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ME203 : HEAT AND FLOW 2
Thermodynamics
Dr Stephanie Ordonez
s.ordonez@strath.ac.uk
www.strath.ac.uk/engineering
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Modified slides originally created by Dr I. Taylor and Dr S. Haeri
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Entropy is an extremely baffling conception. It is
sometimes erroneously treated as if it were simply
a statistical, a probable, or thermodynamic factor,
without any material basis. Its material basis is
almost never defined and is seldom even alluded to
in books on thermodynamics.
Albert Mathews
6. X
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Clausius Inequality
■ Clausius Inequality - important thermodynamic relation.
dQ is heat transfer at a part of the system boundary
T is temperature at that part of the boundary
The cyclic integral of dQ/T is always less than or equal to zero.
0
T
Q
d
0
rev
T
Q
d
0
irrev
T
Q
d
If cyclic integral is applied to
reversible heat engine :
For an irreversible heat engine
(including losses) :
7. X
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Definition of Entropy
■ Consider a cycle consisting
of two reversible
processes, A and B.
■Then :
We have a quantity whose
cyclic integral is zero
P
V
1
2
Process A
Process B
0
rev
T
Q
d
Not process
dependent
Dependent
only on the
end states
In a cycle, the initial and final states are identical.
The cyclic integral of a property will be zero.
8. X
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Definition of Entropy
P
V
1
2
Process A
Process B
For example, using volume as an example
1
2
2
1
2
1
B
proc
B
proc
A
proc dV
dV
dV
0
1
2
2
1
B
proc
A
proc dV
dV
0
dV
Applying to the processes shown (both
reversible)
0
1
2
2
1
B
A
rev T
Q
T
Q
T
Q d
d
d
2
1
2
1
B
A T
Q
T
Q d
d
As the cyclic integral of a property is zero,
the quantity (dQ/T)rev must represent a
property.
9. X
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Definition of Entropy
■ Entropy defined as :
TdS
Q
or
T
Q
dS rev
rev
d
Units : kJ/K
Process represented on a T-S
diagram, the heat transferred
during the process is
represented by the area under
the curve.
Qrev = ∫ TdS (no friction)
(similar to area on P-V diagram
representing work W= ∫ PdV)
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Definition of Entropy
■ Entropy defined as :
TdS
Q
or
T
Q
dS rev
rev
d
Units : kJ/K
What is entropy ?
• Difficult to state exactly or to visualise
• Related to rate of heat transfer
• Often termed a measure of disorder or “chaos”
• Useful to define process direction
• Useful as a measure of irreversibility
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Principle of Increasing Entropy
For irreversible processes, Qrev ≠ ∫ TdS
TdS
E
Q f
where Ef is equivalent heat transferred due to frictional dissipation.
If have frictional heat dissipation, Ef :
0
2
1
T
dE
dQ
S
or
T
dE
dQ
dS
f
f
Note Ef always positive.
For Adiabatic (but still irreversible) process, Q=0 :
0
2
1
S
ie
T
dE
S
or
T
dE
dS
f
f
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Principle of Increasing Entropy
For Adiabatic (irreversible) process, Q=0 :
0
2
1
S
ie
T
dE
S
or
T
dE
dS
f
f
If process is reversible also,
0
0
S
so
Ef
Remember: Isentropic ≡ reversible adiabatic.
Have now shown that
S > 0 or S = 0
13. X
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Principle of Increasing Entropy
A universe consists of a system and its surroundings.
If all heat transfers are internal to the universe, the
universe becomes an adiabatic system.
Have shown that S > 0 or S = 0
i.e. S ≥ 0
“The entropy of an isolated system during a process always increases or, in the
limiting case of a reversible process, remains constant.”
For a system contained within a “universe”
S 0 i.e. Entropy always increases.
work
Universe Surroundings
heat
system
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Importance of Entropy
■ 2nd Law of Thermodynamics : Processes occur in
certain direction only.
■ A process must occur in direction that complies with the
increasing entropy principle.
■ (coffee spontaneously heating : S < 0 )
■ (cold drink spontaneously cooling : S < 0 )
■ Entropy is a non-conserved property.
(Only conserved during reversible processes)
“The entropy of an isolated system during a process always increases or, in the
limiting case of a reversible process, remains constant.”
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Importance of Entropy
■ Thermal efficiency is always less than 100%
■ Thermal efficiency is a maximum when using reversible (ideal)
processes.
■ Cycle using Ideal Processes : Scycle = 0 kJ/K
■ Cycle using processes close to ideal (small losses) :
Scycle small +ve value
■ Cycle using processes not close to ideal (large losses) :
Scycle high +ve value
■ Performance of thermodynamic systems is degraded by
irreversibilities.
■ Entropy can be used as a measure of the irreversibilities associated
with a process.
2
1
rev
T
Q
S
d
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2nd Law of Thermodynamics
Plank’s Statement of the 2nd Law of
Thermodynamics
“Every process occurring in nature proceeds in the
sense in which the sum of the entropies of all bodies
taking part in the process is increased. In the limit,
i.e. for a reversible process, the sum of the
entropies remains unchanged.”
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Other comments on Entropy
■ Only usually concerned about changes in entropy.
■ Can usually define a zero value of entropy
at some arbitrarily selected reference state.
(e.g. Triple point)
■ Third law S = 0 at 0 K.
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Laws of thermodynamics
FIRST LAW :
“Energy can neither be created nor be destroyed, it can only be
transferred from one form to another.”
SECOND LAW :
“The entropy of any isolated system always increases”
“Every process occurring in nature proceeds in the
sense in which the sum of the entropies of all bodies
taking part in the process is increased. In the limit,
i.e. for a reversible process, the sum of the
entropies remains unchanged.”
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Laws of thermodynamics
THIRD LAW :
“The entropy of a perfect crystal at absolute zero is exactly equal
to zero”
“It is impossible for any process, no matter how idealized, to
reduce the entropy of a system to its absolute-zero value in a finite
number of operations”
ZEROTH LAW :
“If two systems are each in thermal equilibrium with a third system, they
are also in thermal equilibrium with each other.”
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Example 5-1 - Entropy
• STOTAL for b) is smaller, this
means that the process is less
irreversible (less losses)
• Expected as b) involves a
smaller temperature difference
T
Q
S rev
Would not usually
calculate entropy using
Example 5-1
A heat source at 800 K losses 2000
kJ of heat to a sink at
(a) 500 K and (b) 750 K.
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Entropy Change for a Process :
Ideal Gas
■ It is not usual to calculate entropy using
■ For a perfect gas, from the energy equation
and the gas law, it can be shown (see notes
for proof) :
T
Q
S rev
A
C
1
2
1
2
1
2 ln
ln
P
P
R
T
T
C
s
s p
B
1
2
1
2
1
2 ln
ln
P
P
C
v
v
C
s
s v
p
1
2
1
2
1
2 ln
ln
v
v
R
T
T
C
s
s v
These relations show that
for a perfect gas, the
entropy change for a
process can be obtained
from the primary
properties.
P, V, T.
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Entropy Change for a Process
For Constant Volume :
1
2
1
2
ln
ln
P
P
C
T
T
C
s v
v from A and C
For Constant Pressure :
1
2
1
2
ln
ln
v
v
C
T
T
C
s p
p from B and C
For Constant Temperature :
1
2
1
2
ln
ln
P
P
R
v
v
R
s from A and B
Can modify these further :
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Entropy Change for a Process
■ Now consider an isentropic process.
s1 = s2 or s = 0
Using C
1
2
1
2
ln
ln
P
P
C
v
v
C v
p
1
2
2
1
v
v
P
P
.
.
2
2
1
1
Const
Pv
Const
v
P
v
P
0
ln
ln
1
2
1
2
1
2
P
P
C
v
v
C
s
s v
p
v
p
C
C
v
v
P
P
1
2
2
1
ln
ln
2
1
1
2
ln
ln
P
P
C
v
v
C v
p
Remember
isentropic
processes from
Week 1.
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Entropy for Liquids and Vapours
■ Cannot use simple equations of state for liquids and
vapours.
■ Use a different approach to that used for perfect gases.
■ Values of entropy given in property tables.
■ Saturated liquids & vapours
■ Superheated vapours
■ For liquid-vapour mixtures, use :
f
g
f s
s
x
s
s
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Example 5-2 - Entropy
Example 5-2
A rigid tank contains 5 kg of steam, initially at 200°C and
150 kPa. The steam is now cooled until its pressure drops
to 70 kPa.
Determine the entropy change of the steam during this
process.
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Example 5-3 - Entropy change.
Example 5-3
A quantity of air is contained within a frictionless, leak proof,
piston-cylinder device occupying a volume of 1.25m3, at 5 bar and
150C. The air is allowed to expand to 1.5 bar, according to
PV1.2=constant. Then, the internal energy of the air is raised by
90kJ, during a constant pressure heating. Find the pressure,
temperature and volume after each process.
These two processes are then replaced by a single reversible
polytropic expansion, which will result in the same final state
being reached. Determine the necessary index of expansion, and
also if the change in entropy for the new process will be greater
than, less than or the same as the entropy change for the original
two processes? ( R, CP, CV in notes )
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Reminder - Importance of Entropy.
■ Process direction can be determined from principle of
increasing entropy.
• Within a “universe”, S ≥ 0.
■ Amount of change in entropy gives a measure of how
close a system is to ideal conditions.
• Zero entropy increase for ideal case.
• Large increase in entropy is due to large losses or
irreversibility.
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Isentropic Efficiencies
■ Actual processes : Losses (irreversibility) reduce
the performance (efficiency) of devices.
■ Entropy will increase for an irreversible but
adiabatic process.
■ Ideal adiabatic reversible process will be
isentropic.
■ Isentropic efficiency is measure of how close
actual process is to the ideal isentropic process.
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Isentropic Efficiencies
■ So far, considered turbines and compressors to be
ideal steady-flow devices, i.e. isentropic.
■ In reality, losses (irreversibility) downgrade the
performance of a device.
■ Can quantify how closely device is to ideal
conditions using Isentropic efficiency.
SFEE : Adiabatic, so use
-w=h.
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Isentropic Efficiency and Entropy.
Inlet pressure, P1
Exit pressure, P2
Enthalpy, h
Entropy, s
1
2s
2a
2b
2c
a → c : Increasing loss
(irreversibility) as move
from process a to c.
S increasing
H (work)
reducing
Less work produced
Larger entropy increase
Work reduction (loss) related to entropy change
Isentropic efficiency used to ascertain how close to ideal
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Isentropic Efficiency - Expansion
■ Expansion process, work is being produced.
■ For given inlet and exit pressure,
• Ideal case produces work, Wout kJ.
• Actual case produces Wout – Ef kJ.
• Hence, less work produced in actual case.
■ Isentropic Efficiency for Expansion
processes defined as:
hs =
Actual work out
Ideal work out
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Isentropic Efficiency - Expansion
■ Irreversibilities mean that less work is produced
than for ideal process.
■ SFEE : Adiabatic, so –w = h.
1
2
1
2
h
h
h
h
h
h
w
w
s
a
s
a
s
a
s
h h2a and h2s are enthalpy at
turbine exit for actual and
isentropic processes respectively.
hs =
Actual work out
Ideal work out
Wideal
Wactual
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Isentropic Efficiency - Compression
■ Compression process, work is being consumed.
■ For given inlet and exit pressure,
• Ideal case consumes work, Win kJ.
• Actual case consumes Win + Ef kJ.
• Hence, more work consumed in actual case.
■ Isentropic Efficiency for
Compression processes
defined as:
hs =
Ideal work in
Actual work in
43. X
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Isentropic Efficiency - Compression
■ Irreversibilities mean that more work is consumed
than for ideal process.
■ SFEE : Adiabatic, so –w = h.
h2a and h2s are enthalpy at
compressor exit for actual and
isentropic processes respectively.
hs =
Ideal work in
Actual work in
h
1
2
1
2
h
h
h
h
h
h
w
w
a
s
a
s
a
s
s
Wideal
Wactual
44. X
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Summary - Entropy
■ Process direction can be determined from principle of increasing
entropy.
• Within a “universe”, S ≥ 0.
■ Amount of change in entropy gives a measure of how close a
system is to ideal conditions.
• Large increase in entropy is due to large losses or irreversibility.
■ Isentropic Efficiency is a practical way of assessing how close a
device is to ideal conditions.
• Uses change in work transfer due to losses
• Amount of loss is related to entropy rise.
s
a
turb
s
w
w
,
h
a
s
comp
s
w
w
,
h → 6 – Carnot
Cycle
45. X
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Example 5-4 - Isentropic Efficiency
Example 5-4
Steam enters an adiabatic turbine steadily at a pressure of 3 MPa
and a temperature of 400C, and exits at a pressure of 150 kPa
and a temperature of 150C. If the power output of the turbine is
2.5 MW, then determine :
a) the Isentropic efficiency of the turbine
b) the mass flow rate of the steam through the turbine.
48. X
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Example 5-4a - Isentropic Efficiency
Example 5-4a
110 kg/s of air, with pressure and temperature of 23 bar and 1250K respectively, is
expanded through a turbine, and exits at a pressure of 9 bar. For the three cases
below, determine the temperature of the air at the turbine exit, the power generated
by the turbine and the entropy generated.
a) Ideal turbine
b) Isentropic efficiency of 91%
c) Isentropic efficiecny of 85%
49. X
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Example 5-4a - Isentropic Efficiency
a) Using P,V,T, Isentropic relations :
0
/
110
1250
900
2300
1
2
1
Q
H
W
Q
s
kg
m
K
T
kPa
P
kPa
P
K
P
P
T
T s 1
.
956
2300
900
1250
4
.
1
1
4
.
1
1
1
2
1
2
MW
T
T
C
m
W s
P
s
49
.
32
1250
1
.
956
005
.
1
110
1
2
12
K
kJ
S
K
kJ
P
P
R
T
T
C
m
S
s
s
P
s
/
0
/
010
.
0
2300
900
ln
287
.
0
1250
1
.
956
ln
005
.
1
110
ln
ln
12
1
2
1
2
12
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I
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Example 5-4a - Isentropic Efficiency
b) Isentropic efficiency = 91%
K
T
T
T
T s
s
a
6
.
982
1250
1250
1
.
956
91
.
0
1
1
2
1
,
2
h
MW
T
T
C
m
W a
P
a 56
.
29
1
1
,
2
1
,
12
K
kJ
P
P
R
T
T
C
m
S a
P
a /
012
.
3
ln
ln
1
2
1
1
,
2
1
,
12
c) Isentropic efficiency = 85%
K
T
T
T
T s
s
a
2
.
1000
1250
1250
1
.
956
85
.
0
1
1
2
2
,
2
h MW
T
T
C
m
W a
P
a 62
.
27
1
2
,
2
2
,
12
K
kJ
P
P
R
T
T
C
m
S a
P
a /
975
.
4
ln
ln
1
2
1
2
,
2
2
,
12
53. X
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Y
O
F
E
N
G
I
N
E
E
R
I
N
G
Example 5-5 - Isentropic Efficiency
Example 5-5
Air is compressed by an adiabatic compressor from 100 kPa and
12°C to a pressure of 800 kPa at a steady mass flow of 0.2 kg/s.
Calculate the exit temperature of the air and the required power
input to drive the compressor.
If the compressor has an isentropic efficiency of 80%, determine
the actual exit temperature of the air and the actual required
power input.
