thermodynamics lecture series - universiti teknologi...

Thermodynamics Lecture Series

Ideal Ideal Rankine Rankine Cycle Cycle ––The The Practical CyclePractical Cycle

Applied Sciences Education Research Group (ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

email: [email protected]://www5.uitm.edu.my/faculties/fsg/drjj1.html

Steam Power PlantExample: A steam power cycle.Example: A steam power cycle.

SteamTurbine Mechanical Energy

to Generator

Heat Exchanger

Cooling Water

Pump

Fuel

Air

CombustionProducts

System Boundaryfor ThermodynamicAnalysis

System Boundaryfor ThermodynamicAnalysis

Second LawSecond LawHigh T Res., TH

Furnace

qin = qH

Low T Res., TLWater from river

qout = qL

Working fluid:Water Purpose:

Produce work,Wout, ωout

Steam Power Plant ωnet,out

An Energy-Flow diagram for a SPP

Second Law – Dream EngineSecond Law – Dream Engine

Carnot CycleP - ν diagram for a Carnot (ideal) power plantP, kPa

ν, m3/kgqout

qin

2

34

1

What is the maximum performance of real engines if it can never achieve 100%??

in

out,net

qnputi equiredroutput desired ω

η ==

revin

outinrev q

qq

−=η

Carnot Principles• For heat engines in contact with the same hot

and cold reservoir P1: η1 = η2 = η3 (Equality)P2: ηreal < ηrev (Inequality)

Second Law – Will a Process HappenSecond Law – Will a Process Happen

revreal ηη ≤

;(K) (K)

H

L

revH

L

TT

qq

=

(K) (K) 11

H

L

revH

Lrev T

Tqq

−=

−=η

Consequence

Processes satisfying Carnot Principles obeys the Second Law of Thermodynamics

Clausius Inequality :• Sum of Q/T in a cyclic process must be zero

for reversible processes and negative for real processes

∫ ≤KkJ ,0

TQδ

Second Law – Will a Process HappenSecond Law – Will a Process Happen

∫ = ,0TQδ

∫ < ,0TQδ

∫ •≤

KkgkJ ,0

Tqδ

reversiblereal

∫ > ,0TQδ impossible

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Isolated systemsFIGURE 6-6The entropy change of an isolated system is the sum of the entropy changes of its components, and is never less than zero.

6-3

Entropy – Quantifying DisorderEntropy – Quantifying DisorderIncrease of Entropy Principle – closed system

The entropy of an isolated (closed and adiabatic) system undergoing any process, will always increase.

Surrounding

System

0≥∆+∆=+=∆ surrsysgenheatisolated SSSSS

)ss(mS 12sys −=∆

( )surr

surroutinsurr T

QQS

−=∆

For pure substance:

( )surrTQ

geninnetssmS ,)( 12 +−=

and

Then

Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance – for any general system

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 + SSgengen = ∆Ssys)


Entropy Transfer

FIGURE 6-61Mechanisms of entropy transfer for a general system.

6-18

Entropy – Quantifying DisorderEntropy – Quantifying DisorderEntropy Balance –Steady-flow device

outinoutin WWQQ••••

−+− kW ,mminout

−

=

••ϑϑ

0SSSS sysgenoutin ==+−•••

∆ inoutgen SSS,So•••

−=

Then:

inmassheat

outmassheatgen SSSSS

+−

+=

•••••

inletin

in

exitout

outgen sm

TQsm

TQS

−−

+= ∑∑

••

••

•



−+− ( ) kW ,m inletexit ϑϑ −=•

( ) kW ,000 34 hhmW out −=−+−••

Assume adiabatic, ∆kemass = 0,∆pemass = 0

where•••

== mmm exitinlet

( )K

kWssmS gen ,00 34 −+−=••

EntropyBalance K

kWsmsmTQ

TQS

in

in

out

outgen ,3344

••••

•

−+−=In,3

Out

Turbine:



−+− kW ,mminletexit

∑∑

−

=

••ϑϑ

Mixing Chamber:

KkW,smsmsm

TQ

TQS 112233

in

in

out

outgen

•••••

•−−+−=

exitinlet mm••

=

kW,hmhmhmWWQQ 112233outinoutin •••••••

−−=−+−

1

3

2where

Vapor Cycle Vapor Cycle Steam Power Plant

External combustionFuel (qH) from nuclear reactors, natural gas, charcoal Working fluid is H2OCheap, easily available & high enthalpy of vaporization hfg

Cycle is closed thermodynamic cycleAlternates between liquid and gas phaseCan Can Carnot Carnot cycle be used for representing real SPP??cycle be used for representing real SPP??Aim: To decrease ratio of TL/TH

Vapor Cycle – Carnot CycleVapor Cycle – Carnot CycleEfficiency of a Carnot Cycle SPP

55.0273374

273151TT1

H

Lrev =

++

−=−=η

627.0273500

273151TT1

H

Lrev =

++

−=−=η

Vapor Cycle –Carnot CycleVapor Cycle –Carnot CycleImpracticalities of Carnot Cycle

Isothermal expansion: TH limited to only Tcrit for H2O.High moisture at turbine exitNot economical to design pump to work in 2-phase (end of Isothermal compression)No assurance can get same xfor every cycle (end ofIsothermal compression)

s3 = s4s1 = s2

qin = qHT, °C

Tcrit

TH

TL

qout = qL

s, kJ/kg°K

Vapor Cycle – Alternate Carnot CycleVapor Cycle – Alternate Carnot CycleImpracticalities of Alternate Carnot Cycle

s3 = s4s1 = s2

qin = qHT, °C

Tcrit

TH

TL

qout = qL

s, kJ/kg°K

Still ProblematicIsothermal expansion but at

variable pressurePump to very high pressure

Can the boiler sustain the high P?

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine Cycle

Overcoming Impracticalities of Carnot Cycle

SuperheatSuperheat the H2O at a constant pressure (isobaric expansion)Can easily achieve desired TH higher than Tcrit.reduces moisture content at turbine exit

Remove all excess heat at condenserPhase is sat. liquid at condenser exit, hence need only a pump to increase pressure Quality is zero for every cycle at condenser exit(pump inlet)


Pum

p

Boiler Turbine

Condenser

qin = qH

ωout

Low T Res., TLWater from river

qout = qLqin - qout = ωout - ωin

qin - qout = ωnet,out

High T Res., THFurnace

Working fluid:Water

ωin

A Schematic diagram for a Steam Power Plant

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleT- s diagram for an Ideal Rankine Cycle

T, °C

s, kJ/kg°K

1

2

Tcrit

TH

TL= Tsat@P4

Tsat@P2

s3 = s4s1 = s2

qin = qH

4

3

PHPL

ωin

ωout

pump

qout = qL

condenser

turbineboiler


FIGURE 9-2The simple ideal Rankinecycle.

9-2


BoilerIn,2 Out,3

qin = qH

Energy AnalysisAssume ∆ke =0, ∆pe =0 for the moving mass, kJ/kg

qin – qout+ ωin – ωout = θout – θin, kJ/kg

qin – 0 + 0 – 0 = hexit – hinlet, kJ/kg

Qin = m(h3 – h2), kJqin = h3 – h2, kJ/kg

( ) kWhhmQ in ,23 −=••

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis Assume ∆ke =0, ∆pe =0 for

the moving mass, kJ/kg


0 – qout + 0 – 0 = hexit – hinlet - qout = h1 – h4,

CondenserOut,1 In,4

qout = qL

( ) kWhhmQ out ,14 −=••

So, qout = h4 – h1, kJ/kg

Qout = m(h4 – h1), kJ

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy Analysis

ωout

In,3

Out,4

Turbineqin – qout+ ωin – ωout = θout – θin, kJ/kg

0 – 0 + 0– ωout = hexit – hinlet, kJ/kg

- ωout = h4 – h3, kJ/kg So, ωout = h3 – h4, kJ/kg

Wout = m(h3 – h4), kJ Assume ∆ke =0, ∆pe =0 for the moving mass, kJ/kg( ) kWhhmW out ,43 −=

••


Pum

pωin

Out,2

In,1

∫∫∫ +=+=2

1

2

1

2

1in,pump dP0dPPd νννω

( ) 1212in,pump hhPP −=−=νω

1P@f12 ννν =≅

So, Win = m(h2 – h1), kJ

( ) kWhhmW in ,12 −=••


0 – 0 + ωin – 0= hexit – hinlet, kJ/kg

ωin = h2 – h1, kJ/kgFor reversible pumps

where


Efficiency

( )23

1423

in

outin

in

out,net

hhhhhh

qqq

q −−−−

=−

==ω

η

( )23

1243

in

inout

in

out,net

hhhhhh

qq −−−−

=−

==ωωω

η

23

1243

hhhhhh

−+−−

==η

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleT- s diagram for an Ideal Rankine Cycle

T, °C

s, kJ/kg°K

1

2

Tcrit

TH

TL= Tsat@P4

Tsat@P2

s3 = s4s1 = s2

qin = qH

4

3

PHPL

ωin

ωout

pump

qout = qL

condenser

turbineboiler

s1 = sf@P1 h1 = hf@P1

s3 = s@P3,T3

s4 = [sf +xsfg]@P4 = s3

h3 = h@P3,T3

h4 = [hf +xhfg]@P4

4P@fg

4P@f3

sss

x−

=

Note that P1 = P4

1P@f12 ννν =≅h2 = h1 +ν2(P2 – P1); where

Vapor Cycle – Ideal Rankine CycleVapor Cycle – Ideal Rankine CycleEnergy AnalysisIncreasing Efficiency

Must increase ωnet,out = qin – qout

Increase area under process cycleDecrease condenser pressure; P1=P4

Pmin > Psat@Tcooling+10 deg C

Superheat T3 limited to metullargical strength of boiler

Increase boiler pressure; P2=P3

Will decrease quality (an increase in moisture). Minimum x is 89.6%.


Lowering Condenser PressureFIGURE 9-6The effect of lowering the condenser pressure on the ideal Rankinecycle.

9-4


Superheating SteamFIGURE 9-7The effect of superheating the steam to higher temperatures on the ideal Rankine cycle.

9-5


9-6

FIGURE 9-8The effect of increasing the boiler pressure on the ideal Rankinecycle.

Increasing Boiler Pressure


FIGURE 9-10T-s diagrams of the three cycles discussed in Example 9–3.

9-8

Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CyclePu

mp

Boiler High

P turbine

Condenser

High T Reservoir, TH

qin = qH

qout = qL

Low

P turbine1

23

4

5

6

qreheat

ωout,1

ωin

ωout,2

Low T Reservoir, TL


FIGURE 9-11The ideal reheatRankine cycle.

9-9

Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleReheating increases η and reduces moisture in turbine

TL= Tsat@P1

ωin

s5 = s6s1 = s2

Tcrit

TH

Tsat@P4

Tsat@P3

s3 = s4

qout = h6-h1

ωout, II

P4 = P5

P6 = P1

61

5

4

qreheat = h5-h4

qprimary = h3-h2 ωoutP3

3

2

T, °C

s, kJ/kg°K

Vapor Cycle – Reheat Rankine CycleVapor Cycle – Reheat Rankine CycleEnergy Analysis

q in = qprimary + qreheat = h3 - h2 + h5 - h4 qout = h6-h1

ωnet,out = ωout,1 + ωout,2 - ωin = h3 - h4 + h5 - h6 – h2 + h1

( )4523

164523

in

outin

in

out,net

hhhhhhhhhh

qqq

q −+−−−−+−

=−

==ω

η

4523

12654321,

hhhhhhhhhh

qq in

inoutout

in

outnet

−+−+−−+−

=−+

==ωωωω

η


s6 = [sf +xsfg]@P6. Use x = 0.896 and s5 = s6where

h6 = [hf +xhfg]@P6

Knowing s5 and T5, P5 needs to be estimated (usually approximately a quarter of P3 to ensure x is around 89%. On the property table, choose P5 so that the entropy is lower than s5 above. Then can find h5 = h@P5,T5.


where s1 = sf@P1 h1 = hf@P1

s3 = s@P3,T3 = s4. h3 = h@P3,T3

1P@f12 ννν =≅h2 = h1 +ν2(P2 – P1); where

P5 = P4.

From P4 and s4, lookup for h4 in the table. If not found, then do interpolation.


Supercritical Rankine CycleFIGURE 9-9A supercriticalRankine cycle.

9-7

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