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Hello and welcome to lecture 7 of the lecture
series on introduction to aerospace propulsion.
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In today's lecture, as I mentioned in my previous
lecture, we shall solve some problems basically
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related to work generated by a system. We
shall mainly focus on the displacement work
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as I mentioned in my previous lecture that
displacement work is of significant interest
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to engineers, because there are lots of engineering
systems which operate on the displacement
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work mode. So, we shall try to solve some
problems which are related to work generation
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through the displacement work mode.
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So, today's lecture will basically be on problem
solving, numerical problem solving related
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to calculation of work done and with more
focus on the displacement work done.
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Let us look at the first problem that we have.
Well, we have a problem here which states
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that hydraulic cylinder has a piston of a
cross sectional area of 25 centimeter square
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and a pressure of 2 mega pascal. If the piston
is moved by 0.25 meters, how much work is
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done? Here we have a question where in there
is a piston cylinder arrangement, the area
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of the piston head is given and the pressure
within the cylinder is also given to us. If
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the piston moves by a certain distance which
is also specified, we need to find out how
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much is the work done during this particular
process.
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Before I start solving this problem what I
would like to emphasize is a systematic approach
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to problem solving; so, it is very important
that we adopt a certain systematic approach
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for any problem solving, whether it is thermodynamics
or any other discipline for that matter. One
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of the important things is to first understand
the problem definition very clearly and to
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be clear as to what is mentioned in the particular
problem statement. The other important thing
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is to make a sketch of the problem that you
have in hand.
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In this example that we are looking, we have
a piston cylinder arrangement and we have
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certain properties of the system which have
been defined. So, it is important for us to
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mark these particular parameters, which have
been defined in this problem on the drawing
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or the sketch of the problem definition. Then
we also need to define the assumptions that
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we are going to make in a particular problem.
In solving a particular problem, if required,
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also justify why these assumptions have been
made. Some of the assumptions are quite obvious
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like in most of the thermodynamics analysis
that we are going to do; we are going to assume
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that the process is carried out quasi-statically.
So, you may or may not mention this as an
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assumption, because it is an obvious assumption
in almost all the thermodynamics analysis
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that the process is quasi-static.
If it is a non quasi-static process, then
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its analysis is entirely different, but we
might as well mention the assumptions that
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we make besides the quasi-static assumption
there could been other assumptions that are
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involved. So, it is also important to clearly
explain the assumptions, which are made in
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the problem and then move towards solving
understanding the problem itself and then
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solving the problem.
If we were in many of the problems that we
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will solve and some of them we will of course
solve today. It would also be advisable to
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draw the process diagram in terms of PV or
little later on in terms of temperature and
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entropy. So, the process illustration along
with the process diagram in terms of PV or
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any other parameters would definitely make
things a lot clear and easier to tackle than
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this directly trying to substitute numerical
values in a formula and getting the answer,
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right? Well, basically it is trying to analyze
a particular problem systematically. Also
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ensure that you do not make any error of judgment,
when you try to understand a particular problem
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statement.
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So, with this in mind let us now look at what
this particular problem is all about. In this
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problem we have a piston cylinder arrangement
as is shown in this diagram. In this piston
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and cylinder arrangement, we have a certain
pressure which is within the system, within
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the cylinder volume, which is in this particular
case given as 2 mega pascals.
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There is a certain pressure, which is exerted
on the piston by the gas which is host within
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the cylinder volumes. Now, the area of the
piston is given as 25 centimeter square; during
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this process it could be it is basically an
expansion process. The piston has moved by
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a certain distance which is specified as 0.25
meters arc or 25 centimeters.
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This is the problem and we are required to
find what the work is done during this particular
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process? One of the assumptions as I already
mentioned is of course, that the process is
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in quasi-equilibrium or it is a quasi-static
process. It is possible for us to integrate
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PdV because it is a quasi-static process.
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To calculate the work done during this process
as we have seen if the basic definition of
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work is that work is force with a displacement.
In this case, we have F is equal to P times
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A pressure times the area. So, work is integral
of F times dx that is force times the displacement.
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Since, force is equal to pressure times area
we have work is equal to pressure into area
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into the distance traveled during this process.
All these parameters are specified in this
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particular problem.
So, we have work is equal to pressure in this
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case it is 2 mega pascals which is equal to
2000 kilopascal into the area which is 25
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centimeter square so, 25 into 10 power minus
4 meter square and delta x which is 0.25 meters.
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Another suggestion that I have when you try
to solve a problem is to try and write the
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corresponding units of each of these parameters,
so that you do not make a mistake while calculating
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by putting in the wrong units. For example,
the area in this problem was given in centimeter
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square; it helps if you specify the units
right next to the property. So that you know
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for sure that you have got the units.
So, you have pressure 2000 kilopascal area
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25 into 10 raise to minus 4 meter square times
delta x which is 0.25 meters. If you were
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to calculate this you would get one 1.25 kilo
joules so, in this particular process the
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work done to move the piston was 1.25 kilo
joules. This is a very simple problem where
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you were to require calculating the work done
during a process.
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If you have been specified the pressure, the
area of the piston and the distance by which
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the piston moves, it is a simple problem where
you can calculate work done by equating work
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is equal to pressure times area times delta
x. In this case the work done to move the
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piston was calculated as 1.25 kilo joules.
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Now, let us look at the next problem that
we have for discussion today, this again is
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a piston cylinder example. A piston cylinder
has 1.5 kilograms of air at 300 kelvin and
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150 kilopascals. It is now heated up in a
two step process. First a constant volume
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to 1000 kelvin, which takes the process to
state 2 as is mentioned in the brackets then
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followed by a constant pressure process 2500
kelvin state 3. Find the final volume and
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the work in the process.
In this case, we have a problem where we have
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a system which carries out a process in two
steps; the first step we have a constant volume
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process which takes it from state 1 to state
2, where state 2 is 1000 kelvin, at state
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1 the system was at 300 kelvin and 150 kilopascals
after the first process at the end of which
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it reaches state 2.
The second process is a constant pressure
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process where in the system is taken to a
temperature of 1500 kelvin, the pressure remaining
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the same. So, we are required to calculate
the volume at the final state that is state
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3 and the work done during this process.
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As I mentioned, one of the first things that
we can do is to draw the process diagram that
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is the process path. In this example, we have
PV diagram and the state was initially at
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the process or the system was initially at
state 1 and state 1 to 2 is a constant volume
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process. Therefore, V 2 is equal to V 1 as
you can see the pressures are not the same
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P 1 is not equal to P 2. The second process
is a constant pressure process from state
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2 to state 3. Since, pressure is constant
we have P 3 is equal to P 2 and what we are
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required to find is the volume at 3 that is
V 3 and the work done during this process.
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So, if you recall from the previous lecture
where we had discussed about work done during
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constant pressure processes and constant volume
processes. If you recall that you can immediately
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see that work done during the first process
will be 0, because it is a constant volume
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process and work done during the second process
is the pressure times the difference in the
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volumes. The total net work done of this obviously
would be equal to work done during the first
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process plus the work done during the second
process.
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So, in this case besides the quasi-static
assumption we shall also use the ideal gas
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approximation for air. In the medium of this
particular process is air and we shall consider
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air to be in ideal, so that is the second
assumption besides the quasi-static assumption.
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At state 1 we know certain properties of the
system some of them are temperature is known
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it is specified in the problem it is 300 kelvin,
pressure is given as 150 kilopascals, then
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we know the mass of the system it is 1.5 kilograms
and we also know the gas constant for air.
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So, gas constant for air is universal gas
constant divided by the average molecular
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weight of air that will be equal to 8.314
kilo joules per kilogram kelvin divided by
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the molecular weight of air which is 29.
If we were to calculate that you would get
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0.287 kilo joules for the gas constant for
air. Since, some of these properties that
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is temperature pressure mass and gas constant
are known and we have approximated air to
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be an ideal gas V 1 that is volume at state
1 will be equal to mRT 1 divided by P 1 this
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is from the state equation that is PV is equal
to mRT therefore, V 1 is mRT 1 divided by
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P 1. So, all these parameters are known 1.5
is kilograms is mass R is the gas constant
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T 1 is temperature and P 1 is the pressure.
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If we were to substitute these values the
volume at state 1 that is V 1 can be calculated
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as 0.861 meter cube, now we look at state
2. We have calculated all the parameters at
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state 1 we know the volume, we know the temperature;
pressure all that is known at state 1, now
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proceed to state 2. We know that the first
process which takes the system from state
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1 to state 2 is a constant volume process.
Therefore, V 2 is equal to V 1.
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If you again use the ideal gas state equation,
then we get P 2 pressure at state 2 is equal
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to P 1 times T 2 by T 1 this is again from
the state equation, because V 2 is equal to
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V 1. So, you get since P 1 is known as 150
kilopascals, temperature T 2 is given as 1000
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kelvin, T 1 is 300 kelvin therefore, P 2 is
equal to 500 kilopascals.
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Now that you know pressure at state 2 we can
proceed to state 3, because the second process
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is a constant pressure process. Therefore,
P 2 is equal to P 3 so that is because the
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second process is a constant pressure process.
From this we can all again use this state
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equation to calculate the volume at the end
of the second process that is at state 3 so,
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V 3 is equal to V 2 times T 3 by T 2 again
from the state equation.
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We know volume at state 2, V 2 is known, T
3 is known and T 2 is also known. T 3 is specified
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as 1500 kelvin, T 2 is already known as 1000
kelvin, and V 2 we had already calculated
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from in the first step of this problem that
is 0.861 meter cube. If you again substitute
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these values you get V 3 is equal to 1.2915
meter cube, so this is answered to one part
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of the question, which was to find the final
volume at state 3 or at the end of the process.
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So, the final volume that is V 3 is 1.2915
meter cube.
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Now, the second part of the question is to
find out the work done during the process.
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I already mentioned at the beginning that
in this particular example, we have two processes
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for which we had calculated the work done
in the previous lecture; one is a constant
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volume process and the second process is a
constant pressure process. Net volume, net
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work done during this process will be work
done during the first process plus work done
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during the second process.
We already know that work done during the
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constant volume process will be 0, because
there is no change in volume, there is no
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displacement. Therefore, PdV work will be
equal to 0 therefore, the net work during
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done during this process is equal to the work
done during the second process which is a
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constant pressure process.
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Let us calculate now the work done during
the second process. So, work done during this
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whole process that is denoted by W subscript
1 3 which denotes work done between states
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1 and 3 is equal to W 1 2 plus W 2 3. Where
the first one is work done during the first
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process between states 1 and 2 plus W 2 3
is work done during the second process between
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states 2 and 3. So, work done during process
1 2 as we know is 0 because it is a constant
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volume process.
Therefore, work done during 1 3 is equal to
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work done during the second process that is
2 3 is equal to P 3 times V 3 minus V 2, which
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is also is equal to P 2 times V 3 minus V
2 because P 3 is equal to P 2 it is a constant
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pressure process. Now, we know the pressure
P 3 or P 2 which is 500 kilopascals, we also
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know V 3 and V 2 from the previous calculation.
So, we just need to substitutes these values,
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so 500 multiplied by o1.2915 minus 0.861,
this comes out to be 215.3 kilo joules.
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So, the total work done during this process
is 215.3 kilo joules, the net work done during
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the process is equal to the second process
that is work done during the second process
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and that was equal to pressure times the difference
in the volumes. Now, I also mentioned in the
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previous lecture when we were discussing about
work that work net work done is essentially
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equal to area under the curve on a PV diagram.
If we now go back to the PV diagram, which
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we had plotted for this particular problem
and then find they are under the curve, it
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will come out to be 215.3 kilo joules because
it is basically equal to the pressure times
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the difference in the specific volumes difference
in the volumes. So, it is P 3 or P 2 multiplied
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by V 3 minus V 2 so that is essentially the
area under the constant pressure process that
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was straight line on the PV diagram horizontal
line on the PV diagram.
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That is essentially the area under the curve
and that is equal to the net work done during
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this particular process. So that was problem
number 2. Now, let us move on to another problem
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which is yet again a piston cylinder example.
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I think I mentioned in the first lecture like
that was lecture 4 that we shall be discussing
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about piston cylinder examples very often
during the thermodynamics course. Now, this
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example is about a piston cylinder device
initially contains 0.4 meter cube of air at
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100 kilopascals and 80 degree Celsius. The
air is now compressed to 0.1 meter cube in
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such a way that temperature inside the cylinder
remains constant. Determine the work done
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during this process.
So, in this piston cylinder example we have
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been specified some initial volume, initial
temperature and pressure. Then the piston
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is compressed or the air with in the cylinder
is compressed by the piston to its final volume
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which is 0.1 meter cube, but this compression
is done in such a way that the temperature
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remains constant. So, it is a constant temperature
that is an isothermal process and we are required
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to find the work done during this process.
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So, the first step to solving this problem
is to draw the process path as well as the
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problem definition itself. Here we have a
piston cylinder arrangement where in the initial
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volume is 0.4 meter cube, pressure is 100
kilopascals, temperature is 80 degree celsius
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which is in this problem a constant.
Now, this piston as it moves down compresses
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the air within the cylinder to a final volume
which is 0.1 meter cube keeping the temperature
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same. So, on a PV diagram this process would
look like this, the process is initially at
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state 1 which corresponds to a volume of 0.4
meter cube and a pressure of 100 kilopascals.
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Now, following a constant temperature that
is an isothermal process, the system is compressed
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00:21:24,290 --> 00:21:28,960
which means that the volume has to decrease
and the pressure has to increase, so this
194
00:21:28,960 --> 00:21:35,530
is an isothermal compression process. Final
state of the system is at state 2 where the
195
00:21:35,530 --> 00:21:42,530
volume is 0.1 meter cube.
Again the assumptions as I mentioned need
196
00:21:42,680 --> 00:21:47,660
to be specifying the process in this case
it is a compression process, it is again in
197
00:21:47,660 --> 00:21:54,660
quasi-equilibrium. At the specified conditions
we will consider air to be an ideal gas and
198
00:21:55,380 --> 00:22:02,030
assuming that the temperatures and pressures
are not close to the critical values for defining
199
00:22:02,030 --> 00:22:09,030
or for assume making the assumptions of an
ideal gas for air. So, this is a problem which
200
00:22:09,440 --> 00:22:16,390
requires us to solve for an isothermal process.
I think I discussed in the previous lecture
201
00:22:16,390 --> 00:22:22,020
also that this would be a process where the
product PV will be equal to constant.
202
00:22:22,020 --> 00:22:29,020
So, PV equal to constant if we were to assume
the ideal gas assumption corresponds to an
203
00:22:29,720 --> 00:22:35,920
isothermal process. We will use PV equal to
constant for this particular solving this
204
00:22:35,920 --> 00:22:42,670
particular problem. Now, let us look at how
we can solve this problem.
205
00:22:42,670 --> 00:22:48,500
For an ideal gas at a constant temperature
here T subscript 0, T naught is remaining
206
00:22:48,500 --> 00:22:54,120
constant here, it is an isothermal process.
So, from the state equation product PV is
207
00:22:54,120 --> 00:23:01,120
equal to mRT naught or T 0 which is equal
to C or a constant which means that P is equal
208
00:23:02,170 --> 00:23:09,170
to C by V where C is a constant.
So, to calculate work is integral 1 to 2 PdV
209
00:23:10,670 --> 00:23:17,670
and which is integral replacing P by C by
V, we get C by V times dV or C times log natural
210
00:23:19,990 --> 00:23:26,990
log V 2 by V 1 which is P 1 V 1 times log
V 2 by V 1. If you had done this exercise
211
00:23:27,690 --> 00:23:34,080
I think I mentioned towards end of the previous
lecture that PV is equal to constant work
212
00:23:34,080 --> 00:23:41,000
done during PV equal to constant process need
to be carried out by as an exercise for you.
213
00:23:41,000 --> 00:23:46,260
If you had done that exercise, you would now
recall that work done you would have calculated
214
00:23:46,260 --> 00:23:53,260
as equal to P 1 V 1 log V 2 by V 1. So, in
this example we can replace P 1 V 1 by P 2
215
00:23:55,380 --> 00:24:02,380
V 2 because it is a isothermal process, P
1 V 1 is equal to P 2 V 2 or since PV is equal
216
00:24:03,500 --> 00:24:10,500
to mRT 0 we can replace it by mRT 0. Also
V 2 by V 1 can be replaced by P 1 by P 2,
217
00:24:12,970 --> 00:24:19,970
because P 1 V 1 is equal to P 2 V 2.
So, if we were to replace make these changes
218
00:24:20,720 --> 00:24:27,010
in the work equation which is P 1 V 1 log
V 2 by V 1, we will get we will basically
219
00:24:27,010 --> 00:24:34,010
have to replace for P 1 V 1 in terms of mRT
because mass is known gas constant for air
220
00:24:34,730 --> 00:24:40,800
is known and temperature is already known.
So, this equation would reduce to work done
221
00:24:40,800 --> 00:24:47,420
is equal to mRT 0 log P 1 by P 2.
222
00:24:47,420 --> 00:24:54,420
If we were to replace those values, we get
the pressure times the volume. This is P 1
223
00:24:54,930 --> 00:25:01,930
times V 1 that is the initial volume times
the pressure and the ratio of the either the
224
00:25:02,580 --> 00:25:09,580
volumes or the pressures in this case volumes
are known so log of V 2 by V 1. So, if you
225
00:25:09,620 --> 00:25:15,870
were to calculate this what you will see is
that you get minus 55.5 kilo joules so you
226
00:25:15,870 --> 00:25:21,160
may wonder what this minus means well this
negative sign essentially indicates that work
227
00:25:21,160 --> 00:25:27,250
is done on the system so there is work input
into the system and the during the last lecture
228
00:25:27,250 --> 00:25:30,490
we had discussed about sign conventions for
heat and work.
229
00:25:30,490 --> 00:25:36,820
Therefore, work done on a system is taken
as negative and in this case it is a compression
230
00:25:36,820 --> 00:25:43,220
process during a compression process always
work is done on the system and therefore,
231
00:25:43,220 --> 00:25:48,630
the work done in this particular example comes
out to be negative minus 55.5 kilo joules
232
00:25:48,630 --> 00:25:55,120
and the negative sign basically indicates
that work is done on the system and for any
233
00:25:55,120 --> 00:25:59,140
compression process work is always done on
the system.
234
00:25:59,140 --> 00:26:06,140
Now, let us look at our the 4th problem that
we have for today's exercise in this example
235
00:26:10,650 --> 00:26:17,650
we have a gas in a piston cylinder assembly
which undergoes an expansion process where
236
00:26:19,160 --> 00:26:26,160
PV raise to 1.5 is equal to constant the initial
pressure is 3 bar and the initial volume is
237
00:26:27,390 --> 00:26:34,390
0.1 meter cube and the final volume is 0.2
meter cube determine the work done for this
238
00:26:35,350 --> 00:26:42,350
process. So this is an example problem which
we have where the process is taking a PV raise
239
00:26:43,340 --> 00:26:49,660
to n equal to constant where n is 1.5 in this
example.
240
00:26:49,660 --> 00:26:53,600
And we have the initial pressure and volume
which have been specified we also have the
241
00:26:53,600 --> 00:27:00,380
final volume we need to find the work done
during this process so PV raise to n if you
242
00:27:00,380 --> 00:27:05,610
recall from our discussion in the last lecture
work done can be calculated by integrating
243
00:27:05,610 --> 00:27:12,410
PdV and because PV raise to n is constant
you can replace for P is equal to C by V raise
244
00:27:12,410 --> 00:27:13,730
to n.
245
00:27:13,730 --> 00:27:20,650
Therefore, determine the expression for calculating
work done during this particular process now
246
00:27:20,650 --> 00:27:25,900
let us now look at the process in terms of
the process diagram and the problem definition
247
00:27:25,900 --> 00:27:32,900
now we have P 1 and V 1 specified P 1 is 3
bar V 1 is 0.1 meter cube.
248
00:27:33,880 --> 00:27:40,880
Now this is an expansion process and so the
system was initially at state 1 which is 0.1
249
00:27:41,920 --> 00:27:48,920
meter cube and by expansion following a process
PV raise to 1.5 equal to constant it reaches
250
00:27:50,350 --> 00:27:57,350
state 2 which is at 0.2 meter cube well some
of the assumptions besides the quasi-equilibrium
251
00:28:01,140 --> 00:28:07,140
assumption are that gas is obviously in a
closed system it is a close system and the
252
00:28:07,140 --> 00:28:14,140
expansion mode is a polytrophic process which
is basically qualified by PV raise to n is
253
00:28:14,340 --> 00:28:21,260
a constant and that PdV work is the only work
mode well this is some of this assumption
254
00:28:21,260 --> 00:28:28,260
are valid also for the previous three examples
which we solved for again PdV work processes.
255
00:28:28,970 --> 00:28:35,970
So to solve this particular problem we will
need to find the expression for work done
256
00:28:37,470 --> 00:28:44,470
during PV raise to n is equal to constant
process I would suggest that instead of memorizing
257
00:28:45,950 --> 00:28:52,690
the expressions for work done in terms of
pressures and 1 minus n and so on it is always
258
00:28:52,690 --> 00:28:57,890
advisable that you derive the expression from
the first principle it is just a two step
259
00:28:57,890 --> 00:29:04,000
process so if you know how to integrate PdV
and for different processes which are PV raise
260
00:29:04,000 --> 00:29:08,210
to n equal to constant and so on.
You would not have to memorize any of these
261
00:29:08,210 --> 00:29:15,210
expression for work done so work done during
this process is W 1 2 integral between 1 to
262
00:29:17,330 --> 00:29:24,330
2 PdV and so we replace P in terms of C by
V raise to n and we get C times V 2 raise
263
00:29:25,110 --> 00:29:31,010
to minus n plus 1 minus V 1 raise to minus
n plus 1 divided by minus n plus 1 which is
264
00:29:31,010 --> 00:29:38,010
basically P 2 V 2 minus P 1 V 1 by 1 minus
n. Now the pressure at state 2 we can find
265
00:29:40,390 --> 00:29:46,450
using the expression P 2 V 2 raise to n is
equal to P 1 V 1 raise to n because we know.
266
00:29:46,450 --> 00:29:53,450
That this is a PV raise to n equal to constant
process so we can find pressure at state 2
267
00:29:53,480 --> 00:30:00,480
from this equation so P 2 in this case equal
to P 1 times V 1 by V 2 raise to n.
268
00:30:02,320 --> 00:30:08,580
So, this if we were substitute the values
you get we have converted here bar into pascal's
269
00:30:08,580 --> 00:30:15,130
so 3 bar is 3 into 10 raise to 5 pascal's
so 3 into 10 raise to 5 pascal's times V 1
270
00:30:15,130 --> 00:30:22,130
which is 0.1 divided by V 2 which is 0.2 raise
to 1.5 so from this we get pressure at state
271
00:30:23,350 --> 00:30:30,350
2 P 2 which is 1.06 into 10 raise to 5 pascal's.
Therefore we substitute for P 2 V 2 and P
272
00:30:31,840 --> 00:30:38,840
1 V 1 and n in the work equation so work as
we now know is P 2 V 2 minus P 1 V 1 divided
273
00:30:38,970 --> 00:30:45,970
by 1 minus n therefore, W is 1.06 into 10
raise to 5 pascal's times 0.2 meter cube minus
274
00:30:47,190 --> 00:30:54,190
3 into 10 raise to 5 pascal's times 0.1 meter
cube divided by 1 minus 1.5 so if we were
275
00:30:54,940 --> 00:31:01,830
to calculate this you will get work done as
plus 17.6 kilo joules so this is the work
276
00:31:01,830 --> 00:31:08,830
done during this particular expansion process
which is following a PV raise to n that is
277
00:31:09,080 --> 00:31:12,400
where n is equal to 1.5 equal to constant
process.
278
00:31:12,400 --> 00:31:18,950
So work done during this process can be calculated
because we know that it is PV raise to n that
279
00:31:18,950 --> 00:31:20,990
is a polytrophic process.
280
00:31:20,990 --> 00:31:26,710
Now we shall make some interesting observations
from this particular problem so work done
281
00:31:26,710 --> 00:31:33,530
as we have calculated from this process is
coming out to be plus 17.6 kilo joules now
282
00:31:33,530 --> 00:31:40,530
again I leave it as an exercise for you to
solve if n was equal to 1 instead of 1.5 if
283
00:31:40,850 --> 00:31:46,700
you had n equal to 1 which makes it an isothermal
process that is PV equal to constant what
284
00:31:46,700 --> 00:31:50,600
is a net work done.
Well, if you if you actually calculate that
285
00:31:50,600 --> 00:31:56,140
you will get the net work done during that
process as plus 20.70 kilo joules.
286
00:31:56,140 --> 00:32:03,140
Now, if you substitute n equal to 0 then you
can calculate the net work done which will
287
00:32:03,280 --> 00:32:08,810
be basically a constant pressure process work
done you will be able to calculate as 30 kilo
288
00:32:08,810 --> 00:32:15,810
joules. So you can notice that as you change
the value of n the work done during the same
289
00:32:16,570 --> 00:32:22,010
system initial state and final state remaining
the same just that the process by which it
290
00:32:22,010 --> 00:32:26,250
is carried out is different.
You have different values of n you get different
291
00:32:26,250 --> 00:32:31,200
values of work done what you will immediately
notice is that as you keep decreasing the
292
00:32:31,200 --> 00:32:38,200
value of n you get higher and higher work
done so please recall from our previous lecture
293
00:32:38,370 --> 00:32:43,040
when we were discussing about the polytrophic
process I had shown a PV diagram for different
294
00:32:43,040 --> 00:32:48,790
polytrophic processes for different values
of n ranging from 3 2 1 and so on up to n
295
00:32:48,790 --> 00:32:53,840
equal to 0 so I will bring that slide here
once again.
296
00:32:53,840 --> 00:32:59,730
That we have here the PV diagram for the for
different polytrophic processes so in this
297
00:32:59,730 --> 00:33:05,420
particular PV diagram you can see as you keep
changing n where as you keep decreasing n
298
00:33:05,420 --> 00:33:12,180
n equal to 3 n equal to 2 1 and so on you
can see that this area under this curve between
299
00:33:12,180 --> 00:33:19,180
1 and 2 will increase so were for n equal
to 1.5 that is W n equal to 1.5 17.6.
300
00:33:20,380 --> 00:33:24,270
You would probably somewhere come somewhere
here this would be the process somewhere in
301
00:33:24,270 --> 00:33:31,270
between n equal to 2 and 1 point and 1 n equal
to 1 is here which is above n equal to 1.5
302
00:33:32,200 --> 00:33:36,950
and therefore, you get a higher work output
n equal to 0 is a constant pressure process
303
00:33:36,950 --> 00:33:42,840
you get the highest work output.
During the constant pressure process, so this
304
00:33:42,840 --> 00:33:49,840
is just to illustrate that as you keep increasing
while decreasing the value of n the work done
305
00:33:50,610 --> 00:33:57,610
by the process will keep increasing so work
done is obviously also a function of the polytrophic
306
00:33:59,710 --> 00:34:04,640
power that is equal to n so as you keep reducing
n you get higher and higher work output the
307
00:34:04,640 --> 00:34:10,379
limit of that is n equal to 0 which is a constant
pressure process were in you get the maximum
308
00:34:10,379 --> 00:34:13,409
work done during the particular process.
309
00:34:13,409 --> 00:34:20,409
Now let us look at the 5th problem we have
for during for discussion during today's lecture
310
00:34:23,079 --> 00:34:29,790
this particular problem is about a piston
cylinder device which contains 0.05 meter
311
00:34:29,790 --> 00:34:36,790
cube of gas initially at 200 kilopascals at
this state a linear spring that has a spring
312
00:34:39,040 --> 00:34:46,040
constant of 150 kilonewtons per meter is touching
the piston but, exerting no force on it now
313
00:34:48,610 --> 00:34:54,800
heat is transferred to the gas causing the
piston to rise and to compress the spring
314
00:34:54,800 --> 00:35:01,270
until the volume inside the cylinder doubles
if the cross sectional area of the piston
315
00:35:01,270 --> 00:35:08,270
is 0.25 meter square determine part a.
The final pressure inside the cylinder the
316
00:35:08,740 --> 00:35:15,740
total work done by the gas and part c the
fraction of this work done against the spring
317
00:35:16,710 --> 00:35:23,710
to compress it so here we have what looks
like a rather complicated problem but, as
318
00:35:25,350 --> 00:35:30,690
we understand the problem little better it
will come that it is not as complicated as
319
00:35:30,690 --> 00:35:33,950
it looks like that though the question looks
a little long.
320
00:35:33,950 --> 00:35:40,950
Well here, we have a piston cylinder device
were in the piston cylinder contains initially
321
00:35:42,630 --> 00:35:49,330
certain amount of gas which is at a pressure
and a volume and then there is a spring which
322
00:35:49,330 --> 00:35:55,390
has a certain spring constant of 150 kilonewtons
per meter it is just touching the piston but,
323
00:35:55,390 --> 00:36:02,140
a spring is not exerting any force on it now
as you heat the gas in the cylinder the piston
324
00:36:02,140 --> 00:36:08,400
will start rising right because the specific
volume of the gas will increase and as the
325
00:36:08,400 --> 00:36:14,460
gas rises that will compress the spring against
the spring constant now so there is a certain
326
00:36:14,460 --> 00:36:19,960
work done by the gas against the atmospheric
pressure and all that and also there is certain
327
00:36:19,960 --> 00:36:25,730
amount of work which the gas has to do against
the spring to compress it.
328
00:36:25,730 --> 00:36:32,730
So what we are required to find is the pressure
inside the cylinder then we also need to find
329
00:36:32,860 --> 00:36:38,060
the work done by the gas that is total work
done and also the fraction of this total work
330
00:36:38,060 --> 00:36:43,220
done which is the work done for compressing
the spring.
331
00:36:43,220 --> 00:36:50,220
So let us now look at the process diagrams
as well as the problem definition in terms
332
00:36:51,250 --> 00:36:58,250
of an illustration so here we have the piston
cylinder arrangement now the piston is enclosing
333
00:36:59,650 --> 00:37:06,650
a certain gas at 0.05 meter cube volume it
has a pressure of 200 kilopascals and area
334
00:37:06,850 --> 00:37:10,270
of the piston is also given as 0.25 meter
square.
335
00:37:10,270 --> 00:37:14,990
So initially there is a spring here which
does not exert any force on the piston spring
336
00:37:14,990 --> 00:37:21,930
constant is given to as now as you start transferring
heat to the system the piston moves up and
337
00:37:21,930 --> 00:37:28,130
it moves till the volume inside the cylinder
doubles that is the final volume would be
338
00:37:28,130 --> 00:37:35,130
equal to 2 into 0.05 that is 0.1 meter cube
so which means that the piston does work against
339
00:37:38,010 --> 00:37:44,620
the cylinder against the spring as well as
work done during the expansion process assumptions
340
00:37:44,620 --> 00:37:50,010
obviously are that the expansion process is
quasi-equilibrium and the spring is linear
341
00:37:50,010 --> 00:37:56,710
in the range of interest.
So the process path here is that initially
342
00:37:56,710 --> 00:38:03,710
we have the system at state 1 which is at
V 1 is equal to 0.05 meter cube and P 1 is
343
00:38:05,380 --> 00:38:12,380
equal to 200 kilopascal and then there is
an expansion process where in the volume increases
344
00:38:13,510 --> 00:38:20,020
to it is double the value initially and to
a pressure which is 300 kilopascals and so
345
00:38:20,020 --> 00:38:26,010
this is how the process is what are indicated
by 1 and 2 will be clear little later I shall
346
00:38:26,010 --> 00:38:31,940
explain what this means that is 1 and 2 in
this diagram.
347
00:38:31,940 --> 00:38:38,650
So, the enclosed volume at the final state
is twice the initial volume therefore, V 2
348
00:38:38,650 --> 00:38:45,390
is equal to 2V 1 and therefore, it is 2 into
0.05 which is 0.1 meter cube.
349
00:38:45,390 --> 00:38:50,470
Now, we need to calculate the displacement
of the piston so how do we calculate the displacement
350
00:38:50,470 --> 00:38:54,970
of the piston we know the change in volume
from the initial in the final states so delta
351
00:38:54,970 --> 00:39:00,510
V is known area of the piston is known so
x which correspondence to the displacement
352
00:39:00,510 --> 00:39:07,510
of the piston will be equal to delta V by
A which is 0.1 minus 0.05 by 0.25 so the net
353
00:39:08,770 --> 00:39:15,770
displacement we get as 0.2 meters.
Now, how do we find the force applied by this
354
00:39:17,080 --> 00:39:22,920
linear spring at the final state while force
is the spring constant times the displacement
355
00:39:22,920 --> 00:39:29,400
so we know the displacement which we have
just calculated as equal to delta V by A and
356
00:39:29,400 --> 00:39:33,850
we also know the spring constant so spring
constant k is given as 150 kilonewtons per
357
00:39:33,850 --> 00:39:39,270
meter.
So k times x will be the force which is been
358
00:39:39,270 --> 00:39:44,640
applied on the spring at the final state so
k times x will be 150 kilonewtons per meter
359
00:39:44,640 --> 00:39:51,210
into 0.25 so that is 30 kilonewtons. So, this
is the force exerted or force applied by the
360
00:39:51,210 --> 00:39:58,210
spring at the final state now we now need
to find what is the final pressure at the
361
00:40:01,700 --> 00:40:04,450
state 2 that is, at the end of the process
what is the final pressure.
362
00:40:04,450 --> 00:40:11,230
Now, pressure as we know is force per unit
area so we have just now calculated the force
363
00:40:11,230 --> 00:40:16,440
that the spring exerts at the final state
so that force divided by the area will give
364
00:40:16,440 --> 00:40:21,260
you the pressure well that is the pressure
which is basically the additional pressure
365
00:40:21,260 --> 00:40:24,000
because of the presence of the spring itself.
366
00:40:24,000 --> 00:40:31,000
Therefore, if you calculate P, P would be
equal to the force that is F divided by A
367
00:40:31,350 --> 00:40:38,350
and so it is 30 kilonewtons divided by 0.25
that is 120 kilopascals; without the spring
368
00:40:39,800 --> 00:40:45,830
the pressure of the gas would remain a constant
because you are allowing the gas to expand
369
00:40:45,830 --> 00:40:49,540
and therefore, the volume is changing but,
the pressure will remain a constant so if
370
00:40:49,540 --> 00:40:55,400
the spring was not there the gas would remain
at 200 kilopascal which is it is initial pressure
371
00:40:55,400 --> 00:41:01,030
while the piston keeps rising but, because
of the presence of the spring the pressure
372
00:41:01,030 --> 00:41:08,030
rises linearly because it is a linear spring
and it rises from 200 kilopascals to 200 plus
373
00:41:09,000 --> 00:41:15,210
120 which is 120 is corresponding to the additional
pressure which is applied by the spring on
374
00:41:15,210 --> 00:41:20,910
the gas at this state. Therefore the final
pressure will be equal to 200 pascal which
375
00:41:20,910 --> 00:41:27,090
is the initial pressure plus 120 which is
the pressure due to the spring which is 320
376
00:41:27,090 --> 00:41:30,740
kilopascals.
Therefore, the final pressure in the cylinder
377
00:41:30,740 --> 00:41:37,170
at the end of the process is 320 kilopascals
so this is the answer to the first part of
378
00:41:37,170 --> 00:41:41,920
the question which was to find the final pressure
at the end of the process at that is at state
379
00:41:41,920 --> 00:41:45,670
2.
So, here the pressure corresponds to two different
380
00:41:45,670 --> 00:41:51,270
components one is because of the initial pressure
itself which would have remained constant
381
00:41:51,270 --> 00:41:58,040
if they were no spring which was 200 kilopascals
and because of the presence of the spring
382
00:41:58,040 --> 00:42:02,810
as the piston expands the spring exerts a
pressure on the piston which we calculated
383
00:42:02,810 --> 00:42:07,080
as 120 kilopascals.
Therefore, the total pressure will be 200
384
00:42:07,080 --> 00:42:14,080
plus 120 that is 320 kilopascals let us know
calculate the work done required or work done
385
00:42:15,030 --> 00:42:20,860
during this process and as well as the work
done against the spring that is the third
386
00:42:20,860 --> 00:42:21,710
part of the question.
387
00:42:21,710 --> 00:42:26,080
So, second part of the question is work done
so how do we find the work done during this
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00:42:26,080 --> 00:42:33,080
process if you go back to our first slide
were we discussed about the process diagram
389
00:42:33,300 --> 00:42:37,880
and also if you recall our earlier discussions
work done during this process will be work
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00:42:37,880 --> 00:42:41,580
area under this curve.
You can clearly see that this is basically
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00:42:41,580 --> 00:42:46,440
a trapezoid so if you calculate area under
this curve because and you know the end points
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00:42:46,440 --> 00:42:52,119
of the trapezoid so it is very easy to calculate
area under this curve which is basically the
393
00:42:52,119 --> 00:42:54,420
work done during the process.
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00:42:54,420 --> 00:43:01,350
So, work done will be equal to area and how
do you calculate the area because it is basically
395
00:43:01,350 --> 00:43:08,160
area of a trapezoid in this case the end points
are pressures and volumes so it will basically
396
00:43:08,160 --> 00:43:15,160
be equal to 200 which is initial pressure
plus 320 final pressure divided by 2 multiplied
397
00:43:16,320 --> 00:43:23,320
by the difference in the volumes 0.1 minus
0.05 therefore, if we were to calculate this
398
00:43:26,230 --> 00:43:31,420
the total work done during this process will
be equal to 13 kilo joules well this is not
399
00:43:31,420 --> 00:43:36,570
the only way you can solve it you can all
solve it in the same way as we solved the
400
00:43:36,570 --> 00:43:42,670
previous problems; that is, if you were to
consider the process which has a pressure
401
00:43:42,670 --> 00:43:47,800
which is varying linearly you can calculate
the work done during that particular process
402
00:43:47,800 --> 00:43:53,210
or you could also calculate separately work
done if it was a constant pressure process
403
00:43:53,210 --> 00:43:57,840
and work done if due to the spring itself
and then add up the two so there are different
404
00:43:57,840 --> 00:44:03,440
ways of calculating this easiest of them being
just you calculate the area under the curve
405
00:44:03,440 --> 00:44:05,880
which corresponds to the particular process.
406
00:44:05,880 --> 00:44:12,720
Therefore, the total work that we have calculated
comes out to be 13 kilo joules now so that
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00:44:12,720 --> 00:44:17,340
is answer to the part 2 of the question part
3 of the question is to find the fraction
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00:44:17,340 --> 00:44:20,750
of the work done against the spring to compress
it.
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00:44:20,750 --> 00:44:27,750
Now, so that, basically represents the region
one here which corresponds to the amount of
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00:44:29,900 --> 00:44:36,900
work that is required to work done against
the spring. So, here in in the the p v diagram
411
00:44:37,090 --> 00:44:42,520
which I had shown, region one corresponds
to the work done against the piston and atmosphere
412
00:44:42,520 --> 00:44:49,350
that is the work done if the work no spring
and region two corresponds to the area, because
413
00:44:49,350 --> 00:44:55,880
of the presence of the spring itself. So,
the work done by the spring is basically the
414
00:44:55,880 --> 00:45:02,880
work done under region two which is half of
320 minus 200 which is difference of pressures
415
00:45:04,260 --> 00:45:09,480
multiplied by the difference in the volumes
that is 0.05 meter cube. So, if we calculate
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00:45:09,480 --> 00:45:13,880
this, you will get 3 kilo joules
So, fraction of this work done against the
417
00:45:13,880 --> 00:45:18,750
spring, to compress it is 3 kilo joules. Well,
you can also calculate the work done by the
418
00:45:18,750 --> 00:45:25,740
spring, using the spring constant which is
given in this case already as 150. So, work
419
00:45:25,740 --> 00:45:32,740
done by a spring if you recall basic mechanics
is half k times x 2 square minus x 1 square
420
00:45:32,910 --> 00:45:39,910
where x 1 and x 2 are initial and final positions
of the spring. So, x 2 in this case is 0.2
421
00:45:40,080 --> 00:45:46,740
meters x 1 is 0. So, if you calculate work
during using this formula, you will also get
422
00:45:46,740 --> 00:45:52,700
the same answer that is 3 kilo joules. So,
this solves our the fifth problem that we
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00:45:52,700 --> 00:45:59,700
have solved today, which was of piston cylinder
assembly and compressing a certain spring.
424
00:46:00,290 --> 00:46:07,290
So, the process consist of two different parts,
one was to calculate the work done by the
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00:46:07,910 --> 00:46:12,490
piston if there were no spring and also to
calculate the work done because of the presence
426
00:46:12,490 --> 00:46:17,820
of the spring. And so we can actually add
up the work done during these two separately
427
00:46:17,820 --> 00:46:24,820
to calculate the total work done during this
particular process. And so we shall now discuss
428
00:46:25,490 --> 00:46:31,490
a few exercise problems which we have, which
I shall discuss with you and which you can
429
00:46:31,490 --> 00:46:37,390
solve later on at your leisure based on what
we have discussed during the previous lectures
430
00:46:37,390 --> 00:46:43,660
as well as based on some of the problems which
we have solved during today's lecture.
431
00:46:43,660 --> 00:46:50,660
So, the first exercise problem which I have
for you is on a fluid contained in a horizontal
432
00:46:52,500 --> 00:46:58,210
cylinder, is continuously agitated using a
stirrer passing through the stirrer cylinder
433
00:46:58,210 --> 00:47:04,710
cover. The cylinder diameter is 0.4 meters
during the stirring process which last for
434
00:47:04,710 --> 00:47:11,710
ten minutes the piston slowly moves out at
distance of 0.485 meters. The net work done
435
00:47:12,050 --> 00:47:17,830
by the fluid during the process is 2 kilo
joules. The speed of the electric motor driving
436
00:47:17,830 --> 00:47:24,830
the stirrer is 840 r p m. Determine the torque
in the shaft and the work out well power output
437
00:47:24,850 --> 00:47:29,190
of the motor.
So this is an example of an exercise problem
438
00:47:29,190 --> 00:47:34,500
for you, were in, we have a a piston cylinder
arrangement and through the head of the cylinder
439
00:47:34,500 --> 00:47:41,500
or the cylinder cover we have a stirrer which
is rotated by an electric motor. And so, we
440
00:47:42,380 --> 00:47:48,859
are basically adding work into the system
using a stirring process. And so, as you add
441
00:47:48,859 --> 00:47:54,180
energy to the system the piston moves because
you are adding energy into the cylinder and
442
00:47:54,180 --> 00:47:59,850
therefore the cylinder, the piston expands
and piston moves to a different state.
443
00:47:59,850 --> 00:48:04,550
And so what we have been given is the net
work done by the system during this process
444
00:48:04,550 --> 00:48:09,890
which is specified as 2 kilo joules and also
we have been specified how much time this
445
00:48:09,890 --> 00:48:14,490
stirring process is done and during which
during this time, how much the piston has
446
00:48:14,490 --> 00:48:20,950
moved. And so, you are required to find what
is the torque in the shaft and the power output
447
00:48:20,950 --> 00:48:26,280
of the motor because work done is specified
and the distances which have been moved have
448
00:48:26,280 --> 00:48:30,080
been given so, you can find the torque in
the shaft and the power output.
449
00:48:30,080 --> 00:48:35,700
So, I have given the answers here you can
verify these answers by calculating it. You
450
00:48:35,700 --> 00:48:42,490
should be getting a torque in the shaft of
0.08 newton meters and power output of the
451
00:48:42,490 --> 00:48:45,670
motor as 6.92 watts.
452
00:48:45,670 --> 00:48:52,670
Well, the second exercise problem we have
today is on again an expansion process p v
453
00:48:54,600 --> 00:49:01,450
diagram process and in this process it is
a two part expansion process. One is a constant
454
00:49:01,450 --> 00:49:08,450
pressure process and consider a two part process
with an expansion from 0.1 to 0.2 meter cube
455
00:49:11,570 --> 00:49:18,570
at constant pressure of 150 kilopascal's followed
by an expansion from 0.2 to0.4 meter cube
456
00:49:21,010 --> 00:49:27,260
with a linearly raising pressure from 150
kilopascal ending at 300 kilopascal. So, show
457
00:49:27,260 --> 00:49:30,760
this process in a p v diagram and find the
boundary work.
458
00:49:30,760 --> 00:49:37,440
So, here you have a problem where the process
is carried out in two parts first, is a constant
459
00:49:37,440 --> 00:49:43,369
pressure process where, the volume changes
from0.1 to 0.2 meter cube and then there is
460
00:49:43,369 --> 00:49:49,760
an expansion process which takes it from 0.2
to 0.4 and this expansion process consists
461
00:49:49,760 --> 00:49:56,760
of a linear variation of pressure from 150
kilopascal's to 300 kilopascal's. So, you
462
00:49:57,190 --> 00:50:01,950
have to find the process. Well, work done
during this process as well as obviously the
463
00:50:01,950 --> 00:50:07,080
to show the process on a p v diagram. So,
the answer for this particular question would
464
00:50:07,080 --> 00:50:13,350
be the work done should be equal to 60 kilo
joules.
465
00:50:13,350 --> 00:50:18,570
And the third problem we have as an exercise
problem for you is a piston cylinder arrangement
466
00:50:18,570 --> 00:50:25,570
which contains water at 500 degree celsius
300 mega pascal's. It is cooled in a polytrophic
467
00:50:27,580 --> 00:50:33,660
process so we have process which cools water
from 500 degree celsius 3 mega pascal's to
468
00:50:33,660 --> 00:50:40,660
200 degree celsius 1 mega pascal. We need
to find the polytrophic exponent and the specific
469
00:50:42,890 --> 00:50:46,450
work in this process that is you need to find
p v raise to n where n you have to have to
470
00:50:46,450 --> 00:50:53,020
find as well as the work done during this
process. So, in this case you will find that
471
00:50:53,020 --> 00:50:59,260
the polytrophic exponent will come out to
be 1.919 and work done during this process
472
00:50:59,260 --> 00:51:03,700
is 155.2 kilo joules.
473
00:51:03,700 --> 00:51:10,700
And the last problem exercise problem for
you is, consider a gas enclosed in a piston
474
00:51:10,990 --> 00:51:16,770
cylinder assembly as a system the gas is initially
at a pressure of 500 kilopascal and occupies
475
00:51:16,770 --> 00:51:23,770
a volume of 0.2 meter cubes. The gas is taken
to the final state where the pressure is 100
476
00:51:24,200 --> 00:51:28,910
kilopascal by the following two different
processes. Calculate the work done by the
477
00:51:28,910 --> 00:51:34,010
gas in each case. So the two processes are
one is, volume of a gas is inversely proportional
478
00:51:34,010 --> 00:51:38,960
to pressure that is v is inversely proportional
to p or in other words p v is a constant it
479
00:51:38,960 --> 00:51:43,190
is a isothermal process.
Second process is, process follows p v raise
480
00:51:43,190 --> 00:51:50,070
to gamma is a constant where gamma is equal
to 1.4. So for this polytrophic processes
481
00:51:50,070 --> 00:51:56,460
you need to find the work done, and so for
the first part is 160.94 kilo joules. For
482
00:51:56,460 --> 00:52:03,460
the second part that is p v raise to gamma
question is 92.15 kilo joules. So, that brings
483
00:52:04,330 --> 00:52:11,000
us to the end of this lecture, where we disused
about problem solving for work done during
484
00:52:11,000 --> 00:52:16,580
different processes and we have discussed
some examples where we calculated work done
485
00:52:16,580 --> 00:52:21,119
during different types of processes. So, what
we are going to do is in the next lecture
486
00:52:21,119 --> 00:52:27,170
is to discuss about the first law of thermo
dynamics as applied to closed system.
487
00:52:27,170 --> 00:52:31,710
So, during this discussion that is during
the next lecture we shall discuss about first
488
00:52:31,710 --> 00:52:37,830
law applied to closed systems, and we shall
discuss about energy balance, we shall discuss
489
00:52:37,830 --> 00:52:43,170
about energy change for a system, we shall
discuss about energy transfer mechanisms,
490
00:52:43,170 --> 00:52:50,170
during for closed systems. And we shall apply
first law for a cycle as well as first law
491
00:52:50,530 --> 00:52:55,790
for a system undergoing a change of state
. So, these are the things that we shall be
492
00:52:55,790 --> 00:53:02,790
discussing during the next lecture that would
be lecture number eight.