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Good morning, I welcome you all to this session. In the last two classes, we have discussed
the analysis of first law of thermodynamics to closed and open systems. In this class,
we will solve some problems to clear our concept about this analysis of first law to closed
and open systems. Through some practical problems, we will see how these analyses are being utilized.
Well, so let us start with the problems.
First problem, example one: This is a problem relating to the application of first law in
a closed system and we start with a very simple problem - straightforward application. A gas
of mass 1.5 kg undergoes a quasi static expansion, you may or may not write the entire problem,
you can write only the pertinent points. For example, a gas the mass of which is 1.5
kg, undergoes a quasi static expansion, already we know that is pdV work, which follows a
relationship p is equal to a plus bv, that is the relationship p and V, p is equal to
a plus bV, where a and b are constants. This is the law of the process displacement work,
where the pressure and volume changes. The initial and final pressures are 1000 kiloPascals
and 200 kiloPascals respectively. The initial pressure is 1000 kiloPascals and the final
pressure is 200 kiloPascals. That means it is an expansion process where the pressure
decreases. The corresponding volumes are 0.2 meter cube - that is the initial volume and
1.2 meter cube - that is the final volume. The specific internal energy of the gas is
given by….I told you earlier that regarding this internal energy, how does it relate with
the pressure volume, this we will see afterwards, but at the present moment we will assume some
relationship which will be given as u is equal to 1.5 pv where V is the specific volume and
p is the pressure minus 85 kilo Joule per kg. This u is expressed in kilo Joule per
kg provided we substitute p in this equation in kiloPascals; these things are very important
utilizing this equation. If you do a mistake, so you will lose marks in examination.
Concept remains the same, but to get the correct answer we have to be very careful. v in meter
cube per kg; that means, if you express p and v in these units, then u will be given
by kilo Joule per kg according to this relationship. These constants - the numerical values - are
adjusted like that. Calculate the net heat transfer and the maximum
internal energy of the gas during the expansion. The problem is clear to you? So, now, how
to solve the problem? First of all, we will have to find out the net heat transfer. So,
this is a problem of a closed system.
For a closed system, we will always write the first law in this fashion, as we know
already what to write; Q1-2, let 1 and 2 be the initial and final state points is equal
to u2 minus u1 plus integral of pdV, that is the work 1 to 2, this is the work transfer
W. That means W1-2 is integral of pdV. It is the very simple problem straightforward
applications. Now we know this p is a liner function of
v, a plus bv; so we can straightforward integral d. But to get a numerical value, we have to
know the numerical values of these constants a and b and it is very simple; it is found
from these two conditions; that means, it is written; when p1 is equal to 1000 kiloPascals,
what is V1? V1 is equal to 0.2 meter cube; this is one set. Another is p2; p2 is how
much? p2 is equal to 200 kiloPascals. What is V2? V2 is equal to 1.2 meter cube. Now
if we substitute this p, not meter cube per kg only meter cube, very good, P is equal
to a plus bv; this is the relation. That means, we will write this 1000 in kiloPascals, I
am writing, accordingly the unit will come. 1000 is equal to a plus b, 0.2 because, I
am not converting into Pascals by multiplying with 10 cube unnecessarily, the number will
be very high, so b in meter cube. Then we write 200 is equal to a plus 1.2b. If you
solve these two equations, you will get a is equal to 1160 kiloPascals. You have done
this problem? Just now you are doing? Very good. The unit of a has to be kiloPascals
because it is pressure. What is the value of b? b is equal to minus
800 kiloPascals per meter cube. Very simple problem that is why you are so prompt. Therefore,
I get av. Now, what is W1-2? Tell me. W1-2 is equal to a into V2 minus V1 plus b by 2
into V2 square minus V1 square. All this students may not catch so fast as for wide spectrum
of the students, so we will go slow. Now we have got the values V2 is equal to
1.2 meter cube, V1 is equal to 0.2 meter cube and ‘a’ is in kiloPascals, ‘b’ is
in kiloPascals per meter cube. If you substitute this value, you get the value of W as 600
kiloJoule. If you just substitute the value a in kiloPascals and V2 minus V1 in meter
cube, so kiloPascals into meter cube will be kiloJoule. Here also, if you substitute
kilopascals per meter cube and this is V2 square; therefore, you get the value 600 kiloJoule.
Now what is Q1-2? Q1-2is equal to u2 minus u1 plus 600 kiloJoule. Now, here W is coming
to be plus; so this sign automatically comes when you integrate this. We put the limit
final minus initial. So, final is the upper limit; initial is the lower limit. Depending
upon the final and initial values of V the sign of this quantity is determined. So it
is determined as a positive quantity which physically signifies that work is coming out
of the system. Now how to find out u2 minus u1? From the property relation, School level
problems in fact. What is this property relation? This is kiloJoule per kg.
If I want to find out capital U, we will have to multiply with the mass; that means, V will
be the volume; this V is also in meter cube per kg. That means, I will write the equation
rather in this fashion. U is equal to 1.5 pV minus 85 into mass of
the gas. This is the relationship between the pressure volume and the total internal
energy of this system. Then the equation becomes U is equal to 1.5 pV minus 85 into 1.5, but
this part is constant, it is not required in finding out the change. So, change will
be given by U2 minus U1 is equal to 1.5 into p2V2 minus p1V1, which is equal to 60 kiloJoule.
Therefore, Q1-2 is equal to 600 plus 60, that means 600 is W1-2 and 60 is u2 minus u1,is
equal to 660 kiloJoule. What is the next part? The maximum internal
energy. How do you do this part? Calculate the maximum internal energy of the gas during
the expansion. How do you do it? Tell me.
Let us first write u. What is u? u is equal to 1.5 pV minus 85 into1.5. So, you express
this in one variable. That is p is a plus bV, then we get u is equal to 1.5 into a plus
bV into V minus 85 into 1.5. It does not matter whatever it is until and unless we are interested
in finding out them. It will be required when we find out the maximum internal energy. Let
us make du by dv. What is du by dv? du by dv is equal to 1.5 into a plus 2bV and said
this is equal to 0 for maximum, but this curve has only a maximum with the values of a and
b, how? d square u by dv square. is negative. What
is d square u by dv square? That is equal to 1.5 into 2b and b is negative in this problem;
minus 800 kiloPascals per meter cube. Therefore, this shows with these values of ab, this curve,
that means the relationship between u and V, shows only a maximum . So, this is the
condition for maximum not a minimum. So that maximum condition is given. So that Vis equal
to minus a by 2b. Therefore maximum internal energy will be
found out umax is equal to 1.5 into a plus b into minus a by 2b into minus a by 2b minus
85 into 1.5. By substituting the values this we can simplify into 1.5 into a by 2 into
minus a by 2b minus 85 into 1.5. So, the minus in the a will get cancelled with the minus
in the b. Ultimately, I do not know the sign of this, but it will be composite. What is
the value? The value will come is 503.2 kiloJoule; this is the change in the maximum internal energy.
Maximum internal energy. It is the maximum internal energy during the
expansion. Any query? It is a very straightforward problem applied to a closed system. Straightforward
application of the law that Q is equal to u2 minus u1 or delta u plus w which is pdV
work incase pdV work it is the integration of pdV.
Now, let us go to the second problem number 2; example 2: A piston cylinder arrangement as shown in figure contains air at 250 kiloPascals
- that means, the initial pressure of the air - and 300 degree Celsius. This is the
piston cylinder. The 50 kg piston, that means, the piston has a weight of 50 kg and it has
a diameter of 0.1 meter and initially pushes against the stops - that means, physically.
What is this? There is a pressure of 250 kiloPascals. This pressure exerts a force which balances
the wet plus the atmospheric pressure and the reaction against these stops, so that
piston is kept fixed with this stop. The atmospheric pressure is 100 kiloPascals and the temperature
is 20 degreeCelsius, which is the initial temperature.
It is atmospheric temperature. . The atmospheric temperature is 20 degree Celsius. Now, the cylinder cools
as heat is being transferred from the cylinder Q to the ambient. At what temperature does
the piston begin to move down? That is part one. How far has the piston dropped when the
temperature reaches ambient? That means 20 degree Celsius.
Now, you see what is the problem? Try to understand the problem physically. Now the piston is
fixed to this . When we cool it, this will be cooling at a constant volume, so pressure
will be reduced. When you take the heat from this at a constant volume the pressure will
be reduced. When the pressure will be reduced, both pressure and temperature will be reduced,
but the volume will remain same, because piston is fixed to this. When pressure will reach
a value, the piston will try to descend downward direction. So, from the mechanics point of
view, what will happen at the point when piston is just trying to descend down?
No.
No.
Pressure plus atmospheric pressure plus.
Will be equal to the atmospheric it is the pressure inside the gas very good, but now
what it is? It is not like that now at this condition
one.
Why?
Normal forces are added, the reaction will be 0, the forces on both the sides except
this reaction forces, that means pressure and the weight and the pressure forces will
be equal to each other. These normal forces here will be 0 . Then, if you reduce the pressure
any more it will automatically descend down. Therefore, we have to find that pressure and
if we know the pressure, we can find out the temperature. This is the problem. This part
is more of mechanics than of thermodynamics. What is this pressure? Calculate. What is
this atmospheric pressure? Let me write in terms of atmospheric with pressure p and all
these things.
Let the weight of the piston be W. I can write this as p0A plus W is equal to p into A, where
A is the area of this piston and p is the pressure when the piston will be in a position
to descend down. So therefore, p is equal to p0 plus W by A; the atmospheric pressure
due to the weight of the piston. The pressure of the air at which the piston will just try
to descend down will be p0 plus W by A. Now substitute the values. What is p0? What is
the value? 100 kiloPascals. This part also has to be expressed in kiloPascals. We get
p0 is equal to 100 kiloPascals plus 50 kg into mass 9.81 Newton into 4 divided by area,
what is the diameter of area? pi into 0.1 square into 1000. Wm into G, that is the weight expressed in Newton,
pi d square by 4 is the area kilopascals; that means, kilo Newton per meter square.
So, this becomes equal to what? 162.45 kiloPascals. This is the pressure when the pressure will be reduced from 250
kiloPascals to 162.45 kiloPascals, the piston will try to descend, but the question is,
at what temperature does the piston begin to move down? Now question is that I know
the pressure, but how can I know the temperature?
Constant volume p1 pi. Therefore, let T be the temperature, then p by t, that is 162.45;
rather I write this way - T by p is 162.45; this is equal to what? The initial temperature
is 300 degree Celcius.
Actually, it is 273.15, but for our practical purpose we will take it as 273 divided by
p. What is p? p is 250 kiloPascals, p is to be in kilo Pascals. This gives T is equal
to what? 99.. Therefore, we get T is equal to 372.33 k which gives t is equal to 99.33
degree Celsius. Now where from you got p1 by T1 is p2 by T2?
Volume, but you are not supposed to know the first law, so that I tell that sometimes it
will be given; if it is not given, then for air I recapitulate the earlier things: you
always know that if p is the pressure, V is the volume, m is the mass and T is the temperature
of a certain mass of gas M. Then for any ideal gas this is the relationship: pV is equal
to mRT. For any ideal gas, pressure, volume, temperature and mass is related by this equation
where R is a constant for that particular gas and is known as characteristic gas constant.
We will see afterwards if this mass is changed to another unit – moles - which is nothing
but the mass scaled with the molecular rate, then this constant is same for all ideal gases.
But if you do not scale the mass by the molecular way rather mass simply has the amount in terms
of kg, then it is the constant which is varying from gas to gas and is known as characteristic
gas constant. For many other problems, you have to use this equation. So you have to
remember one thing - the value of R; this may not be given. Even for the examinations
also because you are supposed to know certain values; so, R is one of such. For air, this
value is 287 Joule per kg, in SI unit it is 287 Joule per kgk. This we will always remember
that pV is equal to 287 mRT where p is in basic unit Newton per meter square, V is in
meter cube, m in kg and T in Kelvin. Now let us come to the next part. How far
has the piston dropped when the temperature reaches ambient how do you find out?
So how to p2V2 correct, but temperature is changing why temperature will be constant,
which is 20 degree Celsius.
V by T is constant.
One solution is that V by T is constant, so you can find out the V, pressure remains constant
that is one. Another person is telling that T will remain constant. T is not constant
because it is given there.
Pressure is remaining constant is the answer for it; that means, we will have to find the
V by making the pressure constant, so that you can use the same equation pV.
If pressure remains constant V by T is constant, you can find out the volume.
Why the pressure will remain constant? Can you tell me those who have told that pressure
will remain constant that will come from the physical concept?
Atmospheric, but inside pressure of the gas, why it will be constant?
Because always we will consider that the piston descends in an equilibrium condition, where
we will neglect the inertia of the piston in a sense that is the acceleration will be
neglected. It is slowly moving. If it moves with a uniform velocity there also there is
an equilibrium of forces, but in this case of thermodynamics all such problems of this
sort will always be considered when the piston descends down or goes up it is always in equilibrium;
the total force acting on both the sides of the pistons are equal. We neglect the acceleration
of the piston. If we make so then atmospheric pressure is constant, then the weight of the
piston is constant; that means, gas pressure has to be same. It is a displacement work
at constant pressure, but we do not have to use the pdV formula; only thing is that we
have to use that now V by t. So, we have to find out the volume.
Let this volume is Vf by Vi is equal to Tf by Ti. So you find out V and what we will
get, but we have to find out what?
What is the value of V? Just a moment.
You know Vi; Vi is not known.
Initial Vi is not known. Just a minute.
So initial Vi is known; therefore, we can find out the Vf. The value of Vf, is 1.544
meter cube, which gives h as 0.01966. How do you get it? So, the drop in height will be 0.25 minus 0.1966 that
is 5.34 centimeter. So, this temperature is which temperature ? Ti is the temperature
at which it starts falling. So we know this temperature, we know the final temperature,
it is told in the problem that it will reach ambient temperature, we can very well use
this, pressure remaining constant. This is another type of problem. There are few more
complicated problems, which I will be solving in tutorial classes. I do not know how much
I can do for these sessions, these audiovisual classes.
Now, I come to another interesting problem. This is a very standard problem for an open
system on control volume. First law for an open system on control volume is known as
problem of charging of a bottle. How it is done? First of all, you see that problem.
Before going through this problem, I should better discuss.
Let us consider a bottle here. This is a typical problem of charging a bottle which is being
charged. Let us consider there is a valve, which is being connected to a pipeline which
contains air. This is the practical problem; there is a main pipeline through which air
flows at a constant pa and Ta, and these are pressure and the temperature of the air which
is flowing. There is a valve and there is a bottle which contains initially a mass mi
and its pressure is pi, its temperature is Ti. If pi is less than pa then what will happen?
If I open this valve, then air will gradually flow in this bottle. Let us consider the entire
thing is adiabatic; that means, insulated or no heat transfer. In thermodynamic sense,
the heat transfer is 0 in either direction. Now what will happen practically? The air
will continue to flow or to charge the bottle. When the mass of air will increase in this
bottle of fixed volume, what will happen? The pressure and temperature both will increase.
This process will automatically stop, naturally stop. This is not a continuous process; so,
it will be automatically stopped when the pressure inside the bottle will reach the
value of pa; no flow will take place in that pressure equalization case; the process will
naturally stop. Parameter of interest to be known is the temperature. At that condition,
what will be the temperature in the bottle, if we do not allow any heat to come out or
go in. Because pressure will be equal to pa, that means what will be the final temperature
of the bottle when the process naturally stops. How to solve this problem?
This is a problem of application of first law to a control volume and unsteady state.
How you would do the problem? You take this bottle as a control volume. There is an inflow
to the control volume but outflow is 0 and you write for a finite time, not in the rate
form, the equation which I develop in the rate form, if we integrate that we will get
in the finite form. The work transfer is also 0; here there is no work interaction.
Let us consider at the end of this process the mass in the bottle is mf, pressure is
automatically pa, and Tf is to be found out. Now what is the energy influx to this control
volume during the entire process is the mass of the air which has come. Let this is ma
times the specific enthalpy of the air associated with this pressure and temperature and this
maha is equal to mf minus mi into ha. I am not writing everything like that. This is
the influx of energy to the control volume. It is a very simple problem mf minus mi into ha will
be equal to the change in the internal energy of the control volume that is mfuf minus miui.
Now you can ask me: sir, you are telling the most practical problem is that engineers are
interested to know what is the temperature? But this equation, where it is temperature
from the property relations, we will know how h relates with the temperature for air
and how u relates with the temperature for air. From this equation we can find out what
is uf knowing mi, ui and ha. So, we can find out uf and from uf we can find out the temperature.
In this connection, I like to tell you that for air if it is not otherwise given.
You will always assume h is equal to cp into T and u is equal to cv into T for air, where
the values of cp is 1.005 kiloJoule per kgK if it is not otherwise given. If the properties
relationships are given, you do not use that, we use the property relationship. Cv is equal
to 0.718 these are the values of cp cv. This comes from the fact that for any ideal gas,
internal energy is a function of temperature; enthalpy is a function of temperature; that
means, they are function of temperatures only not of any other property. That means, one
can write if you follow the definition that du by dt, because u is a function of temperature
there is no question of writing delta u divided by delta t is equal to cv. So you integrate
it, you get delta u is cv delta T plus constant. If you make it delta u is delta T, but if
you just integrate to give u is cvT plus C constant. So if you make the assumption that
at T is equal to 0 internal energy is 0 that is C is 0. In the similar way, we can write
using this equation, dh by dt for ideal gas is cp. I am not writing delta h divided by
delta t at constant pressure, because h is not a function of pressure it is only the
temperature. Therefore, I can write dh by dt and it becomes
cp this will be taught afterwards for an ideal gas. If we integrate it, we get delta h is
cp delta t or h is simply cp into t at T is equal to 0, we consider the h is also equal
to 0, so that h is cpT and u is cvT. If these property relationships are known, I can utilize
this formula to find out the temperature. In case of an evacuated bottle if the bottle
contains 0 mass at 0 pressure; so, this will be 0.
So the application of this equation is there in this problem three.
Now you read this problem which appears to be very simple. An evacuated bottle is fitted
with a valve through which air from the atmosphere, that means the condition of the flowing air
at 1 bar. What is bar? Bar is 100000 Newton per meter square at 25 degree Celsius and
this is precised at the the atmospheric pressure and temperture. Atmospheric pressure is little
more 1.01 bar, however is allowed to flow slowly to fill the bottle.
Now here question comes: sir, why you have neglected kinetic enegry associated with this?
We go slowly, here also I have use that slowly filling up. That means the velocity is neglected,
velocity is so small that the kinetic energy part is much small compare to this enthalpy
part. So that I am only using the enthalpy part as the total energy; kinetic energy is
Air from the atmosphere is allowed to flow slowly to fill the bottle. If no heat is transferred
to or from the air in the bottle that means bottle is totally insulated. What will its
temperature be when the pressure in the bottle reaches 1 bar? The pressure will be same as
that one that is the process will be automatically stop. If the bottle initially contains 0.03
meter cube of air at 400 millimeter f Hg and 25 degree Celsius where first bottle is evacuated.
Here you see the property relationship is like this; this is the value of cv what we
have told T plus 273. So when this relationship is given one student may not know what is
ideal gas? What is cp? What is cv? In ideal gas what is the value of cp? How
it does it relate? He will use this simply as a mathematical formula. That u is equal
to 0.718 into t plus 273 and he will leave simply this as a mathematical formula as far
as the property relationship is concerned. That pv is equal to 0.287, t is equal to 273,
where u is in kiloJoule per kg it is given. These types of equation where these numerical
constants are there are always specified like that. If you substitute this variable, that
is independent variable, in this unit we get the dependent variable in this unit. So this
type of equation are always specified; otherwise there is no meaning to these equations. So
pv is this, where p is in kilo Pascals and v is in meter cube per kg and t is in degree
Celsius. If you add these two we get the h, h is u plus pv that is h is equal to 1.005
into t plus 273.
Please write it. So, how to solve this problem? Just we apply this equation. What is the equation?
That t is equal to 144. Do you have written it? This equation . Now we will be solving;
that means, I write this equation.
mf minus mi into ha is equal to mf uf minus mi ui; very interesting result will come out;
that is, now ha is equal to uf. Therefore, ha is 1.005 as far as the property relationship
are given. Therefore we get 1.005Ta is equal to 0.718Tf. Ta is equal to what?
The air temperature is 25 degree Celsius - 298k. With this I get Tf is 144 degree Celsius.
It is very interesting thing that we know this for air, for any ideal gas, h is cp that
means 1.005 is cp and 0.718 is cv. That means the final temperature, if the bottle is evacuated
will be the ratio of specific heat times the temperature of the air which is greater than
1; the air temperature will be multiplied by the ratio of the specific heat, for air
we know this is 1.41. So, if the ambient temperature is 25 degree Celsius, the inside temperature
will be 1.41 into 298. So, it is the ratio or specific heat which will be multiplied
with the air temperature to get the final temperature of the evacuated is the thermal
rule, but it is true. When we deal with an ideal gas and the bottle is evacuated, but
if the bottle is not evacuated, then we will have to take care of mi and we will have to
find out the mf also. Then you can find out the value of u.
In the second case, the bottle initially contains 0.3 meter cube. This will give the value of
mi because initial pressure is given, initial temperature is given. How can you find out
mi?
Because we know pv is equal to mRT I have already told, so pivi is equal to m into 287
into T. This is Joule per kg. So that these things are to be in consistent unit Newton per meter square meter
cube per kg and Kelvin and m is in kg. This is meter cube not meter cube per kg; this
is meter cube this is kg. Now from this, we can find out the value of mi and we can apply
these equations to find out the Tf. This I am giving you as an exercise with the answer
for Tf. What is the answer for Tf? In that case, what is the problem? I will give you
the answer for this, just wait for sometime. Let me see the answer for it, whether I do
have or not, just wait. This is exercise for you; so, I must give
you the answer for it. Well, just wait. Okay, you do it. I will tell you the answer in the
next class. This is your home exercise. I have misplaced the answer somewhere; so, I
will tell you the answer in the next class.
Which one? Yes, the volume will remain constant. Volume will remain constant. The simple algebraic manipulations then I will
come to problem number four. Time is up? What is the time?
okay Do you have a class now?
Lunch. Just wait because they have some time for them. So, wait.
Problem four: an air line at 300 k and 0.5 MegaPascals as shown is connected to a turbine.
This is the airline, this is connected to a turbine , that exhausts to a closed initially
empty tank of 50 meter cube. Almost the similar problem, but we ask a different quantity.
The turbine operates to a tank pressure of 0.5 MegaPascals. Now what happens this air
line is at 0.5 MegaPascals, that means here the pressure is 0.5 MegaPascals and the turbine
operates to a tank; the turbine operates till the tank pressure is 0.5 MegaPascals, at which
point the temperature is 250 k; the temperature of the tank becomes 250 k that means, this
is the point when the process terminates automatically the same thing. Assuming the entire process
to be adiabatic, both the turbines and the tank along with the pipelines are adiabatic,
determine the turbine work. We have to find out what is the turbine work? Tell me which
one you will consider as a control volume place, any one of you?
Only turbine.
Turbine and tank together. We take turbine and tank together; what will be the equations?
Then in that case also there is an inflow only. It is a definite process over a finite
time.
The mass of air which has come into the enthalpy of air, that is the mass coming in, energy
coming in minus the work done W control volume will be equal to mf uf minus miui because
this is nothing but the change in internal energy of the control volume. If we look back
to our steady flow energy equations integrate it with respect to dt. Then we get these values
that finite quantity of mass times the specific stored energy. Finite quantity of work, finite
quantity of heat and the change in the internal energy. That means, it is written over a finite
time period for definite quantity of matter for a given process. Not in the rate basis.
Therefore here also you can write this that means, if I again draw the picture this is
the turbine, this is the tank. I am considering this as the control volume. Now here one point
is very important, now this is the energy entering to control volume there is no exit
to the control volume. This is the work done by the control volume that means, this is
this minus this must be equal to the increase in internal energy of the control volume,
but turbine part of the control volume does not undergo any change in the internal energy
because it is a steady state device. So, within the turbine things are steady the mass which
is coming in is going out that means energy which is coming in equal to going out in the
form of work and in the form of energy going to with this mass which is coming to this
tank. Therefore, as far as turbine part is considered,
there is no change in the internal energy. Internal energy is changing only in the tank
part which is in unsteady mode. Therefore, if we consider this as a control volume, the
change in the internal energy is contributed only by the tank. This is the only key point
of this problem; so that you can write this any query; you can ask me which will give
the W. Now what is the value in this case? This is an evacuated bottle or the tank is
empty. Therefore this is 0 , you can find out mf and this mf is equal to ma, because
there was no mass. So final mass is the mass which has come during that interval of time
through the control volume; that means, the entire control volume system. So, mf is equal
to ma. You can find out ma,; here ma does not cancel out because work is there. Otherwise,
we have to find out the work per unit mass that is all right, but here ma we can find
out because the final temperature and pressure of the tank is given and volume of the tank
is also given or not 50 meter cube. We can find out the mass by the same formula; for
the air pv is equal to m into RT. You know the mass; you know the internal energy at
the final point, because final temperature is given. When the mass pressure reaches 0.5
megaPascal, temperature is 250 k. We utilize this equation; so if we know uf and ha we
can find out by using this equation where t plus 273 this is given 300 k that means
t plus 273 is already given in that form 300 k. We can find out the value of this. Finally
work comes out to be 42.51 MegaJoule. You do it and you will see this is the final work
done by the turbine that is 42.51 MegaJoule. These are the four problems, where you can
have an idea how this concepts of first law applied to closed system and open systems
are utilized. Well, any query? This is the lunch time so, I think there will be no query.
Problems are simple. Okay then, thank you.