1. Second Low of Thermodynamics

2. Example 1 Heat always flows from high temperature to low temperature.
So, a cup of hot coffee does not get hotter in a cooler room.
Yet, doing so does not violate the first low as long as the energy lost by air is the same as the energy gained by the coffee.

3. Example 2
The amount of Electrical Energy is equal to the amount of energy transferred to the room.
Paddle wheel rotates and stirs the fluid the internal energy of the insulated fluid increase

4. It is clear from the previous examples that..
Processes proceed in a certain direction and not in the reverse direction.
The first law places no restriction on the direction of a process.
Therefore we need another law (the second law of thermodynamics) to determine the direction of a process.

5. Introduction to the 2nd Law
We use the 2nd Law of Thermodynamics to:
Identifie the direction of the processes
Assert that energy has quality as well as quantity
The first law is concerned with the quantity of energy and its transformation from form to another with no regard to its quality
Provide the necessary means to determine the quality as well as the degree degradation of energy during a process.

6. Introduction to the 2nd Law
We use the 2nd Law of Thermodynamics for:
More of high temperature energy can be converted to work and thus it has higher quality that the same amount of energy at lower temperature
Determines the theoretical limits for the performance of commonly used engineering systems such as heat engines and refrigerators.

7. Introduction to the 2nd Law

8. Introduction to the 2nd Law

9. Thermal Energy Reservoir
It is defined as a body to which and from which heat can be transferred without any change in its temperature.
Large thermal energy capacity (mass x specific heat)
If it supplies heat then it is called a source.
If it absorbs heat then it is called a sink.

10. Some obvious examples are solar energy, oil furnace, atmosphere, lakes, and oceans
Another example is two-phase systems,
and even the air in a room if the heat added or absorbed is small compared to the air thermal capacity (e.g. TV heat in a room).

11. Heat Engines
We all know that doing work on the water will generate heat.
However transferring heat to the liquid will not generate work.
Yet, doing so does not violate the first low as long as the heat added to the water is the same as the work gained by the shaft.

12. Previous example leads to the concept of Heat Engine!.
We have seen that work always converts directly and completely to heat, but converting heat to work requires the use of some special devices.
These devices are called Heat Engines and
can be characterized by the following:

13. Characteristics of Heat Engines..
High-temperature Reservoir at TH
receive heat from high-temperature source.
They convert part of this heat to work.
They reject the remaining waste heat to a low-temperature sink.
They operate on (a thermodynamic) cycle. QH HE W QL
Low-temperature Reservoir at TL

14. Part of the heat received by a heat engine is converted to work, while the rest is rejected to a sink.

15. Difference between Thermodynamic and Mechanical cycles
A heat engine is a device that operates in a thermodynamic cycle and does a certain amount of net positive work through the transfer of heat from a high-temperature body to a low-temperature body.
A thermodynamic cycle involves a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid.
Internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution) but not on a thermodynamic cycle. Air is not cooled to the initial temperature but it is replaced by a fresh air and fuel.
However, they are still called heat engines
Best example to fit into the definition of heat engine is the steam power plant (External-combustion Engine)

16. Steam power plant is another example of a heat engine..

17. Thermodynamic cycle:1st Law
Wnet,out = Wout – Win
Qnet = Qin - Qout
First Law of thermodynamics for cycle is Wnet = Qnet or
Wnet = Qin – Qout
(ΔE =0)
Some heat engines perform better than others (convert more of the heat they receive to work).

18. Thermal efficiency Thermal Efficiency < 100 %
= what you gain / what you spend
Thermal Efficiency
= net work output / total heat input
< 100 %

19. Thermal efficiency Thermal Efficiency < 100 %
QH= magnitude of heat transfer between the cycle
device and the H-T medium at temperature TH
QL= magnitude of heat transfer between the cycle
device and the L-T medium at temperature TL
Thermal Efficiency
< 100 %

20. thermal efficiency can not reach 100%
Even the Most Efficient Heat Engines Reject Most Heat as Waste Heat
QH, TH
Automobile Engine 20%
Diesel Engine %
Gas Turbine %
Steam Power Plant 40%
QL, TL

21. Can we save Qout?
Heat the gas (QH=100 kJ)
Load is raised=> W=15 kJ
How can you go back to get more weights (i.e. complete the cycle)?
By rejecting 85 kJ
Can you reject it to the Hot reservoir? NO
What do you need?
I need cold reservoir to reject 85 kJ
A heat- engine cycle cannot be completed without rejecting some heat to a low temperature sink.

22. Example 6-1: Net Power Production of a Heat Engine
Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine.

23. EXAMPLE 6–2 Fuel Consumption Rate of a Car
A car engine with a power output of 50kW has a thermal efficiency of 24 percent. Determine the fuel consumption rate of this car if the fuel has a heating value of 44,000 kJ/kg (that is, 44,000 KJ of energy is released for each 1 kg of fuel burned).
50 KW/0.24 = KW
To supply energy at this rate, the engine must burn fuel at a rate of
Mdot = kW/44,000 kJ/kg = kg/s = 17 kg/hr

24. The Second Law of Thermodynamics: Kelvin-Plank Statement (The first statement)
The Kelvin-Plank statement: Related to Heat Engine
It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work.
Impossible cycle

25. It can also be expressed as:
No heat engine can have a thermal efficiency of 100%, or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace.
Note that 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 idealized and the actual heat engines.

26. Refrigerator and Heat Pumps
Heat can not be transferred from low temperature body to high temperature one except with special devices.
These devices are called Refrigerators and Heat Pumps
Heat pumps and refrigerators differ in their intended use. They work the same.
The working fluid used in the refrigeration cycle called “refrigerant” e.g. R134a, Freon …
They are characterized by the following:

27. Refrigerators QL = QH - W QH W Ref QL High-temperature Reservoir at TH
Objective
Low-temperature Reservoir at TL

28. Heat Engine – Heat Pump/Refrigerator
We have seen that work always converts directly and completely to heat, but converting heat to work requires the use of some special devices Heat Engine.
Heat can not be transferred from low temperature body to high temperature one except with special devices Refrigerator or Heat pump.

29. An example of a Refrigerator and a Heat pump ..

30. Coefficient of Performance of a Refrigerator
The efficiency of a refrigerator is expressed in term of the coefficient of performance (COPR).
COPR > 1 this is not efficiency

31. Heat Pumps QH = W + QL QH W HP QL High-temperature Reservoir at TH
Objective QH
QH = W + QL W HP QL
Low-temperature Reservoir at TL

32. Heat Pump
The objective of a heat pump, is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low-temperature source, such as well water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a
house
The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it.

33. Coefficient of Performance of a Heat Pump
The efficiency of a heat pump is expressed in term of the coefficient of performance (COPHP).

34. Relationship between Coefficient of Performance of a Refrigerator (COPR) and a Heat Pump (COPHP).

35. EXAMPLE 6–3 Heat Rejection by a Refrigerator
The food compartment of a refrigerator, is maintained at 4°C by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance
of the refrigerator and
(b) the rate of heat rejection to the room that
houses the refrigerator.

37. EXAMPLE 6–4 Heating a House by a Heat Pump
A heat pump is used to meet the heating requirements of a house and maintain
it at 20°C. On a day when the outdoor air temperature drops to 2°C,
the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat
pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air

38. (b) The house is losing heat at a rate of 80,000 kJ/h
(b) The house is losing heat at a rate of 80,000 kJ/h. If the house is to be maintained at a constant temperature of 20°C, the heat pump must deliver heat to the house at the same rate, that is, at a rate of 80,000 kJ/h.
the rate of heat transfer from the outdoor becomes

39. The second Law of Thermodynamics: Clausius Statement (the 2nd statement)
The Clausius statement ( related to refrigerators & Heat pump):
It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. ( i.e. compressor driven by external power supply must exist)
Both statements are negative statements!
Impossible cycle

40. Proof that the violation of the Kelvin–Planck statement leads to the violation of the Clausius statement.
Impossible to have 100% HE = Impossible to have HP, R with no Work
No Work from compressor =

41. Equivalence of the Two Statements
High-temperature Reservoir at TH
Net QOUT = QL QH
QH + QL HE
Ref
HE + Ref
W = QH QL
Net QIN = QL
Low-temperature Reservoir at TL

42. Perpetual Motion Machines
Any device that violates the first or second law is called a perpetual motion machine
If it violates the first law, it is a perpetual motion machine of the first type (PMM1)
If it violates the second law, it is a perpetual motion machine of the second type (PMM2)
Perpetual Motion Machines are not possible

43. A perpetual-motion machine that violates the first law of thermodynamics (PMM1).
Wnet, out
No Qin ; Wnet does not equal Qnet (Wnet must =QH-QL)
continuously supplying energy to the outside at a rate of

44. A perpetual-motion machine that violates the second law of
thermodynamics (PMM2).
This way, all the heat transferred to the steam in the boiler will be converted to work, and thus the power plant will have a theoretical efficiency of 100 percent.

45. The second law of thermodynamics state that no heat engine can have an efficiency of 100%.
Then one may ask, what is the highest efficiency that a heat engine can possibly have.
Before we answer this question, we need to define an idealized process first, which is called the reversible process.