1. EGR 334 Thermodynamics Chapter 5: Sections 1-9
Lecture 21:
Introduction to the
2nd Law of Thermodynamics
Quiz Today?

2. 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

3. It does not say anything about the spontaneous direction.
Sec 5.1: Introducing the Second Law
The First Law of Thermodynamics in an energy balance and tells us the magnitudes energy flows.
It does not say anything about the spontaneous direction.
Qout
Qin
Can we spontaneously cool the refrigerator by removing heat from the environment?
Can we spontaneously heat the house by removing heat from the environment?

4. Work needs to be done to move in the non-spontaneous direction
Sec 5.1: Introducing the Second Law
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

5. Sec 5.1: Introducing the Second Law

6. The Second Law of Thermodynamics answers the following
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

7. Kelvin-Planck Statement
Sec 5.2: Statements of the Second Law
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.

8. [ ] entropy within system = + Entropy Statement:
Sec 5.2: Statements of the Second Law
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)
( )
( )
( + or 0 )
Most processes do not operate at Ssys = 0, so generally, Ssys 

9. For a Closed Systems:
Energy Balance:
Energy Rate Balance:
Entropy Balance:
Entropy Rate Balance:

10. Problem 5: 3 Classify the following processes of a closed system as
possible, impossible, or indeterminate.
 Process
Entropy Change
Entropy Transfer
Entropy Production
Possible,
Impossible, or
Indeterminate. A
> 0
if > 0 
Possible B
< 0
if < 0  C
Impossible D E
if > 0  F G
Entropy
change
Entropy
transfer
Entropy
production

11. An alternate way of looking at Entropy:
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

12. Both the system and surroundings can be returned to the initial state.
Sec 5.3: Identifying Irreversibility
Gas
Sand
Gas
Sand
Reversible process:
Both the system and surroundings can be returned to the initial state.
Gas
Sand
One grain of sand is removed.
The grain of sand is replaced.
Irreversible process:
System and surroundings cannot be returned to the original state.
Gas
Rock
Remove rock
Irreversibility can be
internal to the system
external to the system

13. Typical sources of irreversibility: Heat transfer due to a T
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.
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 non- idealities, 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.

14. Kelvin-Planck Statement
Sec 5.4: Interpreting the Kelvin-Planck Statement
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
= 0 : No internal irreversibility
The inequality (<) is associated with irreversibility with the system W

15. Sec 5.6: Second Law Aspects of Power Cycles
Efficiency of a Power Cycle interacting with two reservoirs.
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.

16. As WCycle  0, then  and   ∞
Sec 5.7: Second Law Aspects of Refrigeration and Heat Pumps
COP of refrigeration and heat pumps cycles interacting with two reservoirs
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.

17. The Celsius Temperature Scale:
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

18. Sec 5.8: Temperature Scales

19. Use equation of state to relate T  p
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

20. The driving force for heat transfer is a T
Sec 5.8: Temperature Scales
The driving force for heat transfer is a T
This causes a heat transfer Q  T
Choose
Thus
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.

21. Consider three heat engines , operating reversibly
Sec 5.8: Temperature Scales
Consider three heat engines , operating reversibly 1 2 2 1 3 3 3 2
Therefore

22. Thus, ideally we want to maximize T = (TH - TC)
Sec 5.9 : Maximum Performance
With
For a reversible cycle
The Carnot efficiency
increase TH
decrease TC
To increase η 
Thus, ideally we want to maximize T = (TH - TC)
TH – limited by equipment costs
(high p and high T)

23. What is the maximum efficiency of a power cycle operating between
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
TH = 745 K
and TC = 298 K

24. Sec 5.9 : Maximum Performance
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.
For each of the following cases, determine
whether the cycle operates reversibly,
irreversibly, or is impossible.
Max system efficiency:
a) QH = 500 kW, QC=100 kW
b) QH = 500 kW, Wcycle = 250 kW,
and QC = 200 kW
Impossible: ηa > ηmax
Impossible
c) Wcycle = 350 kW, QC = 150 kW
d) QH = 500 kW, QC = 200 kW
Reversible: ηc = ηmax
Irreversible: ηa < ηmax

25. The rate energy is rejected, in kW
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
The rate energy is rejected, in kW
The lowest theoretical temperature within the refrigerator, in K
The maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings.
W=0.8kW
TH= 20°C= 293 K
TC= ?
=4.5

26. The rate energy is rejected, in kW
Sec 5.9 : Maximum Performance
Example: (5.63) Determine
W=0.8kW
TH= 20°C= 293 K
TC= ?
The rate energy is rejected, in kW
COP of refrigeration cycle
=4.5
Work Energy Balance:
Combining:

27. b) The lowest theoretical temperature within the refrigerator, in K
Sec 5.9 : Maximum Performance
Example: (5.63) Determine
W=0.8kW
TH= 20°C= 293 K
TC= ?
b) The lowest theoretical temperature within the refrigerator, in K
Modeling as Reversible System with Maximum Possible COP = 4.5
=4.5

28. Sec 5.9 : Maximum Performance
Example: (5.63) Determine
W=?
TH= 28°C= 301 K
TH= 20°C= 293 K
c) The maximum theoretical power, in kW, that could be developed from a power cycle operating between the coils and surroundings.
From part a) the rejected heat is
Maximum Possible Power Cycle Efficiency:
Note: Such a power cycle would operate between reservoir temperatures of 20 oC and 28 oC.
Solve for possible Work done:

29. End of Lecture 21 Slides