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[noise]
Welcome back. Ah, in this ah lecture we will
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be reviewing the Energy Analysis Of Closed
System. So, as I already discussed the for
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the closed system [vocalized-noise] you have
ah only interaction with the surrounding is
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through ah [noise] through energy ok, that
means, energy can ah exchange with the surrounding
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[vocalized-noise] and that would ah be either
through ah work or ah heat of work [vocalized-noise].
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So, ah let me first ah describe ah the work
here [vocalized-noise]. So, ah the work can
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be of different types. So, the one classical
definition is ah defined as a force acting
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through a displacement ok where the displacement
is in the direction of the force and thus
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we typically write the work in this form ok
now for such a system of course, physically
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the system may not displace as such ok.
But, if this piston [noise] is is movable;
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that means, a boundary is movable then it
can displace the volume of the gas containing
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it ok [vocalized-noise], but in the other
form there could be other kind of work such
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as electrical work [noise] ok which ah also
will ah supply certain work to the system
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[vocalized-noise].
So, a generic definition is that work is done
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by a system if the sole effect [noise] on
the surrounding could be [noise] the raising
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of a weight [vocalized-noise]. We can take
an example of a battery motor ah which is
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used to run a fan ok and ah the effect of
ah work by the system on the surrounding ok
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that is [noise] if we consider this as a part
of it then this can be understood by the same
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definition by replacing the fan by a pulley
weight system.
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So, while the shaft rotates it is effect is
basically to raise this weight ok through
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this pulley [vocalized-noise]. So, this is
a generic ah [vocalized-noise] definition
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of the work [vocalized-noise] now as I said
the work could be of different type, one is
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a the boundary work if the boundary work is
ah is if the boundary is moveable [noise]
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ok.
Now, how do we define for such a system ah
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that say the boundary work here now a general
definition of mechanical work in this case
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ah using the piston ah cylinder ah device
is simply by F ah F external multiplied by
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the displacement d here, where F external
is nothing, but the force to compress the
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gas or the force felt by the surrounding as
the gas expands [vocalized-noise] ok and you
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can write F external in terms of P external
multiplied by dV, V where V is the volume
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[noise] which get displaced. For the quasi
ah static equilibrium the p external ok [noise]
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here the P external and P internal [noise]
[vocalized-noise] are nearly equal, but one
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is in fact, a similarly larger to accomplish
a net [noise] change in the volume ok.
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For a very slow process you can write the
differential work as simply P dV ok [vocalized-noise]
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and thus you can integrate this in order to
get the in order to get the work you can integrate
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the PdV and if you know the relation between
P and V you can easily get the work out of
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it.
Now, as I said the boundary work is as a simplest
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case here, the piston can be [noise] moved
and thus ah upon applying ah certain ah heat
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ok all sudden changes in the state conditions,
so, the boundary can move and thus can ah
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undergo certain work [vocalized-noise]. So,
for example, for a differential change in
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the volume or the piston location by d s the
differential amount of the work can be written
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as d ah del ah W b is equal to F into d s
and F as I said it could be ah simply the
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pressure multiplied with the cross sectional
area and this can be written as P dV ok.
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So, the integral [noise] form of this would
be w b is equal to integral from state 1 to
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state 2 and P dV [vocalized-noise]. If it
is a quasi equilibrium process one can draw
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a process da path out of it and thus you can
simply integrate for such a system if you
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know the path [vocalized-noise], that means,
you know the relation between P and V [vocalized-noise].
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Now, W b for the sign ah based on the sign
convention is positive for expansion ok and
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negative for compression [noise] ok, that
means, if the work is done by the system and
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surrounding is positive otherwise it is negative
[noise] ok.
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So, let me just [vocalized-noise] use this
ah concept to do a quick example and some
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of you have already done this thing it is
just a recap ah in order to speed up all ah
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the understanding for a more complicated ah
subject or concepts in the later part of this
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course [vocalized-noise]. So, this kind of
example certainly is going to help us [noise]
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later on ok. So, ah this is a simple question
the question is a piston cylinder device which
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has ah air containing at this volume 0.4 meter
cube at ah 100 kilo [noise] Pascal and 80
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degree Celsius, ok [vocalized-noise] and it
is ah compressed to 0.1 meter cube, in such
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a way that the temperature is constant, ok
[vocalized-noise].
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Now, we need to find out the work done during
[noise] the process. So, what we are going
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to assume is that this is a [noise] quasi
equili equilibrium process [vocalized-noise],
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further we are going to assume that the air
is at the air is basically an [noise] ideal
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gas [vocalized-noise] well because of very
simple reason temperature is very high compared
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to it is ah critical point [vocalized-noise]
and at the word the process can be drawn on
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a P V diagram by this green curve [vocalized-noise]
from 1 to 2, such that ah T is equal to constant
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or in other word P V is equal to constant
ok [vocalized-noise].
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Because, V is equal to ah m N R T can be written
directly like this or ah we can write P is
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equal to C some constant ok divided by V where
C is nothing, but N R T [noise] ok [vocalized-noise].
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So, I can now write [noise] the the boundary
work simply as this [noise] this would be
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your C ln V 2 by V 1 [vocalized-noise] and
C is nothing, but P 1 V 1 ok, ln V 2 by V
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1 ok [vocalized-noise]. So, ah this is as
very simple example from here we can get this
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boundary work ok [vocalized-noise].
So, let us move on, in many processes ah the
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actual expansion and compression of the gases
are related [noise] in in a commonly ah relation
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ah which is called poly tropic process where
P and V are often represented by this expression
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ok, P V to the power C to the power n [noise]
is constant.
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So, this is a your poly tropic process ok
ah. So, where n can vary ok, so, from 1 to
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ah of n is a variable for n is equal to 1,
of course, it is a constant temperature for
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particularly ideal gas. So, we can also find
ah boundary work for such a system which is
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a poly tropic process. So, again ah for poly
tropic process we can draw a diagram we can
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draw the path on the P V diagram. So, again
we can write this boundary work in this form
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ok [vocalized-noise]. Now, the P is given
as a some constant C minus n, d V [noise]
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and you can rewrite this expression or you
integrate it you should be able to get [noise]
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the following [noise] ok, where this is 1
to 2 and minus n plus 1 ok [vocalized-noise].
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This can be also written in this form or very
simpler form P 2 V 2 because C V 2 minus n
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is nothing, but P 2 P 2 multiplies by the
remaining term V 2 is [vocalized-noise] this
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minus P 1 [noise] V 1 [noise] and [noise]
1 minus n ok [vocalized-noise]. So, this is
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the case for a generic expression for boundary
work for a poly tropic process [vocalized-noise].
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Now, if you have an ideal gas you can replace
P 2 V 2 by N R T ok and thus you can get N
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R number of moles on the gas constant [noise]
T 2 minus T 1 [noise] 1 minus n and of course,
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n is not equal to 1 in this case [vocalized-noise].
You can also consider specific cases for example,
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ah if n is equal to 1 ok and PV is equal to
C [vocalized-noise].For an ideal gas w b is
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going to be P V [noise] and then V 2 by V
1 which we have derived earlier also [vocalized-noise]
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for a constant [noise] pressure [noise] [vocalized-noise]
W b is going to be [noise] simply P 0 which
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is a constant V 2 minus V 1 ok [vocalized-noise].
What about constant volume? [noise] [vocalized-noise]
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is going to be ah simply [noise] 0, because
the boundary is not changing at all ok. So,
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W b is going to be 0 for constant volume ok.
So, having [vocalized-noise] done this exercise
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for the boundary work, so, let us now ah summarize
the energy ah balance for the closed system
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[vocalized-noise] ok. So, ah as we have already
discussed a generic energy balance is E in
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minus E out ok equal to delta E, E system
where E in minus E out is the net energy transfer
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by heat work and mass [vocalized-noise].
Of course, for closed system there is no mass
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contribution only heat and work is going to
be there whereas, ah for a delta E system
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you will have to include changes in the internal
energy kinetic energy potential energy ok
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for the rate expression you include dot here
which means basically is the rate of energy
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change with respect to time and that is also
included here ok. If a sign convention is
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used the energy balance can be written as
Q net in minus w net out equal to [noise]
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delta E system or in a simpler form Q minus
W [noise] ok, where my definition Q is energy
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supplied to the system and W say as the work
done by the system.
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Now, for a cyclic process ok delta E must
be 0 and in the work W net out is equal to
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Q net in or Q is equal to W ok, where Q net
in is nothing, but Q in minus out and W net
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out is nothing, but W out minus W in ok. Now,
this you can also derive using a generic expression
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here based on this without considering the
sign convention.
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So, let me just now ah do a quick exercise
on our general energy balance for a constant
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pressure expansion and compression process
ok. So, we are going to consider is Q is applied
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to the system and W is ah done ah from the
system. So, in this case ah what will be the
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expression for constant pressure expansion
process, what will be there?
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So, let me start with [noise] E in minus [noise]
E out [noise] delta E system ok. Right now
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this is of course, for that ah energy ah transfer
because of the fact that is a clau closed
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system [vocalized-noise]. So, this will be
net [noise] [vocalized-noise] energy [noise]
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transfer by Q and W [noise] [vocalized-noise].
So, Q minus W [noise] is equal to delta U
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plus delta K E [noise] plus delta P E and
we are going to consider this to be 0 [noise]
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ok [vocalized-noise]. Now, this W could be
[noise] W boundary and [noise] some other
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[noise] work ok [vocalized-noise]. So, in
that case we can write Q minus W other [noise]
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[vocalized-noise] ok and W b for a constant
pressure is P 0 V 2 minus V 1 [noise] and
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that is going to be delta u ok. So, we can
take this [noise] or we can first write here
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is U 2 minus U 1, we can take this to this
side and what remains is [noise] W other [noise]
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[vocalized-noise] U 2 plus P 2 V 2 [noise]
minus U 1 [noise] plus P 0 [noise] V 1 and
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this is nothing, but by definition enthalpy.
So, this will be H 2 and this will be H 1
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[vocalized-noise]. So, for the case of a constant
pressure expansion compression process for
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a closed system [noise] the energy [noise]
valance is going to be simply [noise] H 2
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minus H 1 ok [noise], having derived this
ah now, we can apply it to a example.
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[vocalized-noise]. So, this is our example
of a piston cylinder device which contains
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ah 20 gram of saturated water vapor that is
maintained at a constant pressure [vocalized-noise]
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and that resistance heater within the cylinder
is turned on and it passes a current of 0.2
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ampere for 5 minutes from a 120 volt source,
at the same time a heat of the heat loss of
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3.7 kilojoules occurs and we to we need to
find the final temperature of the system [vocalized-noise]
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ok [vocalized-noise]. So, we can apply that
constant pressure expression. So, ah what
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we have we have ah [noise] again E in minus
E out [noise] is delta E system [vocalized-noise].
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So, ignoring the kinetic energy changes in
the potential energy changes this would be
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delta u and if you just consider the sign
which is given here ah which is known from
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the statement we can write W e [noise] in
electric energy due to resistance heater minus
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Q out [noise] minus whatever the boundary
work which is done ok [vocalized-noise]. So,
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ah if you take out the boundary work to the
other side we know this is going to be delta
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H or in other word m [noise] h 2 minus h 1
ok [vocalized-noise] and we can write W e
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in [noise] minus Q [noise] out ok now ah what
is W e in ok this is nothing, but your I V
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delta t ah. So, small t which is ah 5 minutes
you have to convert it that in 2 seconds ah
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this is going to be 7.2 kilo joules, W ah
Q out is 3.7 [noise] kilojoules and ah M is
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given to you is 0.025 kg [noise] we always
prefer to use SI unit remember that and then
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we need to find out ah from here h 2 what
is h 1? h 1 is ah is a saturated vapor 300
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kilo Pascal [vocalized-noise]. So, this is
going to be h of g at 300 kilo Pascal we look
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at a table of a saturated water table ok ah
particularly the pressure will table [vocalized-noise].
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So, from here the value comes out to be 2724.9
kilo joules per kg [vocalized-noise] we plug
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in here and obtain the value of h 2; h 2 [noise]
is 2864.9 [noise] kilo joules [noise] per
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kg ok [vocalized-noise]. Now, what we have
as a state 2 [noise]? [vocalized-noise] P
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2 is [noise] 300 [noise] kilo Pascal and h
2 is 2864.9 [noise] kilo [noise] joules per
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kg, ok [vocalized-noise]. So, now ah since
ah h 2 is greater than h g, so, it is a super
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heated table. So, you look at the super heated
table and you can find out T corresponding
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to which tons which is very close to the 200
[noise] degree Celsius [vocalized-noise] ok.
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So, super heated table 0.3 mega Pascal you
look at the value you will find this h 2 will
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lie very close to the T you can do the interpolation
or since the value is almost close I am just
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approximating it to 200 degree Celsius [vocalized-noise].
So, this is the example of making use of constant
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pressure energy balance, ok.
So, we can extend this exercise for other
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cases such as un ah restrained expansion ok.
So, we can ah extend this ah understanding
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by applying the concept of energy balance
for the closed system for a case of unrestrained
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[vocalized-noise] expansion. So, ah here what
we have is a rigid tank which is divided into
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2 equal parts ok and initially one side of
the tank contains the water having 5 kg pressure
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at 200 kilo Pascal and temperature is 25 degree
Celsius [vocalized-noise] and other side is
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a vacuum, evacuate the space the partition
is removed water expands in the entire tank
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and the water is allowed to exchange heat
wi with the surrounding, until the temperature
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in the tank returns to 25 degree Celsius ok.
What we need to find is the volume of the
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tank, the final pressure and the heat transfer
for this process [vocalized-noise].
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So, we are going to consider the system a
complete tank ok. So, in that case of course,
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the tank volume is for is fixed, so, the the
boundary work is going to be 0, ok [vocalized-noise].
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So, ah let me just first find ah a couple
of things we need to ah start with a very
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simple idea that what is the volume of this
particular container ah that is ah we can
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start with the ah first part. So, if you look
at the condition is 200 kilo Pascal 25 degree
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Celsius ok. So, by definition this ah water
is in a compressed ah liquid ok and ah thus
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we can approximate the specific volume of
this ah water as we have the specific volume
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of the fluid at 25 [noise] degree Celsius
which is 0.001 meter cube per kg ok [vocalized-noise].
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00:19:12,980 --> 00:19:20,530
In the word, ok, so, we want the volume of
this half volume ah the volume occupied initially
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00:19:20,530 --> 00:19:28,110
by the water is going to be the mass multiplied
by the specific volume [vocalized-noise] which
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00:19:28,110 --> 00:19:35,820
ah since this is 5 kg and you plug in this
value it turns out to be 0.01 meter cube ok.
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00:19:35,820 --> 00:19:43,270
So, now, ah what we need to find out is the
[noise] fa final volume. So, final volume
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00:19:43,270 --> 00:19:49,950
is ah V 2 which is 2 times of course, which
ah 2 times a V 1 ok. So, the what is the sa
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00:19:49,950 --> 00:19:59,090
final specific volume is going to be V 2 divided
by m, so, it turns out to be 0.01 divided
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00:19:59,090 --> 00:20:19,580
by [noise] 5, 0.002 meter cube by kg ok [vocalized-noise].
So, this ah, so, ah this final specific volume
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00:20:19,580 --> 00:20:30,160
of ah state 2 is 0.02 ah meter cube per kg.
So, at [noise] 25 degree Celsius [noise] we
185
00:20:30,160 --> 00:20:38,910
can write down V f and V g and this is needed
in order to find out the quality of quality
186
00:20:38,910 --> 00:20:49,090
of water in state 2. So, this is your 0.001003
straight from the [noise] steam table [noise]
187
00:20:49,090 --> 00:20:58,380
and this is 43.340 meter cube per kg ok. So,
this you can see that you fa these are the
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00:20:58,380 --> 00:21:03,130
values which we are using here [vocalized-noise]
ok [vocalized-noise]. Now, from here we can
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00:21:03,130 --> 00:21:10,270
find out quality because quality is V 2 minus
V f [noise] divided by V g minus V f. So,
190
00:21:10,270 --> 00:21:15,750
this is nothing, but V f g. So, from here
we get 2.3 into 10 to power minus 5.
191
00:21:15,750 --> 00:21:22,460
Now, we write down the expression remember
that also that P sate is ah for this P sate
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00:21:22,460 --> 00:21:33,350
is ah P 2 is P sate [noise] at 25 degree Celsius
which is 3.1698 kilo Pascal. So, we write
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00:21:33,350 --> 00:21:42,060
down the expression [noise] of energy [noise]
balance delta E system and this can be written
194
00:21:42,060 --> 00:21:51,030
as m [noise] u 2 minus u 1 [vocalized-noise]
and Q in is ah simply Q in there is no W b
195
00:21:51,030 --> 00:21:56,170
there is no other thing. So, it is Q in is
equal to delta u which is m u 2 minus u 1
196
00:21:56,170 --> 00:22:00,710
[vocalized-noise]. So, what would be your
ah. So, we need to find out this two terms.
197
00:22:00,710 --> 00:22:09,620
So, what is a U 1? U 1 can be approximated
as u f at 25 degree Celsius ok which is [noise]
198
00:22:09,620 --> 00:22:15,620
104.83 kilo [noise] joules per [noise] kg
ok [vocalized-noise] and thus your U 2 is
199
00:22:15,620 --> 00:22:24,190
your U f [vocalized-noise] and ah yeah. So,
ah for to to calculate U 2 we know already
200
00:22:24,190 --> 00:22:32,190
x. So, we can use U f and U f g. So, again
we can go back here [noise] you we know ah
201
00:22:32,190 --> 00:22:41,480
U f and ah U f g we directly make use of that
[noise] and x we know we we get 104.88 kilo
202
00:22:41,480 --> 00:22:45,559
joules per kg.
We plug in this value two values here in this
203
00:22:45,559 --> 00:22:53,850
equation in order to get [noise] Q in [noise]
ok. So, Q in is this. So, Q in comes out to
204
00:22:53,850 --> 00:23:01,440
be 0.25 kg [noise] 0.25 kilo joules [vocalized-noise]
ok. So, it is a straightforward exercise in
205
00:23:01,440 --> 00:23:08,660
other word what is happening is this in this
process at ah 200 ah kilo Pascal at 25 degree
206
00:23:08,660 --> 00:23:14,890
Celsius. This was the specific volume and
ah this was in a compressed ah liquid region
207
00:23:14,890 --> 00:23:21,160
and in this expansion process ah the pressure
drops and it gets into that saturated region
208
00:23:21,160 --> 00:23:26,920
which is ah, 2 ok [vocalized-noise].
So, this is how you make use of this ah simple
209
00:23:26,920 --> 00:23:31,110
energy balances in order to solve certain
problems ok [vocalized-noise]. So, this is
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00:23:31,110 --> 00:23:35,840
a recap I hope that ah and believe that sa
some of you must have gone through this kind
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00:23:35,840 --> 00:23:41,710
of exercises in your earlier course of thermodynamics.
But, we are trying to just recap some of the
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00:23:41,710 --> 00:23:44,920
concepts.
So, let me just ah move on and I talk about
213
00:23:44,920 --> 00:23:50,980
specific heat here [vocalized-noise] ah specific
heat basically define in order to express
214
00:23:50,980 --> 00:23:58,000
different capacity of the different materials
to ah to store ah heat or to store energy
215
00:23:58,000 --> 00:24:03,640
[vocalized-noise]. So, if you compare.
Let us say two materials such as ah iron and
216
00:24:03,640 --> 00:24:09,460
water. It takes different amount of energy
to raise ah same amount of temperature ok.
217
00:24:09,460 --> 00:24:14,350
It takes less amount of energy for iron to
change it is temperature from 20 to 30 degree
218
00:24:14,350 --> 00:24:19,350
Celsius for a 1 kg ah block of iron [vocalized-noise]
and if you look at the water it takes much
219
00:24:19,350 --> 00:24:26,410
significantly much ah larger amount of ah
energy almost close to ah 10 times more [vocalized-noise].
220
00:24:26,410 --> 00:24:32,510
So, how do you define [vocalized-noise] or
compare this ah kind of ability [vocalized-noise]
221
00:24:32,510 --> 00:24:40,040
hold such a energy [vocalized-noise] and we
define in terms of ah a variable call specific
222
00:24:40,040 --> 00:24:46,440
heat ok and the specific heat is defined as
the energy required to raise the temperature
223
00:24:46,440 --> 00:24:50,880
of a unit mass of a substance by 1 degree
[vocalized-noise] ok and thus it is unit is
224
00:24:50,880 --> 00:24:58,320
ah 5 ah it is it is unit is kilojoules kg
degree Celsius or kilojoules kg Kelvin ok
225
00:24:58,320 --> 00:25:02,930
[vocalized-noise].
Now, you can ah calculate specific heat E
226
00:25:02,930 --> 00:25:08,250
either uh main maintaining the volume constant
or you may you can maintain the ah pressure
227
00:25:08,250 --> 00:25:13,750
constant [vocalized-noise]. If it is ah done
in a way where ah the volume is constant is
228
00:25:13,750 --> 00:25:18,679
call ah specific heat at constant volume which
is the energy required to raise the temperature
229
00:25:18,679 --> 00:25:26,360
for unit mass of a system of a substance by
1 degree as the volume is maintained constant
230
00:25:26,360 --> 00:25:30,110
and the specific heat at constant pressure
is the energy required to raise the temperature
231
00:25:30,110 --> 00:25:35,420
of a unit mass of a substance by 1 degree
as the pressure is maintained constant ok
232
00:25:35,420 --> 00:25:40,179
[vocalized-noise].
So, in other word you can represent ah c v
233
00:25:40,179 --> 00:25:44,490
in terms of the partial derivative u with
respect to T at constant volume and c p a
234
00:25:44,490 --> 00:25:50,410
partial derivative of enthalpy with respect
to temperature ah pressure constant ok. Now,
235
00:25:50,410 --> 00:25:56,000
[vocalized-noise] note that, ah c p is always
greater than c v because this ah. So, in order
236
00:25:56,000 --> 00:26:02,690
to accommodate more energy needed to expand
ah needed for expansion ok [vocalized-noise].
237
00:26:02,690 --> 00:26:08,390
Now, let me just talk about these variables
in terms of for particularly for ideal gas.
238
00:26:08,390 --> 00:26:13,700
For ideal gas ah joules has already shown
that ah the internal energy is going to be
239
00:26:13,700 --> 00:26:17,940
independent of pressure it only depends on
the temperature ok for by simple simple ex
240
00:26:17,940 --> 00:26:23,780
ah experiment [vocalized-noise].
Now, ah by definition enthalpy ah is u plus
241
00:26:23,780 --> 00:26:29,130
P v and ah since u is the only dependent on
temperature for an ideal gas and P v can be
242
00:26:29,130 --> 00:26:35,450
written as R T. So, by definition ah h is
equal to u plus P v and since u is independent
243
00:26:35,450 --> 00:26:39,380
of temperature and P v you can write it for
ideal gas R T [vocalized-noise].
244
00:26:39,380 --> 00:26:44,700
So, as a joules has ah shown that ah find
the ideal gas u is exact differentiation.
245
00:26:44,700 --> 00:26:51,100
So, ah c v now can be written as simply du
by dT and simply similarly you can do the
246
00:26:51,100 --> 00:26:56,710
same thing for c p. In other word you can
write [noise] d u as simply c v T which is
247
00:26:56,710 --> 00:27:05,010
dependent on temperature multiplied by d t
and simply d h is equal to c p dT ok.
248
00:27:05,010 --> 00:27:11,640
So, one can find out now the changes in the
internal energy by just integrating this expression
249
00:27:11,640 --> 00:27:16,340
similarly that for enthalpy. For enthalpy
will be considering c p for internal energy
250
00:27:16,340 --> 00:27:20,860
will be considering c v [vocalized-noise].
What we need is basically now, the relation
251
00:27:20,860 --> 00:27:29,929
of c v as a function of ah temperature so
that relation or that expression would be
252
00:27:29,929 --> 00:27:35,060
needed [vocalized-noise] ok.
Now, if you look at real gases ah real gases
253
00:27:35,060 --> 00:27:41,110
at low temp ah real gases as a very low pressures
and very high temperature ah behaves like
254
00:27:41,110 --> 00:27:45,900
ah ideal gas ok or particularly, when you
consider very low pressure. All at low pressure
255
00:27:45,900 --> 00:27:50,730
all the real gases approaches ideal gas and
only depends on the [noise] temperature ok
256
00:27:50,730 --> 00:27:58,270
and particularly for noble gases ah they are
only dependent on ah or they are almost constant.
257
00:27:58,270 --> 00:28:03,790
On the other hand the molecular fluids they
are linearly dependent on temperature at very
258
00:28:03,790 --> 00:28:08,590
low pressures [vocalized-noise].
So, one can use these ah curves get a ah fit
259
00:28:08,590 --> 00:28:14,640
it into some kind of polynomial expression
and obtain c p ah as a function of temperature
260
00:28:14,640 --> 00:28:20,100
ok [vocalized-noise].
And, this is what often use c p, ah one can
261
00:28:20,100 --> 00:28:25,040
put it into expressions and this can be used
in order to get delta h ok and for an ideal
262
00:28:25,040 --> 00:28:29,840
gas of course, we know that expression of
c p minus c v is equal to r we can also get
263
00:28:29,840 --> 00:28:35,290
from there ah the expression for delta u ok
[vocalized-noise]. Now, ah integrations though
264
00:28:35,290 --> 00:28:41,419
are straightforward ah, but it is rather time
consuming and for many gases ah u and h data
265
00:28:41,419 --> 00:28:46,900
are tabulated ok [vocalized-noise].
So, if you summarize this specific heat based
266
00:28:46,900 --> 00:28:53,799
on let us say ah calculations just we can
take an example of ah u, ah internal energy
267
00:28:53,799 --> 00:28:58,130
you can directly take a make a use of a table
in order to get the changes in the internal
268
00:28:58,130 --> 00:29:02,270
energy or if you have the expression of a
c v as a function the temperature you can
269
00:29:02,270 --> 00:29:08,559
also use that or for a short temperature ah
ah intervals you can just take the average
270
00:29:08,559 --> 00:29:14,630
c v value and obtain delta you ok [vocalized-noise].
So, this depends on what is available and
271
00:29:14,630 --> 00:29:18,940
ah of course, the state conditions [vocalized-noise].
So, with that [vocalized-noise] I will close
272
00:29:18,940 --> 00:29:22,870
ah this lecture and we have just summarized
the the energy balance for the closed systems.
273
00:29:22,870 --> 00:29:28,430
We will take up ah in the next lecture for
the ah case of open system. So, I will see
274
00:29:28,430 --> 00:29:29,180
you in the next lecture [noise].