EGR 334 Thermodynamics
Chapter 5: Sections 1-9
Lecture 21:
Introduction to the
2nd Law of Thermodynamics
Quiz Today?
Today’s main concepts:
• Understand the need for and the usefulness of the 2nd law
• Be able to write and use the entropy balance
• Be able to predict the maximum possible efficiency and COP of
power and refrigeration or heat pump cycles, respectively.
• Be able to provide several different expressions which explain
the 2nd Law of Thermodynamics.
Reading Assignment:
• Reread Chapter 5, Sections 10-11
Homework Assignment:
Problems from Chap 5: 20, 43, 40, 56
Sec 5.1: Introducing the Second Law
3
The First Law of Thermodynamics in an energy balance and tells us the
magnitudes energy flows.
dE CV
dt
 QCV  WCV 

2


V
mh
 gz 
2


It does not say anything about the spontaneous direction.
Qout
Qin
Can we spontaneously cool the
refrigerator by removing heat Can we spontaneously heat the house by
removing heat from the environment?
from the environment?
Sec 5.1: Introducing the Second Law
4
We know intuitively, that the spontaneous direction for heat flow is from
warm to cold.
Qin
Qout
The fridge will warm.
The house will cool.
If TA > TB  TA = TB.
But, we have refrigeration and heating, so what is required to
change the direction of heat flow?
Work needs to be done to move in the non-spontaneous direction
Sec 5.1: Introducing the Second Law
5
Sec 5.1: Introducing the Second Law
The Second Law of Thermodynamics answers the following
• Which direction the process will move spontaneously?
• What is equilibrium?
• What is the maximum efficiency?
• What parameters can be changed to move closer to
the maximum efficiency?
• What is temperature?
• How can we measure u and h?
Since the Second Law answers all these
questions, there is not a single statement
of the second law.
Popular forms of the Second Law include:
•Everything degrades to chaos
•No perpetual motion machine
6
Sec 5.2: Statements of the Second Law
7
Clausius Statement:
“It is impossible for any system to operate in such a
way that the sole result would be an energy
transfer by heat from a colder to a hotter body.”
Kelvin-Planck Statement
“It is impossible for any system to operate in
a thermodynamic cycle and deliver a net
amount of energy by work to its surroundings
while receiving heat transfer from a single
thermal reservoir.”
Q
Thermal Reservoir:
A body where energy exchange as
heat does not effect the temperature.
Kelvin-Planck Statement tells us that
It is not possible to convert heat completely to work.
Sec 5.2: Statements of the Second Law
8
Entropy Statement:
“It is impossible for any system to operate in a way that
entropy is destroyed.”
[ ][ ] [ ]
entropy
within
system
=
Net entropy
transferred
into the
system
+
Entropy
generated
with system
Note: Entropy can be generated (unlike mass)
 S sys
(+ 0 -)


Q
T
(+ 0 -)

 gen
( + or 0 )
Most processes do not operate at Ssys = 0, so generally, Ssys 
9
For a Closed Systems:
Energy Balance:
 E sys  Q sys  W sys
dE sys
Energy Rate Balance:
d E sys
dt
W
Q
dt
 Q sys  W sys
Entropy Balance:
 S sys  
Q
T
Q
  g en
T
Entropy Rate Balance:
d S sys
dt
 
Q
T

d S sys
dt
gen
 g en
10
Problem 5: 3 Classify the following processes of a closed system as
possible, impossible, or indeterminate.
-------------------------------------------------------------------------------------------------------Process
A
B
C
D
E
F
G
Entropy
Change
Entropy
Transfer
>0
<0
0
>0
0
>0
<0
 S SYS
Entropy
change
Entropy
Production
0
if < 0
>0
>0
<0
if > 0
>0
if > 0
<0
<0


Q
T
Entropy
transfer

 SYS
Entropy
production
Possible,
Impossible, or
Indeterminate.
Possible
Possible
Impossible
Possible
Possible
Impossible
Possible
Sec 5.2: Statements of the Second Law
An alternate way of looking at Entropy:
Entropy, S, is a measure of the Disorder within a system
11
Sec 5.3: Identifying Irreversibility
Sand
12
Reversible process:
Both the system and surroundings can be
returned to the initial state.
Gas
Gas
Sand
One grain of sand
is removed.
Sand
The grain of sand
is replaced.
Gas
Irreversible process:
System and surroundings cannot be
returned to the original state.
Irreversibility can be
internal to the system
external to the system
Rock
Gas
Remove
rock
Sec 5.3: Identifying Irreversibilities
Typical sources of irreversibility:
• Heat transfer due to a T
• Unrestricted expansion
• Spontaneous chemical reaction
• Friction
• Electric current producing heat
due to resistance
• Process with hysteresis
• Inelastic deformation.
13
Elastic hysteresis of an idealized
rubber band. The area in the
centre of the hysteresis loop is the
energy dissipated as heat
Since we cannot avoid all nonidealities, most of the time a real
process is irreversible.
In this class we are looking for the theoretical maximum, so
we will model most processes as reversible.
Sec 5.4: Interpreting the Kelvin-Planck Statement
14
Kelvin-Planck Statement
“It is impossible for any system to operate in a thermodynamic
cycle and deliver a net amount of energy by work to its
surroundings while receiving heat transfer from a single
thermal reservoir.”
For a single reservoir:
< 0 : Internal irreversibility present
W cycle  0
= 0 : No internal irreversibility
The inequality (<) is associated with irreversibility with the system
W
Sec 5.6: Second Law Aspects of Power Cycles
15
Efficiency of a Power Cycle interacting with two reservoirs.
 
W cycle
Q in

Q in  Q out
Q in
 1
Q out
 1
Q in
If QC  0, then η  1.0 = 100%
Second Law statements for power cycles
(Carnot Corollaries):
• The thermal efficiency of an irreversible
power cycle is always less than the thermal
efficiency of a reversible power cycle when
each operates between the same two
thermal reservoirs.
• All reversible power cycles operating
between the same two thermal reservoirs
have the same maximum thermal efficiency.
Q Cold
Q Hot
 1
QC
QH
Sec 5.7: Second Law Aspects of Refrigeration and Heat Pumps
16
COP of refrigeration and heat pumps cycles interacting with two reservoirs
 
Q out
W cycle

QC
Q H  QC
&
Q in
 
W cycle
As WCycle  0, then  and   ∞
Such an arrangement would violate the Clausius Statement.
Second Law statements for refrigeration:
• The coefficient of performancee of an
irreversible refrigeration cycle is always less
than the coefficient of performance of a
reversible refrigeration cycle when each
operates between the same two thermal
reservoirs.
• All reversible refrigeration cycles operating
between the same two thermal reservoirs
have the same maximum coefficient of
performance.

QH
Q H  QC
Sec 5.8: Temperature Scales
The Celsius Temperature Scale:
0°C  freezing point of water
(at 1 atm)
100°C  boiling point of water
(at 1 atm)
The Fahrenheit Temperature Scale:
0°F  freezing point of water &
NaCl solution (at 1 atm)
100°F  average normal human body
temperature (at 1 atm)
It is desirable to have a scale that is not
dependent on a single substance.
Phase changes of many substances
allows extension of the scale based on
properties
17
Sec 5.8: Temperature Scales
18
Sec 5.8.2: Gas Thermometer
Gas Thermometer:
Used instead of liquid since it will not
change phase over a large range of
temperatures.
Use equation of state to relate T  p
19
Sec 5.8: Temperature Scales
20
The driving force for heat transfer is a T
This causes a heat transfer Q  T
 QC

Q
 H
Thus


  TC , T H

 reversible
 QC

Q
 H

Choose
 
TC
TH

TC



TH
 reversible
The ratio of the temperatures is equal to the ratio of the
heat rejected and the heat absorbed.
Used to define the Kelvin temperature scale.
 Q
T  273 . 16 
 Q TP



 reversible
Sec 5.8: Temperature Scales
21
Consider three heat engines , operating reversibly
1
QC
  1
QH
  T H , T C
QC
2
 f T C , T I
QI
QH
3
QI
3
2

QH QI

QC Q I
Therefore
QH
QC

2

1
 f T H , T I

3
f T H , T I

f T C , T I 

f T H


f T C 
TH
TC
Sec 5.9 : Maximum Performance
With  Q C 
22
TC



Q 
TH
 H  reversible
The Carnot efficiency  max  1 
To increase η

increase TH
decrease TC
Thus, ideally we want to
maximize T = (TH - TC)
TH – limited by equipment costs
(high p and high T)
TC
TH
For a
reversible cycle
Sec 5.9 : Maximum Performance
Typically, the cold reservoir is the atmosphere and large bodies of water
and thus TC = 298 K, since it will cost too much to use a refrigerated
reservoir.
What is the maximum efficiency of a power cycle operating between
T
298
TH = 745 K
  1 C  1
 60%
and TC = 298 K
TH
745
23
Sec 5.9 : Maximum Performance
24
Example: (5.19) A power cycle operating at steady state receives energy by heat
transfer at a rate of QH at TH=1000 K and rejects energy by heat transfer to a cold
reservoir at a rate QC at TC = 300 K.
Max system efficiency:
For each of the following cases, determine
T
300
 m ax  1  C  1 
 0 .7 0
whether the cycle operates reversibly,
TH
1000
irreversibly, or is impossible.
a) QH = 500 kW, QC=100 kW
a  1
QC
 1
QH
Impossible:
b) QH = 500 kW, Wcycle = 250 kW,
and
100
 0.80
500
c 
QH

W cycle
W cycle  Q C

350
350  150
Reversible: ηc = ηmax
W cycle  Q H  Q C
250 kW
ηa > ηmax

b  ?
c) Wcycle = 350 kW, QC = 150 kW
W cycle
QC = 200 kW
 0.70
500  200  300 kW
Impossible
d) QH = 500 kW, QC = 200 kW
d  1
QC
QH
 1
200
 0.60
500
Irreversible: ηa < ηmax
Sec 5.9 : Maximum Performance
Example: (5.63) The refrigerator shown in the figure operates at
steady state with a coefficient of performance of 4.5 and a power
input of 0.8 kW. Energy is rejected from the refrigerator to the
surroundings at 20°C by heat transfer from the metal coils whose
average surface temperature is 28°C. Determine
(a) The rate energy is rejected, in kW
(b) The lowest theoretical temperature
within the refrigerator, in K
(c) The maximum theoretical power, in
kW, that could be developed from a
power cycle operating between the
coils and surroundings.
TH= 20°C= 293 K
W=0.8kW
=4.5
TC= ?
25
Sec 5.9 : Maximum Performance
26
Example: (5.63) Determine
(a) The rate energy is rejected, in kW
TH= 20°C= 293 K
W=0.8kW
COP of refrigeration cycle
 max 
Q in
W cycle
=4.5
TC= ?
Work Energy Balance:
W cycle  Q H  Q C
Q C  Q H  W cycle
Combining:
 m ax 
Q H  W cycle
W cycle
Q H  W cycle  1   m ax 
  0.8  kW  1  4.5   4.4 kW
Sec 5.9 : Maximum Performance
27
Example: (5.63) Determine
b) The lowest theoretical temperature within the
refrigerator, in K
Modeling as Reversible System with
Maximum Possible COP = 4.5
 m ax 
TC 
TC 
TH= 20°C= 293 K
W=0.8kW
=4.5
TC= ?
TC
T H  TC
 m ax
1   m ax 
4 .5
1  4 . 5 
TH
293   0 . 82 293  
240 K   33  C
Sec 5.9 : Maximum Performance
28
Example: (5.63) Determine
c) The maximum theoretical power, in kW, that
could be developed from a power cycle operating
between the coils and surroundings.
TH= 28°C= 301 K
W=?
From part a) the rejected heat is
Q H  4 . 4 kW
TH= 20°C= 293 K
Maximum Possible Power Cycle Efficiency:
 m ax  1 
TC

W cycle
TH
Note: Such a power cycle would
operate between reservoir
temperatures of 20 oC and 28 oC.
QH
Solve for possible Work done:

T 
W cycle  Q H  1  C 
TH 

293 

 (4.4 kW )  1 
  (4.4 kW )  0.0266   0.117 kW
301 

29
End of Lecture 21 Slides