1. ERT 206 THERMODYNAMICS WAN KHAIRUNNISA WAN RAMLI
ENTROPHY
ERT 206
THERMODYNAMICS
WAN KHAIRUNNISA WAN RAMLI

2. OBJECTIVE
OBJECTIVE
Apply the second law of thermodynamics to processes.
Define a new property called entropy to quantify the second-law effects.
Calculate the entropy changes that take place during
processes for pure substances, incompressible substances,
and ideal gases.
Examine a special class of idealized processes, called
isentropic processes, and develop the property relations for
these processes.
Derive the reversible steady-flow work relations
Introduce and apply the entropy balance to various systems.

3. 2ND LAW OF THERMODYNAMICS CLAUSIUS INEQUALITIES
Involve many inequalities:
Ŋ irrev < ŋrev
COP HP/ R < COP rev
CLAUSIUS INEQUALITIES
VALID FOR ALL CYCLES
Differential over entire cycle
Sum of all differential amounts of heat transfer per T at boundary

4. THE CLAUSIUS INEQUALITY
Energy balance on combined system,
Consider the cyclic device
reversible,
Eliminates δQR,
The combine system undergo cycle,
So, we have CLAUSIUS INEQUALITY
If no irreversibilities & cyclic device is reversible,
The combined system is internally reversible
The system considered in the development of Clausius Inequality
Violates Kevin-Planck of 2nd Law, so Wc cannot be work output (Wc≠+ve)
CLAUSIUS INEQUALITY
The EQUALITY for totally or just internally reversible cycles
The INEQUALITY for the irreversible ones.

5. RELATION TO ENTROPHY? ENTROPYPROPERTY
The entropy change between two specified states is the same whether the process is reversible or irreversible
ENTROPYPROPERTY
The net change in a property (i.e. volume) during a cycle is always zero
Property A quantity which its cyclic integral = 0
Definition of Entrophy
SPECIAL CASE: INTERNALLY REVERSIBLE ISOTHERMAL HEAT TRANSFER PROCESSES

6. REMARKS ON ENTROPHY
Processes can occur in a certain direction only, not in any direction. A process must proceed in the direction that complies with the increase of entropy principle, that is, S gen ≥ 0. A process that violates this principle is impossible.
2. Entropy is a nonconserved property, and there is no such thing as the conservation of entropy principle. Entropy is conserved during the idealized reversible processes only and increases during all actual processes. (Entropy is generated or created during irreversible process due to the presence of irreversibilities).
3. The performance of engineering systems is degraded by the presence of irreversibilities, and entropy generation is a measure of the magnitudes of the irreversibilities during that process. It is also used to establish criteria for the performance of engineering devices.

7. ENTROPHY CHANGE OF PURE SUBSTANCES
Entropy is a property: The value of entropy of a system is fixed once the state of the system is fixed.
Determination of S value = determination of other property (i.e. h)
The entropy of a pure substance
is determined from the tables
(like other properties).
Compressed Liq & Superheated Vapor  straight from data
For saturated mixture, given quality, x
Entrophy change
S used as coordinate on T-s diagram

8. EXAMPLE 1
A rigid tank contains 5kg of R-134a initially at 20°C and 140 kPa. The refrigerant is cooled while being stirred until its pressure drops to 100 kPa. Determine the entrophy change of R-134a during this process.
SOLUTION:
ASSUMPTIONS: the volume of the tank is constant, so v2 = v1
ANALYSIS:
Closed system, no mass crosses the system boundary
Change in entrophy = s2-s1, state 1 is fully specified
Specific volumes remains constant
State 1: at P1=140 kPa, T1 = 20°C s1 = kJ/kg.K, v1 = m3/kg
State 2: P2 = 100 kPa, v2 = v1 = m3/kg  saturated mixture since vf<v2<vg
Determine the quality for state 2 x = (v2-vf)/vfg = 0.859
Thus s2=sf+xsfg = kJ/kg.K
Entrophy change, ΔS = m(s2-s1)= kJ/K

9. ISENTROPIC PROCESS: NO HEAT TRANSFER, AREA = 0
ISENTROPIC PROCESSES
ISENTROPIC PROCESS  A PROCESS WHICH THE ENTROPHY REMAINS CONSTANT.
ISENTROPIC PROCESS: NO HEAT TRANSFER, AREA = 0
During an internally
reversible, adiabatic
(isentropic) process, the
entropy remains constant.
The isentropic process appears as a vertical line segment on a T-s diagram.

10. 3RD LAW OF THERMODYNAMICS
WHAT IS ENTROPHY?
ENTROPHY  A MEASURE OF MOLECULAR DISORDERS OR MOLECULAR RANDOMNESS
As a system becomes more disordered, the entrophy will increase.
The level of molecular
disorder (entropy) of a
substance increases as it melts or evaporates.
S is related to the thermodynamic probability, p (molecular probability)
3RD LAW OF THERMODYNAMICS
A pure crystalline substance at 0 temperature is in perfect order & its entrophy is 0

11. ENTROPHY CHANGES OF LIQUIDS & SOLIDS
Derived from Eq Gibbs equation
Liquids & solids can be approximated as incompressible substances since their specific volumes remain nearly constant during a process
Liquids, solids:
For an isentropic process of an incompressible substance
ISENTROPIC
ISOTHERMAL

12. ENTROPHY CHANGES OF IDEAL GASES
IDEAL GAS PROPERTIES
From the first T ds relation (Eq 7-25)
From the second T ds relation (Eq 7-26)

13. Constant Specific Heats (Approximation Analysis)
For gases whose C vary linearly with T range
Entropy change of an ideal gas on a
unit–mole basis  multiplying by molar mass
Under the constant-specific-heat assumption, the specific heat is assumed to be constant at some average value.

14. Variable Specific Heats (Exact Analysis)
Absolute zero is chosen as the reference temperature & define a function s° as
Thus, entrophy changes between T1 & T2
In unit-mole basis
For gases whose C vary NONlinearly with T range
The entropy of an ideal gas depends on both T and P. The function s° represents only the temperature-dependent part of entropy.
On a unit–mole basis

15. EXAMPLE 2
Air is compressed from an initial state of 100kPa and 17°C to a final state of 600 kPa and 57°C. determine the entrophy change of air during this compression process by using (a) property values from the air table (b) average specific heats
SOLUTION: Air is compressed between 2 specified states. The entrophy change is to be determined by both method
ASSUMPTIONS: Air is an ideal gas
ANALYSIS: the initial & final states are fully specified
Property table for s0 (Table A-17) & substituting into eq. for exact analysis [ kJ/kg.K]
Find Cp values to determine the Cpave at Tave of 37°C (Table A-2b) and solve using approximation analysis [ kJ/kg.K]

16. Isentropic Processes for Ideal Gases
Constant Specific Heats (Approximate Analysis)
For Isentropic (Δs=0) process, setting the above Eqs. equal to zero to get:
The isentropic relations of ideal gases are valid for the isentropic processes of ideal gases only.

17. Isentropic Processes for Ideal Gases
Variable Specific Heats (Exact Analysis)
Relative Pressure & Relative Specific Volume
The use of Pr data
for calculating the
final temperature
during an isentropic
process
Pv=RT
T/Pr is the relative specific volume vr.

18. EXAMPLE 3
Helium gas is compressed by an adiabatic compressor from an initial state of 100 kPa and 10°C to a final temperature of 160°C in a reversible manner. Determine the exit pressure of helium.
SOLUTION:
ASSUMPTION: At specified conditions, helium can be treated as ideal gas.
ANALYSIS:
Find the k value for helium (Table A-2)
Calculate the final pressure of helium using eq.

19. REVERSIBLE STEADY-FLOW WORK
Energy balance for a steady-flow device (internally reversible, +ve for Wout)
but
Yield
For work input,
v , W
For the steady flow of incompressible liquid (v constant) through a device that involves no work interactions (such as a pipe section), the work term is zero.
BERNOULLI EQUATION

20. Proof that Steady-Flow Devices Deliver the Most and Consume the Least Work when the Process Is Reversible
Energy balance for 2 (irrev & rev) steady-flow devices (+ve for Qin, Wout)
Both operate between the same end states,
however,
Gives,
T is absolute T
IRREVERSIBLE
REVERSIBLE
A reversible turbine delivers more work than an irreversible one if both operate between the same end states.
Work-producing devices (turbines )deliver more work, and work-consuming devices (pumps, compressors ) require less work when they operate reversibly

21. ENTROPHY BALANCE
Entrophy transfer + Entrophy generation = Entrophy change
Entropy Change of a System, ∆S system
Energy and entropy
balances for a system
When the properties of the system are not uniform

22. Mechanisms of Entrophy Transfer, Sin & Sout
HEAT TRANSFER
Heat transfer is always accompanied by
entropy transfer in the amount of Q/T,
where T is the boundary temperature.
Entropy transfer by heat transfer:
T ≠ constant
Entropy transfer by work:
No entropy accompanies work as it crosses the system boundary. But entropy may be generated within the system as work is dissipated into a less useful form of energy.

23. Mechanisms of Entrophy Transfer, Sin & Sout
MASS FLOW
Mass contains entropy as well as energy, and thus mass flow into or out of system is always accompanied by energy and
entropy transfer.
Entropy transfer by mass:
s = specific entrophy
When the properties of the mass change during the process

24. Entrophy Generation, Sgen
Entropy generation
outside system
boundaries can be
accounted for by
writing an entropy
balance on an
extended system that includes the system and its immediate
surroundings.
Mechanisms of entropy transfer for a
general system.

25. Closed Systems
The entropy change of a closed system during a process is equal to the sum of the net entropy transferred through the system boundary by heat transfer and the entropy generated within the system boundaries.
ADIABATIC CLOSED SYSTEM
SYSTEM + SURROUNDINGS

26. Control Volumes STEADY FLOW STEADY FLOW SINGLE STREAM
The entropy of a control volume changes as a result of mass flow as well as heat transfer.
STEADY FLOW
STEADY FLOW SINGLE STREAM
STEADY FLOW SINGLE STREAM ADIABATIC
The entropy of a substance always increases (or remains constant in the case of a reversible process) as it flows through a single-stream, adiabatic, steady-flow device.

27. THANK YOU

28. TUTORIAL II
A horizontal cylinder is separated into two compartments by an adiabatic, frictionless piston. One side contains 0.2 m3 of nitrogen and the other side contains 0.1kg of helium.. Both initially at 20°C and 95 kPa. The sides of the cylinder and the helium end are insulated. Now heat is added to the nitrogen side from a reservoir at 500°C until the pressure of the helium rises to 120 kPa. Determine (a) the final temperature of helium, (b) the final volume of the nitrogen, (c) the heat transferred to the nitrogen and (d) the entrophy generation during this process.
ANS: (a) 321.7K, (b) m3, (c) kJ, (d) kJ/K