WEBVTT
Kind: captions
Language: en
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Good morning everybody, today I will start
00:00:30.470 --> 00:00:37.460
with the analytical solution of temperature
distribution. And the last module, last time
00:00:37.460 --> 00:00:44.210
we have discussed the different model and
how it can represent the different types of
00:00:44.210 --> 00:00:49.720
the welding processes, for example, arc
welding process or laser welding process or
00:00:49.720 --> 00:00:54.040
resistant welding process. So, corresponds
to that there are different representation
00:00:54.040 --> 00:01:00.309
of the heat source that we can mathematically
implement in any kind of the heat transfer
00:01:00.309 --> 00:01:04.350
equation for solving the temperature distribution
in case of welding process.
00:01:04.350 --> 00:01:10.540
So, now will start with the analytical solution
of the temperature distribution we generally
00:01:10.540 --> 00:01:17.700
observed in the welding processes. So, any
the basic governing equation here is the
00:01:17.700 --> 00:01:24.930
heat conduction equation. And this he[at]
heat conduction equation by assuming. So,
00:01:24.930 --> 00:01:30.590
many things and we can find out the solution
of this heat conduction if you seen by analytical
00:01:30.590 --> 00:01:38.270
means . So, let us start with the very
basic solution which has been developed earlier
00:01:38.270 --> 00:01:42.860
for the for to estimate the temperature
distribution in case of welding process .
00:01:42.860 --> 00:01:49.630
So, here if we see the governing equation
that that case the thermal conductivity and
00:01:49.630 --> 00:01:57.479
capital T which is also temperature and
Q dot is the internal heat generation and
00:01:57.479 --> 00:02:02.310
we can find out and this is the tangent
variation of the temperature with respect
00:02:02.310 --> 00:02:12.410
to time t. So, this is the typical governing
equation that we need to solve in a very specific
00:02:12.410 --> 00:02:18.820
to the welding process. So, in the first case,
we assume some infinite body and it is also
00:02:18.820 --> 00:02:26.579
mentioned it necessary to mention that in
analytical solution we can it is limited to
00:02:26.579 --> 00:02:32.760
limited with so many assumptions . And we
can get the approximate solution of this equation,
00:02:32.760 --> 00:02:40.720
because this governing equation it is not
possible to exactly exact solution of this
00:02:40.720 --> 00:02:46.750
equation in analytically it is not possible.
So, you need a lot of assumptions that is
00:02:46.750 --> 00:02:52.270
very specific to the welding process.
So, first case suppose the we start with the
00:02:52.270 --> 00:03:00.090
infinite body, we assume the length or dimension
of the hm solution geometry through which
00:03:00.090 --> 00:03:04.590
we are interest to find out the temperature
distribution that is infinite in size in the
00:03:04.590 --> 00:03:10.069
sense that effect of boundary condition can
be neglected in this case. Then we assume
00:03:10.069 --> 00:03:16.439
the instantaneous point heat source, and we
assume the initial temperature equal to 0
00:03:16.439 --> 00:03:20.830
.
So, here the instantaneous point heat source
00:03:20.830 --> 00:03:26.459
means at time t equal to 0, we just applying
the amount of the heat within the domain.
00:03:26.459 --> 00:03:33.040
And then we will be getting the temperature
distribution like this . T as a function of
00:03:33.040 --> 00:03:41.629
a variable space R and as a function of
time a small t that we can find out the solution
00:03:41.629 --> 00:03:47.739
is like that; and it is related to all the
parameters that means, that Q, Q is the I
00:03:47.739 --> 00:03:51.930
thinks supply amount of the heat is applied
here .
00:03:51.930 --> 00:03:58.159
And then rho is the density of the material,
C p is a specific heat, and alpha thermal
00:03:58.159 --> 00:04:07.239
diffusivity alpha. And then we can represent
this in such a way that this estimated
00:04:07.239 --> 00:04:14.910
isotherm contour is a series of spheres with
radius R, so a radial distance equal to R,
00:04:14.910 --> 00:04:21.220
so that means, here we are getting the temperature
distribution as a three-dimensional temperature
00:04:21.220 --> 00:04:28.190
distribution with these assumptions. And these
temperatures varies with respect to time
00:04:28.190 --> 00:04:31.750
also.
Now, if we look into other type of analytical
00:04:31.750 --> 00:04:36.840
solution, then infinite body therefore such
that effect of the boundary condition can
00:04:36.840 --> 00:04:41.750
be neglected . Then next assumption since
the instantaneous line heat source that means
00:04:41.750 --> 00:04:48.470
the representation of the heat source is not
having a one specific point rather it is a
00:04:48.470 --> 00:04:55.840
release of the heat generally happens over
over a length over a finite length of the
00:04:55.840 --> 00:05:01.440
line . So, in this case, it is possible to
estimate the two-dimensional temperature distribution
00:05:01.440 --> 00:05:05.330
and of course, assume the initial temperature
equal to 0.
00:05:05.330 --> 00:05:12.720
So, here we can see that isotherm contours
in the this first solution and for the 2 d
00:05:12.720 --> 00:05:17.860
temperature distribution, it is also a series
of cylinders and we can find out the difference
00:05:17.860 --> 00:05:24.120
the similar kind of expression. But here the
power is here equal to 1 and as compared to
00:05:24.120 --> 00:05:28.889
3 by 2 in the which was in the earlier case
when we try to find out the t-dimensional
00:05:28.889 --> 00:05:36.400
temperature distribution in that case. And
that q by rho c p 4 pi alpha t exponential
00:05:36.400 --> 00:05:42.610
minus R square by 4 alpha t, so that way the
temperature variation will be getting on with
00:05:42.610 --> 00:05:48.139
respect to one origin point .
Now, if we assume the instantaneous plane
00:05:48.139 --> 00:05:53.800
heat source, then our solution dimensionality
actually decreases. So, in this case, if we
00:05:53.800 --> 00:05:57.949
assume the plane heat source then we will
be getting the 1D temperature distribution
00:05:57.949 --> 00:06:03.370
. So, if we look into that we when you try
to apply the point heat source, then we will
00:06:03.370 --> 00:06:07.820
be able to find out the temperature distribution
in three-dimensional. Then when we apply the
00:06:07.820 --> 00:06:11.759
line heat source in the 2D two-dimensional
temperature disturbance we can estimate, but
00:06:11.759 --> 00:06:16.259
when we assume the plane heat source that
mean heat source is that mean heat is released
00:06:16.259 --> 00:06:18.560
assuming that is over a plane define plane
.
00:06:18.560 --> 00:06:23.390
So, in this case, we will be able to estimate
the temperature distribution in one-dimensional
00:06:23.390 --> 00:06:27.810
case. So, that temperature distribution can
be estimate almost similar kind of expression,
00:06:27.810 --> 00:06:34.980
but here the x it is different at the exponential
here it is one by two and remaining sense
00:06:34.980 --> 00:06:40.449
and all these cases we are we will be getting
the hm temperature distribution with respect
00:06:40.449 --> 00:06:47.570
to space as well as with respect to time.
And here we can find out that the solution
00:06:47.570 --> 00:06:54.030
represents the isotherm that means, hm constant
temperature profile that represent the isotherm
00:06:54.030 --> 00:06:56.330
contours and it is a consists of the series
of plane.
00:06:56.330 --> 00:07:04.770
So, in this way with this all assumptions,
we can we just simply find out the solution
00:07:04.770 --> 00:07:10.830
of the temperature distribution in case
of welding process. But if you observe that
00:07:10.830 --> 00:07:16.340
all these three solutions that is no effect
of the moving heat source that means, if heat
00:07:16.340 --> 00:07:21.830
source moves we normally we will find out
in in case of the linear welding process,
00:07:21.830 --> 00:07:26.620
where either laser or arc is moved with some
constant velocity. So, there is in the solution
00:07:26.620 --> 00:07:31.680
there is no effect of the velocity only I
will be able to find out the temperature distribution
00:07:31.680 --> 00:07:36.139
over space and time, but no effect of the
velocity .
00:07:36.139 --> 00:07:42.420
The most widely used that hm analytical solution
of temperature distribution in case of welding
00:07:42.420 --> 00:07:48.770
process that comes is the in the in the
from the Rosenthal's equations . And then
00:07:48.770 --> 00:07:54.940
in this case we assume the moving point heat
source on a semi-infinite body semi-infinite
00:07:54.940 --> 00:08:02.139
one of the dimension of the dimension of
the solution domain is finite and remaining
00:08:02.139 --> 00:08:07.849
becomes infinite. So, in this case, for the
semi-infinite body Rosenthal's equation for
00:08:07.849 --> 00:08:12.210
the steady-state heat flow we can find
out the this is the solution.
00:08:12.210 --> 00:08:18.319
But in this case the assumption was lacked
point heat source and that means we assume
00:08:18.319 --> 00:08:25.039
the applied heat source over a point and then
there is no heat losses. And we will in we
00:08:25.039 --> 00:08:31.060
assume the two-dimensional heat flow in the
welding of a very thin sheet of infinite width
00:08:31.060 --> 00:08:38.020
that means width of the sheet is infinite,
but there is a finite dimension of the thickness
00:08:38.020 --> 00:08:42.070
of the plate. So, in that case, you will
be able to find out the temperature distribution
00:08:42.070 --> 00:08:51.430
like this T equal to T 0 plus Q by twice pi
K h and exponential V x by 2 alpha and
00:08:51.430 --> 00:08:57.569
K 0 V R by 2 alpha.
Here if we see that K 0 is actually modified
00:08:57.569 --> 00:09:03.180
vessels function of the second kind and zero
order. So, you can further expand this
00:09:03.180 --> 00:09:08.000
or approximate this case zero value and we
will be able to find out the temperature distribution.
00:09:08.000 --> 00:09:13.450
And here we can see since it is the steady-state
temperature distribution, so these temperature
00:09:13.450 --> 00:09:20.279
is not a function of time component . And
here you see that the there is a effect in
00:09:20.279 --> 00:09:24.870
this solution there is a effect of the velocity;
that means, if we know the velocity of the
00:09:24.870 --> 00:09:29.430
welding then we will more precisely we can
find out that temperature distribution analytically
00:09:29.430 --> 00:09:34.070
when in case of the linear welding process
.
00:09:34.070 --> 00:09:40.500
And here we see that we assume that this velocity
is actually we have defined the velocity
00:09:40.500 --> 00:09:45.790
that means, moving the welding or in one direction
that is in x-directions for when the heat
00:09:45.790 --> 00:09:52.060
source is moving in x-direction then the solution
of the steady-state solution of temperature
00:09:52.060 --> 00:09:58.850
distribution is valid here . Now, if the welding
direction may be y-directional maybe some
00:09:58.850 --> 00:10:03.830
other direction, so accordingly this variable
should change here .
00:10:03.830 --> 00:10:10.060
And R equal to is the radial distance from
the origin, and we see that it is a we will
00:10:10.060 --> 00:10:15.940
be able to find out that different contour
of the two-dimensional temperature distribution
00:10:15.940 --> 00:10:23.080
on the surface or maybe normal you can find
out the temperature distribution on the on
00:10:23.080 --> 00:10:29.500
the surface in this case. Or basically you
can say that two-dimensional temperature distribution
00:10:29.500 --> 00:10:34.780
can be obtained from the solution .
Next, Rosenthal's other solution that is a
00:10:34.780 --> 00:10:39.250
moving part heat source on a semi-infinite
body , but in this case we will be able to
00:10:39.250 --> 00:10:43.769
find out the three-dimensional equation in
a semi-infinite work piece. And of course,
00:10:43.769 --> 00:10:49.000
in this case also the assumption was that
point heat source and no heat loss, then we
00:10:49.000 --> 00:10:54.570
will be able to find out the solution for
the Rosenthal equation. Here you can see that
00:10:54.570 --> 00:11:00.209
the solution is little bit simplified as
computed earlier because earlier there
00:11:00.209 --> 00:11:05.920
is involvement of the vessels function, but
here is not not such . We can find out the
00:11:05.920 --> 00:11:10.300
t dimension temperature distribution , but
the assumptions says that the welding heat
00:11:10.300 --> 00:11:14.850
source is moving along the x-axis with
the velocity v.
00:11:14.850 --> 00:11:21.070
For in that case, this solution is valid and
T 0 was the initial temperature and the Q
00:11:21.070 --> 00:11:27.180
is the applied heat heat flux in this
case and R is the radial distance case the
00:11:27.180 --> 00:11:31.709
thermal contraction . Alpha is the thermal
diffusivity alpha is basically represented
00:11:31.709 --> 00:11:38.890
by K by rho C p means thermal conductivity
wise then ratio of the thermal conductivity
00:11:38.890 --> 00:11:43.790
and multiplication of the density and a specific
heat that is the alpha .
00:11:43.790 --> 00:11:49.950
And here you can see that the solution
of these Rosenthal's equation although in
00:11:49.950 --> 00:11:55.200
the in case of steady-state solution, and
assuming the point heat source that will not
00:11:55.200 --> 00:12:01.170
be able to find out that solution exactly
at the origin of the quadrant. Because the
00:12:01.170 --> 00:12:07.730
singularity problems also arise from this
at this type of solution of steady-state
00:12:07.730 --> 00:12:17.459
equation . So, this with the with this
assumptions this hm the analytical solution
00:12:17.459 --> 00:12:24.110
has been developed, but there are further
development or so many modifications analytically
00:12:24.110 --> 00:12:31.370
by so with the assumptions and then complicated
equations we can get from this analytical
00:12:31.370 --> 00:12:36.870
solution of the temperature distribution and
that developed over the time .
00:12:36.870 --> 00:12:45.740
But nowadays there is a option to find
out the solution of the temperature distribution
00:12:45.740 --> 00:12:53.279
numerically. And by either this numerical
solution can be obtained some readymade
00:12:53.279 --> 00:12:58.480
available software, commercial software or
can be using some developed code also that
00:12:58.480 --> 00:13:04.260
more is using different a numerical methods
for example, finite element, finite difference
00:13:04.260 --> 00:13:10.760
and finite volume . The more obviously, we
can getting the temperature distribution which
00:13:10.760 --> 00:13:17.430
is who is assembles the more realistic to
the actual welding process. And that means,
00:13:17.430 --> 00:13:22.760
in welding process whatever physics we can
find out, if we follow these things this mode
00:13:22.760 --> 00:13:29.250
is more scientifically defined the problem
which is more realistic with the practical
00:13:29.250 --> 00:13:33.920
what we observe during the welding process
if we follow the in numerical solution than
00:13:33.920 --> 00:13:36.250
with respect to the any kind of analytical
solution.
00:13:36.250 --> 00:13:41.630
But still analytical solution is handy in
the sense the quickly we can estimate the
00:13:41.630 --> 00:13:47.940
temperature distribution and rough estimation
of the cooling rate phenomena also ok estimation
00:13:47.940 --> 00:13:53.029
also possible using the analytical solution
. So, let us see that how we can estimate
00:13:53.029 --> 00:13:59.170
the cooling rate and temperature gradient
from Rosenthal's three-dimensional solution
00:13:59.170 --> 00:14:03.540
of the temperature distribution .
So, you know that this is the temperature
00:14:03.540 --> 00:14:11.410
distribution from the Rosenthal's equation.
And from here we can simply estimate the cooling
00:14:11.410 --> 00:14:20.290
rate. So, first we assume that it is a variable
is the x t v, x t, small t, capital T, temperature,
00:14:20.290 --> 00:14:27.579
time and the space are also variable quantity
here . Now, first if we see that do x the
00:14:27.579 --> 00:14:33.500
partial derivative of distance x with respect
to time t that represents the velocity vector,
00:14:33.500 --> 00:14:38.730
because we assume that in this solution the
heat source is moving along the x-axis. So,
00:14:38.730 --> 00:14:48.090
along that when it is moving along the x-axis,
so this expression that x-axis with that change
00:14:48.090 --> 00:14:54.839
the derivative of x with respect to time
component t that represents actually the velocity
00:14:54.839 --> 00:14:58.270
.
Now, if along the x-axis if we follow along
00:14:58.270 --> 00:15:03.170
the x-axis, then we can put this condition
y equal to 0 and z equal to 0 that represents
00:15:03.170 --> 00:15:09.770
the heat source moving is along x-axis . So,
if we put y equal to 0, z equal to 0, then
00:15:09.770 --> 00:15:19.700
R becomes x, because R was the radial distance.
So, in this equation if we put y equal to
00:15:19.700 --> 00:15:25.750
0 and z equal to 0, then R becomes x. So,
in that sense R becomes x . Now, if R becomes
00:15:25.750 --> 00:15:33.870
x from this 3D solution we can find out
T minus T 0 is equal to Q by twice pi K
00:15:33.870 --> 00:15:39.459
x this expression .
Next, we can find out the temperature gradient
00:15:39.459 --> 00:15:47.290
temperature gradient that means that basically
slope that represents when we keeping the
00:15:47.290 --> 00:15:53.220
variable time is a constant, and this dou
t by dou x that represents the temperature
00:15:53.220 --> 00:16:02.050
gradient, so that temperature gradient diagonally
can be estimated from this equation that minus
00:16:02.050 --> 00:16:11.390
Q by twice pi K x square. And then if we put
it replace the value of variable x this equation,
00:16:11.390 --> 00:16:16.160
we will be able to find out in terms of the
temperature difference and other constant
00:16:16.160 --> 00:16:21.959
parameters, for example, thermal conductivity
and the Q is the heat input.
00:16:21.959 --> 00:16:29.579
So, now the temperature gra[dient] a temperature
sorry that cooling cooling rate, so change
00:16:29.579 --> 00:16:36.480
of temperature with respect to time that
represents that two variable here the dou
00:16:36.480 --> 00:16:43.270
t by dou x first variable another is the dou
x this is the partial derivative of this
00:16:43.270 --> 00:16:49.529
variable. And from the personal derivative
we can express that this cooling rate is
00:16:49.529 --> 00:16:53.061
these two components.
And then if we put these values, we will be
00:16:53.061 --> 00:16:57.709
able to find out the cooling rate can be estimated
like this which incorporated the effect of
00:16:57.709 --> 00:17:06.199
the velocity temperature as well as the heat
input. Now, this is the straightforward
00:17:06.199 --> 00:17:11.500
to estimate the cooling rate when the heat
source is moving along the x-axis so that
00:17:11.500 --> 00:17:16.490
means, the assumptions was that y equal to
z equal to 0. So, in this line this is the
00:17:16.490 --> 00:17:23.159
cooling rate .
So, along one line the along the line which
00:17:23.159 --> 00:17:29.280
in which direction the heat source is moving.
So, in that direction, we can find out the
00:17:29.280 --> 00:17:34.840
cooling rate and that cooling from this cooling
rate we can say that the cooling rate is reduced
00:17:34.840 --> 00:17:41.110
significantly by preheating. Preheating means
if we heat the the T 0 actually the initial
00:17:41.110 --> 00:17:44.260
temperature.
So, normally initial temperature start with
00:17:44.260 --> 00:17:50.090
the ambient temperature, but if the preheater
sample this initial temperature should be
00:17:50.090 --> 00:17:54.620
replaced by the preheat temperature, so
that preheat temperature is very high that
00:17:54.620 --> 00:18:01.200
difference reduces, so in that sense that
the cooling rate is actually reduces. So,
00:18:01.200 --> 00:18:08.110
by simply by doing that preheating process
so that means, the cooling rate cooling
00:18:08.110 --> 00:18:15.000
rate can be reduced significantly by preheating
by the one way to practically doing the some
00:18:15.000 --> 00:18:18.440
preheating process.
Next, if we look into that cooling rate also
00:18:18.440 --> 00:18:24.060
can be decreased with increasing Q by V, Q
by V actually represents the heat input per
00:18:24.060 --> 00:18:29.740
unit length. If you see Q is the heat input
and v is the velocity. So, if there that can
00:18:29.740 --> 00:18:36.110
be represented by some in SI unit joule per
meter ; that means, we can say it is a unit
00:18:36.110 --> 00:18:41.370
of the heat input per unit length. So, that
is the one characteristic parameter and that
00:18:41.370 --> 00:18:47.039
is in specific to the very thin state welding
process. So, from here you can conclude that
00:18:47.039 --> 00:18:54.400
the cooling rate can be decreases with increasing
that Q by V value ok. So, Q by V value
00:18:54.400 --> 00:19:01.840
increasing Q by V value either increasing
heat input more heat input hm or by decreasing
00:19:01.840 --> 00:19:05.970
the value of the velocity.
So, therefore, temperature gradient another
00:19:05.970 --> 00:19:13.049
conclusion we can say the temperature gradient
decreases with increasing with increasing
00:19:13.049 --> 00:19:19.110
Q . So, here we can link this equation and
here you can find out if we increasing Q,
00:19:19.110 --> 00:19:27.440
then temperature gradient also a decreases.
So, this kind of conclusion can be drawn
00:19:27.440 --> 00:19:33.400
simply just by estimating the cooling rate.
And from the analytical solution of the temperature
00:19:33.400 --> 00:19:40.470
distribution. And that analogy can be linked
with the actual process condition or actually
00:19:40.470 --> 00:19:45.520
physically when you try to do some welding
process, we can analyze just looking into
00:19:45.520 --> 00:19:51.190
these kind of parameters say whether we can
increase or decrease in the rate of cooling
00:19:51.190 --> 00:19:54.770
during the welding process. Because this cooling
rate is very much significant parameter in
00:19:54.770 --> 00:20:00.510
this solidification process and final microstructure
is actually influenced by the rate of cooling
00:20:00.510 --> 00:20:05.290
or whether we can say decided by the more
on the rate of the cooling process. So, this
00:20:05.290 --> 00:20:09.700
cooling rate is very important parameter to
analyze the welding process .
00:20:09.700 --> 00:20:17.450
Now, some elemental calculation need
to understand during the heat transfer in
00:20:17.450 --> 00:20:23.790
welding process. So, here you can see that
arc power, so arc power; that means, suppose
00:20:23.790 --> 00:20:29.150
if we assume some arc welding process, so
arc power can be estimated from the arc voltage
00:20:29.150 --> 00:20:34.120
from the machine direct itself which we measured
the arc voltage . And what is the welding
00:20:34.120 --> 00:20:38.500
current we supply based on that we can estimate
the arc power, this is multiplication of the
00:20:38.500 --> 00:20:43.990
voltage into current .
Now, directly we can measure this is in case
00:20:43.990 --> 00:20:50.490
of arc , but in case of laser welding process
directly we can measure the laser power from
00:20:50.490 --> 00:20:56.350
the machine reading itself . And so these
are the input power you actually supplied
00:20:56.350 --> 00:21:02.710
in in case of the welding process, but net
energy which is supplied for the heating to
00:21:02.710 --> 00:21:09.710
the work is that may not be the what is the
input power. So, some uh then then we can
00:21:09.710 --> 00:21:15.470
introduce some efficiency term also that is
called the thermal weld thermal efficiency
00:21:15.470 --> 00:21:18.760
because during to the transmission of the
arc or laser power to the work piece, there
00:21:18.760 --> 00:21:22.950
may be some loss of the heat to the surroundings.
So, if we look into that if you take into
00:21:22.950 --> 00:21:28.860
that account then we can say that Q is the
amount of the energy actually goes to the
00:21:28.860 --> 00:21:36.160
work piece to the material . So, then we use
this efficient term such that efficiency less
00:21:36.160 --> 00:21:45.660
than 1, such that less amount of energy
that is the arc Q 0 is basically the arc power.
00:21:45.660 --> 00:21:50.980
And that and when arc power multiplied by
this weld thermal efficiency that is the
00:21:50.980 --> 00:21:57.460
actual amount of the energy that is actually
goes into the work piece or substrate material
00:21:57.460 --> 00:22:02.590
. Then we can define our other term also that
is called the melting efficiency . So, melting
00:22:02.590 --> 00:22:08.299
efficiency we define that not all the supplied
power is basically utilized for the melting
00:22:08.299 --> 00:22:15.440
purpose w we can divide their total supplied
heat to the work piece is the Q. So, Q may
00:22:15.440 --> 00:22:21.390
be one is the Q 1, another is the amount of
the Q 2 .
00:22:21.390 --> 00:22:27.230
So, Q 1 amount of the heat is basically we
can say that is utilized for the total amount
00:22:27.230 --> 00:22:34.710
of the heat required to melt the weld metal
up to the melting point temperature and that
00:22:34.710 --> 00:22:38.340
also includes the phase transformation from
solid to liquid phase. So, that amount of
00:22:38.340 --> 00:22:45.210
the heat latent heat also in heat accounts.
So, then if we amount if you separate out
00:22:45.210 --> 00:22:50.450
this is the amount of the total amount of
the heat, you could just to melt and that
00:22:50.450 --> 00:22:56.580
Q 2 is the remaining the maybe you can say
it is not necessary the supply heat which
00:22:56.580 --> 00:23:02.740
only just melt the substrate material,
but it achieves some superheated condition
00:23:02.740 --> 00:23:07.299
that means, from melting point to the maximum
temperature that is the superheat temperature
00:23:07.299 --> 00:23:12.280
so that means, liquid overheat temperature
overheated and the part of the heat is also
00:23:12.280 --> 00:23:16.230
conducted air to the surroundings and that
can be considered as a loss also.
00:23:16.230 --> 00:23:22.679
So, the Q 2 this account that overheating
overheating means above melting point temperature
00:23:22.679 --> 00:23:29.400
plus the which amount of the heat is conducted
to the surroundings that accounts that amount
00:23:29.400 --> 00:23:36.370
is accounted by this Q 2 . So, therefore,
based on that we can find out the melting
00:23:36.370 --> 00:23:43.090
efficiency is basically that Q 1 the in this
case the to the substrate material Q 1 is
00:23:43.090 --> 00:23:48.330
actually used just melting the material and
with respect to the what is supplied to the
00:23:48.330 --> 00:23:52.429
work piece Q and that ratio can be defined
as a melting efficiency.
00:23:52.429 --> 00:24:00.300
We can estimate that Q 1 also in another way
that Q 1 suppose the material temperature
00:24:00.300 --> 00:24:05.890
initial temperature was T 0 and it raised
to the melting temperature T m by the application
00:24:05.890 --> 00:24:13.299
of the energy from the arc or laser. So, to
increase this temperature from T 0 to 2 M
00:24:13.299 --> 00:24:19.930
plus to change the phase the latent heat,
so if we that amount of the heat can be estimate
00:24:19.930 --> 00:24:23.419
in this way.
So, that is this amount of the heat just to
00:24:23.419 --> 00:24:27.130
raise the from initial temperature to the
melting point temperature, and this among
00:24:27.130 --> 00:24:35.890
this is this accounts the change of the
phase from solid from from solid phase
00:24:35.890 --> 00:24:41.880
to the liquid phase . And then this is the
amount of this cross sectional area, and the
00:24:41.880 --> 00:24:47.950
over the cross sectional area and normal to
the cross sectional area, V is the velocity
00:24:47.950 --> 00:24:53.909
of the welding source. So, that account the
that where V into A that may be volume
00:24:53.909 --> 00:25:00.409
flow rate. And if we multiply by the density
then it becomes mass flow rate; and that mass
00:25:00.409 --> 00:25:08.159
per unit time this is the this accounts the
mass and the remaining time is the heat accounts
00:25:08.159 --> 00:25:11.890
are increment of the temperature.
So, in this way we can estimate the total
00:25:11.890 --> 00:25:19.919
amount of the energy that is actually utilized
to just melt the substrate material. So,
00:25:19.919 --> 00:25:24.480
in this way just simplified we can separate
out the different amount of energy utilized
00:25:24.480 --> 00:25:28.280
into which purpose; and based on that we can
define different efficiencies for things.
00:25:28.280 --> 00:25:33.060
For example, we can define the well thermal
efficiency we can define the melting efficiency
00:25:33.060 --> 00:25:38.529
also in these two ways.
So, what are we can say that actual heat supplied
00:25:38.529 --> 00:25:56.400
that is the Q and is a arc power equal
to that Q. So, arc power Q 0 and that Q
00:25:56.400 --> 00:26:02.380
from Q to Q 0, so so there is some loss of
the heat and that during when the there is
00:26:02.380 --> 00:26:08.279
a transmission of the arc part to the work
piece and is Q 0. Then Q 0 to some out of
00:26:08.279 --> 00:26:15.539
this key one is utilized for the melting
process. So, what is the actual actual separate
00:26:15.539 --> 00:26:23.279
energy is Q by the from the machines itself
and then to melt the substrate metal we utilization
00:26:23.279 --> 00:26:29.880
of the Q 1, so that means, definitely the
this utilization will be very less as compared
00:26:29.880 --> 00:26:36.590
to the q and that it can be linked between
all this Q, Q 0, Q in terms in terms of defining
00:26:36.590 --> 00:26:42.539
the different efficiency term .
Now, apart from this analytical solution some
00:26:42.539 --> 00:26:49.070
basic elemental calculation of the heat transfer
accounts the amount of the energy transfer
00:26:49.070 --> 00:26:54.529
to the substance material to this thing
. So, finally, we are interested to know that
00:26:54.529 --> 00:26:59.570
what is the temperature distribution in the
in the substrate metal in the work piece material
00:26:59.570 --> 00:27:06.549
. Because that is temperature distribution
actually helps to get the idea that what are
00:27:06.549 --> 00:27:11.900
the cooling rate, we can estimate the cooling
rate also from this temperature distribution.
00:27:11.900 --> 00:27:16.450
And then some kind of methodological point
of view we can explain the we can predict
00:27:16.450 --> 00:27:21.299
the structure solidified structure or maybe
micro structure.
00:27:21.299 --> 00:27:27.020
So, to get this kind of knowledge we need
to know or maybe other way also; that means,
00:27:27.020 --> 00:27:31.840
if we know the heat transfer temperature distribution,
we can we can predict there what was the size
00:27:31.840 --> 00:27:36.460
of the molten pool, and what was the size
of the heat affected zone. All kind all kind
00:27:36.460 --> 00:27:42.660
of this quantitative information is possible
to know only to if we know the temperature
00:27:42.660 --> 00:27:48.289
distribution actually happens during the welding
process, so that is why in this case it is
00:27:48.289 --> 00:27:53.140
necessary to solve this governing equation
basically the heat conduction equation.
00:27:53.140 --> 00:27:57.419
And then we will be able to find out the temperature
isotherm based on the temperature isotherm
00:27:57.419 --> 00:28:01.990
that means, if we for a specific material
if we know the solidus temperature liquidus
00:28:01.990 --> 00:28:07.310
temperature and this temperature. And we if
we know the some phase transforms at temperature
00:28:07.310 --> 00:28:13.919
so that that typical temperature actually
define the different zone in the weld pool
00:28:13.919 --> 00:28:16.830
mathematically.
So, let us look into this that how we can
00:28:16.830 --> 00:28:24.029
do this thing. Just before that some idea
that this this graph actually shows the simulation
00:28:24.029 --> 00:28:29.610
of the laser spot welding that means, laser
spot welding. So, first we define the geometry,
00:28:29.610 --> 00:28:33.799
and the there is application of the laser
source from and with respect to time this
00:28:33.799 --> 00:28:38.770
is a development of the weld pool or and heat
affected zone. So, is a very initial time
00:28:38.770 --> 00:28:44.830
it is a fast heat time is very small then
weld size is small, and gradually continuously
00:28:44.830 --> 00:28:51.270
if there is just supply of the laser energy
to the work piece metal then gradually the
00:28:51.270 --> 00:28:54.549
weld pool actually develop.
So, initial we start is the very small weld
00:28:54.549 --> 00:28:59.970
pool and finally, we are getting the very
big weld pool; at the same time we can estimate
00:28:59.970 --> 00:29:05.149
the heat affected zone by the different
color. So, in this picture, the red color
00:29:05.149 --> 00:29:10.610
actually represents the weld pool dimension;
that means, the red color represents the this
00:29:10.610 --> 00:29:17.679
is a molten state of the work piece material.
And then other colors define there is a solid
00:29:17.679 --> 00:29:23.669
state phase transformation happens and that
that depends on the temperature distribution.
00:29:23.669 --> 00:29:28.610
So, that kind of information from the simulation
also we can find out and that actually resembles
00:29:28.610 --> 00:29:35.559
to the actual welding process .
But what are the typical difficulties or
00:29:35.559 --> 00:29:41.901
maybe issues when you try to mathematically
solve this equation in case of welding process
00:29:41.901 --> 00:29:47.140
. So, governing equation straight forward,
the heat conduction equations with having
00:29:47.140 --> 00:29:53.330
some internal heat generation term, so that
is the very basic governing heat conduction
00:29:53.330 --> 00:29:58.059
equation hm in this case .
But when you try to solve the conduction based
00:29:58.059 --> 00:30:03.630
model we try to develop the conduction based
model inclusion welding process, so we just
00:30:03.630 --> 00:30:09.020
simply solve the heat conduction equation
. Though assumptions was like that convective
00:30:09.020 --> 00:30:14.820
flow of the liquid metal is neglected here
that means, we know that within the liquid
00:30:14.820 --> 00:30:19.500
molten pool is not the liquid molten pool
may not be very still because there are several
00:30:19.500 --> 00:30:24.630
driving force, there is the arc pressure and
there is a there must be some flow of the
00:30:24.630 --> 00:30:29.140
material within the weld pool, so that means,
that convective flow of the liquid metal simply
00:30:29.140 --> 00:30:35.720
we are neglecting in case of conduction based
model in welding process.
00:30:35.720 --> 00:30:41.659
Then sometimes you represents the that
when you discuss that the representation of
00:30:41.659 --> 00:30:45.090
the heat source we can represent the heat
source in terms of the surface heat flux;
00:30:45.090 --> 00:30:50.559
that means, only on the surface the heat flux
is supplied or sometimes we represent the
00:30:50.559 --> 00:30:54.160
heat flux its a volumetric heat flux that
means, heat is applied and the representation
00:30:54.160 --> 00:30:58.540
in the mathematical model that happens over
a defined volume of this thing.
00:30:58.540 --> 00:31:02.950
Now, if it is surface heat flux we can step
for it in the Gaussian follow the Gaussian
00:31:02.950 --> 00:31:07.879
distribution and put through the boundary
condition, we can interact the circuit
00:31:07.879 --> 00:31:12.980
we can put as a boundary condition through
the surface that the surface heat flux .
00:31:12.980 --> 00:31:18.670
But when we assume the volumetric heat heat
flux and if we look into the actual governing
00:31:18.670 --> 00:31:23.600
equation of the heat conduction, here you
can see that Q dot represents the internal
00:31:23.600 --> 00:31:30.480
heat generation . So, this internal heat generation
is practically in welding process, there is
00:31:30.480 --> 00:31:36.919
no internal heat generation , but it can be
utilized that to incorporate the volumetric
00:31:36.919 --> 00:31:42.640
heat through this term and that is a usual
procedure to incorporate the volumetric heat
00:31:42.640 --> 00:31:46.909
through this internal heat generation term
in this equation .
00:31:46.909 --> 00:31:51.870
So, physically internal heat generation term
in that Q dot represents the physically the
00:31:51.870 --> 00:31:58.549
internal heat generation , but in case of
welding simulation we consider as we use this
00:31:58.549 --> 00:32:05.059
term to incorporate the volumetric heat. And
if we solve and then if we solve this equation
00:32:05.059 --> 00:32:10.980
with the proper boundary condition, then finally,
we will be able to get the temperature distribution
00:32:10.980 --> 00:32:16.940
that is shown in this in this figure the
temperature it it simply shows the temperature
00:32:16.940 --> 00:32:20.700
distribution in case of the spot welding;
that means, in case of the stationary heat
00:32:20.700 --> 00:32:24.040
source.
But what are the issues that we have already
00:32:24.040 --> 00:32:28.600
discussed the representation heat source,
the issues is that first this is the difficulty
00:32:28.600 --> 00:32:36.130
in defining the volumetric heat a priori that
means, it is very difficult to define the
00:32:36.130 --> 00:32:42.159
volumetric heat source before the start of
the simulation. So, most of the cases what
00:32:42.159 --> 00:32:47.700
we define the this heat source term is defined
looking in the experimental values and the
00:32:47.700 --> 00:32:53.160
experimentally measure the length and width.
And based on that we can we can map these
00:32:53.160 --> 00:32:58.120
things with the volumetric heat source term
so that is the one way .
00:32:58.120 --> 00:33:03.360
And other way also if we incorporate this
volumetric heat source term in in some adaptive
00:33:03.360 --> 00:33:08.130
manner, so that uh basically if we can do
this thing in volumetric users you know in
00:33:08.130 --> 00:33:15.880
terms of the adaptive way such that over the
time or what the load step there is a mapping
00:33:15.880 --> 00:33:23.440
of the weld dimension with the volumetric
heat source stuff. So, that there is a continuous
00:33:23.440 --> 00:33:27.590
change of the these dimension that means,
the volumetric heat source term, there is
00:33:27.590 --> 00:33:33.640
a continuous updating of the this heat source
term with respect to time and that can be
00:33:33.640 --> 00:33:36.340
linked with the actual simulation of the temperature
distribution .
00:33:36.340 --> 00:33:43.140
So, this mapping can be done in any in in
different types of the geometric shape, in
00:33:43.140 --> 00:33:48.960
case in can be conical it can be ellipsoidal
it can be double ellipsoidal, so all this
00:33:48.960 --> 00:33:55.159
kind of geometric shape this can be mapped.
So, there is a some specific mapping
00:33:55.159 --> 00:34:00.510
techniques is also required to map with this
parameter that more relate to the representation
00:34:00.510 --> 00:34:02.690
of the heat source term that we have already
discussed.
00:34:02.690 --> 00:34:08.510
Now, we tried to look into that other simulation
in case of the autogenous fusion welding process
00:34:08.510 --> 00:34:16.210
that mean without any application of the consumable
electrode. So, in that case, so we generally
00:34:16.210 --> 00:34:20.520
solve the heat transfer heat conduction equation
to find out the temperature distribution.
00:34:20.520 --> 00:34:25.500
So, first figure if you see linear welding
that means, for a moving heat source and in
00:34:25.500 --> 00:34:32.640
this case at any instant of time or maybe
at instant of time, if he represents that
00:34:32.640 --> 00:34:39.609
at a specific point if the heat source at
time t, the heat source is positioned on specific
00:34:39.609 --> 00:34:44.820
point. And that point the temperature distribution
is shows in the first figure.
00:34:44.820 --> 00:34:51.730
Here you can see that we can define the mesh
that means in using the finite because this
00:34:51.730 --> 00:34:56.520
simulation has been done using the finite
element method . So, by defining the mess,
00:34:56.520 --> 00:35:01.130
and at one instant of time that there is application
of the heat flux and this is the typical distribution.
00:35:01.130 --> 00:35:07.460
So, red color basically represents the fusion
zone, and remaining colors represents the
00:35:07.460 --> 00:35:14.280
heat affected zones. The color body is also
defined here that so that 973, 1273, we define
00:35:14.280 --> 00:35:19.599
we assume that this temperature elects to
some kind of phase transformation happens.
00:35:19.599 --> 00:35:25.990
And based that that critical temperatures
or that critical isotherm define certain heat
00:35:25.990 --> 00:35:31.850
affected zone in case of the welding process.
So, the current was 180 amps, voltage was
00:35:31.850 --> 00:35:38.000
was 14.6 volt and speed then means welding
piece was 4.4 millimeter per second . So,
00:35:38.000 --> 00:35:42.710
in that condition, we can find out the temperature
distribution is like this. And we can predict
00:35:42.710 --> 00:35:49.030
the weld pool diamonds and that means, weld
pool size at this welding condition. So, we
00:35:49.030 --> 00:35:54.170
have done the simulation without what we have
done in case of practical cases. And we compare
00:35:54.170 --> 00:36:00.130
with respect the experimentally evaluated
the weld pool dimension and com computationally
00:36:00.130 --> 00:36:04.970
evaluated weld dimension.
So, the red colored fusion zone and this side
00:36:04.970 --> 00:36:10.230
the original that is it is a fusion zone
this represents a fusion boundary so more
00:36:10.230 --> 00:36:17.390
or less it is matching with the experimental
and the numerical model data for a specific
00:36:17.390 --> 00:36:23.359
welding condition. So, this way this is the
way to analyze the actual welding process
00:36:23.359 --> 00:36:30.960
what happens in practically that can be represented
in terms of the some mathematical model, and
00:36:30.960 --> 00:36:35.890
do some computational calculation, we can
represent the similar kind of phenomena what
00:36:35.890 --> 00:36:41.560
we generally observe during the welding process.
So that means by solving the temperature distribution
00:36:41.560 --> 00:36:47.920
only we can define that temperature distribution
with respect to time or temperature distribution
00:36:47.920 --> 00:36:54.960
over the space or and that temperature control
also define the different zones in the weld
00:36:54.960 --> 00:37:00.859
pool weld zone and that and same phenomena
when you do the experiment after doing the
00:37:00.859 --> 00:37:04.950
welding experiment we cut the sample do the
polishing and then after the after the etching
00:37:04.950 --> 00:37:09.800
solution, we will be able to find out that
what was the well dimension, one specific
00:37:09.800 --> 00:37:14.940
welding process parameters, so that welding
process parameters if we try to simulate will
00:37:14.940 --> 00:37:20.119
be getting the almost similar if we
do the very good simulation we will be able
00:37:20.119 --> 00:37:26.301
to find the represents welding condition.
So, the compression represents that if there
00:37:26.301 --> 00:37:31.390
is a more small changes in the process parameters
not necessary to always do the experiment,
00:37:31.390 --> 00:37:37.930
repeat the experiment again. Another we can
rely on the simulation of this process. And
00:37:37.930 --> 00:37:43.370
using simple is a computer simulation, we
can predict the weld pool size, a heat affected
00:37:43.370 --> 00:37:48.380
zone by doing the simple only the thermal
simulation .
00:37:48.380 --> 00:37:54.330
Similarly, this simulation shows that of
the micro laser spot welding process that
00:37:54.330 --> 00:37:59.670
means micro laser spot welding process we
can find out the metal equal to SS304 stainless
00:37:59.670 --> 00:38:08.820
steel. Laser power was used only 75 watt;
laser type was the continuous fiber laser
00:38:08.820 --> 00:38:14.089
practical and experimentally. And thickness
of the sample was only 100 micrometer.
00:38:14.089 --> 00:38:20.550
So, in using the commercial software, uh
in this case the simulation has been performed,
00:38:20.550 --> 00:38:28.770
and we can find out the from the top surface
of the weld pool, weld is the this representation
00:38:28.770 --> 00:38:33.340
of these things . And of course, here we can
we have solved the symmetric problem that
00:38:33.340 --> 00:38:39.540
means, not necessary in computationally
to solve the temperature distribution for
00:38:39.540 --> 00:38:44.720
the whole domain which which can we can reduce
the computational time depending about the
00:38:44.720 --> 00:38:51.579
symmetric nature of the problem. Only half
of this half part of this can be used
00:38:51.579 --> 00:38:57.200
to do the to do the simulation and to avoid
or to reduce the computational time.
00:38:57.200 --> 00:39:02.510
Here you can see that this is the top surface
profile on the weld pool, and this is the
00:39:02.510 --> 00:39:07.990
bottom surface, so that is done from the simulation.
And here the cross sectional profile is like
00:39:07.990 --> 00:39:13.829
that in case of micro laser spot welding process.
And finally, if we compare these things
00:39:13.829 --> 00:39:18.070
we can see there is a comparison between the
experimental value and the numerical value
00:39:18.070 --> 00:39:23.170
this is a good comparison that actually brings
the reliability of the numerical model what
00:39:23.170 --> 00:39:29.200
we have we generally follow.
So, another simulation the thermal simulation
00:39:29.200 --> 00:39:37.560
micro laser spot welding process, but in this
case I think the in initial 75, but in
00:39:37.560 --> 00:39:44.320
laser power was less even for 25 watt . So,
with this laser power , so here we can see
00:39:44.320 --> 00:39:50.440
this is a top surface profile and the thermal
profile that means, a red color is here indicates
00:39:50.440 --> 00:39:56.000
the fusion zone. And here you can see the
cross sectional view also that it is not fully
00:39:56.000 --> 00:40:01.430
penetrated so that means, partially penetrated.
So, that means, here if you use the laser
00:40:01.430 --> 00:40:06.700
25 watt laser for only there is a partial
penetration of the profile. But if we increase
00:40:06.700 --> 00:40:13.910
the laser power, this this profile will increase
a with the seventy 75 watt laser power we
00:40:13.910 --> 00:40:18.800
can get this kind of profile that means, full
depth full depth of penetration can be achieved.
00:40:18.800 --> 00:40:25.070
So, although without doing the analysis
or maybe without doing the experimental things,
00:40:25.070 --> 00:40:29.770
so if we in this case for example, once you
validate on numerical model, we can further
00:40:29.770 --> 00:40:34.930
use it for the simulation of the different
process parameters. And we can predict
00:40:34.930 --> 00:40:40.500
some behavior just simply basing just simply
doing is the simulation. So, here is the advantage
00:40:40.500 --> 00:40:48.079
of the computer simulation .
Now, since we have talked about the heat
00:40:48.079 --> 00:40:52.790
conduction equation. And from the heat conduction
equation, we can incorporate the temperature
00:40:52.790 --> 00:40:57.641
distribution, and that temperature distribution
can be used for the prediction of the weld
00:40:57.641 --> 00:41:02.660
pool dimensions and maybe we can estimate
the cooling rate from this temperature time
00:41:02.660 --> 00:41:06.990
temperature cycle and that cooling that can
be incorporated the microstructure all this
00:41:06.990 --> 00:41:12.450
phenomena.
But often to do this calculation, so we generally
00:41:12.450 --> 00:41:17.990
use most of the cases the heat conduction
based model that means, within the weld pool
00:41:17.990 --> 00:41:24.210
there is a flow of the molten pool and
that flow of the flow of flow velocity
00:41:24.210 --> 00:41:30.360
that means, flow field within the molten pool
that actually modify the the weld pool shape
00:41:30.360 --> 00:41:36.890
and size, so that thing is that velocity
field within the small weld pool we often
00:41:36.890 --> 00:41:40.970
neglect it .
Because because of the reason is because of
00:41:40.970 --> 00:41:47.579
that this incorporation of the metal flow
within the small weld pool is tremendously
00:41:47.579 --> 00:41:54.520
increase the computational cost. But at the
same time if we incorporate the material flow
00:41:54.520 --> 00:42:00.230
within the small weld pool. So, we more accurately
we can predict the temperature distribution
00:42:00.230 --> 00:42:05.020
and although it is computationally expensive
.
00:42:05.020 --> 00:42:10.930
Let us see what is that why the fluid flow
is also important to analyze in case of small
00:42:10.930 --> 00:42:16.140
weld pool and what is the influence of our
impact of this consideration of the metal
00:42:16.140 --> 00:42:22.040
flow specifically in case of the in case
of the welding simulation . So, here you can
00:42:22.040 --> 00:42:28.790
see that that in which you want to solve the
fluid flow field or material flow field within
00:42:28.790 --> 00:42:33.370
the small weld pool into consider transport
phenomena base heat transfer and fluid flow
00:42:33.370 --> 00:42:36.140
model you need to incorporate or we need to
consider .
00:42:36.140 --> 00:42:42.740
So, here extra fluid flow is comes from the
due to the momentum transport due to the three
00:42:42.740 --> 00:42:47.800
military different driving forces. So, one
is the surface tension force that surface
00:42:47.800 --> 00:42:52.460
tension put in entirely depends the material
specific that type of the material whether
00:42:52.460 --> 00:42:57.890
there exist any kind of surface active elements
within the within the material or not . And
00:42:57.890 --> 00:43:02.980
that actually interface between the in arc
welding or maybe in the two different medium
00:43:02.980 --> 00:43:08.349
that means liquid and the gas medium in the
interface, there is a acting of the surface
00:43:08.349 --> 00:43:12.550
tension force.
Next is the and but it is necessary to mention
00:43:12.550 --> 00:43:16.900
that surface tension force is mainly acting
on the over the surface . And other as the
00:43:16.900 --> 00:43:22.990
buoyancy force or an electromagnet, so buoyancy
force can comes into the picture due to the
00:43:22.990 --> 00:43:27.770
hm a temperature difference between the top
and the bottom layer.
00:43:27.770 --> 00:43:31.619
So, the presence of the temperature gradient
actually brings the buoyancy force that buoyancy
00:43:31.619 --> 00:43:37.460
force is actually distributed over the volume
of the molten pool , but there is another
00:43:37.460 --> 00:43:41.900
dragging force which is responsible for the
material movement within the weld pool that
00:43:41.900 --> 00:43:47.760
is the electromagnetic force. But if we use
a electric current especially in case of the
00:43:47.760 --> 00:43:53.140
arc welding process, this electromagnetic
force is more actually impact on the metal
00:43:53.140 --> 00:43:58.930
flow or act as a driving force .
But this electromagnetic force is distributed
00:43:58.930 --> 00:44:04.619
over the volume of the molten weld pool , but
if in case of laser welding normally doing
00:44:04.619 --> 00:44:09.150
we do not find out the electromagnetic force
field. So, in case of laser welding it is
00:44:09.150 --> 00:44:14.210
sufficient to consider only the surface tension
force and the buoyancy force as a driving
00:44:14.210 --> 00:44:21.160
force for the momentum transport within the
molten pool that means that actually decides
00:44:21.160 --> 00:44:25.800
that these two type forces in case of laser
welding actually decides the velocity field
00:44:25.800 --> 00:44:31.160
within the small weld pool .
So, of course, when you in case of transport
00:44:31.160 --> 00:44:34.950
phenomena based heat transfer fluid flow mold,
if we want to use it then it is necessary
00:44:34.950 --> 00:44:41.869
to solve the conservation of mass momentum
and energy equations which is as compared
00:44:41.869 --> 00:44:47.690
to the conduction based model. There was only
to solve the only the energy equations.
00:44:47.690 --> 00:44:52.060
And from the energy equation, we will be getting
the output as a on the temperature field.
00:44:52.060 --> 00:44:57.100
But in this case we will be getting the output
as a velocity field as well as the temperature
00:44:57.100 --> 00:45:02.680
field within the small weld pool as well
as the heat effect regime . But it is it is
00:45:02.680 --> 00:45:08.500
also important to know that these velocity
field will be able to predict in case of the
00:45:08.500 --> 00:45:12.410
welding process within the small weld pool
also, but temperature field will be able to
00:45:12.410 --> 00:45:19.140
predict throughout the whole domain not only
confined into the small weld pool .
00:45:19.140 --> 00:45:24.670
Now, to solve this heat transfer and fluid
flow problem in specific the welding process
00:45:24.670 --> 00:45:35.570
of course, if we try to there is
a limited solution available for the analytically,
00:45:35.570 --> 00:45:42.339
so it is more easy convenient to do the numerical
solution of this problem to get the actual
00:45:42.339 --> 00:45:47.130
effect of this kind of physical phenomena
in case of the welding process, so that is
00:45:47.130 --> 00:45:53.630
why I am directly shifting to the numerical
solution of this problem So, by avoiding
00:45:53.630 --> 00:45:57.930
the and any kind of analytical solution in
case to find out the fluid flow field in case
00:45:57.930 --> 00:46:03.140
of welding process.
So, so definitely first we need to solve the
00:46:03.140 --> 00:46:07.510
conservation of the mass momentum and energy
in case of the heat transfer and fluid flow
00:46:07.510 --> 00:46:12.540
model. So, mass conservation we can represent
this equation. And here we are showing the
00:46:12.540 --> 00:46:17.790
momentum conservation, we can use this equation
that here the small u is actually represent
00:46:17.790 --> 00:46:24.190
the velocity field.
And x the coordinates, x is the coordinate,
00:46:24.190 --> 00:46:28.960
and t is the time component, capital small
t is the time component and P capital P is
00:46:28.960 --> 00:46:35.069
actually represent the pressure, and mu is
the effective is the viscosity. So, that parameter
00:46:35.069 --> 00:46:39.089
is also necessary to define when you try to
solve the fluid flow in case of the welding
00:46:39.089 --> 00:46:44.130
process. And f i represents the body force
through the f i term we can incorporate any
00:46:44.130 --> 00:46:50.109
kind of the body forces here. So, this is
the governing equations for the momentum transfer
00:46:50.109 --> 00:46:55.560
in case of the material flow.
Then we need to solve the energy equation
00:46:55.560 --> 00:47:01.560
also. So, energy equation as compared to the
heat conduction equation the this term is
00:47:01.560 --> 00:47:08.780
extra in this case because this energy equation
or maybe temperature field is modified by
00:47:08.780 --> 00:47:13.550
incorporation of the liquid molten metal,
so that is a convective flow of material we
00:47:13.550 --> 00:47:19.660
can consider here in the by this term . And
we can modify the solution of the temperature
00:47:19.660 --> 00:47:25.170
distribution in case of the energy equation.
So, absence of this term is just simply using
00:47:25.170 --> 00:47:30.589
of the heat conduction equation.
So, here this is a Q dot is the internal heat
00:47:30.589 --> 00:47:35.900
generation term normally when you try to do
the actual phenomena that means, fluid
00:47:35.900 --> 00:47:40.839
flow phenomena in case of the fusion welding
process not necessary to consider the Q dot
00:47:40.839 --> 00:47:47.119
term in in case of the arc welding process.
So, only surface heat flux is sufficient
00:47:47.119 --> 00:47:53.510
that can be incorporated through the boundary
condition hm and that that is that we will
00:47:53.510 --> 00:48:01.750
be able to find out the actual solution of
the temperature distribution as well as
00:48:01.750 --> 00:48:04.890
the velocity distribution in case of the welding
simulation .
00:48:04.890 --> 00:48:10.339
Now, if you look into the boundary condition,
the boundary interaction how we can represent
00:48:10.339 --> 00:48:17.930
here. So, on that surface top surface basically
if you look into this picture, the top surface
00:48:17.930 --> 00:48:22.859
there is a in the heat flux is applied, it
can be either laser, it can be either arc
00:48:22.859 --> 00:48:28.280
welding process. So, that because a simply
Gaussian distributed heat input from the welding
00:48:28.280 --> 00:48:33.619
arc and then on the top surface there is a
interface between the welding arc or may be
00:48:33.619 --> 00:48:40.050
along with the gas, this is shielding gas
as well, and the liquid molten pool . So,
00:48:40.050 --> 00:48:44.920
that interface on the top surface, the gas
medium and the liquid molten pool so on that
00:48:44.920 --> 00:48:50.309
surface there is acting of the surface tension
force . So, that surface tension force can
00:48:50.309 --> 00:48:54.050
be represented in this way.
So, normally we incorporate the surface tension
00:48:54.050 --> 00:48:59.500
force is the temperature coefficients of
the surface tension. This term and f l represent
00:48:59.500 --> 00:49:07.480
the liquid fraction here and to by delta x
that represents the temperature gradient with
00:49:07.480 --> 00:49:13.480
respect to one direction x-axis. And this
is the represent the shear stress basically
00:49:13.480 --> 00:49:18.319
this term so that we estimate on the surface,
the shear stress value two components along
00:49:18.319 --> 00:49:22.320
the x-axis and along the y-axis in this way
.
00:49:22.320 --> 00:49:27.849
And to on this x-axis, y-axis we can consider
the shear stress value and that shear stress
00:49:27.849 --> 00:49:34.880
value in terms of the surface tension coefficients,
temperature surface temperature dependence
00:49:34.880 --> 00:49:41.579
sorry the surface tension coefficient, a temperature
coefficient of the surface tensions and the
00:49:41.579 --> 00:49:48.331
temperature gradient in terms of that.
And then along the z-axis that means, normal
00:49:48.331 --> 00:49:55.420
to the z-axis there is no such that w equal
to 0 means that means, the displacement
00:49:55.420 --> 00:50:00.960
on velocity field normal to the surface equal
to 0 that means, we assume the it is a flat
00:50:00.960 --> 00:50:05.230
surface, and that is why we getting this kind
of boundary conditions.
00:50:05.230 --> 00:50:10.750
And the energy equation mathematically we
can see that energy equation; that means,
00:50:10.750 --> 00:50:16.059
what is the heat conducted a on the surface
and from the surface heat is convey heat convection
00:50:16.059 --> 00:50:20.290
or radiation also happens that means, convective
and radiative heat loss from the surface that
00:50:20.290 --> 00:50:25.760
can be mathematically represented by this
equation. So, that can act as a boundary
00:50:25.760 --> 00:50:30.960
condition here. And the symmetric surface
since it is a symmetric problem. So, on the
00:50:30.960 --> 00:50:36.530
symmetric surface, we can assume the heat
flux equal to 0. And at the solid liquid interface
00:50:36.530 --> 00:50:41.300
with no slip boundary condition, we apply
these are the typical boundary conditions
00:50:41.300 --> 00:50:47.619
for the momentum transfer for the solution
of the momentum and energy equation. So, with
00:50:47.619 --> 00:50:54.760
this solution of this with the boundary condition,
these governing equations and along with the
00:50:54.760 --> 00:50:58.760
energy equation.
If we solve these things we can find out,
00:50:58.760 --> 00:51:04.920
we can implement in finite element method
like this that final final matrix is basically
00:51:04.920 --> 00:51:10.970
from the energy equation it is a kind of A
x equal to B , but in that form . So, a f
00:51:10.970 --> 00:51:17.410
is equal to the right column vector right
side and we are finding the solution for T
00:51:17.410 --> 00:51:24.010
temperature distribution, but this K form
this matrix is a also a function of temperature
00:51:24.010 --> 00:51:27.020
when you try incorporate the temperature dependent
material properties.
00:51:27.020 --> 00:51:31.770
And finally, getting the nodal temperature
as the out that each and every node point
00:51:31.770 --> 00:51:37.870
will be getting the value of the temperature
each and every node point that means, distributed
00:51:37.870 --> 00:51:42.540
value of the temperature in node point. But
when you from the momentum equation we can
00:51:42.540 --> 00:51:49.420
find out the solution of the nodal velocity
in the similar fashion and that finally we
00:51:49.420 --> 00:51:51.839
are making the matrix in the form of A x equal
to B .
00:51:51.839 --> 00:51:57.900
So, they will be getting the velocity field
each and every node point if it is the three-dimensional
00:51:57.900 --> 00:52:02.369
problem will be getting the three component
of the velocity in the in each and every node
00:52:02.369 --> 00:52:09.609
point. And that each and every node part we
can we can that distributed value if we get
00:52:09.609 --> 00:52:14.130
it then we can predict that over the solution
domain what is the velocity field each and
00:52:14.130 --> 00:52:20.270
every node point. So, these are the typical
methodology in in case of finite element
00:52:20.270 --> 00:52:26.069
method, we can solve this equation and we
will be able to find out the temperature
00:52:26.069 --> 00:52:31.390
distribution as well as the velocity field
each and every node point in in in using the
00:52:31.390 --> 00:52:35.420
finite element method.
So, here all the different terms are also
00:52:35.420 --> 00:52:40.750
defined here also and we can find out all
the different term also define that which
00:52:40.750 --> 00:52:46.630
represents of what. So, we each and every
is every component that means, K, K matrix
00:52:46.630 --> 00:52:53.690
and the both the cases the K matrix consists
of the several other matrix component and
00:52:53.690 --> 00:52:58.960
each and having the physically represent the
some physical behavior for example, and in
00:52:58.960 --> 00:53:04.650
the momentum that M is the capital M is the
mass from the mass conservation from taking
00:53:04.650 --> 00:53:09.260
the amount of the mass, and then velocity
dependent convective term is taken care of
00:53:09.260 --> 00:53:15.730
by matrix C, viscous diffusion term is taken
care of by K. And the penalty because in this
00:53:15.730 --> 00:53:20.530
case we use the penalty finite element method
for the solution of the momentum equation,
00:53:20.530 --> 00:53:27.720
so in this case the penalty term is in corporal
bar other K hm K matrix. And b that F sorry
00:53:27.720 --> 00:53:33.910
f that actually represents the body force
and the surface accounts the body force and
00:53:33.910 --> 00:53:38.359
the surface tension force in this in this
form of this equation.
00:53:38.359 --> 00:53:43.349
So, when they solve the linear system of the
equation using some solver, we will we
00:53:43.349 --> 00:53:47.270
can find out the temperature and distribution.
Of course, there are so many strategies to
00:53:47.270 --> 00:53:53.059
follow how to solve all this equation that
is beyond the scope this analysis . So, just
00:53:53.059 --> 00:53:59.320
to get some overall idea how I can solve the
temperature this uh temperature then energy
00:53:59.320 --> 00:54:03.510
equation and momentum equation to get the
solution of the temperature and the velocity
00:54:03.510 --> 00:54:06.400
field.
Here you can see some results also by the
00:54:06.400 --> 00:54:12.069
solution of conservation of mass movement
of energy in case of the linear GTA welding
00:54:12.069 --> 00:54:16.359
process. So, linear GTA welding process if
we look into the right hand side figure first
00:54:16.359 --> 00:54:21.190
here we can see that within the small weld
pool that is red color represents the fusion
00:54:21.190 --> 00:54:26.599
zone . So; that means, there is the liquid
velocity field in the vector form it is represented.
00:54:26.599 --> 00:54:33.090
We can see that one centre point is there
. So, from centre point to outward direction,
00:54:33.090 --> 00:54:39.569
the liquid metal is try to flow from center
to the outer directions . And on the surface
00:54:39.569 --> 00:54:44.270
on the phase that that clears are some circulation
loop also.
00:54:44.270 --> 00:54:48.950
So, in two different cases we have shown that
the temperature distribution , but at the
00:54:48.950 --> 00:54:55.020
same time we the same figure we can see the
also velocity field the temperature distribution
00:54:55.020 --> 00:54:59.530
as well as the velocity field can also be
obtained for the other cases also. Here you
00:54:59.530 --> 00:55:04.410
can see the current equal to 100 amps voltage;
that means, in these two cases the parameters
00:55:04.410 --> 00:55:09.890
are different . For two different parameters
we can find out from the simulation that we
00:55:09.890 --> 00:55:15.210
are getting the a different temperature distribution
that means, different size of the weld pool
00:55:15.210 --> 00:55:23.339
as well as the velocity vector also .
So, if we look into the here in details that
00:55:23.339 --> 00:55:28.760
comparison between the experimental measurement
graph and the computationally evaluated, the
00:55:28.760 --> 00:55:34.240
velocity field also as well as the temperature
field we can find out for stainless steel
00:55:34.240 --> 00:55:42.980
304, we can see that there is a velocity vector
is moves one we look into the it is the clockwise
00:55:42.980 --> 00:55:46.849
direction . Here also we can see the velocity
vector is in such a way the clockwise direction
00:55:46.849 --> 00:55:52.109
so that means, metal flow we observe with
respect to the one centre point, there is
00:55:52.109 --> 00:55:58.750
a the metal flows in the clockwise direction.
Similar, we can see here also, so metal flow
00:55:58.750 --> 00:56:05.920
in this way and other temperature also that
other these this represents that this line
00:56:05.920 --> 00:56:15.500
represent the constant temperature contour.
So, here we can see that it is possible
00:56:15.500 --> 00:56:20.730
to solution of the conservation of the mass
momentum upon energy, we can predict the temperature
00:56:20.730 --> 00:56:25.290
in velocity field and we can compare with
the experimental data. We can see that in
00:56:25.290 --> 00:56:31.849
case of stainless steel the normally the
temperature that flow field is happening flow
00:56:31.849 --> 00:56:41.190
field is following this is in in in this case
all these cases it is the in clockwise direction.
00:56:41.190 --> 00:56:47.740
Similarly we can from this temperature
simulation we can find out the cooling and
00:56:47.740 --> 00:56:53.490
solidification behavior in case of the laser
spot welding. Here we can see that in case
00:56:53.490 --> 00:56:58.619
of the laser power 1 kilowatt laser power,
sheet thickness is 1 millimeter and one time
00:56:58.619 --> 00:57:03.400
equal to 20 millisecond that means, for 20
millisecond laser was found then we can find
00:57:03.400 --> 00:57:08.930
out the time versus temperature graph is like
that at the different fixed basis. So, we
00:57:08.930 --> 00:57:14.819
have defined that different species 1, 2,
3 along the along the surface and the along
00:57:14.819 --> 00:57:20.559
the depth direction the 4, 5 this some
characteristic points has been defined and
00:57:20.559 --> 00:57:26.370
on that point we can find out the temperature
versus time. So, it defined the point in such
00:57:26.370 --> 00:57:29.260
a way having some importance to do the further
calculation .
00:57:29.260 --> 00:57:34.569
So, here you can see that in case of laser
welding process there is a initially there
00:57:34.569 --> 00:57:37.970
is an increment of the temperature is the
peak temperature. We can say that is the heating
00:57:37.970 --> 00:57:44.660
pace and then gradually when the we can
switch up the laser. Then after reaching the
00:57:44.660 --> 00:57:47.640
peak temperature there is a gradually decrement
of the temperature.
00:57:47.640 --> 00:57:53.770
Here you can see reaching the peak temperature
and gradually decreasing and the marked space
00:57:53.770 --> 00:58:02.590
here you can see that there is a kind of
hump is there that actually in this part
00:58:02.590 --> 00:58:07.369
actually represents the if you see the temperature
then there is a phase transformation; that
00:58:07.369 --> 00:58:13.660
means, from the liquid phase to the solid
phase happens and that happens over a narrow
00:58:13.660 --> 00:58:20.780
range of the temperature in case of alloy.
So, here ; that means, in in case of alloy
00:58:20.780 --> 00:58:26.540
the this phase transfer from liquid phase
to the solid phase happens will between the
00:58:26.540 --> 00:58:30.720
solidus and liquidus temperature. So, that
we are getting this kind of harm here. So,
00:58:30.720 --> 00:58:37.980
that actually indication that phase transformation
happens in this during this process, so
00:58:37.980 --> 00:58:42.650
that also actually signify that in this numerical
model also we can incorporate the effect of
00:58:42.650 --> 00:58:48.450
the phase transformation .
Now, right hand side figure shows that
00:58:48.450 --> 00:58:54.480
the temperature and the velocity field distribution
after the switch of the laser sort that means,
00:58:54.480 --> 00:59:01.829
there is a decrement of the weld pool size.
And that first cases it was 20.6 mill second
00:59:01.829 --> 00:59:06.069
there will up to the 20 millisecond we have
applied the heat flux; that means, laser source
00:59:06.069 --> 00:59:12.349
then we switch up the laser source. And then
gradually the size of the molten pool actually
00:59:12.349 --> 00:59:16.690
decreases during the cooling phase.
So, the cooling phase we see 20.6 millisecond
00:59:16.690 --> 00:59:24.480
then 20 millisecond, and then 24 millisecond
also, there is a decrement of the weld pool
00:59:24.480 --> 00:59:31.440
size. But at the same time the velocity field
also decreases gradually during the cooling
00:59:31.440 --> 00:59:36.920
phase. So, all this kind of phenomenological
behavior that we generally observe that actually
00:59:36.920 --> 00:59:42.059
observe in the actual welding processes can
be represented through some mathematical model
00:59:42.059 --> 00:59:46.329
as well .
So, we can see some estimation of the cooling
00:59:46.329 --> 00:59:50.329
in the solidification in case of the laser
welding process . I have already mentioned
00:59:50.329 --> 00:59:55.290
that temperature distribution is and the basically
we can use the temperature distribution in
00:59:55.290 --> 01:00:01.150
case of the welding process. And this temperature
distribution can be used to estimate the different
01:00:01.150 --> 01:00:06.569
cooling and the solidification behavior
possible to predict or cooling rate all these
01:00:06.569 --> 01:00:11.559
parameters can we predict that can be closely
linked with the micro structural phenomena
01:00:11.559 --> 01:00:15.099
in case of the welding process.
So, let us see how we can estimate the
01:00:15.099 --> 01:00:20.040
cooling different solidification parameter.
So, two different solidification parameters
01:00:20.040 --> 01:00:27.299
can be defined here one is the R. R is the
solidification rate or growth rate that this
01:00:27.299 --> 01:00:35.020
R parameter will be will be more explain
the actual kinetics the solidification in
01:00:35.020 --> 01:00:41.940
the the in the chapter that the welding metrology
in in that module we will be explaining
01:00:41.940 --> 01:00:46.670
this phenomenon. But mathematically this can
be calculated here are the solidification
01:00:46.670 --> 01:00:53.690
rate and G is the temperature gradient . These
two parameters can be considered as the solidification
01:00:53.690 --> 01:00:56.609
parameter we can do further calculation using
these two parameters.
01:00:56.609 --> 01:01:04.190
So, either here you can see G, G by R one
is the solidification parameter and GR is
01:01:04.190 --> 01:01:09.050
the another parameter. So, G R is basically
represents the cooling rate, but how can estimate
01:01:09.050 --> 01:01:16.290
the G and R . So, here you can see the R a
change of the dimension with respect to time
01:01:16.290 --> 01:01:21.579
so that means it represented some rate that
means, that is called the solidification rate
01:01:21.579 --> 01:01:24.520
.
And another is the G is the temperature
01:01:24.520 --> 01:01:30.700
gradient. So, change of the temperature with
respect to the distance, so that represents
01:01:30.700 --> 01:01:37.400
the temperature gradient. So, in this case,
so here you can define that the weld typical
01:01:37.400 --> 01:01:42.080
weld pool is like that only the liquidus and
solidus temperature and the is there between
01:01:42.080 --> 01:01:46.690
the liquidus and solidus temperature .
And the distance with respect to the center
01:01:46.690 --> 01:01:54.510
point D 0 ; that means, at angle zero or D
90 along the depth direction that we define
01:01:54.510 --> 01:02:01.260
is the distance D 90 and on the surface
horizontally you can define as the D 0. So,
01:02:01.260 --> 01:02:05.960
this distance parameter can be evaluated
and that actually changes with respect to
01:02:05.960 --> 01:02:15.050
time . So, here if you see to track the changes,
so D 0 and D 90 or R 0 or R 90 having the
01:02:15.050 --> 01:02:21.039
similar dimension here you can see that R
0 and R 90 how it is changes. So, here R 0
01:02:21.039 --> 01:02:29.309
changes is more as compared to the R 90.
That means the solidification rate is more
01:02:29.309 --> 01:02:36.190
in this case the solidification rate
yes solidification rate is more one cases
01:02:36.190 --> 01:02:44.010
is in the with respect to the solidification
time along the on the surface as compared
01:02:44.010 --> 01:02:51.710
to the depth . So, on 90 value is less , but
if you look to the other cases that here
01:02:51.710 --> 01:02:58.099
you can see the G, G is the temperature gradient.
So, temperature gradient is actually is more
01:02:58.099 --> 01:03:04.490
along the depth direction in 90s, G 90 is
more as compared to the G 0.
01:03:04.490 --> 01:03:10.849
So, variation of the G 90, from the very high
value to the low value that happens in case
01:03:10.849 --> 01:03:18.010
of laser spot welding along the depth direction
as compared to the width direction . So, here
01:03:18.010 --> 01:03:24.339
if you if we if we estimate the cooling rate,
so here you can see that cooling at a 90 degree
01:03:24.339 --> 01:03:31.530
at cooling rate 0 degree; that means, the
90 degree means along the depth direction
01:03:31.530 --> 01:03:36.130
the variation of the cooling rate is more
as compared to the variation of the along
01:03:36.130 --> 01:03:44.180
the width direction in case of the laser welding
spotting. So, this kind of mathematical
01:03:44.180 --> 01:03:49.339
calculation of the different solidification
parameter that means, temperature solidification
01:03:49.339 --> 01:03:56.579
rate and the temperature gradient as well
as the cooling rate can also be estimated
01:03:56.579 --> 01:04:02.650
just to know about the time temperature profile
in case of the welding process.
01:04:02.650 --> 01:04:10.230
So, here further the cooling and the solidification
in GTA spot welding process here you can see
01:04:10.230 --> 01:04:17.190
that welding current 140 amps voltage 12 volt
and on-time equal to 3 second. So, here you
01:04:17.190 --> 01:04:23.039
can see that different a similar kind of the
different characteristic point on the along
01:04:23.039 --> 01:04:27.980
the width direction or along the depth direction,
we can find out the temperature distribution.
01:04:27.980 --> 01:04:33.799
This hump also represents that the solid phase
to liquid phase transformation happens over
01:04:33.799 --> 01:04:38.770
between the solidus and liquidus temperature
and that can be clearly captured just simply
01:04:38.770 --> 01:04:45.200
looking into the time temperature simulation.
Here also we can see in GTA spot welding process
01:04:45.200 --> 01:04:49.760
the three second; that means, on-time was
3 second means after 3 second just switch
01:04:49.760 --> 01:04:58.780
up the heat source. Then we can follow the
nature of the weld pool at the cooling phase.
01:04:58.780 --> 01:05:04.220
So, at 3.01 second, 3.09 second and 3.12 seconds
at 3.12 seconds there is a disappearance of
01:05:04.220 --> 01:05:12.050
the weld molten pool.
So, these kind of mathematical calculation
01:05:12.050 --> 01:05:16.440
can also you have to analyze the physical
viewer what is happening within the weld pool
01:05:16.440 --> 01:05:23.980
bottom and this this kind of physical phenomena
its not always possible to capture just simply
01:05:23.980 --> 01:05:28.359
doing the experiment. So, in that way it is
advantageous to analysis these thing.
01:05:28.359 --> 01:05:33.770
So, similar kind of calculation cooling and
solidification GTA spot welding can also be
01:05:33.770 --> 01:05:40.450
done. But if we look into this comparison
that G by R that one solidification parameters,
01:05:40.450 --> 01:05:48.130
and G R and if we look into that, so GTA
spot welding process that G by R is very
01:05:48.130 --> 01:05:55.670
high in case of GTAs as possibly it is high
as compared to the laser spot welding process
01:05:55.670 --> 01:06:05.359
. G by R that means, G by R is also
other cases than at zero degree in case of
01:06:05.359 --> 01:06:09.690
laser it is 40, but in case of GTA spot it
is less.
01:06:09.690 --> 01:06:15.710
But cooling rate is in laser which is significantly
higher as compared to the GTA spot welding
01:06:15.710 --> 01:06:24.849
process . And if we see that at 0 degree that
means, we found out as a 730 Kelvin as compared
01:06:24.849 --> 01:06:31.180
to the, but zero degree 730 Kelvin in case
of GTA welding, but in case of laser it is
01:06:31.180 --> 01:06:37.299
less. So, what we understand that cooling
rate is basically along the width direction
01:06:37.299 --> 01:06:42.859
cooling it is more in case of the GTA spot
welding as compared to the less, but along
01:06:42.859 --> 01:06:47.590
the depth direction the cooling rate is more
in case of the laser spot welding as case
01:06:47.590 --> 01:06:51.270
of an as compared to the GTA spot welding
process.
01:06:51.270 --> 01:06:56.720
So, now we can see the further heat transfer
and the fluid flow in laser micro welding
01:06:56.720 --> 01:07:02.829
processes here you can see the one cases is
that in case of SS304 that only laser power
01:07:02.829 --> 01:07:10.940
was 64 what pulse fiber laser has been used.
Thickness was the 400 micrometer . And we
01:07:10.940 --> 01:07:20.570
can see that uh the temperature distribution
in the within the molten pool as
01:07:20.570 --> 01:07:26.230
well as the velocity distribution. And
we can everywhere we can see the velocity
01:07:26.230 --> 01:07:31.809
distribution it starts from the center center
point to the outwards periphery in that direction
01:07:31.809 --> 01:07:37.410
actually we can observe the flow of the liquid
molten metal happens during the welding process.
01:07:37.410 --> 01:07:42.390
And if we see the comparison between the experimental
and the numerical results and we can see that
01:07:42.390 --> 01:07:52.660
the there is a the there is some circular
loop in this case that means, clockwise
01:07:52.660 --> 01:08:03.190
circulation loop or observed in in case of
the micro laser welding process and . And
01:08:03.190 --> 01:08:08.541
because these are this loop actually depends
on the the which direction the surface tension
01:08:08.541 --> 01:08:14.231
force is acting. So, if there is change of
the surface direction of the surface tension
01:08:14.231 --> 01:08:21.150
force, and then the fluid that this pattern
can also be change. So, we will see later
01:08:21.150 --> 01:08:26.810
on that how what is the possible way to change
the this direction or pattern of the velocity
01:08:26.810 --> 01:08:31.000
field.
Thank you very much for your kind attention.
01:08:31.000 --> 01:08:36.500
The next class will try to discuss the surface
active elements or then fluid flow how
01:08:36.500 --> 01:08:58.790
how the surface active areas actually increases
the material flow.