chapter 4 compressible flows.pdf

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44

COMPRESSIBLE FLOWS KKKJ3123 FLUID MECHANICS

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4.1 IDEAL GAS RELATION

Compressible flow analysis requires: 1. Conservation of mass, 2. Conservation of linear momentum, 3. Conservation of energy, 4. Equation of state for an ideal gas.

The equation of state for an ideal gas is RTp = (4.1)

where p is pressure, is density, R is the gas constant (286.9 J/kgK for air) and T is the absolute temperature (in K).

3


Gas constant R could be obtained from the universal gas constant (8.314 J/molK) and the gas molecular weight (28.97 g/mol for air):

=R (4.2)

For ideal gas, internal energy u( is a function of temperature T and its gradient defines the specific heat at constant volume cv:

dTud

Tuc

Vv

((=

=

Wiht cv almost constant, integrating the above relation yields

= 2121 dTcud v( ( )1212 TTcuu v = (( (4.3) 4


In thermodynamic texts, enthalpy h( is defined as

RTupuh +=+= ((( (4.4)

For ideal gas, enthalpy h( is also a function of temperature T and the gradient defines the specific heat at constant pressure cp:

dThd

Thc

pp

((=

=

With cp almost constant, integrating the above relation yields

= 2121 dTchd p( ( )1212 TTchh p = (( (4.5)

5


The relationship between cp and cv could obtained from Eq. 4.4:

RdT

uddT

hddTRudhd +=+=((((

Rcc vp += (4.6) and, the specific heat ratio k is (also constant for air, k = 1.4)

v

p

cc

k = (4.7)

Combining eqn. (4.6) and eqn. (4.7) yields

1= kkRcp (4.8)

1= kRcv (4.9)

6


In thermodynamic texts, with 1=v( being the specific volume, the relationship for entropy s could be written as

( )1dpudvdpuddsT +=+= ((( (4.10) From the definition of enthalpy:

( ) dpdpudpdudhd ++=

+= 1(((

dpdsThd +=( (4.11)

From eqn. (4.10) and eqn. (4.11), the change in entropy ds is

( )( ) p

dpRTdTcdR

TdTcds pv =+=

11

7


Integration of both equations above yield

2

1

1

212 lnln

RTTcss v += (4.12)

1

2

1

212 lnln p

pRTTcss p = (4.13)

From the Second Law of Thermodynamics, for an adiabatic and frictionless flow, 012 == ssds and is known as isentropic flow, where relationships among variables T, p and could be derived as:

1

2

1

2

2

1

1

2

1

2

1

2

2

1

1

2

lnln1

0lnln1

lnln0lnln

ppR

TT

kkRR

TT

kR

ppR

TTcR

TTc pv

==+

==+

8


Hence, for an isentropic flow,

( )

1

2

1

2

1

1

2

pp

TT

kkk

=

=

(4.14)

constant=kp (4.15)

9


4.2 CHARACTERISTICS OF COMPRESSIBLE FLOW

Compressible flow behaviour could be characterised by the Mach Number (Ma or M) which is defined as

cVM =Ma (4.16)

where V is the flow velocity and c is the speed of sound.

Compressible flow could be classified to three categories: 1. Nearly-Incompressible flow M 0.3

> Nearly symmetrical pressure wave, > Change in density is negligible (< 5%), > Hence, the Bernoulli equation could be used.

10


2. Subsonic flow 0.3 < M < 1.0 > Assymetrical pressure wave, > Change in density becomes more significant (> 5%), > When M 1.0, there exists a plane that separates zone of

silence and zone of action and is called Mach wave.

3. Supersonic flow M 1.0 > Mach wave transforms to Mach cone with angle:

MVc 1sinsin 11 ==

> Change in density is very significant, > Shock wave phenomena could develop when flow is

obstructed by a solid body.

11


Condition when M = 1.0 is called sonic condition.

Figure 4.1 Sound waves from a fixed and moving source

12


Figure 4.1 (continued)

13


Figure 4.1 (continued)

14


15

In aerodynamic texts, two more flow regimes are commonly added: 1. Transonic flow 0.9 M 1.2

> A state of partial subsonic/supersonic condition causing analysis to be more complicated.

2. Hypersonic M > 5 > Fluid molecules begin to experience chemical reactions such

as air ionization, causing analysis to be too complex.


Figure 4.2 Moving and static control volume of a pressure pulse

The relationship for speed of sound c is derived by considering the propagation of a pressure pulse.

16


17

To derive it, lets start from the continuity equation (neglecting higher order terms):

( )( )

cVVcVcc

AVccA

=+=

+=

From the linear momentum equation:

( )( )( ) ( )( )

pVcpccVc

ApppAAVcVccAc

==

+=++2

Combining both relations above yields:

pcpc ==2 (4.17)


For isentropic flow, Eq. 4.17 could be written as

s

pc =

From Eq. 4.15, its derivative yields

0

0

1 =

=

+kk

k

dkpdp

pd

kRTkpddpp

s

===

18


Hence, the speed of sound for an isentropic flow is

kRTkpc == (4.18)

For air in standard condition: ( )

401.1KJ/kg 9.28615K 15.288

==

=

kR

CT

Hence,

km/hr 1225m/s 3.340 ==c

19


EXAMPLE 4.1

An aircraft flying at an altitude of 1000 m, passes directly above an observer. If the aircraft is moving at Mach 1.5 with surrounding ambient temperature of 20C, obtain the time after which the observer is able to hear the sound of the aircraft. Use gas constant R = 287 J/kgK and specific heat ratio k = 1.4.

20


Since Ma > 1.0, a Mach cone is formed. From the equation for Mach cone angle:

=== 8.415.1

1sin1sin 11M

From the diagram above:

Vtz

xz ==tan

( )( )s 17.2

8.41tan15.273202874.15.11000

tantantan

=+=

=== kRTMz

cMz

Vzt

21


4.3 ISENTROPIC GAS FLOW

Main characteristics of an isentropic flow are: 1. No heat transfer across CS adiabatic,

2. No friction between fluid molecules in CV inviscid.

For conservation of mass in one-dimensional flow, the general continuity equation is used:

( )0

0constant

=++==

==

dVAdAVdAVAVdmd

AVm

&

&

Dividing by AV gives:

0=++ VdV

AdAd

(4.19)

22


For conservation of momentum, the general Bernoulli equation is used (assuming steady flow and effect of elevation difference g dz is negligible):

0=+ dVVdp Dividing by V2:

02 =+ VdV

Vdp

(4.20)

Combining Eq. 4.19 and Eq. 4.20 yields:

==

=+

ddpV

Vdpd

Vdp

AdA

Vdp

AdAd

2

22

2

1

0

23


For isentropic flow, ddpc = :

( )222 11 MVdVcVVdVAdA =

= (4.21)

dA > 0 dp > 0 dV < 0

dA > 0 dp < 0 dV > 0

Subsonic flow(Ma < 1)

Supersonic flow (Ma > 1)

dA < 0 dp < 0 dV > 0

dA < 0 dp > 0 dV < 0

Figure 4.3 Flow through diverging and converging ducts

24


Relation between and M could be derived from Eq. 4.19-21:

( ) dMMcVd

cV

VdVM

VdV

VdV

AdAd =

=== 21

( ) ( ) == M MddMdd 0 221221 0 ( )2210221

0

expln MM ==

For density change ( ) %500


In the isentropic flow relationships, static quantities could be related to the stagnation quantities through Eq. 4.15:

k

kk pppp

1

00

0

0 constant

===

From the Bernoulli equation:

( ) ( )( )

[ ] 011

0

0

02

21

11

0

10

0

221

10

10

221

10

102

21

0

0

=+

=+

=+=+

+

V

p

p

kk

Vp

p k

k

k

k

Vk

pp

Vdpdpp

VdpdppVddp

26


012

21

0

0 =

V

ppk

k (4.22)

Using the ideal gas equation (Eq. 4.1):

( ) ( ) 01 22102210 == VTTcVTTkkR

p (4.23)

From the relationship above, total/stagnation enthalpy could be defined:

0h(

pemalar22100 =+== VTcTch pp(

(4.24)

Eq. 4.23 is divided by to yield the relationship for T: kRTc =2

02

11

121

20

2

20 =

=

M

TT

kcV

kRTTT

kkR

27


( ) 2210 11 MkTT += (4.25)

Through Eq. 4.14:

( )

pp

TT

kkk00

10 =

=

Thus, relationships for p and are obtained:

( )[ ] ( )12210 11 += kkMkpp (4.26) ( )[ ] ( )112210 11 += kMk (4.27)

28


For sonic condition (M = 1.0):

( ) ( )11

0

1

00 12,

12,

12

+=

+=+=

kkk

kkpp

kTT

(4.28)

For air, taking k = 1.4: 6339.0,5283.0,8333.0 000 === ppTT

Cross sectional area where sonic condition occurs is known as the critical area A*, where the flow is in a state of choked. From continuity equation:

VV

AAVAAV

==

29


where = kRTV is the velocity of the flow at sonic condition. With kRTcV MaMa == , hence,

( )( )

( )( )0

0

0

0

Ma1

Ma TTTT

kRTkRT

AA

==

( )

( )( )12

1

21

221

11Ma11

Ma1

+

++= k

k

kk

AA

(4.29)

30


Figure 4.4 Relation between the critical area ratio and Ma

At choked flow condition, maximum mass flow rate (M > 1.0): == kRTAVAm xma&

For air (k = 1.4), values from Eq. 4.25-29 are easier to be extracted from the Isentropic Air Flow Table (Appendix A).

31


Eq. 4.29 can be used in the design of nozzles and diffusers for supersonic flows.

Ma < 1 Ma = 1 Ma > 1

Supersonic nozzle

Ma > 1 Ma = 1 Ma < 1

Supersonic diffuser Figure 4.5 Nozzle and diffuser for supersonic flow

The subsonic-supersonic flow transition in a nozzle or diffuser could occur at the throat where M = 1.0.

32


Figure 4.6 Effect of sonic condition to operation of converging nozzle

33


Figure 4.7 Effect of sonic condition to operation of

converging-diverging nozzle

34


35

EXAMPLE 4.2

An air flow with a Mach 3.5 velocity has a static pressure of 304 kPa (abs) and static temperature of 180 K. Calculate:

(a) Stagnation pressure and stagnation temperature, (b) Pressure, temperature and the speed of sound for critical

condition, (c) Flow velocity.

Use gas constant R = 287 J/kgK and specific heat ratio k = 1.4.


(a) From App. A, for M = 3.5: 27.760 =pp , 45.30 =TT . Hence, ( )

( )K 621

18045.3

(abs) MPa 3.22(abs) kPa 23186

30427.76

00

00

===

====

TTT

T

ppp

p

(b) At critical condition, for M = 1.0. From App. A: 893.1*00 == pppp , 2.1*00 == TTTT

Hence,

( )(abs) MPa 2.21

2.23893.11

00

**

=== p

ppp

36


( )

( )( )m/s 564

5182874.1

K 185

6212.1

1

*

00

**

===

===

kRTc

TTTT

(c) Flow velocity:

( )( )m/s 419

1802874.15.3==

== kRTMcMV

37


EXAMPLE 4.3

In a subsonic-supersonic flow through a converging-diverging nozzle, the reservoir pressure and temperature respectively are 10 atm (abs) and 300 K. In the nozzle, there are two positions where 6* =AA ; one in the converging section and the other in the diverging section. Calculate the Mach number, pressure, temperature and flow velocity at those two positions. Use gas constant R = 287 J/kgK and specific heat ratio k = 1.4.

38


In the reservoir, the fluid is in stagnation: K 300(abs) kPa 1013(abs) atm 10 00 === T,p

From App. A, for 6* =AA : c)(supersoni 368.3(subsonic) 097.0 == M,M

For converging section, M = 0.097 (subsonic). From App. A: 006.10 =pp , 002.10 =TT

Hence,

( )( )

( )( )m/s 6.33

2992874.1097.0

K 299300002.11

(abs) kPa 10071013006.11

00

00

====

===

===

kRTMcMV

TTTT

pppp

39


For diverging section, M = 3.368 (supersonic). From App. A: 13.630 =pp , 269.30 =TT

Hence,

( )( )

( )( )m/s 476

8.912874.1368.3

K 1.89300269.31

(abs) kPa 6.01101313.63

1

00

00

====

===

===

kRTMcMV

TTTT

pppp

40


4.4 NORMAL SHOCK WAVE

When a supersonic flow is resisted by a body or an obstruction, shock waves will form.

Subsonic

M < 1.0 V < c

M > 1.0 V > c

Supersonic Figure 4.8 Formation of shock wave

Shock waves are thin (107 m) regions in a supersonic flow where the fluid properties changes abruptly.

41


Crossing the shock wave:

change no decreases,,increases,,,

0

0

hpVM

sTp

(

Normal shock

M > 1

Oblique shock Bow shock

M < 1 M > 1

M > 1 M < 1

Figure 4.9 Typical types of shock waves

42


Figure 4.10 Normal shock wave

43


List of relationships/equations used are (taking A constant): 1. Continuity equation

2211 VV = 2. Momentum equation

( ) ( ) 222221111221 VpVpVVmApp +=+= & 3. Energy equation

222

12

212

11

222

12

212

110 VTcVTcVhVhh pp +=++=+=

(((

4. Speed of sound

pkkRTc ==2

44


From the energy equation relationship above:

2121

21212

222

21

21

222

211

VkcV

kc

VkkRTV

kkRT

+=+

+=+

Taking the transition condition as sonic ( )= cV :

( ) 21121

21

2121212

2222

21

21

22

22

2221

21

Vkcc

kkV

kc

Vkcc

kcV

kc

+=+=+

+=+=+

( ) ( )( ) ( ) 222122122

212

12212

1

11

11

Vkckc

Vkckc

+=+=

45


Dividing the momentum equation with continuity equation yields:

122

22

1

21

1222

2

11

1

222

21

11

1

VVkVc

kVc

VVV

pV

p

VV

pVV

p

=

=

+=+

Combining with the relationship from the energy equation above:

( ) ( ) 121221

2*

12

1222

2*

11

2*

21

21

21

21

21

21

VVVVk

kVV

cVVk

k

VVVkkVckVk

kVck

=++

=+++

46


12

12

1

21

2*

=++k

kVV

ck

k

212* VVc = (4.30)

Eq. 4.30 is known as the Prandtl relation and could be manipulated as follows:

11212

111

12

11

12

1

22

21

22

22

21

21

22

2*

21

2*

+=

+

+

=

+++

+++

=

kkM

kM

kk

Vc

kkk

Vc

k

Vc

Vc

47


( )

( )111

212

1

212

122

+=kkM

MkM (4.31)

The change in density across the normal shock wave could be obtained from the continuity equation:

( )( ) 2121

21

2

21

21

21

2

1

1

2

121

VkcVk

cV

VVV

VV

++====

( )

( ) 2121

1

2

121

MkMk

++=

(4.32)

The change in pressure across the normal shock wave could be obtained from the combination of continuity and momentum equation:

( )211122221112 VVVVVpp ==

48


=

=

=

2

121

1

221

21

1

2

1

211

1

12 111

kMVV

ckV

VV

pV

ppp

( )1121 2112 ++= Mkk

pp

(4.33)

The change in temperature across the normal shock wave could be obtained from the ideal gas equation RTp = and Tch p=

(:

1

2

2

1

1

2

1

2

hh

pp

TT (

(==

( ) ( )( ) 212

121

1

2

112

11

21Mk

MkM

kk

TT

++

++= (4.34)

49


The change in entropy across the normal shock could be obtained from Eq. 4.12 or Eq. 4.13:

1

2

1

2

2

1

1

212 lnlnlnln p

pRTTcR

TTcss pv =+=

If M1 = 1: 012 = ss M1 > 1: 012 > ss M1 < 1: 012


From the entropy equation, compare stagnation quantities between locations (1) and (2):

10

20

10

20

10

2012 lnlnln p

pR

pp

RTT

css p ==

=R

sspp 12

10

20 exp (4.36)

The loss in stagnation pressure compared to static pressure is given by the Rayleigh Pitot-tube formula:

1

10

10

20

1

20

pp

pp

pp = (4.37)

For air (k = 1.4), Eq. 4.31-37 are easier obtained from the Normal Shock Properties Table (Appendix B).

51


Figure 4.11 Normal shock properties

52


EXAMPLE 4.4

Consider a normal shock wave in air, with velocity, temperature and pressure of the flow before crossing the shock of 680 m/s, 288 K and 101.3 kPa (abs). Calculate velocity, temperature, pressure and change in entropy after crossing the shock. Use gas constant R = 287 J/kgK and specific heat ratio k = 1.4.

Let the upstream region be denoted as (1) and downstream region as (2). Speed of sound at (1):

( )( )

0.22.340

680m/s 2.340

2882874.1

1

11

1

====

==

cV

M

kRTc

From App. B, for M1 = 2.0: 5.412 =pp , 687.112 =TT , 5774.02 =M

53


Hence,

( )( )

( )( )m/s 552

4862874.15774.0

K 864288687.1

(abs) kPa 5643.1015.4

22222

11

22

11

22

====

===

===

kRTMcMV

TTT

T

ppp

p

and, the change in entropy:

( ) KJ/kg 100514.1

2874.11

=== kkRcp

( ) ( )KJ/kg 3.69

5.4ln287687.1ln1005lnln1

2

1

212

===

pp

RTT

css p

54


EXAMPLE 4.5

A Pitot tube is stationed on the fuselage of an aircraft to measure its speed. If the static pressure measured is 1 atm (abs), determine the Mach numbers for pressure values on the Pitot tube as follows:

(a) 1.276 atm (abs), (b) 2.714 atm (abs), (c) 12.06 atm (abs).

For each case, check whether flow is subsonic or supersonic. From App. B, for M = 1.0: 893.10 =pp .

( ) (abs) atm 893.11893.100 ===

pppp

(a) Since , flow is subsonic. From App. A, for

< 00 pp276.10 =pp :

6.0=M 55


56

(b) Since , flow is supersonic. From App. B, for

> 00 pp714.2120 =pp :

3.1=M (c) Since , flow is supersonic.

From App. B, for

> 00 pp06.12120 =pp :

0.3=M


4.5 NON-ISENTROPIC GAS FLOW

In this topic, the nonisentropic flows that shall be discussed are: 1. Fanno flow Adiabatic (no heat transfer) flow with friction in

constant cross-section duct, 2. Rayleigh flow Non-adiabatic (with heat transfer) frictionless

flow in constant cross-section duct.

Figure 4.12 One-dimensional adiabatic gas flow with friction

57


For Fanno flows, consider the following one-dimensional flow: List of relationships/equations used are:

1. Continuity equation

0constant =+=VdVdV

2. Momentum equation x-direction

( ) ( ) ( )04 =++

+=+dVV

Ddxdp

VdVVmdxDAdpppA

w

w

&

3. Energy equation

0221002

21 =++===+ dVVdTcVTcTchVh ppp

((

58


4. Ideal gas equation of state

TdTd

pdpRTp +==

5. Darcy friction factor relationship

2812

81

2

8fkpMVf

Vf w

w ===

Using the relationship for speed of sound kRTc = and Mach number cVM = , these relationships could be derived:

( )( )( ) DdxfMkMVdVd

Ddxf

MMkk

pdp

2

2

2

22

12

1211Ma

==+=

59


( )( ) DdxfMMkkTdTDdxfkM

dp

dp

2

4

221

0

0

0

0

121

=

==

( )Ddxf

MMk

kMM

dM2

221

22

2

111

+=

Integration of the above relation yields:

( )[ ]

=+= LM

M DdxfdM

kMMkM

0

1 242

21

2

111

( )

( ) DLf

MkMk

kk

kMM =

++++ 2

2

2

2

121ln

211

(4.38)

60


Other relationships could be derived in a similar manner:

( )( )

=

+= 1 22

2121

111M

M

T

TdM

Mkk

TdT

( ) 2121

Mkk

TT

++= (4.39)

( )

( )21

2

2

121

++=== Mk

MkTTM

kRTkRTM

VV

(4.40)

( )

( )21

2

2

112

++==

Mk

MkVV

(4.41)

= TT

pp

61


( )( )

21

21211

++= Mk

kMp

p (4.42)

=0

0

0

0

pp

pp

pp

pp

( ) ( ) ( )1212

0

0

1121

+

++=

kk

kMk

Mpp

(4.43)

From Eq. 4.38-43, changes in fluid properties with increasing values in f could be summarised as in Table 4.1.

62


63

Table 4.1 Characteristics of Fanno flows

Property Subsonic Supersonic M, V p, p0, 0 T s


Figure 4.13 One-dimensional non-adiabatic frictionless gas flows

64


For Rayleigh flows, consider the following one-dimensional flow: List of relationships/equations used are:

1. Continuity equation

constant2211 == VV 2. Momentum equation x-direction

( ) ( )1221 VVmApp = & 3. Energy equation

( ) 10202121122212 hhmQmQqVhVhmQ ((&&((&& ===+= 4. Ideal gas equation of state

( )1020102022

2

11

1 , TTchhT

pT

pp ==

((

65


Using the relationship for speed of sound kRTc = and Mach number cVM = , the following relationships could be derived:

( ) ( )[ ]

( )222

212

0

0

1

1112

kM

MkMkTT

+++= (4.44)

211

kMk

pp

++= (4.45)

( ) 2

211

++= kM

MkTT

(4.46)

( ) 22

11

MkkM

VV

++==

(4.47)

( ) ( )12

20

0

112

11

++

++=

kk

kMk

kMk

pp

(4.48)

66


From Eq. 4.44-48, changes in fluid properties with respect to increasing values of q could be summarised as in Table 4.2:

Table 4.2 Characteristics of Rayleigh flows

Heating Cooling Property

Subsonic Supersonic Subsonic Supersonic Ma, V T0 T ( )( )1Ma1

1Ma0

k

k

( )( )1Ma11Ma0

k

k

p, p0, 0 s

67


4.6 OBLIQUE SHOCK WAVE

Shock wave in two dimensions is known as oblique shock.

Figure 4.14 Oblique shock wave

68


If a body is (or could be approximated as) a particle, oblique shock angle is equivalent to Mach cone angle (see Fig. 4.1):

MVc 1sin == (4.49)

Consider a wedge-shaped body with deflection angle that produces oblique shock of angle . List of relationship/equations used: 1. Continuity equation

2211 nn VV = 2. Momentum equation normal component

211

22221 nn VVpp =

3. Momentum equation tangential component

( )12110 ttn VVV = 69


4. Energy equation

0222

1222

12

212

1212

11 hVVhVVh tntn =++=++

((

From Fig. 4.14 (and relation 3 above), there is no change in the tangential velocity component:

constant21 === ttt VVV Thus, relations 1, 2 and 4 are similar to the case of normal shock, but V and Ma need to be taken in its normal components:

sin11

11 Mc

VM nn == (4.50)

( ) == sin22

22 Mc

VM nn (4.51)

70


The relationship --M1 could be obtained from Eq. 4.32 by replacing M1 Mn1 and V Vn:

( )( ) ( ) 2

122

1

221

1

2

tantan

sin12sin1

n

n

VV

MkMk ==+

+=

Solving for :

( )( ) 2cos2 1sincot2tan 221

221

++=

kMM

(4.52)

Eq. 4.52 is plotted in Fig. 4.15 or in Appendix D, and it shows 2 solutions for : 1. Weak shock results in M > 1 downstream, 2. Strong shock results in M < 1 downstream.

71


Figure 4.15 --Ma diagram for oblique shock wave

72


Figure 4.16 Supersonic flow past a wedge structure

73


EXAMPLE 4.6

Air at Mach 2 with 70 kPa (abs) pressure, flows through a 10 ramp and forms a weak oblique shock as shown in the figure below. Calculate:

(a) Shock angle , (b) Downstream Mach number M2, (c) Downstream pressure p2, (c) Total pressure downstream p02, measured by a Pitot tube.

10M1 = 2.0

p1 = 70 kPa

74


75

(a) From App. D, for M1 2.0 and p1 70 kPa: 3.39

(b) Normal component of M1: 267.13.39sin0.2sin11 MMn

From App. B: 609.2,707.1,8031.0 120122 ppppM n

Hence, the Mach number downstream:

64.1

103.39sin8031.0

sin2

2

nMM

(c) Pressure downstream:

(abs) kPa 11970707.111

22 pp

pp


76

(d) Total pressure downstream:

(abs) kPa 83170609.211

2020 pp

pp


4.7 PRANDTL-MEYER EXPANSION WAVE

Figure 4.15 Comparison between oblique shock and expansion wave

77


When supersonic flow is deflected from the adjacent flow, an expansion wave is formed.

Crossing this expansion wave: 1. but 12 MM > 121212 ,, TTpp


Figure 4.16 Formation of expansion wave

If d is a small angle between two Mach waves, using sine rule: ( )

( ) ( )

ddVdVV

+=+=+

coscos

2sin2sin

ddVdV

sinsincoscoscos1 =+

79


For small angle ( ) ddd sin,1cos ,

tan11

sincoscos1

ddVdV

==+

Using series expansion (for x < 1) and neglecting high order terms:

L++++=321

11 xxx

x

L++=+ tan11 dVdV

tanVdVd =

80


From Eq. 4.49:

1

1tan1sin2

1

==

MM

Therefore, differential equation for the Prandtl-Meyer flow is obtained:

VdVMd 12 = (4.53)

Integrating the above equation yields:

= 21

2

1

12M

M VdVMd

81


Combining with the relation cVM = and RTc = :

cdc

MdM

VdVcMV +=+= lnlnln

MdM

MkVdVdM

Mk

Mk

cdc

MkccMkTT

cc

22

212

020

20

211

1

211

21

211

211

+=+

=

+=+==

Hence,

+== 2

1

2

1 2

2

12

211

1MM M

dM

MkMd (4.54)

82


The integration above is known as the Prandtl-Meyer function:

( ) + 2

1 2

2

211

1MM M

dM

MkMM

With 01 = , and ( ) 0=M at 11 =M , this yields

( ) ( ) 1tan111tan11 2121 ++= MMkkkkM (4.55) Eq. 4.55 and Eq. 4.49 is given in Appendix C for air ( )4.1=k .

83


EXAMPLE 4.7

A supersonic air flow with properties M1 = 1.5, p1 = 81.5 kPa (abs) and T1 = 256 K passes an expansion angle (see figure below) that deflects the flow through angle 2 = 20. Calculate M2, p2, T2, p02, T02, and the upstream and downstream angles of the Mach lines.

20

M1 = 1.5

84


From App. C, for M1 = 1.5:

=+=+===

91.312091.1181.41,91.11

112

11

From App. C, for 2 = 31.91: 207.2,95.26 22 == M

From App. A, for M1 = 1.5: 45.1,671.3 110110 == TTpp

From App. A, for M2 = 2.207: 974.1,81.10 220220 == TTpp

Flow through an expansion wave is isentropic:

20102010 , TTpp ==

85


86

Therefore,

K 37125645.1

(abs) kPa 2995.81671.3

K 18825645.11975.11

(abs) kPa 7.275.81671.3181.10

1

11

101020

11

101020

11

10

10

20

20

22

11

10

10

20

20

22

TTT

TT

ppp

pp

TTT

TT

TTT

ppp

pp

ppp

95.62095.26angle, downstream lineMach 81.41angle, upstream lineMach

22

1

Chapter 4 Compressible Flows4.1 Ideal Gas Relation4.2 Characteristics of Compressible FlowExample 4.1

4.3 Isentropic Gas Flow Example 4.2Example 4.3

4.4 Normal Shock WaveExample 4.4Example 4.5

4.5 Non-isentropic Gas Flow4.6 Oblique Shock WaveExample 4.6

4.7 Prandtl-Meyer Expansion WaveExample 4.7

Appendix AAppendix BAppendix CAppendix D

chapter 4 compressible flows.pdf

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