CHAPTER 1

EXERGY: A MEASURE OF WORK POTENTIAL

CHAPTER 1 EXERGY: A MEASURE OF WORK POTENTIAL

1.1 EXERGY: WORK POTENTIAL OF ENERGY

1.2 REVERSIBLE WORK AND IRREVERSIBILITY

1.3 SECOND-LAW EFFICIENCY

1.4 EXERGY CHANGE OF A SYSTEM

1.5 EXERGY TRANSFER BY HEAT, WORK AND MASS

1.6 THE DECREASE OF EXERGY PRINCIPLE AND EXERGY DESTRUCTION

1.7 EXERGY BALANCE: CLOSED & OPEN SYSTEM

Energy Source

Waste Energy

Useful Energy

Useful Work

Availability of Energy

Available Energy

A property that used to determine the USEFUL

WORK POTENTIAL of a given amount of energy at some SPECIFIED STATE

The MAXIMUM USEFUL WORK that can be

obtained from a system at the specified state.

Total Energy

Exergy

Unavailable Energy

1.1 EXERGY: WORK POTENTIAL OF ENERGY

Quality of Energy

Comparing the work potential of different energy sources or systems

Amount of energy that can be extracted from an energy source?

EXERGY The WORK POTENTIAL

of the energy in a system at a specified state

1.1.1 TERMINOLOGY INVOLVED WITH EXERGY

EXERGY: The MAXIMUM POSSIBLE WORK as it undergoes a reversible process from the SPECIFIED INITIAL STATE to the state of its environment

(DEAD STATE)

TERMINOLOGY

DEAD STATE SURROUNDINGS

Immediate Surroundings Environment

Affected by the process Not affected by the process

A system that is in thermodynamic equilibrium with the environment

Initial State Final State (Dead State)

EXERGY ANALYSIS Specified: Not Variable

Maximum Work

Output

Reversible Process

100% ENERGY

1.1.2 EXERGY ASSOCIATED WITH KE & PE

Kinetic Energy (KE)

MECHANICAL ENERGY

Potential Energy (PE)

WORK

Exergy of KE Exergy of PE

Example 1-1: A wind turbine with a 12 m diameter rotor is to be installed at a location where the wind is blowing steadily at an average velocity of 10 m/s. Determine the maximum power that can be generated by the wind turbine.

Example 1-2: Consider a large furnace that can transfer heat at a temperature of 1100 K at a steady rate of 3000 kW. Determine the rate of exergy flow associated with this heat transfer. Assume an environment temperature of 25 C.

1.2 REVERSIBLE WORK & IRREVERSIBILITY

2 Quantities Related To The Actual Initial And Final States

ACTUAL ENGINEERING

SYSTEM

Dead State: Hard

REVERSIBLE WORK, Wrev

IRREVERSIBILITY, I (EXERGY DESTRUCTION)

Isentropic Process: Limited to Adiabatic

Process Not only for 2 fix states

EXERGY: The MAXIMUM POSSIBLE WORK as it undergoes a reversible process from the SPECIFIED INITIAL STATE to the state of its environment (DEAD STATE)

T

s

1’

2’

3’

4’1

2

3

4

Ideal

Actual

Pressure drop in the boiler

Irreversibility in the turbine

Irreversibility in the pump

Pressure drop in the condenser

1.2.1 REVERSIBLE WORK & IRREVERSIBILITY

2 Quantities Related To The Actual Initial And Final States

ACTUAL ENGINEERING SYSTEM

REVERSIBLE WORK, Wrev

IRREVERSIBILITY, I (EXERGY DESTRUCTION)

COMPRESSION EXPENSION

Useful Work, Wu

Wu = W a+ Wsurr

Wu = Wa + Po(V2-V1)

Useful Work, Wu

Wu = Wa - Wsurr

Wu = Wa – Po(V2-V1)

Wsurr =0

SURROUNDING WORKS

No Volume Change, ΔV=0

No Moving Boundary

Wu = Wa

T

s

1’

2’

3’

4’1

2

3

4

Ideal

Actual

Pressure drop in the boiler

Irreversibility in the turbine

Irreversibility in the pump

Pressure drop in the condenser

1.2.2 REVERSIBLE WORK & IRREVERSIBILITY

REVERSIBLE WORK, Wrev IRREVERSIBILITY, I (EXERGY DESTRUCTION) The MAXIMUM amount of

USEFUL WORK that can be produced as a system undergoes A PROCESS between the SPECIFIED

INITIAL and FINAL STATES

Initial State Final State

Wrev=Exergy, Final State = DEAD STATE

Any Difference between Wrev and Wu

Irreversibility = Exergy Destroyed

I = Wrev,out - Wu,out I = Wu,in - Wrev,in

Waste work potential @ Lost opportunity to do work

I = 0

Wrev = Wu

Wu >Wrev

Wu<Wrev

Example 1-3: A heat engine receives heat from a source at 1200 K at a rate of 500 kJ/s and rejects the waste heat to a medium at 300 K. The power output of the heat engine is 180 kW. Determine the reversible power and the irreversibility rate for this process.

1.3 SECOND-LAW EFFICIENCY, II

Reversible Thermal Efficiency

(Max), th, rev

EFFICIENCY: System Performance

Thermal Efficiency, th

First Law of Thermodynamics: QUANTITY OF

ENERGY

Performance: MISLEADING

No reference to the best possible performance

Second Law of Thermodynamics:

QUALITY OF ENERGY

Reference to the best possible performance

SECOND-LAW EFFICIENCY, II

THE BEST

HEAT ENGINE

Higher Performance?

II=0.60 II=0.43

1.3.1 OTHER SECOND-LAW EFFICIENCY, II EXPRESSIONS

WORK-PRODUCING

DEVICE:

WORK-CONSUMING

DEVICE:

GENERAL DEFINITION

CYCLIC DEVICES: REFRIGERATOR @

HEAT PUMP

SECOND-LAW EFFICIENCY, II

HEAT ENGINE

1.3.2 OTHER SECOND-LAW EFFICIENCY, II EXPRESSIONS

GENERAL DEFINITION

Exergy Expended: The decrease in the exergy of the heat transferred to the

engine

Exergy Recovered: The net work

output

HEAT ENGINE

Exergy Destroyed: Irreversibility

1.3.3 OTHER SECOND-LAW EFFICIENCY, II EXPRESSIONS

Exergy Expended: The work input

Exergy Recovered:

The exergy of the heat transferred to

the high-temperature

medium (HP) or from the low temperature

medium (REF)

GENERAL DEFINITION

REFRIGERATOR @ HEAT PUMP

Exergy Destroyed: Irreversibility

1.3.4 OTHER SECOND-LAW EFFICIENCY, II EXPRESSIONS

ELECTRIC HEATER

Exergy Expended: The

electrical energy the resistance

heater consume from the resource

Exergy Recovered: The exergy

content of the heat supplied to the room

GENERAL DEFINITION

Exergy Destroyed:

Irreversibility

Q

Heated Space at TH Environment at T0 1st Law: Qe= We

Example 1-4: A dealer advertises that he has just receive a shipment of electric resistance heaters for residential buildings that have an efficiency of 100%. Assuming an indoor temperature of 21 C and outdoor temperature of 10 C, determine the second-law efficiency of these heaters.

1.4 EXERGY CHANGE OF A SYSTEM

The work potential of a system in a specified environment

The maximum amount of useful work that can be obtained as the system is brought to equilibrium with the environment

EXERGY

EXERGY CHANGE

Exergy of a Fixed Mass: Non-flow (or Closed

System) Exergy

Exergy of a Flow Stream: Flow (or Stream or Open

System) Exergy

Energy Balance Equation

Energy Change

Net Energy Transfer

1.4.1 EXERGY OF A FIXED MASS: NON-FLOW (OR CLOSED SYSTEM) EXERGY

Energy Change (Non-flow)

Internal Energy

Sensible Energy:

Molecules KE

Latent Energy: Binding Forces (Phase Change)

Chemical Energy

Nuclear Energy

HEAT

SECOND LAW: Not 100% Heat Work

Work potential of Internal

Energy

How Much Less?

< Internal Energy

1.4.2 EXERGY OF A FIXED MASS: NON-FLOW (OR CLOSED SYSTEM) EXERGY

Stationary Close System

Internal Energy

Sensible Energy

Latent Energy

HEAT

SECOND LAW: Not 100% Heat Work

Work potential of Internal

Energy

How Much Less?

< Internal Energy

Initial: Specified State

Reversible Process

Final: Environment State

Reversible Heat Engine: T T0

1.4.3 EXERGY OF A FIXED MASS: NON-FLOW (OR CLOSED SYSTEM) EXERGY

V

U

S

Reversible Process:

Not allow any heat transfer

EXERGY OF A FIXED MASS (NON-FLOW EXERGY)

KE=0; PE=0:

1.4.4 EXERGY OF A FIXED MASS: NON-FLOW (OR CLOSED SYSTEM) EXERGY

NON-FLOW EXERGY CHANGE

V

U

S

EXERGY OF A FIXED MASS (NON-FLOW EXERGY)

1.4.5 EXERGY OF A FLOW STREAM: FLOW (OR STREAM) EXERGY

Energy Change (Non-flow) Energy Change (Flow)

KE=0; PE=0:

1.4.6 EXERGY OF A FLOW STREAM: FLOW (OR STREAM) EXERGY

FLOW EXERGY CHANGE

EXERGY OF A FLOW STREAM (FLOW EXERGY)

Example 1-5: A 200 m3 piston-cylinder device contains compressed air at 1 MPa and 300 K. Determine how much work can be obtained from this air if the environment conditions are 100 kPa and 300K.

Example 1-6: Refrigerant-134a is to be compressed from 0.14 MPa and -10 C to 0.8 MPa and 50 C steadily by a compressor. Taking the environment conditions to be 20 C and 95 kPa, determine the exergy change of the refrigerant during this process and the minimum work input that need to be supplied to the compressor per unit mass of the refrigerant.

Energy Balance Equation

Energy Change Net Energy Transfer

EXERGY TRANSFER BY HEAT (Q), Xheat

EXERGY TRANSFER BY WORK (W), Xwork

EXERGY TRANSFER BY MASS (m), Xmass

Boundary work: Piston device

Other form of work: Shaft work

1.5 EXERGY TRANSFER BY HEAT, WORK, & MASS (CROSS BOUNDARY)

Mass Flow: Mechanism to transport X, S & E

Reversible:

Conservation of Energy Principle:

Energy cannot be created

or destroyed

The Increase of Entropy Principle: Entropy can be created but cannot be

destroyed

FIRST LAW OF THERMODYNAMICS

SECOND LAW OF THERMODYNAMICS

Actual process:

Alternative Statement for SECOND LAW

The Decrease of Exergy Principle: The exergy of an isolated system during a process always decreases (destroyed) or remain constant for

reversible process

1.6 THE DECREASE OF EXERGY PRINCIPLE

Positive value

Non-Flow Exergy Change

GENERATE ENTROPY

Friction, Mixing , Chemical Reaction, Heat Transfer Through A Finite Temperature Difference, Unrestrained Expansion, Non-quasi Equilibrium Compression Or Expansion

1.6.1 EXERGY DESTRUCTION

DESTROYS EXERGY

Reversible:

Actual process:

IRREVERSIBLE PROCESS ,

Xdestroyed

NO EXERGY DESTROYED

DETERMINATION: 1. IRREVERSIBLE: Xdestroyed > 0 2. REVERSIBLE : Xdestroyed = 0 3. IMPOSSIBLEPROCESS: Xdestroyed < 0

IRREVERSIBILITIES

1.7 EXERGY BALANCE: GENERAL

GENERAL:

Net Exergy Transfer: Heat, Work, Mass

Exergy Destruction

Change in Exergy

1.7.1 EXERGY BALANCE: CLOSED SYSTEM

CLOSED SYSTEM

Example 1-7: Consider steady heat transfer through a 5 m X 6 m brick wall of a house of thickness 30 cm. On a day when the temperature of the outdoors is 0 C, the house is maintained at 27 C. The temperature of the inner and outer surfaces of the brick wall are measured to be 20 C and 5 C, respectively, and the rate of heat transfer through the wall is 1035 W. Determine the rate of exergy destruction in the wall, and the rate of total exergy destruction associated with this heat transfer process.

Example 1-8: A piston-cylinder device contains 0.05 kg of steam at 1 MPa and 300 C . Steam now expands to a final state of 200 kPa and 150 C, doing work. Heat losses from the system to the surroundings are estimated to be 2 kJ during this process. Assuming the surroundings to be at T0=25 C and P0=100 kPa, determine (a) the exergy of the steam at the initial and the final states, (b) the exergy change of the steam, © the exergy destroyed, and (d) the second law efficiency for the process.

1.7.2 EXERGY BALANCE: CONTROL VOLUME (OPEN SYSTEM)

STEADY FLOW SYSTEM

ADIABATIC SYSTEM

W=Wrev Xdestroyed=0

STEADY FLOW SYSTEM

1.8 REVERSIBLE WORK, Wrev

1.9 SECOND-LAW EFFICIENCY, II

GENERAL DEFINITION

STEADY FLOW DEVICES

ADIABATIC TURBINE ADIABATIC COMPRESSOR

Example 1-9: Steam enters a turbine steadily at 3 MPa and 450 C at a rate of 8 kg/s and exits at 0.2 MPa and 150 C. The steam is losing heat to the surrounding air at 100 kPa and 25 C at a rate of 300 kW, and the kinetic and potential energy changes are negligible. Determine (a) the actual power output, (b) the maximum possible power output, © the second-law efficiency, (d) the exergy destroyed, and (e) the exergy of the steam at the inlet conditions.

END OF CHAPTER 1

EXERGY: A MEASURE OF WORK POTENTIAL