Thermo Chap VI
-
Upload
ammar-naveed-bajwa -
Category
Documents
-
view
222 -
download
0
Transcript of 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
-
7/28/2019 Thermo Chap VI
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
=
-
7/28/2019 Thermo Chap VI
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.
-
7/28/2019 Thermo Chap VI
4/34
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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.
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
15/34
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
We now look at example 6-10
REVERSIBLE STEADY FLOW WORK
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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 =
We now look at example 6-17
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
-
7/28/2019 Thermo Chap VI
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
=
= =
= =
-
7/28/2019 Thermo Chap VI
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= =
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
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
+ + =
+ + =
=
=
-
7/28/2019 Thermo Chap VI
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
-
7/28/2019 Thermo Chap VI
33/34
-
7/28/2019 Thermo Chap VI
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
C dT Rso s s dv If C Const
T v
T vs s C R T v
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
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
C dT Rso s s dv If C Const
T v
T vs s C R
T v
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
Tthen s s C R
T
= + = +
= =
= + =
= +
=
= = +
= + =
=
2
1
lnP
P