INSTITUTE OF SPACE AND TECHNOLOGY
PROBLEM BASED LEARNING
Thermodynamics
By
Roshan Raza Khan 230101074
Submitted to:
Dr. Umer Sohail
DATED
10th January 2024
Contents
1. Heat Engine Overview ........................................................................................................ 3
2. Second Law of Thermodynamics ....................................................................................... 3
3. Efficiency and Entropy Generation .................................................................................... 4
4. Comparison with the Carnot Cycle .................................................................................... 4
5. Sources of Irreversibilities ................................................................................................. 5
6. Proposed Enhancements ................................................................................................... 5
7. Example Calculation ........................................................................................................... 5
8. Diagram of the Process ...................................................................................................... 7
9. Conclusions ......................................................................................................................... 8
1. Heat Engine Overview
A heat engine operates between two temperature reservoirs:
• Hot reservoir at temperature 𝑇𝐻 : Supplies heat 𝑄𝐻 .
• Cold reservoir at temperature 𝑇𝐶 : Rejects heat 𝑄𝐶 .
The engine converts part of the heat into useful work 𝑊, following the first law of
thermodynamics:
𝑊 = 𝑄𝐻 − 𝑄𝐶
2. Second Law of Thermodynamics
The second law states:
1. The efficiency of any heat engine is limited by the Carnot efficiency:
𝑇𝐶
𝜂𝐶𝑎𝑟𝑛𝑜𝑡 = 1 −
𝑇𝐻
(Temperatures must be in Kelvin.)
2. Entropy generation (𝛥𝑆𝑔𝑒𝑛 ) is zero for a reversible process and positive for irreversible
processes:
𝛥𝑆𝑔𝑒𝑛 = 𝛥𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + 𝛥𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ≥ 0
3. Efficiency and Entropy Generation
Efficiency of a Real Engine
The efficiency of a real heat engine (𝜂) is:
𝑊 𝑄𝐶
𝜂= = 1−
𝑄𝐻 𝑄𝐻
𝑊 = 𝜂Carnot ⋅ 𝑄𝐻
For a real engine:
• 𝑄𝐶 > 𝑇𝐶 𝛥𝑆 due to irreversibilities (e.g., friction, non-ideal processes).
• 𝜂 < 𝜂𝐶𝑎𝑟𝑛𝑜𝑡 .
Entropy Generation
Entropy generation accounts for irreversibilities:
𝑄𝐻 𝑄𝐶
𝛥𝑆𝑔𝑒𝑛 = − >0
𝑇𝐻 𝑇𝐶
4. Comparison with the Carnot Cycle
Carnot Cycle:-
It provides the maximum possible efficiency a heat engine can achieve when
operating between two temperature reservoirs. While no real engine can operate on the Carnot
cycle, it serves as a theoretical benchmark for efficiency.
The Carnot cycle is an idealized, reversible cycle consisting of:
1. Isothermal Expansion: Heat 𝑄𝐻 is absorbed at 𝑇𝐻 .
2. Adiabatic Expansion: Temperature drops to 𝑇𝐶 without heat exchange.
3. Isothermal Compression: Heat 𝑄𝐶 is rejected at 𝑇𝐶 .
4. Adiabatic Compression: Temperature rises back to 𝑇𝐻 .
The Carnot efficiency sets the upper limit, with no entropy generation (𝛥𝑆𝑔𝑒𝑛 = 0).
Comparison:
• Real engines have lower efficiency due to irreversibilities like:
o Non-adiabatic processes.
o Friction and turbulence in mechanical components.
o Heat transfer through finite temperature differences.
5. Sources of Irreversibilities
1. Friction: Causes energy loss as heat.
2. Finite Heat Transfer: Heat transfer between finite temperature gradients generates
entropy.
3. Non-ideal Working Fluids: Deviations from ideal gas behavior reduce performance.
4. Leakages: Loss of working fluid.
6. Proposed Enhancements
1. Minimize Temperature Gradients: Use better heat exchangers to reduce
irreversibilities.
2. Reduce Friction: Use lubricants and low-friction materials.
3. Optimize Cycle Design: Improve the compression and expansion processes to approach
reversibility.
4. Regenerative Heat Exchangers: Recover heat from the exhaust to preheat incoming
fluid.
5. Use High-Temperature Reservoirs: Increase 𝑇𝐻 to improve efficiency while keeping 𝑇𝐶
low.
7. Example Calculation
Problem Statement:
A Carnot heat engine operates between two temperature reservoirs: a hot reservoir at 𝑻𝑯 =
𝟔𝟎𝟎 K and a cold reservoir at 𝑻𝑪 = 𝟑𝟎𝟎 K. The engine absorbs 𝑸𝑯 = 𝟏𝟓𝟎𝟎 J of heat from the
hot reservoir. Analyze the engine's performance and determine:
The efficiency of the Carnot engine.
The work output.
The heat rejected to the cold reservoir.
The entropy changes in the system and reservoirs.
Step 1: Carnot Efficiency
The Carnot efficiency is given by:
𝑇𝐶
𝜂Carnot = 1 −
𝑇𝐻
Substitute the given temperatures:
300
𝜂Carnot = 1 − = 1 − 0.5 = 0.5 (50%)
600
Thus, the maximum theoretical efficiency is 50%.
Step 2: Work Output
The work output of the engine is related to the heat input and the efficiency:
𝑊 = 𝜂Carnot ⋅ 𝑄𝐻
Substitute the values:
𝑊 = 0.5 ⋅ 1500 = 750 J
The work output is 750 J.
Step 3: Heat Rejected
The heat rejected to the cold reservoir is given by:
𝑄𝐶 = 𝑄𝐻 − 𝑊
Substitute the values:
𝑄𝐶 = 1500 − 750 = 750 J
The heat rejected is 750 J.
Step 4: Entropy Analysis
For a Carnot cycle, the entropy change in the system is zero since it is a reversible process.
However, the entropy changes in the reservoirs must be analyzed.
• Entropy change of the hot reservoir:
𝑄𝐻
𝛥𝑆𝐻 = −
𝑇𝐻
Substitute the values:
1500
𝛥𝑆𝐻 = − = −2.5 J/K
600
• Entropy change of the cold reservoir:
𝑄𝐶
𝛥𝑆𝐶 =
𝑇𝐶
Substitute the values:
750
𝛥𝑆𝐶 = = 2.5 J/K
300
The net entropy change:
𝛥𝑆total = 𝛥𝑆𝐻 + 𝛥𝑆𝐶 = −2.5 + 2.5 = 0
As expected, the entropy change for the Carnot cycle is 0, confirming that it is a reversible
process.
Summary of Results
Carnot efficiency: 𝜂Carnot = 50%
Work output: 𝑊 = 750 J
Heat rejected: 𝑄𝐶 = 750 J
8. Diagram of the Process
The diagram includes:
9. Conclusions
In this example, we analyzed the performance of a Carnot heat engine operating between two
reservoirs: a hot reservoir at 𝑇𝐻 = 600 K and a cold reservoir at 𝑇𝐶 = 300 K. By applying the
fundamental principles of thermodynamics, we were able to calculate the engine’s efficiency,
work output, heat rejection, and entropy changes.
1. Carnot Efficiency: The Carnot efficiency, which represents the maximum theoretical
efficiency of a heat engine operating between two temperature reservoirs, was
calculated as 50%. This value is derived from the temperature difference between the
hot and cold reservoirs, and it sets an upper bound on the efficiency of any real heat
engine. In this case, with the given temperatures of 600 K and 300 K, the Carnot
efficiency represents the most efficient energy conversion possible under these
conditions.
2. Work Output: The engine's work output was found to be 750 J. This value is derived
from the heat input (𝑄𝐻 = 1500 J) and the efficiency of the engine. The work done by
the engine is the portion of the heat input that is converted into useful work, with the
remaining portion being rejected as heat to the cold reservoir.
3. Heat Rejected: The amount of heat rejected to the cold reservoir was calculated to be
750 J, which matches the work output, ensuring energy conservation. The heat rejected
is the portion of the energy absorbed by the engine that cannot be converted into work
due to the inherent limitations set by the second law of thermodynamics. In the Carnot
cycle, the amount of heat rejected is equal to the amount of work output, and this heat
is expelled to the cold reservoir at a lower temperature.
4. Entropy Change: The total entropy change for the Carnot cycle was found to be zero,
𝛥𝑆total = 0 J/K. This result is a direct consequence of the fact that the Carnot cycle is a
reversible process. In an ideal Carnot engine, there is no net generation of entropy
during the operation, which is a hallmark of its theoretical efficiency. While the hot
reservoir undergoes a decrease in entropy (𝛥𝑆𝐻 = −2.5 J/K), the cold reservoir
experiences an equal and opposite increase in entropy (𝛥𝑆𝐶 = 2.5 J/K), leading to no
overall entropy generation.
