Chapter Three

Two-Phase Flow Contents 1- Introduction. 2- Notations and relations. 3- Flow patterns. 4- Homogeneous flow model. 5- Separated flow model. 6- Examples. 7- Problems; sheet No. 3 1- Introduction: Background Two-phase flow is widely seen in the nature: Rain, snow, smog, dust-flow, etc. Two-phase Flow: Is a flow of two-phase mixture of a substance with the same chemical composition. (e.g., Water-steam flow, water-ice flow, etc.) Two-Component Flow: Is a flow of two-component mixture with different chemical composition. (e.g., air-water flow, oil-water flow, air-dust flow, etc.) Two- phase flows are related to the phase-change Phenomena. (e.g., boiling, condensation, freezing, melting, solidification, crystallization). 2- Notation and relations: Subscripts

f = fluid g= gas l= liquid h= enthalpy

Void

Thu

And

Mas

Tota

Volu

Tota

Qua

And

And

d fraction

us

d

ss flow ra

al mass fl

umetric fl

al volume

ality (or m

d

d in a therm

n α :

ate Wg, W

low rate W

flow rate Q

etric flow

mass quali

modynami

Wf:

W:

Qg, Qf:

rate Q:

ity) x:

ic equilibrrium,

Mas

And

Ther

Volu

Thu

Volu

Rela

ss flux (or

d

refore

umetric fl

us

umetric q

ative velo

r superfic

flux (or su

quality β:

city

cial veloci

uperficial

ty) j:

velocity) j:

3- Flow patterns: Vertical flow patterns

(a), (d), (f), (h) and (i): Flow patterns frequently observed in large diameter tubes. (b), (c), (e) and (g): Flow patterns specially appear in capillary tubes.

Horizontal flow patterns

Flow pattern maps

(1) Vertical flow [Hewitt & Roberts, 1969]

1

(2) Horizontal flow [Baker, 1954]

(3) Generalized map [Taitel & Dukler, 1976]

a+b : F vs. X c : K vs. X d : T vs. X

.

.

.

.

4- H Con

Mom

dzdp

−

dd

−

Ener

dzdi

For

dd

−

dF

wτ

=wτ

Homogen

ntinuity eq

mentum E

dzdp

zp

⎠⎞

⎜⎝⎛−=

Adzdp

=1

rgy Equat

(dzd

+21

dw=0, the

(1vdz

dp

h

=

pdZwτ=

w : wall sh

⎜⎝⎛=

21

hhf ρ

neous flo

quation:

Equation:

ZF dzdp⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

gdzdF

+

tion:

( ) gu h + s2

e energy e

21(

dZdE

+

hear stress

⎟⎠⎞2

hhu

ow mode

aZ dzdp⎟⎠⎞

⎜⎝⎛−

vg

h

+θsin

dd

=θsin

equation be

( )2udzd

h

s ,p:tube p

el:

dzdu

AW h+

dzdw

dzdq

−

ecomes:

sin θg+

perimeter

h

)θ

hf : f For

−

⎜⎝⎛−

d

Get

McA

Cicc

Duk

hμ =

For

For

friction fa

a tube of d

dzdp

F)( =−

=⎟⎠⎞ 1

F Adzdp

=2

t

Adams

μ1

chitti

hμ =

kler

ff

jjμ +=

laminar fl

hf =

turbulent

actor

diameter,

dzdF

A1

( ) =1w p

Aτ

DvGf hh

22

gh

xμ

+=1

(gxμ +=

gg ufj

=+

low:

hRe16

=

flow (Bla

D

⎜⎝⎛

21

hfAp ρ

f

xμ−

+1

) fx μ−1

[ ggh x μυρ

asius corre

⎟⎠⎞2

hhuρ

f

( )g x−+ 1

elation):

) ]ffv μ

When using the McAdams Correlations,

1

11−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

g

fg

f

fg

foF

xvv

xdzdp

dzdp

μμ

Laminar flow

41

11−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

g

fg

f

fg

foF

xvv

xdzdp

dzdp

μμ

Turbulent Flow

Thus

(Laminar Flow)

41

2 11−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

g

fg

f

fgfo x

vv

xμμ

φ (Turbulent flow)

Acceleration Pressure Drop

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛− h

h

a

vA

WdzdG

dzdu

Gdzdp

Where

( )[ ]fgh vxxv

dzd

dzdv

−+= 1

= ( )dzdxv

dpdv

xdpdv

xdzdp

gffg +⎥⎦

⎤⎢⎣

⎡−+ 1

1

2 11−

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+=

g

fg

f

fgfo x

vv

xμμ

φ

⎟⎠⎞

⎜⎝⎛+=

AdzdWGv

dzdvG h

h 12

And

( ) }1)1(2

dzdA

Axvv

dpdv

xdpdv

xdzdp

dzdxvG

dzdp

gfffg

gfa

+−⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡−++=⎟

⎠⎞

⎜⎝⎛−

Gravitational Pressure Drop

fgfZ xvvg

dzdp

+=⎟

⎠⎞

⎜⎝⎛−

θsin

Finally

( ) ( )

( ) ⎥⎦

⎤⎢⎣

⎡−++

+++−++

=⎟⎠⎞

⎜⎝⎛−

dpdv

xdpdv

xG

xvvg

dzdA

AxvvG

dzdxvGxvv

DGf

dzdp

fg

fgffgfgffgf

h

a 11

sin12

2

222 θ

5- Separated flow model: Pressure Drop Correlations:-

aZF dzdp

dzdp

dzdp

dzdp

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−=−

Where:

dzdF

Adzdp

F1)( =−

( )[ ] θαρρα sin1 gdzdp

gfZ

+−=⎟⎠⎞

⎜⎝⎛−

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

+=⎟⎠⎞

⎜⎝⎛−

αα 111 222

fg

a

vxvxAdz

dA

Wdzdp

For a circular tube of diameter, D

22

2 2)( fo

ffofo

foF

F DvGf

dzdp

dzdp φφ =⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

22

2 2)( go

ggogo

goF

F DvGf

dzdp

dzdp φφ =⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

2

222 2

)( ggg

gg

FF D

vxGfdzdp

dzdp φφ =⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

Where gfgofo φφφφ ,,, are the two-phase frictional multipliers related to each other as follows:

( )fof

goggo

fof

ggg

fo

fffo fv

fvfvfv

xff

x 222222 1 φφφφ ==−=

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−

−+=⎟

⎠⎞

⎜⎝⎛−

αα 111 222

fg

a

vxvxAdz

dA

Wdzdp

( ) ( )

( ) ⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

−⎟⎠⎞

⎜⎝⎛∂∂

+⎭⎬⎫

⎩⎨⎧

−

−−= 2

2

2

22

1

1.

1122

ααα

ααgf

p

fg vxvxx

vxxvdzdxG

( ) ( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

+⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

++ 2

2

2

2222

11

.11

ααα

ααgf

x

fg vxvxpdp

dvxdpdvx

dzdpG

( )⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−

−+−

αα 11 222

fg vxvxdzdA

AG

( ) 222

2 12)( f

fff

fF

F DvxGf

dzdp

dzdp φφ

−=⎟

⎠⎞

⎜⎝⎛−=⎟

⎠⎞

⎜⎝⎛−

Finally,

2

1

HH

dzdp

=−

Where

( ) ( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

−⎟⎠⎞

⎜⎝⎛+

⎭⎬⎫

⎩⎨⎧

−−

−+= 2

2

2

222

2

1 11

.11222

ααα

ααφ gffg

foffo vxvx

dxdvxxv

dzdxG

DvGf

H

( ) ( )( ) θρααρ

ααsin1

11 222

gvxvx

dzdA

AG

fgfg −++

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−

−+−

and

( ) ( )( ) ⎥

⎥⎦

⎤

⎢⎢⎣

⎡

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

−⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−−

++= 2

2

2

2222

2 11

.111

ααα

ααgffg vxvx

dpd

dpdvx

dpdvx

GH

6- Examples: 1- Using the homogeneous flow pressure drop method, calculate the two phase pressure drop for up flow in vertical tube of 10mm internal diameter that is 2m long .The flow is adiabatic ,the mass flow rate is 0.02 kg/s and the vapor quality is 0.05 .The fluid is R-123 at a saturation temperature of

3οC and saturation pressure of 0.37 bar ,whose physical properties are : ρl= 1518 kg/m3 , ρg= 2.6 kg/m3 , µg= 12.6×10-6 Pa.s , µl= 58.56×10-5 Pa.s Solution:

1

11 0.9685

1 50.3

∆ 50.3 2 9.81 90 ∆ 987

1 10.000557 .

4571

0.079

. 0.00961

∆2

4953

Total pressure drop =∆ ∆ 5.94 .

University of Technology Sheet No. 3 Mechanical Engineering Dep. Two-Phase Flow Fluid Mechanics II (3 rd year) 2011/2012 1- Find the flow pattern when 4 kg/s of steam –water mixture of quality 20% at 20 bar flows in a circular tube of internal diameter 0.1 m: a) When the flow is vertically upward. b) When the flow is horizontal. The physical properties required are: ρl= 850 kg/m3, ρg = 10 kg/m3 , μl= 128*10-6 Pa.s , μg= 16*10-6 Pa.s 2- A steam –water mixture of quality 0.1 at 5 bar flow through a smooth vertical round tube of diameter 0.05m.The total flow rate is 0.6 kg/s .Calculate: a) The homogeneous void fraction. b) Homogeneous gravitational pressure gradient. c) Homogeneous frictional pressure gradient. The physical properties required are: ρl= 915 kg/m3, ρg = 2.67 kg/m3 , μl=180*10-6 Pa.s , μg= 14*10-6 Pa.s

[0.97 ; 255 N/m3 ; 547 N/m3]

3- Consider mixture of water (Wl=0.42 kg/s) and air (Wg=0.01 kg/s) flowing upward in a vertical pipe (D=25mm ,L=45cm). Given friction factor of (f=0.079/Re0.25).Find the total pressure drop ,the volumetric flow rates ,void fraction ,and mean water and air velocity. Using homogeneous model. the physical properties required are: ρl= 915 kg/m3, ρg = 1.2 kg/m3 ,μl= 1*10-3 Pa.s , μg= 1.8*10-5 Pa.s

[3169 N/m2 ; 0.00833m3/s ; 0.0042m3/s ; 0.952 ; 17.65m/s]