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[noise]
Welcome back, in this lecture we will continue
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the review of the properties continue from
the previous lecture on the properties. So,
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we will start with the ideal gas equation
of state, now the ideally from the ah ah perspective
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of thermodynamics we would prefer to have
analytical equation of state, where it represent
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the state in terms of various different thermal
properties [noise] such as PVT and one of
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the classical example is basically ideal gas
equation of state.
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So, let me just first formally define what
is any equation of state? An equation that
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relates the pressure temperature and specific
volume, so this essentially will be [noise]
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your equation of state. So, equation of state
is nothing, but some function which relates
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P V T and the simplest and the best known
equation of state is ideal [noise] gas equation
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of state, in principle at low pressure and
high temperature all gases approach ideal
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gas behavior.
So, typically a gas will behave like idle
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we call this conditions the gas it does not
have any effective interaction between themselves
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and essentially there is no energy; that means,
it can have a kinetic energy, but it [noise]
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cannot it does not have any potential energy
ok.
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So, such a of course, gross approximation
for gases does wonderfully work in certain
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range selective regions and something which
we will be talking about it. So, what is the
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equation of state for ideal gas is nothing,
ah but PV is equal to NRT, where n is of code
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number of moles R is a gas constant and temperately
is the temperature now of course, we cannot
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represent properties of many [noise] different.
Substances by simply ideal gas equation of
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state and thus for real fluids there are many
equations states which has been developed
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one of the classical 1 due to wonder wall.
So, wonder Waal equation of state in early
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nineteen thirties the famed equation included
the terms related to attractive forces between
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the molecules and effectively the volume of
the molecules. So, this due to the attractions
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it was this volume and thus if you consider
natural gas this is going to be 0 a is equal
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to 0 b is equal to 0 and you're going to get
the same equation as an idle gas equation
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of state.
Now, in order to obtain the parameters a b
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for different fluids 1 can consider PV plot,
and we know that at critical point there is
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inflection point or critical point is represented
by the inflection point. Thus the criticality
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or the critical temperature and pressure can
be used to evaluate the properties or valid
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this constant a and b.
So, using these 2 equation for the critical
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isotherms, we can obtain the values a and
b and those you can use equation of state
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for many different substances, if you know
the critical point of the substances you can
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directly get a l b for a specific fluid or
gas, and this you can find out the equation
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of state control equation state for a specific
gas.
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Now, of course, including only this attractive
force and this size is not good enough and
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that is the accuracy of the wonder Waal equation
the state is often inadequate, and this is
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the reason that many different equations of
state have been developed and this the listed
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are just like Burt Leigh, Rutledge Kwan and
so forth.
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Peng-robinson equation of state also and,
so I will not describe in all our details
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something which we can use it and some aspect
we will be using in the later part, but there
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are many different equations of stated some
are popular and some are very specific (Refer
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Time: 04:22) is an equation of state is a
motion one of the most popular equation of
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state for chemical systems.
Particularly using idle gas, so this is an
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example where the gauge pressure of an automobile
tire is measured to be 200 and 10 kilo Pascal
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before a trip, and 220 kilo Pascal after the
trip at a location, where the atmosphere pressure
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is 95 kilo Pascal so; that means, the total
pressure P 1 is equal to 210 plus 95 kilo
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Pascal, and P 2 is 220 plus 95 kilo Pascal
and what is being said that assumed the volume
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of the tire remains constant and the air temperature
before the trip is 25 ok.
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So, it is a 25 plus 273.15 Kelvin, now what
we need to find out is T 2, now this temperature
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25 degree Celsius is way above the critical
point, and it can be considered an ideal gas.
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So, considering a as an ideal gas we can simply
use P 1 V 1 is equal to and T 1 P 2 V 2 is
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equal to N R T 2 and thus since the volume
is constant P 1 by P 2 is equal to T 1 by
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T 2 and you can find T 2 from this simple
relation.
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You can plug these values in order to get
the T 2. So, of course, ideal gas equation
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makes life easier in terms of its application,
but it works out to be quite well even if
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you use equation of state very, very complex
a to create constant equations 9, constraint
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equation 5, constant equations the value may
not change for this particular example.
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Ok, but maybe it may change if the conditions
are watch well below temperature settings,
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where it can be let us say saturated system
and so forth. So, as I said our intention
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is to have some kind of equation of state,
but many times we fail and for complex molecules
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is extremely difficult to have form a formulation
with a function or the equation which relates
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P v T for all conditions, and thus property
table is more popular for very commonly used
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fluid such as water refrigerant ok.
So, therefore, the properties are frequently
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presented in the form of tables now there
are some properties which can be measured
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each easily, but others such as entropy and
so forth. You need to use a some kind of relation
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from the measurable quantities in order to
evaluate such a function. So, you know usually
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what we do is we measure and calculate all
these properties, some are easily measurable,
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some are dependent on the measurable quantity
and present in the table which in the convenient
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form, one of the properties enthalpy which
is often encountered in control volume cases.
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So, typical table will look like this this
is a ka table of saturated water, where we
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say is a pressure table the reason for that
is that the first column is a pressure corresponding
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table is saturation temperature of this because
the saturation water. So, saturation temperature
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and then you have saturated liquid vapour
saturated liquid internal energy such a vapor
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into energy and the difference of that. So,
U g minus U f is U f g and similarly that
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for enthalpy ok.
So, to give you an example here, you have
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a case where the tank contains a saturated
liquid at 90 degree Celsius. So, essentially
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on the T v diagram you since, T 90 degree
Celsius is less than critical point and its
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saturated. So, it will be somewhere below
here, but what would be the corresponding
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pressure for T 90.
So, we will look at specifically saturated
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water temperature table where the temperature
is the first in the column. So, corresponding
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pressure will be your saturated pressure.
So, here is a temperature and the corresponding
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pressure is 90 70.183 and this is where the
system is because its saturated liquid it
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is not saturated vapor liquid otherwise it
will have it would have been somewhere in
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between, and it is not certainly the saturated
vapor otherwise it should have been here ok.
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So, as I said for a given pressure and volume
table or temperature column table these 2
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points at the end one corresponds to saturated
liquid, 1 correspond to saturated vapor for
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the case of or let us say 100 kilopascal,
you have a saturated liquid here and saturated
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vapor of course, will have a different volume
which will be substantially larger then the
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1 contains by such was a liquid. So, for a
case of 100 Pascal you are looking at a pressure
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table.
So, the 500 Pascal this will be your saturated
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Vf. So, V f is this and saturated Vg saturated
vapor get a volume is going to be vg which
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will be 1.6941 meter cube per kg.
So, this is how we try to make use of the
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table in order to evaluate the volumes and
other properties similarly, you can also find
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the corresponding Uf here, Ug here, Hf here,
Hg here from the same table ok.
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So, what about when the system is at 2 phase
mixture; that means, the vapor and liquid
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both so; that means, within the saturated
regions. So, we in order to define such a
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case we need to find out the quality of it.
So, quality is basically the relative amount
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of liquid and vapor and the way it is represented
is the mass ratio of the mass of the vapor
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in the system; that means the ratio of the
mass of the vapor and total mass ok. So, the
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total mass will be your m liquid plus vapor
mf plus mg m x is equal to m vapor ok.
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So, you one can show clearly that having this
definition, where m total is equal to m fluid
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plus m gas and x is m gas by m total, one
can show that the average volume of the system
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is nothing, but Vf plus x multiplied by VG
minus Vf, and similarly for internal energy
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similarly for enthalpy.
So, where s x in terms of the volume is going
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to be the Vg minus Vf divided by V fg. Now
from the point of view of a plot if you look
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at P or T versus V you can use the diagram
to find out x also for example, if b is at
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which our system is then we can use the length
from a to b and a to see in order to get x
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here. So, this is directly from this expression
ok.
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Now, our in general you can write any property
y average as Y f which means the property
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y of the fluid plus x multiplied by Y f g
or the difference between the property Y in
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the gas phase a saturated gas phase and minus
saturated liquid phase. So, in general Y average
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is going to be in between Y f and Yj and I
could be Vu h. So, that is what we say here.
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So let me just make use of this understanding
to do a simple example, widget tank contains
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10 kg of water at 90 degree Celsius. If 8
kg of water is in the liquid form and the
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rest is in the vapor form, determine the pressure
in the tank and the volume of the time. So,
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what we have is information is that it is
at 90 degree Celsius and the total mass is
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given as 10 kg 8 kg is in the liquid form
so; that means, the mass of the vapor is 2
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kg. So, x is 2 kg by 10 kg this is going to
be point 2 ok.
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Now, we have been given ah ah ninety degree
and certainly it is a saturated system. So,
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we are going to look at saturated water temperature
table, which gives you the temperature and
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the corresponding pressure is this soapy set
of the tank at 90 degrees Celsius is 70.183
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kilo Pascal right. Now what is the volume
[noise] of the tank, so volume of the tank
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we can simply use this information that the
Vf and Vg is this. So, V is V f plus x V g
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minus V f that is of course, V f g. So, we
can write down here the point 0 0 1 0 3 6
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plus 0.2 multiplied by 2.3593 ok.
So, that will get V here in terms of meter
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cube per kg, now in order to get the volume
we need to multiply V by 10 kg. So, this is
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nothing, but V average right. So, V average
by into 10 kg and that will give us the volume
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of the tank. So, tank volume, so a simple
way of using the tables to get the information
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for this case ok.
So, let us continue further with another example
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a mass of 100 gram of saturated liquid water
is completely vaporized, at a constant pressure
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of 100 kilopascal and we have to determine
the volume change and the amount of energy
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transferred to the water. So, what we have
is a mass which is point 2 kg and the pressure
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is given here.
So, now essentially since if the pressure
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is here, we are going to make use of the saturated
water constant pressure table and since it's
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already informed that there is a complete
vaporization. So, it is going to be saturated
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system, so on a PV diagram it looks like this,
so its 100 kilo Pascal and its being said
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that its completely vaporized from here to
here, and we have been asked to find out the
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change in the volume from this to this and
as well as the amount of energy transferred
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within this so; that means, what will be the
difference here also.
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So, we can directly take a look at the tables,
the tables are given here 100 is given here
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pressure kilopascal corresponding Vfe is here.
So, of course, a Vfg is 1.6941 minus this
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and the internal energy is given the difference
between the gas and liquid is already given
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here, so we want to directly use this ok.
So, that means, the volume is point 2 multiplied
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by 1.6, let me say again 1.6941 minus point
0 0 1 0 4 3. So, that is going to be the volume
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of the system whereas, U is going to be 2408.2,
which is nothing, but U fg multiplied by point
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2, and that will be your kilojoules. So, that
is the delta U transferred in the system and
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this will be your delta V of the change in
the volume.
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So, this is a example, of using the saturated
water tables now of course, many times we
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obtain phases where the it is a superheated,
which is basically this region now superheated
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region is in the right side of the saturated
vapor line.
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So, the way the way to represent or to find
out this operated whether the system is at
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superheated let us say at a constant pressure
is by looking at is property, they should
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be greater than the saturated property if
this is the saturated vapor point, then the
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corresponding that is enthalpy is a g then
if the system which is at superheated condition
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the enthalpy of that is room is going to be
greater than sg. So, for this also we have
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tables are typically in terms of V U H and
temperature for a given pressure. So, that
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is how we make use of such a superheated tables
in order to solve problems related to superheated
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conditions.
So, let us look at an example, so here we
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have a system which is at state where the
pressure is defined and enthalpy is given,
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and we need to find out the temperature now
since we don't know whether the system is
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at superheated or saturated or compress we
need to first we find out what is the possible
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state of that.
So, the best way to do these to make use of
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first the condition of saturated water pressure
table where we look at the condition here
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0.5 mega Pascal, which is 500 kilo Pascal
and look at the corresponding enthalpy of
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the vapor ok. Now, 2890 is certainly more
than 2748.1 that is h is greater than sg and
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this it is in the super-heated vapour condition.
So that is we should directly, take superheated
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water tables. So, there are many values we
have to pick up a condition, where we should
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get P is equal to 0.5 mega Pascal. So, we
take a values corresponding to P is equal
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to 0.5 mega Pascal and.
Here you have v u h, so this is like v u h
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at the entropy. So, we will talk about that
later, so within this we look at the value
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h is equal to 2890, and you can see that h
is equal to 2890. So, this is our h here lies
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between temperature 200 and to 50, because
2890 is in between here so; that means, temperature
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is in between this now in this case in order
to find precisely the temperature what we
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are going to do is we are going to do a linear
interpolation.
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So, very simple way to do that is we consider
and the difference in the this 2 temperature
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250 and 200, and the slope of the linear line
from 250 to 200 will be the same as that in
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the temperature in between to 200, and that
is what we are going to use that delta h divided
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by the temperature, and from here we rearrange
this to get this temperature which comes out
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to be 216.3 degree Celsius ok.
So, what we are looking at is simply is the
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slope of the line from this 2 prime point.
So, this is a common way of solving problems,
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where the data's are not given for each point
and so we make use of linear interpolation.
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So in addition to the superheated vapor which
is on the right side of this saturated vapor
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line you have a compressed liquid which is
on the left side of a saturated liquid line
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on the P or T versus V plot. So, a compressed
liquid line usually what we consider is for
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a given point, somewhere let us say here which
is on a P is equal to 500 kilopascal at temperature
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75 the property, we try we approximate to
that of the liquid saturated liquid at the
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75 degree Celsius. So, usually we approximate
because the compressed liquid property depend
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on temperature more than on pressure.
So, in this case the pressure variations are
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usually neglected, but for enthalpy for example,
the more accurate relation is even here. So,
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this is something which 1 can derive using
thermodynamic relations and which we will
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be doing it in the later part of this course.
So, what are the conditions for the compressed
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liquid? So, for corresponding to 75 degree
Celsius the saturated point is here and at
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this point. So, this will be your P sat add
that temperature of this system this is a
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P sat at the temperature of the system which
is 75. So, basically [vocalized-noise] the
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system is at 75 degree Celsius and 500 kilopascal
ok.
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So, if you look at it the pressure P here
is greater than P sat at T. So, that is the
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condition which we have P should be greater
than [noise] P set at a given temperature,
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in addition the temperature if you look at
it for a given pressure that the T sat for
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this P is 151.83 and 75 is less than 151.83.
So, the temperature is less than T sat at
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a given pressure. So, in general these 2 conditions
will conclude that it is a compressed liquid,
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but in general all the properties such as
your specific volume internal energy enthalpy
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will be less than the corresponding properties
of the fluid at a given pressure or temperature
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so.
That is how we define compressed liquid. So,
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let me now talk about the deviation of the
properties from the ideal gas behavior, as
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we said that you know we are interested much
to represent the properties or the state of
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the fluid in terms of some kind of equational
state.
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So, one way to look at is how much is a deviation
from the ideal class is it possible to quantify
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the (Refer Time: 24:47), non-ideality in terms
of very simple parameters and that is what
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the compressibility factor talks about it.
So, now, the compressibility factor can be
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represented in terms of Pv is equal to ZRT
well this is a molar volume. So, Z basically
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represent deviation from non-ideality for
ideal gas that is equal to 1 for real guess
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it could be less than 1 or greater than 1,
if you decrease the pressure such that it
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is 0, where the particles are far apart you
to start behaving ideal gas.
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So, what is the definition in general for
low pressure it need not be like you know
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going to 0 or absolute 0, but in general Z
is unity, when the gases are at low pressure
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or the high temperatures so, but what is the
definition of low pressure at high temperature
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it is with respect to its critical point.
So, the low pressure high pressure of a gas
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is high or low related to its critical temperature
of pressure.
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So, in practicality or in general are you
on a TV diagram this region would behave like
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idle gas behavior because the extremely low
pressure, because yes it is extremely low
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pressure right. So, the pressure will decrease
here all right and as well as this pressure
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because the temperature is quite high.
So, these 2 regions will start behaving more
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or less like ideal gas now the other important
thing is right, now it was noted that if you
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reduce the properties such as your pressure
temperature and volume by is critical point
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and many substances who are similar type sizes,
they start behaving similarly in a reduced
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00:26:44,289 --> 00:26:51,690
plot and this is the principle of corresponding
state ok. So, if you reduce [noise] pressure
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by it is by the critical point of the fluid,
and get a variable P R and similarly TR and
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VR.
We can plot Z as a function of P are four
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different TR, and it appears that the many
different substances which is here written
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here ethylene ethane and so forth. They all
fall on almost similar on the same curve this
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is what ah is you know beautiful because essentially
tells you that if you take out the characteristic
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property of the molecules in the form of the
critical point then all behave similarly.
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So, thus you can map this on a ah single curve
which we call it compressibility chart, and
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00:27:41,040 --> 00:27:46,179
which is nothing, but the Z here as a function
of P R, now you can clearly see here that
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can be less than 1 as well as a certain condition
it would be more than 1 ok.
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00:27:51,000 --> 00:27:56,340
So, if you want to find out Z just using the
compressibility chart its quite easy for a
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00:27:56,340 --> 00:28:02,730
given fluid for example, in this case any
of this you need to find out for a given P,
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00:28:02,730 --> 00:28:12,820
and let us say for given T you can get P R
and T R using his critical point and then
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00:28:12,820 --> 00:28:19,360
for given T R of y given P R and T R, you
know exactly this line and then you can take
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00:28:19,360 --> 00:28:25,210
Z and from there you can find out other properties,
or for that matter if the Z is given and a
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specific an inner temperature is given that
say you can use that in order [noise] to get
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the other unknowns. So, this is how the compressibility
chart is used; now this is a simpler compressibility
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chart this there is more complete compressibility
chart for.
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00:28:41,020 --> 00:28:46,070
Many different systems this is something which
we call in then cell about generalized compressibility
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00:28:46,070 --> 00:28:52,670
charts. [noise] And it is designed such that
it has a specific chart for low pressure and
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high pressures, and here you can find out
that it can go the Z can also deviate from
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1. Now in addition to PR and TR you have also
VR.
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In this case in order to simplify this exercise
of course, this say something which is a commonly
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used in industry and 1 can also do an ah ah
exercise on it or which I leave it to you.
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So, this was a very quick introduction to
properties particularly the phase diagram
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utility of the tables and equation of state
we will continue our discussion in the next
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lecture, see you in the next lecture.