ENGR 2213 Thermodynamics

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ENGR 2213 Thermodynamics

• F. C. Lai
• School of Aerospace and Mechanical
• Engineering
• University of Oklahoma

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Second Law of Thermodynamics
First law of Thermodynamics
Energy
Second law of Thermodynamics
Entropy
Unlike energy, entropy is a non-conserved
property.
Clausius Inequality
The cyclic integral of dQ/T is always less than
or equal to zero.
This inequality is valid for all cycles,
reversible or irreversible.
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Clausius Inequality
For reversible heat engines
0
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Clausius Inequality
For irreversible heat engines
From Carnot principle
Wrev gt Wirrev
QL, rev lt QL, irrev
QL, irrev QL, rev Qdiff
lt 0
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Example 1

• A heat engine receives 600 kJ of heat from a
  high-
• temperature source at 1000 K during a cycle. It
• converts 150 kJ of this heat to work and rejects
  the
• remaining 450 kJ to a low-temperature sink at
• 300 K. Determine if this heat engine violates
  the
• 2nd law of thermodynamics on the basis of
• the Clausius inequality.
• (b) the Carnot principle.

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Example 1 (continued)
(a) Clausius inequality
- 0.9 kJ/K
lt 0
(b) Carnot principle
?th lt ?rev
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Second Law of Thermodynamics
Internally Reversible Processes
A process is called internally reversible if
no irreversibilities occur within the boundaries
of the system during the process.
Clausius Inequality
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Entropy
If a quantity whose integral depends only on the
end states and not the process path, then it is a
property.
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Entropy
Isothermal Processes
Q T0 ?S
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Increase-in-Entropy Principle
Clausius Inequality
S1 S2
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Increase-in-Entropy Principle
The entropy change of a closed system during an
irreversible process is greater than the
integral of dQ/T evaluated for that process.
For an adiabatic process, Q 0
In the absence of heat transfer, entropy change
is due to irreversibilities only, and their
effect is always to increase the entropy.
(?S)adiabatic 0
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Increase-in-Entropy Principle
(?S)adiabatic 0
A system plus its surroundings constitutes an
adiabatic system, assuming both can be enclosed
by a sufficiently large boundary across which
there is no heat or mass transfer.
(?S)total
(?S)system (?S)surroundings
0
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Increase-in-Entropy Principle
gt 0 irreversible processes
0 reversible processes
Sgen (?S)total
lt 0 impossible processes
Causes of Entropy Change
? Heat Transfer
? Irreversibilities
Isentropic Process
A process involves no heat transfer (adiabatic)
and no Irreversibilities within the system
(internally reversible).
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Remarks about Entropy
1. Process can occur in a certain direction
only. A process must proceed in the
direction that complies with the
increase-in-entropy principle.
2. Entropy is a non-conserved property. There
is no such thing as the conservation of
entropy principle.
3. The quantity of energy is always preserved
during an actual process (the first law),
but the quality decreases (the second law).
The decrease in quality is always
accompanied by an increase in entropy.
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What is Entropy?
Entropt can be viewed as a measure of
molecular disorder, or molecular randomness
From a statistical point of view, entropy is a
measure of the uncertainty about the position of
molecules at any instant.
The entropy of a pure crystalline substance at
absolute zero temperature is zero since there is
no uncertainty about the state of the molecules
at that instant.
- the 3rd Law of Thermodynamics
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Second Law of Thermodynamics
Heat is, in essence, a form of disorganized
energy and some disorganization (entropy) will
flow with heat.
Work instead is an organized form of energy, and
is free of disorder or randomness and thus free
of entropy.
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Example 1

• Saturated water at 100 ºC is contained in a
  piston-
• cylinder assembly. The water undergoes an
• internally reversible heating process to the
• corresponding saturated vapor state. Find
• the work per unit mass for the process.
• (b) the heat transfer per unit mass for the
  process.

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Example 1 (continued)
(a)
(101.4)(1.673 0.001044)
170 kJ/kg
(b)
2257 kJ/kg
(373.15)(7.3549 1.3069)
2257 kJ/kg
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Example 2
Steam at 7 MPa and 450 ºC is throttled through a
vavle to 3 MPa. Find the entropy generation
through the process.
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Example 2 (continued)
T1 450 ºC p1 7 MPa
h1 3287.1 kJ/kg s1 6.6327 kJ/kg K
Table A-6
Throttling Process h2 h1
p2 3 MPa h2 3287.1 kJ/kg
s2 6.9919 kJ/kg K
Table A-6
sgen ?s
6.9919 6.6327
0.3592 kJ/kg K
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Second Law of Thermodynamics
Carnot Cycle in T-S Diagram