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In thermodynamics, energy is a fundamental quantity that can neither be created nor destroyed, only transformed from one form to another. However, not all energy is equally useful. Some energy can be fully converted into work, while other forms cannot. This leads us to the concept of exergy, which represents the maximum useful work obtainable from a system as it comes into equilibrium with its surroundings.
Unlike energy, which is conserved, exergy can be destroyed due to irreversibilities such as friction, unrestrained expansion, mixing, and heat transfer across finite temperature differences. Understanding exergy helps engineers evaluate the quality of energy and identify where losses occur in real systems, enabling better design and optimization.
In this section, we will explore the definition, physical meaning, and calculation of exergy, its relation to thermodynamic laws, and practical applications in engineering systems such as power plants, heat exchangers, and refrigeration cycles.
Exergy is defined as the maximum useful work that can be extracted from a system as it is brought into thermodynamic equilibrium with a specified reference environment, commonly called the dead state. The dead state represents the surroundings at ambient conditions where no further work can be extracted.
To understand this, consider a hot cup of tea placed in a room. The tea has energy due to its temperature, but not all this energy can be converted into work. As the tea cools down to room temperature, some energy is lost as heat to the surroundings. The portion of energy that can be converted into work before reaching ambient conditions is the exergy.
Energy can be divided into two parts:
The dead state is characterized by the ambient temperature \( T_0 \), pressure \( P_0 \), and chemical composition of the environment. When a system reaches this state, it has no potential to perform work.
Just as energy balances are fundamental in thermodynamics, exergy balances help us track the useful work potential through a system. The exergy balance for a system can be expressed as:
Here, exergy destruction \(\dot{E}_d\) is a measure of irreversibility in the process. It is always positive or zero, reflecting the second law of thermodynamics. The more irreversible a process, the greater the exergy destruction and the lower the system's efficiency.
graph TD Exergy_Input --> System System --> Exergy_Output System --> Exergy_Destruction[Exergy Destruction (Irreversibility)] System --> Exergy_Loss[Exergy Loss to Environment]
In real engineering systems, minimizing exergy destruction is key to improving performance and reducing waste.
Physical exergy refers to the exergy associated with a system's temperature and pressure relative to the environment. For a fluid at state \((T, P)\) with specific enthalpy \(h\) and entropy \(s\), the physical exergy per unit mass \(\psi\) is given by:
The dead state properties \(h_0\) and \(s_0\) correspond to the fluid at ambient temperature \(T_0\) and pressure \(P_0\). The formula essentially subtracts the unavailable energy portion (related to entropy) from the total energy difference to find the maximum useful work.
| Variable | Description | Typical Units |
|---|---|---|
| \(T\) | Temperature of the system state | K (Kelvin) |
| \(P\) | Pressure of the system state | kPa or MPa |
| \(h\) | Specific enthalpy at system state | kJ/kg |
| \(s\) | Specific entropy at system state | kJ/kg·K |
| \(T_0\) | Ambient (dead state) temperature | K |
| \(P_0\) | Ambient (dead state) pressure | kPa |
| \(h_0\) | Specific enthalpy at dead state | kJ/kg |
| \(s_0\) | Specific entropy at dead state | kJ/kg·K |
The concept of exergy is deeply connected to the Second Law of Thermodynamics, which introduces the idea of irreversibility and entropy generation. While the first law deals with energy conservation, the second law explains why some energy cannot be converted into work.
The Clausius inequality states that for any real process:
This inequality implies that entropy generation is always positive or zero, leading to exergy destruction. Thus, exergy analysis provides a quantitative measure of the second law inefficiencies in a system.
Exergy analysis is a powerful tool used in various engineering fields:
By focusing on exergy rather than just energy, engineers can pinpoint inefficiencies that traditional energy analysis might miss.
Besides physical exergy, other important calculations include:
Exergy analysis bridges the gap between the quantity of energy and its quality, providing a clear picture of where and how useful work potential is lost in thermodynamic systems. By mastering exergy concepts and calculations, students and engineers can design more efficient, sustainable, and cost-effective mechanical systems.
Step 1: Identify given data and convert temperatures to Kelvin.
Ambient temperature \(T_0 = 30 + 273 = 303\,K\).
Step 2: Use physical exergy formula:
\[ \psi = (h - h_0) - T_0 (s - s_0) \]
Step 3: Substitute values:
\[ \psi = (3215 - 125) - 303 \times (6.7 - 0.44) = 3090 - 303 \times 6.26 \]
\[ \psi = 3090 - 1896.78 = 1193.22 \, \text{kJ/kg} \]
Answer: The physical exergy of the steam is approximately 1193 kJ/kg.
Step 1: Write the exergy balance for the heat exchanger:
\[ \dot{E}_{in} = 200 \, \text{kW}, \quad \dot{E}_{out} = 150 \, \text{kW} \]
Step 2: Calculate exergy destruction:
\[ \dot{E}_d = \dot{E}_{in} - \dot{E}_{out} = 200 - 150 = 50 \, \text{kW} \]
Step 3: Calculate exergy efficiency:
\[ \eta_{ex} = \frac{\dot{E}_{out}}{\dot{E}_{in}} = \frac{150}{200} = 0.75 \text{ or } 75\% \]
Answer: Exergy destruction is 50 kW and the heat exchanger operates at 75% exergy efficiency.
Step 1: Recall that the thermal efficiency of an ideal Otto cycle is:
\[ \eta_{thermal} = 1 - \frac{1}{r^{\gamma - 1}} \]
where \(r\) is compression ratio and \(\gamma\) is specific heat ratio. However, for exergy efficiency, we use the second law efficiency concept.
Step 2: The maximum possible work (exergy input) is related to the Carnot efficiency:
\[ \eta_{Carnot} = 1 - \frac{T_0}{T_{max}} = 1 - \frac{300}{2200} = 0.8636 \]
Step 3: Assuming the Otto cycle thermal efficiency is 0.6 (typical value), exergy efficiency is:
\[ \eta_{ex} = \frac{\eta_{thermal}}{\eta_{Carnot}} = \frac{0.6}{0.8636} = 0.695 \]
Answer: The exergy efficiency of the Otto cycle engine is approximately 69.5%.
Step 1: Calculate exergy input and output:
Exergy input = compressor work = 5 kW
Exergy loss = heat rejection = 3 kW
Step 2: Exergy destruction is:
\[ \dot{E}_d = \dot{E}_{in} - \dot{E}_{out} = 5 - 3 = 2 \, \text{kW} \]
Step 3: Major exergy destruction occurs in the expansion valve and condenser due to irreversibility in throttling and heat rejection.
Answer: The major source of exergy destruction is the throttling process and heat rejection in the condenser, accounting for 2 kW of lost work potential.
Step 1: Calculate useful exergy output before and after improvement:
Before: \(1000 \times 0.40 = 400\, \text{kW}\)
After: \(1000 \times 0.60 = 600\, \text{kW}\)
Step 2: Calculate fuel input required to produce 600 kW useful exergy at old efficiency:
\[ \text{Fuel input} = \frac{600}{0.40} = 1500 \, \text{kW} \]
Step 3: Fuel saved due to efficiency improvement:
\[ 1500 - 1000 = 500 \, \text{kW} \]
Step 4: Annual energy saved:
\[ 500 \times 8000 = 4,000,000 \, \text{kWh} \]
Step 5: Annual cost savings:
\[ 4,000,000 \times 5 = Rs. 20,000,000 \]
Answer: The plant saves Rs.2 crore annually by improving exergy efficiency from 40% to 60%.
When to use: At the start of any exergy problem to avoid confusion and errors.
When to use: During all numerical problem solving.
When to use: When checking results for physical feasibility.
When to use: When interpreting results and comparing different systems.
When to use: In problems involving heat exchange between system and surroundings.
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