WEBVTT
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good morning everybody myself ritwik maiti
research scholar from department of mechanical
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engineering iit kharagpur so this is a tutorial
class on adiabatic two phase flow and flow
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boiling in microchannel this class has a two
section in the first section mister alex koshy
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will discuss some numerical problems related
two phase flow and it is based on the classes
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which have been conducted and this is a part
of assessment of this course so he will discussed
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one problem related to normal type and one
problem related to calculation of pressure
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drop in in a typical two phase flow when it
flow through a conduit and in the second section
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mainly content with the computation of fluid
dynamics in two phase flow and this is a not
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part of this assessment of this course in
this part mister vinmay dhor will discussed
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in c f d and c f d in two phase flow he will
also discussed in briefly the stimulation
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of two phase flow with the help of open source
like jets and then mister abhir chakravarthi
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will also discuss simulation of two phase
flow to find out the flow pattern and pressure
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drop with the help of commercially used package
like console now i am just hand over this
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class to master alex koshy thank you
hello everyone i am alex koshy i am the the
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t f of the course adiabatic two phase flow
and flow boiling in microchannel so in this
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tutorial i am going to explain few a numerical
problems that are related to multiphase flow
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so the first problem as mister ritwik maiti
has already given introduction about what
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my session will be the first problem is a
nomenclature related problem and the second
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one is a typical multiphase flow problem that
d s with the pressure drop calculation that
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is involved when a two phase mixture flow
through a micro channel
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so i am going straight to first problem so
i will read it once derive the relations to
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express x the mass quality in terms of volumetric
quality beta and phase densities rho l and
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rho g
mass quality of air in the system suppose
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it is an air water system the mass quality
air in the system can be represented as w
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a by w l plus w a where a is the w a is the
mass flow rate of air and w l is the mass
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flow rate of liquid and it can be written
as q a rho a by q a rho a plus q l rho l ok
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where q a and q l are the monometric chlorides
of air and liquid phase and rho a and rho
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l are the phase densities in the similar way
we can write one minus x that is a mass quality
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of liquid phase as q l rho l by q a rho a
plus q l rho l all right here now this is
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equation one and this is equation two so we
are [multiply/dividing] dividing one by two
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therefore we will get any equation like x
by one minus x is equal to q a rho a by q
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l rho l
now i am dividing both the numerator and denominator
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of this equation of this term with q a plus
q l ok now if you see we are having q a divided
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by q a plus q l term here in the numerator
and q l by q a plus q l term term here
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in the denominator and this is nothing other
than beta that is volumetric as a volumetric
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ratio volumetric no ratio at the inlet and
here q l by q a plus q l is one minus beta
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ok
now i am simplifying that equation by cross
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multiplication
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ok now after further simplification we will
get the value of x as we will we can express
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x as thus we have expressed x the mass quality
in terms of beta and phase densities now we
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go to the second part that is the slip ratio
k it need to be expressed as a function of
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a alpha and beta only as we know k is the
ratio of incentive velocities of gas phase
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divided by incentive velocities of liquid
phase and this can be written as
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q g by a g divided by q l by a l where q g
and q l are the volumetric chlorides of gas
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phase and liquid phase and a g and a l
suppose we are having a conduit we are having
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a annular flow this is the a l region and
this is the gas region and this is the liquid
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region the area the fraction of area i am
sorry the area obtained by the gas region
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is turned as a g and the area obtained by
the liquid region is termed as a l this a
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g can be written as a into alpha and a l can
be written as a into one minus alpha and this
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a and this a will be cancelled out thus we
get k as q g by alpha divided by q l by one
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minus alpha and i divide both numerator and
denominator with q g plus q l all right and
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now if we see this term is as we have done
in the as we got in the first problem this
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term can be written as beta and this term
can be written as one minus beta
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therefore we got k this slip ratio as beta
by alpha divided by one minus beta by one
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minus alpha and it can be further simplify
to beta by one minus beta into one minus alpha
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by alpha so now we go to the second problem
i will read the problem once air and water
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flow through a horizontal conduit of a diameter
four hundred micro meter and length four hundred
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centimeter assuming homogeneous isothermal
flow at room temperature calculate the frictional
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pressure drop across the conduit
so we have been provided with certain data's
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that are the superficial velocities of
liquid and gas phase that is sixty millimeter
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per second and ten millimeter per second and
we have been provided with the phase densities
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as well as phase viscosities now before going
into the frictional pressure drop estimation
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i would like to tell you something
the pressure drop across a conduit can be
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categorized into three components first one
is gravitational pressure drop and second
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one is frictional pressure drop and the third
one is accelerational pressure drop
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so here in this case we are neglecting gravitational
pressure drop because first of all it's
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a horizontal conduit and secondly the amount
of liquid that is contained in the conduit
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is very very small so we can neglect the gravitational
pressure drop here and we will be calculating
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frictional pressure drop and talking about
the accelerational pressure drop we can neglect
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the accelarational pressure drop if there
is no phase change as well as there is no
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compressibility effects involved in the system
so suppose if the relative change in specific
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volume across the specific volume happening
across the system is greater than five percent
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we have to consider acceleration pressure
drop if it is less than five percent we will
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neglect accelerational pressure drop so for
the time being i am neglecting accelarational
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pressure drop i am assuming that there is
no accelarational pressure drop is involved
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because mainly there is no heating of
conduit heating of conduit occurring there
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is no phase change occurring
so i am taking only frictional pressure drop
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as of now and at the end i will cross check
whether of my assumption of [vocalized-noise]
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neglecting acceleration pressure drop is right
or wrong so now going to the frictional pressure
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gradient calculation it can be calculated
as darcy weisbach equation and the equation
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is shown in the screen diagram while looking
in the screen we can find out that there
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are plenty of what do say plenty of unknowns
involved here one is f t p one is g t p and
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one is rho t p and d h is a hydraulic diameter
here as it is circular channel we can take
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the hydraulic diameter as they are diameter
of the conduit and g t p we need to find out
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the g t p as a total mass flux of the two
phase mixture across a conduit
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it is a sum of g l and g g that can be written
as rho l j l plus rho g j g well j g and j
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l are the superficial velocities phase velocities
and rho l rho g are the phase densities we
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know all these values these values have been
given and we got the values of g t p as sixty
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point zero one two two five kilogram per meter
square second i am writing the value straight
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away because i am having a time restriction
that's why i am not doing the calculation
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i have already calculated it and
now rho t p it can be found out as rho g into
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alpha plus rho l it one minus alpha alpha
is a incentive volume factor as we are assuming
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it is a homogeneous flow model we can take
alpha as beta that is a in a volumetric fraction
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and it can be written as q g divided by q
l plus q g by q g and q l are the phase volumetric
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chlorides it can be written as a into j g
by a into j l plus j g with j g j g and j
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l r are the superficial velocities of phases
thus we can cancel out a and a from numerator
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and denominator thus it becomes j g by j l
plus j g we got the values of j g and j l
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thus we got the value of alpha as point one
four three and one minus alpha as point eight
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five seven in we substitute the value of alpha
and one minus alpha here we got the value
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of rho t p as eight fifty seven point one
seven five kilogram per meter cube
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so now we got g t p we got rho t p if we look
at the equation we see that only one unknown
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is remaining that is f t p so for the estimation
of the f t p where f where f is the infraction
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factor to phase mixture ok suppose it was
hm a single phase rho case what we would have
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done we would have found out the reynolds
number r e associated with the process and
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if r e comes below to one zero zero that is
the laminar regime we would have taken sixteen
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by n r e as sorry sixteen by r e as in the
fan infraction factor value and if r e is
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greater than to two one zero zero that is
turbulent regime we would have either took
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the value of f as a constant that is zero
point zero zero five or we would have use
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the co relation zero point zero seven nine
r e raised to minus point two five or you
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would find the value of r e from moodys plot
here i have already found out the value of
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r e and the value of r e is twenty four i
mean value of r e t p two phase reynolds number
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is twenty four and that's comes in the laminar
regime but we cannot use this equation because
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this equation does not does not take into
account the wa wald interactions the wald
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effects that are involved in the flow process
so we are not taken this equation instead
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of that we are going to another equation which
is termed as the churchill equation and this
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churchill equation is applicable only for
n x x value that ranges from zero to zero
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point two five x value is the the mass
mass fraction value mass quality ok
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so we need to calculate the mass quality first
of all whether to see if it is come in the
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in this range zero to zero point two five
so mass quality x can be found out as w g
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by w l plus w g
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here we are cancelling out the a value thus
all the remaining value we have already hm
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it's already given j g value is given rho
g rho l j l all are given thus we got the
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value of e as two point zero four into to
ten raised to minus four and if we see this
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value it comes in the range zero point two
five and zero so we can use churchill equation
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for our calculation for the estimation
of the f t p if we look at the equation there
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are plenty of unknowns here there is an a
there is a b there is r e t p that are we
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have already found out hm and there is
a c one for circular tubes or value of c one
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given as eight and for r e t p estimation
we know that r e t p is equal to d into g
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t p by mu t p we know the value of g d we
know the value of g t p we need to find out
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the value of mu t p and mu t p can be found
out using equation three here if we see
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here if we see we have got the values of
mu g we have got the values of mu l and we
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got the value of x
so substituting all these values we will get
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the values of mu t p and if we substitute
the values of mu t p here we will get the
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value of r e t p as twenty four ok and this
comes in the laminar range and we substitute
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the value of r e t p here we need to find
out a and b value and find out a and b value
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we got equations that is given under equation
four and b value we can straight from the
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r e value that we have got and a value for
that we need to find out hm one by root c
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t and for circular tubes the value one by
root c t as given as two point four five seven
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and we know the value of r e here there is
a there is a term that includes hm component
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known as epsilon d that is the roughness factor
involved in the process
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here we are assuming that it is a smooth conduit
so we take the value of epsilon d as zero
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thus hm hm the whole term zero point two seven
epsilon d by d cancel out so we know the value
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of r e we know the value of one by root c
t thus we got the value of a and b as
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ok and now i substituted all the values here
a b r e t p have got i substitutes all the
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values and i got the value of fan infraction
factor f t p as to two point six seven and
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i have i have substituted the value of fan
infraction factor here in the darcy weisbach
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equation and i got the pressure gradient d
p by d z f t p as five six zero nine zero
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point seven nine pascal per meter and now
we need to find out the pressure drop value
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from the pressure gradient value it can be
founded using this equation using integration
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across the conduit length and this is equating
minus d p by d z f t p into l l we know it
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is the value of the l is four centimeter that
is zero point zero meter and by doing the
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calculation we get the value of delta p pressure
drop as two two four three point six three
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pascal
now we got the value of pressure drop using
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homogeneous flow model now initially i told
you that we have neglected the acceleration
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pressure drop value so we need to cross check
whether our assumption to neglect acceleration
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pressure drop value was right or wrong so
in order to find out that i assumed that the
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inlet pressure is p i and the inlet specific
volume is v i and the inlet temperature is
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the room temperature that is two ninety eight
kelvin the outlet temperature is the same
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because is the isothermal process it is two
ninety eight kelvin outlet pressure it is
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the atmospheric pressure it is one zero one
three two five pascal outlet volume we need
19:45.910 --> 19:49.740
to calculate we need to calculate v zero we
need to calculate v i and we need to check
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whether this relative variation is greater
than five percent or less than five percent
19:56.650 --> 20:07.390
so we have calculated the value of p v i and
v zero from ideal gas flow
20:07.390 --> 20:18.600
value of r we know it is eight point three
one four joule per kelvin mole and now substituting
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all the values i got the value of v i as zero
point zero two three nine meter cube i got
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the value of v zero as zero point zero two
four four five and hm i forgot to mention
20:33.510 --> 20:39.630
i got the value of p i by adding the outlet
pressure that's the pressure drop and the
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values p i is one zero three five six eight
point six three pascal
20:46.640 --> 20:52.230
now we have the v i value we have the v zero
value we need to find out the relative change
20:52.230 --> 21:07.500
in volume specific volume delta v by v i and
i got it in two point three percent that's
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obviously less than five percent thus we are
neglecting any kind of compressive effects
21:13.820 --> 21:18.380
and thus we have assume that our assumption
to neglect the acceleration pressure drop
21:18.380 --> 21:27.720
was right
and now we have found out the pressure drop
21:27.720 --> 21:32.470
using homogeneous flow model and i am repeating
the same method means repeating the same process
21:32.470 --> 21:37.020
that is finding out the pressure drop across
the conduit using another method that is the
21:37.020 --> 21:46.500
method that uses martinelli parameter this
is approach two here martinelli parameter
21:46.500 --> 21:54.470
chi square is usually phi g square phi l square
where phi g and phi l are two phase multipliers
21:54.470 --> 22:10.140
that can be written as minus d p by d z f
l and g this term is the pressure drop that
22:10.140 --> 22:16.429
is involved when the liquid flows along through
the conduit at a mass flow rate at at a mass
22:16.429 --> 22:22.440
flux of g l and this happens when gas flow
alone through the conduit at a mass flex of
22:22.440 --> 22:29.700
g g and we use the same procedure that is
the darcy weisbach equation darcy weisbach
22:29.700 --> 22:36.750
equation plus the f t p estimation churchill
equation we use the same thing for the
22:36.750 --> 22:42.860
calculation of single space pressure drops
pressure gradients and we got the single phase
22:42.860 --> 23:04.350
pressure gradient values as
23:04.350 --> 23:10.169
we substitute these values here and we got
the value of chi square as three zero two
23:10.169 --> 23:19.780
point nine five ok and now we have a correlation
known as a chisholms correlation there we
23:19.780 --> 23:24.580
substitute the value of chi square and chi
and we get the value of phi l square that
23:24.580 --> 23:30.410
is the value of two phase multiplier and here
also we got an unknown that is c for viscous
23:30.410 --> 23:35.330
viscous cases that is the case that we have
encountering now because hm the renoylds number
23:35.330 --> 23:40.490
involved when liquid flows alone through the
pipe as well as conduit as well as gas flows
23:40.490 --> 23:44.830
alone through the conduit both comes under
the laminar region the laminar regime
23:44.830 --> 23:51.100
so we take viscous viscous case here we take
the value of c as five and we substitute the
23:51.100 --> 24:10.029
values in this equation
24:10.029 --> 24:23.900
we got the value of phi l square as one point
two nine one ok now what is phi l square phi
24:23.900 --> 24:29.370
l square is minus d p by d z f t p that is
the chisholms pressure drop involved when
24:29.370 --> 24:35.020
two phase mixture flows through the conduit
by chisholms pressure drop involved when liquid
24:35.020 --> 24:40.950
phase flow involved to the conduit at a mass
flux g l and here we need to find out these
24:40.950 --> 24:46.950
value this is our pressure drop pressure gradient
and this value we already know we already
24:46.950 --> 24:55.450
found it out and the value was found out we
four eight zero six zero we substitute the
24:55.450 --> 25:01.049
value here and we multiply four eight zero
three zero that is this value under this value
25:01.049 --> 25:13.750
and we got the two phase frictional pressure
drop value as
25:13.750 --> 25:23.760
ok and from this value we got the pressure
drop value as we found out earlier as two
25:23.760 --> 25:29.570
four eight one point eight two pascal and
this is the pressure drop value that is calculated
25:29.570 --> 25:37.840
using chisholms correlation
we found out earlier that the value of pressure
25:37.840 --> 25:54.559
drop calculated choosing homogeneous method
was hm if we see we can find out a small different
25:54.559 --> 26:00.049
between the these two values these two values
were obtain form two different methods
26:00.049 --> 26:05.020
and this difference is basically due to fact
that in chisholms correlation it is assumed
26:05.020 --> 26:09.130
that the there is no interaction between the
phases at the interphase and in homogeneous
26:09.130 --> 26:13.700
flow model it is assume that complete interactions
occurs at the phase interphase complete in
26:13.700 --> 26:19.409
ration occurs between the phase at the interphase
ok and these two are the extremes i will show
26:19.409 --> 26:25.400
you a graph this has been taken from a literature
and the top curve it denotes the phi l square
26:25.400 --> 26:30.750
that is two phase multiplier value for chisholms
correlation method and the bottom curve denotes
26:30.750 --> 26:37.059
the two phase multiplier value of obtain from
a homogeneous method and in this case at low
26:37.059 --> 26:41.179
mass quality values we can see a deviation
and in this deviation is due to the fact that
26:41.179 --> 26:47.710
i have explained earlier and in reality the
the pressure drop value will be somewhere
26:47.710 --> 26:55.740
in between these two values
so before winding up i will just first summarize
26:55.740 --> 27:00.230
what i have all done in the second problem
i have adopted two methods in order to calculate
27:00.230 --> 27:05.029
the pressure drop of two phase flow through
a conduit and first method is homogeneous
27:05.029 --> 27:09.130
flow model method and second one is chisholms
correlation method and in the homogeneous
27:09.130 --> 27:14.660
flow model method i found out the values of
g t p rho t p f t p but f t p value is found
27:14.660 --> 27:19.340
out using churchill equation and i have substituted
the values in the darcy weisbach equation
27:19.340 --> 27:24.559
to get the value of two phase pressure gradient
and thus the two phase pressure drop and then
27:24.559 --> 27:28.630
i cross check whether the assumption to cancel
out acceleration pressure drop was right or
27:28.630 --> 27:35.340
wrong by checking if the relative change in
specific volume was less than five percent
27:35.340 --> 27:39.549
and in chisholms correlation method i found
out the value of martinelli parameter using
27:39.549 --> 27:45.620
values of single phase that is water and air
pressure gradient and from the value of martinelli
27:45.620 --> 27:49.659
parameter using chisholms correlations i found
out the value of two phase multiplier phi
27:49.659 --> 27:55.550
l square and from there from the values
of the two phase multiplier as well as liquid
27:55.550 --> 28:00.390
phase pressure drop i found out the two phase
pressure gradient and pressure drop that's
28:00.390 --> 28:05.990
all from me now mister will be explaining
the basis of c f d analysis of two phase flow
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using software
thank you