thermodynamics lecture series
DESCRIPTION
Thermodynamics Lecture Series. Assoc. Prof. Dr. J.J. Entropy – Quantifying Energy Degradation. Applied Sciences Education Research Group (ASERG) Faculty of Applied Sciences Universiti Teknologi MARA. email: [email protected] http://www5.uitm.edu.my/faculties/fsg/drjj1.html. Quotes. - PowerPoint PPT PresentationTRANSCRIPT
Thermodynamics Lecture Series
email: [email protected]://www5.uitm.edu.my/faculties/f
sg/drjj1.html
Applied Sciences Education Research Group (ASERG)
Faculty of Applied SciencesUniversiti Teknologi MARA
Entropy – Quantifying Entropy – Quantifying Energy DegradationEnergy Degradation
QuotesQuotes
• “The principal goal of education is to create men and women who are capable of doing new things, not simply repeating what other generations have done” Jean Piaget
• “What we have to learn to do, we learn by doing” Einstein
How to relate changes to the cause
Dynamic Energies as causes (agents)
of change
SystemE1, P1, T1, V1
ToE2, P2, T2, V2
Properties will change indicating change of
state
Mass out
Mass inWin
Wout
Qin
Qout
Review - First LawReview - First Law
Energy Balance
Ein – Eout = Esys, kJ orein – eout = esys, kJ/kg or
Energy Entering a system
-
Energy Leaving a system
=
Change of system’s energy
kW,EEE sysoutin
Review - First LawReview - First Law
Mass Balance
min – mout = msys, kg or
Mass Entering a system
-
Mass Leaving a system
=
Change of system’s
mass
s/kg,mmm sysoutin
Review - First LawReview - First Law
Esys= 0, kJ; esys= 0 , kJ/kg, Vsys= 0, m3; msys= 0 or min = mout , kg
Energy Balance – Control Volume
Steady-Flow
kg/s ,0msys
Steady-flowSteady-flow is a flow where all propertiesall properties within boundary of the system remains constant with timeremains constant with time
kg/s ,mm or 0mm outinoutin
Review - First LawReview - First Law
Mass & Energy Balance–Steady-
Flow: Single Stream
outinoutin WWQQ
qin – qout+ in – out = out – in, kJ/kg
Energy balance kJ/s ,EE So, .0E outinsys
kg/s , mm So, .0m outinsys
Mass balance
kW ,mminout
Review - First LawReview - First Law
Second LawSecond Law
Steam Power Plant
High T Res., TH
Furnace
qin = qH
net,out
Low T Res., TL
Water from river
An Energy-Flow diagram for a SPP
qout = qL
Working fluid:
Water
qin - qout = out - in
net,out = qin - qout
Purpose:
Produce work,
Wout, out
qin = net,out + qout
Second LawSecond Law
Thermal Efficiency for steam power plants
in
out,net
qnputi equiredr
output desired
in
out,net
q
in
outin
q
in
out
in
in
q
q
q
q
in
out
q
q1
H
L
q
q1
Second LawSecond Law
Refrigerator/ Air Cond
High T Res., TH, Kitchen room / Outside house
qout = qH
net,in
Low TemperatureRes., TL, Inside fridge or house
An Energy-Flow diagram for a Refrigerator/Air Cond.
qin = qL
Working fluid:
Ref-134a
qout – qin = in - out
net,in = qout - qin Purpose:
Maintain space
at low T by
Removing qL
net,in = qH - qL
Second LawSecond Law
Coefficient of Performance for a Refrigerator
in,net
inR
q
nputi equiredr
output desiredCOP
in,net
inR
qCOP
inout
in
q
in
in
in
out
1
1qq
1
in
out
1qq
1
L
H
Second LawSecond Law
Heat Pump
High TemperatureRes., TH, Inside house
qout = qH
net,in
Low TemperatureRes., TL, Outside
house
An Energy-Flow diagram for a Heat Pump
qin = qL
Working fluid:
Ref-134a
qout = net,in + qin
net,in = qout - qin
Purpose:
Maintain space
at high T by
supplying qH
net,in = qH - qL
Second LawSecond Law
Coefficient of Performance for a Heat Pump
in,net
outHP
q
nputi equiredr
output desiredCOP
in,net
outHP
qCOP
inout
out
q
out
in
out
out
1
out
in
1
1
H
L
1
1
Second Law – Energy DegradeSecond Law – Energy Degrade
Factors of irreversibilities• less heat can be converted to work
– Friction between 2 moving surfaces– Processes happen too fast– Non-isothermal heat transfer
What is the maximum performance of real engines if it can never achieve 100%??
Second Law – Dream EngineSecond Law – Dream EngineCarnot Cycle• Isothermal expansionIsothermal expansion
Slow adding of Q resulting in work done by system (system expand)
Qin – Wout = U = 0. So, Qin = Wout . Pressure drops.
• Adiabatic expansionAdiabatic expansion0 – Wout = U. Final U smaller than initial U.T & P drops.
Second Law – Dream EngineSecond Law – Dream EngineCarnot Cycle• Isothermal compression
Work done on the systemSlow rejection of Q- Qout + Win = U = 0. So, Qout = Win .Pressure increases.
• Adiabatic compression0 + Win = U. Final U higher than initial U.T & P increases.
Second Law – Dream EngineSecond Law – Dream EngineCarnot Cycle
P - diagram for a Carnot (ideal) power plantP, kPa
, m3/kgqout
qin
2
34
1
Second Law – Dream EngineSecond Law – Dream EngineReverse Carnot Cycle
P - diagram for a Carnot (ideal) refrigeratorP, kPa
qout
qi
n
3
4
2
1
, m3/kg
Second Law – Dream EngineCarnot Principles• For heat engines in contact with the
same hot and cold reservoir All reversible engines have the same
performance.Real engines will have lower
performance than the ideal engines.
(K)
(K)
L
H
revL
H
T
T
q
q
Second LawSecond Law
Steam Power Plants
High T Res., TH
Furnace
qin = qH
net,out
Low T Res., TL
Water from river
An Energy-Flow diagram for a Carnot SPPs
qout = qL
Working fluid:
Not a factor
P1: 1 = 2 = 3
P2: real < rev
(K)
(K)
H
Lrev T
T1
H
Lreal q
q1
Second LawSecond Law
Rev. Fridge/ Heat Pump
High T Res., TH, Kitchen room / Outside house
net,in
Low TemperatureRes., TL, Inside fridge or house
An Energy-Flow diagram for Carnot Fridge/Heat Pump
Working fluid:
Not a factor
revH
LHP
1
1COP
rev
1TT
1COP
L
HRrev
1qq
1COP
L
HR
H
LHP
1
1COP
H
LHP
TT
1
1COP
rev
1qq
1COP
revL
HRrev
qin = qL
qout = qH
Carnot Principles• For heat engines in contact with the
same hot and cold reservoir P1: 1 = 2 = 3 (Equality)P2: real < rev (Inequality)
Second Law – Will a Process HappenSecond Law – Will a Process Happen
Processes satisfying Carnot Principles obeys the Second Law of Thermodynamics
revreal
Clausius Inequality :• Sum of Q/T in a cyclic process must be
zero for reversible processes and negative for real processes
K
kJ ,0
T
Q
Second Law – Will a Process HappenSecond Law – Will a Process Happen
,0T
Q
,0T
Q
Kkg
kJ ,0
T
q
reversible
impossible
real
,0T
Q
Steam Power Plant
Source
qin = qH
net,out
Sink
qout = qL
0T
Q
T
Q
T
Q
L
L
H
H
0T
Q
T
Q
L
L
H
H
0T
Q
Q
T
Q
T
T
Q
L
L
H
H
H
H
H
H
0T
T
Q
Q1
L
H
revH
L
0T
T
T
T1
L
H
H
L
011
Second Law – Will a Process HappenSecond Law – Will a Process Happen
Processes satisfying Clausius Inequality obeys the Second Law of Thermodynamics
Carnot SPP
Steam Power Plant
Source
net,out
Sink
revintT
QdS
2
1
12 dSSSS
Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy
Entropy Change in a process
2
1 revint12 T
QSSS
H
H
H
outsource T
Q
T
QS
H
H
H
insys T
Q
T
QS
qout = qL
qin = qH
L
L
L
outsys T
Q
T
QS
L
L
L
inksin T
Q
T
QS
• EntropyQuantitative measure of disorder or chaosIs a system’s property, just like the othersDoes not depend on process pathHas values at every state
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
Entropy Quantifies lost of energy quality Can be transferred by heat and mass or generated due to irreversibilty factors:
Frictional forces between moving surfaces. Fast expansion & compression. Heat transfer at finite temperature difference.
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
Increase of Entropy PrincipleThe entropy of an isolated (closed and
adiabatic) system undergoing any process, will always increase.
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
Surrounding
System
0SSS surrsysisolated
)ss(mS 12sys
surr
surroutinsurr T
QQS
For pure substance:
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Clausius Inequality
1
2 revint
2
1T
Q
T
Q
T
Q
1
2 revint
1
2
21 T
QdSSS
So:
0SST
Q21
2
1
Then:
2
1
12 T
QSS
0
EntropyChange
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Then entropy change for the closed system:
2
1
12sys T
QSSS
rev. , 2
1
12sys T
QSSS
irrev. , 2
1
12sys T
QSSS
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Then entropy change for the closed system:
2
1
12sys T
QSSS
genS heat12sys SSSS
gen
2
1
12sys ST
QSSS
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Then entropy change for the closed system:
genS heat12sys SSSS
gen S 0SSS 12sysadiabFor adiabatic process:
genS 12 SS
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Then entropy change for the closed system:
genS heat12sys SSSS
0 0SSS 12sysiso
For adiabatic reversible system:
12 SS
Isentropic or constant entropy process
Increase of Entropy Principle Proven– Consider the following cyclic process containing an
irreversible forward path and a reversible return path
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
irreversibl
e reversible
21
Then entropy change for the closed system:
genS heat12sys SSSS
0S0S gensysiso
For isolated (adiabatic & closed) system:
0SSS surrsysgen
T – S diagram
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
Area of curve under P – V diagram represents total work done
Area of curve under T – S diagram represents total heat transfer
revintT
QdS
TdSQ revint
Recall
Then 2
1
2
1
revint TdSQQ
Hence total heat transfer is
Area under T- S diagram is amount of heat in a process
T – s diagram
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
•Adding all the area of the strips from state 1 to state 2 will give the total area under process curve. It represents specific heat received for this process
2
1
in
2
1
qTdsdAA
The infinite area dA = area of strip = Tds
s, kJ/kgK
2
1
qin=qH
T, C
TH
T
TL
ds
Factors affecting Entropy (disorder)
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
Entropy will change when there is
Heat transfer (receiving heat increases entropy)
Mass transfer (moving mass changes entropy)
Irreversibilities (entropy will always be generated)
Entropy Balance
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
For any system undergoing any process,
Energy must be conserved (Ein – Eout = Esys)
Mass must be conserved (min – mout = msys)
Entropy will always be generated except for reversible processes
Entropy balance is (Sin – Sout + Sgen = Ssys)
Entropy Balance –Closed system
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
pekeuqq outinoutin
sysgenoutin ssss
sysgenoutmassheatinmassheat ssssss
inheatoutheatsysgen 0s0sss
sysin
in
out
outsys12gen T
q
T
q)ss(ms
Energy Balance:
Entropy Balance: Kkg
kJ,
Kkg
kJ,
Kkg
kJ,
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
outinoutin WWQQ
kW ,mminout
0SSSS sysgenoutin
inoutgen SSS,So
Then:
in
massheat
out
massheatgen SSSSS
inletin
in
exitout
outgen sm
T
Qsm
T
QS
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
outinoutin WWQQ
kW ,m inletexit
Nozzle:
kW ,kehkehm0000 1122
Assume adiabatic, no work done,pemass = 0
where
mmm exitinlet
K
kW,ssm00S 12gen
Entropy BalanceK
kW,smsm
T
Q
T
QS 1122
in
in
out
outgen
InInState 1State 1
OutOutState 2State 2
AA22 << A << A11
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
outinoutin WWQQ
kW ,m inletexit
Turbine:
kW ,hhmW000 12out
Assume adiabatic, kemass = 0,
pemass = 0
where
mmm exitinlet
K
kW,ssm00S 12gen
EntropyBalance
K
kW,smsm
T
Q
T
QS 1122
in
in
out
outgen
In
Out
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
outinoutin WWQQ
kW ,m inletexit
Heat exchanger: energy balance; 2 cases
kW,hmhmhmhm0 11223344
Assume kemass = 0,pemass = 0
where
mmm exitinlet
kW,hmhmQQ 1122outin
1 2
4
3Qin
kW,hmhm0Q 1122in
Case 1
Case 2
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
K
kW,smsm
T
Q
T
QS 1122
in
in
out
outgen
Heat exchanger:
mmm exitinlet
K
kW,smsmsmsm00S 11223344gen
Entropy Balance
where
1 2
4
3Qin
Case 1
Case 2
Entropy Balance –Steady-flow device
Entropy – Quantifying DisorderEntropy – Quantifying Disorder
outinoutin WWQQ
kW ,mminletexit
K
kW,smsmsm
T
Q
T
QS 112233
in
in
out
outgen
Mixing Chamber:
exitinlet mm
where
kW,hmhmhmWWQQ 112233outinoutin
1
3
2