Evaluating entropy changes

2

2 11 int rev

QS ST

• Since entropy is a property, the entropy change between two specified states does not depend on the process (reversible or irreversible) that occurs between the two states.• However, to evaluate entropy change by the above definition a reversible process has to be imagined connecting the initial and final states and integration has to be carried along its path.

revQ TdS rev

Q dST

2

rev1

Q TdS

T

S

1 2

Using entropy change and thermodynamic temperature to graphically interpret the heat supplied in reversible processes

V

P

revW PdV

2

rev1

W PdV

Qnet,in

Showing a Carnot cycle on a T-s diagram

T

s

2

4

1

3

Isotherms: horizontal linesIsentropes:vertical lines

Application:Carnot engine

Isobars and isochores on the T-s diagram for an ideal gas (Sec. 7.9)

11 1

~ ln lnavgv

T vs s c RT v

11 1

~ ln lnavgp

T Ps s c RT P

Isochores: 11~ exp avg

v

s sT Tc

Isobars: 11~ exp avg

p

s sT Tc

T

s

1

1

vv TT

v

Isochores:1

1

PP TT

P

T-s diagram of a cycle: try yourself

• The Brayton cycle used in a gas turbine engine consists of following steps:– isentropic compression (1-2). – isobaric heat addition (2-3). – isentropic expansion (3-4). – isobaric heat rejection (4-1).

Plot this process on a T-s and P-v diagram.

Use of the Tds relations in calculating entropy changes

On an intensive (per unit mass) basis:

du PdvdsT T

dh vdPdsT T

dH VdPdST T

dU PdVdST T

First TdS relation

Second TdS relation

2 2

2 11 1

dh vdPs sT T

2 2

2 11 1

du Pdvs sT T

Integrating between initial (1) and the final state (2):

Even though the Tds equations are derived using an internally reversible process, the change in entropy during an irreversible process occuring between the same two equilibrium states can be also calculated using the integrals on the left.

Entropy change of liquids/solids (Sec. 7.8)

•Liquid/solid can be approximated as incompressible substances (v=constant): i.e. specific volume ( or density) remains constant even when other properties change (dv=0) Internal energy of a solid/liquid is a function of temperature alonecp=cv=c (one specific heat)

2

2 11

cdTs sT

2 2

2 11 1

du Pdvs sT T

from First Tdsrelation

For solids/liquids, the isentropic process is also isothermal.

22

11

ln TdTc cT T

if temperature variation of specific heat (c)can be neglected.

du cdT

Second law for a closed system undergoing a process

2

1 21

0b

Q S ST

2

2 11 b

QS ST

0b

QT

˜Clausius inequality:

> for irreversible process 1-2= internally reversible process 1-2

b: boundary

2 1

1 2 int rev

0b

Q QT T

Tb=T foran internallyreversible process

Second law for an isolated system

2

1 2 2 11

QS S ST

Also, across the boundaries of an isolated system no energy (heat/work) is transferred (by definition)

1 2

isolated 0S

The universe (any system + its surroundings) can be considered an isolated system. “The total entropy of the universe is increasing.”

Second law for an adiabatic process undergone by a closed system

2

1 2 2 11

QS S ST

Second law for a system undergoing adiabatic (irreversible/reversible) process(Q=0)

1 2 2 10S S S

The entropy balance for closed systems

• Sgen>0 if irreversibilities present inside the system• Sgen=0 for no irreversibilities inside the system• The value of Sgen is a measure of the extent of

irreversibilities within the system. More irreversibilities Sgen ↑

2

2 11

genb

QS S ST

( 0)genS

entropy generationwithin thesystem

entropy transfer accompanying heat transfer

entropy change

Entropy generation (Sgen>0) within the system is due to irreversible processes within the system boundary

Irreversible processes occurring within the system boundary result in entropy generation.Examples:• Friction (solid-solid, solid-fluid) may be present

between parts of the system. • Hot and cold zones may be present within the

system which may be interacting irreversibly through heat transfer.

• Non-quasi-equilibrium compression/expansion occuring between parts of the system.

• Other: mixing between substances having different chemical composition, chemical reaction.

Example of entropy generation due to an irreversible process in an isolated system

Objective: • To show that the process is irreversible.

Remove separator

Initial state (1)

Pf,Tf, (m1+m2)

Final state (2)

2

2 11

genb

QS S ST

Isolated system has Q=0

2 1to show 0S S

To show a process is irreversible

) to show Sgen>0

Example (contd.): The final state of the process

P1,V1, T, m1 P2,V2, T, m2

• From the first law Uf=Ui.

• At the final state: .

Pf,Tf, (m1+m2)

1 2 1 21 2

1 2 1 1 2 2 1 1 2 2

( )/ / / /f

m m RT m mm m RTV V m RT

PP m RT P m P m P

High pressure Low pressure

1 2 1 2)( v f v v fm C T mC T m C T T Tm Final temperature:

Example of entropy generation due to an irreversible process in an isolated system

P1,,,V1, T, m1 P2,V2, T, m2 Pf,Tf, (m1+m2)

1 2 1 2)( v f v v fm C T mC T m C T T Tm High pressure Low pressure

isolated genf iS S S S Entropy balance

1 1 2 1 1 2 2 1 2 212 )( ) ( ) ( ) ( )(isolated f f f fS S S m s m s m s m s sS sm m s

Entropy change of the system is the entropy change of its parts

Example of entropy generation due to an irreversible process in an isolated system

P1,,,V1, T, m1, P2,V2, T, m2 Pf,Tf, (m1+m2)

1 1 12( ( ))isolated f fS m s s m s s

Second law (entropy balance)

11

1ln ln lnff p

f

p pTs sT p

C R Rp

from integration of Tds=dh-vdP

22 lnf

f

R ps sp

Similarly

1 2 1 2

1

21 2

1 2

1 1 2

1

2

( ) ln

/ /

m m m mpm m

m mm

pR

P m P

Example of entropy generation due to an irreversible process in an isolated system

P1,,,V1, T, m1, P2,V2, T, m2 Pf,T, (m1+m2)

1 2

1 2

2)

1 2

1 1 2

(

2

1ln

/ /

m m

gen m m

pSm m

m P m P

pR

2 21 1 1 2( ) ( )gen f fS S S m s s m s s

1GM HM0 process is irreversiblegenS

1 2

1 2

21 2

1 2 1( )ln ln ln m

f f f

m m

m

pm R p p pmp p

R Rp

• Reversible cycles develop the maximum work.• Reversible cycles consist of reversible processes

through each device.• At every stage of the internally reversible process:

– qrev-wrev=dh +d(ke)+d(pe)– –

• Combining: wrev=-vdP-d(ke)-d(pe)• Integrating:

Calculating work done in a reversible steady flow process

revq Tds Tds dh vdP

2 222 1

2 11( )

2revw v vvdP g z z

First law for control volumes

if changes in KE and PEcan be neglected (e.g. inturbines, compressor andpumps, but not in nozzles)

Graphical representation of reversible steady flow work on a p-v diagram

2

1rev vw dP

Reversible steady flow work for a pump (PUMP HANDLES LIQUIDS)

1 2 1( )pumprevw v p p

2

1

pumprev vw dP

Since specific volume of a liquid is nearly independent of pressure:

1 2~ ~v v v

isenpumpw (since pump is an adiabatic device)

1 1 2 2 2 1 1 2,( ) , )(revpump pump

isen sw w h h hP P s s h

similar to compressor

Alternatively

In steam power plants, how big is “turbine work out” compared to “pump work in”?

v

4

3

turbinrev

ew vdP Specific volume of vapors are orders of magnitude larger than thespecific volume of liquid (e.g. vf' 10-3 m3/kg vg' 2 m3/kg at 100 kPa))

2

2 11)(pump

rev vdP v Pw P

pumprev re

compre orv

ssw wGas turbine pumps consume a significant part of work developedin turbines to run compressor