Thermo Chap VI

7/28/2019 Thermo Chap VI

1/34

ENTROPYThe Second Law by its connotations has showed that processes could be

reversible or irreversible.

A reversible process although not possible in nature , is the ideal process

possible between two states.

An irreversible process between the same two states would be less than ideal.

This has also been pointed out by the efficiency and COP of Carnot devices

( ) ( )

( ) ( )

th thCarnot Heat Engine IRR Heat Engine and

COP Carnot COP IRR Device

>

>

Now we need to determine the amount or degree of irreversibility of

processes, and to do that we define a new property called as ENTROPY

This is a developed property , and is a result of the

CLAUSIUS INEQUALITY which is 0dQ

T


2/34

For a Carnot Cycle we have

2 2 2 1

1 1 2 1

1 1

0

H L L L H L L Lth

H H H H H H

Q Q Q T T T Q T so

Q Q T T Q T

Q T Q Q

so for our cycle above or Q T T T

= = = = =

= =

Now if this cycle was an irreversible cycle then

2 2 2 2 2 1 1 2

1 1 1 1 2 1 1 2

1 1Q T Q T Q Q Q Q

or or or veQ T Q T T T T T

< > > =

so 0 Re 0Q Q

for v Cycle and for IRR CycleT T

=


3/34

Lets take a reversible cyclic device taking Heat from a Thermal Reservoir. The

rejected heat is given to a system which produces work

( )

( . )R C C

C R System

C

R R

RC C

For the combined system Q W dE F Law where

W Net work of combined system W W

and dE change of energy of Combined system

Q TSince the Device A is rev and cyclic so

Q T

Tso then W Q dE T

=

= = +

=

=

=

Now let us allow the combined arrangement to complete a cycle , with the

Rev. Device undergoing several cycles.Then for combined system dEC=0

C R

Qor W T net work by combined system in a cycle

T

= =

This violates the Second Law as there is no Heat Rejection.

So Wc is not +ve but can be zero or negative. TR is positive

So the only possibility is that 0Q

T

This is valid for all Thermodynamic cycles both reversible or irreversible.

0

Q

T

=for internally reversible cycle and

DeviceA

RW

HIGH TEMPERATURE RESERVOIR at TR

System at T

(due to a process)

Dotted Line encloses thecombined system.


4/34


5/34

2

2 1

1

2

1

intREV

IRR

QOn the reversible path we egrate s s

T

QREMEMBER IS NOT ENTROPY CHANGE

T

=

INTERNALLY REVERSIBLE ISOTHERMAL HEAT TRANSFER PROCESS

If we have an internally Reversible Isothermal Process then we can have

2 2

1 1

1

REV o o

Q QS Q

T T T

= = =

only for Int. Rev. Iso.process

For such processes heat transfer can be negative or positive depending upon

the direction of heat flow from or to the system. Then

if Q is +ve it will mean that S is positive

if Q is ve it will mean that S is negative

We now look at Example 6-1


6/34

PRINCIPLE OF INCREASE OF ENTROPY

Let us look at a cycle which is made up of two processes

1.2 Maybe be Reversible or Irreversible Process

2-1 Reversible Process

( )

( )

( )

( )

2 1 2

1 2

1 2 1

2

2 1

1

2

2 11

2

2 1

1

0

0 0

1 2

1 2

REV

IRR

QFrom Clausius Inequality we have or

T

Q Q Qor S S or

T T T

QS S

T

Q

Now if is reversible then S S T

Qso if is irreversible then S S

T

+ +

=

So we say

THE ENTROPY CHANGE OF A CLOSED SYSTEM DURING AN

IRREVERSIBLE IS GREATER THAN

2

1 IRR

Q

T

FOR THE PROCESS

IF THE PROCESS IS REVERSIBLE THEN

2

1 REV

Q

T

IS THE ENTROPY

CHANGE

P

V

2

REV

1

Dotted line is the process from 1 to 2


7/34

Now if we have an irreversible process then we can find S2- S1 by generating

a reversible process between the two states and evaluating

2

1

Q

T

Now we have

( )2 2

2 1

1 1

2

1

IRR IRR

IRR

Q QS S or dS

T T

QWe call the term as ENTROPY TRANSFER T

>

Thus in an irreversible process , the entropy change is greater than the

entropy transfer during the process of the system.

We thus write this as

( )

2

2 1

1Generated

IRR

QS S S

T

+

ENTROPY IS GENERATED DUE TO IRREVERSIBILITIES

SGENERATED IS ALWAYS +VE OR Zero. It cannot be negative.

SGENERATED IS NOT A PROPERTY. IT DEPENDS UPON THE PROCESS AND

IS A PATH FUN CTION.

The more the irreversibilities the more will be the generation of entropy

If in a process there is no

2

1

Q

T

then S = SGen


8/34

Now

( )

( )

2

2 1

1

2 10 0

IRR

ISOLATED

QS S so if we have an isolated system

Tthen Q or S S

=

So it is zero if the processes in the isolated system are reversible

It is positive if processes in the isolated system are irreversible

ENTROPY OF AN ISOLATED SYSTEM NEVER DECREASES. IT CAN

EITHER REMAIN THE SAME OR INCREASE

This is known as the Increase of Entropy Principle.

So lets say we have several sub systems operating in an isolated envoirnment

0Z

Total n

n A

S S if processes are irreversible

=

= >

As the sub systems interact with each other , the sum of there entropy

changes is never less than Zero if they have irreversible processes.

Also in the absence of Heat Transfer the change of entropy will be due to

other irreversibilities.

If we have a system and its surroundings then it can be represented as

A CB

ISOLATED


9/34

0Gen Total Sys Surrounding

S S S S = = + Entropy Change Entropy Change

of System of surrounding

If we take the UNIVERSE as an isolated system , then because of

IrreversibilitiesSUniverseis always increasing

SUniversemeans an increase of disorder in the universe. So the disorder is

increasing day by day .

In fact there will be a collapse of the universe when its Entropy reaches a

maximum.

So lets remember some facts

SGen for IRR process is always positive

SGen for REV process is always ZERO

SGen < Zero is not possible

The Entropy change(S2- S1) on the other hand can be negative , positive or

zero.

System

Surrounding

Isolated

We can assume the system andsurroundings as two sub systemsinteracting with each other


10/34

0Gen Sys Surrounding

S S S= +

Sys

Surrounding

S may be positive or negative and

S may be positive or negative

But there sum may either Zero or positive

REMARKS ON ENTROPY

A. Processes can occur only in a certain direction .

It must obey 0GenS

B. Entropy is non conserved

0 , 0Isolated Isolated

For REV S For IRR S = >

Entropy of Universe is constantly increasing

C. Increase of Entropy means presence of Irreversibilities. Thus a process

is good ifIsolated

S is as close to Zero . Higher values of

IsolatedS means irreversibilities are more.

ENTROPY CHANGE OF A PURE SYSTEM.

Entropy is a property , so it is used to define the state of a system.

By utilizing the two property rule Entropy will be useful to define the state of a

system.

We can also find CHANGE OF ENTROPY with the help of other properties.

Entropy is generally calculated against a REFERENCE STATE.

For Steam Sf ( Entropy of Saturated Liquid at 0.01oC) is taken as zero. So at

-5oC the entropy will be negative

For Ref-134 Sf (Entropy of Saturated Liquid at 40oC) is taken as Zero. So at

-45oC the entropy will be negative.


11/34

( )

( )

@ @

2 1

SAT

f g f

T and P f T

For Liquid Vapor region s s x s s

For Compressed Liquid s s

also S m s s

= +

=

=

Entropy is also used for property diagram

ISENTROPIC PROCESS

Constant Pressure Lines

Pressure Increases

Constantspecificvolume Lines

Specific volume Increases

P

s


12/34

2 1

2 1 0 Isentropic Process

This means 0 0

Or process is Adiabatic and Reversible = Isentropic

The other possibility is that 0 (

REV

REV

REV

QBy defination S S

T

Now if S S Then we call it

Qso either Q

T

Q

T

=

=

= =

=

)

so not only Adiabatic Reversible is Isentropic but other processes

can be Isentropic as well if 0REV

Due to Integration

Q

T

=

Ex 6-5

Process is Reversible and Adiabatic hence Isentropic KE = PE = 0

Find Wx

( )

( )

* * *

2 1

*

2 1 1 1

1 2 2 2

*

3316.2 / 6.8186 /

2966.6 / 1.4

3316.2 2966.6 349.6 /

x

o

x

x

Q W m h h KE PE or

w h h h kJ kg s kJ kg K

Now s s so at s we get h kJ kg at P MPa

w kJ kg

= + +

= = =

= = == =

PHYSICAL CONCEPT OF ENTROPY

P1

= 5Mpa

T1

= 450 oC

P2

= 1.4 Mpa

s

1

2

T


13/34

We have developed Entropy fromREV

dQ

T . Lets now look at its physicalsignificance

The value of entropy is a means of judging MOLECULAR DISORDER or

MOLECULAR RANDOMNESSWhen entropy increases , we go towards a state where more disorder is

present. Thus in

SOLIDS: we have low value of entropy, since the molecules have positional

stability , even though they vibrate.

LIQUIDS: have more entropy, because the positional stability is less stable

than solids. The disorder is higher.GASES: have high value of entropy because the molecules move in a random

manner , collide with each other and we cannot predict their position. There is

great degree of CHAOS

HIGH MOLECULAR CHAOS = HIGH ENTROPY

Lets look at a few examples

If we put a paddle wheel into a gas container

If we have a shaft with a weight attached with frictionless bearing

We find that although the molecules havehigh KE , they cannot rotate the paddlewheel , because they move in randomdirections and do not push in an orderedway

As the weight drops the shaft rotates inone direction, hence the molecules of theshaft are moving in an organized manner.There is no disorder. If we raise the weightthen the shaft will rotate in the oppositedirection.

W


14/34

Work is generally free of disorder and in the absence of friction does not

increase entropy.

Thus work in most cases in nearly reversible.

Here energy conversion is efficient.

But what if we have a rotating shaft in a gas container.

Heat is a form of disorganized energy and thus it causes disorganization.

THE VALUE OF ENTROPY IS DEFINED BY THE THIRD LAW OF

THERMODYNAMICS

ENTROPY OF A PURE CRYSTALLINE SUBSTANCE AT ABSOLUTE ZERO

IS ZERO.

This provides the absolute reference and it is important that the substance be

in crystalline form.

As the shaft rotates , the paddle wheelrotates and work is converted into heat ,because the gas is also rotated andchurned. The temperature will rise andmolecular disorder is increased. Thisallows the entropy to increase and theprocess is highly irreversible.

GAS

W

HOT BODYCOLD BODY

Q

HIGH ENTROPYHIGHLY DISORDERED

AS A RESULT OF

HEAT TRANSFER THE

ENTROPY

DECREASES

LOW ENTROPYBEING CONVERTEDINTO HIGH ENTROPY

AS A RESULT OF

HEAT TRANSFER THE

ENTROPY

INCREASES


15/34


16/34

H-S diagrams

For water the Mollier diagram looks like this

The temperature lines are nearly horizontal because superheated steam

behaves like ideal gas at Low pressures

Tds Relations

T

s

21 All processes are reversible.s

2= s

3and s

1= s

4

4 3

QH

QL

H

s

2

1Such a diagram is useful for processesinvolving steady flow devices.

H-S diagrams are also called MollierDiagrams

P=900 kPa

P=150 kPa

T= 220 oC

T= 80 oC

x = 0.9

x = 0.96

x = 1.0

H

s


17/34

If we have a reversible process of a closed system then from first Law

( )

REV REV REV

REV

Q W dU where Q TdS and

W PdV if there is no shaft work

so TdS dU PdV REVERSIBLE PROCESS

If we divide by mass Tds du Pdvh u Pv so dh du Pdv vdP

or dh Tds vdP or Tds dh vdP

= ==

= +

= += + = + += + =

Both Equations were generated by Reversible Processes but

T,S,H,U,P and V are properties , so the change of property is

independent of path.

BOTH EQUATIONS ARE VALID FOR IRREVERSIBLE PROCESSALSO.

These equations are also written as

du Pdv dh vdP ds and ds

T T T T = + =

Relations are for simple systems and can be applied to open and closed

systems.

ENTROPY CHANGE OF SOLIDS AND LIQUIDS

Solids and liquids are incompressible dv is nearly 0. Thus

2 2

22 1

11 1

2

2 1 2 1

1

ln

Pr

0 ln

P v

du Pdv du CdT ds

T T T T

C C C For Solids and Liquids

Tdu CdT or s s C T T T

And for Isentropic ocess of these

Ts s C so T T

T

= + = =

= =

= = =

= = =

Isentropic Process of Liquids and Solids are ISOTHERMAL

Example 6-7

ENTROPY CHANGE OF IDEAL GASES


18/34

2 2

2 1

1 1

2 2

2 1

1 1

.

ln ln

v

v

v

v

v

C dTdu Pdv R ds dv

T T T v

P RSince u C dT and

T v

C dT Rso s s dv If C Const

T vT v

s s C R T v

= + = +

= =

= + =

= +

2 2

2 1

1 1

2 2

2 1

1 1

ln ln

P

P

PP

P

Also for an ideal gas dh C dT

C dTdh vdP R so ds dP

T T T P

C dT Rso s s dP If C Const T P

T Pthen s s C R

T P

=

= = +

= + =

=

BOTH RELATIONS ARE VALID FOR REVERSIBLE AND IRREVERSIBLE

PROCESSES

If CP and Cv are not constant then we use

( ) 2 22 11 1

ln lnv AvgT v

s s C R T v

= +

( ) 2 22 11 1

ln lnP Avg

T Ps s C R

T P =

If we want to find per Molar Mass then divide the equations by M.

If the values of CP and Cv are variable and cannot be averaged then we use

other technique. For this we have to the reference values.

We calculate the entropy change with reference to Absolute Zero and define

( )1

1

0

T

o

P

dTS C as function T

T

=

This is calculated at various temperatures and tabulated


19/34

Now if we look at the original equation

( )( )

( )

1

2

2

2

2 1

11

10

2

0

2

2 1

1

22 1 2 1

1

22 1 2 1

1

ln

ln

ln

P

T

o

P

T

o

P

o o

P

o oo

o o

u o

Ps s C dT R and if

P

dTS C as function T and

T

dTS C as function T

T

Then C dT S S

P kJs s S S R P kg K

On a molar basis

P kJs s S S R

P kmol K

=

=

=

=

=

=

We now look at example 6-9

ISENTROPIC PROCESSES OF IDEAL GASES


20/34

Constant Specific Heats; For an ideal gas we have

2 22 1

1 1

2 2 2 22 1

1 1 1 1

1

2 2 2 2 2

1 1 1 1 1

ln ln

ln ln ln ln

1

1 11

1ln ln ln (1)

P

P

P

P v

pv

P P v

P

k

k

P

T Ps s C R

T P

T P T P Rif s s then C R or

T P T C P We have R C C for ideal gas thus

CCRnow we define k or

C C C

R kthen

C k k

T P P T P R kso or

T C P k P T P

Now al

=

= = =

=

= =

= =

= = =

2 22 1

1 1

2 2

2 11 1

2 2

1 1

1 1

2 2 1

1 1 2

11

2 1

1 2

2 1

1 2

ln ln

ln ln

ln ln 1

(2)

1 2

v

v

P v

v v v

k k

kk

k

k

k

T vso s s C R

T v

T vif s s then C R or

T v

T v C C R Rbut k

T C v C C

T v vso

T v v

P vif we equate and we get

P v

P vor or Pv co

P v

= +

= =

= = =

= =

=

= =

1

1

tan

tan tank

k k

ns t

also Tv cons t and TP cons t

= =

Isentropic Process with Variable Specific Heats


21/34

( )

( )

2

2 1

1

2

2 1 2 1

1

2

2 1

1

2

2 1

1

2 2

2 1

1 1

ln

Pr

0 ln

ln . Pr

.

ln

oS

o o R

oS

R

o o

o

o o

o

o o

S S

o o R

P kJNow s s S S R

P kg K

so if we have an Isentropic ocess

P kJS S R or

P kg K P

S S R Now if we do not know the essure RatioP

then we have a problem

P P eNow S S R or e

P Pe

Th

=

=

= +

= = =

2

2 2

1 1

2

1 1 2 2 2 2 1 2 1 2

11 2 1 1 2 1 2

1

2

1

Pr Pr

Pr

Pr

Pr Pr

PrPr

Pr

oS

R

r

r

rIsentropic

is term e is called relative essure

Pso

P

TPv P v v T P T

Now or TT T v T P T

Tis a function of temperature only and is called

vvrelative volume v so

v v

=

= = = =

=

1


REVERSIBLE STEADY FLOW WORK


22/34

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2

1

0

rev rev

rev

rev

rev

rev

rev

rev

For a steady flow device we can say that

q w dh d ke d pe

but q Tds when Tds dh vdP

so q dh vdP and

w dh d ke d pe dh vdP

w d ke d pe vdP which on Integration

W ke pe vdP

If ke pe then W vd

= + +

= =

=

= + + +

= + +

= + +

= = =

( ) ( ) ( )

2

1

2 1

tan

rev

P

If subs ce is incompressible then

W ke pe v P P = + +

IRREVERSIBLE WORK AND HEAT OF STEADY FLOW DEVICES.

( ) ( )

0

irr irr rev rev

irr irr rev rev

rev irr rev irr rev

rev irr irr

rev irr irr irr

If we carry out the above process irreversibly then

q w dh d ke d pe q w so q w q w

w w q q but q Tds so

w w Tds q

w wq qor ds but ds

T T T

so

= + + = =

= =

=

= >

0rev irr rev irr

w wor w w

T

> >

Thus work producing devices like Turbines produce more work if operated

reversibly and work consuming devices such as pumps and compressors

require less work when operated reversibly.

We look at examples 6-10 and 6-12


23/34

ISENTROPIC EFFICIENCIES OF STEADY FLOW DEVICES

Irreversibilities occur in most processes and their effect is to degrade the

performance of the devices which carry out the process.

For Heat engines we compare the Ideal Carnot Cycle with real cycles, by

looking at th. Carnot cycle has the best efficiency

Similarly for Heat Pump and Refrigerators we know that COPReversed Carnot is the

best.

So how can we compare the efficiencies of Devices which do not operate in a

cycle. There are several types of devices , but we will limit ourselves to Steady

Flow Devices like Turbines , Compressors and Nozzles.

All these steady flow devices operate in Adiabatic Condition. So it will be

necessary that we compare them as

Adiabatic Irreversible Process vs Adiabatic Reversible Process (Isentropic)

So we need to look at these devices as they operate actually and compare it

with their operation Isentropically.

We call this comparison as ISENTROPIC OR ADIABATIC EFFICIENCY

If a device has an Isentropic Efficiency of 1.00 then it behaves ideally, so

.

ISEN

ISEN

Actual Performance

Isentropic Performance

is defined seperately for each device

=

W e will now look at these cases individually

ISENTROPIC EFFICIENCY OF TURBINE


24/34

In a turbine a steady flow device produces external work .

Here the inlet and outlet pressure is the same for actual and Isen process

( ) ( ) ( )

( )

* * * 2 2

2 1 2 1 2 1

* *

1 2

1

2actual

actual

actualISEN

Isentropic

x a

x a

WActual Performance

Isentropic Performance W

For a turbine

Q W m h h V V g z z

Now ke pe o for a turbine then

W m h h

If the process is done isentropically t

= =

= + + = =

=

( )* *

1 2

1 2

1 2

Isen

Turbine

ax s ISEN

s

hen

h hW m h h so

h h

= =

At the exit, state will be different for Ideal and Actual cases. P2 will remain

same.

The exit entropy drop is more for the actual process.

2 1 2 2a Gen a ss s s and s s = > For most turbines

Isentropic Efficiency is between 70 to 90 %. Lets see example 6-16

ISENTROPIC EFFICIENCY OF COMPRESSORS AND PUMPS

InletState 1

OutletState 2

Wx

1

2s

2a

P2

P1

Ws

Wa

T

s

P1


25/34

Here a steady flow device receives external work .

Here the inlet and outlet pressure is the same for actual and Isen process

( ) ( ) ( )

( )

* * * 2 2

2 1 2 1 2 1

* *

2 1

1

2

/

actual

actual

Isentropic

ISEN

actual

x a

x a

WIsentropic Performance

Actual Performance W

For such devices

Q W m h h V V g z z

Now ke pe o for Comp Pump then

W m h h

If the process is done isen

= =

= + + = =

=

( )/

* *2 1

2 1

2 1

Isen

Comp Pump

sx s ISEN

a

tropically then

h hW m h h so

h h

= =

At the exit, state will be different for Ideal and Actual cases. P2 will remain

same.

The exit entropy increase is more for the actual process.

2 1 2 2a Gen a ss s s and s s = > For most compressors

Isentropic Efficiency is between 75 to 85 %.

A pump is mostly used for liquids so for a pump

InletState 1

OutletState 2

Wx

2s

2a

P1

P1

Ws

Wa

s1

P2


26/34

( )2 1

2 1

Pump

a

v P P

h h

=

Most compressors get heated , as the gas gets to a high temperature due to

high pressure. Thus Compressors are required to be cooled.

SO COMPRESSORS ARE NOT ADIABATICSo for cooled compressors , we compare the actual process with a

REVERSIBLE ISOTHERMAL PROCESS. And so

ReCooled Comp

Isothermal versible Work

Actual Work =


ISENTROPIC EFFICIENCY OF NOZZLE

A nozzle is used to accelerate a fluid from high pressure to low pressure. So

Nozzle

Actual Kinetic Energy at Nozzle Exit

Isentropic Kinetic Energy at Nozzle Exit =

( ) ( ) ( )

( ) ( )

( )

* * *2 2

2 1 2 1 2 1

* *2 2

2 1 2 1 2 1

2 2

1 2 2 1 2 1

2 2

1 2 2 1 2 2

1 2

1 2

1

2

10 ,

2

1

2

1 1

2 2

. 6 18

x

x

a a s s

aNozzle

s

Q W m h h V V g z z

Q W z z so h h V V

or h h V V Generally V V

so h h V and h h V

h hLets see exampleh h

= + +

= = = =

= >>

= =

=

ENTROPY BALANCE

T

s

2a 2s


27/34

Suppose we have any system then we can say

TotalSystem

S Total Entropy Transfer Total Entropy Generated = +

Now if the entropy enters and leaves the system then the balance is given by

TotalSystem

S Entropy In Entropy out Total Entropy Generated = +

This is known as Entropy Balance

The entropy change of a system during a process is equal to the net entropy

transfer through the system boundary , and the entropy generated within the

system due to Irreversibilities.

We will discuss the various terms in the above relation.

Entropy Change of system.

TotalSystem

S Entropy at end of process Entropy at start of process =

Thus if we have an open steady system , like Nozzles, Compressors ,

Turbines then for such devices

0

int

TotalSystem

System

S during steady flow operation

and if properties are not uniform then

S s m s dV egrated over entire volume

where V volume and density

=

= =

= =


28/34

Entropy Transfer

Entropy can be transferred by either

a. Heat b. Mass Flow

When Heat is given to a system its entropy increases, and when heat is

removed then entropy decreases.

2

1

k

tan

tan

Heat Transfer through the boundary at temp T at location k

Heat

kHeat

k

k

QEntropy Transfer by Heat S when T cons t

T

This ratio is called Entropy Transfer due to Heat

Q QWhen T Cons t then S

T T

Q

= = =

= =

=

When two systems are in contact with each other and share heat , then

500 5001.25 1.25

400 400

Hot Body Cold Body

Outside Inside

Heat Heat

Heat Heat

S S so at boundary

S and S

=

= = = =

THERE IS NO ENTROPY GENERATION AT THE BOUNDARY

WORK on the other hand is entropy free.Also WORK does not transfer entropy. So SWork = 0

So energy interaction with entropy transfer is HEAT

Energy interaction without entropy transfer is WORK.

This definition differentiates HEAT and WORK.

System

Qk

= 500J

Tk

= 400 oK1.25

k

kHeat

k

QS

T= =


29/34

Now work can be transferred across a system , but it does not transfer entropy

, but how work is utilized in the system , may generate Entropy

Entropy Transfer by mass flow

Mass contains Energy as well as Entropy , so it also transfers entropy. Both

energy and entropy are proportional to mass or Smass= m(s).

In a closed system entropy cannot be transferred by mass as no mass flows

In open system flow of mass in brings entropy and the flow of mass out

removes entropy.

The rate of entropy transfer by mass depends upon mass flow rate.

So for steady flow it will depend upon the value of s of mass at inlet and exit,

even though mass flow is constant.

So if the properties of the mass changes then we do the following:

( )*

/ sec

int

mass n c c

n c

mass

S s V dA where A c tional area of Flow

V normal velocity at A

and S s m egrated during a time period

= =

=

=

As fan rotates , the frictionconverts work energy into heatenergy . This energyconversion to Heat causesEntropy Generation


30/34

ENTROPY GENERATION

Entropy generation is due to the result of IRREVERSIBILITIES

For a reversible process the

Re0

0

( )

vGEN

GENIRR

IRR

IRR

QS for reversible process

T

QBut our defination of S S for all processes

T

Qwhere Entropy Transfer

T

QIf a process is Adiabatic and Irreversible then

T

Entropy Transfer for such a process is zero

Als

= =

= +

=

=

0exp

mass

System in out Gen

in out

o S ms for closed systemSo we ress the entropy change of any system as

S S S S

whereS S Net Entropy Transfer by Heat and Mass

= =

= +

=

If we take in terms of rate then it is

* * * *

** * *

System in out Gen

Heat Mass

in out gen system

S S S S

Qwhere S and S ms

T

if the mass flow is steady then

s s s s

= +

= =

+ =

The above equations account only for SGen within the system. It is taking place

inside the boundary. It does not cater for the SGen in the surroundings.

Entropy TransferDue to Massgoing in

Entropy TransferDue to Heatgoing in

Entropy TransferDue to Heatgoing out

Entropy TransferDue to Massgoing out

Sin SoutSGEN


31/34

* * * *

* *

0 intSystem

Surrounding

Surrounding System

Gen

Net System Surrounding

in out Gen

Gen Gen

Now if S Then the process is ernally reversible

If we cater for system and surroundings then

S S S

S S S S

If S and S are both zero then the p

=

= +

= +

rocess

is totally reversible

Let us apply this on a closed system. Since there is no mass transfer then

Entropy transfer is by Heat Transfer. So we can say that

2 1

,

0 0

( )

Gen System

Gen Adiabatic System

Gen System Surr Isolated System

Q

S ST

And if process is adiabatic then S S

If we combine system and surroundings

as an Isolated System then for such an

isolated system Q and W and

S S S

Qm s s

+ = =

= =

= +

+

. .

** * * *

. .

*

. .

** * *

. .

0

Surr

Surr

in in out out Gen C V

kin out Gen C V in out

C V

kout inGen out in

T

If we have an open system or C V then we have

Qm s m s S S

T

Qor m s m s S S

T

For Steady flow devices S

QS m s m s

T

+ + =

+ + =

=

=


32/34

If flow is having one stream and is steady but irreversible then

*

* * *k

out inGen out in

QS m s m s

T=

If flow is having one stream , is adiabatic and is steady but irreversible then

* * *

out inGen out inS m s m s=

If flow is reversible then

* * *

0 out inGen out inS so m s m s= =

And if flow is reversible and steady with adiabatic conditions

out inso s s=We now look at Example 6-19


33/34


34/34

2 2

2 1

1 1

2 2

2 1

1 1

2 2

2 1

1 1

2

2 1

1

.

ln ln

ln

v

v

vv

v

P

P

PP

P

C dTdu Pdv R ds dv

T T T v

P RSince u C dT and

T v


T v

T vs s C R T v



T T T P

C dT Rso s s dP If C Const

T P

Tthen s s C R

T

= + = +

= =

= + =

= +

=

= = +

= + =

=

2

1

lnP

P

2 2

2 1

1 1

2 2

2 1

1 1

2 2

2 1

1 1

2

2 1

1

.

ln ln

ln

v

v

v

v

v

P

P

P

P

P

C dTdu Pdv R ds dv

T T T v

P RSince u C dT and

T v


T v

T vs s C R

T v



T T T P

C dT Rso s s dP If C Const

T P

Tthen s s C R

T

= + = +

= =

= + =

= +

=

= = +

= + =

=

2

1

lnP

P

Date post:	03-Apr-2018
Category:	Documents
Upload:	ammar-naveed-bajwa
View:	222 times
Download:	0 times

Thermo Chap VI

Documents