Thermodynamics · Chapter 8

Exergy

Exergy measures work potential relative to a specified environment. Energy is conserved, whereas exergy is destroyed by irreversibility, making exergy analysis a practical way to locate improvement opportunities without promising that every loss can be recovered.

Learning objectives

Define physical and flow exergy relative to a stated dead state.
Distinguish reversible work, useful work, irreversibility, and energy transfer.
Apply exergy balances to closed systems and steady control volumes.
Calculate heat, work, and mass exergy transfers with consistent signs.
Audit exergy destruction, second-law efficiency, and uncertainty using entropy generation.

8–1

Exergy: Work Potential of Energy

Environment-dependent value

A system at \(T_0,P_0\) with no kinetic, potential, chemical, or other disequilibrium is at the dead state and has zero physical exergy. Exergy therefore depends on both state and reference environment.

Physical meaning

Two streams can carry equal energy but unequal ability to produce work. Exergy assigns value to temperature, pressure, motion, and elevation differences that can drive change.

8–2

Reversible Work and Irreversibility

Gouy–Stodola relation

\[X_{dest}=I=T_0S_{gen}\ge0\]

Reversible work is the maximum output or minimum input for the defined end states and environment. It is not automatically shaft work; subtract atmospheric displacement work when defining useful work for a closed system.

8–3

Second-Law Efficiency

Second-law efficiency compares desired exergy product with required exergy fuel. State numerator and denominator explicitly: a compressor, heat exchanger, and engine do not share one universal ratio. Thermal efficiency and COP remain first-law measures and must not be substituted.

8–4

Exergy Change of a System

Physical exergy forms

\[\phi=(u-u_0)+P_0(v-v_0)-T_0(s-s_0)+\frac{V^2}{2}+gz\] \[\psi=(h-h_0)-T_0(s-s_0)+\frac{V^2}{2}+gz\]

Use \(\phi\) for a closed mass and \(\psi\) for a flowing stream. Property differences must come from one internally consistent authoritative source and reference convention.

8–5

Exergy Transfer by Heat, Work, and Mass

Transfer mechanisms

\[X_Q=\int\left(1-\frac{T_0}{T_b}\right)\delta Q,\qquad X_m=m\psi\]

Work transfers exergy one-for-one after the useful-work convention is applied. Heat crossing below the environment can carry negative exergy under the sign convention; use the actual boundary temperature distribution.

EXAMPLE 1Work potential of kiln heat

A batch kiln supplies 480 kJ of heat through a boundary maintained at 750 K to equipment in a 300 K environment. Determine the exergy accompanying that heat transfer.

Recognition strategy

Heat transfer at a specified boundary temperature and a stated environment signals the Carnot factor. Exergy is not the heat amount; it is the maximum useful-work potential relative to the dead state.

Given information

Closed receiving-system viewpoint for one transfer event; no mass crosses the selected boundary.
Boundary and reference temperatures are absolute: \(T_b=750\ \text{K}\), \(T_0=300\ \text{K}\).
Given heat into the system is \(Q=+480\ \text{kJ}\); useful work out is positive.
No property table is needed; heat is treated at uniform boundary temperature, with KE and PE irrelevant.

Model and interpretation

Use \(X_Q=(1-T_0/T_b)Q\). The relation is exact only for heat crossing at the stated uniform temperature; a varying boundary requires integration.

Step-by-step solution

1. Define reference. Set the dead-state environment at the supplied \(300\ \text{K}\), against which work potential is measured.

2. Formulate. The exergy fraction is \(1-T_0/T_b\), because the remainder must be rejected to the environment.

3. Substitute. \(1-300/750=0.600\), a dimensionless availability factor multiplying \(480\ \text{kJ}\).

4. Calculate. The heat-transfer exergy is therefore \(X_Q=0.600(480)=288\ \text{kJ}\).

Verification

Units: A dimensionless Carnot factor multiplying heat leaves the required energy unit, kJ.

Signs and direction: Heat entering gives positive exergy transfer; reversing the same transfer reverses its sign under the stated convention.

Plausibility: The result lies between zero and 480 kJ, and tends to zero as the boundary temperature approaches the 300 K environment.

Final result: \(X_Q=288\ \text{kJ}\) of maximum work potential accompanies the heat.

Engineering interpretation

The remaining 192 kJ is energy without work potential relative to this environment, not destroyed energy by itself. Actual equipment will deliver less useful work because finite temperature differences, friction, and other irreversibilities generate entropy.

8–6

The Decrease of Exergy Principle

Exergy of an isolated system cannot increase: it remains constant only in a reversible process and falls by \(T_0S_{gen}\) otherwise. Exergy can increase inside a nonisolated system through transfers, so define the boundary before invoking this principle.

8–7

Exergy Balance: Closed Systems

Closed-system balance

\[X_{in}-X_{out}-X_{dest}=X_2-X_1\]

Expand transfers by heat and useful work, retain closed-system \(u\) and \(P_0v\), and verify the associated first-law energy balance independently.

8–8

Exergy Balance: Control Volumes

Steady-flow exergy balance

\[0=\sum_k\left(1-\frac{T_0}{T_k}\right)\dot Q_k-\dot W_{useful}+\sum_i\dot m_i\psi_i-\sum_e\dot m_e\psi_e-\dot X_{dest}\]

First close mass and energy, then entropy, then exergy. A negative destruction result cannot be repaired by interpretation; revisit signs, environment, states, heat temperatures, and property references.

EXAMPLE 2Exergy destruction in an insulated turbine

Air flows steadily through an insulated turbine at 1.50 kg/s. Supplied inlet and exit data are h1=900, h2=620 kJ/kg and s2-s1=0.120 kJ/(kg·K). Find power and exergy destruction at T0=298 K.

Recognition strategy

A one-inlet, one-exit adiabatic turbine calls for energy and entropy balances. The supplied state properties avoid inventing table values; their entropy difference directly reveals irreversibility.

Given information

Steady turbine control volume with equal inlet and outlet mass rates.
Supplied absolute environment temperature \(T_0=298\ \text{K}\); pressures are understood absolute in the source property evaluation.
Given authoritative-input placeholders for this exercise: \(h_1=900\), \(h_2=620\ \text{kJ/kg}\), \(s_2-s_1=0.120\ \text{kJ/(kg}\!\cdot\!\text{K)}\).
Adiabatic; shaft work out positive; negligible kinetic and potential energy changes; mass basis and real, irreversible component.

Model and interpretation

Steady energy gives \(\dot W_{out}=\dot m(h_1-h_2)\). With no heat transfer, steady entropy gives \(\dot S_{gen}=\dot m(s_2-s_1)\), then Gouy–Stodola gives destruction.

Step-by-step solution

1. Mass and energy. Steady one-stream flow conserves mass and gives \(\dot W_{out}=1.50(900-620)\ \text{kW}\).

2. Power. The enthalpy decrease is \(280\ \text{kJ/kg}\), so actual shaft output is \(420\ \text{kW}\).

3. Entropy generation. \(\dot S_{gen}=1.50(0.120)=0.180\ \text{kW/K}\), which is nonnegative as required.

4. Exergy destruction. \(\dot X_{dest}=T_0\dot S_{gen}=298(0.180)=53.6\ \text{kW}\).

Verification

Units: Mass flow times kJ/kg gives kJ/s, while kelvin times kW/K gives kW.

Signs and direction: Positive work output follows an enthalpy drop; positive entropy generation produces positive, never negative, exergy destruction.

Plausibility: Destruction is smaller than shaft output and vanishes in the isentropic limit \(s_2=s_1\); mass, energy, entropy, and exergy signs are consistent.

Final result: \(\dot W_{out}=420\ \text{kW}\) and \(\dot X_{dest}=53.6\ \text{kW}\).

Engineering interpretation

The destroyed exergy identifies improvement potential, not guaranteed recoverable output from a retrofit. Release analysis must retrieve h and s from traceable real-gas software at measured absolute states and include heat leakage and velocity changes.

EXAMPLE 3Exergetic performance of a plant compressor

A steady adiabatic compressor handles 2.20 kg/s of gas. Supplied flow-exergy rise is 145 kJ/kg and measured shaft input is 390 kW. Determine exergy destruction and second-law efficiency.

Recognition strategy

A work-consuming control volume with a supplied stream-exergy increase is best audited by the steady exergy balance. Compare the desired exergy rise with actual shaft exergy input.

Given information

Steady compressor control volume; equal mass flow at inlet and outlet.
Supplied \(\psi_2-\psi_1=145\ \text{kJ/kg}\), already evaluated from authoritative h and s data at absolute states relative to specified \(T_0,P_0\).
Actual shaft work input is \(390\ \text{kW}\); heat transfer, KE, and PE changes are neglected.
Work out is positive, so compressor work is negative in the balance; reported input magnitude is positive. Real component, mass-rate basis.

Model and interpretation

For an adiabatic compressor, shaft-work exergy in equals stream-exergy rise plus destruction. The second-law efficiency is desired stream-exergy rise divided by actual shaft input.

Step-by-step solution

1. Desired exergy. Compute \(\dot m(\psi_2-\psi_1)=2.20(145)=319\ \text{kW}\) as the useful product.

2. Exergy balance. Write \(\dot W_{in}=\dot m(\psi_2-\psi_1)+\dot X_{dest}\) for this adiabatic boundary.

3. Destruction. Thus \(\dot X_{dest}=390-319=71\ \text{kW}\), implying positive entropy generation.

4. Efficiency. The exergetic efficiency is \(\eta_{II}=319/390=0.818\), or \(81.8\%\).

Verification

Units: Every exergy-rate term is kW; the ratio of two rates is dimensionless.

Signs and direction: Positive destruction and an efficiency below unity agree with an actual work input exceeding the reversible minimum.

Plausibility: The balance closes exactly: 390 = 319 + 71 kW, and \(\dot S_{gen}=71/T_0\ge0\) for every positive absolute environment temperature.

Final result: \(\dot X_{dest}=71\ \text{kW}\) and \(\eta_{II}=0.818\) (81.8%).

Engineering interpretation

About 18.2% of shaft exergy is lost inside the stated boundary. Decisions require a defined dead state, uncertainty in flow and power, vendor compressor maps, off-design operation, cooling losses, and safety review.

Problem-solving strategy

Recognize whether the boundary encloses a fixed mass, one steady component, or an entire cycle.
Identify every state with absolute pressure and temperature, phase, and an authoritative property source.
Write mass, energy, entropy, and when relevant exergy balances before inserting component models.
Separate ideal benchmark states from actual states and iterate when properties depend on the unknown temperature.
Audit units, signs, cycle closure, entropy generation, and sensitivity to supplied data.

Chapter recap

Always state the dead state because exergy has no meaning without an environment.
Use flow exergy with enthalpy for streams and closed-system exergy with internal energy.
Link every destroyed-exergy value to nonnegative entropy generation and an independent energy audit.

Before moving on

You should be able to construct exergy balances, define a meaningful second-law efficiency, and explain what additional evidence is needed before claiming recoverable savings.

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