1. THE SECOND LAW OF THERMODYNAMICS
ERT206
THERMODYNAMICS
THE SECOND LAW OF THERMODYNAMICS
MISS WAN KHAIRUNNISA WAN RAMLI

2. OBJECTIVE Introduce the second law of thermodynamics.
Identify valid processes as those that satisfy both the first and second laws of thermodynamics.
Discuss thermal energy reservoirs, reversible and irreversible processes, heat engines, refrigerators, and heat pumps.
Describe the Kelvin–Planck and Clausius statements of the second law of thermodynamics.
Discuss the concepts of perpetual-motion machines.
Apply the second law of thermodynamics to cycles and cyclic devices.
Apply the second law to develop the absolute thermodynamic temperature scale.
Describe the Carnot cycle.
Examine the Carnot principles, idealized Carnot heat engines, refrigerators, and heat pumps.
Determine the expressions for the thermal efficiencies and coefficients of performance for reversible heat engines, heat pumps, and refrigerators.

3. INTRODUCTION TO SECOND LAW 1ST LAW OF THERMODYNAMIC
NO VIOLATION OF 1ST LAW
Heat IN? NO
Conservation of Energy
Heat OUT? YES
Processes proceed in certain direction,
not in reverse direction.
2nd LAW OF
THERMODYNAMIC

4. 1st Law 2nd Law MAJOR USES OF THE SECOND LAW
The second law may be used to identify the direction of processes.
The second law also asserts that energy has quality as well as quantity. The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality. The second law provides the necessary means to determine the quality as well as the degree of degradation of energy during a process.
The second law of thermodynamics is also used in determining the theoretical limits for the performance of commonly used engineering systems, such as heat engines and refrigerators, as well as predicting the degree of completion of chemical reactions.
PROCESS OCCURENCE
2nd Law
1st Law

5. THERMAL ENERGY RESERVOIRS
SOURCE:
Reservoirs which supplies energy in the form of heat
Bodies with relatively large thermal masses can be modeled as thermal energy reservoirs.
SINK : Reservoirs which absorbs energy in form of heat.
THERMAL ENERGY RESERVOIR  A hypothetical body with a relatively large thermal energy capacity (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature
In practice, large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or thermal masses.

6. HEAT ENGINE Heat Work Work Heat The devices that convert HEAT  WORK.
They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.).
They convert part of this heat to work (usually in the form of a rotating shaft.)
They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.).
They operate on a cycle.
Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid.

7. CONSIDER AS CLOSED SYSTEM,
STEAM POWER PLANT
A portion of the work output of a heat engine is consumed internally to maintain continuous operation.
CONSIDER AS CLOSED SYSTEM,
ΔU = 0

8. PERFORMANCE OF HEAT ENGINE (THERMAL EFFICIENCY, ŋth )
The fraction of heat input that is converted to net work output
Some heat engines perform better than others (convert more of the heat they receive to work).
Schematic of a heat engine.
THERMAL EFFICIENCY, ŋth of Heat Engine is relatively low, even with the most efficient Heat Engine available. Almost half of Energy supplied were rejected as waste

9. EXAMPLE 1
6-21
An automobile engine consumes fuel at a rate of 22 L/h and delivers 55 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg an a density of 0.8 g/cm3. Determine the efficiency of this engine. [Ans: 25.6 %]
SOLUTION:
Assumptions The car operates steadily.
Properties & given values
Volume flow rate, V = 22 L/h
Power, Wnet, out = 55 kW
The heating value, qhv = 44,000 kJ/kg,
ρ = 0.8 g/cm3,
Analysis
Draw HE diagram & label given values
Find the mass consumption flow rate, m
Calculate heat supply, Q
Calculate ŋ

10. CAN WE SAVE Qout?
Can we just take the condenser out of the plant and save all that waste energy?
In a steam power plant, the condenser is the device where large quantities of waste heat is rejected to rivers, lakes, or the atmosphere.
A heat-engine cycle cannot be completed without rejecting some heat to a low-temperature sink. NO
Because without a heat rejection process in a condenser, the cycle cannot be completed.
Every heat engine must waste some energy by transferring it to a low-temperature reservoir in order to complete the cycle, even under idealized conditions.

11. KEVIN-PLANCK STATEMENT
A heat engine that violates the Kelvin–Planck statement of the second law.
KEVIN-PLANCK
STATEMENT
It is impossible for any device that operates on a cycle to receive heat from a single reservoir & produce a net amount of work
Heat engine MUST exchange heat with a low-T SINK & a high-T SOURCE to keep operating
No heat engine can have a thermal efficiency of 100 %,
The impossibility of having a 100% efficient heat engine is not due to friction or other dissipative effects. It is a limitation that applies to both the idealized and the actual heat engines.

12. TO REMOVE HEAT (QL) FROM THE REFRIGERATED SPACE
REFRIGERATORS
The transfer of heat from a low-temperature medium to a high-temperature one requires special devices called refrigerators.
Refrigerators, like heat engines, are cyclic devices.
The working fluid used in the refrigeration cycle is called a refrigerant.
The most frequently used refrigeration cycle is the vapor-compression refrigeration cycle.
Basic components of a refrigeration system and typical operating conditions.
OBJECTIVE :
TO REMOVE HEAT (QL) FROM THE REFRIGERATED SPACE
Desired output = ?
Required input = ?

13. Coefficient of Performance, COPR
The efficiency of a refrigerator is expressed in terms of the coefficient of performance (COP).
It requires a work input of W net, in to remove QL
COPR can be > 1

14. EXAMPLE 2 REFRIGERATOR 6-42
A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine:
The electric power consumed by the refrigerator [0.83kW]
The rate of heat transfer to the kitchen air [110 kJ/min]
HEAT PUMP
6-49
A heat pump is used to maintain a house at a constant temperature of 23°C. The house is losing heat to the outside air through the walls and the windows at a rate of 60,000 kJ/h while the energy generated within the house amounts to 4000 kJ/h. For a COP of 2.5, determine the required power input to the heat pump. [6.22 kW]

15. SOLUTION:
6-42
Assumptions The refrigerator operates steadily.
Properties & given values
COP = 1.2, Q? = 60 kJ/min
Analysis
Draw Refrigerator diagram & label given values
Calculate power input, Win, net using COP eqn
Calculate heat rejection, Q? using energy balance
6-49
Assumptions The heat pump operates steadily.
The rate of heat loss, the rate of internal heat gain, and the COP of a heat pump are given.
Draw HP diagram & label given values
Calculate heating load, Q of HP

16. HEAT PUMPS Coefficient of Performance, COPHP
DIFFERENT OBJECTIVE FROM REFRIGERATORS :
TO SUPPLY HEAT QH INTO A TH MEDIUM
Heat pumps operate on the same cycle as refrigerators.
Coefficient of Performance, COPHP
for fixed values of QL and QH
Can the value of COPHP be lower than unity?
What does COPHP=1 represent?

17. A refrigerator that violates the Clausius statement of the second law.
It is impossible to construct a device that operates in a cycle & produces no effect other than the transfer of heat from a lower temperature body to a higher temperature body
It states that a refrigerator cannot operate unless its compressor is driven by an external power source
The net effect on the surroundings involves the consumption of some energy in the form of work, in addition to the transfer of heat from a colder body to a warmer one.

18. EQUIVALENCE OF THE TWO STATEMENTS
The Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the second law of thermodynamics.
Any device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa.
Proof that the violation of the Kelvin–Planck statement leads to the violation of the Clausius statement.

19. PERPEPTUAL-MOTION MACHINE
PERPEPTUAL-MOTION MACHINE  Any device that violates the 1st or 2nd law of thermodynamics.
PMM1  A device that violates the 1st law (by creating energy)
PMM2  A device that violates the 2nd law (exchanging heat with 1 reservoir only)
A perpetual-motion machine that violates the second law of thermodynamics (PMM2).
A perpetual-motion machine that violates the first law (PMM1).

20. REVERSIBLE & IRREVERSIBLE PROCESSES
A process that can be reversed without leaving any trace on the surroundings.
An idealization to actual processes
A process that cannot reverse spontaneously to restore its initial state
Processes in nature are irreversible
Why are we interested in reversible processes?
Reversible Processes
Easy to analyze
Can be viewed as theoretical limits for corresponding irreversible processes.
Leads to the definition of 2nd-Law Efficiency  degree of approximation to the corresponding reversible processes
Reversible processes deliver the most and consume the least work.

21. EXAMPLE  Friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions.
IRREVERSIBILITIES
The factors that cause a process to be irreversible
The presence of any of these effects renders a process irreversible.
(a) Heat transfer through a temperature difference is irreversible, and (b) the reverse process is impossible.
Friction renders a process irreversible.

22. INTERNALLY & EXTERNALLY REVERSIBLE PROCESSES
INTERNALLY reversible process  No irreversibilities within the boundaries of the system during the process. [Eg: Nonquasi Equilibrium process]
EXTERNALLY reversible process  No irreversibilities outside the system boundaries.
TOTALLY reversible process  Involves no irreversibilities within the system or its surroundings. [No heat transfer through a finite temperature difference, No nonquasi-equilibrium changes, and No friction or other dissipative effects.]
A reversible process involves no internal and external irreversibilities.
Totally and internally reversible processes.

23. THE CARNOT CYCLE  best known reversible cycle
ISOTHERMAL
ADIABATIC
EXPANSION
Wout
COMPRESSION
Win
Reversible Isothermal Expansion (process 1-2, TH = constant)
Reversible Adiabatic Expansion (process 2-3, temperature drops from TH to TL)
Reversible Isothermal Compression (process 3-4, TL = constant)
Reversible Adiabatic Compression (process 4-1, temperature rises from TL to TH)

24. THE REVERSED CARNOT CYCLE
P-V diagram of the Carnot cycle.
THE REVERSED
CARNOT CYCLE
Area under curve  W out
P-V diagram of the reversed Carnot cycle.
Area under curve  W in
Area enclosed by both curve ( )  Wnet done during cycle
The Carnot heat-engine cycle is a totally reversible cycle, it can be reversed  Carnot refrigeration cycle.
Reversed carnot  differ in Q & W direction

25. THE CARNOT PRINCIPLES
The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.
The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.
Ŋirrev < ŋrev between the same TH & TL
Ŋrev1 = ŋrev2 between the same TH & TL
Proof of the first Carnot principle.
The Carnot principles.

26. THE THERMODYNAMIC TEMPERATURE SCALE
thermodynamic temperature scale: A temperature scale that is independent of the properties of the substances that are used to measure temperature
Based on principle 2, Ŋrev is independent of working fluid, execution of the cycle, or type of reversible engine used
The arrangement of heat engines used to develop the thermodynamic temperature scale.
Carnot principle 2:
All reversible heat engines operating between the same two reservoirs have the same efficiency.

27. KELVIN SCALE TH & TL  Absolute T
For reversible cycles, the heat transfer ratio QH /QL can be replaced by the absolute temperature ratio TH /TL.
KELVIN SCALE
TH & TL  Absolute T
On Kelvin scale, T ratios depend on the ratios of heat transfer, Q between reversible heat engine & reservoirs.
Absolute T using the magnitude of Kelvin:

28. THE CARNOT HEAT ENGINE
A hypothetical Heat Engine operates on reversible Carnot cycle
Any heat engine (Rev/ Irreversible):
For Carnot Heat Engine known as Carnot Efficiency:
Carnot Efficiency  highest efficiency for HE operating between TH & TL
No heat engine can have a higher efficiency than a reversible heat engine operating between the same high- and low-temperature reservoirs.

29. EXAMPLE 3
6-81
A heat engine operating on a Carnot cycle has a thermal efficiency of 75%. The waste heat from this engine is rejected to a nearby lake at 15°C at a rate of 14kW. Determine the power output of the engine and the temperature of the heat source, in °C. [42kW, 879°C]
SOLUTION:
Assumptions The heat engine operates steadily. Heat losses from the
working fluid at the pipes and other components are negligible.
Properties & given values
ŋ = 0.75, Tl = 15°C, Ql= 14 kW
Analysis
Draw HE diagram & label given values
Find Qh using ŋ eqn. & calculate power output, Wnet, out
Calculate the temperature of high-T reservoir, Th

30. THE CARNOT REFRIGERATORS & HEAT PUMPS
A refrigerator & heat pump operates on reversed Carnot cycle
Coefficient of Performance, COP for any refrigerator/ heat pump (Rev/ Irreversible):
For Reversible Refrigerator/ Heat pump:
It is the highest COP value for Refrigerator & heat pump operating between TH & TL
No refrigerator can have a higher COP
than a reversible refrigerator operating
between the same temperature limits.

31. EXAMPLE 4
6-92
A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. It keeps a space at 26°C by consuming 4.25 kW of power. Determine the temperature of the reservoir from which heat is absorbed and the heating load provided by the heat pump [265K, 37 kW]
SOLUTION:
Assumptions The heat pump operates steadily.
Properties & given values
COP = 8.7, Th = 26°C, Power, Wnet, in = 4.25 kW
Analysis
Draw HP diagram & label given values
Find the temperature of low-T reservoir, Tl
Calculate heating load, Q

32. THANK YOU ^^

33. TUTORIAL I
A heat pump with refrigerant-134a as the working fluid is used to keep a space at 25°C by absorbing heat from geothermal water that enters the evaporator at 60°C at a rate of kg/s and leaves at 40°C. Refrigerant enters the evaporator at 12°C with a quality of 15 percent and leaves at the same pressures as saturated vapor. If the compressor consumes 1.6 kW of power, determine (a) the mass flow rate of the refrigerant, (b) the rate of heat supply, (c) the COP and (d) the minimum power input to the compressor for the same rate of heat supply.