Chapter 2. Fluid Statics

(1)

Chapter 2. Fluid Statics

Fluid mechanics (유체역학) – 단위조작의 기본 분야

관, 펌프에서의 유체이동, 교반, 반응

열전달, 확산, 물질전달 등 분리조작의 기초

“Engineering science dealing with the behavior of fluids”

↓

liquid, gas, vapor Fluid Statics (유체 정역학)

: treats fluids in the equilibrium state of no shear stress Fluid Dynamics (유체 동역학)

: treats fluids when portions of the fluid are in motion

Incompressible fluids (비압축성 유체) – density changes are neglected ( = const.) Compressible fluids (압축성 유체) – density changes are significant

ρ

(2)

Hydrostatic equilibrium

정지 유체의 압력: 높이에 따라 다름.

지면과 평행한 단면에서는 동일 All forces on the small volume of fluid Æ 0

(그렇지 않은 경우엔 “유동”이 존재)

F₁ (the force acting in an upward direction) : pS

F₂ (the force acting in a downward direction) : (p+dp)S

F₃ (gravitational force):

= 0

∴ ∑ F

_i

dZ S gρ ⋅

(3)

Å upward direction: positive (+) --- (2.2)

For incompressible fluids ( = const.)

--- (2.4) 즉,

- Head of a liquid

(액체의 높이압)

: the relationship between pressure and the height of a liquid column (Z: head)

g p Z = / ρ

0 )

( + − =

−

∴ pS p dp S g ρ SdZ

= 0 + g dZ dp ρ

) (

_a _b

a

b

p g Z Z

p − = ρ − Z g p = ∆

∆ ρ

ρ

(4)

- Barometric equation

(기압 방정식) For an ideal gas,

Eq. (2.2) becomes

Integrating between levels a & b, (assuming that T is constant)

: barometric eq’n (2.7)

RT

= pM ρ

) 0 (ρ ≠

) / ( m V M RT

nRT m

pV = = ρ =

= 0

+ dZ

RT gM p

dp

⎟ ⎠

⎜ ⎞

⎝

⎛ − −

= RT

Z Z

gM p

p

_b _a

a

b

( )

exp

(5)

- Rotating centrifuge

(원심분리기)

Pressure drop over any ring of rotating liquid, (회전하는 원심분리기 속 액체의 압력강하)

2 2

2

,

ω θ ω

ω ω

r dt v

vd dt

a dv r a

adm dF

ma F

dm r

dF

=

←

=

∴

=

←

=

dr r b dF

rdr dS

b dS dm

b rdr

rbdr dm

2

2 높이 ) (

면적 ), (

2 2

ω πρ

π ρ

π πρ

=

∴

=

⋅

=

←

=

∆p

(6)

Since

r₁에서 r₂까지 적분하면,

Assuming is constant,

--- (2.8)

rdr rb dp dF

ρ ω

π

2

2 =

=

2 ) (

₂² ₁²

2 1

2

r p r

p − = −

∴ ω ρ

= ∫

−

²

1

2 1

2

r

rdr

p

p ω ρ

ρ

(7)

- Manometer

(압력차 측정기)

여기서는 U-tube manometer의 원리

Æ --- (2.10) 관련 문제: Prob. 2.1

만일 를 무시할 수 있으면 으로 됨

b

B m

A m

B m m

a a

p

gZ p

p

gR p

p

p p

R Z

g p

p

p p

=

−

=

+ +

ρ ρ

ρ

4 5

3 4

2 3

2 1

: ) 5 pt.

at (

: ) 4 pt.

at (

) 3 pt.

at (

) (

: ) 2 pt.

at (

: ) 1 pt.

at (

( )

[

_m _m _B _m _A _m _B

]

_b

a

g Z R R Z p

p + + − − =

∴ ρ ρ ρ

(

_A _B

)

m b

a

p gR

p − = ρ − ρ

∴

R

m

g p = ∆

∆ ρ ρ

B

U관 마노미터

(8)

(Ex. 2.1) Pressure drop across an orifice (orifice meter)

streamline

Liquid B (brine)

1,260 kg/m³ Liquid A (mercury)

13,590 kg/m³

= ?

R

m

Fig. 8.17 (p. 228)

mmHg

− 250

b

= p bar

14 .

= 0 p

a

(gauge pressure)

(9)

From Eq. (2.10):

cf.) Absolute p = Atmospheric p + Gauge p (절대압 = 대기압 + 게이지압) 절대압으로는