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로드 중.... (전체 텍스트 보기)

전체 글

(1)

 Viscous Stress and Force

 The Navier-Stokes Equation

 Equation of motion

 Boundary/initial conditions

 Applications of the Navier-Stokes Equation

 Couette flow

 Poiseuille flow

 Combined Couette-Poiseuille flow

 Creeping (Stokes) flow

 Unsteady laminar flow

A Quick Look

(2)

 Fluid stress is the sum of

Pressure stress Viscous stress (in normal direction) (proportional to )

) shear (

tangential normal

n

p  

  

Translation Rotation

(rigid body)

Shear Compression

yield viscous stress

Fluid Stress

(3)

 Viscous stress is proportional to the rate of motions

Fluid element V

x

y x

xy

yx

yx

xy

Fluid element V

y

xy xy

yx

yx

y V x

xy ?

 ?

direction surface

 

 

 

 

 

 

 

gradient V μ

fluid stress

shear

Viscous Stress

(4)

D erivation

dy

y

z dx dz x

o

xz

xy xx

xz xx

xy dy

y

xy

?

 ?

x dx

xx

?

 ?

z dz

xz

?

 ?

constant ,

0 V

if

) z , y , x k

( V

i

f k 2 k

 

3. Obtain the Navier-Stokes equation as V g

1 p t

D V

D   2

 

  p * ν V

ρ V 1

t V

V    2

 

Navier - Stokes Equation

1. Start from the Euler equation + viscous term

 

 

 f

g 1 p

t D

V

D 

 

2. Determine the viscous force per unit mass

(5)

Fluid-solid boundary condition: no slip

Immiscible fluid-fluid boundary condition

solid

V V  

2 1

2 1

2 1

τ τ

) n p ( )

n p ( inviscid If

stress σ

σ

 

n  2

n  1

1

2

p 2

p 1

Fluid 1

Fluid 2

Boundary Conditions

(6)

Navier-Stokes Equation

x x

z y

y

h

V

p

w

w

xy xy

V xy

* 1 p

t D

V

D  2

?

?

0 0

 Boundary condition

p x

x ( 0 ) 0 , V ( h ) V

V  

x p w

p x

h i μ V τ

h V y ) y ( V

 

2 V V

2 h V W

Q

p p

Plane Couette Flow

(7)

x y

z

high

p h low

p

w

w

y=h

y=0

V x

L

w

w

p (x) p (x+L)

C.V.

2 1

2 x

2

C y C x y

d

* p d 2 μ V 1

ν V

* ρ p

1 t

D V D

 

Plane Poiseuille Flow

Navier-Stokes Equation

(8)

 Boundary condition

  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  

 

 

 

x d

* p d 2

) h (

* p ) (

* L p 2 τ h

x i d

* p d 2

i h x d

* p d μ

2 μ h y i

μ V τ

3 V 2 x

d

* p d 12

h h W V Q

x d

* p d 12

h W

Q

x d

* p d μ

8 h 2

h x

d

* p d μ

2 1 2

y h V V

) y h ( x y

d

* p d μ

2 y 1 x d

* p d μ 2 y h

x d

* p d μ 2 V 1

W

x x

x 0 y x W

max x 2

3

2 2 x

max x

2 x

L x x

 

0 ) h ( V 0

) 0 (

V xx

 Results

Plane Poiseuille Flow

(9)

x y

z

high

p h low

p

w

w

y=h

y=0

V x

V p

V

p

) y h ( x y d

* p d 2 μ

) 1 y ( V : Poiseuille plane

h V y ) y ( V : Couette plane

x

p x

 

 

 

Superposition of the two solutions

   

x x

p W PP

W PC W

3 p

PP PC

p x

x i d

* p d 2

i h h μ V τ

τ τ

x d

* p d μ

12 h 2

h V W

Q W

Q W

Q

) y h ( x y

d

* p d μ

2 1 h V y ) y ( V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Combined Plane Couette + Poiseuille Flow

Individual solutions

(10)

z r

R w

w

) r ( V z

z 0 0 V

θ V

0 V 0

V

z z

θ r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

D μ V τ 8

z or d

* p d 4

L π D ) π DL τ (

z d

* p d μ

32 D A

V Q

z d

* p d μ

128 π D dr π r 2 V Q

) r R z (

d

* p d μ

4 V 1

z dr d

* p d μ 2 dV r

W 2

W

2 R

0

4 z

2 2 z

z

 

 

 

z

2 z

z

r 2 2 θ θ θ 2

r θ θ

θ 2 2

r r 2 2

θ r

r

ν V r

* p ρ V 1

t V V

θ V r

2 r

V V θ ν

* p ρ

1 r

V V V

t V V

θ V r

2 r

V V r ν

* p ρ

1 2

V V t V

V

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Circular Poiseuille Flow

Navier-Stokes Equation

 Boundary condition

 Results

(11)

?

)

V F

R

a

V R

V j

j

t V D

V D

V

* 1 p

t D

V D

2

2

 

 



 

Results

 

 

    j

 

 

  

j

 j

3 3 3 3 R

R 4

a 5 R

4 a 1 3

sin V

V

R 2

a R

2 a 1 3

cos V

V

a V 6

F

R 4

a 5 R

4 a 1 3

R sin V

R 2

cos aV

p 3

* p

3 3 R

2

*

 j 

 

 

 

  

j

 

j

 

Creeping Flow (having a very small Re)

Navier-Stokes Equation

(12)

const

x with slowly

changes a

pipe converging

x d

p d

) x ( a a

x x 2

r x

x

x V

x p 1 r

V V x

V V t

V   ?

?

?

 

?? 

?

?? ?

?

?

?

 ?

?

 ?

?

?

) steady (

0 x

p

?

 ?

x p

?

 ?

) a r

x ( p 1 4 V 1

. C . B apply

r r V d x dr

p r

2 2

x

x

 

 

 

 

 

 

 

r

a o a Q x

x=0 p=p

1

steady

x=L p=p

2

Converging Pipe

Navier-Stokes Equation

Solution

(13)

Volume flow rate

Pressure drop

4

4

a 0

2 a 2

0 x

a Q 8 x

p

x p 1 8

a

dr ) a r

( x r

p 1 4 dr 2

r 2 V Q

 

 

 

 

 

 

 

 

 

 

  

 

 

 

 

 

 

 

) 1 1

( 1 a

Q p 8

Δ

L x 1

a ) x ( a

3 4

o o

Converging Pipe

(14)

Rayleigh problem (1-D)

  V V 1 p * V

t

V 2

0

 

 

 

 

 

x y z

V 

U

? ) 0 t ( V

0 ) 0 t ( V

 

Infinite large plate ( no x,y dependence) No asymptotic profile!

- momentum diffuses into fluid via viscosity

- if the fluid were inviscid there would be no fluid motion

t > 0 ( time ↑)

V z

Unsteady Flow - Example

Navier-Stokes Equation

(15)

x z

y fluid z ≥0

V x (z, t)

U 0 at t=0

x-comp. Navier-Stokes eq.

x 2

x z x

y x

x x

x V p 1

z V V

y V V

x V V

t V

 ?

 

?

?

 

?

 ?

?

 ?

?

 ?

?

?

0 0 0

0

very large plate

(continuity)

(no external pressure applied)

mass

everywhere 0

V

z 0 V y

V x

V V

z

y z x

? 

 ?

?

 ?

?

 ?

?

? 

0 0

thus diffusion eq .

z V t

V

2 x 2

x

 

 

Rayleigh Problem

Solution

(16)

initial condition boundary condition variables change 0

) 0 t , z (

V x  

0 ) 0 t , z

( V

U )

0 t , 0 z ( V

x x

t z

 

V 0 2 V

V t

1 z

and V V

t 2 t

now V

x 2

x 2

2 x 2

t 2

x 2 x

z x

 

 

 

 

 

 

 

 similarity

solution

U V x

t z

 

All solutions are similar so that they are combined into 1 solution.

Rayleigh Problem

f

 V x

 

 

 

  

1 2 V x

exp 4 C

f

 

 

 

 t

erf z t U U

) t , z ( V x

Let

then Apply IC + BC to finally obtain

the desired velocity profile as

(17)

 Head Loss in Pipes & Ducts

 Straight pipes:  h given Q; Q given h

 Straight noncircular ducts

 Loss from area change, bend, fittings, etc.

 Total head loss

 Head Changes in Systems with Pumps and Turbines

 Matching pumps & hydroturbines to flow systems

 Complex Networks

 Serial components

 Parallel components

A Quick Look

(18)

 Head (incompressible, 1-phase flow)

 Straight Pipes

 viscous shear on a pipe causes a Δp ;

) gz p

R g p

* p g (

2 V g

* p g

R g g

2 V g

p

2

2

      

 

 

 

    

D V L

2 p 1

p

*in *out 2

 

  

f

 If there is no change in velocity  Dimensionless loss coefficient

g 2 V D

L g

p

p

*in *out 2

out

in

f

f

 

   

g 2 K V D ,

K L thus

V g 2 g 2 / K V

2 2

2

f f

f

f  

 

 

Head Loss in Pipes and Ducts

(19)

bulent tur

a 1 r 49 60 V

V

laminar

a 1 r

V 2 V

1/7 z

2 z

 

  

 

 

 

 

 

a g

z V V

2 . 49 1

60 V

V V 2 V : 0 r

tur max, z

lam max, z

 

 

 

 

 

Velocity Profiles

(20)

 Basic Formula:

 laminar:

 turbulent:

) Re (

64

D

2300 Re

f

D

) D / , Re ( D

f f

) 2300 Re

f (

f  

D

 

  

Re

D

51 . 2 7

. 3

D log /

0 . 1 2

0.008

10

3 2300

10

4

10

5

10

6

10

7

10

8

0.10

0.04

0.01

0 D / 

10

6

D / 

10

-5

10

-4

10

-3

10

-2

V

D Re

D

f

(Darcy)  / D

① Critical Zone, Discontinuous

smooth pipes

rough pipes

① Transition Zone

① Laminar Flow Zone

① Complete Turbulence Zone

The Pipe Darcy-Weisbach Friction Factor

(21)

 Turbulent Skin Friction and Drag

 Wall shear stress is much larger than in a laminar flow

 Much less dependent upon the flow Re number

 Even independent of Re in some circumstances

 Turbulent Pipe Flow

 Friction factor

 Including roughness 51 . 2

f log Re

0 . 2 f

1  D

 

 

  

f Re

51 . 2 3.7

log /D 2 f

1

D

10

-2

10

-4

10

-6

ε/D=0 log f

4 6 8

-1.5 -2.0 -2.5

Pipe Flow

(22)

x

y z

 

 

 

 

2 2 2 2

x 2 2

z V y

μ V dx

* 0 dp

: direction x

ν V

* ρ p

1 t

D V

D  

) 1 ( ρ V

2 1

D ) dx /

* dp f (

Perimeter Area D 4

: D

2 h h

h

 

 

; factor friction

Darcy

diameter hydraulic

l

; l dimension stic

characteri

 

 

 

 

 

 

x d

* p d μ V

D Re 2

f

gives )

1 (

) 2 ( ρ V

2 1 D f L x L

d

* p p d

Δ

h D

2 h

h

; difference pressure

Poiseuille Flow in Noncircular Ducts

(23)

a

b

a

b b

2a

h D

2 2

D D

b 2

L V b

/ D

Re

* f p ) 2 ( from Then

) tube circular

a for ( 64 Re

f from 25%

than more

by differ not

does Re

f

h

h h

 

b/a

Ellipse Rectangle Isosceles Triangle

D

h

Re

f D h / b f Re D

h

D h / b f Re D

h

D h / b

1.0 64.00 1.0 56.91 1.0 52.61 0.828

0.5 67.29 1.298 62.19 1.333 50.49 0.944 0.2 74.41 1.497 76.28 1.667 48.63 0.990 0.1 77.26 1.548 84.68 1.818 48.31 0.998

Friction Factor for Noncircular Ducts

Friction factor & hydraulic diameter for noncircular tubes

(24)

 Note the discontinuity at

2 T

2 L

D

10 728 . 4

10 783 . 2 2300

Re

f f

Actually the increase in f is spread over a range of Re near 2300. This numerical

discontinuity seldom causes trouble in pipe flow calculation. In most of the computer codes, they have continuous Darcy friction factor curves interpolated in the critical zone.

 Head loss for given Q:

 Calculate

 Calculate or depending on

 Calculate

2 2

D Q V 4

4 V D

Q     

 

 

D Q 4 D

Re

D

V

f

L

f

T

Re

D

 2300 or Re

D

 2300

f f

f

f        

 

 

 , p * g

g 2 K V D ,

K L

2

The Pipe Darcy-Weisbach Friction Factor

(25)

 8 inch Φ Commercial Steel Pipe, ε = 5×10 -5 m

gpm 1000 Q  km 2

m 22

m 100

Tank Storage

 

 

 

 

 

 

f

f

5

5

5 D 6

2

3 2

10 95 . 3

51 . 2 )

2032 . 0 ( 7 . 3

10 log 5

0 . 1 2

turbulent :

10 95 . 10 3

1

) 2032 . 0 ( ) 944 . 1 Re (

s / m 944 . D 1

Q V 4

s / m 10 303 . 6 gpm 1000 Q

m 2032 . 0 D . 1

2 -2

10 1.616 iterations

more 3

after obtain

finally to

10 1 assuming

by Begin .

2

f f

     

 

g z

z z

z g p

p p

p

64 . ) 30 807 . 9 ( 2

) 944 . 1 10 ( 59 . g 1

2 K V

10 59 . 2032 1 . 0 10 2000 616

. D 1 f L K . 3

out in

out in

out in

* out

* in out

in

2 2

2 f f

2 2

f

 

 

 

Pipe Friction Factor - Example

(26)

A

l

A

S

V

S

Sudden Contraction

A

l

A

S

Sudden Expansion A 0

when A 4

. A 0

1 A 4 . 0

K c s    s

 

 

l l

A 0 when A A 1

1 A

K s

2

e s   

 

 

l l

 Total Head Loss

 

 

i

 

i

i i

f i

j

i j i

f i

 

 

 

 

 

 

 

  

 

g p

p

D f L where

g 2 K V

* out

* in

2

 Head Changes in System with Pumps / Turbines

t p

in

out    

     l    

Loss from Area Changes

(27)

 A Heat Collecting Circuit for a Solar Panel

K e = 1.0

K c = 0.4 K b = 1.0

header diameter

>> pipe diameter

D = 1 cm L = 22 m Q = 4 l /min

70 . 53 ( 10 K K K )V 3 . 01 m

g 2 K V

D h L

2 2

 

 

 

 

 

 

 

 

 

j j

l

f

489 , 8 /

D V Re

s / m 8489 .

0 A / Q V

m 10 854 . 7 4 / D A

s / m 10

667 . 6 Q

D

2 5 2

3 5

 Choose 10 6 m

2

D

10 205 . 3

Re 51 . 2 7

. 3

D log /

0 . 1 2

 

 

  

f

f f

Loss from Area Changes - Example

(Turbulent)

(28)

max power

Q max

/

 Q

) pump

 (

system piping

0 0.2 0.4 0.6 0.8 1.0

0.2 0.4 0.6 0.8 1.0

h max

h

  x

max

max h

Q

h Q

 

 x

? ?

?

?

??

?

?

?

  

0 h

Q Q at

h h

) .

e . i ( 0 Q at

pump

max max

off colsed

discarge

Matching Pumps to Flow Systems (1)

(29)

Q will be determined for given pump and the piping system .

 

Q as h

Q as h p

l

h g

Q h

g

m     

 Procedure

 Ideal power delivered to fluid stream passing through the pump [W]

Power required to through the pump:

/ p

h g

Q  

 where  p : pump efficiency

 Pump are designed such that is reached at i.e. the maximum power.

max ,

 p ( Q  h ) max

Matching Pumps to Flow Systems (2)

(30)

 Pump:

 Assume

 Calculate Q 20

reservoir water

80 m pump Q ?

? ?

?

?

? ?

?

?

? ?

?

? ?

 ?

?

2 p

5

) gpm ( 1000

) gpm ( 1 Q

) m 150 ( h

m 10 5

. in 6 D

km 5 . 1 L

) loss head

viscous (

h m

80

h

p

  

l

m 20 h 

l

4 2

3

10 043 . L 3

D h g

Re 2  

 

l

f

10

5

292 . 2

Re 51 . 2 7

. 3

D log /

Re 0

. 2 Re

 

 

  

f f

 

1 435 / 1000121 . 6 m ( 80 20 ) m

150 h

, gpm 4 435

Re

Q

20

  D   

p

 

2

  

 Now assume  h l  40 m , Q 40 = 626 gpm

 

1 626 / 100091 . 22 m ( 80 20 ) m

150

h

p

 

2

  

 Iterate more in this way, Q  527 . 7 gpm and  h p  108 m

Matching Pumps to Flow Systems - Example

(31)

 Hydroturbine

s / m 659 . D 5

Q V 4

2

 

turbulent :

10 488 . D 8

Re

D

V  

6

 

366 . Re 9

51 . 2 7

. 3

D log /

0 .

1 2   

 

  

f

f

281 . D 2

K L  

 

f

f

0 . 1 K

K , 4 . 0 K

K

inlet

C

outlet

e

m 010 . 2 6

) V K

K K

2 ( K V h

2 out

in

2

     

 

 

l

f

m 99 . 93 h

100 h

Now 

t

  

l

kW 501 h

g Q )

Power Turbine

(     

) 85 . 0 (

turbine 

t

h = 100 m

D = 1.5 m

m 10 1 ?

4

 L = 300 m

Q = 10 m

3

/s

Matching Pumps to Flow Systems - Example

(32)

pipe A

pipe B

Q Q A Q B Q

Q A = Q B = Q

Series Parallel

Q

A

Q Q

pipe B pipe A

Q

B

Q

A

= Q

B

= Q

B , l A

,

l

h

h  

 If Q is known

 determine h A & h B

 then h

l

h A h B

 If is known  h

l

 assume h A ,

 get Q A & Q B

A

B h h

h    

l

 find that makes h A Q A Q B

 If is known

 find

 get

 If Q is known

 guess  h

l

 calculate Q A , Q B h

l

B A & Q Q

B

A Q

Q

Q  

B

A Q

Q

Q  

until

Complex Networks

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