Exergy Change Calculations — Compute exergy change for simple systems (intro) - Thermodynamics

SECTION 1

Historical Context & Motivation

The first law of thermodynamics tells us that energy is conserved, but it says nothing about the quality of that energy or how much of it can actually be converted into useful work. A hot reservoir and a lukewarm lake may hold comparable amounts of internal energy, yet one is far more capable of driving a heat engine than the other. This observation—that not all energy is equally useful—motivated the development of exergy (also called availability), a property that combines the first and second laws to quantify the maximum useful work obtainable from a system as it reaches equilibrium with a reference environment. The concept evolved over more than a century, from early reflections on engine efficiency to a mature analytical framework used in modern energy engineering.

1824 • Carnot's Ideal Engine

Sadi Carnot established the theoretical maximum efficiency of heat engines, implicitly recognizing that a temperature difference relative to the surroundings governs work potential—a precursor to the exergy concept.

1873 • Gibbs and Available Energy

Josiah Willard Gibbs introduced the notion of 'available energy' in his foundational papers on thermodynamic equilibrium, formalizing how free energy functions measure the capacity of a system to perform work.

1956 • Rant Coins the Term 'Exergie'

Zoran Rant proposed the term 'exergy' (from Greek ex, 'out of,' and ergon, 'work') to unify various names—availability, essergy, work potential—under a single internationally recognized label.

1980s–present • Exergy Analysis in Engineering

Exergy methods became standard tools for evaluating power plants, refrigeration cycles, and chemical processes, enabling engineers to pinpoint where and why thermodynamic losses occur.

The central question that exergy analysis answers is deceptively simple: Given a system that is not in equilibrium with its environment, how much of its stored energy can, in principle, be converted to useful work? This lesson introduces the mathematical machinery for computing exergy change in simple closed and open systems—an essential skill before tackling full exergy destruction and second-law efficiency analyses.

SECTION 2

Core Principles & Definitions

Before diving into calculations, it is essential to establish the foundational ideas that underpin exergy analysis. Unlike energy, exergy is not conserved; it is destroyed whenever irreversibilities are present. The following core concepts form the scaffolding for all exergy change computations.

Dead State

The reference condition (T₀, P₀) at which a system is in complete thermodynamic equilibrium with its environment. At the dead state, the system has zero exergy because no further work can be extracted.

Exergy (Availability)

The maximum useful work obtainable as a system transitions from its current state to the dead state via reversible processes. It quantifies the work potential of energy rather than its quantity.

Exergy Destruction

Work potential that is irreversibly lost due to friction, heat transfer across finite temperature differences, mixing, and other irreversibilities. By the Gouy–Stodola theorem, exergy destruction equals T₀ times the entropy generated.

Closed vs. Flow Exergy

For a closed (non-flow) system, exergy involves internal energy, while for a steady-flow system it involves enthalpy. Kinetic and potential energy contributions appear in both forms.

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KEY TAKEAWAY

Think of exergy like the purchasing power of currency rather than the face value of a banknote. A $100 bill in a country where everything costs $100 is worthless in practical terms—the bill is at its 'dead state.' Similarly, a system at the same temperature and pressure as its surroundings still possesses internal energy, but none of it can be converted to work. Exergy measures how far the system is from this equilibrium and, therefore, how much useful work is theoretically available.

SECTION 3

Visual Explanation — Exergy on State Diagrams

Visualizing exergy on a familiar thermodynamic diagram helps build geometric intuition. The diagram below depicts a T–s (temperature–entropy) representation for a closed system undergoing a state change, with the dead-state properties marked for reference. The shaded regions illustrate the portions of energy that correspond to exergy versus the unavailable energy that cannot be converted to work.

On a T–s diagram, the area beneath the process curve but above the dead-state temperature line T₀ represents the exergy (cyan region), while the area below T₀ represents the unavailable energy (violet region). Moving from State 1 to State 2 changes both regions; the exergy change is the difference in the cyan-shaded areas between the two states.

The diagram makes a crucial point visually: as entropy increases (moving rightward), the unavailable-energy rectangle grows, which means the exergy of the system diminishes. Irreversibilities within the system generate entropy and thereby shift useful work potential into the unavailable category. Conversely, removing heat from a system at a temperature above T₀ reduces both entropy and unavailable energy, thereby increasing the fraction of stored energy that is exergy.

SECTION 4

Mathematical Framework

The general expression for the specific exergy (also called specific availability) of a simple compressible system can be derived by imagining a reversible process that brings the system from an arbitrary state to the dead state while exchanging heat solely with the environment at T₀ and doing boundary work against the atmosphere at P₀. The two principal forms are the non-flow (closed-system) exergy and the flow (open-system) exergy.

Closed-System (Non-Flow) Exergy

NON-FLOW SPECIFIC EXERGY

ϕ = (u − u₀) + P₀(v − v₀) − T₀(s − s₀) + V²/2 + gz

where u = specific internal energy, v = specific volume, s = specific entropy, V = velocity, g = gravitational acceleration, z = elevation, and subscript 0 denotes the dead-state property.

The first three terms represent the thermomechanical exergy: the combination (u − u₀) + P₀(v − v₀) − T₀(s − s₀) captures both the internal energy difference and the adjustments for boundary work against P₀ and the entropy 'tax' imposed by the second law. The kinetic and potential energy terms are fully convertible to work, so they enter the expression undiminished.

Flow (Steady-State) Exergy

FLOW SPECIFIC EXERGY

ψ = (h − h₀) − T₀(s − s₀) + V²/2 + gz

where h = specific enthalpy. This form applies at each inlet and exit of a control volume. Note that the P₀v terms are absorbed into enthalpy, simplifying the expression compared to the closed-system form.

Exergy Change Between Two States

CLOSED-SYSTEM EXERGY CHANGE

Δϕ = ϕ₂ − ϕ₁ = (u₂ − u₁) + P₀(v₂ − v₁) − T₀(s₂ − s₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)

Dead-state properties cancel when taking the difference. For stationary systems with negligible KE and PE changes, the last two terms vanish.

FLOW EXERGY CHANGE

Δψ = ψ₂ − ψ₁ = (h₂ − h₁) − T₀(s₂ − s₁) + (V₂² − V₁²)/2 + g(z₂ − z₁)

This is the form most commonly used for turbines, compressors, heat exchangers, and nozzles operating at steady state.

ℹ

Sign Convention

A positive Δϕ (or Δψ) means the system's exergy increased—its work potential grew. A negative value means exergy was either transferred out (e.g., as work) or destroyed by irreversibilities. In any real (irreversible) process, the total exergy of the universe decreases.

SECTION 5

Detailed Breakdown — Components of Exergy Change

It is instructive to decompose the exergy change into its constituent parts and examine each one's physical significance. The diagram below separates the closed-system exergy change into its thermomechanical, kinetic, and potential contributions, showing how each term relates to the overall work potential of the system.

The exergy change breaks into three additive components. The thermomechanical term captures internal energy, atmospheric boundary work, and the entropy penalty. The kinetic and potential terms are pure mechanical energy contributions that are entirely convertible to useful work.

The entropy penalty term −T₀(s₂ − s₁) deserves special attention. When entropy increases (s₂ > s₁), this term is negative, meaning the exergy decreases—entropy generation has eroded work potential. This is the mathematical expression of the second law's mandate: irreversibilities always reduce the capacity to do useful work. In contrast, the P₀(v₂ − v₁) term accounts for the fact that any volume change of the system does work against (or receives work from) the atmosphere at pressure P₀, and that portion of work is not 'useful' to the engineer.

Term	Physical Meaning	Positive When…
u₂ − u₁	Change in internal energy stored in the system	System gains internal energy (e.g., heated)
P₀(v₂ − v₁)	Work exchanged with the atmosphere during volume change	System expands (pushes atmosphere back)
−T₀(s₂ − s₁)	Entropy 'penalty' — work potential lost due to disorder	System entropy decreases (becomes more ordered)
(V₂² − V₁²)/2	Change in kinetic energy (100% exergy)	System speeds up
g(z₂ − z₁)	Change in gravitational potential energy (100% exergy)	System moves to higher elevation

Summary of each term in the closed-system exergy change expression.

SECTION 6

Worked Example — Steam in a Closed Tank

Consider a rigid, insulated tank containing 2 kg of steam that undergoes an irreversible process. We wish to calculate the change in exergy. The dead-state environment is at T₀ = 25 °C (298.15 K) and P₀ = 100 kPa. The tank is stationary, so kinetic and potential energy changes are zero, and because the tank is rigid, v₁ = v₂.

Exergy Change of Steam in a Rigid Tank

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Step 1 — State the Given Information

Mass m = 2 kg. State 1: T₁ = 400 °C, P₁ = 600 kPa → from superheated steam tables, u₁ = 2962.1 kJ/kg, s₁ = 7.7086 kJ/(kg·K), v₁ = 0.5137 m³/kg. State 2: T₂ = 250 °C (pressure drops to ≈ 220 kPa at same specific volume) → u₂ = 2731.4 kJ/kg, s₂ = 7.7956 kJ/(kg·K), v₂ = v₁ = 0.5137 m³/kg. Dead-state: T₀ = 298.15 K, P₀ = 100 kPa.
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Step 2 — Write the Exergy Change Expression

Since the tank is rigid (v₂ = v₁) and stationary (ΔKE = ΔPE = 0), the exergy change per unit mass simplifies to: Δϕ = (u₂ − u₁) + P₀(v₂ − v₁) − T₀(s₂ − s₁). The volume term P₀(v₂ − v₁) = 0 because the specific volume does not change.
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Step 3 — Calculate Each Term

Internal energy change: u₂ − u₁ = 2731.4 − 2962.1 = −230.7 kJ/kg. Entropy penalty: −T₀(s₂ − s₁) = −298.15 × (7.7956 − 7.7086) = −298.15 × 0.0870 = −25.94 kJ/kg.

Δu = −230.7 kJ/kg; −T₀Δs = −25.94 kJ/kg
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Step 4 — Sum the Terms for Specific Exergy Change

Δϕ = −230.7 + 0 + (−25.94) = −256.6 kJ/kg.

Δϕ = −256.6 kJ/kg
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Step 5 — Total Exergy Change for the System

Multiply by mass: ΔΦ = m × Δϕ = 2 × (−256.6) = −513.2 kJ. The negative sign confirms that the system's work potential decreased. Part of this decrease is exergy destruction due to the irreversible process inside the insulated tank.

ΔΦ = −513.2 kJ (exergy decreased)

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Interpretation

Both the internal-energy decrease and the entropy increase contribute to the exergy loss. Even though the tank is insulated (no heat transfer out), the irreversible process inside generated entropy, which further reduced the system's ability to do useful work. This is a concrete example of exergy destruction.

SECTION 7

Energy Analysis vs. Exergy Analysis

Students sometimes ask why exergy analysis is needed when energy balances already provide a complete accounting. The table below highlights the distinct strengths and limitations of each approach, making clear that exergy analysis provides information about the quality and usefulness of energy that first-law-only analysis cannot.

Criterion	Energy (1st Law) Analysis	Exergy (2nd Law) Analysis
Conservation	Energy is always conserved	Exergy is destroyed by irreversibilities
Quality of energy	Treats all energy forms equally	Distinguishes high-quality (work) from low-quality (waste heat)
Locating losses	Identifies where energy exits the system	Pinpoints where and how much work potential is destroyed
Reference environment	Not required	Dead-state (T₀, P₀) must be specified
Efficiency metric	Thermal (first-law) efficiency: η₁ = W_net / Q_in	Second-law (exergetic) efficiency: η₂ = exergy recovered / exergy supplied

Energy analysis vs. exergy analysis comparison

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KEY TAKEAWAY

Energy analysis is like tracking total dollars flowing through a business—it tells you the revenue and expenses balance. Exergy analysis is like tracking profit and loss: it reveals where value is being wasted, even when the books balance. An engineer who performs only energy analysis might conclude a boiler is 90% efficient, while an exergy analysis reveals that 50% of the work potential of the fuel was irreversibly destroyed due to large temperature gradients during combustion.

SECTION 8

Connection to Advanced Exergy Topics

The exergy change calculations introduced here form the foundation for more advanced analyses encountered in upper-division courses and graduate research. Understanding how the introductory material connects to these advanced topics helps place the current lesson within the broader discipline.

Introductory Concept (This Lesson)	Advanced Extension
Specific exergy ϕ or ψ at a single state	Exergy balance equation for control volumes with multiple inlets/exits, heat transfer at varying T, and shaft work
Exergy change Δϕ between two states	Gouy–Stodola theorem: X_destroyed = T₀ · S_gen, linking exergy destruction to entropy generation
Dead state defined by T₀, P₀ only	Chemical exergy: dead state includes chemical equilibrium with the environment; accounts for fuel combustion potential
Second-law efficiency η₂ (simple ratio)	Thermoeconomics (exergoeconomics): assigning monetary costs to exergy streams to optimize plant economics

From introductory to advanced exergy analysis

In future coursework, you will encounter the full exergy balance equation (analogous to the energy balance but with an exergy-destruction term), which enables systematic optimization of multi-component systems such as combined-cycle power plants, cogeneration systems, and chemical reactors. The change-of-exergy calculations practiced here are the building blocks for those analyses: once you can compute ψ at each state point, assembling the full balance is a matter of careful bookkeeping.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain, in your own words, why the exergy of a system is zero at the dead state even though the system still possesses internal energy. Why can no useful work be extracted?

Show answer

At the dead state (T₀, P₀), the system is in complete thermomechanical equilibrium with its surroundings. There is no temperature difference to drive a heat engine and no pressure difference to drive expansion work beyond what the atmosphere resists. While the system retains internal energy, none of it can be channeled into useful work because every potential driving force—temperature gradient, pressure gradient—has vanished. Exergy measures the departure from this equilibrium; at the dead state the departure is zero, hence exergy is zero.

Question 2: BASIC CALCULATION

Air (ideal gas,

c_v

= 0.718 kJ/(kg·K),

c_p

= 1.005 kJ/(kg·K), R = 0.287 kJ/(kg·K)) in a rigid tank is cooled from T₁ = 500 K to T₂ = 350 K. The dead-state conditions are T₀ = 300 K, P₀ = 100 kPa. Neglect KE and PE. Compute the specific exergy change Δϕ.

Show answer

For a rigid tank, v₂ = v₁, so the P₀(v₂ − v₁) term is zero. For an ideal gas with constant specific heats: u₂ − u₁ =

c_v

(T₂ − T₁) = 0.718 × (350 − 500) = −107.7 kJ/kg. The entropy change at constant volume is s₂ − s₁ =

c_v

ln(T₂/T₁) = 0.718 × ln(350/500) = 0.718 × (−0.3567) = −0.2561 kJ/(kg·K). The entropy penalty is −T₀(s₂ − s₁) = −300 × (−0.2561) = +76.83 kJ/kg. Therefore Δϕ = −107.7 + 0 + 76.83 = −30.87 kJ/kg. The exergy decreased by about 30.9 kJ/kg, reflecting both the loss of internal energy and the partial offset from reduced entropy.

Question 3: INTERMEDIATE

Steam enters a well-insulated turbine at P₁ = 4 MPa, T₁ = 500 °C (h₁ = 3445.3 kJ/kg, s₁ = 7.0901 kJ/(kg·K)) and exits at P₂ = 100 kPa with a quality of x₂ = 1.0 (h₂ = 2675.5 kJ/kg, s₂ = 7.3594 kJ/(kg·K)). Neglect KE and PE changes. With T₀ = 298.15 K, compute the change in flow exergy Δψ between inlet and exit.

Show answer

The flow exergy change is Δψ = (h₂ − h₁) − T₀(s₂ − s₁). Enthalpy change: h₂ − h₁ = 2675.5 − 3445.3 = −769.8 kJ/kg. Entropy change: s₂ − s₁ = 7.3594 − 7.0901 = 0.2693 kJ/(kg·K). Entropy penalty: −T₀(s₂ − s₁) = −298.15 × 0.2693 = −80.3 kJ/kg. Therefore Δψ = −769.8 + (−80.3) = −850.1 kJ/kg. The flow exergy decreased by 850.1 kJ/kg. Part of this decrease appeared as turbine work output, and the remainder (proportional to T₀ × entropy generated) was destroyed by irreversibilities inside the turbine.

Question 4: APPLIED

A piston–cylinder device contains 0.5 kg of refrigerant R-134a at State 1: P₁ = 800 kPa, T₁ = 50 °C (u₁ = 263.9 kJ/kg, v₁ = 0.02846 m³/kg, s₁ = 1.0309 kJ/(kg·K)). The refrigerant is cooled at constant pressure to State 2: T₂ = 20 °C (u₂ = 227.8 kJ/kg, v₂ = 0.02567 m³/kg, s₂ = 0.9105 kJ/(kg·K)). The dead state is T₀ = 25 °C (298.15 K), P₀ = 100 kPa. Compute the total exergy change ΔΦ for this process.

Show answer

Because this is a closed system with changing volume: Δϕ = (u₂ − u₁) + P₀(v₂ − v₁) − T₀(s₂ − s₁). Internal energy: u₂ − u₁ = 227.8 − 263.9 = −36.1 kJ/kg. Boundary work against atmosphere: P₀(v₂ − v₁) = 100 × (0.02567 − 0.02846) = 100 × (−0.00279) = −0.279 kJ/kg. Entropy penalty: −T₀(s₂ − s₁) = −298.15 × (0.9105 − 1.0309) = −298.15 × (−0.1204) = +35.90 kJ/kg. Specific exergy change: Δϕ = −36.1 + (−0.279) + 35.90 = −0.479 kJ/kg. Total: ΔΦ = 0.5 × (−0.479) ≈ −0.24 kJ. The exergy change is small and negative, indicating that most of the exergy lost via internal energy decrease was nearly offset by the entropy decrease (the system became more ordered as it cooled).

Question 5: CRITICAL THINKING

A student argues: 'If I heat a gas in a rigid tank from T₀ to some higher temperature T₁, the exergy change must equal the heat added Q, since all the heat raises the system's work potential.' Critically evaluate this claim using the exergy change formula, and identify the flaw in the reasoning.

Show answer

The claim is incorrect. While the heat Q does increase the internal energy by

c_v

(T₁ − T₀), the exergy change also includes the entropy penalty −T₀(s₁ − s₀). For a rigid-tank ideal gas, Δϕ =

c_v

(T₁ − T₀) − T₀

c_v

ln(T₁/T₀). The logarithmic term is always larger than zero for T₁ > T₀, so the exergy gained is strictly less than the heat added. Physically, the heat transfer raises both the energy and the entropy of the system; the second-law 'tax' of T₀ΔS accounts for the fraction of added energy that becomes unavailable. Only if the heat were delivered reversibly by a Carnot-like process at infinitesimally varying temperatures would every increment of heat transfer fully convert to exergy—a practically impossible scenario. The ratio Δϕ/Q can be shown to equal 1 − T₀ ln(T₁/T₀)/(T₁ − T₀), which is always less than 1 for finite temperature differences, confirming that heat and exergy are not equivalent.

SUMMARY

Lesson Summary

Summary

Exergy (availability) quantifies the maximum useful work a system can deliver as it reaches the dead state (T₀, P₀). For a closed system, the specific exergy is ϕ = (u − u₀) + P₀(v − v₀) − T₀(s − s₀) + V²/2 + gz, while for a steady-flow system it is ψ = (h − h₀) − T₀(s − s₀) + V²/2 + gz. Computing the exergy change between two states is straightforward: form the difference Δϕ = ϕ₂ − ϕ₁ (or Δψ = ψ₂ − ψ₁), noting that dead-state properties cancel. The entropy penalty −T₀Δs captures how irreversibilities erode work potential, embodying the essence of the second law.

Unlike energy, exergy is not conserved—it is destroyed whenever real (irreversible) processes occur. This makes exergy analysis a powerful diagnostic tool for identifying and minimizing thermodynamic losses in engineering systems. Mastering these introductory calculations prepares you for full exergy balance equations, second-law efficiency calculations, and ultimately thermoeconomic optimization of complex energy systems.