3-Isentropic Flow With Area Variation

7/30/2019 3-Isentropic Flow With Area Variation

1/30

GasDynamics 1

Isentropic Flow with Area Variation


2/30

GasDynamics

Simple convergent nozzle

2

If the flow is subsonic


3/30

GasDynamics

Isentropic Mach number relations

3

+=

20

2

11 M

T

T

120

2

11

+=

M

P

P

1

1

20

2

11

+=

M

( )121

2

2

*1

1

1

21 +

+

+

+

=

M

MA

A

2

1

20

2

11

+= M

a

a


4/30

GasDynamics

Example (Duct Flow)

4


5/30

GasDynamics

Analysis:

5


6/30

GasDynamics 6


7/30

GasDynamics 7

The maximum velocity is found by converting all the thermal energy to kinetic energy. Taking

zero thermal energy to correspond to absolute zero (despite the fact that air would not be a gas at

this point) one could estimate


8/30

GasDynamics

Sonic Properties(Let * denote a property at the sonic state)

8


9/30

GasDynamics

Effect of Area Change

9

In the isentropic limit the mass, momentum, and energy equation for

a compressible ideal gas reduce to:


10/30

GasDynamics 10

Substitute energy then mass into momentum:


11/30

GasDynamics 11

equation singular when M2 = 1 if M2 = 1, one needs dA = 0 area minimum necessary to

transition from subsonic tosupersonic flow!!

can be shown area maximumnot relevant


12/30

GasDynamics 12

Substitute from earlier-developed relations and get:


13/30

GasDynamics

Reference Stagnation and Critical Conditions

for Isentropic Flow

13


14/30

GasDynamics

Area versus Mach number for a calorically

perfect ideal gas

14


15/30

GasDynamics

Choking

15

Consider mass flow rate variation with pressure difference

small pressure difference gives small velocity, small mass flow

as pressure difference grows, velocity and mass flow rate grow

velocity is limited to sonic at a particular duct location

this provides fundamental restriction on mass flow rate

can be proven rigorously that sonic condition gives maximum mass flow

rate

A flow which has a maximum mass flow rate is known as

choked flow. Flows will choke at area minima in a duct.


16/30

GasDynamics

Isentropic Flow in a Converging Nozzle

16


17/30

GasDynamics 17

1. Pb = Po, Pb /Po = 1. No flow occurs. Pe = Pb, Me=0.

2. Pb > P* or P*/Po < Pb /Po < 1. Flow begins to increase as

the back pressure is lowered. Pe = Pb, Me < 1.

3. Pb = P* or P*/Po = Pb /Po < 1. Flow increases to the choked

flow limit as the back pressure is lowered to the criticalpressure. Pe = Pb, Me=1.

4. Pb < P* or Pb /Po < P*/Po < 1. Flow is still choked and

does not increase as the back pressure is lowered below the

critical pressure, pressure drop from Pe to Pb occurs outsidethe nozzle. Pe = P*, Me=1.

5. Pb = 0. Results are the same as for item 4.


18/30

GasDynamics

Choking Example

18

Given: Air with stagnation conditions Po = 200 kPa To = 500 K flows through a

throat to an exit with Mach number of 2.5. The desired mass flow is 3.0 kg/s,

Find: a) throat area, b) exit pressure, c) exit temperature, d) exit velocity, and e)

exit area.


19/30

GasDynamics

Analysis:

19


20/30

GasDynamics 20


21/30

GasDynamics

Discharge Example

21


22/30

GasDynamics 22


23/30

GasDynamics

Isentropic Flow in a

Converging-Diverging Nozzle

23


24/30

GasDynamics 24


25/30

GasDynamics 25

PA = Po, or PA/Po = 1. No flow occurs. Pe = Pb, Me = 0.

Po > PB > PC > P* or P*/Po < PC/Po < PB/Po < 1. Flow begins

to increase as the back pressure is lowered. The velocity

increases in the converging section but M< 1 at the throat; thus,

the diverging section acts as a diffuser with the velocity

decreasing and pressure increasing. The flow remains subsonicthrough the nozzle. Pe = Pb and Me < 1.

Pb = PC = P* or P*/Po = Pb/Po = PC/Po and Pb is adjusted so

that M=1 at the throat. Flow increases to its maximum value at

choked conditions; velocity increases to the speed of sound atthe throat, but the converging section acts as a diffuser with

velocity decreasing and pressure increasing. Pe = Pb, Me < 1.


26/30

GasDynamics 26

PC > Pb > PE or PE/Po < Pb/Po < PC/Po < 1. The fluid that achieved sonic velocity

at the throat continues to accelerate to supersonic velocities in the diverging section

as the pressure drops. This acceleration comes to a sudden stop, however, as a

normal shock develops at a section between the throat and the exit plane. The flowacross the shock is highly irreversible. The normal shock moves downstream away

from the throat as Pb is decreased and approaches the nozzle exit plane as Pbapproaches PE. When Pb = PE, the normal shock forms at the exit plane of the

nozzle. The flow is supersonic through the entire diverging section in this case,

and it can be approximated as isentropic. However, the fluid velocity drops to

subsonic levels just before leaving the nozzle as it crosses the normal shock.

PE > Pb > 0 or 0 < Pb/Po < PE/Po < 1. The flow in the diverging section is

supersonic, and the fluids expand to PF at the nozzle exit with no normal shock

forming within the nozzle. Thus the flow through the nozzle can be approximated

as isentropic. When Pb = PF, no shocks occur within or outside the nozzle. When

Pb < PF, irreversible mixing and expansion waves occur downstream of the exitplane or the nozzle. When Pb > PF, however, the pressure of the fluid increases

from PF to Pb irreversibly in the wake or the nozzle exit, creating what are called

oblique shocks.


27/30

GasDynamics

Example : Converging-Diverging Area

27

A converging-diverging nozzle has an exit-area-to-throat area ratio of 2.Air enters this nozzle with a stagnation pressure of 1000 kPa and a

stagnation temperature of 500 K. The throat area is 8 cm2. Determine the

mass flow rate, exit pressure, exit temperature, exit Mach number, and exit

velocity for the following conditions:

Sonic velocity at the throat, diverging section acting as a nozzle.Sonic velocity at the throat, diverging section acting as a diffuser.


28/30

GasDynamics 28

For A/A* = 2, Table A-32 yields two Mach numbers, one > 1 and one < 1.When the diverging section acts as a supersonic nozzle, we use the value for M> 1.

Then, for AE/A* = 2.0, ME = 2.197, PE/Po = 0.0939, and TE/To = 0.5089,

P P kPa kPa

T T K K

E o

E o

= = =

= = =

0 0939 0 0939 1000 939

08333 0 5089 254 5

. . ( ) .

. . (500 ) .

C kRT

kJ

kg K

K

m

s

kJkg

m

s

E E=

=

=

14 0 287 254 5

1000

319 7

2

2

. ( . )( . )

.

Analysis:

r

V M Cm

s

m

sE E E= = =2197 319 7 702 5. ( . ) .


29/30

GasDynamics

The mass flow rate can be calculated at any known cross-sectional area where the

properties are known. It normally is best to use the throat conditions. Since the flow

has sonic conditions at the throat, Mt = 1, and

*0.8333

0.8333 0.8333(500 ) 416.6

T Tt

T To o

T T K K t o

= =

= = =

21000

21.4(0.287 )(416.6 )

409.2

V C kRT t t t

m

kJ sKkJkg K

kg

m

s

= =

=

=

*0.528

0.528 0.528(1000 ) 528

P Pt

P Po o

P P kPa kPat o

= =

= = =

tP

RT

kPa

kJ

kg KK

kJ

m kPa

kg

m

= = =

=

**

*

(528 )

( . )( . )

.

0 287 416 6

4 416

3

3


30/30

GasDynamics

&

. (8 )( . )( )

.

m AV

kg

mcm

m

s

m

cm

kg

s

t t t=

=

=

4 416 409 2100

1446

3

22

2

When the diverging section acts as a diffuser, we use M< 1. Then, for

AE /A* = 2.0, ME = 0.308, PE /Po = 0.936, and TE /To = 0.9812,

P P kPa kPa

T T K K

E o

E o

= = =

= = =

0 0939 0 936 1000 936

08333 0 9812 490 6

. . ( )

. . (500 ) .

C kRT

kJ

kg K

K

m

s

kJkg

m

s

E E=

=

=

14 0287 490 6

1000

444 0

2

2

. ( . )( . )

.

Since M= 1 at the throat, the mass flow rate is the same as that in the first

part because the nozzle is choked.

r

V M Cm

s

m

sE E E

= = =0 308 444 0 136 7. ( . ) .

3-Isentropic Flow With Area Variation

Documents

Transcript of 3-Isentropic Flow With Area Variation