This document discusses the second law of thermodynamics in three main points: 1) The second law is about the direction of heat transfer and that heat cannot spontaneously flow from a colder to a hotter body without work. 2) For heat engines, it is impossible to convert all the heat absorbed from a hot reservoir into work - some heat must be exhausted to a cold reservoir. 3) For refrigerators, work must be done to accomplish the transfer of heat from a colder to a warmer body.

1. Second Law of Thermodynamics
S.Gunabalan
Associate Professor
Mechanical Engineering Department
Bharathiyar College of Engineering & Technology
Karaikal - 609 609.
e-Mail : gunabalans@yahoo.com

2. Second law of thermodynamics
• The second law of thermodynamics about the
direction of heat transfer
• Second law can be visualized in terms of the
waterfall
– You can not send water up with out energy

3. Second law of thermodynamics
• Second Law for Heat Engines
It is impossible to extract heat QHot from a hot
reservoir and use it all to do work W . Some amount of
heat QCold must be exhausted to a cold reservoir.
Called Kelvin-Planck statement

4. Second law of thermodynamics
• Second Law for Refrigerator
• It is not possible for heat to flow from a colder body
to a warmer body without any work having been
done to accomplish this flow.
Called Clausius statement

5. Second law of thermodynamics
Second Law: Entropy
A measure of the amount of energy which
is unavailable to do work.
A state variable whose change is defined
for a reversible process at T where Q is the heat
absorbed
A measure of the disorder of a system.

6. First Law efficiency
• Efficiency = output energy of device / input
energy of device
– Irrespective of form of energy
– Availability of energy at different temperature

7. Second Law efficiency
• Efficiency =
• Defined as the available energy for the work

8. Exergy Balance For a Closed System
• Exergy balance for a closed system can be
developed by combining the energy and
entropy balances for a closed system.
• Energy Balance
∫ = ∆ + ----------(1)

9. Exergy Balance For a Closed System
• ∫ = ∆ + −−−− −(1)
• Entropy Balance
∫ / = ∆ --------------(2)
• Multiplying the second equation by T0 and
subtracting it from the first one yields
• ∫ = . ∆

10. Exergy Balance For a Closed System
• ∫ = ∆ + −−−− −(1)
• ∫ = . ∆
• Subtract ------------------------------------
• ∫ 1 − = ∆ + − . ∆
• E = u + v2/2+gZ

11. Exergy Balance For a Closed System
• ∫ 1 − = ∆ + − . ∆
• E = u + v2/2+gZ
1 − = 2 − 1 + (
2
2 − 1
2
2
) + 0( 2 − 1) − ( 2 − 1)

12. Reference
• http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html
• Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett
Publishers, Sudbury, Mass.
• Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New
Delhi.