WEBVTT
Kind: captions
Language: en
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So, welcome back. Now, what we were discussing
in the last class is about the homogeneous
00:00:30.620 --> 00:00:35.610
flow model and there what we have done, we
have just reduced the single phase flow model
00:00:35.610 --> 00:00:41.750
with the mixture velocity, mixture viscosity
and mixture density and, we said that the
00:00:41.750 --> 00:00:50.220
dou P by dou x minus is what it is going to
be the contribution of the friction. So, it
00:00:50.220 --> 00:01:00.640
will be P upon a and tau instead of w mixture
w it means on the mixture composition plus
00:01:00.640 --> 00:01:11.310
it is going to be rho m into G plus it is
going to be rho m into V mixture into d V
00:01:11.310 --> 00:01:15.280
m upon dx.
So, what we have done we have reduce everything
00:01:15.280 --> 00:01:20.500
in terms of the mixture velocity and tau m
w is nothing, but we have written in terms
00:01:20.500 --> 00:01:30.520
of tau m w we have written in terms of it
will be rho m into u or u I will write it
00:01:30.520 --> 00:01:36.729
as a V m square by 2 multiplied by the friction
factor which will be based on the mixture
00:01:36.729 --> 00:01:42.110
velocity and Reynold number for this f m will
be based on the mixture Reynolds number and
00:01:42.110 --> 00:01:50.009
that will be nothing, but V m rho m d upon
mu m and what we have said that the mu m is
00:01:50.009 --> 00:01:53.990
going to be the problem because we do not
know the mixture viscosity though there are
00:01:53.990 --> 00:01:58.820
some simplified correlations available, but
most of the time we have to go and check
00:01:58.820 --> 00:02:04.259
mixture viscosity experimentally.
Now, so, what we have learned that this is
00:02:04.259 --> 00:02:09.200
a very tedious process and if you want to
calculate the delta P across a pipeline which
00:02:09.200 --> 00:02:14.510
will require a huge experimental effort as
well as you have to solve the equations to
00:02:14.510 --> 00:02:22.400
get the accurate delta P. Now, in industry
most of the time what we do we use some empirically
00:02:22.400 --> 00:02:27.769
developed correlations to calculate the delta
P in a pipeline or across a pipeline.
00:02:27.769 --> 00:02:33.540
Now, this correlation those they are developed
empirically are very very useful because it
00:02:33.540 --> 00:02:39.129
can give you a very quick result and you do
not require these much information and with
00:02:39.129 --> 00:02:44.629
a very limited information you can actually
find the delta P. So, to get the first hand
00:02:44.629 --> 00:02:50.290
idea it is these correlations are very very
important and very good to have a calculations
00:02:50.290 --> 00:02:55.349
instead of doing the numerically you can use
some empirical based equation and calculate
00:02:55.349 --> 00:02:59.879
the delta P.
Now, though again please note it down that
00:02:59.879 --> 00:03:06.329
those these correlations are accuracy suffers
the accuracy they do not provide you hundred
00:03:06.329 --> 00:03:12.410
percent accurate result, but they are very
good to get the first hand idea and with a
00:03:12.410 --> 00:03:19.209
certain percentage of kind of modification
or extra pipeline or extra core correction
00:03:19.209 --> 00:03:24.629
you can have a delta P calculation which will
be very quick and today what we are going
00:03:24.629 --> 00:03:31.530
to learn is about that one kind of a correlation
which was one of the first correlation developed
00:03:31.530 --> 00:03:37.150
to calculate the delta P in multi phase flow
or in a gas liquid pipeline or multi-phase
00:03:37.150 --> 00:03:41.150
flow pipeline.
Now, the name of that correlation is called
00:03:41.150 --> 00:03:57.579
Lockhart Martinelli correlation, Lockhart
Martinelli
00:03:57.579 --> 00:04:03.180
Lockhart Martinelli correlation. So, this
is the very very classical correlation. It
00:04:03.180 --> 00:04:16.440
is one of the oldest correlation and developed
in 1949 to calculate delta P in horizontal
00:04:16.440 --> 00:04:30.580
pipe. So, pipe for multiphase flow and I will
say it is not for multiphase flow, but for
00:04:30.580 --> 00:04:40.460
two-phase flow and mainly the two phases,
gas and liquid, ok.
00:04:40.460 --> 00:04:46.690
So, this is one of the oldest correlation
in 1949 this correlation has been developed
00:04:46.690 --> 00:04:51.419
where the computational approaches was not
this was very limited even the computer was
00:04:51.419 --> 00:04:57.120
not available. So, doing a hard code three
dimensional CFD simulations was very very
00:04:57.120 --> 00:05:03.280
tough. To calculate the delta P, we need to
have certain correlation these correlations
00:05:03.280 --> 00:05:08.520
are developed based on the empirical rules
and what are those rules we will try to learn
00:05:08.520 --> 00:05:14.430
uh. But, even after say more than 70 - 80
years these correlations are still very very
00:05:14.430 --> 00:05:21.550
popular and widely used in industry to calculate
the delta P and why, because we learn that
00:05:21.550 --> 00:05:26.449
instead of solving the three dimensional navier-stokes
equation along with the continuity equation
00:05:26.449 --> 00:05:30.920
for the two phase flow or based on the mixture
viscosity and mixture density and keep on
00:05:30.920 --> 00:05:37.420
calculating the mixture density and mixture
viscosity always you can get delta P very
00:05:37.420 --> 00:05:43.490
quickly in a very short time and that approximation
will not be a bad approximation and we will
00:05:43.490 --> 00:05:49.490
see that we will get that how to calculate
the delta P in two phase flow and very quickly.
00:05:49.490 --> 00:05:55.270
So, this is what is the good thing is in this
flow is that whatever we have done till
00:05:55.270 --> 00:06:00.870
now, we need definitely the information of
the holdup or the volume fraction or we need
00:06:00.870 --> 00:06:06.039
to assume that there is no slip condition,
so, hold up and volume fraction is equal.
00:06:06.039 --> 00:06:21.889
In this case no volume fraction requirement
is there is needed. So, you do not need any
00:06:21.889 --> 00:06:32.009
volume fraction information ok. These correlation
is very simple, but not that accurate and
00:06:32.009 --> 00:06:38.110
that you have to keep in mind that these correlations
are very very simple, but not that accurate,
00:06:38.110 --> 00:06:42.120
but still a very good to get the first hand
idea, ok.
00:06:42.120 --> 00:06:49.569
So, it is based on a multiplier and how the
multiplier has been decided based on that
00:06:49.569 --> 00:06:54.210
we will actually develop some empirically
correlations are available. So, what does
00:06:54.210 --> 00:07:01.289
Lockhart Martinelli correlation says that
the pressure drop in two phase flow will be
00:07:01.289 --> 00:07:08.139
equal to the pressure drop in single phase
flow multiplied by some multiplier ok. So,
00:07:08.139 --> 00:07:18.220
suppose dP by dx in two phase flow I am writing
is TP or two phase flow will be equal to dP
00:07:18.220 --> 00:07:29.199
by dx of single phase liquid, ok or it will
be equal to the dP by dx of single phase gas,
00:07:29.199 --> 00:07:47.080
ok; so, single phase liquid pressure drop,
single phase gas
00:07:47.080 --> 00:07:55.350
pressure drop. So, these are equal, but proper
this you multiply it with a multiplier and
00:07:55.350 --> 00:07:59.400
these multiplier is phi square L and phi square
G.
00:07:59.400 --> 00:08:05.819
So, if I know that multiplier value. So, Lockhart
Martinelli correlation says simply that two
00:08:05.819 --> 00:08:13.969
phase flow pressure drop is equal to the single
phase flow pressure drop of liquid or of gas
00:08:13.969 --> 00:08:20.180
multiplied with certain multiplier and for
if suppose, you are using that single phase
00:08:20.180 --> 00:08:26.180
liquid the multipliers would be phi square
L, if you are using single phase gas the multiplier
00:08:26.180 --> 00:08:32.360
will be phi square G. So, if I calculate these
two multiplier if I know this these two multiplier
00:08:32.360 --> 00:08:42.950
I can calculate the two phase flow pressure
drop. Flow pressure drop we can calculate.
00:08:42.950 --> 00:08:48.750
This is as simple as this the whole correlation
and that is why this correlation is very simple
00:08:48.750 --> 00:08:53.690
and very popular because you can easily calculate
the single phase flow pressure drop by using
00:08:53.690 --> 00:08:59.100
the conventional method. If you just calculate
this phi square L and phi square G you will
00:08:59.100 --> 00:09:02.780
able to find it out that what will be the
two phase flow pressure drop.
00:09:02.780 --> 00:09:08.680
So, how these multipliers are defined if you
just see this equation you can say that phi
00:09:08.680 --> 00:09:21.660
square L is nothing, but dP by dx of two phase
divided by dP by dx of single phase liquid,
00:09:21.660 --> 00:09:33.400
phi square G will be dP by dx of two phase
divided by dP by dx of single phase gas, ok.
00:09:33.400 --> 00:09:38.230
So, this is the way the multiplier has been
defined and multiplier if you know the single
00:09:38.230 --> 00:09:43.220
phase flow liquid pressure drop or single
phase flow gas pressure drop you just multiply
00:09:43.220 --> 00:09:47.860
these two values you will get that what will
be your two phase flow pressure drop ok. So,
00:09:47.860 --> 00:09:52.980
that is what is the this things.
Now, to calculate the multiplier another multiplier
00:09:52.980 --> 00:09:58.530
has been defined another parameter has been
defined which is called two phase flow multiplier
00:09:58.530 --> 00:10:07.680
and it is called defined by the symbol psi
and this is defined as psi square psi is equal
00:10:07.680 --> 00:10:15.331
to or psi square is equal to phi square G
upon phi square . So, to calculate the multiplier
00:10:15.331 --> 00:10:20.830
value a parameter has been defined which is
called Lockhart Martinelli parameter. This
00:10:20.830 --> 00:10:36.350
is called Lockhart Martinelli parameter which
is the ratio of phi square G upon phi square
00:10:36.350 --> 00:10:39.340
L.
Now, if you do that phi square G upon phi
00:10:39.340 --> 00:10:48.710
square L phi square G will be dP by dx of
two phase flow, of two phase flow divided
00:10:48.710 --> 00:10:57.730
by dP by dx of single phase gas this will
be phi square L will be dP by dx of two phase
00:10:57.730 --> 00:11:06.600
flow divided by dP by dx of single phase liquid,
this two will be cancelled out. So, what you
00:11:06.600 --> 00:11:14.500
will get, psi square will be equal to phi
square G upon phi square L and this will be
00:11:14.500 --> 00:11:24.020
dP by dx of single phase liquid divided by
dP by dx of single phase gas.
00:11:24.020 --> 00:11:30.190
So, you can calculate this psi square value.
So, if you know the psi square value you can
00:11:30.190 --> 00:11:35.430
find the phi square value and then phi square
value you can calculate that what will be
00:11:35.430 --> 00:11:41.680
the delta P in two phase flow. Now, the problem
is this psi square and phi square G are correlated
00:11:41.680 --> 00:11:47.110
and phi square L are correlated it means you
need certain other correlation to calculate
00:11:47.110 --> 00:11:51.940
the phi square G and phi square L value. Because,
if you want two phase flow pressure drop you
00:11:51.940 --> 00:11:56.700
need to have phi square G and phi square L
and the ratio of phi square G and phi square
00:11:56.700 --> 00:12:03.160
L is defined as a psi square. Psi square can
be calculated with the ratio of delta P upon
00:12:03.160 --> 00:12:08.790
x or dP by dx for single phase liquid to the
dP by dx of single phase gas. So, you need
00:12:08.790 --> 00:12:12.210
another correlation where you can calculate
the phi square G.
00:12:12.210 --> 00:12:17.190
Now, once the phi square G and phi square
L because these are two variable one equation
00:12:17.190 --> 00:12:23.820
you need another equation to calculate that.
So, the phi square G has been found experimentally,
00:12:23.820 --> 00:12:33.460
these all are experimentally fitted correlation
is C psi plus psi square, this is the phi
00:12:33.460 --> 00:12:42.640
square G and phi square L has been defined
as 1 plus C upon psi plus 1 upon psi square,
00:12:42.640 --> 00:12:47.090
ok.
Now, many value this value says zeta many
00:12:47.090 --> 00:12:51.880
value say the x cross. So, there different
symbols available whatever symbol we are using
00:12:51.880 --> 00:12:58.010
is the psi square. Now, the value is this
will be the value of psi square G and psi
00:12:58.010 --> 00:13:03.411
square L and if you know this what is the
problem now is to calculate the value of C.
00:13:03.411 --> 00:13:09.230
So, if you know this value of C ideally what
you can do you can calculate the single phase
00:13:09.230 --> 00:13:14.320
flow pressure drop in liquid if only liquid
is flowing; single phase flow pressure drop
00:13:14.320 --> 00:13:19.910
if only gas is flowing. The ratio will give
you the size square. From size square you
00:13:19.910 --> 00:13:24.450
can calculate the value of phi square G and
phi square L. The moment you have phi square
00:13:24.450 --> 00:13:29.230
G and phi square L you can calculate the two
phase flow pressure drop by using this correlation.
00:13:29.230 --> 00:13:34.900
Now, to calculate the phi square G and phi
square L the only problem now left is to calculate
00:13:34.900 --> 00:13:40.750
the value of C. Now, several experiment has
been conducted Lockhart Martinelli had done
00:13:40.750 --> 00:13:45.640
several experiments. They have performed the
experiments by changing the velocity of the
00:13:45.640 --> 00:13:50.330
gas velocity of the liquid.
And, they have come up with the table and
00:13:50.330 --> 00:13:58.370
that table phase that only based on the liquid
and gas velocity and again I am saying only
00:13:58.370 --> 00:14:03.160
based on the liquid and gas velocity it means
we will assume that only once only liquid
00:14:03.160 --> 00:14:08.350
is flowing, then we will assume only gas is
flowing. We will calculate the Reynolds number
00:14:08.350 --> 00:14:19.240
and based on that Reynolds number if both
the flow is turbulent
00:14:19.240 --> 00:14:25.810
both the flow is turbulent then the C value
is found to be 20, if one laminar say liquid
00:14:25.810 --> 00:14:41.730
is laminar, gas is turbulent you get the value
12, C value 12. If liquid turbulent gas laminar
00:14:41.730 --> 00:14:57.940
you get the value 10 and if both the laminar,
sorry, if both are laminar you get the value
00:14:57.940 --> 00:15:00.740
5.
So, now what you have the C value with you
00:15:00.740 --> 00:15:06.490
for the different condition. If both the flow
is turbulent you will get the C value 20,
00:15:06.490 --> 00:15:13.120
if liquid is laminar gas is turbulent the
value will be 12. If both one liquid is turbulent
00:15:13.120 --> 00:15:18.620
gas is laminar the value will be 10 and if
both are laminar the value will be 5. So,
00:15:18.620 --> 00:15:22.860
based on that what you will have now? You
will have the C value. So, by using these
00:15:22.860 --> 00:15:32.950
correlation which says that phi square L is
nothing, but dou P by dou x of two-phase flow
00:15:32.950 --> 00:15:41.690
divided by dou P by dou x of single phase
liquid or phi square G which says dou P by
00:15:41.690 --> 00:15:50.440
dou x of two-phase flow divided by dou P by
dou x of single phase liquid.
00:15:50.440 --> 00:15:58.760
Another equation psi square which is nothing,
but phi square G or phi square L ratio of
00:15:58.760 --> 00:16:06.610
this or you can say it is the ratio of dou
P by dou x of single phase liquid to dou P
00:16:06.610 --> 00:16:14.600
by dou x of single phase gas and then using
the correlation that phi square G is equal
00:16:14.600 --> 00:16:26.860
to 1 plus C psi plus psi square or phi square
L is 1 plus C upon psi plus 1 upon psi square
00:16:26.860 --> 00:16:31.900
you can calculate the multiplier the C value
will be this and you can calculate the two
00:16:31.900 --> 00:16:37.230
phase flow pressure drop. So, this whole method
is being developed by the Lockhart Martinelli
00:16:37.230 --> 00:16:41.930
and that is why the correlation came name
came as a Lockhart Martinelli correlation.
00:16:41.930 --> 00:16:47.570
And, you can see that by using just performing
four step you can calculate the delta P. You
00:16:47.570 --> 00:16:51.820
do not need to numerically discretize it,
even in the single phase flow equation you
00:16:51.820 --> 00:16:56.310
do not need to calculate the mixture viscosity,
you do not need to need the information of
00:16:56.310 --> 00:17:00.320
the volume fraction of phase holdup, you do
not need the information that whether the
00:17:00.320 --> 00:17:06.089
flow is in no slip condition or not. Without
having all these information just by solving
00:17:06.089 --> 00:17:11.600
the four step you can calculate that what
will be the delta P in a horizontal pipe in
00:17:11.600 --> 00:17:16.270
which gas and liquid are flowing together
and that is the beauty of this correlation.
00:17:16.270 --> 00:17:22.929
Though it suffer accuracy, it has some error,
but it will just solving the four step which
00:17:22.929 --> 00:17:29.029
you can solve with your calculator or even
by your hand you can find the delta P in a
00:17:29.029 --> 00:17:34.830
pipeline of any length, because this is dP
by dx. So, any length you just multiply with
00:17:34.830 --> 00:17:39.700
the length you will get the delta P in that
length. So, whether it is a 100 kilometre
00:17:39.700 --> 00:17:44.159
pipe, 1000 kilometre pipe, 10000 kilometre
pipe, does not matter you can calculate the
00:17:44.159 --> 00:17:50.429
delta P within sometime within some few minutes
and that is the beauty of this correlation
00:17:50.429 --> 00:17:56.150
and why this correlation is. So, popular even
now to calculate the delta P, ok.
00:17:56.150 --> 00:18:01.940
So, now if I just do the gas guess if I just
take some simplified values to find it out
00:18:01.940 --> 00:18:06.350
that how the two phase flow pressure drop
will change. So, let us assume that my flow
00:18:06.350 --> 00:18:11.860
is turbulent, it means both gas and liquid
is in turbulent flow psi value is say one
00:18:11.860 --> 00:18:18.389
the what will be the phi square G value. The
phi square G value will be equal to 21, ok,
00:18:18.389 --> 00:18:27.240
not 21, it will be 22 ok. So, this will be
1 plus 20 plus 1. So, that will be equal to
00:18:27.240 --> 00:18:32.920
22, it means what if I see this equation it
means two phase flow pressure drop will be
00:18:32.920 --> 00:18:38.789
22 times of the single phase flow gas pressure
drop, ok.
00:18:38.789 --> 00:18:44.860
So, you can simply calculate it. If this both
the force are turbulent and that is the mostly
00:18:44.860 --> 00:18:50.409
used case where both the phases are actually
flow in turbulent to have a higher throughput
00:18:50.409 --> 00:18:56.620
you will see that the two phase flow pressure
drops. So, dou P by dou x of two phase will
00:18:56.620 --> 00:19:04.900
be equal to 22 times of dou P by dou x of
single phase flow gas. So, your pressure drop
00:19:04.900 --> 00:19:10.850
will enormously increase and that is the region
that one need to calculate the delta P before
00:19:10.850 --> 00:19:16.120
designing any pipelines system for flow of
two phase flow, because the flow pressure
00:19:16.120 --> 00:19:22.200
drop can have enormously high and that is
the way the Lockhart Martinelli has done they
00:19:22.200 --> 00:19:27.139
have taken a pipeline they have flowed the
two phase flow they have done it for the different
00:19:27.139 --> 00:19:32.519
velocities and they developed these four correlations
and then based on this four correlation they
00:19:32.519 --> 00:19:37.519
found that you can measure the delta P in
two phase flow problem in any two phase flow
00:19:37.519 --> 00:19:40.870
problem.
Now, how to use these correlations, we will
00:19:40.870 --> 00:19:45.970
see it with a simple example that how to use
this correlation and this is very important,
00:19:45.970 --> 00:19:49.960
that is why I am going to solve a numerical
problem we will have a assignment on this
00:19:49.960 --> 00:19:55.210
so that you can solve more numerical problems
to understand the methodology that how to
00:19:55.210 --> 00:20:00.909
use this kind of a flow .
So, let us assume that I have a horizontal
00:20:00.909 --> 00:20:06.940
pipe in which two phase flow is flowing both
gas and liquid is flowing inside and it is
00:20:06.940 --> 00:20:22.200
given that volume fraction of air or holdup
of air air is 45 percent, ok. That is given,
00:20:22.200 --> 00:20:32.179
it is given that the mixture flow rate it
means gas plus liquid flow rate is 10 kg per
00:20:32.179 --> 00:20:40.429
second and the diameter is pipe is specified
as say 0.05 meter. So, these all information
00:20:40.429 --> 00:20:51.850
is given gas is air and liquid is water. So,
the gas density is 1.2 and so rho of air at
00:20:51.850 --> 00:21:00.870
standard condition we have taking 1.2 and
rho of water we are taking it 1000, ok. And,
00:21:00.870 --> 00:21:10.049
what is the other property we are going to
take is the mu value. So, mu value of water
00:21:10.049 --> 00:21:17.870
a air is given as 10 is to the power minus
5, let me see the data. So, that 1.7 into
00:21:17.870 --> 00:21:25.549
10 is to the power minus 5 and mu of water
is given as 10 is to the power minus 3, ok
00:21:25.549 --> 00:21:30.669
and this unit is kg per meter per second,
ok.
00:21:30.669 --> 00:21:37.669
So, these information is given, now what we
need to do we need to calculate the delta
00:21:37.669 --> 00:21:44.940
P in this pipeline. The overall length of
the pipe is given say 1000 meter. So, we need
00:21:44.940 --> 00:21:52.160
to calculate that how much delta P will take
place in this pipeline. So, what we are going
00:21:52.160 --> 00:21:57.660
to do? We are going to use the Lockhart Martinelli
correlation. So, to use that first what we
00:21:57.660 --> 00:22:02.820
are going to do we are going to calculate
that what is the velocity, mixture velocity
00:22:02.820 --> 00:22:07.740
inside. To calculate the mixture velocity
what I need? I need a volumetric flow rate
00:22:07.740 --> 00:22:11.879
and the flow rate is given in terms of the
mass flow rate kg per second it is given.
00:22:11.879 --> 00:22:17.190
So, first I need to calculate the mixture
density. So, how the mixture density will
00:22:17.190 --> 00:22:26.820
be calculated? Rho m is going to be the epsilon
of air into rho of air plus epsilon of water
00:22:26.820 --> 00:22:36.840
into rho of water. Now, epsilon of air is
given 0.45, 45 percent means 0.45 into 1.21
00:22:36.840 --> 00:22:44.390
plus epsilon of water, we know that epsilon
air plus epsilon water will be equal to 1
00:22:44.390 --> 00:22:50.809
because it is a two phase flow. So, epsilon
water is going to be equal to 1 minus epsilon
00:22:50.809 --> 00:22:57.749
air and that is going to be 0.55, because
this is 0.45, 1 minus 0.45 you will get 0.55.
00:22:57.749 --> 00:23:05.419
So, this will be 0.55 into 1000.
Now, if you calculate this, if you add it
00:23:05.419 --> 00:23:09.509
you will get the rho m value, I have already
did the calculation, but you can do it in
00:23:09.509 --> 00:23:18.330
your calculator that will be 550.54 kg per
meter cube. So, that will be the mixture density.
00:23:18.330 --> 00:23:23.139
Now, if you have the mixture density, I can
calculate the mixture volumetric flow rate
00:23:23.139 --> 00:23:33.049
and that will be what m naught upon rho m.
So, this will be m naught is in kg per second,
00:23:33.049 --> 00:23:38.179
this is in kg per meter cube. So, what you
are going to get? You are going to get meter
00:23:38.179 --> 00:23:44.000
cube per second ok. So, you will get the volumetric
flow rate and that is what and not value is
00:23:44.000 --> 00:23:50.779
given 10, this value rho m value will get
five 550.54. So, what you are going to get
00:23:50.779 --> 00:24:02.649
the Q m value, that is, equal to if you divide
it you will get 0.182. 0.182 meter cube per
00:24:02.649 --> 00:24:08.890
second is the volumetric flow rate.
Now, I have the volumetric flow rate. I can
00:24:08.890 --> 00:24:13.270
find it out the volumetric flow rate of mixture;
I can find it out the volumetric flow rate
00:24:13.270 --> 00:24:16.650
of air and water because I know the volume
fraction.
00:24:16.650 --> 00:24:25.740
So, the volumetric flow rate of air will be
what epsilon of air into Q of mixture. So,
00:24:25.740 --> 00:24:39.559
that will be 0.45 into 0.0182 and that correspond
to 8.19 into 10 raise power minus 3 meter
00:24:39.559 --> 00:24:46.369
cube per second, ok. Similarly, I can find
the volumetric flow rate of water. So, Q of
00:24:46.369 --> 00:24:57.940
water will be equal to epsilon of water into
Q of mixture, so, 0.55 into 0.0182 if you
00:24:57.940 --> 00:25:07.220
do that we will get 0.00999 meter cube per
second. If you add these two, you will get
00:25:07.220 --> 00:25:10.220
0.0182 that is the overall volumetric flow
rate.
00:25:10.220 --> 00:25:15.110
So, I know the volumetric flow rate, now what
I need to calculate I need to calculate the
00:25:15.110 --> 00:25:20.619
pressure drop in single phase flow liquid
pressure drop one single phase flow gas, it
00:25:20.619 --> 00:25:25.809
means assuming that only liquid is flowing
inside and only gas is flowing inside. So,
00:25:25.809 --> 00:25:32.960
to calculate that delta P what I need, I need
to calculate the velocity and for velocity
00:25:32.960 --> 00:25:40.129
I need to calculate the area and area of this
will be equal to pi by 4 D square and this
00:25:40.129 --> 00:25:47.299
will be pi by 4 into 0.05 whole square and
this if you will do that you will get the
00:25:47.299 --> 00:25:52.059
area is 1.96 into 10 raise to power minus
3 meter square.
00:25:52.059 --> 00:25:57.919
So, if you do that what you can calculate,
you can calculate the velocity of the air.
00:25:57.919 --> 00:26:11.850
So, velocity of air or I will say superficial
velocity, because we are assuming only air
00:26:11.850 --> 00:26:19.440
is flowing inside. So, I will say V a this
will be equal to Q of air divided by area
00:26:19.440 --> 00:26:28.080
and that will be coming as 8.19 into 10 is
to the power 3 three divided by 1.96 into
00:26:28.080 --> 00:26:36.290
10 is to the power minus 3, you will get it
the value as equal to 4.178 meter per second.
00:26:36.290 --> 00:26:44.830
So, that you will get the V of a.
Similarly, we can calculate the superficial
00:26:44.830 --> 00:27:01.309
velocity velocity of water and that is V w
it will be Q w, Q of water divided by area
00:27:01.309 --> 00:27:10.280
and that is 0.00999 upon 1.96 into 10 is to
the power minus 3 and that you will get V
00:27:10.280 --> 00:27:16.600
w superficial velocity of the water. Let me
see the calculation you will get this will
00:27:16.600 --> 00:27:25.240
be equal to 5.097 meter per second. So, generally
try to keep after least 3 digit after the
00:27:25.240 --> 00:27:29.259
decimal of four digit after the decimal to
get the accuracy.
00:27:29.259 --> 00:27:34.170
Now, what we need to do we need to calculate
the pressure drop for calculating the pressure
00:27:34.170 --> 00:27:39.269
drop I need friction factor, to get the friction
factor value that what will be the friction
00:27:39.269 --> 00:27:42.140
factor value we will be using the Reynolds
number.
00:27:42.140 --> 00:27:47.249
So, we need to calculate Reynold number of
air which will be what which will be V of
00:27:47.249 --> 00:27:55.570
air rho of air into diameter of the pipe upon
mu of air. So, if you do that V of air we
00:27:55.570 --> 00:28:08.340
have calculated 4.178, rho of air is 1.2 and
into 0.05 upon 1.7 into 10 to the power minus
00:28:08.340 --> 00:28:19.090
5. If you do that Reynold number of air will
come 14868. So, this is greater than 2100,
00:28:19.090 --> 00:28:27.759
so, flow is turbulent. So, we get that that
one flow is turbulent.
00:28:27.759 --> 00:28:35.869
Now, we will calculate Reynold number of water
and that will be V of water into rho of water
00:28:35.869 --> 00:28:46.399
into D upon mu of water. Now, V of water we
have got 5.097 into 1000 into 0.05 upon 10
00:28:46.399 --> 00:28:55.760
raise to the power minus 3. If we do that
we will get the number 254850. Now, again
00:28:55.760 --> 00:29:03.309
because this is the pipe flow if Reynold number
is greater than 2100 it means this flow is
00:29:03.309 --> 00:29:09.090
turbulent.
So, we have we came to know that both the
00:29:09.090 --> 00:29:14.059
flow is turbulent. Now, what we need to do
we need to calculate the delta P for the single
00:29:14.059 --> 00:29:20.519
phase flow liquid, delta P for the single
phase flow gas. Now, because the flow is horizontal
00:29:20.519 --> 00:29:25.989
the gravity term will be 0, if the flow is
steady state and velocity is nine velocity
00:29:25.989 --> 00:29:37.450
is fully developed it means the flow is dV
by dx is equal to 0 fully developed flow
00:29:37.450 --> 00:29:42.710
it means dV by dx is going to be 0. So, what
we are going to have we will see the dP by
00:29:42.710 --> 00:29:51.809
dx is only because of P upon A tau w ok and
if I am doing the dP by dx. So, this P upon
00:29:51.809 --> 00:29:58.919
a for the cylindrical pipe will be what P
will be pi D area will be pi by 4, D square
00:29:58.919 --> 00:30:05.010
into tau w. So, you will see that it will
be 4 tau w upon D, ok.
00:30:05.010 --> 00:30:14.039
Now, tau w, we have written as what is friction
factor into rho u square or V square upon
00:30:14.039 --> 00:30:19.190
2.
So, if you replace it the dP by dx for the
00:30:19.190 --> 00:30:28.179
single phase flow will be equal to if you
do that here, so, you will get 2 of f into
00:30:28.179 --> 00:30:35.820
rho into V square upon D. So, you will get
this as single phase flow pressure drop. Now,
00:30:35.820 --> 00:30:43.999
if I say dP by dx for the single phase flow
gas it means this will be 2 f for the gas
00:30:43.999 --> 00:30:49.120
friction factor for the gas, rho of gas V
of gas square upon D.
00:30:49.120 --> 00:30:54.620
Now, friction factor of the gas, how to calculate
the friction factor? We know that f will be
00:30:54.620 --> 00:31:03.769
equal to 16 by R e, if the flow is laminar
and if I assume that the flow is turbulent
00:31:03.769 --> 00:31:11.190
and using the Blasius correlation I can say
that f will be equal to 0.79 upon R e raised
00:31:11.190 --> 00:31:20.590
to the power 0.25 by using the Blasius correlation.
So, because the flow is turbulent we will
00:31:20.590 --> 00:31:25.869
use the Blasius correlation.
So, f we need to calculate for the air. So,
00:31:25.869 --> 00:31:39.440
f of air will be equal to 0.79 upon R e and
R e of air was 14868 raised to the power 0.25.
00:31:39.440 --> 00:31:46.769
If you do this calculation you will get that
f value for the air is 7.154 into 10 raised
00:31:46.769 --> 00:31:57.639
to the power minus 3. Similarly, f for water
you can calculate f water is equal to 0.079
00:31:57.639 --> 00:32:06.539
upon Reynold number of water which is 254850
raised to the power 0.25 and if you do this
00:32:06.539 --> 00:32:12.269
calculation you will get it this value 3.516
into 10 raised to the power minus 3.
00:32:12.269 --> 00:32:18.220
So, now, I have everything to calculate the
single phase flow pressure drop for the gas,
00:32:18.220 --> 00:32:23.450
single phase flow pressure drop for the liquid.
So, let us do that calculation dP by dx for
00:32:23.450 --> 00:32:31.980
the single phase flow of gas that will be
2 into f of gas or f of air that will be 7.154
00:32:31.980 --> 00:32:41.799
into 10 raised to the power minus 3 into rho
G, 1.21 and the V G, V G was 4.178 square
00:32:41.799 --> 00:32:50.789
divided by 0.05 which is D and if you calculate
that this value will come 6.044, clear.
00:32:50.789 --> 00:32:59.789
Similarly, we can calculate dP by dx for water.
So, dP by dx for water that will be equal
00:32:59.789 --> 00:33:08.769
to 2 into f; f is 3.516 into 10 raised to
the power minus 3 into rho of water that will
00:33:08.769 --> 00:33:17.289
be 1000 into velocity square 5.097 square
divided by the diameter 0.05 and what you
00:33:17.289 --> 00:33:26.259
will calculate you will get the dP value and
dP value will be 3653.7 if you calculate that.
00:33:26.259 --> 00:33:32.130
So, I have two phase single phase flow pressure
drop for water single phase flow pressure
00:33:32.130 --> 00:33:38.419
drop for the gas. So, what I can do, I can
calculate the size square value the parameter,
00:33:38.419 --> 00:33:45.929
Lockhart Martinelli parameter value and that
is dP by dx of single phase flow liquid divided
00:33:45.929 --> 00:33:55.830
by dP by dx of single phase flow gas. So,
in this case what it will be? It will be liquid
00:33:55.830 --> 00:34:08.159
is 3653.7 gas it is 6.044. So, the psi square
value you are going to get is 24 psi value
00:34:08.159 --> 00:34:15.940
actually will be 24.59, ok. So, this will
be the value of psi which you will get here
00:34:15.940 --> 00:34:20.850
and if you know the value of psi what we can
do we can calculate the value of phi square
00:34:20.850 --> 00:34:31.030
L and phi square G phi square L will be 1
plus 1 plus C upon psi plus 1 upon psi square
00:34:31.030 --> 00:34:36.470
phi square G value will be 1 plus C psi plus
psi square.
00:34:36.470 --> 00:34:48.180
Now, we know that that C value is what because
both the flow is turbulent we can write here,
00:34:48.180 --> 00:35:00.880
the flow is turbulent C will be equal to 20.
So, phi square L you can calculate 1 plus
00:35:00.880 --> 00:35:15.700
20 upon 24.59 plus 1 upon 24.59, you will
get the value is equal to 1.8146. Similarly,
00:35:15.700 --> 00:35:20.230
you can find the phi square G value also.
What you have to do, the phi square G value
00:35:20.230 --> 00:35:25.530
you have to multiply by 20 into 24 plus 24
square, you will get the huge value.
00:35:25.530 --> 00:35:32.570
Now, what you can say the dP by dx, I am calculating
based on the liquid for the two-phase flow
00:35:32.570 --> 00:35:41.400
is nothing, but phi square L into dP by dx
for single phase flow liquid. So, it means
00:35:41.400 --> 00:35:51.260
what you are going to get 1.8146 into dP by
dx of single phase liquid that is equal to
00:35:51.260 --> 00:36:05.850
1.8146 into 3653.7. Now, if you do that you
will get 6630.2 Pascal per meter dP by dx
00:36:05.850 --> 00:36:12.010
for the two phase flow. So, it is 1.81 times
higher than the liquid phase flow and more
00:36:12.010 --> 00:36:16.380
than thousand times higher for single phase
gas flow.
00:36:16.380 --> 00:36:22.450
Now, if you do that because we have said that
the pipe length is thousand meter I can find
00:36:22.450 --> 00:36:32.760
the dP for the two phase flow is 6630.2 into
1000, it is Pascal. So, you will get this
00:36:32.760 --> 00:36:41.850
much of your value will be 6630.2 into 10
raised to the power 6 Pascal will be the dP
00:36:41.850 --> 00:36:48.290
by dx value in the two phase flow. So, you
can calculate the delta V by dx or delta P
00:36:48.290 --> 00:36:55.310
value for the two phase flow in a pipeline
of 1000 meter long or 1 kilometre long even
00:36:55.310 --> 00:37:01.340
if it is 10000 kilometre long or 10000 meter
long this calculation is going to be the same
00:37:01.340 --> 00:37:06.040
and within the fraction of time what we have
done we have calculated that what will be
00:37:06.040 --> 00:37:11.800
the delta P in two phase flow.
So, though again coming back to the same though
00:37:11.800 --> 00:37:17.070
it may suffer the accuracy, but it gives a
very quick result and you can see within 15
00:37:17.070 --> 00:37:21.840
minute of time we have calculated that what
will be the delta P in a pipeline in a two
00:37:21.840 --> 00:37:27.810
phase flow pipeline, where the gas and liquid
are flowing. Now, being it 10000 kilometre
00:37:27.810 --> 00:37:33.530
pipeline being it even more bigger pipeline
the calculation does not take even extra time
00:37:33.530 --> 00:37:39.110
because you have to just multiply with the
length of the pipe at the end of the calculation.
00:37:39.110 --> 00:37:46.010
You can do the calculation very quickly to
find that what will be the delta P in a two
00:37:46.010 --> 00:37:50.040
phase flow.
So, this is Lockhart Martinelli correlation.
00:37:50.040 --> 00:37:55.100
This is the method how to use the Lockhart
Martinelli correlation. What we do to summarize
00:37:55.100 --> 00:38:00.780
again, we take single phase flow Lockhart
Martinelli says that two phase flow pressure
00:38:00.780 --> 00:38:06.720
drop will be equal to the single phase flow
pressure drop of liquid or gas multiplied
00:38:06.720 --> 00:38:14.320
by some multiplier and that multiplier can
be calculated by using some empirical correlations
00:38:14.320 --> 00:38:19.390
developed by the Lockhart Martinelli which
is in terms of the Lockhart Martinelli parameter
00:38:19.390 --> 00:38:25.180
psi square psi square is nothing, but phi
square G upon phi square L. And, then phi
00:38:25.180 --> 00:38:33.010
square G has been defined as 1 plus C x plus
phi square C psi plus psi square of phi square
00:38:33.010 --> 00:38:37.760
L has been defined as 1 plus C upon psi plus
1 upon psi square.
00:38:37.760 --> 00:38:44.020
The value of C can be found by finding that
individual flow if assume that only gas is
00:38:44.020 --> 00:38:49.720
flowing or only liquid is flowing what is
your Reynold number? If both the Reynold number
00:38:49.720 --> 00:38:56.770
is turbulent the value is 20. If one is
laminar say liquid is laminar gas is turbulent
00:38:56.770 --> 00:39:02.900
value is 12, if liquid is turbulent gas is
laminar value is 10, if both are laminar value
00:39:02.900 --> 00:39:06.370
is 5.
So, what we do we calculate the pressure drop
00:39:06.370 --> 00:39:12.150
for single phase flow assuming that only that
phase is flowing inside we calculate the Reynold
00:39:12.150 --> 00:39:18.010
number. We calculate the f value, we calculate
the pressure drop, we calculate the Lockhart
00:39:18.010 --> 00:39:24.300
Martinelli parameter psi square, then we calculate
the multiplier phi square G of phi square
00:39:24.300 --> 00:39:28.650
L and we know the calculate that two phase
flow pressure drop. So, that is the Lockhart
00:39:28.650 --> 00:39:34.410
Martinelli correlation widely used in many
industries and to get the first hand idea
00:39:34.410 --> 00:39:39.600
of pressure drop in two phase flow which you
will get if you do the advanced CFD simulation
00:39:39.600 --> 00:39:45.470
to get a accurate result. It will take huge
time and will depend on the length of the
00:39:45.470 --> 00:39:52.500
pipe here you can do it very quickly.
But, the limitation is the accuracy limitation
00:39:52.500 --> 00:39:57.160
is we have not considered any bends, we have
not considered any joints; we have not considered
00:39:57.160 --> 00:40:02.372
any inclination. If you do those things, your
calculation will be keep on increasing. It
00:40:02.372 --> 00:40:08.310
will be tedious. You have to incorporate the
losses due to the fittings, the losses due
00:40:08.310 --> 00:40:12.660
to the bends and all which has not been done
by the Lockhart Martinelli correlation.
00:40:12.660 --> 00:40:18.230
So, what you need to do you have to find it
out the losses in the bend as L e by D it
00:40:18.230 --> 00:40:23.780
means assuming that how much length of the
straight pipe line is required to account
00:40:23.780 --> 00:40:29.500
for that bend loss. Like I have told you that
in pneumatic conveying the bend losses is
00:40:29.500 --> 00:40:36.070
equal to the 7.5 meter of the straight pipe
line. So, similarly for each bend each fittings
00:40:36.070 --> 00:40:40.650
you have to calculate those kind of a correlation,
you have to develop that kind of a correlation
00:40:40.650 --> 00:40:47.320
L e upon D value or equivalent length value,
then you can add the length in that much amount
00:40:47.320 --> 00:40:52.430
to get the delta P in two phase flow if you
are using the Lockhart Martinelli correlation.
00:40:52.430 --> 00:40:59.140
So, with this today, chapter is over. Now,
what we are going to see in the next class
00:40:59.140 --> 00:41:05.690
is, to see that how to write the equation
for two phase flow, it means once the phases
00:41:05.690 --> 00:41:10.750
are not homogeneously makes they are not a
mixture model they are either separated or
00:41:10.750 --> 00:41:30.650
flowing in annular flow, how to write the
equations and how to solve those equations.
00:41:30.650 --> 00:41:38.850
Thank you .