This document discusses availability and irreversibility in transient and steady flow systems. It provides definitions for available energy, unavailable energy, dead state, and effectiveness. It examines availability in terms of the first and second laws of thermodynamics. Equations are given for calculating maximum work, availability, and irreversibility for various thermodynamic processes involving changes in temperature, pressure, and volume. Two sample problems are included and solved to demonstrate calculating work, availability change, and irreversibility for specified thermodynamic systems.

1. Irreversibility and Availability in
Transient Systems
S.Gunabalan
Associate Professor
Mechanical Engineering Department
Bharathiyar College of Engineering & Technology
Karaikal - 609 609.
e-Mail : gunabalans@yahoo.com
Part - 3

2. Available Energy is the maximum work output
obtainable from a certain heat input in a cyclic
heat engine.
Unavailable energy is the minimum energy that
has to be rejected to the sink by the second law
Dead state is defined as the system with zero
velocity and minimum potential energy.
Effectiveness is defined as the ratio of actual
useful work to the maximum useful work.
third law of thermodynamics states “When a
system is at zero absolute temperature, the
entropy of system is zero

3. AVAILABLE ENERGY REFERRED TO A CYCLE
Q1 = A.E. + U.E.
Wmax = A.E. = Q1 – U.E.
the reversible efficiency,
For a given T1,
ηrev - increase with the decrease of T2.
The lowest practicable temperature of heat rejection
is the temperature of the surroundings, T0.

6. Availability in Transient flow
Transient flow is a flow where the velocity and
pressure changes over time.
Transient flows usually occur during the starting
or stopping of a pump, the opening or closing of
a tank, or simple changes in tank levels.
Transient flow usually refers to surge or water
hammer. The main reason transient flow can be
a problem is it can cause pressure that would
exceed the limits of pipes, fittings, etc.

7. AVAILABILITY IN NON-FLOW SYSTEMS
= − 0 1 − 0

8. The heat supplied to the engine (Q)is equal to
the heat rejected by the fluid in the cylinder.
Wfluid = (u1-u0) – Q
Wfluid + = (u1−u0) – + − 0 1 − 0
Wfluid + = (u1−u0) − 0 1 − 0
Work done on atmosphere = p0 ( 0 – 1)
maximum work available Wmax
= (u1 – u0) – T0 (s1 – s0) – p0 ( 0 – 1)
Wmax = (u1 + p0 1 – T0s1) – (u0 + p0 0 – T0s0)
Wmax = (h1 – T0s1) – (h0 – T0s0)

9. AVAILABILITY IN STEADY FLOW
SYSTEMS
the function ‘b’ , ‘a’ is a composite property of a system and its environment ; this is
also known as Keenan function.

10. IRREVERSIBILITY
The actual work which a system does is always less
than the idealized reversible work, and
the difference between the two is called the
irreversibility of the process.
Irreversibility, I = Wmax – W
This is also referred as ‘degradation’ or ‘dissipation’.

11. For a non-flow process between the equilibrium
states, when the system exchanges heat
only with environment, irreversibility (per unit
mass)

12. Irreversibility for steady flow-process
The same expression for irreversibility applies to both flow and
non-flow processes.

13. Prob-1
A rigid tank of volume 2.5m3 contains air at
200kPa and 300K. The air is heated by supplying
heat from the reservoir at 600K until the
temperature reaches 500K. The surrounding
temperature is at 100kPa and 300K. Determine
the maximum useful work and the irreversibility
with the process.

14. volume = 2.5m3
P1 = 200kPa
T1 = 300K
T of reservoir = 600K
T2 = 500K.
Surrounding temperature
P0 = 100kPa
To = 300K.
Determine the maximum useful work and the irreversibility with the process.

15. Max work or reversible work or
availability of air at 1
. 1 1 0 = 1 + 0 1 − 0 1 − ( 0 + 0 0 − 0 0)
. 1 1 0 = 1 − 0 + 0 1 − 0 − 0 1 − 0
To – is the surrounding
temperature

16. Max work or reversible work or
availability of air at 2
. 2 2 0 = 2 − 0 + 0 2 − 0 − 0 2 − 0
. 2 1 = . 2 2 0 - . 1 1 0
. 2 1
=
2 − 1 + 0 2 − 1 − 0 2 − 1

19. Prob-2
Two kg of air at 500 kPa,80℃ expands
adiabatically in a closed system until its volume
is doubled and its temperature is equal to that
of the surroundings which is at 100 kPa and 5℃.
For this process, determine the maximum work,
the change in availability and the irreversibility.
Assume for air cv = 0.718 kJ/kgK. And R = 287
J/kgK.

20. m = 2 kg
P1 = 500 kPa,
T1 = 80℃
expands adiabatically in a closed system
until its volume is doubled ie v2 = 2 v1
P2 = 100 kPa
T2 = 5℃.
Determine
1. the maximum work,
2. the change in availability and
3. The irreversibility.
Assume for air cv = 0.718 kJ/kgK. And R = 287 J/kgK.

23. Reference
• Rathakrishnan, E. 2005. Fundamentals of engineering thermodynamics.
Prentice Hall, New Delhi.
• Rajput, R. K. 2010. Engineering thermodynamics. Jones and Bartlett
Publishers, Sudbury, Mass.
• Nag, P. K. 2002. Basic and applied thermodynamics. Tata McGraw-Hill, New
Delhi.