Good morning. Last class we discussed about
what is meant by gas power cycles and air standard cycles? And why it is necessary to
study the air standard cycles? I just repeat a few words again and air standard cycle is
a reversible cycle that all processes are reversible. Working fluid is air and where,
heat is added from outside source is not like that, that it is generated internally though
heat is not generated in fact this is a wrong word but the internal energy is generated
inside which boost up the temperature of the working fluid.
The temperature of the working fluid increases because of a heat addition from the outside
source why I am telling this because in much practical gas power plant the temperature
is generated inside because of the generation of internal energy so the if working fluid
gets heated. When you study an ideal cycles that is an air standard cycle we consider
heat is coming from outside and heat is rejected also to the an outside sink source and sink
concept and throughout the cycling processes air is the fluid which behaves as an ideal
gas with constant specification. Today we will first study about the Carnot
cycle We have studied Carnot cycle in few last occasions,
so we know what is a Carnot cycle. If you plot the PV diagram for Carnot cycles we will
see the Carnot cycle consists of two isotherms and two isentropic so these are the two isentropic
line air s is constant s is constant and this is e is T constant so this line is constant
temperature line so these two lines is T is constant, T is constant so it consist of two
isothermal and two isentropic. We can start from any point as 1 it goes in
this direction and isentropic expansion 2 where the specific volume is increased per
unit mass I am drawing the cycle. This we have drawn earlier also in some other occasions
then there is an isothermal compression that is 2 to 3 process then again 3 to 4 is an
isentropic compression this is isentropic process and four to one is isothermal expansion.
These three processes are there which constitute the Carnot cycles if we draw this diagram
first and foremost task is to draw the diagram in PV and TS plane and we will have to develop
this habit. Now 1 to 2, 2 to 3 we start with that 1, 2 is an isentropic process so an isentropic
process obviously will be a line like this so this is 1, 2.
Similarly 2, 3 is isothermal process so this will be I am drawing by a black so this 2,
3. Again 3, 4 is an isentropic process, this is 3, 4 and 4, 1 is again the isothermal process.
Let me stop drawing by different pen it will take unnecessary time therefore we see that
Before calculating the cycle efficiency net work output there are many things which may
be up to calculate to calculated. Net work output, cycle efficiency what is the output?
What is the power output in process 1, 2? What is the gross we added many things we
can calculate. Best process is to do it that you make a table
like that process where you write the different process 1, 2 that means this process 2, 3
there are four process 3, 4 and you go on computing this delta u change in internal
energy of that process work and heat quantity. One thing is very important in analyzing this
type of air standard cycle two information are very important one information is the
working fluid which is ideal gas which is ideal gas air? Air is the working fluid which
behaviors as an ideal gas ideal gas is a broad thing air is the working fluid. Next important
thing which is very much needed is this what type of process is performed by the gas? That
means whether it is a Non flow process or flow process? The answer to this is non flow
process that means process performed in a closed system this two are extremely important
while evaluating the parameters like work done heat added all these things non flow
process. Though change in internal energy is not required
it is a point function because it is difference of the property as you know for an ideal gas
it is CV times delta T but for WNQ it is much important whether the process is a non flow
or flow process. Accordingly things will change afterwards we will study another air standard
cycles Brayton cycles where the processes are flow processed study flow process. It
is extremely important to know what type of process? it is a non flow process being performed
by the working fluid air during the cyclic operations. Non flow process means process
executed by a closed system closed system process.
Now the process 1, 2 you start from process 1, 2 is an isentropic process so what happens
reversible adiabatic process. First of all you see whether any of these three columns
there is a special case or not so that I can put 0 or something like that so for this process
is there any of these three columns would be 0? This one is 0 so better you do that
first. What is the technique? this is the way if you do it will be very clear I am telling
books may not tell that then what you do you write the first law of thermodynamics? Which
is Q which is valid for the closed system in this fashion del u plus W. Next is that
first you search whether there is 0 for these two column next is the del u always shutting
your eyes you can just write for any process without seeing any cycle that if process is
1, 2 then its change is internal energy is T2-T1 where it comes from the concept that
ideal gas. In this column you can write without seeing
even the shape of the cycle but from the particular process you can write here Q is 0 so when
Q is 0 work done is minus delta e here I am writing only work done so plus minus sign
will indicate whether work is done or work is being done on the process. W is minus delta
u that means this will be simply Cv, T1-T2. Next comes to the process 2, 3 one thing is
that in this process T1 is equal to T4 one has to write that T1 is equal to T4 let this
is equal to TA. T1 is T4 that means this TA line isothermal similarly T2 is equal to T3
is equal TB let this line is TB lines that means there are two isotherms TA and TB and
two isentropic lines isentropes this is the Carnot cycle here also this is TA this is
TB. Next part is 2, 3 you tell me 2, 3 is there
any column where I can write 0? Which column? Delta u the first column. You have studied
thermodynamics well then if delta u is zero Q is equal to W if any of the two quantity
is found that means we can write for both this how to find out? Which one is simple
to find? How to find out Q?. T into delta h T in to delta h. Why T into delta? It is
Q where from you get you have to find out from the basic why Q into delta it is T is
a reversible process so what so what is the equation which gives you the heat transfer
in this process 2, 3 is Q into delta h how do you know?
You have to find out W, W is what? this is a non flow process it is pdv if it is non
flow process per unit mass that’s why small v it could not be pdv in that case work done
is equal to something else we have to go for the study flow energy equations very careful
at this stage do not do a blunder at this stage which you can calculate from equations.
pdv is the work done reversible non flow process only work done is pdv there is no other work
expect the displacement mode over. If you evaluate pdv you see that pv is equal to what
RT2 or RT3 because this is same pv is equal RT2 or RT3 that means during this process
temperature remains constant 2, 3 which is TB. I can write RTB T2, T3,TB RTB pv RTB by
v that means this will be ln v3 by v2 so this the work done.
Therefore I write RTBln v3 by v2 since delta u is 0 from the first law Q is equal to W
so Q is RTBln v3 by v2. Now 3, 4 one thing you see before that this is an isothermal
process where volume is reduced that is a compression process this was explained physically
earlier so in this process both work and heat transfer is there. If you consider this term
work done v3 is less than v2 that means this is negative that means work is done on the
fluid if you see this term RTBln v3 by v2 this is also negative the same term that means
heat is being rejected by the fluid that means isothermal compression process. Where heat
is being rejected by the fluid and fluid is being compressed.
What happens? Just you imagine in a piston cylinder arrangement this was explained earlier
if there is an air if you want to compress the air at a constant temperature. What will
happen you want to compress the air? That means you give the work so when the temperature
is constant for the air the temperature has to be constant it behaves as an ideal gas
and more over it is isothermal process that means if temperature is constant means the
internal energy will ah remain constant that means you want to make an isothermal compression.
Isothermal means temperature remains same and temperature remain same means for an ideal
gas internal energy remains same that means when you compress if the work of compression
which is being added to the working fluid has to come out in the form of heat so that
heat has to be rejected so that the temperature remain same. This is the way one can understand
also one can understand without these heat and work that if I compressive it its temperature
will increase to make the temperature constant I will have to take heat away so that it is
cooled and temperature is maintained constant. Therefore this is a process where heat is
being rejected and work is being done that means this is the process where heat rejection
takes place. Let this is QR and work is being done on this process. Work done W I am just
writing one W but in this process there is an work but there is no heat there is an work
W12 this is W23 this QR is Q23 I am writing as QR because heat is rejected.
Heat transfer processes are only 1, 4, 4, 1and 2, 3 because these two processes are
reversible adiabatic process there will be no heat transfer. Whenever you write these
things in the table you complete one row different columns you fill then immediately physically
understand that what the for example this process there is no heat transfer because
it is reversible adiabatic is an isentropic process. There is a work done which is positive
work is coming out and there is a change in internal energy. Now I come to the process
three 4, 3 it is synonymous to 1, 2 in respect that this is 0. What is this Cv T4 minus T3?
What is this Cv T3 minus T4 negative of this because Q0 W is minus delta u. You see there
is T4 is more than T3 or T3 is more than T4 here you cannot see it here T4 is more than
T3. Therefore there is an increase in internal energy but there is a negative work that means
work is being W3, 4 given to the system there is no heat transfer.
4, 1 process again 4, 1 process which will be 0 it is the synonymous to the 2, 3 process
so that means this is delta u 0. W is RTA in this case it will be RTAln final volume
means v1 divided by v4 similar is the thing that for the work because delta u 0 RTAln
v1 minus v4. If you recognize this you see there is no internal energy change just the
reverse get an expansion so heat has to be added because it is expanded means its temperature
is reduced so cost and temperature means it has to be added and if temperature remains
constant internal energy remains constant that means the work which is being developed
by the expansion has to be nullified by the amount of heat which has to be supplied so
that summation of these two is 0 earlier expansion there is no change in internal energy.
Therefore v1, v4, v1 is more than v4 so therefore this is positive work is done so in this case
work is done W4, 1 and the heat quantity you see similar way v1 is more than v4 that means
this is the quantity so that heat is being added to this system Qa. Qa is the heat added
which is the isothermal heat addition 4 1 and Qb is the heat rejected that means this
is Qa heat is added this was discuss earlier also in Rankine cycle the same cycles is QA
and this is the heat rejections QR and OR quantities are involved in all the four processes.
In a closed system work quantity will become zero only if there is in process with constant
volume dv is 0 isobaric process constant volume processes otherwise all processes work transfer
is there. I think this is clear this table after you complete this table you check the
first law thermodynamics for cyclic process. What is that sigma of delta u has to be 0
and it is 0 you see T2 is T3 and T1 is T4, T2 is T3 and T1 is T4 so they cancelled up
it become 0. If you sum it up 0 because T2 is T3, T1 is T4, T2 is T3, T1 is T4.
Now delta W, if delta u is 0 this cyclic process delta W must have must be equal to delta Q
delta W and delta Q are same. If you do that T2 is T3, T1 is T4 that means these two terms
cancels out. Therefore this plus this is equal to this plus this so therefore the sum of
this column is sum of this column. If one can do these things but he can calculate anything
whatever is asked for but what is the most important thing that is asked for is thermal
efficiency? Now thermal efficiency it is always better
without going for calculating this sigma W until and unless it is asked for what is the
net power output of the cycle or work transfer for a particular process then you will cote
this result. If only the thermal efficiency asked we should write from this what is thermal
efficiency? Heat1 minus heat rejected by heat added that means heat added by heat rejected,
heat divided by heat added. That means one minus heat rejected by heat added because
here heat rejected term is already negative. Which is the heat rejected term this one and
what is the heat rejected term is this one which is already negative so that I can write
1 plus RTBln v3 by v2 divided RTAln v1 by v4. With a minus sign I can write eta is 1
minus R, R cancels TB by TA into ln v2 by v3 so that make it positive by ln v1 by v4.
Immediately my commonsense dictates that since the efficiency of a Carnot cycle is 1 minus
TB by TA so these two things automatically will cancel how can you cancel v2 by v3 as
v1 by v4? You have to write now any manipulations you will have to do now with the process calculations.
For this isentropic expansion process 1, 2 what can we write? We can write this if we
before writing this I just tell you that these equations will be very important for isentropic
process. pv to the power gamma is equal to constant
that we know for isentropic process performed by an ideal gas pv to the power gamma is constant
where gamma is cp by cv it is the very important relation this will never be giving in examination.
At the same time we know pv is equal RT, pv and TR related for an ideal gas like that
at R is a constant characteristic gas constant this has been told so if you want to know
pT and vT relationship we will have to substitute one either p or v which is not required from
this equation. If I want to know the vT relationship p is
substituted that means RT by v into v to the power gamma is constant or i can write T by
v to the power one minus gamma is constant or one can write v by T to the power 1 by
1 minus gamma is constant. We can write v by T to the power 1 by 1 minus gamma is constant.
If I use this relation v by T to the power 1 by 1 minus gamma constant then it will be
easier for me because here I want to relate this v. That means I can write v1 by v2 is T1 by T2
to the power one by one minus gamma. v is directly proportional to T to the power one
by one minus gamma. You will have to work out this this you do not have to keep in memory.
Similarly I can write for this process v4 by v3 is equal to T4 by T3 to the power 1
by 1 minus gamma. Now T1 is equals to T4 this is an isothermal process. Similarly T2 equals
to T3 this is an isothermal process that means v1 by v2o is v4 by v3 that means v1 by v2,
v4 by v3 which means that v1 by v4 is equal to v2 by v3 that means which I want to prove
that this equals to this so these two cancels out and I get eta is 1 minus TB by TA which
is the efficiency of a Carnot cycle that means 1 minus the temperature of heat rejection
divided by the temperature of heat addition. Another interesting feature that when you
draw the pv diagram what does this area of the closed loop represent? This represents
the work done. Total heat added net heat added net heat added so the area these two are they
equal? yes sir yes this because delta u is 0 and the basic first law of thermodynamics
is that algebraic sum of the work is equal to algebraic sum of the heat that is the conservation
of energy because the system comes to the initial stage where that there is no change
in the internal energy this is the Carnot cycle.
Now I will come to Otto cycle. What is an Otto cycle? Next is an Otto cycle it is named by Carnot
as we have as Sadi Carnot and French engineer Otto cycle is German Otto cycle. Otto cycle
is another air standard cycle now I will explain afterwards why a Carnot cycle is not used
for any gas power plant but air standard cycle Otto cycle is a theoretical cycle of gas power
plant like the Rankin cycle was a theoretical cycle for a steam power plant. Similarly Otto
cycle is followed for an automobile engineering automotive engines or automobile engines using
petrol as the working fluid that is the petrol engine. You simply call that is a petrol engine
and it is known as reciprocating IC engine. Therefore without going in to any details
of reciprocating IC engines reciprocating spark ignition IC engine this is a terminology.
We like to know that what of the operations that a reciprocating IC engine performs spark
ignition engine performs so that we can understand the cycle just like the way we appreciated
the different components of a steam power plant and what are the operation?. For that
reason only we just go probably you will know this thing from your popular knowledge that
a reciprocating internal combustion engine which work on spark ignition process this
is like this. Let me draw this first the figure then I will
explain this is a cylinder piston arrangement and a working fluid performance a non flow
process. Let me explain this thing this performance a reciprocating motion this is the center
line and this is a connecting rod and crank mechanism. Now there are two extreme positions
between which it now thing is like this. This is a piston cylinder arrangement connected
with two valves I will explain this afterwards. This is the piston which makes a reciprocating
motion within this cylinder and this reciprocating motion is being generated by the use of a
rotary motion through a crank and connecting rod mechanism. This is a connecting rod I
am not writing everything this is the crank so I can write. This is the mechanical engineers
will read connect this is one of the classical mechanism by which the rotary to reciprocating
or a reciprocating rotary conversion may be made.
It reciprocates within the cylinder between two limits that means the inner most point
and inner most plane to which it can go without heating the head of the cylinder is known
as inner dead center position IDC or TDC top dead. So top dead center position the work
top is justified when it is a vertical engine that mean it is the top one but IDC goes for
both the case horizontal and vertical that means the inner dead center these two are
known as dead center inner dead center or top dead center in case of a vertical configuration.
Similarly this is ODC Outer Dead Center or Bottom Dead Center if it is a vertical engine
so this is the bottom dead center that means between the inner most and the outer most
that IDC and ODC the piston makes a reciprocating motion and this length is known as stroke
not this length the volume actually this one movement from one dead center to another center
is one stroke. When the piston moves from one top IDC or
Top Dead Center one dead center to other dead center either in this direction piston makes
one stroke movement that is known as stroke. Now what happens I am telling that, if you
start from this piston is at the piston is continuously moving like this in a continuous
operation and we get the work done from here. Now what happens I start from one particular
position the piston is at the IDC or top TDC in that means what happens this valve is open
this valve is made in such a way that this is opened in this direction. This valve is
open and this valve is known inlet valve this is made to open by a mechanism known as cam
which is attached to a shaft known as camshaft. Valve is made to open it take some time for
a complete opening of the valve. So valve starts opening before the piston reaches the
top dead center or inner dead center position that is true.
Before it reaches the inner dead center position valve starts opening but we consider as a
when the piston has reached is undead center position valve is wide open. Then what happens
when the piston moves down the gas or air. If you consider an air standard cycle it will
be air but all usually it is a gas in practice expands which was almost at an atmospheric
pressure. That will come back when you will complete the cycle. An atmospheric pressure
gas expands because of the increase in volume due to the descending motion of the piston
what happens the pressure inside when the piston descending and come here so this pressure
inside will be less than the atmospheric pressure because some gases at atmospheric pressure
is expanded so its volume is increased so pressure is decreased less than the atmosphere
so there is a suction pressure inside the cylinder.
The fresh air from outside this is connected by a pipe line known as intake manifold this
is intake manifold these are the usually terminologies so air comes through this intake manifold
to this cylinder this is very simple. Most idealized case I am telling there are many
things they are filter these that which will be reading in IC engine class so air comes
in. When the piston descends down from the IDC due to the suction created within the
cylinder the air comes in till the piston goes to the outer dead center position. Therefore
this is the condition where you see the pressure goes below and again must comes so pressure
is getting ah really equalized try to understand physically this is very important that when
the piston goes down the suction is created immediately air comes piston goes down suction
is created expansion and air comes. Air continuously comes as the piston goes
down this is approximated in theory to be a process of induction of air it is a process
of induction of air at constant pressure that means it as expansion process from some volume
which corresponds to the volume of this that is when the piston was IDC to a volume at
constant pressure why at constant pressure? Practically it may not be at constant pressure
there will be a change in pressure because whenever there is a suction air is coming
that means fall in pressure is being utilized by the addition of the fresh mass that is
why this is being concept as a be constant pressure expansion and in theory it is like
this 0-1. In practice it is not like that in practice
the process is like this initially the pressure goes down initially the suction pressure fall
in suction pressure is much more then the induction of the air is in such a way of the
pressure is not being utilized and then slowly slowly when the air masses increase inside
this the pressure takes up to the atmospheric pressure. It ends up to the atmospheric pressure
but we we we approximate this in an ideal cycle or air standard cycle as if a constant
pressure 0 one 0 is the starting point here that i will you will standard why I start
with 0 not with one 0 one as the expansion process. v1 corresponds to the volume of the
cylinder volume of the gas inside the cylinder when the piston is at the outer dead center
position. Now you consider a situation where the valve
is closed it is not closing much before there but I am not going in to this practical things
there are certain time for closing and opening the valves which are known as valve timing
which is being done to get the maximum possible power needed that means heat the iron when
it is what type of thing that means you have to open the valve. When valve opening is much
more utilized you have to close the valve when the valve closing is much more utilized
from view point of the operation of the IC engines that i am not going to describe.
In ideal case let us consider as a as soon as the piston reaches here instantaneously
the valve is closed this valve is closed throughout so therefore now the two valves is closed
these valves is closed throughout so far we have discussed these two valves are closed
that means at this position piston has inducted somewhere that means the cylinder is full
of air whose volume is v11. Now piston when ascends up what happens is
air is getting compressed it is a spark ignition engine this is not air fuel vapor. This is
not the air this was diesel engine air is being sucked. It sucks an air fuel mixture
which is known as charge which mean that the mixture or air fuel vapor which is being inducted
to the engine during this descending motion of the piston.
This is one stroke as I have told this trouble is known as suction stroke that before explaining
this in this diagram pv diagram I first better explain the what happens actually in this
engine when it reaches here after inducting this charge the mixture of air and fuel vapor.
When it is full of air and fuel vapor this valve gets closed and this valve is also closed
then what happen? Piston goes on ascending and what happen the fixed amount of mass which
is the air and mixture of air and fuel vapor known as charge which is getting compressed.
When the piston comes to this inner dead center position then some compressed air and fuel
vapor mixture is there within this portion of the cylinder.
Whose pressure and temperature is high this is because the compression the piston is insulated
during this but no heat is going out or is not allowed to go out so that the charge attains
some high pressure and temperature because of the compression by the upward movement
of the piston or inward movement of the depending upon the cylinder piston configuration. Then
what happens at these pressure and temperature this charge mixture of air and fuel vapor
cannot ignites spontaneously this stroke length and the design of the cylinder the cylinder
it is designed in such a way that this compression cannot make the temperature as high as required
for this spontaneous ignition or burning of fuel vapor with air. Even in the absence of
any ignition ignite of source even in the absence of any initiation.
Therefore initiation is requiring by a igniter this is an igniter so which ignites the fuel
vapor and air mixture for which the burning takes place. Now this burning is an is an
exothermic reaction the fuel is actually an hydrocarbon compound which rapidly reacts
with air which is a rapid exothermic reaction oxidation reaction because of which there
is a huge pressure and temperature created within this discharge. This thing will be
made very clear when I will take the reactive thermodynamics afterwards. This is an exothermic
reaction though loosely many people tell that which generates heat or liberates heat that
is the wrong terminology which I will make clear afterwards. So exothermic reaction does
not mean that liberates heat because the question of heat does not come here.
This is the closed system which is insulated so that no question of heat is there because
heat is then energy which is in transit because of a temperature different so therefore this
is a exothermic reaction which makes the products of combustion to attain a very high temperature
and pressure because of this burning that means the rapid exothermic oxidation reaction
of fuel vapor and air. Therefore a high pressure and temperature
is created which gives a high thermal energy this is at form of internal energy form the
energy conservation point of view this comes from the chemical energy of the fuel vapor
all these things will be made clear in the subsequent chapter which I will describe in
this chapter. That is the thermodynamics of reactive system so these high temperature
and pressure of the gas pushes the piston downwards. Therefore the third stroke the
second stroke is the compression stroke. Third stroke while the piston goes downward or goes
outwards expanding this high temperature and pressure mass of the products of combustion
products of combustion this the pushes the piston and the piston moves downward this
is known as power stroke this stroke we get the power. When the piston comes at the bottom
dead center position or outer dead center position then the temperature and pressure
both of these charge which was here after burning is reduced pressure is reduced. The
volume is increased and temperature is reduced because of the expansion and the work done.
What happens when it is at this position immediately this valve opens and we consider an instantaneous
opening in the ideal cycle but it gradually opens it is starts opening even before the
piston reaches this outer dead center position or top dead center position these details
I am not teaching here. We consider theoretically or ideally as if the valve opens when the
piston reaches this ODC that outer dead center or bottom dead center position then what happens
some gases are expelled through this manifold or through this duct that means the exhaust
duct which is known as exhaust manifold to atmosphere. Then what happen the piston again
ascends upward during this upward movement of piston bodily pushes the remaining charge
or relatively high temperature and pressure as compared to the atmospheric temperature
and pressure through this port through this exhaust manifold to the air.
When piston goes to this portion the pressure air is almost atmospheric pressure there is
an equalization of pressure. In fact when the valve is open immediately some charge
is expelled and obviously the pressure equalization takes place and when the piston gradually
moves up because of the mass going out what happens the pressure inhales the atmospheric
pressure. We can consider this as an atmospheric constant pressure at atmospheric pressure
discharge so therefore when the piston again comes to this a inner dead center position
or top dead center position. It comes to its initial condition that means this volume which
is full of charge that is burned products of combustion then again the piston moves
downward to start the another cycle. This way the cycle of events repeat so therefore
is see one cycle composites of four strokes that means four movements of the reciprocating
motions of the piston from the IDC to ODC and then again the piston when starts descending
downward what happens it inducts the fresh charge that is air mixture of air and fuel
vapor along with that this charge products of combustion also is there which gets expanded
and that causes some other problems which I will not describe here. Now this is the
operation which takes place the primary operations which take place in an petrol engine in an
internal combustion engine working with petrol this is known as the petrol engine or spark
ignition engine. This is because the ignition is initiated by virtue of a spark here.
How do we simulate in ideal cycle?. Ideal cycle what we consider we do not consider
air and fuel vapor mixture because when you consider air and fuel vapor mixture you see
the first two strokes that is downward and upward stroke the system is the air and fuel
vapor mixture but after burning the system constituents change from that of air and fuel
vapor to the products of combustion so the system composition changes for the remaining
two strokes that means the power stroke, downward stroke and the upward stroke or the inward
stroke that is the exhaust stroke in during these strokes two strokes of the cycle the
working fluid is the products of combustion. Therefore this does not truly define a thermodynamic
cycle so an ideal thermodynamic cycle define like this, as if air is being sucked by the
piston and it is being sucked during the downward or outward stroke during which the pressure
remains constant as I explain in the last the class this is because as it goes down
the pressure inside falls below the atmospheric pressure. At the same time the air must comes
in so these two things equalize the pressure to make it constant at atmospheric pressure
and that is being represented by zero one line but in actually case the process is like
this by dotted line. Where initially the pressure goes below the atmospheric pressure but the
latter part the suction stroke the pressure increases to atmospheric pressure but is simulate
this as a constant pressure induction of air. When the air is pushed in the compression
stroke from the outer dead to inner dead center position and when the piston is insulated
the ideal process should be isentropic because when the piston is insulated no heat is allowed
to flow out side because of the rise in temperature due to compression. This is adiabatic and
moreover if you consider the process to be free from any or fluid friction less absence
of viscous fluid friction or solid friction then we consider the process to be isentropic.
Therefore this part of the curve represents the isentropic s is equal to constant so this
is in pv diagram I am drawing the ideal cycle. This point2 represents the state when the
piston reaches the top dead center position then when the burning takes place this is
very important. When the igniter burns this mixture air and fuel vapor are so thoroughly
mixed. Before it is being admitted to the chamber which is done in a device known as
carburetor and the fuel which is used is petrol. It is so highly volatile that it is easily
made vapor and can be thoroughly mixed with air.
The physical contact is very good and immediately after ignition it releases the heat which
is very colloquial term. It makes the combustion complete and the pressure and temperature
is instantaneously raised. This type of almost instantaneous combustion which takes place
in a petrol engine which is known as a premixed combustion a fuel vapor and air is mixed before
and thoroughly and immediately after ignition almost instantaneously the mixture burns.
Sometimes colloquial heat is almost instantaneously heat is being generated but heat is not generated
but it is immediately burned and raises the pressure and temperature so this process is
so instantaneous. We consider this as a constant volume process
in ideal cycle as if with the limitations that during this process of completing the
combustion and raising the temperature and pressure of the charge piston gets a little
ah downward movement the time for which the downward moment of the piston is very little
so that the process is almost complete within a constant volume and moreover we consider
in ideal cycle where there is no fuel vapor mix there is air that means this particular
burning almost instantaneous burning process is simulated as if piston is fixed at the
out inner dead center position and air is heated some from outsource air is heated to
raise that temperature at constant volume. This is the reason for which this burning
process in a petrol engine is simulated by an heating process of the air which is the
cyclic fluid at a constant volume where heat is being added from outside QA this part is
clear to everybody this is the most important part in this cycle. Obviously the power stroke
which is the descending stroke or outward stroke from IDC or TDC to ODC or BDC which
has to be also an isentropic process 3, 4 ideal case this should be an isentropic process
but it is not the isentropic process. This is because the temperature is so high after
the burning if you allow the piston to move downward temperature ah decreases relatively
less during this descending motion so that the piston material the surface material cannot
withstand that temperature. Effective cooling arrangement is made by passing
water through water jacket surrounding this cylinder tube surface from outside this is
the actual case but when ideally we will consider this system should not reject any heat to
the outside fluid so that we can get the maximum power. It is simulated by an ideal process
at s is equal to an isentropic process then what happens again the important part is here
that at when it reach is at 4 similar to this outer dead inner dead center position. When
it reaches that ODC that means that straight port immediately the valve is open you consider
that valve is opened instantaneously. When instantaneously the valve is open we
consider all the hot air which was there after the expansion is expelled through this so
that this process is instantaneous and the piston almost remain at the outer dead center
position. So this is replaced by the 4, 1 and then we consider as open this instantaneously
some hot mass of air immediately is expelled to make this pressure same to the atmosphere
pressure that means pressure equalization you make the valve open this pressure difference
causes the air to flow so immediately the pressure equalization takes place the air
here is at atmospheric pressure and then the piston moves upward to push the remaining
part of the air going out of this exit manifold or outlet manifold. This thing happens similar
way is simulate in the similar way of the suction stroke or the induction stroke that
at constant atmospheric pressure why because when you push this there is a compression
so pressure should rise but this an open system that means this is open system means this
is going out so have you understand exactly this is an open system the mass is going out
so that if we consider it as a close system we can consider it as a constant pressure
exhaust that means constant pressure discharge of the air.
This is because the increase in pressure is being nullified by a this is by the exit or
air flux or masses the way the decrease in pressure was nullified by the intake of mass
so that the inlet admission of the air that is inlet stroke is taking place in this direction
zero to one at atmospheric pressure. Similarly the exhaust of the air after the point1when
the piston reaches the outer dead center position or bottom dead center position is also simulated
by the process one zero they cancel each other. Ultimately we are remaining with 1, 2, 3,
4 process that remain that defines a cycle this is a constant volume I am not writing
this is a pv diagram so you see here heat is rejected that means heated is added at
constant volume heat is rejected. In actually practice what happens when the
piston descends at this position we keep a sink here where heat is being rejected and
this air is coming to point. This thing that means the simulation is like that first of
all it is air is being taken in and this is the process 1, 2 the air is taken in here.
When we simulate the ideal cycle we just remove this air induction part and the discharge
job air part because they two cancel each other so now you think in this way there is
a cycle which starts like this there is some amount of air in a closed system it is a close
system cycle. Which is first isentropically compressed then after isentropic compression
to a volume two that means here I do this way 1, 2 to then at this point keeping the
piston fixed at this point we bring a source and heat it to a temperature 3. Then from
3 we allow the piston to come down or come out so that an isentropic expansion is there
when it come. This is 3, 2 to3 is the heating then when it comes to 4 then what we do we
keep. Then we remove the insulation we keep a heat sink and make an instantaneous heat
rejection process at constant volume or it may not be instantaneous it may be a heat
rejection process which is made at constant volume because in reversible heat rejection
or reversible heat addition will not be instantaneous it will take time but in actual case it is
instantaneous so that we can simulate it at a constant volume but when we will explain
it with respect to an ideal system with air as the working fluid. We will make that heat
sink is there with which heat is being rejected making the volume constant so 4 to 1 so therefore
1, 2, 3, 4 is an ideal cycle. If you have got any query you tell me and
this cycle is known as Otto cycle so Otto cycle is this 1, 2, 3, 4 so I could have started
like this there is another way of starting. Let us ah read the Otto cycle or analyze the
Otto cycle which compresses of these four processes 1, 2 isentropic compressions in
this direction 2, 3 a constant volume heat addition. Where the pressure and temperature
is increased and 3, 4 is a isentropic is an isentropic expansion and 4, 1 is a constant
volume heat rejection and the working fluid is air so this way I could have started the
definition of Otto cycle but I tell you that that this Otto cycle is a theoretical representation
or a theoretical cycle of an actual spark ignition or petrol engine. How do you conceive
this Otto cycle from the actual operation of the spark ignition engine? Rest part is
now a routine part. Now I again draw the pv diagram of the Otto
cycle before do our routine job so this is like this so doing our routine job so this
is 1- 2, 2 -3, 3- 4. This is s is equal to constant this is s is equal constant and we
see that heat transfer takes place only in these two process that means in this process
heat is added Qa small a also the give and this process Qr heat is rejected. Now we can
draw the Ts diagram also because this practice we have developed in Ts diagram what you do
1-2 is the isentropic process which is like this 1, 2 then 2 to 3 is the constant volume
process which looks like this in the Ts diagram then 3 to 4 is again an isentropic process
and 4 to 1 is the constant volume process. Therefore this is constant volume, this is
constant volume, these are isentropic process therefore heat addition is there only in two
process this process heat added and this process heat rejected.
The area under 2, 3 this area 2, 3, a, b is the heat added and 1, 4, a, b is the heat
reject these are all well known thing so that we can represent this cycle in pvts diagram
here hs diagram is not necessary because the ideal gas air is an ideal gas as the working
fluid. So enthalpy is nothing but cp in to t. Therefore Ts diagram is enough because
there is no difference between enthalpy and temperature it is only a function of temperature
like cp in to T directly proportional to T. Therefore enthalpy entropy diagram is not
required. Now you go one by one 1, 2 this is s is equal to constant 1, 2 this is the
diagram. Now you tell me one by one during 1, 2 process that means the isentropic compression
is any one zero which column is zero? Third column then what is did now this is a closed
system so again this thing is that closed system that is non flow process this is very
important and air as the ideal gas non flow process then we have to write the corresponding
first law of thermodynamics that Q is equal to delta u plus W.
Then delta u when Q is 0 W is minus delta u so you have the find out delta u delta uI
have told always you write for any process 1, 2 it is cv T2 minus T1 there is no doubt
so long it is an ideal gas the change in internal energy is cv T2 minus T1. Then what is work
done so work done is minus delta u so work done is cv T1 minus T2 so T1 is greater than
T2 so T1 is less than T2 so it is stated as T1 is less than T2 so work is done this negative
work is done on this on this system. Then work constant volume process can you tell
anything is 0 in this column W0 so when W is 0 Q is delta u so delta u is same. What
is delta u cv T3 minus T2 and T3 minus T2, T3 is more than T2 that means there is an
increase in internal energy and since W is 0 Q is delta u. So Q is cv T3 and that is
positive that means heat is added during this process.
3, 4 is synonymous to 1, 2 this is 0 and this will be cv T4 minus T3 final minus initial
and therefore work done is cv T3 minus T4 and here you see T3 is more than T4. Therefore
this is positive that means work is done on the system. The work is done by the system
and there is a change in internal energy which is a decrease because T4 is less than T3 the
internal energy is decreased. 4, 1 is synonymous to process 2, 3 constant volume why W is 0?
pdv work done is 0. What is cv? This is always T1 minus T4 this is very simple a child can
do that is the final minus initial so this is the process in this direction cvT1 so heat
added is equal to delta u that mean cv same thing T1minus T4.
T1 is less than T4 this is negative so heat is added to the system. If you can make this
chart everything will be clear so you prove this delta u 0 and delta W is equal to delta
Q. Then this part is clear next part we will find out the efficiency of this cycle. Efficiency
of this cycle can be defined as 1 plus why I am writing plus because i am using this
value so what is heat added? Heat is added during this and heat is rejected during this
cv. So this is already negative cv T1 minus T4 divided by cv T3 minus T2 or I can write
1 minus T4 minus T1 divided by T3 minus T2. T4 minus T1 if I did do then this is a positive
quantity and represents the heat rejection that is why I am writing with a minus sign
now so 1 minus heat rejected by heat added so heat rejected is taken in its sign algebraic
sign that is why I have written here plus or one can tell that cv T4 minus T1 is the
heat reject to that is the amount of heat rejected.
We will have to derive this expression in terms of one important parameter which is
known as the compression ratio I think time is up for today which I will do for the next
class. From here I will derive this expression eta so we will have to make process calculation
to express in terms of the compression ratio. I told you because this is an air standard cycle so here you consider
air as an ideal gas, If we consider the air fuel mixture I cannot consider the air fuel
mixture throughout the cycle so the constituents changes. It does not define the cycle definition.
When you come back even if your pressure and temperature remains the same but if you start
from this point you come back but the constituent elements that change so you have to do it
only one single fluid and air standard cycle does that with a single fluid air as an ideal
gas so that is the reason. How it is model this is beyond the scope of
this thermodynamics class. Modeling is not done in thermodynamics. We are studying the
air standard cycles only how the air mixture is modeled? but how the air mixture is made
in the carburetor? that is the different story. You have to understand that in petrol engine
the air and fuel vapor are mixed before the engine and then it is being directed to the
engine cellular so this is the practice but I am not going in to any detail of this modeling.
We have simulating this cycle with an air standard cycle Otto cycle. Where air is the
working fluid there is no air fuel vapor mixture nothing and air is being heated from outside
the heat is being rejected with heat transfer to the outside that is the air standard cycle.
