Chapter 17-compressible fluid flow

8/12/2019 Chapter 17-compressible fluid flow

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Chapter 17

Compressible Flow

Study Guide in PowerPoint

to accompany

Thermodynamics: An Engineering Approach, 5th edition

by Yunus A. engel and Michael A. Boles


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Stagnation Properties

Consider a fluid flowing into a diffuser at a velocity , temperature T , pressure P , andenthalpy h, etc. Here the ordinary properties T , P , h, etc. are called the staticproperties; that is, they are measured relative to the flow at the flow velocity. Thediffuser is sufficiently long and the exit area is sufficiently large that the fluid isbrought to rest (zero velocity) at the diffuser exit while no work or heat transfer isdone. The resulting state is called the stagnation state.

V

We apply the first law per unit mass for one entrance, one exit, and neglect thepotential energies. Let the inlet state be unsubscripted and the exit or stagnationstate have the subscript o.

q h V

w h V

net net oo

2 2

2 2


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Since the exit velocity, work, and heat transfer are zero,

h hV

o

2

2

The term ho is called the stagnation enthalpy (some authors call this the totalenthalpy). It is the enthalpy the fluid attains when brought to rest adiabatically whileno work is done.

If, in addition, the process is also reversible, the process is isentropic, and the inletand exit entropies are equal.

s so The stagnation enthalpy and entropy define the stagnation state and the isentropicstagnation pressure, P o. The actual stagnation pressure for irreversible flows will besomewhat less than the isentropic stagnation pressure as shown below.


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Then2

2

2

2

2

3003264.5

21000

3309.5

oV

h h

kJ mkJ kg s

mkg s

kJ

kg

and

= = 7.4670okJ

s skg K

h h P so o o ( , )


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We can find P o by trial and error (or try the EES solution for problem 3-27 in the text).The resulting stagnation properties are

3

1.16

422.2

13.640

o

oo

oo

P MPa

T C

kg v m

Ideal Gas Result

Rewrite the equation defining the stagnation enthalpy as

h hV

o

2

2For ideal gases with constant specific heats, the enthalpy difference becomes

C T T V

P o( ) 2

2


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Conservation of Energy for Control Volumes Using Stagnation Properties

The steady-flow conservation of energy for the above figure is

Since

h hV

o

2

2


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For no heat transfer, one entrance, one exit, this reduces to

If we neglect the change in potential energy, this becomes

For ideal gases with constant specific heats we write this as

Conservation of Energy for a Nozzle

We assume steady-flow, no heat transfer, no work, one entrance, and one exit andneglect elevation changes; then the conservation of energy becomes

1 1 2 2

in out

o o

E E

m h m h


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But m m1 2

Thenh ho o1 2

Thus the stagnation enthalpy remains constant throughout the nozzle. At any crosssection in the nozzle, the stagnation enthalpy is the same as that at the entrance.

For ideal gases this last result becomes

T T o o1 2Thus the stagnation temperature remains constant through out the nozzle. At anycross section in the nozzle, the stagnation temperature is the same as that at theentrance.

Assuming an isentropic process for flow through the nozzle, we can write for theentrance and exit states


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So we see that the stagnation pressure is also constant through out the nozzle forisentropic flow.

Velocity of Sound and Mach Number

We want to show that the stagnation properties are related to the Mach number M ofthe flow where

M V

C

and C is the speed of sound in the fluid. But first we need to define the speed of

sound in the fluid.

A pressure disturbance propagates through a compressible fluid with a velocitydependent upon the state of the fluid. The velocity with which this pressure wavemoves through the fluid is called the velocity of sound, or the sonic velocity.

Consider a small pressure wave caused by a small piston displacement in a tubefilled with an ideal gas as shown below.


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It is easier to work with a control volume moving with the wave front as shown below.


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Apply the conservation of energy for steady-flow with no heat transfer, no work, andneglect the potential energies.

h C

h dh C dV

h C

h dh C CdV dV

2 2

2 2 2

2 2

22

2

( ) ( )

( ) ( )

Cancel terms and neglect ; we havedV 2

dh CdV 0Now, apply the conservation of mass or continuity equation to the controlvolume.

m AV

AC d A C dV

AC A C dV Cd d dV

( ) ( )

( )

Cancel terms and neglect the higher-order terms like . We haved dV

C d dV 0


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Also, we consider the property relationdh T ds v dP

dh T ds dP

1

Let's assume the process to be isentropic; then ds = 0 and

dh dP 1

Using the results of the first law

dh dP C dV 1

From the continuity equation

dV C d

Now


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ThusdP d

C

2

Since the process is assumed to be isentropic, the above becomes

For a general thermodynamic substance, the results of Chapter 12 may be used toshow that the speed of sound is determined from

where k is the ratio of specific heats, k = C P /C V .

Ideal Gas Result

For ideal gases


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Example 17-3

Find the speed of sound in air at an altitude of 5000 m.

At 5000 m, T = 255.7 K.

C kJ

kg K K

m s

kJ

kg m s

14 0 287 255 71000

3205

2

2

. ( . )( . )

.


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Notice that the temperature used for the speed of sound is the static (normal)temperature.

Example 17-4

Find the speed of sound in steam where the pressure is 1 MPa and the temperatureis 350 oC.

At P = 1 MPa, T = 350 oC,

1 s

s

P P C

v

Here, we approximate the partial derivative by perturbating the pressure about 1MPa. Consider using P 0.025 MPa at the entropy value s = 7.3011 kJ/kg K, to findthe corresponding specific volumes.


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What is the speed of sound for steam at 350 oC assuming ideal-gas behavior?

Assume k = 1.3, then

C kJ

kg K K

m

skJ kg

m s

13 0 4615 350 2731000

6114

2

2

. ( . )( )

.

Mach Number

The Mach number M is defined as

M V

C M

1 flow is supersonic


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Example 17-5

In the air and steam examples above, find the Mach number if the air velocity is 250m/s and the steam velocity is 300 m/s.

M m sm s

M

m s

m s

air

steam

250

32050780

300

60550 495

..

..

The flow parameters T o/T , P o/P , o/ , etc. are related to the flow Mach number. Let'sconsider ideal gases, then

T T V

C T T

V C T

o P

o

P

2

22

12


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butC

k k

R or C

k kR P P

1

1 1

T

T

V

T

k

kR

o 12

12 ( )

andC kRT 2

soT T

k V C

k M

o

1

12

1 1

2

2

2

2

( )

( )

The pressure ratio is given by


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We can show the density ratio to be

See Table A-32 for the inverse of these values ( P /P o, T /T o, and / o) when k = 1.4.

For the Mach number equal to 1, the sonic location, the static properties are denoted

with a superscript *. This condition, when M = 1, is called the sonic condition.When M = 1 and k = 1.4, the static-to-stagnation ratios are


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Effect of Area Changes on Flow Parameters

Consider the isentropic steady flow of an ideal gas through the nozzle shown below.

Air flows steadily through a varying-cross-sectional-area duct such as a nozzle at aflow rate of 3 kg/s. The air enters the duct at a low velocity at a pressure of 1500 kPaand a temperature of 1200 K and it expands in the duct to a pressure of 100 kPa.The duct is designed so that the flow process is isentropic. Determine the pressure,temperature, velocity, flow area, speed of sound, and Mach number at each pointalong the duct axis that corresponds to a pressure drop of 200 kPa.

Since the inlet velocity is low, the stagnation properties equal the static properties.

T T K P P kPao o 1 11200 1500,


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After the first 200 kPa pressure drop, we have

P

RT kPa

kJ kg K

K

kJ m kPa

kg m

( )

( . )( . )

.

1300

0 287 11519

3932

3

3


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A m

V

kg s

kg m

m s

cmm

cm

( . )( . )

.

3

3 9322 310 77

10

2455

3

4 2

2

2

C kRT kJ

kg K K

m s

kJ kg

m s

14 0 287 115191000

680 33

2

2

. ( . )( . )

.

M V

C

m sm s

310 77

680 33

0457.

.

.

Now we tabulate the results for the other 200 kPa increments in the pressure until wereach 100 kPa.


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Summary of Results for Nozzle Problem

V Ste p

P kPa

T K m/s

kg/m 3

C m/s

A cm 2

M

0 1500 1200 0 4.3554 694.38 0

1 1300 1151.9 310.77 3.9322 680.33 24.55 0.457

2 1100 1098.2 452.15 3.4899 664.28 19.01 0.681

3 900 1037.0 572.18 3.0239 645.51 17.34 0.886

4 792.4 1000.0 633.88 2.7611 633.88 17.14 1.000

5 700 965.2 786.83 2.5270 622.75 17.28 1.103

6 500 876.7 805.90 1.9871 593.52 18.73 1.358

7 300 757.7 942.69 1.3796 551.75 23.07 1.709

8 100 553.6 1139.62 0.6294 471.61 41.82 2.416

Note that at P = 797.42 kPa, M = 1.000, and this state is the critical state.


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Now let's see why these relations work this way. Consider the nozzle and controlvolume shown below.

The first law for the control volume is

dh VdV

0The continuity equation for the control volume yields

d dA

A

dV

V

0

Also, we consider the property relation for an isentropic process

Tds dh dP

0


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and the Mach Number relationdP d

C V

M 2

2

2

Putting these four relations together yields

dA A

dP V

M

221( )

Lets consider the implications of this equation for both nozzles and diffusers.

A nozzle is a device that increases fluid velocity while causing its pressure to drop;thus, d > 0, dP < 0.V

Nozzle ResultsdA A

dP V

M

221( )

Subsonic

Sonic

Supersonic

: ( )

: ( )

: ( )

M dP M dA

M dP M dA

M dP M dA

1 1 0 0

1 1 0 0

1 1 0 0

2

2

2


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To accelerate subsonic flow, the nozzle flow area must first decrease in the flowdirection. The flow area reaches a minimum at the point where the Mach number isunity. To continue to accelerate the flow to supersonic conditions, the flow area mustincrease.

The minimum flow area is called the throat of the nozzle.

We are most familiar with the shape of a subsonic nozzle. That is, the flow area in asubsonic nozzle decreases in the flow direction.

A diffuser is a device that decreases fluid velocity while causing its pressure to rise;thus, d < 0, dP > 0.V

Diffuser ResultsdA A

dP V

M

221( )

Subsonic

Sonic

Supersonic

: ( )

: ( )

: ( )

M dP M dA

M dP M dA

M dP M dA

1 1 0 0

1 1 0 0

1 1 0 0

2

2

2


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To diffuse supersonic flow, the diffuser flow area must first decrease in the flowdirection. The flow area reaches a minimum at the point where the Mach number isunity. To continue to diffuse the flow to subsonic conditions, the flow area mustincrease.

We are most familiar with the shape of a subsonic diffuser. That is the flow area in asubsonic diffuser increases in the flow direction.

Equation of Mass Flow Rate through a Nozzle

Let's obtain an expression for the flow rate through a converging nozzle at anylocation as a function of the pressure at that location. The mass flow rate is given by

m AV The velocity of the flow is related to the static and stagnation enthalpies.

V h h C T T C T T

T o P P o 2 2 2 10 0( ) ( ) ( )


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and

Write the mass flow rate as

m AV oo

We note from the ideal-gas relations that

oo

o

P RT


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What pressure ratios make the mass flow rate zero?

Do these values make sense?

Now let's make a plot of mass flow rate versus the static-to-stagnation pressure ratio.

0.00 0.20 0.40 0.60 0.80 1.000.00

0.02

0.04

0.06

0.08

0.10

0.12

0.14

0.160.16

P*/P o

m [

k g / s

]

P/P o

Dia.=1 cm

T o =1200 K

P o =1500 kPa


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This plot shows there is a value of P /P o that makes the mass flow rate a maximum.To find that mass flow rate, we note

The result is

So the pressure ratio that makes the mass flow rate a maximum is the same pressure

ratio at which the Mach number is unity at the flow cross-sectional area. This valueof the pressure ratio is called the critical pressure ratio for nozzle flow. For pressureratios less than the critical value, the nozzle is said to be choked. When the nozzle ischoked, the mass flow rate is the maximum possible for the flow area, stagnationpressure, and stagnation temperature. Reducing the pressure ratio below the criticalvalue will not increase the mass flow rate.

What is the expression for mass flow rate when the nozzle is choked?


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Using

The mass flow rate becomes

When the Mach number is unity, M = 1, A = A*

Taking the ratio of the last two results gives the ratio of the area of the flow A at a

given Mach number to the area where the Mach number is unity, A*.


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Then

0.0 0.5 1.0 1.5 2.0 2.5 3.00.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

M

A / A *

From the above plot we note that for each A/ A* there are two values of M : one forsubsonic flow at that area ratio and one for supersonic flow at that area ratio. Thearea ratio is unity when the Mach number is equal to one.

ff f k l h h l


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Effect of Back Pressure on Flow through a Converging Nozzle

Consider the converging nozzle shown below. The flow is supplied by a reservoir atpressure P r and temperature T r . The reservoir is large enough that the velocity in thereservoir is zero.

Let's plot the ratio P /P o along the length of the nozzle, the mass flow rate through thenozzle, and the exit plane pressure P e as the back pressure P b is varied. Let'sconsider isentropic flow so that P o is constant throughout the nozzle.


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1 P P P /P 1 N fl P P M 0


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1. P b = P o, P b /P o = 1. No flow occurs. P e = P b, M e=0.

2. P b > P * or P */P o < P b /P o < 1. Flow begins to increase as the back pressure islowered. P e = P b, M e < 1.

3. P b = P * or P */P o = P b /P o < 1. Flow increases to the choked flow limit as the backpressure is lowered to the critical pressure. P e = P b, M e=1.

4. P b < P * or P b /P o < P */P o < 1. Flow is still choked and does not increase as theback pressure is lowered below the critical pressure, pressure drop from P e to P b occurs outside the nozzle. P e = P *, M e=1.

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

Consider the converging-diverging nozzle shown below.


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L t' l t th ti P /P d th M h b l th l th f th l th


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Let's plot the ratio P /P o and the Mach number along the length of the nozzle as theback pressure P b is varied. Let's consider isentropic flow so that P o is constantthroughout the nozzle.

P A = P o, or P A/P o = 1. No flow occurs. P e = P b, M e = 0.

P o > P B > P C > P * or P */P o < P C/P o < P B/P o < 1. Flow begins to increase as the backpressure is lowered. The velocity increases in the converging section but M < 1 at thethroat; thus, the diverging section acts as a diffuser with the velocity decreasing andpressure increasing. The flow remains subsonic through the nozzle. P e = P b and M e < 1.

P b = P C = P * or P */P o = P b/P o = P C/P o and P b is adjusted so that M =1 at the throat.Flow increases to its maximum value at choked conditions; velocity increases to thespeed of sound at the throat, but the converging section acts as a diffuser withvelocity decreasing and pressure increasing. P e = P b, M e < 1.

P > P > P P /P < P /P < P /P < 1 Th fl id th t hi d i l it t


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P C > P b > P E or P E/P o < P b/P o < P C/P o < 1. The fluid that achieved sonic velocity atthe throat continues to accelerate to supersonic velocities in the diverging section asthe pressure drops. This acceleration comes to a sudden stop, however, as a normalshock develops at a section between the throat and the exit plane. The flow acrossthe shock is highly irreversible. The normal shock moves downstream away from thethroat as P b is decreased and approaches the nozzle exit plane as P b approaches P E.When P b = P E, the normal shock forms at the exit plane of the nozzle. The flow issupersonic through the entire diverging section in this case, and it can beapproximated as isentropic. However, the fluid velocity drops to subsonic levels justbefore leaving the nozzle as it crosses the normal shock.

P E > P b > 0 or 0 < P b/P o < P E/P o < 1. The flow in the diverging section is supersonic,and the fluids expand to P F at the nozzle exit with no normal shock forming within thenozzle. Thus the flow through the nozzle can be approximated as isentropic. WhenP b = P F, no shocks occur within or outside the nozzle. When P b < P F, irreversiblemixing and expansion waves occur downstream of the exit plane or the nozzle. WhenP

b > P

F, however, the pressure of the fluid increases from P

F to P

b irreversibly in the

wake or the nozzle exit, creating what are called oblique shocks.


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Example 17-6

Air leaves the turbine of a turbojet engine and enters a convergent nozzle at 400 K,871 kPa, with a velocity of 180 m/s. The nozzle has an exit area of 730 cm 2.Determine the mass flow rate through the nozzle for back pressures of 700 kPa, 528kPa, and 100 kPa, assuming isentropic flow.

The stagnation temperature and stagnation pressure are

T T V

C o P

2

2


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For air k = 1.4 and Table A-32 applies. The critical pressure ratio is P */P o = 0.528.

The critical pressure for this nozzle is P P

kPa kPa

o* .

. ( )

0528

0 528 1000 528

Therefore, for a back pressure of 528 kPa, M = 1 at the nozzle exit and the flow ischoked. For a back pressure of 700 kPa, the nozzle is not choked. The flow rate willnot increase for back pressures below 528 kPa.

For the back pressure of 700 kPa


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For the back pressure of 700 kPa,

P P

kPakPa

P P

B

o o

7001000

0700.*

Thus, P E = P

B = 700 kPa. For this pressure ratio Table A-15 gives

M E 0 7324.

T T

T T K K

E

o

E o

09031

0 9031 0 9031 416 1 375 8

.

. . ( . ) .C kRT

kJ kg K

K

m s

kJ

kg m s

V M C m

sm s

E E

E E E

14 0 287 37581000

3886

0 7324 388 6

284 6

2

2

. ( . )( . )

.

( . )( . )

.

EP kPa kJ( )700


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E E

E

P RT

kPakJ

kg K K

kJ m kPa

kg

m

( )

( . )( . )

.

700

0 287 3758

6 4902

3

3

Then

. ( )( . )( )

.

m A V

kg m

cm m

sm

cm

kg s

E E E

6 4902 730 284 6100

1348

32

2

2

For the back pressure of 528 kPa,

P P

kPakPa

P P

E

o o 528

1000 0528.

*

This is the critical pressure ratio and M = 1 and P = P = P * = 528 kPa


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This is the critical pressure ratio and M E = 1 and P E = P B = P = 528 kPa.

T T

T T

T T K K

E

o o

E o

*

.

. . ( . ) .

08333

0 8333 0 8333 4161 346 7

And since M E = 1,V C kRT

kJ

kg K

K

m s

kJ kg

m s

E E E

14 0 287 346 71000

3732

2

2. ( . )( . )

.

E P

RT kPa

kJ kg K K

kJ m kPa

kg m

**

*

(528 )

( . )( . )

.

0 287 346 7

53064

3

3

m A V


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. ( )( . )( )

.

m A V

kg m

cm m

sm

cm

kg s

E E E

5 3064 730 373 2100

144 6

32

2

2

For a back pressure less than the critical pressure, 528 kPa in this case, the nozzle ischoked and the mass flow rate will be the same as that for the critical pressure.Therefore, at a back pressure of 100 kPa the mass flow rate will be 144.6 kg/s.

Example 17-7

A converging-diverging nozzle has an exit-area-to-throat area ratio of 2. Air entersthis nozzle with a stagnation pressure of 1000 kPa and a stagnation temperature of500 K. The throat area is 8 cm 2. Determine the mass flow rate, exit pressure, exit

temperature, exit M ach 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.


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For A/ A* = 2, Table A-32 yields two M ach 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, M E = 2.197, P E/P o = 0.0939, and T E/T o = 0.5089,

P P kPa kPa

T T K K E o

E o

0 0939 0 0939 1000 93 9

0 8333 0 5089 254 5

. . ( ) .

. . (500 ) .

C kRT

kJ kg K

K

m

skJ kg

m s

E E

14 0 287 254 51000

319 7

2

2

. ( . )( . )

.

V M C m m

2 197 319 7 702 5( )


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V M C s s E E E

2 197 319 7 702 5. ( . ) .

The mass flow rate can be calculated at any known cross-sectional area where theproperties are known. It normally is best to use the throat conditions. Since the flow

has sonic conditions at the throat, M t = 1, andT T

T T

T T K K

t

o o

t o

*

.

. . (500 ) .

08333

0 8333 0 8333 416 6

V C kRT

kJ kg K

K

m s

kJ kg

m s

t t t

14 0 287 416 61000

409 2

2

2

. ( . )( . )

.

P P

P P

P P kPa kPa

t

o o

t o

*

.

. . ( )

0528

0528 0528 1000 528

P kPa kJ** (528 )


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t P

RT kPa

kJ kg K

K

kJ m kPa

kg

m

**

(528 )

( . )( . )

.

0 287 416 6

4 416

3

3

. (8 )( . )( )

.

m AV

kg m

cm m

sm

cm

kg s

t t t

4 416 409 2100

1446

32

2

2

When the diverging section acts as a diffuser, we use M < 1. Then, for AE / A* = 2.0, M E = 0.308, P E /P o = 0.936, and T E /T o = 0.9812,

P P kPa kPa

T T K K E o

E o

0 0939 0 936 1000 936

0 8333 0 9812 490 6

. . ( )

. . (500 ) .

C kRT E E


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kJ kg K

K

m s

kJ kg

m s

E E

14 0 287 490 61000

444 0

2

2. ( . )( . )

.

V M C m

sm s E E E

0 308 444 0 136 7. ( . ) .

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

Normal Shocks

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 thediverging section experiences a normal shock . The normal shock causes a suddenrise in pressure and temperature and a sudden drop in velocity to subsonic levels.Flow through the shock is highly irreversible, and thus it cannot be approximated asisentropic. The properties of an ideal gas with constant specific heats before(subscript 1) and after (subscript 2 ) a shock are related by


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We assume steady-flow with no heat and work interactions and no potential energychanges. We have the following

Conservation of mass

1 1 2 2

2 2 2 2

AV AV

V V

Conservation of energy


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gy2 2

1 21 2

1 2

1 2

2 2

:

o o

o o

V V h h

h h

for ideal gases T T

Conservation of momentum

Rearranging Eq. 17-14 and integrating yield

1 2 2 1( ) ( ) A P P m V V

Increase of entropy

2 1 0 s s

Thus, we see that from the conservation of energy, the stagnation temperature is

constant across the shock. However, the stagnation pressure decreases across theshock because of irreversibilities. The ordinary (static) temperature rises drasticallybecause of the conversion of kinetic energy into enthalpy due to a large drop in fluidvelocity.

We can show that the following relations apply across the shock.


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g pp y2

2 12

1 2

21 12

21 2 2

22 12 2

1

1 ( 1) / 21 ( 1) / 2

1 ( 1) / 2

1 ( 1) / 2

2 /( 1)2 /( 1) 1

T M k T M k

M M k P P M M k

M k M

M k k

The entropy change across the shock is obtained by applying the entropy-change

equation for an ideal gas, constant properties, across the shock:2 2

2 11 1

ln ln pT P

s s C RT P

Example 17-8

Air flowing with a velocity of 600 m/s, a pressure of 60 kPa, and a temperature of 260K undergoes a normal shock. Determine the velocity and static and stagnationconditions after the shock and the entropy change across the shock.

The Mach number before the shock is

1 1V V M


55/57

55

11 1

2

2

600

10001.4(0.287 )(260 )

1.856

M C kRT

m s

mkJ s K

kJ kg K kg

For M 1 = 1.856, Table A-32 gives

1 1

1 1

0.1597, 0.5921o o

P T P T

For M x = 1.856, Table A-33 gives the following results.

2 22

1 1

2 22

1 1 1

0.6045, 3.852, 2.4473

1.574, 0.7875, 4.931o oo

P M

P P P T

T P P

From the conservation of mass with A2 = A1.


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56

2 1

2 2 1 1

1

2 2

1

600245.2

2.4473

V V

mV m sV

s

22 1

1

2

2 11

60 (3.852) 231.1

260 (1.574) 409.2

P P P kPa kPa

P

T T T K K

T

11 2

1

1

260439.1

0.5921o o

o

T K T K T

T T

11

1

1

60 375.60.1597o

o

P kPa P kPa P P

2

2 1

1

375.6 (0.7875) 295.8oo oo

P P P kPa kPa

P

The entropy change across the shock is


57/57

57

py g

2 22 1

1 1

ln ln P T P

s s C RT P

2 1 1.005 ln 1.574 0.287 ln 3.852

0.0688

kJ kJ s s kg K kg K

kJ kg K

You are encouraged to read about the following topics in the text:

Fanno lineRayleigh lineOblique shocksChoked Rayleigh flowSteam nozzles

Chapter 17-compressible fluid flow

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Transcript of Chapter 17-compressible fluid flow