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On this topic, we (wikitec Team) have tried to explore the complete analysis of the thermal engineering portion or, you may say, thermodynamics with all of its applied formulas in competitive or semester-wise examinations. Dear aspirant, please take a moment to read the full articles, which will be your appreciation and trust in us as we help you achieve your goals.
.png "Carnot Cycle | Third law of Thermodynamics | Analysis of Entropy")
Table of Contents:
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Carnot Cycle: Reversible Isothermal
Expansion Reversible Adiabatic
Expansion Reversible Isothermal
Compression Reversible Adiabatic
Compression P-V diagram
(Pressure-Volume) for Carnot Cycle: Carnot Principle
(Carnot’s Theorem): Third Law of
Thermodynamics: Entropy: Clausius’s Theorem: Entropy Change: The principle of
Entropy: Applications of Entropy
Principle: Conclusion: |
Carnot Cycle:
Today we have
discussed one of the most interesting topics about Carnot Cycle, which is the
part of second law of thermodynamics.
We know that
heat engines are cyclic in nature and its working fluid of heat engine basically
returns to its initial state during the end of each cycle.
Here, we can
observe two most important things that work is done by and work is done on.
That’s mean
work done executes by the working fluid in one most part and work done executes
on the working fluid in another part. But here, the question is that the
difference between two part is the net work being delivered by heat engine.
If we will analyse
about the efficiency of heat engine, then we will get how the individual
process that helps to execute the cycle. So, the total work done, and the efficiency
should be maximized during the process with least amount of work and should be
delivered most by the help of reversible processes. A reversible cycle Is an
ideal hypothetical in nature, where all processes execute the cycle are
reversible. Thus, Carnot Cycle is a reversible cycle.
Here, we will
follow up some of our most important process.
Carnot cycle works
under four reversible cycles. Now, we will discuss about two isothermal and two adiabatic processes.
Here, we have discussed the four individual process one by one.
Let suppose,
we have a closed system, enclosed by some gases under adiabatic piston-cylinder
arrangement. You may see figures with the explanation for each process.
Reversible Isothermal Expansion:
Form the above figure, we have 1-2 process of our
piston-cylinder arrangement.
Here, Tg is the temperature of the gas at the source. In this above figure, we
see that the cylinder head is supposed to contact with source at Tg. If we allow to expand the gas gradually, then we see that some work
done execute on the surrounding. One thing here, we will notice due to the
temperature of the gas decreases, there is small change of temperature drop by dT, and
the heat transfer occurs from the reservoir into the gas which results
temperature raise at source Tg. Such process seems to be reversible heat transfer,
the total heat transfer to the gas during this process Qnet, still it continues until the piston touches position 2.
Reversible Adiabatic Expansion:
We have
already discussed 1-2 process earlier, now we will discuss next 2-3 process during
adiabatic expansion under piston-cylinder arrangement. So, let’s talk about it.
From the
above figure, the process 2-3, there is the change or drop the corresponding
temperature Tg to Ts. Where Ts is the sink temperature. When the reservoir tends to
contact with the cylinder head at position 2, after we remove or replace by
making insulation, then the system becomes adiabatic in nature. The gas will
expand gradually and will do some works on surrounding until the temperature
drops from Tg to Ts
at position 3.
Here, we consider the piston seems to be frictionless and quasi-equilibrium. Thus, the process is reversible as well as adiabatic.
Reversible Isothermal Compression:
Here, we have talked about reversible isothermal process 3-4.
During such process, we keep sink temperature Ts is constant. Now at position 3, if we remove
insulation at our cylinder head, ultimately the cylinder comes into contact
with sink temperature Ts, then the piston moves inward direction by external
force by doing work on the gas. Which will result temperature raise due to compression
of gas inside it and simultaneously small amount of change or drop of dT, result will be the heat transfer occurring from gas
to sink portion.
Thus, the temperature of gas remains constant in
nature at Ts. There is the difference between the gas and sink
have both limit up to dT. During Such process of heat transfer still
continues, until the piston rod reaches at position 4 with the amount of heat Qs exhausted from the gas.
Reversible Adiabatic Compression:
Under reversible adiabatic compression process, that’s
mean 4-1 process. From this above figure, here the interesting part is that the
temperature raises from Ts
to Tg. In this case, if we remove sink temperature Ts reservoir and keep insulation on the cylinder head,
then the gas will compress in reversible manner, where the gas tends to restore
at its initial position 1. Due to the temperature increases from Ts to Tg during such process, result will be the completion of entire
cycle.
P-V diagram (Pressure-Volume) for Carnot Cycle:
From this above P-V diagram, we have the area
under the curve during the process is supposed to be work under quasi-equilibrium
process. Here, one thing, we have to observe the area under the curve 1-2-3
is the work done by the gas during expansion and the area under the curve 3-4-1
work done on the gas during compression of the cycle.
Now, we will get from the area under the curve above
two cycles 1-2-3 and 3-4-1, there is the significant difference between
the expansion as well as compression both indicating the net work done during
the cycle under area 1-2-3-4.
So, being a reversible cycle, Carnot cycle is the most efficient between two specified temperature limits.
Carnot Principle (Carnot’s Theorem):
According to the second law of thermodynamics, there are some limits for operation of cyclic devices by both Kelvin-Planck and Clausius rules. So, no any heat engine work by exchanging heat with a single reservoir and even no any refrigerator work without supply of energy from an external source.
From this above figure, we have three reversible heat engines. Here, all heat engines are operating between both specified constant source temperature and sink temperature at T1 and T2 respectively.
Note: -
- The efficiency of an irreversible heat engine is always less than the efficiency of a reversible heat engine operating at same two reservoirs.
- All reversible heat engines having their efficiencies are same operating under same two reservoirs.
- If we decrease the sink temperature T2, then ultimately increase thermal efficiency of Carnot cycle which is more than the source temperature T1.
Third Law of Thermodynamics:
 |
| Crystalline Substance at T=0K |
Let’s consider a pure crystal see the above figure having
its temperature T=0K; where Entropy=0.
Here, we should know that the entropy is the
measurement of molecular disorder. If we will increase the entropy of a system
then there will be the molecular uncertainty or you may say randomness of
molecular probability of a system ultimately increases. That’s mean the
molecules of a substance in solid state is continually oscillating and making their
position uncertain. Due to such oscillation, the temperature decrease, then the
result become no molecular motion occurs at absolute zero temperature.
Hence, the entropy of a pure crystal at absolute zero temperature is zero since the molecules at that instant become uncertain. This statement leads to the third law of thermodynamics, which is only helpful for entropy of an absolute reference point as known as absolute entropy.
Now, we will talk about entropy.
Entropy:
Especially, the second law of thermodynamics leads to Entropy.
No body able to define entropy significantly. It is very difficult to point out the physical significance without its microscopic study of the system. Only option is left at our hand by taking commonly used in engineering fields.
We have already discussed about the second law of
thermodynamics earlier topic; you may follow our previous article. which often
deals with the inequalities. For the best example is irreversible heat engine, which
is not more efficient than a reversible engine at same temperature limits.
Now, you may say any kinds of irreversible
refrigerator or heat pump have lower value of co-efficient of performance than a
reversible at same temperature limits.
But here, entropy is a useful property or you may say
important tool in our second law of thermodynamics.
Entropy has considered as a measure of molecular disorder
or randomness. If we make disorder or displace the entire system, then
molecular position will be very difficult to predict. Thus, entropy of the
system increases.
Here, we should remember one thing always that under
solid state and gas state, entropy of a substance lowest and highest
respectively.
Here, we can see this below figure:
 |
Molecular Disorder at Different states |
Thus, the entropy of a system related to the total
number of microscopic possible states of a system is called Boltz-mann relation
has expressed as
S=k In p
Where, the value of k= 1.3X 10-23 (Boltz-mann Constant)
N.B.: - Entropy is an extensive property. But the entropy per unit mass is an
intensive property and its unit’s kJ/kg dot K
Clausius’s Theorem:
As per the Clausius’s
Theorem in thermodynamics, the cyclic integral is always less than or equal to
zero. Such inequality condition will be valid for all cycles, reversible and
irreversible processes.
∮δQ/T≤ 0
The heat
transfer occurs to or from a system, which will result differential amounts of heat
transfer. The summation of such differential amounts of heat transfer divided
by the temperature should be the cyclic integral of ∮δQ/T.
Here, we have
some conditions for Clausius’s Inequality for reversibility
or irreversibility.
∮δQ/T≤ 0
If the cycle is in reversible; then ∮δQ/T= 0
If the cycle is in irreversible and process;
then ∮δQ/T< 0
Similarly, for the impossible cycle; ∮δQ/T> 0
N.B. :- The above discussion, we conlude that for internally reversible cycles,equality under clausius inequality and inquality for irreversible cycles.
Next, we will discuss about the change in entropy.
Entropy Change:
The change in entropy in
reversible process is a point function and exact differential. When the system
tends to be equilibrium state, the entropy is supposed to be zero.
Here, the change in
entropy of a system during the process should be expressed by the integration
between initial and final states of the process.
We have discussed for
your better understanding by taking under T-S (Temperature-Entropy) diagram.
 |
| Change in Entropy 1-2 process |
dS=S2-S1 = integrating of process 1-2 of (δQ/T) for internally reversible process
For reversible process; ds=δQ/T
Where ds= Change in entropy, δQ=Heat Transfer, T= Absolute Temperature.
For any process by a system ∮δQ/T≤ ds
Where δQ/T is the value of entropy change and its integral of δQ/T is irreversible path.
Thus, the chnge of entropyof a system is due to heat transfer and internal irreversibility for the entropy generation.
S2-S1=integrating of process 1-2 of (δQ/T) due to heat transfer + s due to internal irreversibility.
Here, we should know that entropy generation is a path function and its positive value.
The principle of Entropy:
Let’s consider a cycle, having its two processes. Here, we have considered the processes of 1-2 and 2-1. You may see it below.
 |
Cycle under both reversible and irreversible |
Important points:
Form this
above figure, we conclude that under 1-2 process for arbitrary conditions i.e. either
reversible or irreversible. Similarly, 2-1 process for internally reversible.
We have discussed
earlier about the Clausius’s Inequality
From this Clausius’s Inequality, we have ∮δQ/T≤ 0.
Or
we can express as it like integral
of process 1-2 of (δQ/T)
+ integral of process 2-1 of (δQ/T) ≤ 0.
Hence, we
have second integral relation integral of process 1-2 of (δQ/T) for internally
reversible process
Where, the change in entropy S2-S1
= dS
We can also write this integral
of process 1-2 of (δQ/T)
+ S1-S2≤ 0
i.e. S2-S1
≥ integral of process 1-2 of (δQ/T)
or you can write as dS ≥ δQ/T
For the reversible process, the
change in entropy (dS)
is equal to integral of process 1-2 of (δQ/T) integral of
process 1-2 of (δQ/T).
For the irreversible process,
for closed system the change in entropy (dS) is always greater than integral of
process 1-2 of (δQ/T) integral of process 1-2 of (δQ/T).
So, due to this irreversibility, the entropy is generated during this process,
that is called entropy generation (Sgen).
From the above expression, dS ≥ δQ/T
If we consider for isolated system dSiso ≥0
Similarly, for both reversible and irreversible
process dSiso=0 and dSiso >0 respectively.
Applications of Entropy Principle:
Case:1 Heat transfer occurs through a finite temperature
difference.
Case:2 The mixture of two fluids by the help of adiabatic
condition.
Case:3 The maximum work gainable from two finite bodies at different
temperature limits.
Case:4 The maximum work gainable from a finite body and TER.
Case:5 The processes under isothermal and adiabatic dissipation
of work.
Combined Equations of First Law and Second Law:
There are
two equations, which are valid for both reversible and irreversible of both
open as well as closed system.
TdS= dU+pdV
TdS = dH-Vdp
Entropy Change during Phases:
Here,
we have already known as TdS= dU+pdV
From
this above equation, we can also express as dS=
dU/T+pdV/T
Basically, the liquids and solids phases are
incompressible in nature and their specific volumes are remains constant; for
such condition dV=0.
Now, we can write from the above
equation as dS = dU/T = c dT/T
For incompressible substances, Cp = Cv = C =
Specific Heat and also dU = c dT
Then the
entropy changes during a process for both liquids and solids
[S2-S1=integral of the process
1-2 of c(T) dT/T = Cavg lnT2/T1]
Now, we have
discussed the entropy change of Ideal gases.
Entropy
Change for Ideal Gases:
From the
above equation, we have the relation dS= dU/T + pdV/T
For the
entropy change of an ideal gas dS=Cv dT/T + R dV/V
N.B.:
-The change in specific entropy during isothermal process of a system for ideal
gas is dS = Cp- CvInV2/V1
Here, some points to be noted that all of these above expressions are specifically applied in every numericals or you may say short-notes, which will be very helpful for your quick revision.
Conclusion:
In this post-discussion, we are emphasizing the point-to-point bullet lines, or, concisely, the topic of Carnot cycle, Carnot principle and third law of thermodynamics, only with most relevant and important explanations, as well as its formulas to help our readers with their revision and enable them to move confidently towards success. We wish you to ace your path significantly.
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