Concepts of Thermodynamics
Prof. Suman Chakraborty Department of Mechanical Engineering
Indian Institute of Technology, Kharagpur
Lecture – 60
Diesel Cycle In the previous lecture, we discussed about
Otto cycle and today, we will discuss about Diesel cycle. From the name itself, it is
clear that this cycle, thermodynamic cycle, is used to model the thermodynamic performance
of diesel engines and that is why the name Diesel cycle. So, we will draw the p v and T s diagram and
try to identify what is the difference between this cycle
and the Otto cycle. So, process 1 to 2: reversible adiabatic compression.
This is same for both Otto cycle and Diesel cycle. The major difference between the Otto
cycle and diesel cycle, and perhaps the only difference, is that the heat addition in diesel
cycle is considered to take place at constant pressure, instead of constant volume.
So, 2 to 3: constant pressure heat addition. So, constant pressure heat addition means
it will be a horizontal line here. So, after the constant pressure heat addition—so,
this is p equal to constant, there will be the expansion or the power stroke. So, 3 to
4: reversible plus adiabatic expansion. And then, from 4 to 1, you have constant volume
heat rejection. See, the heat rejection is not done at constant pressure for the reason
that it will increase the stroke volume to an extent that it will be a bulky cylinder
for—and too bulky for internal combustion engine purposes. So, although the heat addition
is constant pressure, but heat rejection for automobiles is traditionally constant volume.
Here, I have purposefully drawn this line steeper than this because in a T s diagram,
the constant volume lines are steeper than constant pressure lines. Now, we are going
to analyse the efficiency of this cycle and compare that with Otto cycle; that is one
of the tasks that we will do. So, what is q H here? Without referring to, you know,
individual systems for writing the first law of thermodynamics, we will simply now—we’ll
write that the heat addition is from 2 to 3.
So, q H is u 3 minus u 2 plus w 23, right? Straight away, this we are writing. So, one
of the major assumptions in the air standard cycle which, in the previous lecture, I forgot
to mention, but you also add it, that you neglect changes in kinetic energy and potential
energy of the working fluid. Because, we are writing these expressions neglecting the changes
in kinetic energy and potential energy, although that is not what we explicitly tell always,
for thermodynamic purposes, the thermal energy contribution is usually much more significant
than the kinetic energy, potential energy contribution.
So, this will be C v into T 3 minus T 2 and what is w 23? So, this is integral pdv from
2 to 3. So, p into V 3 minus V 2; pressure is constant, okay? So, the fuel injection
process which is typical to the diesel engines is such that, during the fuel injection—so,
the difference in paradigm, try to understand. In the previous cycle, we were trying to model
the combustion in a already mixed air and—premixed air and fuel, in the Otto cycle. So, here
we are considering that the air is highly compressed and the fuel is injected.
So, once you do that, the fuel injection process during which this chemical energy of the fuel
is input to the highly compressed air, you will have roughly a constant pressure heat
addition process. So, that is why this difference between the heat addition processes in Otto
cycle and diesel cycle. So, now, p V 3, this is as good as RT 3 and p V 2, this is as good
as RT 2. So, you can write this as C v plus R into T 3 minus T 2. So, that is nothing
but C p into T 3 minus T 2, okay? What is q L? q L
is nothing but—now, it is a constant volume process, so C v into T 4 minus T 1.
Remember, heat transfer—sign-wise, this is negative heat transfer, but when you are
writing q L, for the purpose of calculating efficiency, it’s already the mod. It’s
already the magnitude. So, we have to write T 4 minus T 1 and not T 1 minus T 4. The efficiency—we
will calculate the efficiency in this side. Is equal to 1 minus q L by q H. So, this is
1 minus C v into T 4 minus T 1 by C p into T 3 minus T 2. C v by C p is 1 by gamma. Just
like the previous example of Otto cycle, we will extract T 1 by T 2 here. We will write
T 4 by T 1 minus 1 by T 3 by T 2 minus 1. Now, T 4 by T 1, we can write T 4 by T 3 into
T 3 by T 2 into T 2 by T 1, right? Now, T 1 by T 2, or T 2 by T 1, but it is T 2 by
T 1, v 1 by v 2 to the power gamma minus 1. So, that is R to the power gamma minus 1.
And, what is T 3 by—and what is T 4 by T 3? Is equal to v 3 by v 4 to the power gamma
minus 1. v 3 by v 4 is v 3 by v 2 into v 2 by v 4. v 2 by v 4 is same as v 2 by v 1,
right? Because v 4 is same as v 1. So, this is—now, before that, what is v 3 by v 2?
v 3 by v 2, we call as cut off ratio or r c. So, this is a new definition. So, r c to
the power gamma into 1 by r to the power gamma minus 1, where r c is v 3 by v 2 is equal
to cut off ratio. Why this is called as cut off ratio? Because this is the volume ratio
at which the heat addition is cut off, otherwise it’d have indefinitely continued.
So, then, let us write this T 4 by T 1 as T 4 by T 3 is r c to the power gamma minus
1 into 1 by r to the power gamma minus 1. What is T 3 by T 2? p 2 v 2 by T 2 is equal
to p 3 v 3 by T 3, right? And p 2 is same as p 3. So, T 3 by T 2 is equal to v 3 by
v 2, which is again equal to r c. So, this becomes T 3 by T 2 is equal to r c and T 2
by T 1 is r to the power gamma minus 1. So, T 4 by T 1 becomes r c to the power gamma.
Therefore, the efficiency expression… What is the efficiency expression? 1 minus
1 by gamma into T 1 by T 2, T 1 by T 2 is 1 by r to the power gamma minus 1, into T
4 by T 1 minus 1. So, r c to the power gamma minus 1 divided by T 3 by T 2 is r c, right?
So, essentially the efficiency expression is the Otto cycle efficiency expression multiplied
by a correction factor. Right? With this basis, now let us compare the Otto cycle with this
cycle. So, for the Otto cycle, we have this term multiplied by—so, eta is equal to 1
by r to the power gamma minus 1; this multiplier is not there. So, let us see whether this
multiplier is greater than 1 or less than 1 or what it is. So, for that, you just let r c minus 1 is
equal to x, just take it. So, r c is equal to 1 plus x. So, r c to the power gamma minus,
so, not—so this minus 1 is full minus 1, not r c to the power gamma minus 1. So, r
c to the power gamma minus 1 is equal to 1 plus x to the power gamma minus 1. 1 plus
x to the power gamma is, you can expand in the binomial series. So, 1 plus gamma x plus
gamma into gamma minus 1 bi factorial 2 x square, like that it goes on, minus 1.
So, this is gamma x plus higher order terms. So, r c to the power gamma minus 1 by gamma
into r c minus 1 is equal to gamma x plus whatever by gamma x, right? So, this is this
plus whatever term, which is greater than 0, right? So, this multiplier is a multiplier…
Greater than 1. Oh, sorry—greater than 1, not 0. Okay, greater
than 1, right? So, 1 plus something, not greater than 0, so, greater than 1. So, then, this
multiplier being greater than 1, so, what is the efficiency that we expect? You are
subtracting a number magnified by a factor greater than 1. So, this should be less as
compared to what is there for the Otto cycle. So, the conclusion is for the same r, eta
Otto should be greater than eta diesel, but this is very misleading, you know. Why is
this very misleading? This is misleading because same r is not a good basis for comparison.
Same compression ratio will not be the case for Otto cycle and diesel cycle. Because you
are solely compressing air and not air fuel mixture in the diesel cycle, you can go up
to very large compression ratio, and at very large compression ratios, the diesel cycle
might have an efficiency which is quite large. So, same r may not be the same basis for comparison,
but hypothetically, if it’d have been same r, then Otto is greater than diesel. So—but
this mathematical way of comparison can also be substituted by a more physical way of comparison.
So, I will try to give you two examples of comparison of Otto cycle and diesel cycle
before we call it a day today. Okay? So, the first—so, for any comparison, there
should be a basis for comparison. You should not compare apple with an orange. So, for
common basis, let us say, for the same r and same Q H. When heat transfer is given as a
basis, it is better to use the T s diagram, because heat transfer information is directly
available from T s diagram. Work done information is directly available from the p v diagram.
But interestingly, the area under the p v diagram is same as the area under the Ts diagram
for a cycle. Because cyclic integral of heat equal to cyclic integral of work, and area
under the Ts diagram is cyclic integral of Tds which is cyclic integral of del q. So,
area under the p v diagram is same as the area under the T s diagram is same as the
net work and is same as the cyclic integral of heat, all these things are equal if it
is an internally reversible cycle. Because if it is not an internally reversible cycle,
you cannot write del q equal to Tds. So, area under the T s diagram is equal to
integral of del q—cyclic integral of del q, that you can write only for an internally
reversible cycle. Similarly, area under the p v diagram is same as the net work done for
an internally reversible or quasi equilibrium process or a collection of quasi equilibrium
processes. So, these things we must remember. So, same r and same QH. Same r means if you
start with the point 1, the state point 2 will be the same, because T 2 by T 1 is governed
by only R and gamma, right? Then—so, with white line, I will draw the—let us say,
the Otto cycle. This is—what? This is constant volume. And with a red colour, I
will draw the diesel cycle. So, from 2, you will have a constant pressure
line. So, if you have a constant pressure line, constant volume line is steeper than
constant pressure line. So, constant pressure line from 2 will go like this, and heat addition
being the same, the area under 2 to 3 represents the heat addition process in the T s diagram.
So, from 2 to 3, whatever is the area under that, that is the heat transfer, during heat
addition for the Otto cycle. For the diesel cycle, because this line is
going below to compensate for this loss of area, it should go further towards the right,
to make the area under 2 to 3 same as area under 2 to 3 prime. And then—so if you reject
heat with constant volume, then it is a basic continuation of 1 to 4 diagram. Then you see
for the same QH, which is having more q L? For the Otto cycle, this is the q L and for
the diesel cycle, you have this additional q L, right?
So, for same Q H, you have more q L for the diesel cycle, that means its efficiency is
less. So, the mathematical form, that for same r, the efficiency of Otto will be more
than efficiency of diesel, we can justify this just physically from the T s diagram.
But this may not be a good basis for comparison as I have already mentioned, that same r may
not be very practical way to look into it. So, let us say, we give a constraint: same
p max and T max. If you have same p max T max—see, p max and T max is occurring at
state point 3, right? So, if p max and T max is occurring at state point 3, and if you
specify these two states, V max is also specified because p v equal to RT, that means, for same
p max T max, that is, the combination of p max T max, you could, in principle, fix up
the state point 3 as unique state point for both Otto cycle and diesel cycle.
So, let us—so, this is a Otto cycle; with the white line, we have drawn it. So, with
the same state point 3…Now, so—with the same state point 3, the constant volume line
is steeper than constant pressure line, right? So, the constant pressure line will be less
steep and it will go something like this. So, this is p equal to constant. It will go
less steep than constant volume line and then, if you keep 4 and 4 prime and 1 and 1 prime
same, the point 2 prime will above the point—be above the point 2, right? The point 2 prime
will be above the point 2 because of it being—being less steeper.
So, area—the heat rejection being the same, heat addition is area under the T s diagram
from 2 to 3. Because 2 dash is above 2, you have an extra area representing extra heat
addition for the diesel cycle. So, for same heat rejection, if you have more heat addition,
you have more efficiency. So, this is a more logical way of comparing the two cycles and
here, you can see that you have eta diesel greater than eta Otto, because the practical
engine constraints are most of the times dictated by what is the maximum pressure it can withstand,
what is the maximum temperature it can withstand and so on, and not intrinsically by the compression
ratio. So, if somebody asks you a question, does
Otto cycle have more efficiency than diesel cycle or does diesel cycle have more efficiency
than Otto cycle? The first return question that you must ask: what is the basis for comparison?
What are the parameters that you are keeping fixed? So, depending on what you are keeping
fixed, you will have different answers. Right? In this case, you have eta Otto greater than
eta diesel, but in the case two, you have eta diesel greater than eta Otto.
Now, finally, I will give you a small note. What does a real internal combustion cycle—internal
combustion engine cycle look like? Is it Otto, is it diesel? In practice, the heat addition
is neither a constant pressure heat addition nor a constant volume heat addition. So, it
could be perceived as a combination of constant volume and constant pressure heat addition
and there is a cycle that takes that into account. I will not get into the details of
that cycle, but I will just draw the—maybe p v and T s diagram of that cycle; that is
called as a dual cycle. So, dual cycle is a combination of Otto cycle
and diesel cycle, in a sense that part of the heat is added at constant pressure and
part of the heat is added at constant volume. So, if you draw the p v diagram, so you have
compression, then heat addition, partly constant volume and partly constant pressure, but heat
rejection definitely at constant volume. And the T s diagram: so first, constant volume
curve which is steeper and then, less steep constant pressure curve, then reversible adiabatic
expansion and then heat rejection. Okay? So, to summarise, we have discussed two important
air standard cycles: one is the Otto cycle, another is the diesel cycle and we have compared
their efficiencies. But air standard cycle is not just restricted to cycles for internal
combustion engines. We could have other applications where air standard cycle can be applied. One
such application is the gas turbine cycle for which we consider the Joule cycle or the
Brayton cycle. That, we will study in the next lecture.
Thank you.
