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[noise]
Welcome back. In this lecture we are going
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to summarize mass energy balance for open
system.
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So, we start with the conservation of mass
principle. For a simple control volume as
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depicted here the total or the conservation
of mass principle says that the net mass transfer
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to and from a control ah ah volume during
a interval time delta t is equal to the net
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change in the total mass within the control
volume during delta t or the time change time
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interval.
So, in a simpler form one can simply write
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down in expression the total mass entering
the control volume during a certain time [noise]
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interval minus total mass leaving the control
volume during delta t and that should be same
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as or that should be equal to net change in
mass within the control volume del delta t.
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Or, [vocalized-noise] in other word m in minus
m out is equal to delta m control volume,
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for in the case of rate expression you will
be considering dots here and this will be
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your dm by dt.
For a steady-flow process, the mass of the
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control volume will remain constant and hence
this expression this will be 0, this will
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turned out to be 0 for steady flow process
and thus you will [noise] have m in minus
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is equal to m out for a many inlets and outlets
this would be a generic expression for the
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case of steady flow process.
Now, in many systems of course, there are
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many inlets and outlets. So, one for simple
steady flow process an example would be let
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us say if you have two inlets you have m 1
and m 2 and if this outlet is 1, then based
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on simple your m in is equal to m out your
m in is m 1 plus m 2 and that should be same
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as m 3.
But, in many engineering devices such as nozzles,
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diffusers, turbine, compressors and pumps
they involve only a single stream. In such
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case you have m 1 is equal to m 2 or the m
1 dot is equal to m 2, and this you can represent
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in terms of density volumetric flow rate and
the cross sectional area as well ok.
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So, since we are talking about energy balance
also so and we know that for generic ex expression
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the net energy balance would be through heat
work as well as from us. So, me just talk
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about component which we have not discussed
earlier.
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Now, for a flow system and for a steady state
system say for a steady-flow system the fluid
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continuously flows through the inlet and exit
through the outlet and in order to maintain
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the control volume when the steady state there
will be work which is needed to continuously
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flow. Now, there will be work which we need
to push the certain mass here before the inlet
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and similarly, there will be work which will
have to be pushed out in order to maintain
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the continuity.
So, ah, let us consider again ah, so, this
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is the pressure of volume of the element of
the flow which need to be pushed in and this
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push this work which say is work of or flow
work or flow energy. So, be this is before
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the entering and it is after the entering
and so, consider this that this pushing this
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work is being done by some imaginary piston,
ok, there is a certain force which is being
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applied and this is basically the steady state
there is no acceleration and the constant
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force is being applied and this mass gets
in.
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So, the question is what is basically the
work for such a case, ok. So, me just take
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a simpler view of it.
So, this is again the force and this this
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element here have to be pushed in by a certain
distance L and this force in absence of acceleration
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is equal to the force applied on the piston.
So, this would be equal to the opposite force
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here assuming this to be a strow slow process
this P into A is going to be the force, ok.
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Now, this work would be the force multiplied
by that is an L and that would be PAL, that
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will be P into volume. Now, in terms of a
kilo joules this will be P into specific volume.
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So, this will be the work of flow of course,
the PV value here and PV value in the outlet
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may be different, because that will depend
also on the pressure of the outlet and the
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specific volume of the outlet. So, that means,
there is a some PV work here which is associated
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with it and there is also PV associated with
the outlet.
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Now, we can summarize this concept in terms
of comparing both the case of non flowing
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and a flowing fluid for the non flowing fluid
you have energy of the fluid as internal energy
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or energy or the fluid contains internal energy
kinetic energy potential energy as shown here.
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So, this internal energy per unit mass the
kinetic energy and the potential energy; for
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the case of a flowing fluid in addition to
that internal energy kinetic energy and the
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potential energy it also poses a flow energy
because it moves inside and certainly outside.
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So, it contains that energy and we use that
energy as a part of the energy of the flowing
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fluid there is PV or in other word we can
also make use of the basic definition this
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u plus PV comes in the flowing case and this
can be written as simply h ok.
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So, again we go back to the generic energy
balance that is e in minus e out is equal
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to e e dE system by dt and we are considering
the rate expression considering is a flow
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process. Now, in this case the net energy
transfer is heat work and due to heat work
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and mass ok; for the case of a steady state
of course, the e in is equal to e out , so,
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if we know the directions of various different
interactions we can write this expression
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where Q in plus Q out this part of E in plus
whatever the contribution due to the flow
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ah, that is, mass that could be your enthalpy
plus kinetic energy plus g z potential energy
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and that should be equal to Q out W out plus
the energy contained by the flowing mass.
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So, that would be m h.
Now, this is summation is for each inlet and
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summation here is for each outlet. So, this
is for the case where we know the direction
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here.
We can write also in terms of very generic
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expression that would be Q net in this will
be Q net in and Q net W net out this is equal
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to the summation for the case of outlet, that
means, it is summation of m dot theta minus
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summation of m dot theta this is for in and
this will be out ok. The change in the energy
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due to the mass flowing mass for a single
stream of course, you are going to consider
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only h 2 minus h 1, most of the time we will
be ignoring this velocity and kinetic energy
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kinetic energy and potential energy and only
for some cases we have to use that.
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So, for the case of where we ignore this two
terms will have q minus w in terms of per
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unit mass q where if you divide this q here
is going to be small q is q dot divided by
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m dot and thus you can write a very simpler
case here which is q minus w is equal to h
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2 minus h 1,. So, a very simple expressions
for energy balances, you can apply this for
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many devi defice devices. I am not going to
go through in detail for all the devices and
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I expect that some of the exercises you have
done that in earlier, but we will just recap
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how to do this for few cases.
So, for example, nozzles and diffusers are
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typically used to change the fluid velocity
and kinetic energy where is there is no heat
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interaction and the work there the body does
not change. So, that is your Q and W's are
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zeros, delta p e are going to be also considered
to be 0. So, you can apply this basic definition
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and you get E in minus is equal to E out or
in the other word m h 1 plus V 1, V 2, V 1
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square by 2 is equal to m h 2 plus V 2 square
by 2. Now, you can do this analysis knowing
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that inlet and outlet conditions and the ones
un unknown variables you can find out by using
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simple algebra ok.
So, the this is the final expression of that
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and you can do the analysis subsequently.
Now, the similar kind of exercises you can
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do for turbine which is a work producing device
for compressor fans and pumps again where
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the work is needed in order to change the
pressures. So, you can do this exercises again
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based on very simple logics ok.
For example, this is the case for compressor
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where air is being compressed from 100 kilopascal
to 600 kilo Pascal whereas, the flow rate
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of mass is given here 0.02 kg per second and
what we need to find out is the work which
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is required the rate of work which is required
and in this process the heat also lost.
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So, you can start with again the basic definition
of energy balance here , where is it this
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is a steady state then you can consider this
and then you can write this E m i is equal
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to E out. Now, knowing the direction here
already we can simply write W in plus the
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flow contribution m h 1, this is going to
be equal to Q out plus m h 2 ok. So, Q out
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is here this part plus m h 2 is due to this.
So, you can rewrite this expression in this
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form.
Now, this you can find it from the tables
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or you can use C p dT this if you know the
expression in terms of C p in terms of temperature
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and Q out also of course, we know that. Now,
this kind of analysis is of course, applicable
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to other cases throttling valves, mixing,
heat exchanger and so forth. So, I am not
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again going to in details of that you can
apply and revisit some of the example to hone
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your skills again.
Now, I will just talk about the case where
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we have unsteady flow process which is also
very common. These are the processes where
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the changes in the within the control volume
with time such processes are called a transient
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flow process or as I said unsteady flow, ok.
The examples are discharge of the fluid from
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pressurized vessel, inflating tires or the
balloon. So, for example, here in this case
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during this process of the filling of the
control volume this piston can also move ok.
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So, let me just quickly explain how to work
with this kind of system. So, most steady
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flow processes can be represented by uniform
flow process where we assume that the fluid
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flow at any inlet or exit is uniform and steady
and thus the fluid properties do not change
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with time or position over the cross section
of an int inlet or exit and if they do they
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are averaged and treated as constant for the
entire process.
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So, for such a case we can simply write mass
balance as m in minus m out is equal to delta
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m system, where delta m system is [noise]
m final minus m initial sometimes also we
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write m i inlet and exit is equal to m 2 minus
m 1 of the control volume ok. So, this is
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our mass balance.
What about the energy balance? so, we will
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write again here energy balance E in minus
E out is equal to delta E system. Now, delta
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E system is not going to be 0, in this case
delta E system is m 2 e 2 minus m 1 e 1, where
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this contains only the internal energy, potential
energy, kinetic energy as shown here whereas,
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E in is Q in plus Q w in plus summation of
m theta where theta contains h because it
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is a flowing fluid and k e plus p e, whereas
for the case of outlet is Q out plus W out
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plus summation m theta for the outlet.
Now, assuming p e and k e changes are negligible,
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you can write down this generic expression
in this form , ok, q where q is net heat minus
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W is net work out and this equal to summation
m h m h summation, this equal to the contribution
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due to the changes in the enthalpy from the
outlet and inlet and as well as the changes
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due to the internal energy of the system.
Now, the question is how do you apply this?
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So, we will take up one example and this we
will end the lecture with this example. This
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is an example where you have an insulated
8 meter cube rigid tank which contains air
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at 600 kilo Pascal, 400 at 4 400 Kelvin and
the wall is connected to the tank which is
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now opened. Air is allowed to escape until
the pressure inside the draw inside the drops
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to 200 kilo Pascal.
The air temperature during the process is
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maintained constant. So, this is maintained
and by how it is maintained by providing this
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electrical resistance. So, there is a work
electrical work is applied to the system.
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What we need to find is basically the electrical
energy supplied to the air during this process
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ok. During the process, where the pressure
is changed from 600 to 200 Kelvin, keeping
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the temperature constant.
So, the first thing is since the valve is
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opened certain mass has escaped. So, we will
be applying the mass energy mass [noise] balance
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first followed by energy balance. So, m in
or m in minus m exit is equal to m 2 minus
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m 1 of control volume. Now, this there is
no inlet, so, this m exit is nothing, but
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m 1 minus m 2, of control volume.
Now, E in minus E out is delta E system ok.
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What is E in W e in here and what is E out
because only the mass flow then we can simply
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write m e h e ignoring the kinetic energy
and potential energy. So, and as well as since
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it is insulated, so, q delta e delta K E and
delta P E are going to be 0, and this is m
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2 u 2 minus m 1 u 1.
Now, V 1 is constant ok. P 1 we know P 2 we
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know, now the thing is it is air ok.
So, we can consider air to be idle gas and
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thus we can make use of the equation idle
gas equation to obtain m 1 which will be your
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ma molar mass P 1 V 1 by RT 1 and this itself
is 0.287 kilo Pascal meter cube per kg Kelvin
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and if you plug in the P 1 and V 1, P 1 is
600 kilo Pascal, T 1 is and T 1 and T 2 is
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constant, V 1 and V 2 is also constant is
V 41.81 kg ok.
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Similarly, you can get m 2 M P 2 V 2 by RT
2 you get 13.94 kg. Thus, you can obtain m
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e which is m 1 minus m 2 27.87 kg. So, now
knowing this m e we can now plug in in this
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equation ok. So, so, you can plug in this
information m e and what about h e?.
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So, you can plug in 2; 2 will become W in
ah and you can rearrange here this is m e
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h e plus m 2 u 2 minus m 1 u 1 ok. The reason
where you have taken again this is because
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this is the delta E of the system [noise]
which is non flowing is only of the control
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volume and this u since this idle gas is your
only depends on the temperature which is 400
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Kelvin you can use the table given in any
book and this comes out to be 286.16 kilojoules.
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Similarly, h e which h e is basically the
enthalpy , which can be approximate because
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this can be enthalpy of the air at 400 Kelvin,
ah ah ok, because enthalpy is the temperatures
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is a fixed over there and this from the table
also we can get there. So, with this information
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and with the value m e which we have calculated
we know m 2 m 1 as well this value W in comes
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out to be 3200 kilojoules ok.
So, with this we complete this lecture and
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we have just covered the energy balance for
the open system. [noise] So, we will continue
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our reviewing process of basic engineering
thermodynamics first. So, with that I will
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end this lecture and I will see you in the
next lecture.