2nd Law of Thermodynamics
Two classical statements of the Second Law of Thermodynamics:
| Kelvin & Planck "No (heat) engine whose working fluid undergoes a cycle can absorb heat from a single reservoir, deliver an equivalent amount of work, and deliver no other effect" |
| Clausius "No machine whose working fluid undergoes a cycle can absorb heat from one system, reject heat to another system and produce no other effect" |
Both statements of the Second Law constrains the First Law of Thermodynamics by identifying that energy goes downhill. The Second Law is concerned with Entropy (S) which is produced by all processes and associated with the loss of ability to do work. The Second Law states that the entropy of the universe increases.
For energy to be available there must be a region with high energy level and a region with low energy level. Useful work must be derived from the energy that flows from the high level to the low level.
- 100% energy can not be transformed to work
- entropy can be produced but never destroyed
Efficiency of a heat machine
The thermodynamic efficiency of a heat machine working between two energy levels is defined in terms of absolute temperature and can be expressed as
η = (Th - Tc) / Th = 1 - Tc / Th (1)
where
η = efficiency
Th = temperature high level (K)
Tc = temperature low level (K)
As a consequence, to attain maximum efficiency - Tc has to be as cold as possible. For 100% efficiency Tc should equals to 0 K. This is in real life impossible so the efficiency will always be less than 1 (100%).
| Change in entropy > 0 irreversible process |
Change in entropy = 0 reversible process |
Change in entropy < 0 impossible process |
Entropy is used to define the unavailable energy in a system and entropy defines the relative ability of one system to act on an other. As energy moves toward lower energy levels where one is less able to act upon the surroundings, the entropy is said to increase.
- For the universe as a whole - the entropy is increasing
- Entropy is not conserved like energy
Go to Thermodyamics key values internationally agreed, Standard state and enthalpy of formation, Gibbs free energy of formation, entropy and heat capacity and Standard enthalpy of formation, Gibbs energy of formation, entropy and molar heat capacity of organic substances for listing of values for a lot of inorganic and organic substances.
Thermodynamic Entropy
Change of entropy in a thermodynamic system can be expressed as
dS = dH / Ta (2)
where
dS = change in entropy (kJ/kg K)
dH = change in enthalpy or internal energy (kJ/kg K)
Ta = average temperature (K)
|
Carnot Heat Cycle The sum of (dH / T) values for each step in the Carnot cycle equals 0. This only happens because for every positive H there is a countering negative H, overall. |
Example - Entropy Heating Water
A process raises 1 kg of water from 0 to 100 oC (273 to 373 K) under atmospheric conditions.
Specific enthalpy at 0 oC (hf) = 0 kJ/kg (from steam tables) (Specific - per unit mass)
Specific enthalpy of water at 100 oC (hf) = 419 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / Ta
= 1.297 kJ/kgK
Example - Entropy Evaporation Water to Steam
A process changes 1 kg of water at 100 oC (373 K) to saturated steam at 100 oC (373 K) under atmospheric conditions.
Specific enthalpy of steam at 100 oC (373 K) before evaporating = 418 kJ/kg (from steam tables)
Specific enthalpy of steam at 100 oC (373 K) after evaporating = 2675 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / Ta
= ((2675 kJ/kg) - (418 kJ/kg)) / (((373 K) + (373 K)) / 2)
= 6.054 kJ/kgK
The total change in specific entropy from water at 100 oC to saturated steam at 100 oC is the sum of the change in specific entropy for the water, plus the change of specific entropy for the steam.
Example - Entropy Superheated Steam
A process superheats 1 kg of saturated steam at atmospheric pressure to 150 oC (423 K).
Specific total enthalpy of steam at 100 oC (373 K) = 2675 kJ/kg (from steam tables)
Specific total enthalpy of superheated steam at 150 oC (423 K) = 2777 kJ/kg (from steam tables)
Change in specific entropy:
dS = dH / Ta
= ((2777 kJ/kg) - (2675 kJ/kg)) / (((423 K) + (373 K)) / 2)
= 0.256 kJ/kgK
Related Topics
• Thermodynamics
Work, heat and energy systems.
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