of the second of these equations gives answer (a) Ma2 = 0.145.
Writing the following expression for pressure ratios yields for (b), P2 P2* P *
P2 = P1 * * 1
P2 P P
1
1  v v IX16  eText Main Menu  Textbook Table of Contents  Study Guide 1
P2 = (1.5)
Ma2 ( k + 1)
Ma1 2 + ( k − 1) Ma12 2 + ( k − 1) Ma 2 (1) 1 k +1 2
1/2 1/2 1/2 1
2.4 0.05 2 + ( 0.4) 0.052 P2 = (1.5) (1) 1 0.145 2 + (0.4 ) 0.1452 2.4 1/2 = 0.516 Application of the temperature ratios yields answer (c), T1* T2* T2
2 + (k − 1) Ma12
2 + ( 0.4) 0.052
T2 = T1
= 400
= 399
2 = 400
T1 T1* T2*
2 + ( k − 1) Ma 2
2 + ( 0.4) 0.1452
This example demonstrates what happens when the flow at the inlet to the duct is
subsonic, the Mach number increases as the duct gets longer. When the inlet flow
is supersonic, the Mach number decreases as the duct gets longer. A plot of the
specific entropy of the fluid as a function of the duct Mach number (length) is
presented in Figure 4 for both subsonic and supersonic flow. Figure 4 These results clearly illustrate that the Mach number in the duct approaches unity
as the length of the duct is increased. Once the sonic condition exists at the duct
exit, the flow becomes choked. This figure also demonstrates that the flow can
never proceed from subsonic to supersonic (or supersonic to subsonic) flow, as this
would result in a violation of the second law of thermodynamics.  v v IX17  eText Main Menu  Textbook Table of Contents  Study Guide Other compressible flows in constant area ducts such as isothermal flow with
friction and frictionless flow with heat addition may be analyzed in a similar
manner using the equations appropriate to each flow. Many of these flows
also demonstrate choking behavior. Oblique Shock Waves
Bodies moving through a compressible fluid at speeds exceeding the speed of
sound create a shock system shaped like a cone. The halfangle of this shock cone
is given by = sin−1 1
Ma This angle is known as the Mach angle . The interior of the shock cone is called
the zone of action. Inside the zone of action, it is possible to hear any sounds
produced by the moving body. Outside the Mach cone, in what is known as the
zone of silence, sounds produced by the moving body cannot be heard.
An oblique shock wave at angle
with respect to the approaching compressible
fluid whose Mach number is supersonic is shown in Figure 5. Observe that the
streamlines (parallel to the velocity vector) have been turned by the deflection
angle by passing through the oblique shock wave. Figure 5  v v IX18  eText Main Menu  Textbook Table of Contents  Study Guide This flow is readily analyzed by considering the normal velocity components
Vn 1 = V1 sin
and Vn2 = V2 sin ( − ) and the tangential components
Vt 1 and Vt 2 . Application of the momentum principle in the tangential
direction (along which there are no pressure changes) verifies that
VV
=
tt
12 Vt 1 = Vt 2 By defining the normal Mach numbers as Man1 = Vn1
= Ma1 sin
a1 and Man2 = Vn 2
= Ma2 sin ( −
a2 ) The simultaneous solution of the conservation of mass, momentum, and energy
equations in the normal direction along with the ideal gas equation of state are the
same as those of the normal shock wave with Ma1 replaced with Man1 and Ma2
replaced with Man2. In this way, all the results developed in the normal shock
wave section can be applied to twodimensional oblique shock waves.
Oblique Shock Example:
A twodimensional shock wave is created at the leading edge of an aircraft flying
at Ma = 1.6 through air at 70 kPa, 300 K. If this oblique shock forms a 55o angle
with respect to the approaching air, what is (a) the Mach number of the flow after
the oblique shock (this is not the normal Mach number) and (b) the streamline
deflection angle ?
The velocity of the fluid upstream of the oblique shock wave is V1 = Ma1 a1 = Ma1 k R T = 1.6 (1.4)( 287)( 300) = 556 m / s
whose components are Vn 1 = V1 sin = 5 5 6 s i n 5 5 455 m / s
= Vt 1 = Vt2 = V1 cos = 556cos55 319 m / s
=  v v IX19  eText Main Menu  Textbook Table of Contents  Study Guide The upstream normal Mach number is then Man1 = Ma1 sin = 1 . 6 s i n 5 5 1.311
= and the downstream normal Mach number is
2 ( k − 1)Man 1 + 2 Man2 = 2 2 k Man 1 − ( k − 1) 1/2 1/2 (0.4 )(1.311) 2 + 2 =
2 2 (1.4) (1.311) − 0.4 = 0.780 and the downstream temperature is
2 2 k Man1 − ( k − 1)
2
T2 = T1 [(k − 1) Man1 + 2] (k + 1)2 Man21 2 (1.4) 1.3112 − 0.4 2
T2 = 300 [ (0.4)1.311 + 2] = 359 K
( 2.4) 21.3112 Now, the downstream normal velocity is Vn 2 = Man 2 a2 = Ma n2 k R T2 = 0.780 (1.4)(287)( 359) = 296 m / s and the downstream fluid velocity is V2 = Vn22 + Vt 22 = 296 2 + 319 2 = 435 m / s
and the downstream Mach number is Ma2 = V2
435
=
= 1.15
a2
(1.4 )( 287)(359) According to the geometry of Figure 5,  v v IX20  eText Main Menu  Textbook Table of Contents  Study Guide = − tan −1 Vn 2
296
= 55 − tan −1
= 12.1
Vt 2
319 Other downstream properties can be calculated in the same way as the
downstream temperature by using the normal Mach numbers in the normal
shock relations. PrandlMeyer Expansion Waves
The preceding section demonstrated that when the streamlines of a supersonic flow
are turned into the direction of the flow an oblique compression shock wave is
formed. Similarly, when the streamlines of a supersonic flow are turned away
from the direction of flow as illustrated in Figure 6, an expansion wave system is
established. Unlike shock waves (either normal or oblique) which form a strong
discontinuity to change the flow conditions, expansion waves are a system of
infinitesimally weak waves distributed in such a manner as required to make the
required changes in the flow conditions. Figure 6 The Mach waves that accomplish the turning of supersonic flows form an angle
−1
with respect to the local flow velocity equal to the Mach angle = sin (1 / Ma)
and are isentropic. Application of the governing conservation equations and
equation of state to an infinitesimal turning of the supersonic flow yields  v v IX21  eText Main Menu  Textbook Table of Contents  Study Guide k + 1
− ( Ma) = (Ma) = k − 1 2 −1 ( k − 1)( Ma − 1)
tan k +1 1/2 1/2 − tan −1 ( Ma 2 − 1) 1/2 where (Ma) is the PrandlMeyer expansion function . The overall change in the
flow angle as a supersonic flow undergoes a PrandlMeyer expansion is then ∆ ( Ma1 ) − ( Ma2 ) = where 1 refers to the upstream condition and 2 refers to the downstream condition.
The flow through a PrandlMeyer expansion fan is isentropic flow. The
isentropic flow equations can then be used to relate the fluid properties
upstream and downstream of the expansion fan.
Example:
Air at 80 kPa, 300 K with a Mach number of 1.5 turns the sharp corner of an airfoil
as shown here. Determine the angles of the initial and final Mach waves, and the
downstream pressure and temperature of this flow. The initial angle between the flow velocity vector and the PrandtlMeyer fan is the
Mach angle.
−1
1 = sin 1
1
= sin −1
= 41.8 0
Ma1
1.5  v v IX22  eText Main Menu  Textbook Table of Contents  Study Guide The upstream PrandtlMeyer function is ( Ma1 ) = k + 1 k − 1 2
−1 ( k − 1)( Ma1 − 1) tan k +1 1/2 1/2 2
1/2 2.4 tan −1 ( 0.4)(1.5 − 1) ( Ma1 ) = 0.4
2.4 1/2 − tan −1 ( Ma12 − 1) 1/2 − tan−1 (1.52 − 1) 1/2 ( Ma1 ) = 11.90 0
The downstream PrandtlMeyer function is then ( Ma2 ) = ( Ma1 ) − ∆ = 11.90 − 100 = 1.90 0 Solving the PrandtlMeyer function gives the downstream Mach number
Ma2 = 1.13 . The downstream Mach angle is then 2 = 62.2 0 . According to
the geometry of the above figure,
2 = 2 − ∆ = 62.20 − 10 0 = 52.2 0 Since T0 and P0 remain constant, the isentropic flow relations yield k −1
2
Ma12
T01 T2
1 + 0.2(1.5)
2
T2 = T1
= T1
= 300
= 346 K
k −1
2
T1 T02
1 + 0.2(1.13)
1+
Ma2
2
1+ 1 + k − 1 Ma 2 1
P01 P2 2
P2 = P1
= P k −1
1 2
P P02
1
1+
Ma2 2 k / ( k −1) 1 + 0.2(1.5)2 = 80 1 + 0.2(1.13) 3.5 = 132 MPa Students are encouraged to examine the flow visualization photographs in Ch 9.  v v IX23  eText Main Menu  Textbook Table of Contents  Study Guide ...
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 Spring '08
 Sakar

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