03 part3 availability irreversibility

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

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

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.

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.

AVAILABILITY IN NON-FLOW SYSTEMS

푤푚푎푥 = 푄 − 푡0 푆1 − 푆0

The heat supplied to the engine (Q)is equal to the heat rejected by the fluid in the cylinder.

Wfluid = (u1-u0) – Q Wfluid +

푤푚푎푥 = (u1−u0) – 푄 + 푄 − 푡0 푆1 − 푆0

Wfluid +푤푚푎푥 = (u1−u0) − 푡0 푆1 − 푆0

Work done on atmosphere = p0 (푣0 – 푣1) maximum work available Wmax = (u1 – u0) – T0 (s1 – s0) – p0 (푣0 – 푣1) Wmax = (u1 + p0푣1 – T0s1) – (u0 + p0 푣0 – T0s0)

Wmax = (h1 – T0s1) – (h0 – T0s0)

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.

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’.

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

Irreversibility for steady flow-process

The same expression for irreversibility applies to both flow and non-flow processes.

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.

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.

Max work or reversible work or availability of air at 1

푊푚푎푥푎푡. 1표푟1푡표0 =푢1 + 푝0푣1 − 푡0푠1 − (푢0 + 푝0푣0 − 푡0푠0)

푊푚푎푥푎푡. 1표푟1푡표0

= 푢1 − 푢0 + 푝0 푣1 − 푣0 − 푡0 푠1 − 푠0

To – is the surrounding temperature

Max work or reversible work or availability of air at 2

푊푚푎푥푎푡. 2표푟2푡표0= 푢2 − 푢0 + 푝0 푣2 − 푣0 − 푡0 푠2 − 푠0

푊푚푎푥푎푡. 2푡표1 =푊푚푎푥푎푡. 2표푟2푡표0

- 푊푚푎푥푎푡. 1표푟1푡표0

푊푚푎푥푎푡. 2푡표1

=푢2 − 푢1 + 푝0 푣2 − 푣1 − 푡0 푠2− 푠1

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.

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.

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.