6. Kompresible_1

8/6/2019 6. Kompresible_1

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6. COMPRESSIBLE FLOW6. COMPRESSIBLE FLOW6. COMPRESSIBLE FLOW

Review of thermodynamics

Equation of state

Where is constant for each gas

is universal gas constant, 8314 . /

(1544 . / . ) and is molecula

u

m

u u

m

pv RT

R

R R

M

R R N m kgmole

ft lbf lbmole R M

=

=

=

r mass of the gas.

The ideal gas, the internal energy of a substance may expressed

as ( , ) then

The specific heat at constant volume is defined as

v T

v

u u v T

u udu dT dvT v

uc

T

=

∂ ∂ = + ∂ ∂

∂ ≡∂

so that

for the ideal gas ( ) consequently

v

v

T

v

udu c dT dv

v

u u T

du c dT

∂ = + ∂ =

=



( )

The enthalpy of a substance is defined as

for an ideal gas and hence

We express in its most general form as

,

thenv T

h u pv

p RT h u RT

h

h h p T

h hdh dT dpT p

ρ

= +

= = +

=

∂ ∂ = + ∂ ∂

for an ideal gas

The specific heats for an ideal gas have been shown to be function

of T only.

P

T

P

hdh c dT dp

p

dh c dT

h u RT

dh du

∂= + ∂

=

= +

= +

.

then

The ratio of specific heat is defined as

P v

P v

P

v

RdT

c dT c dT R dT

c c R

ck

c

= += +

=

2 2

1 1

2 2

1 1

2 1 2 1

2 1 2 1

For an ideal gas, the specific heats are functions of temperature.

( )

( )

u T

v v

u T

h T

P P

h T

u u du c dT c T T

h h dh c dT c T T

− = = = −

− = = = −

∫ ∫

∫ ∫



2 2

2 1 2 12 1 2 1

2 1

The first law thermodynamics

2 2 s

p p v vq w u u gz gz ρ ρ

− = − + − + − + −

T

s

qin

qout

1

23

4

elirreversib prosesT

dqss

reversibel prosesT

dqss

2

1

rev12

2

1

rev12

∫ >−

∫ =−

1

2

1

2v12

1

2

1

2 p12

lnR T

Tlncss

p plnR

TTlncss

ρρ

−=−

−=−



12 2 2

1 1 1

1

12 2

1 1

2 2

1 1

Equation isentropic prosessconstant

pc k

R k

k

k

k

p T T

p T T

T

T

p

p

pv

ρ

ρ

ρ

ρ

−

−

= =

=

=

=

T1

= 440 oK

p1

= 188 kPa

v1

= 210 m/s

= 0,15 kg/s

0Q <

24

311

1

3

25

1

11

21222111

222111

VA

m10.79,4

sm210.

m

kg49,1

skg

15,0

V

mA

m

kg49,1

K 440.kgK

Nm287

m

N10x88,1

RT

p

AAA;AVAVm

0AVAV

0Ad.VdVt

−==ρ

=

===ρ

==ρ=ρ=

=ρ+ρ−

=ρ+ρ∂∂

∫ ∫

kgK

kJ282,0

10.88,1

10.13,2ln

kgK

kJ287,0

440

351ln

kgK

kJ1

p

plnR

T

Tlncss

5

5

1

2

1

2 p12 −=−=−=−



=

3

5

1

o

K

p

2

=

2

1

3

k

P

a

kg

kJ89K )440351(

kgK

kJ1)TT(cdTchhh

2

1

T

T12 p p12 −=−∫ =−==−=∆

∫ −=−=−==−=∆2

1

T

T

12vv12kg

kJ8,63)440351(

kg

kJ717,0)TT(cdTcuuu

kgK

kJ282,0

10.88,1

10.13,2ln

kgK

kJ287,0

440

351ln

kgK

kJ1

p

plnR

T

Tlncss

5

5

1

2

1

2 p12 −=−=−=−

Propagation of soundwaves

x

c

ρ+dρp+dp

dVx

ρp

Vx=0

y

x

ρ+dρc-dVx

p+dp

ρc

p



a. Continuity Equation

. 0

Assumption : 1) Steady Flow

2) Uniform Flow at each section

then

( )( ) 0

0

neglecting the prod

CV CS

x

x x

dV v dA

t

cA d c dV A

or

cA cA dV A d cA d dV A

ρ ρ

ρ ρ ρ

ρ ρ ρ ρ ρ

∂+ =

∂

− + + − =

− + − + − =

∫ ∫

uct of differentials

0 x AdV cAd ρ ρ − + =

b. Momentum Equation

.

Assumption : 3. 0

4. = 0

The surface forces acting pn the infinitesimal CV are

- ( )

wher

x

x

Sx Bx rfx x x xyz

cv

B

rfx

S x

F F a dV V dV V V dAt

F

a

F dR pA p dp A

ρ ρ ρ ∂

+ − = +∂

=

= + +

∫ ∫ ∫

e represents all forces applied to the horizontal portion of CS,

Consider only a portion of moving sound wave, 0

-

substituting into the basic equation

x

x

x

S

dR

dR

F Adp

=

=

{ } { }{ }

{ } { }{ }

gives

- - ( ) ( )( )

using the continuity equation, reduces this to

- - ( )

-

or

1

x x

x

x

x

Adp c cA c dV d c dV A

Adp c cA c dV cA

Adp cAdV

dV d c

ρ ρ ρ

ρ ρ

ρ

ρ ρ

= + − + −

= + −

=

=



2

combining continuity and momentum

1 x

S

cdV dp d

c

dpcd

pc

ρ ρ ρ

ρ

ρ

= =

=

∂=

∂

c kRT =

For an ideal gas, the pressure and density in isentropic

Flow are related by .

ln ln ln .

0

k

pcons

p k cons

dp d k

p

p pk kRT

ρ

ρ

ρ

ρ

ρ ρ

=

− =

− =

∂= =

∂

v=0

v=0

c.2∆t

c.3∆t

v<c

c.2∆t

c.∆t

12

3

123

v=c v>c

1

23

α

1

The cone angle can be related to the Mach number

at which the source moves

1sin

or

1sin

c

V M

M

α

α −

= =

=



dx

Rxu

x

Stream line

V = 0

p = po

T = To

ρ + dρVx+dVx

A+dA

p+dp

T+dT

1

a. Continuity equation

. 0

Assumptions : 1). Steady Flow

2). Uniform flow at each solution

( )( )( )

VA VA

x x x

dV v dAt

V A d V dV A dA

ρ ρ

ρ ρ ρ

∂+ =

∂

= + + +

∫ ∫

Momentum Equation

.

Assumptions : 3) 04) 0

5) Frictionless

x

x

s Bx rfx x x xyz

B

rf

F F a dV V dV V V dAt

F a

ρ ρ ρ ∂

+ + = +∂

==

∫ ∫ ∫

2

( ) 02 xdp V

d ρ

+ =

The average pressure and the area component in direction x

( ) ( )( )2

sx

dp F p dA pA p dp A dA dpA= + + − + + = −

substituting into the Momentum Equation,

( ) ( )( )( )( ) x x x x x xdpA V V A V dV d V dV A dA ρ ρ ρ − = − + + + + +

2

obtain

( )

2

x x x

V dp V dV d ρ ρ = − = −

1 12

For an ideal gas became

2k k

v dpd p C dp

ρ

− − = = −

o p

p

k

1k

k /12

p1k

k C

2

v

−=

−1

211

2

k

o k v

p k p

ρ −

−= +



12

2

1

12

The isentropic stagnation properties of

an ideal gas as follows

11

2

11

2

11

2

k

k o

o

k o

p k M

p

T k M

T

k M

ρ

ρ

−

−

− = + −

= +

− = +

1 2

p1 = 350 kPa

T1 = 60oC

V1 = 183 m/s

p02 = 385 kPa

T02 = 350K

M2 = 1,3

s

m366K )60273(kgK

Nm287x4,1kRTC 11 =+==

5,0366

183

C

VM

1

11 ===

186,1M2

1k 1

p

p 1k

k

21

1

1o =

−

+=−



338,1M2

1k 1

T

T

kPa139kPa77,2

385

77,2

p p

77,2M2

1k 1

p

p

K 350T05,1T

05,1M2

1k 1

T

T

22

2

2o

2o2

1k

k

22

2

2o

11o

21

1

1o

=−

+=

===

=

−

+=

==

=−

+=

−

K 262338,1

K 350

338,1

TT 2o

2 ===

T

∆s

po2po1

p1

T1

p2

T2

To1 = To2 =350K

s

kgK

kJ0252,0

50,3

39,1ln

kgK

kJ287,0

333

262ln

kgK

kJ1

p

plnR

T

TlnCss

1

2

1

2 p12 =−=−=−

577,12

1k 1

2,12

1k 1

T

T

4,1k ;893,12

1k 1

p

p

1k

1

*

*o

*

*o

1k

k

*

*o

=

−

+=ρ

ρ

=−

+=

==

−

+=

−

−



* *

*1

*

*

*

1*

1

*

* *

** *

* *

Critical Condition M=1 or V

11 1,893 ; 1,4

2

11 1,2

2

11 1,577

2

2

1 112

2

1

k

k o

o

k o

o

o

c

p k k

p

T k

T

k

c kRT

T

T T k k

k v c RT

k

ρ

ρ

−

−

=

− = + = =

−= + =

− = + =

=

= =− ++

= =+

*

o

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6. Kompresible_1