Multiphase Flow Questions - NPTEL · Multiphase Flow Questions with Answers Chapter 2 1) With the...
Transcript of Multiphase Flow Questions - NPTEL · Multiphase Flow Questions with Answers Chapter 2 1) With the...
Multiphase Flow Questions with Answers
Chapter 2
1) With the help of neat sketches discuss the flow patterns observed in vertical and horizontal
heated tubes.
Ans: Refer to Fig 2.3 & Fig 2.4 along with the associated discussion.
2) Also discuss the probable reasons for the differences in flow patterns in a horizontal heated
tube as compared to (i) vertical heated tube (ii) horizontal unheated tube.
Ans: (i) Influence of gravity which leads for asymmetric phase distribution and stratification of
the two phases.
(ii) (a) Departure from hydrodynamic and thermal equilibrium
(b) Presence of radial temperature profile
3) Gas‐liquid stratified flow in a horizontal pipe encounters a vertical T junction. How do you
anticipate the flow pattern to change after the T?
Ans : As the stratified gas‐liquid mixture encounters a vertical T, some amount of gas will enter
the side branch. This reduces the relative proportion of gas in the main arm after the T and is
expected to cause gas‐liquid bubbly flow after the two phase mixture encounters the junction.
4) What is film inversion? When is it encountered in two phase flow?
Ans : The phenomena of film inversion occurs when a two phase mixture under stratified flow
encounters a return bend. This is illustrated in Fig. 2.11 (a). Under this condition, the film on the
inner walls while travelling through the 1800 bend flows along the outer wall after the bend. The
film inversion of blue kerosene is evident in Fig. 2.11 (a).
5) With the help of neat sketches describe a mixed flow pattern (neither separated nor dispersed)
for (a) gas‐liquid flow (b) liquid‐liquid flow.
Ans : Gas‐liquid flow – Slug Flow
Liquid‐liquid flow – Three layer flow with lighter liquid on the top, heavier liquid on the bottom
and a mixture of droplets of the two liquids in the central portion. Refer to Table 2.1.
6) A distillation column in a petroleum refinery plant operates with a mixture of crude oil and
natural gas as feed which is fed by a pipeline into the feed tray of the column. Due to some
reasons, the refinery is provided with an increased supply of raw materials (crude and gas). In
order to handle the increased supply, the plant manager had two options – (a) t install a
distillation column with a capacity to handle large quantity feed and (b) to install another
column along with the existing one, introduce the previous quantity of the feed in the old
column and divert the extra quantity to the new column. From cost considerations, the
management decided to go for the second option. With that option, they just needed to
required quantity of feed into the new column. These installations would cost less and require
less time to start the operation. Accordingly, the necessary changes were made. The new
column was installed in the plant site and the total feed (crude+gas) was divided into two parts
by a T junction and each part was fed into the respective distillation column. It was expected
that the plant would run smoothly after this. However, after a few days it was noticed that
neither of the two columns were operating property and there was a drastic fall in efficiency of
the existing column which was operating fine before the new venture was taken up. You as a
new management trainee of the company was called upon to look into the problem since you
had a course of multiphase flow in your college. You were asked to investigate the problem and
propose a possible solution to it. You found out that the columns were not getting the
composition of the feed they were used to handle and this occurs because there is
misdistribution of the two phases at the T. You were aware of this feature of the T junction from
your knowledge of multiphase flow. Accordingly, you suggested a revised design of the pipeline
so that the problem can be alleviated. Present schematics of the original and revised pipeline
layout in order to ensure trouble free operation of the plant.
Ans :
7. Using relevant flow pattern maps (Baker et al for horizontal tube and Hewitt and Roberts for vertical
tube) evaluate the most likely flow pattern occurring for a steam‐water system flowing in a 2.54cm
diameter ( vertical (b) horizontal pipe where the system pressure is 70 Bar, the mass qualities are 1%
and 50 % and the mass fluxes are (a) 500 Kg/m2 –sec and (b) 2000 Kg/m2 –sec respectively.
Solution:
For Vertical tube:
For 1st Case
1 21 2
W WWG G G
A A
Here
In 70 Bar, we get saturation temperature from stem table = 255.90 c and vg = 0.027382 m3/kg
and vf = 0.0013514 m3/kg and 3 336.520kg/m and 739.98kg/mg g
1 1 1 11
1 1 1 1 1 1
2 2 21 1 1
22 1
1 11
22
22 2
3
1
495 / sec331.1238142 / sec
739.98 /
s
ff f
f
W u AG
A Au u j
G j
Gj
Gor j
kg mkg m
kg m
Similarly
21 2
2
2
22
21
500 / sec
11% 0.01
1000.01
0.01 500 5 / sec
500 5 495 / sec
G G G kg m
Gx
GG G
G kg m
G kg m
22 222
3
2
5 / sec
36.520 /
0.684556407 / sec
gg g
g
kg mGj
kg m
kg m
Therefore from figure we get flow pattern is Bubble – slug.
For 2nd Case:
For G = 2000 kg/m2 –sec and x = 50% = 0.5 we get
G2 = xG = 0.5 ×2000 = 1000kg/m2-sec
and
G1 = (1-x)G = 0.5 ×2000 = 1000 kg/m2-sec
Now
Therefore from figure we get flow pattern is Annular
For Horizontal tube:‐
1st case
For pressure 70 bar from figure we get 4.8and =2.3 for G = 500 kg/m2-sec and x = 1%
= 0.01
G2 = Gg = 495 kg/m2-sec
2
495 2.3
1138.5 / sec
fG
kg m
222 2 2
2 22
2
2
222
2
2
1000
36.520
27382.2563 /
1000and
739.98
1351.387875 / sec
gg g
ff f
f
GGj j
kg Sec m
Gj
kg m
and 22 51.04166 / sec
4.8gG G
kg m
There from figure we get flow pattern is slug
2nd Case:
For G = 2000 kg/m2-sec and x = 50% = 0.5
We get G1 = Gf = 1000 kg/m2 and
G2 = Gg = 1000 kg/m2 –sec
21000 2.3 2300 / secfG kg m
21000208.333 / sec
4.8fG kg m
Therefore from figure we get flow is Dispersed Bubbly flow.
Chapter‐3
1.) Derive a relation to express (a) x, the mass quality in terms of volumetric quality β and the
density ρ1 and ρ2 of the phases.(b) slip ratio in terms of α and β only (c) j, the overall volumetric
flux in terms of inlet velocity of saturated liquid (ulo) to a evaporating tube, the mass quality x
and the phase densities ρl and ρv.
(1a)
1 1 1
11
G L
L G
L
G
GG g L L
L G L
vx
x v
xx
x x x x
(1b)
1 1
11
g
g g
LL L
QU QA
QU QA
(1c)
0
0
0
1
1
1
1
1
1
1 1
L G
G G G GL L L L
L G
L G
L G
L G
GL
L
GF
L
GF
L
GF
L
j j j
Q W W vQ W W v
A A A A A A
x v xvW
A A A
Wx v xv
AW
x v xvA
xvWvx
A v
vU x
v
vU x
v
vU x
v
(2) Estimate α in a heated tube in which evaporation os occurring in terms of x (mass quality), ulo (inlet
velocity of liquid to be evaporated), gju (drift velocity), phase densities ρl and ρv and C0 the distribution
coefficient. Assume the drift flux model to be applicable for the vapor‐liquid mixture.
1 1
1 .1 1 1
11 1
11
11
G G G G
L L L L
G G G
L L L
G G
L L
GG
GL
G
L
Q W v vx
Q W v x v
v v vx x x
x v x v x v
v vx xor
x v x v
xvvxx vx v
vxx v
1
1L G
L
x v xv
x v
1
1
11
G
L G
L
G
xv
x v xv
x vx v
(3) Find in a heated tube in which evaporation is occurring in terms of x (mass quality), 0Lu (inlet
velocity of liquid to be evaporated), gju (drift velocity), ,L G (phase densities) and 0C (distribution
coefficient) assuming drift flux model to be applicable for the two phase vapor – liquid mixture.
Solution : From drift flux model , 0
g
g gj
ju C j u
Dividing by j throughput , 0
gg gjju u
Cj j j
Or, 0
ggjju
Cj j
Or,
0gju
Cj
(1)
In equation (1)
11 1
G
G G G G L L
G LG L L G G L Gx
G LG L G L
W xW xQ x x
W x xW W xQ Q x x xW
1 1 1G GL L LL G
L G L G L G L G
W xQ WQ W QWx W x x x xj j j
A A A A A A A A
At inlet, mixture is all liquid with ,L LW W Q Q
Substituting individual terms in equation (1) we get :
0 0
1
11
1
LL
G LG L G
gj gjL
L G LOLO
LG
G
xxx xx
u uxC C xu
u x xx x
1
L
G
L
G
x
x x
0 0 0
1
gjL
G LO
L
G
uC C x C x
u
x x
1 1LO L L LLO
L G G
Au xx x u x
A
0 0
0 0
1
1
L
G
gjL
G LO
G
L L
gjG
L LO
G
x
uC x C
u
x
uC x C
u
0 0
0 0
1 gjG G
L Lo L
gjL G G
L Lo L
xu
C x Cu
xu
C x Cu
4) Water at 10 atm enters a straight evaporator tube. If the velocity ratio of water to vapor is
constant at 2.0 and mass flux is 2X105 kg/hr (m2 ) , estimate the void fraction and momentum flux at a
quality of 0.1. Assume water = 903.18 kg/m3 and vapor = 3.6142 kg/m
3 .
2 2 2 2
1 1 1
1
1
2
(1 )(1 )
1
1
TP
TP
G xu Q A
G xu Q A
x
x
5 22
1
3 3
10 , 2.0 2 10 / ( )
0.1
903.2 / 3.6 /
TP
w ater steam
uP atm G kg hr m
u
x
kg m kg m
Substituting the values of 2
1
,u
xu
and 1
2
, we get from the following equation
0.1 903.2 12.0
0.5 3.6
Momentum flux = 1 1 2 2 1 21G u G u G x u xu
11
1
1
1
g xQu
A
22
Gxu
Momentum flux
2 22
1 2
1
1
x xG
Chapter 5
1) A vertical tubular test section is installed in an experiment high pressure water loop. The tube is
1.016 cm ID and 2.134m long and is uniformly heated with 100 KW power. Saturated water
enters at the base at (400) psia with a flow rate of 450 kg/hr. Calculate the total pressure drop
and compare with the measured value of 8 psia.
Heat flux 100
DL
1000 68.0P psia pqv
From steam table
Mass flow rate = 450 /kg hr
31
32
12
312
0.001347 /
0.02795 /
1531.34 /
0.026603 /
v m kg
v m kg
h KJ kg
v m kg
dh dh DW D
dz dz W
1
12 12
100 100. 0.2432
4531531.34 2.134
3600
dx dx dh D Dm
dz dh dz Wh Wh DL
2
22
4531552.095 /
1.016 10 36004
WG kg m s
A
12
12 12
12
100
100 1000.5189 1.0
4531531.34
3600
he
e
DL DLDL W e x
Wh Wh DL
xWh
2
1 12
2 f
F
C Gdpv xv
dz D
0.005fC Re 87.1231.1LOl
DG
2
2
2
2
2 0.005 1552.0950.001347 2.134
1.016 10
2.13441745.397 / . 0.2432 0.0266
26.0621
Fp
kg m s
psi
1 12
2
9.81
0.001347 0.24322 0.026603
9.81 0.001347 0.006469 2.134ln
0.006469 0.001347
3670.208 0.5329
g
dp g
dz v xv
kgpsi
ms
212
ace
dp dxG v
dz dz
2
2
1552.095 0.026603 0.2432 2.134
33260.2 4.829
acep
kgpsi
ms
11.424total
p psi
Since totalp is small compared to 1000psia assumption of constant fluid properties are justified.
Check that exit steam is not superheated by calculating ex . Check for compressibility and flashing effects
by evaluating the terms , ,g f
n
dv dv x
dp dp p
3 33.8 10 / 0g fdv dvft lb psi
dp dp
at 1x
at 0x
at 0.5x
2. Derive the expression of pressure gradient for a homogeneous two phase mixture for appreciable
effect of kinetic energy.
Considering unit mass of fluid, the energy balance equation for single phase flow is as follows.
21( sin ) ( )
2d u dU d zg d pv dq dw (i)
Kinetic energy internal energy gravitational pressure energy heat work content energy where u, U, z, p,
v and θ refer to the velocity, internal energy, length of flow, pressure, volume and angle of elevation
respectively.
Now, there is no work content 0dw and internal energy change can be given by
dU dq dF pdv (ii)
Where dF is the irreversible frictional energy loss per unit mass of fluid.
Substituting in eqn (ii) in eqn (i) we get,
4 1
4 1
0
3.9 10
2 10
h
x
p
psi
psi
2
2 ( sin )
1( sin )
2
12
dF d g z
d u dq dF Pdv d g z Pdv vdP dq
vdp d u
(iii)
Dividing both sides by dz we get,
21( sin )
2
1sin
dp d dF dv u g z
dz dz dz dz
dp u du dF g
dz v dz v dz v
(iv)
Now, 1
v
Thus substituting 1
v we get,
s indp du dF
u gd z d z d z
(v)
This corresponds to
a cc f g
d p d p d p d p
d z d z d z d z
In two phase flow according to HOMOGENEOUS MODEL u and ρ in eqn (v) should be substituted with
Mu and M .
sinM M M M
dp du dFu g
dz dz dz (vi)
Where gravitational pressure gradient is
2 1
1 121 12
sin sin / [ (1 ) ]
sinsin / ( )
( )
Mg
dpg g xv x v
dz
gg v xv
v xv
and 1v and 2v are the specific volume of phase 1 and 2 respectively
Frictional pressure gradient
M
Mf
dp dF
dz dz
According to Fanning, head loss due to friction fh
2
42
ff
PL u Fh f
D g g g
Where F is energy loss due to friction per unit mass of fluid.
24
2
fL uF
D
For unit length of pipe
24
2
f uF
D
Now, for two phase flow u = uM
24
2
M MM
f uF
D
Now, M
MM
Gu
2
222 12
4 4(1 )
2
M M MM
M
f G fF G x v x v
D D
(vii)
Differentiating both sides with respect to z we get,
22
2 1
22 1
2 1 122
22 1
1 12 12
21 12 2 1
121 12
4(1 )
2
42 (1 ) (1 )
2
4( ) (1 )
( )4(1 )
( )
M M M
M M
M
M M
M M MM
dF f G dxv x v
dz D dz
dv dvf G dxx v x v x x v
D dz dz dz
dv dvf G dxxv xv x x v
D dz dz dz
v xv dv dvdF f G dxx x v
dz D v xv dz dz dz
(viii)
Now, variation of specific volume v with respect to z is due to pressure.
Therefore,
22 1
12
4(1 )
M M MM
dv dvdF f G dP dP dxx x v
dz D dP dz dP dz dz
Pressure gradient due to acceleration
MMMM M
acc M
dudP Wu u
dz dz A
2
2 1 2
22 2 1
12 1 12
2 2 112
1 1
(1 )
(1 ) ( )
(1 )
M M M M
M M M
M M M MM M
M
MM
M
du d W W d W dA
dz dz A A dz A dz
du W W d W dAu xv x v
dz A A dz A dz
dv dv dx G dAG x x v v xv
dz dz dz A dz
dv dvdp dp dxG x x v
dp dz dp dz dz
2
1 12( )MG dA
v xvA dz
(ix)
Now combining three pressure gradients terms we get,
222 1 2 1
12 121 12
2
1 12
sin 4(1 ) (1 )
( )
M MM
M
dv dv dv dvdp g f G dp dp dx d dp dXx x v G x x v
dz v xv D dp dz dp dz dz dp dz dPp dz dz
G dAv xv
A dz
2 22
12 12 1 121 12
222 1 2 1
sin 4( )
41 (1 ) (1 )
M M
M MtotalM
g f G dx dx G dAv G v v xv
v xv D dz dz A dzdp
dz dv dv dv dvf Gx x G x x
D dp dp dp dp
When there is too much flashing then
( , )
| |
| |
P h
P h
x x h p
dx xdx dh dp
dh Pdx dx dh dx dp
dz dh dz dp dz
So , by substituting we get,
2 22
12 12 1 121 12
2222 1 12 2 1
12
sin 4| | ( )
441 (1 ) | (1 ) |
M MMP P
MM MtotalMh h
g f G dx dh dx dh G dAv G v v xv
v xv D dh dz dh dz A dzdp
dz dv dv fG v dv dvf G dx dxx x G x x v
D dp dP D dp dp dp dp
Now, 12
1dx
dh h where molal enthalpy of vapourisation
For horizontal evaporator gravity pressure gradient is absent, then,
2 2212 12
1 1212 12
2222 1 12 2 1
12
4( )
441 (1 ) | (1 ) |
M M MM
M MM MtotalMh h
v vf G dh dh dh G dAG v xv
D h dz dz h dz A dzdP
dz dv dv f G v dv dvf G dx dxx x G x x v
D dp dp D dp dP dP dP
3. Liquid evaporated from an inlet conduction at saturated temperature 0x to a vapour liquid
mixture having a mass quality x . For a linear change of x over length L dxconst
dz
, derive an
expression of the pressure drop over length L. As derived in chapter‐5,
2
21 12 12
1 12
2 sinTPf Gdp dx gv xv G v
dz D dz v xv
or
2
21 12 12
1 120 0
2 22
1 12 121 12
2 22 12
1 1212 1
2 sin
2 2 sin
2 2 sinln 1
2
L LTP
TP TP
TP TPn
f Gdp gdz v xv dz G v kdz dz
dz D v xv
f G f G g dzv L v kzdz G v kdz
D D v xv
f G f G vx g Lv L v L G v kL x
D D xv v
4. Saturated water at a rate of 300 kg/hr (m2) enters the bottom of a vertical evaporator tube 2.5 cm diameter and 2.0 m long. The tube receives a heat flux of 2x105 BTU/hr ft2 and there are no heat losses. Calculate the pressure drop for inlet pressures of 350 psia. Assume homogeneous flow with a constant friction factor of 0.005.
Assume homogeneous flow theory, proceed similarly as problem 1
Equating rate of heat addition /unit length to heat flux,
edqD
dz heat flux
12h
edq dW h
dz dz
dh D
dz w
D
G
2
4
4
GDD
(For constant W)
For small pressure changes
12 12 12
1 1 4 4,
dx dhor dx dz
dz h dz h GD h GD
2
21 12 12
1 12
2 cosdP fG dx gv xv G v
dz D dz v xv
Neglectingdp
dz, area change, flashing, compressibility effects
22
121 12 1 12 0
0 0 0
2 221 12
121 120 0 0
cos2
2 2cos
LL L L
TP
L L LTP TP
g dxG v dzf Gdp
p dz v xv dz v xv dzdz D
f G v f G v dz dxL xdz g G v dz
D D v xv dz
212
4 4x z c x z
hG GD h GD
C=0 from B Cs
212 1.3064 /v ft lb 20.01912 /lv ft lb
12 794.7B u
hlb
1.02p psi
p small compared to pressure , so assumption of constant fluid properties justified.
Check for compressibility effects and confirm exit steam is not superheated.
12
40.605e
zx
Gh D
1 12
2 11
h h xh
xh x h
1
12
h hx
h
2 2 221 12
1212 120 0
112
221 12 12 12 12
12 12 12 12
2 2 44cos
42
2 cos2 4 4ln 1
4
L LTP TP
TP LTP
f G v f G G vL dzP v g dz
D D h GD h GDv zh GD
f G v L v v g Gh vL L Lf G G
D D h h D v Gh D
For compressibility & flashing effects ,x x h p
0 0h
xx
p
43.9 10 4 12 10 psi at x=0.5
At 0.6x
xp
lg
1g l
g l l
l
v xv x v
x v v v
xv v
l g l
lg l g l
h h h
v v x v v
vv xv v x v v
p p p p
lgg
h h h
vv xv x
p p p
5 An air‐water mixture flows through a circular pipe of cross‐section A1. It has a nozzle of cross‐section A2 at the centre. The pressure at the upstream, throat and downstream sections are p1, p2 and p3 respectively. Assuming incompressible homogeneous flow, derive the following equations where G is the mass flux and ρ1 and ρ2 are the densities of water and air respectively:
2
1 1 11 2
1 2 2 2
1 1 1A AG
p p xA A
2
21 1
1 31 2 2
1 1 12
AGp p x
A
Clearly state any additional assumptions made
Chapter 6
Problem 1 Air‐water at individual mass flow rates of 50kg/hr flow through a 3 cm diameter pipe at 270 C
and 1.2 atm pressure. What is the overall volumetric flux. If the drift flux is 3m/s, what are the average
velocities of the phases ?
Solution :
21 23j u j …. (1)
1
11 1
Wu
A
……..(2)
22
2
Wu
A ..….(3)
1 2 1 21 2
1 2
Q Q W Wj j j
A A A A
23 u j ….(4)
From equations (3) and (4), find and 2u . Substitute in equation (2) to find 1u
Problem 2 Show that drift flux is independent of the motion of the observer during concurrent flow of
two fluids .
Solution :
Let fluids 1 and 2 move at velocities u1 and u2. Let the observer move at velocity u in direction of fluid
motion.
Therefore, fluid velocity is observed by observer =
21 2 1 2 1
2 1 2 1
1 ' 1 1 '
1 ' ' 1
j j j u u
u u u u
Similarly for observer moving at velocity u opposite to direction of fluid motion, 2 2'u u u and
1 1'u u u . In this case also 21 211j u
1 2
2
50 /
0.034
W W kg hr
A
2 2
1 1
'
'
u u u
u u u
0
31
27
1.2
1000 /
T C
P atm
kg m
Problem 3
Sketch 2j vs. as a function of 1j and 1j vs. as a function of 2j and identify the condition for
flooding , co‐current & counter current flows
Given 12 1nj
u
.
Solution :
2 21 2 21
2
1j j j j
j j j
21 2j j j ……..(1)
21djj
d
Again, 21 . 1n
j u ….(2)
121 . . 1 . 1
1 11
n n
n
dju n u
dn
u
For flooding, curve of eqn(1) is tangent to curve of eqn (2).
Or,
Condition for flooding
1 212
1 21
1 1
j jj y mx c
j j j j
21 21
1 2
. 1dj dj
d d
1 . 1 11
n nju
Problem 4
For what value of n will be linear? How does drift velocity of ‘2’ depends on ?
Solution:
For
Now,
Drift velocity of component 2 = terminal velocity .
Thus,
Or,
Chapter 7
Problem1: Derive the expression to estimate pressure drop for flow of boiling water in straight pipes
(no area change in pipe) where the water enters the pipe under saturated conditions and water and
steam flows under stratified conditions.
Solution: From eqn (7.27), the equation for pressure gradient is:
1 2 1 2 1 1 2 2
1sin 1 W W
dp dg F F W u W u
dz A dz
2 1nj
u
0n 0221
jj u
u
22
ju u
2u const f
2
2 210
0
0
u
j j u uC
j j
= accfg dz
dp
dz
dp
dz
dp
….. (7.27)
The frictional pressure gradient can be expressed in terms of two phase multiplier as:
22
0 0
22
0
2.
2 1
TP
L L
lo loF l
Llo
lol
dp Gdz f dz
dz D
f G Ldz
D L
Since quality increases linearly with distance (derivation below), x=0 at z=0 and
22
0 0
2 1
TP
L xlo
lof l
f G Ldpdz
dz D x
The relationship between length and quality can be expressed by considering the enthalpy balance of
the flow situation.
Let be the heat flux. Therefore the rate of heat addition per unit length is
edqD
dz
From energy equation
2
2
2
22
22
22
4,
.
3.
4 1. .
2
2.
2. .
TP
lo
TP
lo
lo
f
lo
f
wF
loF
lo w
lo lol
lo lol
lo lol
dpdzdpdz
dpor
dz D
dp
dz
A
Gf
D
Gf
D
Gf
D
,
edq dW h
dz dzdh D
ordz W
And 1 12h h xh Or, 12p
dhh
dx
Now,
From the aforementioned expression of dx
dz , it is clear that quality increases linearly with distance.
The acceleration pressure gradient can be expressed a:
Now since A is constant, G is constant and
12
2 1212
.
1.
4
4
dx dx dh
dz dh dzdh
dh dzdxD
Wh
D
DGhD Gh
1 1 2 2
2 21 2
1 1 2 2
22 2 2
1 2
1
1
11
1
dW u W u
A dz
W Wd
A dz A A
W xd W x
A dz A A
1 1 2 2
22 2
21 2
2 22
1 2
1 1 2 2
0
2 22
1 20
2 22
1 20
1
1
1
1
1
1.
1
1
1
1
L
L
L
dW u W u
A dz
xW d x
A dz
xd xG
dz
dW u W u dz
A dz
xd xG dz
dz
x xG
Substituting the expressions for individual components of pressure gradient, the pressure drop for
length L is
1 1 2 2 1 2
00 0 0
22 22 2
1 21 20 00
11 sin.
1.2 1sin 1
1
LL L L
w
Lx L
lolo
l
dW u W u gSdp
P dz A dzdz A
xf G L xdz G g dz
D x
Problem 2. Two incompressible fluids are flowing under separated flow through a nozzle in horizontal
orientation. Express the pressure drop of the two phase system .T PP in terms of 1P and , the
pressure drop encountered by either of the fluids if they would be flowing alone through the nozzle.
Solution: For separated flow the momentum balance equation for flow of the individual components
per unit volume under steady state condition can be written as:
11 1 1 1
22 2 2 2
...........( )
...........( )
du dpu b f i
dz dzdu dp
u b f iidz dz
2P
Since the two component separated flow is accelerated rapidly through nozzle, body force and
frictional components can be neglected as compared to inertia term. Thus
From eqn (1)
11 1
dudpu
dz dz
And from eqn (2)
22 2
dudpu
dz dz
Or on integration,
2 2 2 21 1 2 2
.1 22 2T P
u uor P
2 2 2 2 2 21 1 1 2 2 2
. 2 21 1 2 2
2 21 2
. 2 2 2 21 2
2 21 2
2 21 2
2 21 2
2 21 2
1 1
2 2
1
2 (1 )
( / ) ( / )1 1
2 (1 ) 2
1 1
2 (1 ) 2
T P
T P
u A u Aor P
A A
W Wor P
A A
W A W A
G G
Now if either of fluid flows alone through the nozzle,
21
11
1
2
GP
And
22
22
1
2
GP
1 21 1 2 2
du dudpor u u
dz dz dz
11. 2T PP
21
21 12
1.2
1
22
22 222.
22
12
(1 ) ..........( )12 (1 )
12
..........( )12
T P
T P
G
Piii
GP
GP
ivGP
From equation (iii) we get
2 1
.
1/ 2
1
.
(1 )
(1 ) ..........( )
T P
T P
P
P
Por v
P
And from equation (iv) we get
2 2
.
1/ 2
2
.
..........( )
T P
T P
P
P
Por vi
P
Adding equation (v) and equation (vi) we get
1/ 2 1/ 21 2
1/ 2.
1/ 21/ 2 1/ 21 2 .
1/ 2 1/ 2 2. 1 2
1/ 2 1/ 2
1 1
. .
( ) ( )
( )
( ) ( )
[( ) ( ) ]
(1 )
1T P
T P
T P
T P T P
P P
P
or P P P
or P P P
P PP P
or
Problem 3 Develop the separate cylinders model for stratified gas‐liquid flow assuming turbulent flow
and a constant friction factor for both phases.
Solution: Let gas flow in a cylinder of effective radius gr and liquid flow in a cylinder of effective radius
lr
2
2
o
g
r
r
2
2
1o
l
r
r
2/5
221
o
ggggg
g
g
g r
jfj
r
f
dz
dp
dz
dp
Where gf is the constant friction factor for the gas phase
2/5
2 1
gasonly
g
dz
dpdz
dp
or
22/5 1
g
Similarly
2/5
2
1
1
l
or
5/2
2
11
l
Thus
5/2
2
1
g
,
5/2
2
11
l
Therefore,
5/2
2
5/2
2
11
1
lg
Or 111
5/2
2
5/2
2
lg
Problem 4
For flow through a packed bed of spheres with diameter‘d’ and void fraction , deduce the values of
and and using the Carman‐Kozeny equation for the frictional pressure drop
during viscous flow through void space between the solid
2
2 3
180 1f fo
F
jp
z d
is the fluid flux relative to the particles and is the liquid fraction in the3 solid‐liquid system.
Subscripts f and s refer to the fluid and solid respectively.
Solution:
In one dimensional form the momentum equation per unit volume of individual phase is
1 11 1 1 1
u u pu b f
t z z
2 22 2 2 2
u u pu b f
t z z
Now, for steady state condition
1 20, 0,u u
t t
If we neglect the inertial effect then 1 21 20 0
u uu and u
z z
If we neglect body force then
1 0b and 2 0b
Then from momentum equation we get,
1 2F
p pf f
z z
………(1)
foj
1ff f 2sf f
The force 2f on particles is made up of two parts, one due to fluid , 21f and the other due to the
particles .
.
Thus
But force on fluid is only due to particle
So, 1 12f f
Therefore, from equation (1) we get
2
12 1 2 3
1180 f foj
f fd
…….(2)
2
2 21 22 2 3
180 1f fojf f f
d
……….(3)
Now, 12F = equivalent of 12f per unit volume of whole flow.
12
12
1 f
f
21F equivalent of 21f per unit volume of whole flow
=
21
211
f
f
12F and 21F forces are entirely due to mutual hydrodynamic drag and since action and reaction are
equal and opposite we have.
22f
2 21 22f f f
1f
12 21
12 21
1 21
21 1
2
2 3
2 2
, 1
, 1
, .1
180 1.
1
180 1
f fo
f fo
F F
or f f
or f f
or f f
j
d
j
d
Now, from equation (3) we get
22 2 21
2
2 3 2 2
2 2
2 3
2 3
2 3
180 1801 1.
1 2180
180 1
180 1
f fo f fo
f fo
f fo
f fo
f f f
j j
d d
j
d
j
d
j
d
Problem 4 Using the results of the previous problem, deduce the fluid flux necessary to cause
fluidization in a bed with void fraction and estimate the pressure gradient through the bed in this
case.
Solution:
In a fluidized bed particle are supported by an upward flow of fluid around them and inter particle force
are negligible. Mathematically
For steady state flow, neglecting the inertial term, the momentum equation per unit volume of the
individual phase becomes,
1 1 0p
b fz
……….(1)
And 2 2 0p
b fz
…. (2)
22 0f
where 1 1 2 2,b g b g
2
1 12 2 3
2 21 2 2
1180
180 1
f fo
f fo
jf f
d
jf f
d
Therefore eqns (1) and (2) become,
2
1 2 3
1180 .f fojp
gz d
= 0 ….(3)
2 2 2
1180 . 0f fojdp
gdz d
….(4)
Subtracting eqn (2) from equation (1) we get
2 1 2 2
2 32 1
180 1 1. 1
, .180 1
f fo
fof
jg
d
d gor j
And from equation (1) we get,
1 1
2
1 2 3
2 2 32 1
1 2 3
1 2 1
1 2 1
1 2 1 1
1 2
180 1
180 1. . .
180 1
1
1 1
1
1
f fo
f
f
pb f
z
jg
d
d gg
d
g
g
g
g
Applying Lockhart Martinelli assumption to annular flow
2
2
2
4
41
D
D
D
D
A
A ll
l
(1)
or
2
1
lD
D (2)
Substituting the value of in eqn (1) we get:
5)2(2
22 1
n
l
n
l
nl D
D
D
D (3)
Chapter 8
Problem 1 : Air water mixture flows in a 3m long 5cm diameter pipe and discharges at 94.7 psia.
Assume bubbly flow under turbulent conditions at 270c, calculate inlet pressure for a volumetric flow of
Jl=0.15 m/s and Jg=4.5 m/s at atmospheric pressure & temperature.
Solution : 21 gjj U
2 21
2
2
1 2
1
(1)
TP
gj
j j
j j
j
j j U
As per changes down the pipe, 2j will change
Assuming isothermal expansion, 2 2( ) | (2)pa
paj j
p
Check for sonic flow ‐‐‐‐‐> con. Choking
Then exit atmp p
Acc. 2 12 1
acc
du dudpG G
dz dz dz
2 1,u u from equation (1) in equation 1 21 21
j jU U
2 1 2
1 2 21 1
1 2
gj
j
j
U j j U
j j UU j
j U
Only 2j changes down duct
2 1 2 1
1
1 22 1
1 2
g
gj
acc j
dU dj du dj j
dz dz dz dz j U
j djdpG G
dz j U dz
For small intervals, 2dj
dz found by differentiating eqn. (2)
22 2
12 1 2 2
1 2
( )
( )
pa
paacc j
dj pa dpj
dz p dz
jdp pa dpG G j
dz j U p dz
Friction
1 1 2 2
1 2
2 1
2 1
2 1 1 1 2 2 2
2 2
2
0.005
( )
2 / (1 )
1 /( )( )
1
TP j
f
TP
g
TP
j pa
acc
f Gdp
dz D
f
G j j
j j j
dpg
dz
f Gj D gdp
dz paG G j j U j
p
dpa for a ve
dz
Flow supersonic
Not permissible.
exitP adjust until choking is reached & 1a
Condition of choking
2 12 2 1
1 2
( )c Paj
jP Pa j G G
j U
Problem 2 : For annular flow pattern (Gas Core + Liquid film), deduce ( )fn for 2 ( )l fn
Solution: Annular flow can be analyzed by separated flow model.
A very simple model of separated flow can be developed by assuming that the two phase flow, without
interaction, in two horizontal separate cylinders and that the areas of the cross sections of these
cylinders add up to the cross‐sectional area of the actual pipe. The pressure drop in each of the
imagined cylinder is the same as in the actual flow, is due to frictional effects only, and is calculated
from single phase flow theory.
Therefore,
TPF liquid cylinder gascylinder
dp dp dp
dz dz dz
2
4l lA D cross‐sectional area of liquid cylinder
2
4g gA D cross‐sectional area of gas cylinder
Where l gA A A cross‐sectional area of main cylinder
and are the shape factor.
2 ,4
A D D
Diameter of main cylinder
Now
,
2
2
2
2 2
2
2
2
2
2
2
2
2
2Re
Re
l
l
f liquid cylinder
fll l
l
fl ll
l l l
fl l l
l l l
fl l
l l l
n l
l l l
l l l
l
dp
dz
CU
D
C W
D A
C W
D A
C W
D A
WK
D A
U DNow
2
2
Re
2
l
l l l l
ll l
l
l l
l l
n
l l l
liquid cylinder l l l l l
Now W U A
WU
A
W D
A
W D Wdpk
dz D A A
2
2
2
22 22
22 1 4 2
2
22 5
2
2
.2. .
.( ) .
.2. .
. . . .4
2.
.4
2.
.4
T
n nl l
n nl l l l
n nl l
nn nl n
l l l
n nn n nl l
ln
l
n nn nl l
ln
l
Fl
W Dk
D A
W Dk
DD
kWD
kWD
dpdz
Now
.
2 22 (1 ). .
P
l
l
l
F
F l
dpdz
dp G xor f
dz D
2
2
2
2
2
2
2
2
2
22
2, (1 )
2 (1 ). (1 )
2. . .
.2. .
. .
2. . .
4
l
n l le
l
n
l l
l l
n
l l
l l l
n nln n
l l
n nnl l
nl
W Wk R G x
D A A
W WG xk G x
D A A
W D Wk
D A A
W Dk
D A
Wk D
DD
21 4 2
2
25
2
2 .
4
2 .
4
n nn nl l
n
l
n nnl l
n
l
k WD
k WD
. .2
22 5
2
25
2
52 2
2.
4
2.
4
.
T P
l
l
Fl
F
liquid cylinder
F
n nn nl l
ln
l
n nnl lln
l
nn l
l
dpdz
dpdz
dpdz
dpdz
kWD
kWD
D
D
In the expression of 2l , n is coming from expression of lf
For laminar flow n=1 If we take =shape factor=1, then
1 5 42
2222
22
2
2
2
2
4
4
1 1
1
1
l ll
ll
ll
l
D D
D D
DD
D D
A
A
A
A
Problem 3 Estimate the rise velocity of air bubbles in H2O for equivalent radius of 0.25, 1.5, 0.85 cm
given
21 / , 0.01 981 / .l lg cc poise g cm s
Solution: All the bubbles lie in the Stokes region
2
2
1/
18
611624.72 /
3.823 / 0.125
15.29 / 0.25
550.46 / 1.5
176.76 / 0.85
b g bU d g
d m s
m s r cm
m s cm
m s cm
m s cm
Problem 4 Find relation between Q1 and Q2 for flooding in a pipe of radius 25 cm assuming bubbly
flow with n=2 and
Solution:
For Flooding Condition,
7 0 /dyne cm
1.5 /u m s
2
2
21
2
2 21
0 1 2
1 2
. 1 1 11
2, 1.5 1 1 1
1
, 3 1 1 2
, 3 1 3
1,
3
1
0.0478
3.140.5 0.19625
4
0.0054 0.4226 0
n
n
u n
or
or
or
or
j u
A
Q Aj
C Q Q
Q Q
Problem 5 :
What is the velocity of sound in a hydrogen‐water mixture at 6894.7KPa, 210C, and with mean density
640.7 kg/m3? Repeat the problem with the pressure as 3.44KPa and the radius of the bubble as 1 mm.
Solution :
Mean density =640.7 kg/m3
3
3
640.72
1000 /
281.4 /
g f
f
g
kg m
kg m
Taking 0.5
The velocity of sound in bubbly mixture can be calculated from the following equation.
2
2 2
1
11
;
g fg g f f
f g g f
C
C C
C C
The above case be approximated as
1
2
1
g
f
g
CC
Now, 2g gC r RT for a rapid compress
1 1
2
2
8.314
294
1.4 for diatomic gas
1.4 8.314 294
58.49
g
g
g
R Jmol K
T K
r H
C
C
1 1
2 2
58.49
10000.5 0.51
281.4
62.05 / .
g
f
g
CC
C m s
At low pressure the sonic velocity can be obtained by following equation
12
12 0
21
3 41
g
f
g
CC
PR
3
3
70 10 /
3.44
10
61.64 /b
N m
P KPa
R m
C m s
Problem 6 : In a 10 cms diameter Counter current flow bubble column, it was observed that flooding occurred for
the following mass fluxes of carbon dioxide and water in (Kg/hr/m2)
fW 73305 48870 24435
gW 185.7 205.54 234.57
Compare the results with theoretical values at 15Psia, 021 C.
Solution Let us consider bubble rise velocity through infinite medium u is independent of bubble
size i.e. , in region 4
0.25
1.18
0.1909 /
f
gu
m s
Now, 21 1n
j u
Again,
21 2
2
2
1
1
n
n
j j j
j j u
j j u
Now, 32
101.325 441.823 /
8.314 294
PMkg m
RT
We have
3
2 2
3 2
185.7
1.823 3600 .
0.0282 / .
mj
m s
m m s
Similarly, 3 21
733050.0203 / .
1000 3600j m m s
3 20.0485 /j m m s
22 185.7 /G Kg m hr
Now, 2 1n
j j u
1.530.0485 1 0.1909 0.0282 0j
After solving this equation we get,
0.1412
Now, we know that at flooding point,
21 21 2
21
121
1
1
1 1
1 11
1 1
n
n n
n
n
dj j j
d
j u
dju n
dn
u
u n
21 2 21
2121 2
2121 2
1
2
2
1 1 1
11 1
1
1
11
n n
n
n
n
j j dj
ddj
j jd
djj j
d
j u u n
nu
u
j u
1
122
1
1n
n
j u n
211 21
1
1
1 1 1 1
1 1 1
1
n n
n n
n
djj j
d
u u n
u u n
u
1
1 1n
n
u n
So, far
Similarly,
3 21
21
0.11856 / .
426834.35 / .
j m m s
G Kg m hr
So , far other cases theoretical mass fluxes are shown in table
1
2/ .
G
kg m hr 426834.35 389878.22 338020.6
2
2/ .
G
kg m hr 35.70 47.429 68.56
0.1412 0.1650 0.2007
Problem 7:
A contain silicone fluid has a viscosity of 5000 CP a surface tension of 21 dynes/cm, and a density of 1
g/cm3
. What is the rise velocity of slug – flow bubble in stationary liquid in vertical pipes with diameters 0.25
cm, 1.2 cm, 12.5 cm, and 24 cm.
Solution :
2 1.53 1
2
3 3 2
22
0.1412
0.1909 0.1412 1.53 1 0.1412
5.4406 10 / .
35.70 / .
j
m m s
G kg m hr
5000f CP 21 /dynes cm 31 /f gm cm
Neglecting gas density in comparison to liquid density
Eotvos number
2 2
2.916 3.37f g f
EO
gD gDN
Inverse viscosity
122
13 2 2
13 2
2
0.25 980 10.0782 2
5000 10
f g f
ff
f
f
D gN
D g
Archimedes No.
3 1 32 2 2
31 2 22 2
21 11.229 10
50 980
fAr
f
Ng
Properly group,
2
1661375.66
Ar
YN
Now,
1 00.345 1 exp 0.01 / 0.345 1 exp 3.37 /K Nf NE m
For, 0.25 cm diameter pipe
1
1122
1
0.345
5.4 /f f
K
u K gD cm s
Diameter
d cm 0.25 1.2 12.5 24
Eotvos
EON 2.9167 67.2 37.2917 10 26880
Inverse Viscosity 0.0783 0.823 27.6699 73.6199
fN
1K 51.4315 10 0.0075 0.1903 0.3042
u __ 0.2572 21.0624 46.6528
(cm/s)
m 25 25 21.58 15.32
Problem 8 :
When a long bubble rises in a tube closed at the bottom, the value of j ahead of the bubble is not zero
because of expansion of the gas in the hydrostatic pressure gradient . A bubble 0.00016 m3 in volume
is injected into a column of water 30.5 m high in a 0.0254 m pipe. If the temperature is 210 C and the
pipes is closed at the bottom and open to the atmosphere at the top, how long does it take after
release before the bubble breaks the surface.
Solution :
Bubble volume = 30.000lbm
Ltube =30.5 m
D=0.0254m
Now, initially when bubble rises from bottom of the liquid column,
2
2
2 4 0.000
3.14 0.0254
Bubblevolume D lb
Tubevolume D
Where film thickness
22
0.315D
D
Now, considering a single bubble rises through a stationary water column from bottom to the upward
direction. When it gets a long shape after rising certain height we can write :
22 TB
TB IS
LBubblevolume D
Tubevolume D L L
For one unit cell,
film thickness is very small in comparison to D
0.315
0.315 30.5 9.6075
TB
TB
L
LL m
Column
1 ,CP 31 /f gm cm 70 /dynes cm
Neglecting gas density.
Eotvos No.
2
13 2 2
4
3 12 2
12 2
112
90.3224
1.267 10
187082.86
12.857 10 10
f
EO
f
ff
fAr
f
Ar
gDN
D gN
Ng
Y mN
0.01 3.370.345
1 0.345 1 1
0.3198
15.95 /
f EON N
mK e e
u cm s
So, time take for bubble to break
29.6075 10
15.9560.23
1min 23
S
s
Problem 9 :
What is the minimum tube size in which large bubbles of air will rise in stationary water as 300 C
a) On earth b) in a spaceship for which ‘g’=0.003048cm/s2
Solution : Bubble will not rise when there is surface tension dominating effect
i.e. 2
3.37
3.37
EO
f g
N
gD
On earth, g=980cm/s2
Neglecting g
2
2
2
3.37
3.37
3.37 70
9800.49
fgD
Dg
D
D cm
On space, 20.003048 /g cm s
2
2
3.37
3.37 70
0.003048278.19
fgD
D
D cm
So, minimum tube size on earth = 0.49cm
Minimum tube size on space=278.19cm
Problem 10 :
A liquid metals 3300 / ; 5 / , 0.02fdynes cm g cm poise fills a 0.0095m diameter
horizontal pipe. It is derived to blow gas through the pipe to the metal and solidify it as a uniform film 0.0127cm thick on the walls. What gas flow rate should be used ?
Solution :
300 /dynes cm 35 /f g cm 0.02 poise 0.0095D m 0.0127cm
The available flow area for gas in the tube
2
220.95
3.14 0.0127 0.6712 2bDA cm
Area of tube A=2 2
23.14 0.950.708
4 4
Dcm
227
0.7081.0558
0.671
0.947
0.022.80 10
0.95 5 300
b
b
b
f
f
v A
j A
jv
D
From figure
3
3
3.2 10
48 /
48 1.0558 50.64 /
35.90 /
f
b
g
j
j cm s
v cm s
Q cm s
Chapter 10
1. Discuss a commonly used technique for measuring volume average void fraction of a flowing vapor‐liquid mixture? What are the drawbacks of the technique? Quick closing value technique
Drawbacks –
i) Finite Time required to close down the values. This may cause changes in system
hydrodynamics.
ii) Finite Time required to bring system back to steady state. So not possible for a continuous
operating plant.
ii) Not suitable for transient measurements.
2. How can area average and chordal average measurements of void fraction be converted to
volume average values? Area average measurements of void fraction is the volume average value for infinitesimal length
of the test section. So several area average values at different axial lengths gives the volume
average value. When void fraction does not vary with length (fuliy developed flows), both are
equal.
Chordal average value is connected to area average value either by mathematical manipulation
or by the use of multiple beams.
3. Why does one need to know the voidage profile in addition to the average void fraction of a gas‐
liquid mixture? Voidage profile gives an estimate of the distribution of voids in the flow passage. This
adequately describes the structures of the flow field and identifies sites of active transport and
reactions.
4. What are the drawbacks of photographic methods of flow pattern estimation?
Refer to section 10.3, part 1
5. Suggest a suitable arrangement (shown diagrammatically) of the conductivity probes for
distinguishing between (a)bubbly and annular flow in vertical pipes, (b) plug and slug flow in horizontal pipes, (c) stratified and annular flow in horizontal pipes,
The probe signal for bubbly and annular flow are as follows
6. State any three limitations of the radiation attenuation technique for estimation of void fraction and suggest ways of minimizing them. Refer to section 10.2.1
7. State any two limitations of the conductivity probe technique for gas‐liquid systems.
i) Needs a priori knowledgeof flow pattern
ii) Does not work for gas continuous patterns
8. State the principle and the specific application of (a) infra red absorption method (b) Electromagnetic flow metering technique for estimation of void fraction. a) IR absorption technique based on the differential amount of absorption of IR ray by the two
phases. Specifically suitable for high void fraction flows.
b) Electromagnetic flow metering based on principle of independent measurement of average
liquid velocity ( )Lu from which can be calculated as (1 ) L
L
j
u for low quality flows.
Specific application for liquid metal system.
9. For measurement of two phase pressure drop when are gas filled lines preferred to liquid filled
ones?
For low offset value at zero p
10. What are the advantages of liquid filled lines in general? Less chances of gas ingress since liquid tends to meet the manometer lines and pumping action
is less severe since it is incompressible.
11. What are the advantages and disadvantages of differential pressure transducer over absolute
ones? Refer to text, page
12. Define liquid holdup for gas‐liquid systems. Refer to chap 4
13. Discuss briefly the different ways of expressing liquid holdup and one conventional technique to
measure each of them. Refer to chap 4
14. With the help of a schematic (i) show the location of the different probes and (ii) mention the characteristics of the probe signals which distinguish between the following gas‐liquid flow patterns: wavy annular and bubbly in vertical flow (b) churn and bubbly in vertical flow
(b)
15. How can the following methods be used for flow pattern detectors:
Average pressure gradient (b) transient pressure signal
Refer to section 10.3 subsection 3
16. (ii) State a suitable technique to measure the in‐situ composition of two phase flow under the following conditions: (a) High quality steam water flow (b) water content of margarine a) Infra red absorption
b) Microwave absorption
17. Discuss the PSDF analysis for flow pattern identification. What are the different PSDFs obtained
for gas‐liquid two phase flow in vertical tubes.
Refer to section 10.3 subsection 3
18. What are the problems of using differential pressure transducers for measurement of pressure
drop in two phase system. i) All problems associated with ambiguity in tapping line content as discussed for manometers
ii) Slight fluctuations due to rig vibration can alter readings
iii) Pressure smaller than offset cannot be measured
19. How can we measure the void fraction during sodium liquid / vapor flow?
Electromagnetic flow metering
20. Discuss the acoustic method of void fraction measurement and state its problems. Refer to section 10.2.7