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So, today will discuss about density of sea
water. Now, density is important from the
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point of view of ocean circulation. So, primarily
if you want to have look at density, so
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density determines depth to which the water
will settle or depth to which water particle
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00:01:56,291 --> 00:02:20,210
will settle in equilibrium, so this is the
defining parameter for density. So, here you
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00:02:20,210 --> 00:02:30,319
will
find
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00:02:30,319 --> 00:02:44,480
the least dense water, present on top, so
this is called natural present on top, and
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00:02:44,480 --> 00:03:06,749
densest at
ocean bottom. Now, mixing is easiest, mixing
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of seawater is easiest in waters
having same density.
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So, this point is to be noted, because maps
of density surfaces, so these are called
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00:03:50,709 --> 00:04:05,830
isopycnals, so these are surfaces having the
same density, iso means same, that means
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another term for density, the density are
the same. So, these are mapped and these maps
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00:04:16,090 --> 00:04:36,670
are used
to depict circulation pattern, so density
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is a governing parameter for ocean
circulations, so depicts of circulation patterns.
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00:04:47,240 --> 00:04:48,240
..
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00:04:48,240 --> 00:04:59,070
Now, how density is given, so density it is
given in terms of this parameter called rho
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and rho for rho is equal to 1000 Kg per meter
cube at 0 degree centigrade, and no salt.
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Now, see sea water density is 1021.00 Kg per
meter cube, this is at sea surface and
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normally this sea surface is designated with
p equal to 0, that pressure equal to 0 or
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00:05:45,870 --> 00:06:01,460
atmospheric pressure. So, 1021 Kg at sea surface,
and this becomes 1070 Kg per meter
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00:06:01,460 --> 00:06:19,510
cube, this is at a pressure of 10000 decibar.
So, this is almost at 10 kilometers from the
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00:06:19,510 --> 00:06:29,650
ocean surface, so density there is a variation
in density, now common method of describing
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00:06:29,650 --> 00:06:42,970
density is by this parameter called sigma.
So, sigma s stands for salinity, t is temperature
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and p is pressure, so this is given in terms
of density above 1000 Kg per meter, so in
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00:06:54,741 --> 00:07:00,890
ocean graphic literature you find this
parameter called sigma s t p. So, this is
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00:07:00,890 --> 00:07:30,030
called in Situ density, in Situ density calculated
at sea surface, so this is how the density
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00:07:30,030 --> 00:07:36,430
is defined and there is another equation,
which is
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00:07:36,430 --> 00:07:54,370
to be noted that is density is also given
as an equation of state
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00:07:54,370 --> 00:08:07,930
for seawater.
So, this is given, this are all writing in
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Situ density you simply write rho, in terms
of
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salinity temperature pressure and this will
be rho in terms of salinity, temperature and
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pressure you take it 0. That means, at the
sea surface divided by these factors 1 minus
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p
divided by K, K is called the bulk modulus,
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00:08:38,459 --> 00:08:49,660
so this is another expression for density.
So,
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00:08:49,660 --> 00:08:57,290
salinity temperature and pressure, at any
pressure on the below the sea surface, it
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is
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00:08:58,290 --> 00:09:09,279
.linked with rho s, t, 0 at c surface 1 minus
the pressure at that point K (s, t, p) transfer
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your bulk modulus.
Now, this equation also can be return in this
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00:09:15,569 --> 00:09:35,100
form, now ((Refer Time.) K hear it
stands for bulk modulus, now here actually
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this is slightly different formula, but this
is
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also nowadays is used. So, this stands for
density at any point, so this is your rho
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00:09:55,630 --> 00:10:06,019
0 plus
you can represent this in more mathematical
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form, so T minus T 0 I will give you what
this T 0 rho 0 stands for and alpha and beta
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are constants, S minus S 0, so this is your
equation. Now, alpha is the density gradient,
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so this is delta rho divided by delta T, so
amount of change of density with respect to
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T stands for four temperatures.
And beta stands for what, this will stands
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for change of remember your parameter is your
densities, so this is again delta rho divided
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by delta S that is change of density with
respect to salinity. So, alphas stands for
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change of density with respect to temperature,
beta stands for change of density with respect
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to salinity, and this rho 0, T 0 and S 0,
so
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this represents the density temperature than
salinity, you can take these are at any point
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are arbitrary constants. But, normally you
can take them at the sea surface, and rho,
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T, S
these are mean values
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for the region.
Now, if you take the special region in the
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ocean obviously, it will vary according to
the
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your x and y values, where you just take the
mean values. So, this gives you rho that is
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the density at that particular location with
reference to some rho 0, T 0 and S 0 were
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you
normally you can take this at this sea surface,
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if you write are at other any point.
So, this gives us the equation for state,
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linking density, temperature and salinity,
so this
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equation of state actually models these three
terms or links, so this is vital for defining
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your density links, density with temperature
and salinity. So, that is why this equation
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is
more preferable to the one that having that,
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bulk modulus K, so this is the equation of
state and with this we finish our discussion
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on density, now there is a problem on this
current.
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..
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So, last time we discussed about geostropic
current, now in the Northern Hemisphere
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you come across this huge current, which is
called the gulf stream meander this is called
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gulf stream meander; so this is quite famous
current, this is called gulf stream meander.
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Now, this current actually goes from the eastern
coast of US to the united coast of United
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State kingdom, so UK is somewhere here, so
this is called the gulf stream meander, it
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flows from the Northern Hemisphere like this
and it goes towards the UK coast. So, this
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is one of the main reason for why you get
on the eastern shores of UK you get the warm
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current, but whereas on the west coast you
get a very cold climate.
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So, this is the reason of called the gulf
stream meander, so this occurs in the
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Northern
Hemisphere, now let us try to calculate the
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velocity of this current. Now, in the problem
this is given as pressure change across, you
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have to calculate this pressure change, so
pressure change
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across gulf stream
is the result of an increase in sea level
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value is 1
meter. Now, this take place over a distance,
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this distance across gulf
stream of 100
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kilometer, so pressure change results in a
sea surface elevation of 1 meter.
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So, in brackets you can write this as delta
z, over a distance across 100 kilometers,
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so
this is your delta x, now you calculate the
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geostropic current. So, first you get the
hydrostatic balance equation or simply write
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down the hydrostatic equation, so what is
the hydrostatic equation, so this is given
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by delta p is equal to rho g delta z. So,
this is
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your hydrostatic also or sometimes this is
called a hydrostatic balance, and this is
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.actually supporting the weight that is why
this is called hydrostatic balance. Now, in
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the
Northern Hemisphere what is happening your
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flow, let us see have a look at it direction
of the flow.
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.
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So, this is your low pressure region, so these
the pressure of low pressure, so these are
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your pressure or lines of equi pressure
and right has a bottom you have high pressure.
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So,
in which direction their flow will be directed,
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so flow is always directed from high
pressure, so this is a region of high pressure.
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So, flow is always directed from high
pressure to low pressure region, so that is
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your driving force, your flow is going to
go
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like this, so this is called the gulf stream
meander.
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And by towards a right, so your velocity vectors
keep on increasing, now till at a certain
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magnitude you find the flow is becoming horizontal,
now at this region you find your
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pressure that is given by F p g, your pressure
force will balance the Coriolis force F c
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f.
So, in bracket I am writing pressure, pressure
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force is going to balance your this is F c
f
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is the Coriolis force, then you get the velocity
of the flow this is given has V g, so
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pressure equation I have written.
Now, you write down the equation for geostropic
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current, this is called geostropic
current; now in the Northern Hemisphere I
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have told you the flow is always directed
towards the right, you are throwing a particle
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from the equator to the poll. So, that will
have it same velocity plus vector will be
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add the rotation of the earth, which is the
from
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.left to right, so your particle will always
get deflected towards the right. So, this
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is a
result of the Coriolis force or Coriolis acceleration,
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now in this case you are the
geostropic current, you have to calculate
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from V g this is equal to 1 by rho f.
Now, f is a this is a Coriolis factor multiplied
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by the pressure gradient, so del p over del
x, so this is your pressure gradient and this
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value is given, and f you will take it has
twice
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omega sign phi. Now, omega is your earth angular
velocity, so this is a geostropic
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current or Coriolis it as to be linked to
with the velocity of the earth or earths angular
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velocity. So, this is equals to 7.29 into
10 to the power minus 5 radius per second,
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so this
is the angular velocity of the earth. Now,
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this phi is the latitude, anyway from this
equation written calculate the value of f,
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now for calculation V g this value of f is
already
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given.
.
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Now, V g is equals to 1 by rho f into del
p over del x, now you substitute how much
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is
the value of del p, we have find out del p
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from the hydrostatic balance equation and
what
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is that value. So, this is equal to 1 by rho
f del p is nothing but rho g delta z, now
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you
divide this by rho f and what else you have
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got delta x, so ultimately you can see we
are
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getting delta z over delta x, so rho will
cancel out g divided by f. So, the final equation
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you will get has g over f multiplied by delta
z over delta x, now value of g is in
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centimeters this is 980 centimeters per second.
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.And f is given as 10 to the power minus 4,
so this is radians per second, so at that
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particular latitude what is the angular velocity,
now what is the value of delta p, delta p
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we are given this is delta z, now delta z
is given as 1 meter. So, 1 meter you are doing
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it
in centimeters, so obviously, this would be
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100 centimeters and this is the elimination
over how much delta x value is 100 kilometers,
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so you convert that into centimeters. So,
this will be 10 to the power 7, now you reduce
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this will be approximately this will give
you 100 centimeter per second, (()) approximately
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coming here I am writing this.
So, this 100 centimeter per second, so geostropic
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current velocity V g is 100 centimeter
per second, approximately you are although
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you are getting this has 98, so this is the
gulfstream velocity 100 centimeter per second,
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you see the units right or wrong, so this
is
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100 square. So, this is 5 minus 4 is much
you are getting almost in the this 98 centimeter
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per second, almost 100 centimeter per second,
so that is your gulfstream velocity. So,
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with this let us finish about the properties
of seawater.
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.
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Now, next let me start about the linear wave,
so how one gets this wave equation, so
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your wave equation, if you calculate the wave
equation this comes from the equation of
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continuity. So, what is this equation of continuity,
so this is given as del u or rather you
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can write it del u x over x plus del u y over
del y plus del u z and del z equal to 0, so
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this
equation continuity is what, what is the significance
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of equation of continuity. So, I am
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.not going in detail this you ask your hydrodynamic
teacher, how he has got the equation
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of continuity is coming from conversation
of mass.
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So, this regression of continuity, now you
substitute this values of u x and u y and
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u z, a
u x is how much it is del phi over del x,
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u y is del phi over del y and u z is del phi
over
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del z, and you substitute these values in
the equation of continuity and see what you
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get.
Now, you substitute this you will get the
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Laplace equation are getting it, so you
substitute this, so this will be del square
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phi del x square plus del square phi over
del y
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square plus del square phi over del z square,
so this is equal to 0. Now, this in
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00:30:23,039 --> 00:30:33,850
mathematics that term this has number of square
phi equal to 0, so this is your famous
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Laplace equation.
.
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Now, the linear wave or sometimes this is
called the airy wave is nothing but a solution
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of this equation, so a linear wave or an airy
wave is the solution of Laplace equation.
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So,
you remember this, solution of Laplace equation
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and what is this equation, equation in
short is written as number of square phi equal
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to 0. Now, if you look at the solution, you
find the solution is written in this form
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has beta, beta x t, beta is wearing with the
position x and the time t. So, this is given
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00:31:44,510 --> 00:31:55,799
as a sin of omega t minus K x, so this is
your
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00:31:55,799 --> 00:32:15,450
surface region four linear wave, now if you
draw a graph this will look like this, now
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later on at the end of the class, I mean next
class have look at non-linear waves.
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00:32:23,860 --> 00:32:33,230
.But, before that whereas, expose this and
what we had to be careful about is, the
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applications regime is of the linear non-linear
waves, so this is what this is
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00:32:51,360 --> 00:32:58,240
one wave
period, now the horizontal axis can either
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the t or x, since your surface derivation
beta,
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00:33:04,500 --> 00:33:12,710
beta is called as surface derivation at any
point of time. Now, what is your a, a is your
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00:33:12,710 --> 00:33:47,820
amplitude
and T is called wave period, L is the wavelength,
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now since I have selected
one horizontal axis, you can debit as T and
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L, then what is wave weight and H is your
wavelength.
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00:34:04,230 --> 00:34:18,309
So, that is the distance between the stuff
and impressed, so these parameters are the
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00:34:18,309 --> 00:34:36,069
physical parameters of the airy wave, now
you calculate the speed of propagation. Now,
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this is not a standing wave, it is a propagating
wave now propagating wave has certain
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velocity. Speed of propagating wave and if
you want to calculate this, then this angle
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00:35:04,309 --> 00:35:14,470
that
is omega t minus K x is referred to as phase
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00:35:14,470 --> 00:35:21,599
angle, now there is a simple method of
calculating the propagating wave velocity.
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You simply take the partial derivative of
this phase angle to be 0, that is the phase
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00:35:36,349 --> 00:35:43,329
angle
remains constant, this is equal to partial
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00:35:43,329 --> 00:35:48,329
derivative of omega t minus K x is equals
to 0.
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00:35:48,329 --> 00:36:04,069
So, this implies that phase angle remains
constant obviously, the partial derivative
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00:36:04,069 --> 00:36:11,012
of any
remains constant, that is the derivative or
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00:36:11,012 --> 00:36:13,710
a constant would always will give you 0.
.
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00:36:13,710 --> 00:36:22,609
Now, what is you differentiate this with respect
to time how much we get, so this will be
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00:36:22,609 --> 00:36:38,340
del omega t over del t minus del del x, first
you differentiate these two with this respect
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00:36:38,340 --> 00:36:48,010
.x, then you multiplied this by d x d t, so
this is equal to 0. Now, from this we get
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00:36:48,010 --> 00:36:56,310
omega
minus k d x d t, so this is equal to 0, so
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now you can find out the velocity, so d x
d t is
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00:37:01,000 --> 00:37:08,130
how much, so that is how the linear velocity
of the wave. So, this is we are getting this
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as
omega by K, now omega is how much, omega is
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00:37:16,520 --> 00:37:28,059
called the omega is equals to 2 phi,
omega is a circular frequency and K is the
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00:37:28,059 --> 00:37:43,800
wave number.
So, omega is the circular frequency fine,
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00:37:43,800 --> 00:37:51,460
and K is 2 phi over in this case it is L,
so this is
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00:37:51,460 --> 00:38:04,180
called wave number. So, remember the expression
for this omega and K, omega is 2 phi
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00:38:04,180 --> 00:38:12,950
by T, T is your weight period and K is the
wave number is 2 phi by L, L is the wave
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00:38:12,950 --> 00:38:26,160
length, so omega is K and how much it comes.
So, this is 2 phi by T over 2 phi by L, so
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00:38:26,160 --> 00:38:40,900
this comes as a very neat relationship of
L by T, and sometimes this is given as C is
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00:38:40,900 --> 00:38:45,330
the
propagating velocity, so this is called speed
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00:38:45,330 --> 00:38:59,740
of propagation.
So, we have got the equation for this surface
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00:38:59,740 --> 00:39:08,579
that is the airy wave, that is given by eta
and we are got the, see there is a velocity
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00:39:08,579 --> 00:39:23,740
of propagation, now you find out what is the
particle velocity, now your surface relation
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00:39:23,740 --> 00:39:32,040
equation is eta x, t. So, this is we are using
a
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00:39:32,040 --> 00:39:42,301
sin wave, a sin that is omega t minus K x,
so actually there are two variables and one
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00:39:42,301 --> 00:40:07,740
variable is t, and another variable is x.
Now, you have to define velocity potential,
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00:40:07,740 --> 00:40:11,450
now
this velocity potential have to satisfy the
216
00:40:11,450 --> 00:40:22,720
Laplace equation for this surface profile.
Now,
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00:40:22,720 --> 00:40:49,760
((Refer Time.) this is your velocity potential,
cos hyperbolic K d plus z divided
218
00:40:49,760 --> 00:41:14,700
by sin hyperbolic K d cos omega t minus K
x, so this is the velocity potential that
219
00:41:14,700 --> 00:41:18,080
is to be
used.
220
00:41:18,080 --> 00:41:19,080
..
221
00:41:19,080 --> 00:41:42,050
This should satisfy your Laplace equation,
satisfy number of square phi equal to 0 the
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00:41:42,050 --> 00:41:59,510
Laplace equation, now you find out particular
velocity, now we are considering the wave
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00:41:59,510 --> 00:42:10,660
in two dimensional that is x and z. So, particle
velocity or rather you can write water
224
00:42:10,660 --> 00:42:37,369
particle velocity, u x horizontal
water particle velocity and the other one
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00:42:37,369 --> 00:42:50,920
is u z. So, this is
your vertical, vertical water particle relationship,
226
00:42:50,920 --> 00:43:00,040
what is the expression for u x, now u x
you will get if you differentiate the velocity
227
00:43:00,040 --> 00:43:14,260
(( )) phi with respect to x, and u z you
calculate from del phi over del z.
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00:43:14,260 --> 00:43:21,980
Now, later on you find, if you want to calculate
the pressure term, you will require u x
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00:43:21,980 --> 00:43:31,660
and u z, your ultimate goal is to find out
the pressure term has an a ocean engineer
230
00:43:31,660 --> 00:43:42,351
anyways, so that will come later on. Now,
you find out what is this value of u x, u
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00:43:42,351 --> 00:43:48,630
x is
del phi over del x, and phi we have got the
232
00:43:48,630 --> 00:43:57,349
expression phi, phi is this is expression
((Refer Time.), so you differentiate this
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00:43:57,349 --> 00:44:10,270
with respect to x. And see how much will
get, so u x will omega a into there will be
234
00:44:10,270 --> 00:44:18,849
two minus coming from that minus K u minus
minus will cancel out, and K will also cancel
235
00:44:18,849 --> 00:44:23,050
out.
So, you get ((Refer Time.) this expression
236
00:44:23,050 --> 00:44:41,140
cos hyperbolic of K d plus z and this is
divided by sign hyperbolic of K d, and you
237
00:44:41,140 --> 00:44:50,730
differentiate cos you will get, so the
differentiation will be over x, so cos will
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00:44:50,730 --> 00:44:59,740
be sign, sign omega t minus K x, so this is
your
239
00:44:59,740 --> 00:45:20,440
del phi over del x. Now, next to find out
u z, so this is del phi over del z, so intermittent
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00:45:20,440 --> 00:45:45,040
this expression, so this will be omega a sin
hyperbolic of K multiplied by d plus z and
241
00:45:45,040 --> 00:45:57,440
.denominator will be sin in hyperbolic of
K d. And the other term will be you differentiate
242
00:45:57,440 --> 00:46:13,070
how much phi over del z, so phi is this differentiate
with this is a cos, this will be u z,
243
00:46:13,070 --> 00:46:23,569
this term will be cosine of omega t minus
K x.
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00:46:23,569 --> 00:46:44,090
So, this is your expression, now you find
in deepwater, in deepwater what happens that
245
00:46:44,090 --> 00:47:00,930
is
when K d is infinity, now this omega a cos
246
00:47:00,930 --> 00:47:17,450
hyperbolic this term, that is K d plus z over
sin hyperbolic K d will a approaches certain
247
00:47:17,450 --> 00:47:32,530
value. So, this is omega a e raised to the
power K z, now the another term was approaches
248
00:47:32,530 --> 00:47:35,500
the same value.
.
249
00:47:35,500 --> 00:47:58,530
So, this you find out from hyperbolic expressions,
so omega a sin hyperbolic K d plus z
250
00:47:58,530 --> 00:48:11,599
over sin hyperbolic K d, so this will also
approached the same value, so omega a e raised
251
00:48:11,599 --> 00:48:25,299
to the power K z. So, now, you find out the
resulting velocity u, u is root over of u
252
00:48:25,299 --> 00:48:48,160
squares x plus u squares z, so this is omega
a e raise to the power K z, so at surface
253
00:48:48,160 --> 00:48:56,370
you
put z equals to 0, velocity of water u equals
254
00:48:56,370 --> 00:49:19,060
to simply omega a. Now, this is your
velocity equations, you find out water particle
255
00:49:19,060 --> 00:49:40,130
path, now once you get this part you will
find, the shape of the or the trajectory of
256
00:49:40,130 --> 00:49:56,640
the water particles.
Now, how to find this
257
00:49:56,640 --> 00:50:16,820
you obtained path by integrating velocity
questions, so what is
258
00:50:16,820 --> 00:50:31,089
actually velocity u x, u x is d x over d t,
so you will get x if you integrate d x by
259
00:50:31,089 --> 00:50:51,829
d t. So, u
x we have got us omega a cos hyperbolic K
260
00:50:51,829 --> 00:51:06,720
d plus z divided by sin hyperbolic, which
is
261
00:51:06,720 --> 00:51:18,880
K d, and the other term is this sin expression
omega t minus K x, now you try to integrate
262
00:51:18,880 --> 00:51:29,079
this, so this one equation. So, the equation
number 1 and the other u z is velocity is
263
00:51:29,079 --> 00:51:30,079
how
264
00:51:30,079 --> 00:51:47,299
.much d z over d t, so u z we have got the
expression as omega a
265
00:51:47,299 --> 00:52:02,760
sign hyperbolic K d plus
z over sin hyperbolic K d. And the other term
266
00:52:02,760 --> 00:52:12,730
your got is cosine, cos omega t minus K x,
this is equation number 2, now integrate these
267
00:52:12,730 --> 00:52:16,250
two, you integrate these two you will get
x
268
00:52:16,250 --> 00:52:24,759
and z; and if you integrate you find come
across the interesting conclusion.
269
00:52:24,759 --> 00:52:25,759
.
270
00:52:25,759 --> 00:52:35,790
So, next class will do this, and you will
find that the water particles part is elliptic,
271
00:52:35,790 --> 00:52:45,740
you
will get an ellipse you try to do this, and
272
00:52:45,740 --> 00:53:07,910
this is your wave profile and you will find
ellipse coming, so this is our z, this is
273
00:53:07,910 --> 00:53:27,690
our x. So, water particle path is elliptic,
so we will
274
00:53:27,690 --> 00:53:45,320
explore this we try to find equation of ellipse,
this is very easy from this expression you
275
00:53:45,320 --> 00:53:55,619
can find out the equation for ellipse. And
you find out equation of ellipse you have
276
00:53:55,619 --> 00:54:16,270
to
find major and minor axis minor axis of ellipse,
277
00:54:16,270 --> 00:54:21,740
now in shallow water you would be find,
shallow water now there will be two distinct
278
00:54:21,740 --> 00:54:37,740
cases or rather create distinct cases.
Number 1 is the find for deepwater, next you
279
00:54:37,740 --> 00:54:49,390
find intermediate waters
and the last
280
00:54:49,390 --> 00:54:59,930
category you will be shallow water, now these
are three cases will come across, for all
281
00:54:59,930 --> 00:55:19,869
these you find path of particles. So, next
class we will do this, but before this we
282
00:55:19,869 --> 00:55:35,619
find out
the equation of the ellipse, so this is a
283
00:55:35,619 --> 00:55:40,600
discussion on the linear wave theory, and
after this
284
00:55:40,600 --> 00:55:59,049
will go to the normal linear waves.
285
00:55:59,049 --> 00:55:59,049
.