Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 05
Properties of Pure Substances: Example Problems Today, we will look into various aspects of
identifying state points, considering the property tables of pure substances. The best
way in which this can be understood is through solution of problems, and we will consider
a few problems which we will work out. So, the first problem, which is problem 1.1
is written here, described here, the as per the custom I will go through the problem description,
then draw a schematic of this problem, and then we will proceed towards solving the problem.
A cylinder has a thick piston initially held by a pin, as shown in the figure below. The
cylinder contains carbon dioxide at 150 kilo Pascal and at and ambient temperature of 290
Kelvin. The metal pistons density is given and the atmospheric pressure is given. The
pin is now removed, allowing the piston to move and after a while the gas returns to
ambient temperature. So, you can see that there is a piston, there is a pressure imbalance
across the two sides of the piston. This is the first important information from the problem.
So, because there is a pressure imbalance the piston has a tendency to either move up
or down it depends on you know which side pressure is more.
So, to arrest the piston from moving further there is a pin which is holding the piston.
If the pin is removed, the piston will freely move and it will come to a location where
it is dictated by the equilibrium of forces from all sides, which is called as mechanical
equilibrium. Question is: that is the piston against the stops or not. So, to work out
this problem, so let us draw a schematic of this in the board. So, I will draw a schematic
of this problem in the board. So, you have a pin, this is the height of
the piston H is equal to 100 millimeter. Then you have again this as 100 millimeter and
this as 50 millimeter. The system here is carbon dioxide which we show by this dotted
line. So, whatever is given let us note it down it may help us.
So state 1: 150 kilo Pascal, this is P1, ok. Now what will be the equilibrium pressure
at which the piston will float if the pin is removed; let us calculate that. So, what
is p equilibrium? P equilibrium means, if the piston is in equilibrium under all forces
then what would be the corresponding pressure inside. So for that, we draw the free body
diagram of the piston, free body diagram of the piston.
So, this is atmospheric pressure, right, A is the area of the piston. Then, you have
mass of the piston into g which is the weight of the piston. And then from this side you
have p equilibrium into A right. So, if the piston is moving in a quasi equilibrium process
or quasi static process these forces are always in equilibrium. So, if the volume is changing
this is changing, but locally it is always in equilibrium; that means, resultant of these
forces is zero. So, we can write p equilibrium into A is equal
to p0 into A plus mg. So, p equilibrium is equal to p 0 plus m g by A ok. The mass of
the piston you can calculate: the density into volume. So, rho piston into A into H,
this is the mass, so A is not required. Density of the piston what is given? So, let us mark
out these values. Density of the piston is 8000 kg per meter cube. H is 100 millimeter;
that means 0.1 meter, g is 9.1 81. And what is p0? p0 is this one; p0 is given as 101
kilo Pascal. So, if you substitute the values in most of
the problems that I will be solving in the class, I have made some pre-calculations.
And I will not waste time by you know clicking the calculator to work this out. I will give
you the value if it is available because in thermodynamics depending on this value the
solution of the problem will be there. That is why thermodynamics is so interesting. So,
depending on the specific value the physics of the answer will be has to be set. It does
not have a specific rule that whatever is the value the answer to the state point is
unique; the answer to what is the state depends on what is the value which is coming. So,
this value is very important not that we can say that whatever is the value based on which
a general answer will come. So, this value is 108.8 kilo Pascal. So, if
this pressure is more than. So, what is p1? p1 is 150 kilo Pascal. That means, at state
1 this force in the upper direction is greater than the force in the lower direction. That
means, had the pin not been there the piston will have a tendency to go up. Piston wants
to go up and the pin actually restraints it from doing that, ok.
Now, what happens when the pin is removed? When the pin is removed this piston will have
a tendency to go up. Question is how far will it go up, right? So, it says that the piston,
the pin is now removed allowing the piston to move and after a while that the gas returns
to ambient temperature. So question is: does the piston hit the stops. So, hypothetically
assume that the piston hits the stops, ok. If that
be the case then the final volume will be whatever is the volume plus this much up to
the stops. So, we can write p1 V1 by T1 is equal to pstop in to Vstop by Tstop.
What is the assumption? Assumption is that we can treat carbon dioxide at such low pressures
as ideal gas. Remember a real gas can be approximated as an ideal gas at reasonably low pressure
or high temperature. So assuming ideal gas behavior. What is given is that at state point
2 which is assumed to be the stop here, the gas returns to the ambient temperature. It
started with the ambient temperature at what is that T1 is 290 Kelvin, and it is returning
also to the ambient temperature. So, if it is returning to the ambient temperature and
starting with the ambient temperature then T1 is Tstop, right.
Remember, that T1 is equal to Tstop that does not mean that the process from one to stop
is isothermal process. It says that the piston goes to whatever it may be stop or before
the below the stop and then the system is allowed to come to ambient temperature by
maybe transferring heat. So, T1 is equal to Tstop, but in between all the state points
were not same as T1. So, an isothermal process means that all the intermediate state points
will be at the same temperature, not just that the final and initial temperature at
the same. So, this is a classical example where final
and initial temperature at the same, but it is not an isothermal process. In between the
temperature was not guaranteed to be same as T1. So now, you can calculate from here
pstop is equal to p1 into V1 by Vstop, okay. So what is V1 by Vstop? V1 by Vstop is 100
by 100 plus 50; so this is 100 by 150, into what is p1? Is 150, so this is 100 kilo Pascal,
ok. Now look into this equilibrium pressure, when the when the piston is moving its pressure
can it fall below equilibrium pressure: the answer is no. It’s It will move in a quasi
equilibrium process so, its pressure will be equilibrium pressure. If it hits the stops,
the pressure inside will build up. Why? because it wants to go further but the stops is preventing
that. So, the pressure the final pressure has to
be greater than equal to the equilibrium pressure. It is equal to the equilibrium pressure, if
it is not hitting the stops, but if it is hitting the stops the pressure will be greater
than the equilibrium pressure. So, this conceptual understanding you have to develop that, in
the thermodynamics that we are studying in this course if the piston is moving it is
moving in a quasi equilibrium process. If it is moving in a quasi equilibrium process;
that means, the pressure here is always the equilibrium pressure. The resistance pressure
is not changing, so the equilibrium pressure is also not changing. When it hits here, the
pressure inside will build up. So, none of this is happening, that means,
actually this hypothesis is wrong. That means, the piston does not hit the stops ok. So,
that is the answer to this problem. The problem what is asked is that is the piston against
the stops? The answer is no, the piston is not against the stops. We will work out another
problem. So, we will look into the next problem, problem
1.2. So, the diagram of the problem is given here and again I will read out the problem.
A cylinder is fitted with a 10 centimeter diameter piston that is restrained by a linear
spring; linear spring means the spring force is proportional to the displacement as shown
in the figure below. The spring force constant is 80 kilo Newton per meter and the piston
initially rests on the stops, with a cylinder volume of 1 litre.
Now, there is a air supply line. When the air supply line is opened the piston begins
to rise, because the additional air will pressurize the air from the below and the piston which
was earlier resting on the stops will experience more pressure from bottom than the top, and
then it will go up. In this process the cylinder pressure rises to 150 kilo Pascal, when the
valve is closed the cylinder volume is 1.5 litre and the temperature is 80 degree centigrade.
Question is: what is the mass of air inside the cylinder. So, just like the previous problem
here also we will draw a schematic. So the system is air, and then what is asked
is to calculate; that what is the mass of air inside when the valve is closed. So, let
us try to note the property data diameter of the piston 10 centimeter, this is 10 into
10 to the power minus 2 meter. The spring constant is 80 kilo Newton per meter. V1 is
1 litre that is 1 into 10 to the power minus 3 meter cube. Then p equilibrium
1, the pressure at which the equilibrium pressure at which the piston starts rising is 150 kilo
Pascal. V2 is 1.5 litre is 1.5 into 10 to the power minus 3 meter cube. And T2 is 80
degree centigrade. Question is: what is m2 ok.
So, can you tell: what is the difference between the previous problem and this problem? The
difference is that because the spring force varies as the piston is moving the equilibrium
pressure also varies as the piston is moving. In the previous problem there was no spring,
so equilibrium pressure remain constant; now here the equilibrium pressure will vary. So,
we draw the free body diagram of the piston. So, we will identify the forces. So, first
is let us identify the forces due to pressure. So, you have p0 into A which is the atmospheric
pressure the pressure is p0; here it is p equilibrium into A. And then, the spring force,
it is very important to figure out what is the direction of the spring force. So, when
the air is being supplied the pressure here is increasing. So, if the pressure here is
increasing the piston will have a tendency to go up. So, the spring will give a restoring
force downwards. So, you have a kx like this. So, p equilibrium into A is equal to p0 into
A plus kx. So, p equilibrium is equal to p0 plus kx by A, right.
So, p equilibrium 2 state 2 minus p equilibrium state 1 is equal to k into x2 minus x1 by
A, right. And x2 minus x1 is V2 minus V1 by A, so this becomes V2 minus V1 by A square.
So, what is A? A is pi d square by 4, it is a circular piston. So, k is given, A is given,
V2 is given, V1 is given, all this data are given, so if you calculate this and p equilibrium
1 is also given. So, this equation will give you what is p equilibrium 2. So, p equilibrium
2 is equal to, I mean some value I do not have the value with me exactly, but you can
substitute all these, so p equilibrium 2 value is known.
Sir, why did not you consider the weight of the piston?
So, the weight of the piston here is not given; so the important question is why don’t we
calculate, why don’t we consider the weight of the piston we can consider the weight of
the piston and still that will not be important. So, we can add the weight of the piston; you
will see that it gets cancelled out. So, actually you can consider mpg along with this, so plus
mpg. So, we will see that I mean whatever is the weight of the piston it will not matter
if you find the difference in pressure. Ideally, you are correct you should consider the weight
of the piston; I mean somehow I omitted that in the free body diagram, but you it is very
easy and straight forward to include the weight of the piston.
So, if you include the weight of the piston, you have an additional term, but this p0 plus
mg by A this gets cancelled out, so you will get p equilibrium 2. Once you get p equilibrium
2 it is very easy because m2. So, p equilibrium 2 into V2 is equal to m2 R T 2. What is this
R? Again coming from school students confuse, this is not the universal constant. This is
the universal constant divided by molecular weight ok. So, this is the specific gas constant.
So, this for air, what is the substance here? The, air, right this is air. So, R for air
is 0.287 kilo Joule per kg Kelvin. This you divide the universal constant divided by the
molecular weight ok. So, once you do that, this is R air, this is T2. Remember that when
you write this equation this temperature has to be in Kelvin. So, this is 273.15 plus 80
Kelvin, V 2 is known, p equilibrium 2 is known, so from here you will get what is m2. The
answer to this problem is m2 is 0.012 kg ok. I hope I could illustrate two very typical
problems in which one case there is a piston cylinder without spring, another case piston
cylinder with spring, and how to calculate the equilibrium pressures under that condition.
Let us stop here for this lecture, we will continue with other problems on properties
of pure substances in the next lecture. Thank you.
