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Availability and exergy

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Introduction to Availability and Exergy

In thermodynamics, understanding how energy can be converted into useful work is crucial for designing efficient engineering systems. While the First Law of Thermodynamics tells us that energy is conserved, it does not tell us how much of that energy can actually be used to perform work. This is where the concepts of availability and exergy come into play.

Availability refers to the maximum useful work that can be extracted from a system as it comes into equilibrium with its surroundings. Exergy is a quantitative measure of this useful work potential. Unlike energy, exergy is not conserved; it can be destroyed due to irreversibilities such as friction, unrestrained expansion, or heat transfer through a finite temperature difference.

To analyze exergy, we must define a reference environment or dead state. This is a hypothetical state representing the surroundings with which the system eventually equilibrates. The dead state is characterized by fixed temperature, pressure, and chemical composition, typically those of the ambient environment.

Understanding availability and exergy helps engineers identify where and how energy quality is lost in processes, enabling improvements in efficiency and cost savings. For example, in power plants or refrigeration cycles, exergy analysis reveals the true potential for work and highlights irreversibilities that reduce performance.

Availability and Exergy Definitions

Let's start by defining these key terms clearly:

  • Availability: The maximum useful work obtainable from a system as it is brought into equilibrium with the environment by reversible processes.
  • Exergy: The measure of a system's potential to cause change or do useful work relative to the environment.
  • Dead State: The state of the system when it is in complete equilibrium with the environment - no further work can be extracted.

Imagine a hot cup of tea in a room. The tea has thermal energy, but only part of that energy can be converted into work (like running a small engine). As the tea cools to room temperature (the dead state), its ability to do work disappears. The difference between the tea's current state and the dead state represents its exergy.

System Environment (Dead State) Exergy Flow

This diagram illustrates a system exchanging energy with its environment. The arrow represents the exergy flow - the useful work potential that can be extracted before the system reaches the dead state.

Exergy Balance Equations

To quantify exergy changes in a system, we use exergy balance equations derived from the First and Second Laws of Thermodynamics. These balances account for exergy entering, leaving, and being destroyed within the system.

For a closed system (no mass crossing the boundary), the exergy balance is:

\[ \Delta E_x = E_{x,in} - E_{x,out} - E_{x,destroyed} \]
Change in exergy equals exergy input minus output and destruction

Here, \( E_{x,destroyed} \) represents exergy lost due to irreversibility - energy that cannot be converted into useful work.

For an open system (with mass flow), the exergy balance includes exergy carried by mass entering and leaving, as well as heat and work interactions:

graph TD    A[Exergy In]    B[Exergy Out]    C[Exergy Destruction]    D[Change in System Exergy]    A --> D    D --> B    D --> C

The exergy destruction term is directly related to entropy generation \( S_{gen} \) by:

\[ E_{x,destroyed} = T_0 \times S_{gen} \]
Exergy destroyed equals environment temperature times entropy generated

This shows that irreversibility (entropy generation) reduces the useful work potential of a system.

Types of Exergy

Exergy can be categorized based on the form of energy involved:

Type of Exergy Definition Typical Engineering Applications
Physical Exergy Exergy due to temperature and pressure differences from the environment. Steam turbines, heat exchangers, refrigeration cycles.
Chemical Exergy Exergy due to chemical composition differences relative to environment. Fuel combustion, chemical reactors, fuel cells.
Kinetic Exergy Exergy due to velocity of the system relative to surroundings. Jet engines, wind turbines, fluid flow systems.
Potential Exergy Exergy due to elevation or position in a gravitational field. Hydroelectric power, pumped storage systems.

In many practical problems, physical and chemical exergy dominate, but kinetic and potential exergy must be considered when velocities or elevations are significant.

Formula Bank

Formula Bank

Physical Exergy
\[ e = (h - h_0) - T_0 (s - s_0) \]
where: \( e \) = physical exergy (kJ/kg), \( h \) = enthalpy of system (kJ/kg), \( h_0 \) = enthalpy at dead state (kJ/kg), \( T_0 \) = environment temperature (K), \( s \) = entropy of system (kJ/kg·K), \( s_0 \) = entropy at dead state (kJ/kg·K)
Exergy Balance for Closed System
\[ \Delta E_x = E_{x,in} - E_{x,out} - E_{x,destroyed} \]
where: \( \Delta E_x \) = change in exergy, \( E_{x,in} \) = exergy input, \( E_{x,out} \) = exergy output, \( E_{x,destroyed} \) = exergy destroyed
Exergy Efficiency
\[ \eta_{ex} = \frac{Useful\ Exergy\ Output}{Exergy\ Input} \]
where: \( \eta_{ex} \) = exergy efficiency
Exergy Destruction
\[ E_{x,destroyed} = T_0 \times S_{gen} \]
where: \( E_{x,destroyed} \) = exergy destruction (kJ), \( T_0 \) = environment temperature (K), \( S_{gen} \) = entropy generated (kJ/K)

Worked Examples

Example 1: Calculating Physical Exergy of Steam Medium
Calculate the physical exergy of steam at 400°C and 2 MPa relative to an environment at 30°C and 100 kPa. Given steam properties: enthalpy \( h = 3214 \) kJ/kg, entropy \( s = 6.5 \) kJ/kg·K. At dead state (30°C, 100 kPa), \( h_0 = 125 \) kJ/kg, \( s_0 = 0.43 \) kJ/kg·K. Environment temperature \( T_0 = 303 \) K.

Step 1: Write the physical exergy formula:

\( e = (h - h_0) - T_0 (s - s_0) \)

Step 2: Substitute values (ensure temperature in K):

\( e = (3214 - 125) - 303 \times (6.5 - 0.43) \)

\( e = 3089 - 303 \times 6.07 = 3089 - 1838.21 = 1250.79 \) kJ/kg

Answer: The physical exergy of the steam is approximately 1251 kJ/kg.

Example 2: Exergy Destruction in a Heat Exchanger Hard
A counterflow heat exchanger transfers heat from hot oil at 150°C to water at 30°C. The oil mass flow rate is 2 kg/s with specific heat \( c_p = 2 \) kJ/kg·K, cooling to 90°C. Water mass flow rate is 3 kg/s, heating to 70°C with \( c_p = 4.18 \) kJ/kg·K. Environment temperature is 30°C. Calculate the exergy destruction and exergy efficiency of the heat exchanger.

Step 1: Calculate heat transfer from oil:

\( Q_{oil} = m_{oil} c_p (T_{in} - T_{out}) = 2 \times 2 \times (150 - 90) = 240 \) kW

Step 2: Calculate heat gained by water:

\( Q_{water} = m_{water} c_p (T_{out} - T_{in}) = 3 \times 4.18 \times (70 - 30) = 501.6 \) kW

Note: Heat gained by water is higher due to different mass flow and specific heat; assume ideal mixing for exergy calculation.

Step 3: Calculate exergy change for each fluid using physical exergy approximation:

Exergy change for oil:

\( \Delta e_{oil} = c_p [ (T - T_0) - T_0 \ln \frac{T}{T_0} ] \)

Calculate at inlet (150 + 273 = 423 K) and outlet (90 + 273 = 363 K), \( T_0 = 303 \) K:

\( \Delta e_{oil,in} = 2 \times [ (423 - 303) - 303 \ln \frac{423}{303} ] \)

\( = 2 \times [120 - 303 \times 0.343] = 2 \times (120 - 104) = 32 \) kJ/kg

\( \Delta e_{oil,out} = 2 \times [ (363 - 303) - 303 \ln \frac{363}{303} ] \)

\( = 2 \times [60 - 303 \times 0.180] = 2 \times (60 - 54.5) = 11 \) kJ/kg

Exergy lost by oil per kg: \( 32 - 11 = 21 \) kJ/kg

Total exergy lost by oil: \( 21 \times 2 = 42 \) kW

Similarly, calculate exergy gained by water:

Water inlet: 30 + 273 = 303 K, outlet: 70 + 273 = 343 K

\( \Delta e_{water,in} = 4.18 \times [ (303 - 303) - 303 \ln 1 ] = 0 \)

\( \Delta e_{water,out} = 4.18 \times [ (343 - 303) - 303 \ln \frac{343}{303} ] \)

\( = 4.18 \times [40 - 303 \times 0.125] = 4.18 \times (40 - 37.9) = 8.7 \) kJ/kg

Total exergy gained by water: \( 8.7 \times 3 = 26.1 \) kW

Step 4: Calculate exergy destruction:

\( E_{x,destroyed} = Exergy_{lost} - Exergy_{gained} = 42 - 26.1 = 15.9 \) kW

Step 5: Calculate exergy efficiency:

\( \eta_{ex} = \frac{Exergy_{gained}}{Exergy_{lost}} = \frac{26.1}{42} = 0.62 \) or 62%

Answer: Exergy destruction is 15.9 kW and exergy efficiency is 62%.

Example 3: Exergy Analysis of a Gas Turbine Hard
A gas turbine operates with air entering at 1 bar, 300 K and leaving at 0.1 bar, 900 K. The mass flow rate is 5 kg/s. Calculate the exergy destruction and lost work if the environment is at 300 K and 1 bar. Assume ideal gas with \( c_p = 1.005 \) kJ/kg·K and \( R = 0.287 \) kJ/kg·K.

Step 1: Calculate inlet and outlet enthalpy and entropy:

Inlet enthalpy \( h_1 = c_p T_1 = 1.005 \times 300 = 301.5 \) kJ/kg

Outlet enthalpy \( h_2 = 1.005 \times 900 = 904.5 \) kJ/kg

Entropy change for ideal gas:

\( s_2 - s_1 = c_p \ln \frac{T_2}{T_1} - R \ln \frac{P_2}{P_1} \)

\( = 1.005 \ln \frac{900}{300} - 0.287 \ln \frac{0.1}{1} = 1.005 \times 1.0986 - 0.287 \times (-2.3026) = 1.104 + 0.661 = 1.765 \) kJ/kg·K

Assuming dead state at \( T_0 = 300 \) K, \( P_0 = 1 \) bar, entropy \( s_0 = s_1 \) (reference), so \( s_0 = s_1 \).

Step 2: Calculate physical exergy at inlet and outlet:

At inlet, \( e_1 = (h_1 - h_0) - T_0 (s_1 - s_0) = 0 \) (since system at environment state)

At outlet, \( e_2 = (h_2 - h_0) - T_0 (s_2 - s_0) = (904.5 - 301.5) - 300 \times 1.765 = 603 - 529.5 = 73.5 \) kJ/kg

Step 3: Exergy destruction is difference between input and output exergy minus work output:

Assuming no work output given, exergy destruction \( E_{x,destroyed} = m (e_1 - e_2) = 5 \times (0 - 73.5) = -367.5 \) kW

Negative value indicates exergy loss (destruction) of 367.5 kW.

Step 4: Lost work equals exergy destruction, so 367.5 kW.

Answer: Exergy destruction and lost work in the gas turbine are 367.5 kW.

Example 4: Cost Implication of Exergy Loss Medium
An industrial process has an exergy destruction rate of 500 kW. The cost of fuel is Rs.5 per kWh. Estimate the monetary loss per hour due to exergy destruction.

Step 1: Convert exergy destruction to kWh:

1 kW = 1 kJ/s, so 500 kW = 500 kJ/s

Energy lost in 1 hour = \( 500 \times 3600 = 1,800,000 \) kJ

Convert kJ to kWh: \( \frac{1,800,000}{3600} = 500 \) kWh

Step 2: Calculate cost of lost exergy:

Cost = 500 kWh x Rs.5/kWh = Rs.2500

Answer: The monetary loss due to exergy destruction is Rs.2500 per hour.

Example 5: Exergy Balance for an Open System Medium
Air enters a compressor at 100 kPa, 300 K and leaves at 500 kPa, 450 K. The mass flow rate is 1.5 kg/s. Calculate the exergy input, output, and destruction. Environment conditions are 100 kPa and 300 K. Assume ideal gas with \( c_p = 1.005 \) kJ/kg·K and \( R = 0.287 \) kJ/kg·K.

Step 1: Calculate inlet and outlet enthalpy and entropy:

Inlet enthalpy \( h_1 = c_p T_1 = 1.005 \times 300 = 301.5 \) kJ/kg

Outlet enthalpy \( h_2 = 1.005 \times 450 = 452.25 \) kJ/kg

Entropy change:

\( s_2 - s_1 = c_p \ln \frac{T_2}{T_1} - R \ln \frac{P_2}{P_1} \)

\( = 1.005 \ln \frac{450}{300} - 0.287 \ln \frac{500}{100} = 1.005 \times 0.4055 - 0.287 \times 1.609 = 0.408 - 0.462 = -0.054 \) kJ/kg·K

Step 2: Calculate physical exergy at inlet and outlet:

At inlet (dead state): \( e_1 = 0 \)

At outlet:

\( e_2 = (h_2 - h_0) - T_0 (s_2 - s_0) = (452.25 - 301.5) - 300 \times (-0.054) = 150.75 + 16.2 = 166.95 \) kJ/kg

Step 3: Calculate exergy input and output rates:

Exergy input \( E_{x,in} = m \times e_1 = 1.5 \times 0 = 0 \) kW

Exergy output \( E_{x,out} = 1.5 \times 166.95 = 250.4 \) kW

Step 4: Since work is done on the system, exergy destruction is:

\( E_{x,destroyed} = E_{x,in} + W_{in} - E_{x,out} \)

Assuming work input \( W_{in} \) equals enthalpy increase (ideal compressor), \( W_{in} = m (h_2 - h_1) = 1.5 \times (452.25 - 301.5) = 225.4 \) kW

Therefore, \( E_{x,destroyed} = 0 + 225.4 - 250.4 = -25 \) kW (negative indicates exergy destruction is zero or negligible, actual irreversibility is small)

Answer: Exergy input is 0 kW, output is 250.4 kW, and exergy destruction is approximately zero, indicating an efficient compression process.

Tips & Tricks

Tip: Always define the dead state/environment conditions clearly before starting calculations.

When to use: At the beginning of any exergy problem to avoid confusion.

Tip: Use entropy generation to quickly find exergy destruction instead of detailed energy balances.

When to use: When entropy data is available and irreversibility needs to be quantified.

Tip: Remember that exergy is zero at dead state; use this to simplify calculations.

When to use: When setting reference conditions or checking answers.

Tip: Convert all units to metric (SI) before calculations to maintain consistency.

When to use: Always, especially in competitive exams.

Tip: Relate exergy loss to cost by multiplying lost exergy with fuel cost per kJ for practical insights.

When to use: In problems involving economic analysis of thermodynamic systems.

Common Mistakes to Avoid

❌ Confusing energy with exergy and treating them as the same.
✓ Understand that energy is conserved but exergy is destroyed due to irreversibility.
Why: Students often overlook the second law implications and focus only on energy.
❌ Using incorrect or inconsistent environmental (dead state) conditions.
✓ Always specify and use consistent reference environment parameters for all calculations.
Why: Environmental conditions affect exergy values significantly; inconsistency leads to errors.
❌ Ignoring exergy destruction term in exergy balance equations.
✓ Include exergy destruction to account for irreversibility and lost work.
Why: Neglecting this leads to overestimation of system performance.
❌ Mixing units, especially temperature in Celsius instead of Kelvin.
✓ Convert temperatures to Kelvin before using in formulas involving entropy and exergy.
Why: Thermodynamic formulas require absolute temperature scales.
❌ Forgetting to include kinetic and potential exergy when significant.
✓ Include all relevant exergy components based on problem context.
Why: Omission can lead to inaccurate exergy accounting.
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