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In the last lecture we were discussing about
the theory of electrochemical machining I
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elaborately told about the Faraday’s law
of electrolysis with the help of which we
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are able to calculate linear material removal
rate volumetric material removal rate mass
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material removal rate
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Then we also discussed about the evaluation
of the inter electrode gap with the help of
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parabolic equation that is applicable for
zero feed rate and an implicit equation that
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is applicable for finite feed rate both these
equations are used in the calculation or theoretical
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electrochemical machining but the equation
with finite feed rate has certain drawbacks
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it takes so many iterations before one can
arrive at the correct inter electrode gap
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And in case of finite element method finite
difference method the large number of points
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at which these equations are to be evaluated
so single equation applicable for both the
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cases and taking very less time had been proposed
that was discussed then we discussed how to
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evaluate electrochemical equivalent of an
alloy two methods were discussed percentage
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weight method and total charge required method
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And after that today let us discuss the maximum
permissible feed rate for the given machining
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conditions and that is very important if the
feed rate is very low then productivity will
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be low accuracy will be also low if it is
very high although material removal rate will
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be high but there are all possibilities that
short circuit may take place and that short
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circuit will result in the damage of the tool
as well as the work piece
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So let us see how to evaluate the maximum
permissible feed rate for the given machining
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conditions Theoretically there is no upper
limit for feed rate in electrochemical machining
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as we have seen in case of self regulating
feature that has the inter electrode gap keep
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decreasing because of the higher feed rate
then material removal rate linear material
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removal rate keeps on increasing to maintain
the equilibrium condition
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Still there is always a limit for the maximum
permissible feed rate that is what we are
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going to evaluate today for this purpose during
actual electrochemical machining electrolyte
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flow rate should be such that it is able to
carry away the heat produced during ECM and
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electrolyte boiling does not take place
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Now here there is a question mark why we want
that boiling of the electrolyte should not
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take place and heat generated in the inter
electrode gap should be taken away as quickly
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as possible because if the heat produced during
in the inter electrode gap during ECM is not
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taken away it is going to change the temperature
of the electrolyte as a result of that although
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conductivity will change more important is
whatever you have designed or expecting the
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work piece shape or anode shape that will
not be obtained because you are not able to
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control the conductivity of the electrolyte
in process
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Secondly boiling should not take place if
the electrolyte boiling takes place then vapor
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generation will take place and once vapor
generation takes place those vapors or bubbles
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will be added to the electrolyte and that
is again going to change the conductivity
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of the electrolyte that simply means again
you are not able to get the predicted anode
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shape or the work piece shape and you are
not able to control the conductivity of the
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electrolyte
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Now in the derivation of this particular equation
for maximum permissible feed rate certain
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assumptions has been made to simplify the
analysis the basic assumption and foremost
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assumption is heat produced only due to ohmic
heating is considered and heat produced by
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other sources is neglected
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Heat produced due to the viscous flow of electrolyte
also there are certain chemical reactions
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that take place they also produce the heat
they are the exothermic reactions All these
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sources of heat production during ECM are
neglected
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Maximum permissible temperature of electrolyte
is always less than or equal to the electrolyte
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boiling temperature for the reasons I have
just mentioned Specific heat and density of
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electrolyte do not change with its temperature
Electrolyte temperature is going to change
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from the room temperature to say 90 degree
centigrade or so and room temperature varies
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from say 5 to 10 degree in winter to the 45
degree centigrade or more in summer so there
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is a large variation
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And above this the temperature can further
go during the ohmic heating and it maybe as
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high as 90 degree centigrade or so and at
such a large scale of temperature variation
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the specific heat of the electrolyte is going
to change and to some small extent the density
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of the electrolyte may also change but for
present modeling or for present development
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of the equation for maximum permissible feed
rate both these changes are neglected or ignored
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So we are going with the assumption that heat
generated due to the ohmic heating is equal
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to heat carried away by the electrolyte
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Now this is another very important assumption
that no conduction of heat to the anode and
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cathode is taking place as you know already
that if you recall earlier figures and here
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you can see tool and work piece both are in
contact with the electrolyte where heat is
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generated so obviously some heat will be conducted
away to the tool and some heat will be conducted
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away to the work piece but for this simple
calculation here we are assuming that heat
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conducted away to the tool and work piece
both is negligible
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The thermal conductivity of the tool is very
high however for simplification of the problem
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we are considering it negligible also we are
neglecting here the losses due to the radiation
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because the temperature developed is not very
high so we can easily neglect the heat loss
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due to the radiation
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Now they are various subscripts being used
in the following equation that you can see
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here b is used for boiling d for density e
for electrolyte h for inter electrode gap
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i for initial condition also capital I not
small i for current small m for mass p for
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permissible and capital P for power r for
resistance capital T for temperature U subscript
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e for volumetric flow rate small t for time
and rho s for electrical resistivity
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So we can see here that total heat H is equal
to m e c e T b minus T i so here m is the
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mass of the electrolyte c e is the specific
heat of the electrolyte T b is the boiling
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temperature of the electrolyte that is the
highest permissible temperature in this case
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and T i is the initial temperature that the
temperature at which electrolyte is entering
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into the inter electrode gap
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Now this mass m e can be written in terms
of the volumetric flow rate or volume of the
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electrolyte flowing in the given period of
time and rho e is the density of the electrolyte
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so we have this modified equation Now divide
both sides of this equation by the electrochemical
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machining time t then you get H by t is equal
to U e over t rho e C e multiplied by T b
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minus T i
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Now this volume flow U e when divided by time
of flowing then you get volumetric flow rate
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so you can write and H divided by T total
heat generated divided by time then you get
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power P So you can write this equation equal
to 4.186 U dot e rho e C e T b minus T i here
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U dot e this indicate the volumetric flow
rate of the electrolyte and 4.186 is the joules
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constant and this can be written as equal
to I p suffix square R where I p is the permissible
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current that can flow in the inter electrode
gap
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And its value is determined on the basis such
that it does not permit electrolyte to boil
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or it does not permit electrolyte temperature
to go above the boiling temperature as we
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have already made the assumption and R is
resistance of the inter electrode gap that
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is given over here I p square R
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Now simplification of this equation will give
us I p that is the permissible current in
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the circuit equal to under root 4.186 U e
rho e C e T b minus T i divided by rho s h
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over A Now this R has been substituted by
rho s h over A rho s h over A is the resistance
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and is nothing but the resistance that we
already know rho that is the rho s is resistance
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h is the length and A is the cross-sectional
area through which the current is flowing
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Now rho S resistance that can be written as
1 over k that is the electrical conductivity
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of the electrolyte
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We already know from the this first lecture
of theory of ECM that m dot that is the mass
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mass removal rate is equal to IE over F Now
if we want to calculate specific material
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removal rate in terms of gram per ampere second
then you divide this particular equation by
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the current I that is amperes then we get
m dot s s indicates the specific material
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removal rate is equal to E over F because
I in the numerator and denominator cancel
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is equal to A dash over Z multiplied by 1
over A I have already explained in the earlier
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lecture that A dash is atomic mass Z is valency
of dissolution A dash divided by Z gives you
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the chemical equivalent E and F is the Faraday’s
constant
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Now specific volumetric material removal rate
represented as MRRsv suffix in cubic millimeter
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per ampere second this is given by m dot s
divided by rho w because when mass of the
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material removed you divide by its density
you get the volume of the material removed
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and here s indicates specific so you this
can be written as A dash over Z divided by
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96500 that is the Faraday’s constant multiplied
by 1 over rho w that is the density of the
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work piece material and zeta is the machining
efficiency which you can assume 1 for practical
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purposes
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We also know that the f p that is the permissible
feed rate is equal to current density multiplied
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by specific material removal rate you can
see the balance of the units current density
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is given by ampere per millimeter square and
specific material removal rate is given in
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terms of cubic millimeter per ampere second
when you simplify this you get millimeter
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per second that is the unit of the feed rate
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So the equation shown in the previous slide
can be written as I p over A where here A
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indicates the cross-sectional area through
which the current is flowing multiplied by
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A dash over Z into 1 upon 96500 zeta upon
rho w Now substitute the value of I p and
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simplify this equation because I p we have
already calculated as given here that we calculated
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in the earlier slide that 4.186 U e rho e
C e multiplied by T b minus T i divided by
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rho s h over A whole under root
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Now substitute this value of I p then we get
permissible feed rate f p is equal to zeta
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A dash over Z multiply by 1 upon 96500 multiplied
by under root whatever is the term for I p
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that is 4.186 U dot e rho e C e multiplied
T b minus T i bracket closed divided by rho
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s h A Now here it should be rho s h divided
by A not multiplied by A as written over here
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in the last equation for F p so please correct
it
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This equation that we have just derived can
also be used to evaluate the change in temperature
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for the given feed rate because feed rate
is going to control the temperature of the
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electrolyte because we know that feed rate
will control the inter electrode gap and depending
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upon the inter electrode gap the length of
this or the value of h will change that is
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the that is controlling the conductivity of
the electrolyte as a result of that the temperature
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will also will change
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So for the given machining conditions let
us calculate what is going to be the change
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in the temperature change in temperature is
nothing but T b minus T i that you can represent
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as delta T so substitute T b is equal to T
f now here when we are finding out the change
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in temperature it is not necessary that every
time the temperature of the electrolyte reaches
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to the boiling point so in place of this T
b you can replace it by T f that is the final
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temperature of the electrolyte in the previous
equation
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And then you simplify this and final equation
that you get is T f that is the final temperature
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of the electrolyte minus T i that is the initial
temperature of the electrolyte and this is
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nothing but delta T that is equal to 2.23
into 10 raise to power 3 multiply by rho s
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h A divided by U e dot rho e C e again multiplied
by f is the feed rate Z over A dash multiplied
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by rho w over n whole square
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We know that electrical conductivity k of
the electrolyte changes with various parameters
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namely temperature of the electrolyte sludge
contamination or reaction product of the electrolyte
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that or reaction product that get mixed up
in the electrolyte vapors or gas bubbles that
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are formed and get mixed up with the electrolyte
they also change the conductivities of the
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electrolyte
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Now the question is how to represent all these
effects in the form of mathematical equation
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we will see and another point is the temperature
gradient how the temperature is varying along
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the electrolyte flow direction suppose the
electrolyte is flowing from here towards this
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direction so what is the temperature over
here and how this temperature is varying in
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this direction is it increasing it is decreasing
or remaining the constant that also we should
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know and current density that is the J because
once k is varying and current density J is
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the function of the electrical conductivity
of the electrolyte so once K is varied how
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J is varying that also should be known to
us so let us derive these equations for the
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present purpose
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Electrical conductivity of the electrolyte
depends on many parameters sludge contamination
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gases evolved during electrochemical machining
and change in temperature that is capital
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T we know the dependence of electrolytes electrical
conductivity k is given as k is equal to k
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0 multiplied by 1 plus alpha T minus T0 bracket
close Now here k 0 is the electrical conductivity
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at the inlet point here the electrolyte is
entering at this particular point so whatever
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is the electrical k value here that is represented
as K0 the temperature at this particular point
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is represented as T0
And the temperature at the outlet that is
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here that is represented either as TF or in
this particular equation it has been written
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as just T alpha we already know let us see
this is the conductivity at any desired point
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Conductivity at the entry to the inter electrode
gap alpha is the temperature co-efficient
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of specific conductance
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Now in the previous equation we have not accounted
for the effect of gases evolved during ECM
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or void fraction so let us see accounting
for voids in the electrolyte the above equation
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can be modified as k is equal to k 0 1 plus
alpha delta T that is the change in temperature
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multiplied by 1 minus alpha V whole raise
to power n where alpha V is the void fraction
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and its value is always less than or equal
to 1 and n is the exponent
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Temperature gradient along the electrolyte
flow direction can also be evaluated from
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the fundamental equation as follows from the
law of conservation of energy we can write
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heat generated is equal to heat dissipated
that we have already seen now let us see this
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particular figure here tool work piece inter
electrode gap represented by value of Y and
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w small w represents the width of the tool
and also the width of the work piece and V
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is the electrolyte flow velocity
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So heat generated we have already calculated
that is given by I square R now I can also
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be represented is equal to V over R so I square
becomes V square over sorry yes V square over
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R square so using this particular equation
we can write this as V minus delta V square
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over R where delta V is the over potential
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Now important point here is that R of tool
and work piece are neglected it is assumed
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that the conductivity of the tool material
and work piece material are very high so that
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the resistance electrical resistance of the
tool and work piece are negligible so the
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heat generated due to the resistance of the
tool and the work piece are negligible and
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that are neglected here and we have taken
only the heat generated due to the ohmic heating
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of the electrolyte
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So this can be written like V minus delta
V whole square k divided by y over A now here
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note it that A is the cross-sectional area
through which the current is flowing in the
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electrolyte Now heat dissipated we have calculated
earlier also that is rho e w is the width
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of the path through which the electrolyte
is flowing y is the gap and v is the velocity
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of the electrolyte that is given in terms
of centimeter per second or meter per second
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C is the specific heat and delta T is the
change in temperature from this we can calculate
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the heat dissipated Now note here that we
are not considering any heat going to the
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tool any heat going to the work piece by conduction
or any heat loss due to the radiation
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So that equation can be written like this
V minus delta V whole square k w dx divided
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00:26:14,000 --> 00:26:26,670
by y is equal to rho e w y v C e delta T now
dx is the small moment in X direction so this
208
00:26:26,670 --> 00:26:33,110
equation can be written as dt over dx simplification
of the above equation will give you dt over
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00:26:33,110 --> 00:26:42,490
dx is equal to V minus delta V square k divided
by rho e y square v C e is equal to A k
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00:26:42,490 --> 00:26:49,940
Now please note here that V is constant it
is not varying during the process delta V
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00:26:49,940 --> 00:26:56,090
is also assumed to be constant that is the
over potential and that does not vary during
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00:26:56,090 --> 00:27:03,740
the ECM process that is assumption rho e remains
constant that is the electrolyte density inter
213
00:27:03,740 --> 00:27:10,140
electrode gap is assumed to remain constant
for a short period of time and at a short
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00:27:10,140 --> 00:27:18,440
length of the flow direction V is the velocity
of the electrolyte and C is the specific heat
215
00:27:18,440 --> 00:27:21,580
both of them are again considered to be small
216
00:27:21,580 --> 00:27:29,300
Let us take it like this that this is the
tool and this is the work piece and this we
217
00:27:29,300 --> 00:27:41,020
are assuming that at a small distance that
is dx these parameters are constant specially
218
00:27:41,020 --> 00:27:47,010
the inter electrode gap and flow velocity
they are considered to be constant that means
219
00:27:47,010 --> 00:27:53,170
only k remains as variable and k is electrical
conductivity of the electrolyte so we can
220
00:27:53,170 --> 00:27:59,470
write it as A dot k where A is constant now
please do not confuse this A with the cross
221
00:27:59,470 --> 00:28:12,850
sectional area this is not the cross sectional
area some constant is there
222
00:28:12,850 --> 00:28:20,140
If you see here w is shown over and y is also
shown both are shown over there you can see
223
00:28:20,140 --> 00:28:33,470
this is the w and this is the y
224
00:28:33,470 --> 00:28:40,970
Now as in the previous equation temperature
gradient is a function of K that is the conductivity
225
00:28:40,970 --> 00:28:47,650
whose equation has been derived in the previous
slides without accounting for the void fraction
226
00:28:47,650 --> 00:28:56,370
later on we accounted for the void fraction
also so this temperature gradient can be now
227
00:28:56,370 --> 00:29:04,080
be represented in place of k you can write
down k 0 multiplied by 1 plus alpha T minus
228
00:29:04,080 --> 00:29:07,350
T0 parenthesis closed bracket closed
229
00:29:07,350 --> 00:29:15,690
So this is the equation which can be used
to calculate the temperature gradient to find
230
00:29:15,690 --> 00:29:22,020
the final solution of this you can simplify
this equation like dt is equal to A k 0 multiplied
231
00:29:22,020 --> 00:29:32,560
by the value of the k dx now you can integrate
both sides from T0 to TI or TF or T and dx
232
00:29:32,560 --> 00:29:40,540
side you can integrate from 0 to x now integration
and then you have to evaluate the constant
233
00:29:40,540 --> 00:29:48,190
of integration and if you substitute the values
then you find this as T minus T0 is equal
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00:29:48,190 --> 00:29:58,390
to 1 over alpha multiplied by exponent alpha
A k 0 x A is the constant that I derived in
235
00:29:58,390 --> 00:30:03,640
the last slide note the cross sectional area
x minus 1
236
00:30:03,640 --> 00:30:09,630
Now from this equation you can find out what
is the temperature variation along the electrolyte
237
00:30:09,630 --> 00:30:15,100
flow and from the earlier equation dt over
dx you can find out what is the temperature
238
00:30:15,100 --> 00:30:22,840
gradient along the electrolyte flow path
239
00:30:22,840 --> 00:30:30,230
Now current density can also be evaluated
by considering the variation in the electrical
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00:30:30,230 --> 00:30:37,740
conductivity of the electrolyte and we all
know this equation we can derive after substituting
241
00:30:37,740 --> 00:30:46,240
the value of k in this equation V minus delta
V divided by y multiplied by k 0 that is the
242
00:30:46,240 --> 00:30:56,700
initial electrical conductivity of the electrolyte
multiplied by 1 plus alpha T minus T0 substitute
243
00:30:56,700 --> 00:31:03,820
the value of T minus T0 in this equation as
derived earlier then if you write this if
244
00:31:03,820 --> 00:31:08,880
you substitute the equation of T minus T0
you get final equation as like this J is equal
245
00:31:08,880 --> 00:31:20,250
to V minus delta V k 0 divided by y multiplied
by exponent alpha A k 0 into x
246
00:31:20,250 --> 00:31:27,060
From this equation we can finally calculate
the current density at a particular point
247
00:31:27,060 --> 00:31:31,980
after accounting for the variations in the
temperature as well as the void fraction
248
00:31:31,980 --> 00:31:47,310
Now this indicates this figure indicates that
how the current density is varying along the
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00:31:47,310 --> 00:31:53,950
electrolyte flow path you can see work piece
is shown there tool is shown and the shape
250
00:31:53,950 --> 00:31:59,490
of the work piece is continuously changing
from the inlet to the outlet it is not remaining
251
00:31:59,490 --> 00:32:07,370
as the straight line and as the electrolyte
flow path or you move along the electrolyte
252
00:32:07,370 --> 00:32:14,620
flow path the current density may decrease
because of the higher inter electrode gap
253
00:32:14,620 --> 00:32:22,500
as shown there the arrow moving showing downward
along with the capital J So this indicates
254
00:32:22,500 --> 00:32:30,310
that current density is varying continuously
along the electrolyte flow path
255
00:32:30,310 --> 00:32:36,870
Now with this introduction to the basic theory
of electrochemical machining let us attempt
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00:32:36,870 --> 00:32:48,250
to solve some of the numerical problems in
electrochemical machining cases Let us take
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00:32:48,250 --> 00:32:57,630
first problem in ECM process pure iron is
taken as a work piece material and the desirable
258
00:32:57,630 --> 00:33:04,330
volumetric material removal rate is 5 cubic
centimeter per minute determine the current
259
00:33:04,330 --> 00:33:08,500
required to achieve this particular material
removal rate
260
00:33:08,500 --> 00:33:17,570
The data given are volumetric material removal
rate as 5 cubic centimeter per minute from
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00:33:17,570 --> 00:33:24,830
the data book or a book from electrochemical
machining you can find out the density of
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00:33:24,830 --> 00:33:33,690
the pure iron as 7.85 grams per cubic centimeter
and the chemical equivalent of it is taken
263
00:33:33,690 --> 00:33:44,500
as 28 gram we already know that Faraday’s
constant is 96500 coulombs we also know this
264
00:33:44,500 --> 00:33:50,020
equation with the help of which we can calculate
the material removal rate in terms of gram
265
00:33:50,020 --> 00:33:58,740
per second and this is given by rho a multiplied
by MRRv where MRRv is volumetric material
266
00:33:58,740 --> 00:34:00,580
removal rate
267
00:34:00,580 --> 00:34:07,660
So we can write it like this IE over F is
equal to rho a multiplied by MRRv substitute
268
00:34:07,660 --> 00:34:16,300
because we already know that MRRV required
is 5 cubic centimeter per minute so substitute
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00:34:16,300 --> 00:34:26,710
that value rho a is known 7.85 coulombs Faraday’s
constant that is 96500 coulombs and we get
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00:34:26,710 --> 00:34:37,000
we have to convert this per minute into per
second that we divide by 60 So finally we
271
00:34:37,000 --> 00:34:45,360
get the current as 2254.53 ampere
272
00:34:45,360 --> 00:34:50,899
Now please note that all parameters should
be taken in the same units whether millimeter
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00:34:50,899 --> 00:34:55,190
then for everything you have to account for
millimeter or centimeter then everything should
274
00:34:55,190 --> 00:34:58,570
be in terms of centimeter or meter
275
00:34:58,570 --> 00:35:12,570
Now let us take next problem the composition
percentage by weight of Nimonic 75 super alloy
276
00:35:12,570 --> 00:35:19,240
is given as follows now here we find that
there are various constituents of the alloy
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00:35:19,240 --> 00:35:26,620
like nickel chromium iron titanium silicon
manganese and copper and it gives the percentage
278
00:35:26,620 --> 00:35:36,480
by weight composition of each element in the
alloy that is 72.5 percent nickel 19.5 percent
279
00:35:36,480 --> 00:35:44,220
chromium and so on the density of each element
that is nickel 8.9 gram per cubic centimeter
280
00:35:44,220 --> 00:35:50,440
chromium 7.19 gram per cubic centimeter and
so on are given over there and the atomic
281
00:35:50,440 --> 00:36:01,470
weight of each of them is given like 58.71
51.99 and so on
282
00:36:01,470 --> 00:36:08,250
We have to calculate material removal rate
in terms of cubic centimeter per minute that
283
00:36:08,250 --> 00:36:13,910
is volumetric material removal rate when a
current of thousand ampere is used Use the
284
00:36:13,910 --> 00:36:19,520
lowest valency of dissolution for each element
Here you are not given with the valency of
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00:36:19,520 --> 00:36:24,850
different constituents of the alloy that is
nickel chromium etc that you have to find
286
00:36:24,850 --> 00:36:30,760
out from the handbook or the book on the electrochemical
machining or advanced machining processes
287
00:36:30,760 --> 00:36:37,140
as I have already told you from there you
can find out these values of the valency of
288
00:36:37,140 --> 00:36:39,060
various elements
289
00:36:39,060 --> 00:36:45,680
Other data that are given here is the current
that is 1000 ampere chemical equivalent E
290
00:36:45,680 --> 00:36:52,720
is equal to A by Z that we have already discussed
A is the atomic mass of the element and Z
291
00:36:52,720 --> 00:36:57,450
is the valency at which it is dissolving and
you are given that you have to take the lowest
292
00:36:57,450 --> 00:37:03,920
valency of dissolution of any element and
F we already know Faraday’s constant
293
00:37:03,920 --> 00:37:11,950
Now volumetric material removal rate we know
is given by I E a over rho a F here please
294
00:37:11,950 --> 00:37:21,860
note that E a is the chemical equivalent of
the alloy and rho a is the density of the
295
00:37:21,860 --> 00:37:30,080
alloy we have to find out chemical equivalent
of the alloy that is not given and we also
296
00:37:30,080 --> 00:37:35,390
have to find out the density of the alloy
while we are given the density of the individual
297
00:37:35,390 --> 00:37:36,740
element
298
00:37:36,740 --> 00:37:42,430
We can find out the chemical equivalent of
the alloy from one of the methods that we
299
00:37:42,430 --> 00:37:49,780
have already discussed that is E a is equal
to 100 divided by summation X i Z i over A
300
00:37:49,780 --> 00:37:56,780
i Now as I have mentioned you have to find
out the Z i from the handbook or the book
301
00:37:56,780 --> 00:37:58,820
on the advance machining processes
302
00:37:58,820 --> 00:38:09,510
Now taking those values you can see here E
a is equal to 100 divided by 72.5 into 2 by
303
00:38:09,510 --> 00:38:16,530
58.71 so here 72.5 is the atomic mass of the
element of the alloy that is the nickel 2
304
00:38:16,530 --> 00:38:26,950
is its valency and so 58.71 is the that is
the atomic mass of the nickel so using these
305
00:38:26,950 --> 00:38:34,640
values for different elements we can calculate
the value of chemical equivalent of the alloy
306
00:38:34,640 --> 00:38:40,420
as 27.68 gram
307
00:38:40,420 --> 00:38:48,700
Same way we have the equation for calculating
the density of the alloy that is 100 divided
308
00:38:48,700 --> 00:38:58,970
by summation X i over rho i rho i is the density
of the I F element X i is its percentage contribution
309
00:38:58,970 --> 00:39:06,200
to the alloy substitute the values and take
the density from the previous table that is
310
00:39:06,200 --> 00:39:12,900
given in the problem you can substitute these
values and you get this one and by simplification
311
00:39:12,900 --> 00:39:21,020
of this you will get the density of the alloy
as 8.187 gram per cubic centimeter
312
00:39:21,020 --> 00:39:25,840
You need not to take it to 4 decimal place
or 5 decimal place normally you should take
313
00:39:25,840 --> 00:39:35,870
it to only 2 decimal places that is good enough
so we can now calculate the volumetric material
314
00:39:35,870 --> 00:39:41,060
removal rate that is I E a over rho a over
F we have already calculated the value of
315
00:39:41,060 --> 00:39:50,340
E a as 27.68 gram and rho a as 8.187 gram
per cubic centimeter substitute these values
316
00:39:50,340 --> 00:39:58,790
in this particular equation I is 1000 ampere
and this is a constant and 60 is the conversion
317
00:39:58,790 --> 00:40:02,770
of minutes into the seconds
318
00:40:02,770 --> 00:40:14,800
You get volumetric material removal rate as
2.102 cubic centimeter per minute now if you
319
00:40:14,800 --> 00:40:21,230
want to find out the material removal rate
in terms of grams then you can divide this
320
00:40:21,230 --> 00:40:30,110
by the density of the alloy you will get the
gram per minute as the mass material removal
321
00:40:30,110 --> 00:40:32,350
rate
322
00:40:32,350 --> 00:40:42,020
Let us take the third problem in an electrochemical
machining operation 10 volt direct current
323
00:40:42,020 --> 00:40:51,750
power supply is used the conductivity of the
electrolyte used is 0.2 per ohm per centimeter
324
00:40:51,750 --> 00:40:55,050
and the feed rate used is 1 millimeter per
minute
325
00:40:55,050 --> 00:41:04,040
The work piece is of pure iron calculate the
equilibrium gap that is Y e take over voltage
326
00:41:04,040 --> 00:41:18,869
equal to 1.5 volt Here let us assume that
atomic mass of the iron is 55.85 gram valency
327
00:41:18,869 --> 00:41:30,520
at which the iron is dissolving is 2 and density
is 7.86 gram per cubic centimeter The data
328
00:41:30,520 --> 00:41:37,960
that are given in the problem are conductivity
is 0.2 per ohm per centimeter we can calculate
329
00:41:37,960 --> 00:41:44,619
chemical equivalent A by Z from these two
values that comes out that you can calculate
330
00:41:44,619 --> 00:41:46,710
and density is already given over here
331
00:41:46,710 --> 00:41:55,070
We already know Faraday’s constant feed
rate is given there as 0.1 millimeter per
332
00:41:55,070 --> 00:42:05,300
minute and equilibrium inter electrode gap
we have to calculate Now we know the equation
333
00:42:05,300 --> 00:42:15,300
of equilibrium inter electrode gap is equal
to V minus delta V zeta k E over rho F f now
334
00:42:15,300 --> 00:42:23,830
this F is the feed rate Substitute the values
of this because this is the 10 volt given
335
00:42:23,830 --> 00:42:29,590
over there this is the over potential other
values like efficiency note it here efficiency
336
00:42:29,590 --> 00:42:36,430
has been taken as 100 percent that is 1 unless
it is given you can always take it as 1
337
00:42:36,430 --> 00:42:48,020
k is 0.2 and E is 55.85 and you can take this
as the density Faraday’s constant this is
338
00:42:48,020 --> 00:42:58,650
the feed rate now substituting all these values
we get Y e is equal to 0.375 millimeter So
339
00:42:58,650 --> 00:43:06,380
this gives you the equilibrium inter electrode
gap now in this way you can solve various
340
00:43:06,380 --> 00:43:20,380
real life problems using the theory that we
have discussed in these two lectures So thank
341
00:43:20,380 --> 00:43:25,110
you very much