Chapter 15 Compressible Flow - home.sogang.ac.krhome.sogang.ac.kr/sites/thermal/menu4/Lists/b8/Attachments/31/... · Fundamentals of Thermodynamics Chapter 15 Compressible Flow

Fundamentals of Thermodynamics

Chapter 15

Compressible Flow

Thermal Engineering Lab.

reversible adiabatic deceleration

V = 0

- isentropic stagnation state

흐르는유체가 을

통해 가 되었을때의상태

2

02

V- h h stagnation enthalpy

15.1 Stagnation properties

2

Chapter 15. Compressible Flow


Ts

h

vdPdhTds

P

3


Thermal Engineering Lab. 4

Ex. 15.1 Air flows in a duct at a pressure of 150 kPa with a velocity of 200 m/s.

The temperature of the air is 300 K. Determine the isentropic stagnation

pressure and temperature.



0

x

C.V.

x e ex i ix

d mVSSSF

dt

F m V mV

에서

C.V.에 가해지는 힘

volumecontrol : -)(

mass fixed : )(

..

exeixix

vc

x

xx

VmVmFdt

mVd

Fdt

mVd

15.2 The momentum equation for a control volume


5


- C.V.(유체)에가해지는 힘

x ex ixF m V V


6


Ex. 15.2 On a level floor, a man is pushing a wheelbarrow (Fig. 15.3) into which

Sand is falling at the rate of 1 kg/s. The man is walking at the rate of 1

m/s, and the sand has a velocity of 10 m/s as it falls into the wheelbarrow.

Determine the force the man must exert on the wheelbarrow and the

force the floor exerts on the wheelbarrow due to the falling sand.



R: external force C.V. 에의해 에가해지는 힘

0

0

0

0

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

( ) 0

x x

x x

x x x

x i i e e t i e xx

i i e e i e x x

i i o i e e e x x

t x

F P A P A P A A A R

P A P A P A A R

P A P A P A P A R

P A

15.3 The momentum equation for a control volume

8



Ex. 15.3 A jet engine is being tested on a test stand (Fig. 15.5). The inlet area to

the compressor is 0.2 m2, and air enters the compressor at 95 kPa, 100

m/s. The pressure of the atmosphere is 100 kPa. The exit area of the

engine is 0.1 m2, and the products of combustion leave the exit plane at a

pressure of 125 kPa and a velocity of 450 m/s. The air-fuel ratio is 50

kilograms of air to one kilograms of fuel, and the fuel enters with a low

velocity. The rate of air flow entering the engine is 20 kg/s. Determine

the thrust, Rx, on the engine.



e i i i e e

i i e e

- m = m = AV = AV

AV AV

2 2

2 2

02 2

02 2

i ei i e e

e ie i e i

- first law

V Vq w h gz h gz

V Vh h g z z

10


15.4 Adiabatic, one-dimensional, steady-state flow of an incompressible fluid through a nozzle


.

e i

e i

e

e i e ii

- second law

s s

rev adiabatic s s

Tds dh vdP dh vdP

h h vdP v P P

Bernoulli eq.

2 2

02 2

e ie i e i

V Vv P P g z z

11



Ex. 15.4 Water enters the diffuser in a pump casing with a velocity of 30 m/s, a

pressure of 350 kPa, and a temperature of 25℃. It leaves the diffuser

with a velocity of 7 m/s and a pressure of 600 kPa. Determine the exit

pressure for a reversible diffuser with these inlet conditions and exit

velocity. Determine the increase in enthalpy, internal energy, and entropy

for the actual diffuser.



17.4 Ex 보충설명

P

ieieie

i

ePie

iseie

ieie

eiie

ieiePie

i

ePie

P

P

P

P

c

PPvhhTT

T

Tcssleirreversib

PPvhh

ssTTisentropic

VVhh

PPvTTchh

vdPTdsdh

T

Tcss

dTT

cds

T

vibleincompress

dPT

vdT

T

cds

)()(

ln:

)(

,:

22

)()(

ln

,0

,

22

Ex. 15.4

13



- pressure disturbance : sonic velocity

- sound wave : very small pressure disturbance

의 전파속도

14


15.5 Velocity of sound in an ideal gas



2

2 2

0

0

0, 0

,

2

s

s

i) cA= +d c dV A= m

cd dV

c dVcii) h+ = h+dh +

2 2

dh cdV

dP dPiii) isentropic Tds = dh dh =

dP dP cdcdV c

dP Pc c

d

Pc

과정에서 이므로

1

2

Continuity와 열역학 1, 2법칙을 적용



2

( ) ( )

0

( ) ( ) ( )

s

Ac d A c dV

cd dV

PA P dP A m c dV c Ac c dV c

dP cdV

cddP c

dP Pc c

d

Continuity와 momentum eq.을 적용



Ideal gas sonic velocity의

kPv const

ln lnP k v C

2

1 1

0

dP dv+k = 0

P v

v dv d

dP dk

P

미분

이므로

2

dP kP PP = RT = RT

d

dPkRT c kRT c kRT

d

에서


Ex. 15.5 Determine the velocity of sound in an air at 300 K and at 1000 K.




15.6 Reversible, adiabatic, one-dimensional flow of an ideal gas through a nozzle

0

0

0

0

2Vh+ = const

2

dh+VdV = 0

continuity AV m const

d AV dA V dV A

d dA dV

A V

property relation

Tds = dh vdP rev. adiabatic isentropic

dh vdP

dPdh

미분하면

에서

1

2

3



dPdh VdV

dPdV

V

에서

2

2

2 2

1 1

1 1

dA d dV

A V

d dP dP

dP V

dP

VdP

d

dP

c V

에서

2 3

1

와



2

2 2

2

2

1

1

:

dP V VM

V c c

dPM

V

VMach number M

c

M 1 : super sonic

M 1 : sub sonic

M 1 : sonic



- : 0

; 1 0

nozzle dP

subsonic M dA

; 1 0supersonic M dA

- : 0

; 1 0

diffuser dP

subsonic M dA

supersonic ; M 1 dA 0

1 0M dA throat

Nozzle

Diffuser

2

21

dA dPM

A V



0

0

0

0 0

0

0

1

P

P

P

P v

P

v

CC R

k

k C kR

C C R

Ck

C

0

2

0

2 00 0

22 2 0

22 0

2

20

2

2 2 2 11

21

1

21

1

11

2

P

Vh h

TkRT- V h h C T T

k T

Tc- c kRT V

k T

TVM

c k T

kTM

T

이므로



11 1

0 0 0 0

11 1

2 20 0

*

0

1* *1 1

0 0

,

1 11 1

2 2

1

2

1

2 2

1 1

kk k

kk k

kk k

T P T-

T P T

k kP- M M

P

- throat M

T

T k

P

P k k

이므로

에서 인 경우



15.7 Mass flow rate of an ideal gas through an isentropic nozzle

0

0

2

0

0

1

0 2 12

0

1

0 2 12

1

11

2

11

2

11

2

k

k

k

k

m PV- V

A RT

TPV k PV k

c R T TkRT RT

PM k kM

RT

P k M

RT kM

P k M

RT kM

0

121

12

kk

PP

kM



*

0

1*

0 2 1

1

2 12

*

1

2

, 1,

1

1 2 11

1 2

k

k

k

k

k

- At throat M A A

Pm k

A RT

A k- M

A M k



( )- Effect of back pressure converging nozzle



( )- Effect of back pressure converging diverging nozzle


Ex. 15.6 A convergent nozzle has an exit area of 500 mm2. Air enters the nozzle

with a stagnation pressure of 1000 kPa and a stagnation temperature of

360 K. Determine the mass rate of flow for back pressures of 800 kPa,

528 kPa, and 300 kPa, assuming isentropic flow.



Ex. 15.7 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 360 K. The throat area is 500 mm2. Determine

the mass rate of flow, exit pressure, exit temperature, exit Mach number,

and exit velocity for the following conditions:

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

(corresponds to point d in Fig. 15.13.)

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

(correspints to point c in Fig. 15.13.)




15.8 Normal shock in an ideal gas flowing through a nozzle

: &shock wave extermely rapid abrupt change of state

22

;2 2

;

2 : 0

yxx y ox oy

x x y y

x y y x

y x

VVh h h h

mV V

A

A P P m V V

nd law S S

v

v

v

v

Fanno line

Rayleigh line

Rankine Hugoniot line





22

2 2

1 1 1

2

0

1 1

oy y y x

ox x x y

x x x y y y

y x y x

x y

y x

Fanno line

h h V

h h V

Rayleigh line

P V P V

Rankine Hugoniot line

h h P P

- s s x y

M M

이므로 로 일어남

로 일어남

v

v

v



Normal shock governing equation 의 Fanno - line

2 2

2

2

1 11 , 1

2 2

11

2 - 17.491

12

ox oy ox oy P

oyoxx y

x y

xy

xy

x x y y

yxx y

x y

- T T h h assume const C

TT k k- M M

T T

kMT

-kT

M

- V V

PP

RT RT

에서

그런데

또한 이므로

5



2 2

- 17.50

y y yy y y y y y yx

x y x x x x x x x x x

y y y y y y

x x x x xx

P M TT P P V P M c

T P P V P M c P M T

T P M T P M

P M T P MT

2

2

11

2: - 17.51

11

2

x xy

x

y y

kM M

P- Fanno - line

P kM M

5

5



2 2

2 2

2 2 2 2

x y y x y y x x

x x x y y y

x x x x y y y y

xx

Rayleigh - line

m- P P V V V V

A

- P V P V

- P M c P M c

P- P

R

xT

2

xM k R xT y

y

PP

R

yT

2

yM k R yT

2 2

2

2

1 1

1; - 17.52

1

x x y y

y x

x y

P kM P kM

P kMRayleigh - line

P kM

2

2

2

17.51 , 17.52

2

1 ,2

11

x

y y x

x

MkM M f M k

kM

k

에서

5


Ex. 15.8 Consider the convergent-divergent nozzle of Example 15.7, in which the

diverging section acts as a supersonic nozzle (Fig. 15.16). Assume that a

normal shock stands in the exit plane of the nozzle. Determine the static

pressure and temperature and the stagnation pressure just downstream of

the normal shock.




Ex. 15.9 Consider the convergent-divergent nozzle of Examples 15.7 and 15.8.

Assume that there is a normal shock wave standing at the point where

M = 1.5. Determine the exit-plane pressure, temperature, and Mach

number. Assume isentropic flow except for the normal shock (Fig. 15.18).




:

17.9 Nozzle and Diffuser Coefficients

N

nozzle efficiency

- actual flow velocity coefficient

discharge coefficient

- nozzle efficiency

actual kinetic energy at nozzle exit

kinetic energy at nozzle exit with

isentropic f

low to same exit pressure

15.9 Nozzle and diffuser coefficients



V

V

- Velocity coefficient : C

Actual velocity at nozzle exit C =

Velocity at nozzle exit with isentropic flow to same exit pressure

ev N

s

VC

V



D

D

: 1

- Discharge coefficient : C

Actual mass rate of flowC =

Mass rate of flow with isentropic flow

s

a

s

choked flow throat M mm

m 에서 기준



3 1 3 1

2

1 1 1 2 1

:

2

sD

o o

h h h h hDiffuser efficiency

V h h h h



0

0

3 1

1

3 1 1

2

12 1

12

2 2 2 3 211 1 1 1

1 1

12

12

1

2

, , ,1

1

D

o

P

kk

oP

kk

o

D

T TT

T T T

VT T

C

T PckRC T V M c

k kR T P

cP

P

kR

2 2

1 1M c

2kR

1

2

1

2

1

1

1

2

1

kk

oP

P

k M

k



1

2 21

1

2

1

11 1

2

1

2

kk

o

o

D

PkM

P

kM

11 1

2 1 2

1 1 1

kk kkk k

o o o

o

P P P

P P P

1

211

1

11

2

k kk

oP kM

P

1k1k

k2

1

11

2

kM



15.9 Nozzle and orifices as flow-measuring devices

2

2 222 2 2 2

12 11 2 1 2

1 2

2 2

2

1

,

2 2

2

1

incompressible reversible

AV V

AV Vv P P v P P

v P PV

AA



ΔP ideal gas: Incompressible 가작은경우 로계산 가능

0

2 2

2 2

1 1

1

2

2 2

2

1 1

1 11 1

2

i ei e

e ii e P i e

k kk k

e e i i

i i i i

kk

i i

i i

V V- h h

V Vh h C T T

T P T T P P

T P T P

T P

T P

T k P k

T k P k

0 0 0

2

2

2 2

1

11

2

i

i i

e i i i iP i P i P i

i i i

P

P

T k P

T k P

V V T T T k PC T C T C T

T T k P

k

1

R

k

1k

ki

i

i

PT

P

v P



Pitot tube



2

0 0

0

2

2

incompressible

Vh h v P P

V v P P

인 경우

0 0

0

2

00 0

1

0

12120

2

12

1

1 11 1

2 2

P P

kk

P

kk kk

compressible

TVh h C T T C T

T

PC T

P

P k k VM

P c

인 경우

그런데



0 0

2

0

2

0

22 2

0

22

0

2 2

2 22200

2 2 2

2

2 1 1

2 1 1

221

1 1

2 11

2 2

P P

VC T C T

V kR kRT T

k k

cV c

k k

cc

V k V k

c V kcc k

V V V



1

22

0

2

0

)1(2

2

2

)1(1

kk

kVc

Vk

P

P

12

0

1

2

0

22

0

1

222

0

22

0

1

22

0

2

0

2

)1(1

4

)}1(2{2

2)1()}1(2{2

)}1(2{2

)1(2

2

2

)1(1

1

kk

kk

kk

kk

c

Vk

c

kVc

VkkVc

kVc

kVc

VkP

P



2 1

0 0

2 4

0 0 0

2 4

00 0 0

0 0

11

2

12 8

2 8

kk

P k V

P c

P k V k V

P c c

kP kV k VP P P P

c c

2

0

2

P V

k 0

k

RT

22

0

08

P V V

c k

2

0

2

0

0

22 2

0 00

0

22 2

0 00

0

2

0

0

0

2

2

2 8

1

2 2 4

11

4

1 for incompressible flow

V

V

RT

V V VP

c

V V VP P

c

P P V

c

P P



:참고자료 Fanno, Rayleigh, Rankine-Hugoniot Line 작도

22

222

2

222

2

2

)

2 2

1

2 2

1

2 2

2

yxx y

x x y y

y y yx x x xy x x x x

y x y y x x y

yx xP y x x

x y

yx xP x y x

x y

x

i Fanno - line

VVh h

V V

RT T TP P PV V V V V

RT P P T T P

TV PC T T V

T P

TV PC T T V

T P

VC

2

2 2 2

12 2

2 2

2 2

1

2

2 2

ln

xP x y y x y

x

x xx y x y

x xy

x P x y x P x y

y P y x x

y y

y x P

x x

PT T P V T

T

P PV T V T

T TP

V C T T V C T T

h C T T h

T PS S C R

T P



2 2

)

yx

x x x y y y x y

x y

x

xx x y y y x

y

ii Rayleigh

PPP V P V P P

RT RT

P

RV V V V

x

y

T

P

R

22

2 2

22 2

2

2

2

2 2

1 1

x y

x x

x y

y

y yx xx x y x

x y x y

x xx y x x y

x x y

x y xy x y x

x x x

P TV V

T P

T

P TP PP V P V

RT RT T P

P PP P V V T

RT R T P

RT P PT P P V

P V RT



2 2

2 2

)

2 2

1 1

2 2

1 1

2 2

1 1 1

2

2

12

x xx y

y x x y x y x y

x y y x

y x

y x x y y x x y

y x

x y

yxP y x y x

x y

P y x

y

iii Rankine - Hugoniot line

V Vh h

h h V V V V V V

P P A m V V

P P A A Ah h V V P P V V

m m m

P P

TTC T T R P P

P P

C R P PP

에서

2

2

12

xy P x y x

x

xP x y x

xy

P y x

y

TT C T R P P

P

TC T R P P

PT

C R P PP



Ex. 17.8 작도 계산예

2

2

93.9 : 1.4, 0.287 /

183.2 1.005 /1

2.197

596.1 /

271.3 /

1

2

1

2

x

x P

x

x

x

x xy

x

y y

Fanno - line

P kPa air k R kJ kg K

kRT K C kJ kg K

k

M

V m s

c m s

kM M

P

P kM M

512.7

339.7

?

0.547

y

y

y

y

P kPa

T K

V

M

Ex. 15.8



22

22

2

2

1.4 0.287 1000 200 283.5 /

2 2

2 2 1005 183.2 200 596.1 567.1 /

567.12.000

283.5

1.4 12.197 2.197

2.197 0.982521.2067

2.000 0.89441.4 12.000 2.000

2

y

y y

yxx y

y p x y x

y

y

x

T

c kRT m s

VVh h

V C T T V m s

M

P

P

S

i) 가정(200K)

200

1.005ln 0.287 ln 1.2067 0.03425 /183.2

y xS kJ kg K



2

2

1.4 12.197 2.197

2.197 0.982521.2067

2.000 0.89441.4 12.000 2.000

2

2001.005ln 0.287 ln 1.2067

183.2

0.03425 /

y

x

y x

P

P

s s

kJ kg K



150

160

170

180

183.2

339.7

340

350

400

x x x yP T V T



22

22

)

1 1.4 2.19711.1754

1 1 1.4 2.000

2001.005ln 0.287 ln 1.1754

183.2

0.04180 /

y x

x y

y x

ii Rayleigh line

P kM

P kM

s s

kJ kg K



40000

60000

80000

93900

100000

120000

500000

512700

520000

yP



0

0

00

0

0

0

1ln

P

TP TP

T

air table

reversible process

dPdh vdP C dT RT

P

C sdP dT P dTC

P R T P R T R

2 2

1 1

22 1

1

r

r

r

r

P P

P P

PP P

P

:r

P relative pressure

Documents

Chapter 15 Compressible Flow - home.sogang.ac.krhome.sogang.ac.kr/sites/thermal/menu4/Lists/b8/Attachments/31/... · Fundamentals of Thermodynamics Chapter 15 Compressible Flow