gas dynamics Unit 3 Clear Notes.208-248

7/25/2019 gas dynamics Unit 3 Clear Notes.208-248

1/41

02

11280.522

* 2160.91T

From Rayleighs flow table for =1.4and02*T

=0.522

M2=.4

2

2*

1.961P

P

02

02*

1.961P

P

2

2

0.615*

T

T

2 2* 1.196

1.759 1.196

P P

= 3.44 x10 5N/m2

T2=T2* x.615

=1801.18K

P02=3.849 x105N/m2

Stagnation pressure loss

P0=P01-P02

=4.11 x10 5-3.84 x105

=.261 x10 5N/m2

Heat supplied Q=CP(T02-T01)

=1005(1128-376)

=755.7 x10 3J/kg

Normal Shocks

When there is a relative motion between a body and fluid, the disturbance is created

if the disturbance is of an infinitely small amplitude, that disturbance is transmitted


2/41

through the fluid with the speed of sound. If the disturbance is finite shock waves are

created.

Shock Waves and Expansion Waves Normal Shocks

Shocks which occur in a plane normal to the direction of flow are called normal

shock waves. Flow process through the shock wave is highly irreversible and cannot

be approximated as being isentropic. Develop relationships for flow properties before

and after the shock using conservation of mass, momentum, and energy

In some range of back pressure, the fluid that achieved a sonic velocity at the

throat of a converging-diverging nozzle and is accelerating to supersonic velocities in

the diverging section experiences a normal shock. The normal shock causes a

sudden rise in pressure and temperature and a sudden drop in velocity to subsonic

levels. Flow through the shock is highly irreversible, and thus it cannot beapproximated as isentropic. The properties of an ideal gas with constant specific

heats before (subscript 1) and after (subscript 2) a shock are related by

Assumptions


3/41

1. Steady flow and one dimensional

2. dA = 0, because shock thickness is small

3. Negligible friction at duct walls since shock is very thin

4. Zero body force in the flow direction

5. Adiabatic flow (since area is small)

6. No shaft work

7. Potential energy neglected

Intersection of Fanno and a Rayleigh Line

Fanno and Rayleigh line, when plotted on h-s plane, for same mass velocity

G, intersect at 1 and 2.as shown in fig. All states of Fanno line have same stagnation

temperature or stagnation enthalpy, and all states of Rayleigh line have same stream

thrust F / A. Therefore, 1 and 2 have identical values of G, h 0and F / A. from 1 to 2

possible by a compression shock wave without violating Second Law

Thermodynamics. A shock is a sudden compression which increases the pressureand entropy of the fluid but the velocity is decrease from supersonic to subsonic.

A change from states 2 to 1 from subsonic to supersonic flow is not possible in

view Second Law Thermodynamics. (Entropy can not decrease in a flow process)


4/41

Governing Equations

(i) Continuity

x ym m

x y

AV AV

x x y yV V (Shock thickness being small Ax= Ay)

Gx= Gy(Mass velocity)

Mass velocity G remains constant across the shock.

(ii) Energy equation

SFEE: 2 22 1

sh 2 1 2 1

V Vq h h g z z

2

2 22 1

2 1

V Vh h

2 2

Across the shock

2 2y xy x

V Vh h

2 2

ox oyh h

ox oyT T

Toremains constant across the shock.

(iii) Momentum

Newtons second law

xx xx xx xxcv out inF mV mV mVt

x y xx xxout inP A P A 0 mV mV

y x y xm V V AV V V 2 2y y x xAV AV


5/41

Momentum gives

2 2x y y y x xP A P A AV AV

2 2x x x y y yP A AV P A AV

x yF F

Impulse function remains constant across the shock.

Property relations across the shock.

(1)y

x

T

T

Energy

ox oyh h

ox oyT T (1)

For the isentropic xox

2oxx

x

T K 11 M

T 2

2ox x x

K 1T T 1 M

2

.......... (2)

Similarly

2oy y y

K 1T T 1 M2

............ (3)

Combining (1), (2) and (3)

2 2x x y y

K 1 K 1T 1 M T 1 M

2 2


6/41

2x

y

2xy

K 11 MT 2

K 1T1 M

2

(ii)y

x

P

P

Momentum

Fx= Fy

2 2x x x y y yP A AV P A AV

22y yx x

x yx y

VVP 1 P 1

P P

2 2x x y yP 1 KM P 1 KM

2y x

2x y

P 1 KM

P 1 KM

(iii)y

x

Equation of state P = RT

x x xP RT

y y yP RT

x

x yxx

y x yy

y

PP TRT

T PP

RT

22

xyx2

2y xy

K 11 M1 KM 2

K 11 KM 1 M2


7/41

(iv)y

x

V

V

Continuity equation

x x y yV V

y x

x y

V

V

Equation of state

P = RT

x x xP RT

y y yP RT

22

xy y

22x xy

K 11 MV 1 KM 2

K 1V 1 KM 1 M2

(v) oy

ox

PP

For the isentropic xox

K

K 12oxx

x

P K 11 M

P 2

For the isentropic (yoy)

K

K 1oy 2y

y

P K 11 M

P 2

K

K 12oy y y

K 1P P 1 M

2


8/41

K

K 12y

oy y

2ox xx

K 11 MP P 2

K 1P P1 M

2

But2

y x2

x y

P 1 KM

P 1 KM

K

K 122 y

oy x2

2ox yx

K 11 MP 1 KM 2

K 1P 1 KM 1 M2

(vi) Entropy change (S)

y xS S S

y y oy oy oyp p

x x ox ox ox

T P T P PC ln R ln C ln R ln R ln

T P T P P

K

K 122

yxy x 2

2yx

K 11 M

1 KM 2S S R lnK 11 KM 1 M

2

oy oxT T

(vii) Relation between 2xM and2yM

Prove that

2x

2y

2x

2M

K 1M2K

M 1

K 1

We have

(1)

2x

y

2xy

K 11 MT 2

K 1T1 M

2


9/41

(2)2

y x2

x y

P 1 KM

P 1 KM

(3)y yx x

x y

P MP M

T T

Equation (3)

2 2y y y

2 2xx x

P M T.

TP M

22 22 xyx

22xyy

K 11 MM1 KM 2

K 1M1 KM

1 M2

2 2

2 2 2 2 2 2x y y y x x

K 1 K 11 KM M 1 M 1 KM M 1 M

2 2

L.H.S. = 2 2 4 2 2x x y yK 1

1 2KM K M M 1 M2

22 2 2 4 2 2 2 4 2

y x x y x y x y

K K 1K 1M 1 2KM K M M K K 1 M M M M

2 2

22 2 2 2 4 2 4 2 4 4 4

y x y x y y x y x y

K K 1K 1M 2KM M K M M M K K 1 M M M M

2 2

Similarly R.H.S.

22 2 2 2 2 2 4 2 4 4 4

x x y y x x y x y x

K K 1K 1M 2K M K M M M K K 1 M M M M

2 2

2 2 2 4 2 4 2 4 4 2 4 2 4y x x y y x y x x y y xK 1

M M K M M M M M M K K 1 M M M M2

2 2 2 2 2 2y x y x x yK 1

M M 1 M M KM M 02

Since y xM M

So 2 2y xM M cannot be equal to zero


10/41

Hence 2 2 2 2y x x yK 1

1 M M KM M2

2 2 2 2

y x x y

K 1

M M KM M 12

2 2 2 2y x x y2

M M KM M 1K 1

2 2 2 2y x y x

2K 2M M M M

K 1 K 1

2 2 2y x x

2K 2M 1 M M

K 1 K 1

2x

2y

2x

2M

K 1M2K

M 1K 1

Property relations in terms of incident Mach Number Mx.

(1) Provey 2

xx

P 2K K 1M

P K 1 K 1

Momentum Fx= Fy

2 2x x y yP 1 KM P 1 KM

2y x

2x y

P 1 KM

P 1 KM

but

2

x2y

2x

2M

K 1M2K

M 1K 1

2y x

2xx

2x

P 1 KM

2PM

K 11 K2K

M 1K 1

2 2x x2 2x x

2K1 KM M 1

K 1

2K 2M 1 K M

K 1 K 1

(1)


11/41

Dinominator 2 2r x x2K 2

D M 1 K MK 1 K 1

2x

2K 2K M K 1K 1 K 1

2x

K 1 K 1KM K 1 K 1

2xK 1

1 KMK 1

substitute in equation (1).

y 2x

x

P 2K K 1M

P K 1 K 1

(2) Prove that

2 2x x

y2

x 2x

K 1 2K 1 M M 1

T 2 K 1T K 1

M2 K 1

Energy

ox oyh h

ox oyT T

2x

y

2xy

K 11 MT 2

K 1T1 M

2

2x

2y

2x

2M

K 1M2K

M 1K 1

2x

y

2xx

2x

K 11 MT 2

2TM

K 1 K 112K2

M 1K 1

2 2x x

2 2x x

K 1 2K 1 M M 1

2 K 1

2K K 1 2M 1 M

K 1 2 K 1

Denominator, 2 2r x x2K K 1

D M 1 1 MK 1 2


12/41

2 2

2 2 2x x x

4K K 1 K 12K K 1M M M

K 1 2 2 K 1 2 K 1

Hence

2 2x x

y

2x 2

x

K 1 2K 1 M M 1T 2 K 1

T K 1M

2 K 1

(3) Prove that

K

K 1 12x K 1oy 2

x2oxx

K 1MP 2 K 12 KM

K 1P K 1 K 11 M

2

For the isentropic xox,

K

K 12oxx

x

P K 11 M

P 2

K

K 12ox x x

K 1P P 1 M

2

K

K 12oy y y

K 1P P 1 M

2

K

K 12y

oy y

2ox xx

K 11 MP P 2

K 1P P1 M

2

......... (1)

Buty

x

PP

is obtained from momentum as

2y x

2x y

P 1 KM

P 1 KM

But

2x

2y

2x

2M

K 1M2K

M 1

K 1


13/41

2y x

2xx

2x

P 1 KM

2PM

K 11 K2K

M 1K 1

y 2x

x

P 2K K 1M

P K 1 K 1

2x

22xy

2 2x x

2M

K 1 K 112KK 1 2

M 11 MK 12

K 1 K 11 M 1 M2 2

2

2x

2 2x x

K 1M

2 K 1

K 1 2K 1 M M 12 K 1

Equation (1) becomes

K2 K 1

2x

oy 2x

2 2oxx x

K 1M

P 2 K 12K K 1M

K 1 2K P K 1 K 11 M M 1

2 K 1

K

K 12x

2x

2 2x x

K 1M

2K K 1 2MK 1 2K K 1K 1 K 1

1 M M2 K 1 K 1

K

K 1K 21

xK 12

x 2x

K 1M

2K K 1 2MK 1K 1 K 1 1 M

2

K

K 11 2xK 1oy 2

x2oxx

K 1MP 2K K 1 2M

K 1P K 1 K 11 M

2


14/41

(4) Prove that

2x

y

2xx

K 11 MV 2

K 1VM

2

22 xx 2

x

VM

C

2 2 2x x xV M C

2 2 2y y yV M C

2 2y y y

2 2xx x

V M T

. TV M ......... (1)

But

2x

y

2xy

K 11 MT 2

K 1T1 M

2

But

2x

2y

2x

2M

K 1M2K

M 1K 1

Substituting and simplifying

2 2x x

y

2x 2

x

K 1 2K 1 M M 1

T 2 K 1

T K 1M

2 K 1

............ (2)

Substituting (2) in (1), we have

2x

22 2

2x x2 x

y

2 2 2x x 2

x

2M

K 1K 1 2K 2K 1 M M 1M 1V 2 K 1K 1

V M K 1M

2 K 1


15/41

2 2x x

2

4x

2 K 1M 1 M

K 1 2

K 1M

2 K 1

2 22 2x x

2 2 44 xx

2 K 1 K 11 M 1 M

K 1 2 2

K 1 K 1 MM

2 K 1 4

2x

y

2x

x

K 11 MV 2

K 1VM

2

(5) Prove that

K 1

K 1 K 1oy 2 2x x

x

P K 1 2K K 1M M

P 2 K 1 K 1

(This is known as Rayleigh supersonic Putot tube formula)

We have,

K

K 1oy 2y

y

P K 11 M

P 2

y 2x

x

P 2K K 1M

P K 1 K 1

K

K 1oy 2 2y x

x

P K 1 2K K 11 M M

P 2 K 1 K 1

K

K 12

x 2x

2x

2

MK 1 2K K 1K 11 M2K2 K 1 K 1

M 1K 1

K

K 12 2x x

2x

2x

2K K 1M 1 1 M

2K K 1K 1 2 M2K K 1 K 1

M 1K 1


16/41

K

K 12x

2x

2

x

2K K 1M

2K K 1K 1 2M

2K K 1 K 1M 1

K 1

K2 2 K 1

x

2x

2x

K 1 M

2 K 1 2K K 1M

2K K 1 K 1M 1

K 1

K2

K 1x

2xK

K 12x

K 1 M2 2K K 1

MK 1 K 1

2K K 1M

K 1 K 1

K 1

2 K 1 K 1oy x 2x

x

P K 1 M 2K K 1M

P 2 K 1 K 1

(6) Show that shocks are possible only in supersonic flow:

2 2p

1 1

T PS C ln R ln

T P

Fro a shock

y yp

x x

T PS C ln R ln

T P

But oS S

oy oy oyp

ox ox ox

T P PC ln R ln R ln

T P P

1

oy

ox

PSln

R P


17/41

K

K 11 2xK 12

x2x

K 1M

2K K 1 2ln MK 1K 1 K 1

1 M

2

A plot of this equation is shown in the Figure.

Fro Mx < 1 the entropy change is negative as shown. Since e shocks are

highly irreversible in nature, second law insists that the entropy should be produced

across a shock. Therefore shocks are possible only in supersonic flow.

Alternatively,V

S

C

can be obtained by expanding the logarithmic tersm

interms of 2xM 1 and considering terms up to third powers of 2xM 1 , we get,

32x2

v

2K K 1SM 1

C 3 K 1

From the above equation for Mx > 1,V

S

C

becomes positive. Therefore

shocks are possible only in supersonic flows.

Prandtl-Meyer relationship

*2x yV .V C or

2 2x yM M 1

Governing relations for a normal shock

(1) Continuity

x x y yV V ....... (1)

(2) Energy

ox oyh h

22yx

x y 0

VVh h h

2 2 ............ (2)


18/41

(3) Momentum

xx xx xx xxCV out inF mV mV mVt

x y y xP A P A m V V

x y y xP P G V V

x y y xP P V V V .............. (3)

From (2) we have

2x

x 0V

h h2

2x

p x p o

VC T C T

2

2x

x 0

VKR KR T T

K 1 2 K 1

22 2

0x x CC VK 1 2 K 1

2 2 2x 0 x

K 1C C V

2

.......... (4)

Similarly 2 2 2y 0 yK 1

C C V2

............. (5)

Equation (3) x y y xP P V V V

x y y xP P

V VV

yx y xx x y y

PPV V

V V

yx y xx y

RTRTV V

V V


19/41

yx y xx y

KRTKRTK V V

V V

22yx

y xx y

CC K V VV V

............. (6)

Using (4) and (5) in (6)

2 2 2 20 x 0 y y xx y

1 K 1 1 K 1C V C V K V V

V 2 V 2

20 y x y xx y

1 1 K 1C V V K V V

V V 2

20 y x

y x y xx y

C V V K 1V V K V V

V V 2

20

x y

C K 1K

V V 2

20 x y

K 1

C V V2

.......... (7)

*2 * *

20 00

C KRT T 2

KRT T K 1C

20

T K 11 M

T 2

0*

T K 1

2T

2 *2

0

K 1

C C2

............ (8)

Substitution (8) in (7) we have

*2x y

K 1 K 1C V V

2 2

*2x yC V V


20/41

x y

*2

V V1

C

for a shock, * * *

x yC C C

yx* *x y

VV1

C C

* *x yM .M 1 By definition

*

*

VM

C

THE RANKINEHUGONIOT

EQUATIOS

DENSITY RATIO ACROSS THE SHOCK

We know that density,P

RT

x

x x

For upstream shock

P

x R T

y

y

y y

For down stream shock

P

R T

y

y y y

x

x x

P

R TPx

R T

X

Y

T

T

y y

x

p

x p

We know that,


21/41

22YX

X

PM

P

22 YX

X

PM

P

2 2

22

21 1

2

X X

Y

X X

we know that

M MT

T M

2

21 1

2

Y Y

X XY

XY

X

P P

P PT

T P

P

2

3 2

1 1Y Y

X XY

XY

X

P P

P PTT P

P

3

Y

X

PTaking out

P

3 2

2 2

3

Y Y

X X

Y

X

P P

P P

P

P

2

2 2

Y Y

X X

Y

X

P P

P P

P

P


22/41

2

2

Y Y

X X

Y

X

P P

P P

PP

2

2

Y Y

X X

Y

X

P P

P P

P

P

2

Y Y

X X

Y

X

P P

P P

P

P

2

2

Y Y

X X

Y

X

P P

P P

P

P

1Y Y

X XY

Yx

X

P P

P PT

PT

P

1

Y Y

x XY

X Y

X

T P

T PP

P P

P


23/41

1

Y

xY X

YX Y

X

P

TP P

PP T

P

y xY

x X Y

We have

TP

P T

1

Y

y X

Yx

X

P

P

P

P

1 Y

y X

x Y

X

P

P

P

P

1 Y

y X

x Y

X

P

P

P

P

1y Y Y

x X X

P P

P P

1y y Y Y

x x X X

P P

P P


24/41

1y yY Y

x X x X

P P

P P

1y yY

x X x

P

P

1y yY

x X x

P

P

1

.

y

xY

yX

x

P

P

theabove eqn s is knows Rankine Hugoniot equations

Strength of a Shock Wave

It is defined as the ratio of difference in down stream and upstream shock pressures(py- px) to upstream shock pressures (px). It is denoted by .

Substituting forpy/px


25/41

From the above equation;

The strength of shock wave may be expressed in another form using Rankine-

Hugoniot equation.


26/41


27/41

From this equation we came to know strength of shock wave is

directly proportional to;


28/41

PROBLEMS

Que.1

The state of a gas (=1.3,R =0.469 KJ/KgK.) upstream of normal shock wave is

given by the following date:

Mx =2.5 , Px =2 bar. Tx =275 K calculate the Mach number

,pressure,temperatureand velocity of a gas down stream of shock: check the

calculated values with those given in the gas tables.

Take K =

22x

2y

2 2x

222.5M

12.921.3 1K 1M 0.2432K 2 1.3 53.19

M 1 2.5 1K 1 1.3 1

0.4928YM

22YX

X

PM

P

22 1.3 1.5


29/41

7.065 0.130 6.935Y

X

P

P

6.935YP bar

2 2x x

y

2x 2

x

K 1 2K 1 M M 1

T 2 K 1

T K 1M

2 K 1

2 2

22

1.3 1 2 1.31 2.5 2.5 1

2 1.3 1

1.3 12.5

2 1.3 1

2

031 6.25

1.9372

55.1042.36.25

2 0.3

y

x

T1.869

T

YT 1.869

3 1 0.30.269

1.3 1 6.25 2.3

Y

x

C

C

0.269 .0269Y x x xC C M a

0.269Y x xC M KRT

0.2692.5 1.3 469 275Y xC

,

0.4928 1.3 469 513.975

275.16 / .

y y y

y

Alternately C M KRT

C m s


30/41

A gas( =1.3,R =0.287KJ/KgK) at a Mach number of 1.8 P =0.8 bar and

T=373K pass through a normal shock .Compare this value in an isentropic

compression through the same pressure ratio.

530.8 10 0.747 /

287

x

x

PKg m

RT

1.4 1.8xFrom normal shock table for at M

3.613 , 1.532yY

x x

TP

P T

X

Y

T

T

y y

x

p

x p

3.6132.358

1.532

y

x

32.358 /y Kg m

1

y y

x

For isentropic flow

p

x p

1

3.613 2.5y

x

32.5 /y Kg m

It is noted that the final density of the isentropic process is grater than in the shock

process.

Q.A jet of air at 270 K and 0.2 bar has an intial Mach number of 1.9.If the process

through a normal shock wave. Determine the following for the down stream of the

shock.

1.Mach number.= My

2. Pressure. = Py

3. Temperature = Ty.


31/41

4.Speed of sound .= ay

5. Jet velocity = Cy

6. Density .= y

Given.

Tx=270 K

Px =0.7 bar =0.7105 N/m2

From normal shock table for Mx=1.9 and =1.4

0.596

4.045

1.608

Y

Y

x

y

x

M

PP

T

T

4.045

4.045

y xP P

2.831

1.608

1.608

y

y x

P

T T

Speed of at down of the shock

ya YRT

1.4

ya 417.66 /m s

yay Y

Jet velocity at downstream of the shock

C M


32/41

0.596

248.93 /yC m s

32.83 2.27 /287

y

y

y

PDensity kg m

RT

Q.An Aircraft flies at a Mach number of 1.1 at an altitude of 15,000 metres.The

compression in its engine is partially achieved by a normal shock wave standing at the

entry of the diffuser. Determine the following for downstream of the shock.

1. Mach number

2. Temperature of the air

3. Pressure of the air

4. Stagnation pressure loss across the shock.

1.1, 15,000

x

Given

MAltitude Z m

1.

2.

3.

y

y

y

To find

M

T

P

04. P Ox OyP P

Re , 15,000fer gastables for Altitude Z m

216.6xT K

0.120xP bar


33/41

5 20.120 10 /xP N m

Re 1.1 1.4xfer Normal shocks gastables for M and

0.911yM

1.245y

x

P

P

1.065y

x

T

T

0

0

0.998y

x

P

P

02.133

y

x

P

P

5 2

1.24 5

1.24 10 /

y xP P

N m

5 20.149 10 /yP N m

1.067

1.065 216.6

y xT T

230.67yT K

05

2.133

2.133 10y x

P P

5 2

0 0.259 10 /yP N m

0

00.998

y

x

PP

5

0

0.259 10

0.998xP


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5 2

0 0.2564 10 /xP N m

0

Pr

POx Oy

essure loss

P P

5 50.2564 10 0.259 10

2

0P 50 /N m

1) When a pitot tube is immersed in a supersonic stream a normal shock is

formed ahead of the Pitot tube mouth. After the shock the fluid stream

decelerates isentropically to the total pressure of the entrance to the pitot

tube. A pitot tube travelling in a supersonic wind tunnel gives values of

15Kpa and 70Kpa for the static pressure upstream of the shock and the

pressure at the inlet of the tube respectfully. Find the Mach no. of the

tunnel if the stagnation temperature is 575K. Calculate the static

temperature and the total (stagnation) pressure upstream and the

downstream of the tube.

Px= 15 103

pa = 15 103N/m

2

Poy = 70 10

3

N/m

2

T0= T0x= T0y = 575 K

Refer Normal shock tables for0

4.67y

x

P

P ,and = 1.4

30

3

70 104.375

15 10

y

x

P

P

Bow shock

Mx My

Pitot tube


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1.8

0.1616

x

y

M

M

3.613y

x

PP

1.532y

x

T

T

0

0

0.813y

x

P

P

3

0 78 100.1813

xP

Pox = 0.86 105N/m

2

we know that,

20 112

TM

T

20 112

xx

x

T MT

2575 11 1.82xT

348.9xT K

From table,

1.532y

x

TT

1.532 348.9 534.51yT K

Result:

Mx = 1.8

My = 0.616

Tx = 348.9K


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Ty = 534.51K

P0x = 0.86 105

N/m2

P0y = 70 10

3

Supersonic nozzle is provided with a constant diameter circular duct at its exit. The

duct diameter is same as the nozzle diameter. Nozzle exit cross ection is three times

that of its throat. The entry conditions of the gas (=1.4, R= 287J/KgK) are P 0 =10

bar, T0 =600K. Calculate the static pressure, Mach number and velocity of the gas in

duct.

(a) When the nozzle operates at its design condition.

(b)

When a normal shock occurs at its exit.(c) When a normal shock occurs at a section in the diverging part where

the area ratio, A/A* =2.

Given:

A2=3A*

Or A2/A*=3

= 1.4

R= 287 J/KgK

P0= 10 bar = 106Pa

T0= 600K

For nozzle

Throat (*)

Entry (1) Exit (2)

M1

Solution:

Case (i)

Refer isentropic flow table for A2/A*=3 and = 1.4

M2= 2.64

T2/T02= 0.417


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p2/ p02= 0.0417

{Note: For A2/A* =3, we can refer gas tables page no.30 and 36. But we

have to take M > 1 corresponding values since the exit is divergent

nozzle}

i.e. T2= 0.417 xT02

= 0.417 x 600 {sinceT0=T01=T02}

T2= 250.2 K

P2= 0.0471 x p02

= 0.0471 x 10 x 105 {since p0=p01= p02}

p2= 0.471 x 105N/m2

We know

C2= M2 a2

= 2 2M RT

= 2.64 (1.4 287 250.2).5

C2=837.05 m/s

Case (ii)

Normal shock occurs at exit. Shock

wave

Entry Exit

My

M1 Mx

A2/A* = Ax/Ax* =3 [since in this case A2=Ax]

Refer isentropic table for Ax/Ax* =3 and =1.4.

Mx=2.64


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Tx/Tx* =0.417

Px /P0x=0.0417 [From gas tables page no. 36]

i.e. Tx=0.417 T0x

= 0.417 600

i.e.Tx= 250.2 K [Since T0=T0x]

So px= 0.0417 P0x

= 0.0417 10 105 [Since p0=p0x]

i.e. px= 0.417 105N/m2

Refer Normal shock table for Mx= 2.64 and = 1.4

My= 0.5

Py /Px=7.965

Ty/Tx=2.279

i.e. Py= 7.695 px

=7.695 0.471 105

So py= 3.75 105N/m2

Also Ty=2.279 Tx

= 2.279 250.2

i.e. Ty= 570.2 K

Also Cy=Myay

= Y YM RT

= 0.5 (1.4 287 570.2)

Cy= 239.32 m/s


39/41

Case (iii)

Area ratio A/A* =2

i.e . Ax/Ax* =2

Refer isentropic flow table for Ax/Ax* =2 and = 1.4

Mx=2.2 [from gas tables page no. 35]

Refer Normal shocks table for Mx= 2.2 and = 1.4

My =0.547

P0y/P0x = 0.628

P0y= 0.628 P0x

= 0.628 10 105

P0y= 6.28 105N/m2

Throat (*)

[A*=Ax*] [ Mx, Ax, M>1] [ My, Ay, M


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Refer isentropic table for A2/Ay* = 1.8825= 1.871 and = 1.4

M2= 0.33 [From gas tables page no. 29]

T2/T0y =0.978

P2/p0y=0.927

i.e. T2= 0.978 T0y

= 0.978 600

or T2= 586.8 K

So P2= 0.927 p0y

=0.927 6.28 105

or P2= 5.82 105N/m2

C2= M2 a2

= 2 2M RT

= 0.33 (1.4 287 586.8).0.5

C2= 160.23 m/s

RESULT

Case (i): p2=0.471 105

N/m2

, M2= 2.64, c2= 837.05 m/s

Case (ii): py= 3.75 105N/m2, My=0.5, cy= 239.32 m/s

Case (iii): p2= 5.82 105N/m2, M2= 0.33, c2= 160.23 m/s

The following refers to a supersonic wind tunnel Nozzle throat area- 200 cm2

Test section cross section -337.5 cm2 Working fluid is air. Determine the test section

mach no. and diffuser throat area if normal shock is located at test section?


41/41

Given:

Nozzle throat area = Ax*

= 200cm

2

= 20010-4 m2

Test section area = Ax = 337.5 cm2 = 337 10-4m2

To find:

1. Test section Mach no. Mx

2. Diffuser throat area, Ay*

Solution:

-4x

-4

x

A 337.5101.6875

A* 20010

Refer isentropic table =1.4

Then M = Mx = 2

Refer normal shock table for Mx= 2 and =1.4

0

0

0.721y

x

P

P

We know that 0 0* *x x Y yA P A P

0

0

* * xY xy

PA A

P

-4 1200100.721

2* 0.0277YA m

Result:

Test secton Mach number = 1.97

Diffuser throat area = .0272 m2

gas dynamics Unit 3 Clear Notes.208-248

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Transcript of gas dynamics Unit 3 Clear Notes.208-248