1. 2nd Law of Thermodynamics
Topic 3
2nd Law of Thermodynamics

2. 3.1 Statement of Second Law
3.2 New Concepts
3.2a Heat Engine
3.2b Heat Reservoir
Practical Heat Engine
Reversed Heat Engine:
Refrigeration
3.2c The concept of Entropy

3. 3.1 Statement of 2nd Law of Thermodynamics
For a closed system
undergoing a cycle,
Net Work Done < Gross Heat Supplied

4. “It is impossible to construct a system operating in a cycle, which takes an amount of heat from the surroundings and producing an equal amount of net work on the surroundings.”
Mathematically,this can be written as:

5. 2ND Law of Thermodynamics
For reversible cycle

6. 3.2: NEW CONCEPTS
3.2a Heat engine
is a closed system which operates in a cycle, receiving an amount of heat and producing a net amount of work. This engine is depicted
below :

7. W = net work produced by the heat engine
Heat engine diagram
QH = heat supplied to the system from the hot reservoir (source with temperature TH)
QC = heat removed from the system to the cold reservoir (sink with temperature TC)
W = net work produced by the heat engine

8. 3.2b HEAT RESERVOIR
Refers to the part of the surroundings which exchanges heat with a thermodynamic system due to temperature difference between the system and the reservoir. If heat comes out of the reservoir and enters the system, the reservoir is called a SOURCE, otherwise it is called a SINK.

9. From the 1st Law
(∑δW)cycle-net = (∑δQ)cycle-net
i.e.
W = QH - QC
From 2nd Law,
W < QH
The bigger the portion of QH, is changed to net work the more efficient is the engine.
This brings us to the idea of engine efficiency which is defined as below:

10. But W = QH - QC  = (QH - QC)/QH = 1 -QC/QH Net work done by engine
Net work done by engine
Efficiency =  =
Gross heat given to engine
= (W/QH) x 100%
But W = QH - QC
 = (QH - QC)/QH = 1 -QC/QH

11. Example
A steam power plant can be represented by a heat engine which produces 120 MW of net power. If the heat engine is operated at an efficiency of 30%, calculate the rate of heat (a) Received by the engine and (b) rejected by the engine.

12. Heat received:= QH =W/ =120/0.3 MW =400 MW
Solution
Heat received:= QH =W/
=120/0.3 MW
=400 MW
Heat rejected:= QC = QH – W
= MW
= 280 MW

13. PRACTICAL HEAT ENGINES
An example of a practical heat engine is the steam power cycle as applied in the electric power generation station. The cycle essentially consists of four main processes, Boiler, Turbine, evaporator and heat pump. {The real cycle has a lot more additional equipment such as reheaters, superheaters and regenerators

14. Reversed Heat Engine
A device such as a refrigerator or air conditioner, designed to remove heat from a cold region and transfer it to a hot region, is essentially a heat engine operating in reverse, as shown:

15. Efficiency of Reversible Heat Engine (RHE)
The efficiency of an RHE is normally stated in terms of a performance index, or coefficient of performance (c.o.p), i.e.
  Heat removed from system into hot reservoir
(c.o.p.) =
Net work supplied from surroundings to system
= QH/W
= QH/(QH - QC)

16. (c.o.p.)ref = 1/ = QC/(QH - QC) = [ - (QH - QC) + QH]/(QH - QC)
Heat taken by system from cold reservoir =
Net work from surroundings to system
= QC/(QH - QC)
= [ - (QH - QC) + QH]/(QH - QC)
= QH/(QH - QC)

17. EXAMPLE
In a reversed heat engine, the net work done on the system is 75 kJ and the heat transferred to the system from the cold reservoir is 220 kJ. Calculate the heat transferred from the hot reservoir, and determine the coefficient of performance as refrigerator.

18. QH = W + QC
= = 295 kJ
(c.o.p.)ref = QC/W = 220/75 = 2.93

19. 3.2c The concept of Entropy
Entropy is a corollary (consequence) of the second law of thermodynamics
It is one of 3 derived and immeasureable property of thermodynamics
“There is a thermodynamic property of a closed system whose change of value is equal to [] for a reversible process between state 1 and state 2.”