Thermodynamics Lecture Series

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


Mass Balance

min – mout = msys, kg or

Mass Entering a system

-

Mass Leaving a system

=

Change of system’s

mass

s/kg,mmm sysoutin


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


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


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


Thermal Efficiency for steam power plants

in

out,net

qnputi equiredr

output desired

in

out,net

q

in

outin

q

qq

in

out

in

in

q

q

q

q

in

out

q

q1

H

L

q

q1


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


Coefficient of Performance for a Refrigerator

in,net

inR

q

nputi equiredr

output desiredCOP

in,net

inR

qCOP

inout

in

qq

q

in

in

in

out

qq

qq

1

1qq

1

in

out

1qq

1

L

H


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


Coefficient of Performance for a Heat Pump

in,net

outHP

q

nputi equiredr

output desiredCOP

in,net

outHP

qCOP

inout

out

qq

q

out

in

out

out

qq

qq

1

out

in

qq

1

1

H

L

qq

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


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


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

qq

1

1COP

rev

1TT

1COP

L

HRrev

1qq

1COP

L

HR

H

LHP

qq

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


,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


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.


Increase of Entropy PrincipleThe entropy of an isolated (closed and

adiabatic) system undergoing any process, will always increase.


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


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




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




irreversibl

e reversible

21


2

1

12sys T

QSSS

genS heat12sys SSSS

gen

2

1

12sys ST

QSSS




irreversibl

e reversible

21


genS heat12sys SSSS

gen S 0SSS 12sysadiabFor adiabatic process:

genS 12 SS




irreversibl

e reversible

21


genS heat12sys SSSS

0 0SSS 12sysiso

For adiabatic reversible system:

12 SS

Isentropic or constant entropy process




irreversibl

e reversible

21


genS heat12sys SSSS

0S0S gensysiso

For isolated (adiabatic & closed) system:

0SSS surrsysgen

T – S diagram


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


•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 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


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


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


outinoutin WWQQ

kW ,mminout

0SSSS sysgenoutin

inoutgen SSS,So

Then:

in

massheat

out

massheatgen SSSSS

inletin

in

exitout

outgen sm

T

Qsm

T

QS



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



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



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



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



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

Thermodynamics Lecture Series

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