3_heat&work

8/6/2019 3_heat&work

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

Applied Sciences Education Research Group(ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

Dynamic Energy TransferDynamic Energy Transfer Heat, Work and MassHeat, Work and Mass

Thermodynamics Lecture

Series

CHAPTER

2

Properties of PureSubstances- A Review

Pure substance

QuotesQuotes

"Good judgment comes from experience.Experience comes from bad judgment

- Anonymous

"The roots of education are bitter, butthe fruit is sweet." -Aristotle

"The roots of education are bitter, butthe fruit is sweet." -Aristotle

Example: A steam power cycle.Example: A steam power cycle.

Steam

TurbineMechanical Energy

to Generator

Heat

Exchanger

Cooling Water

Pump

Fuel

Air

Combustion

Products

System Boundary

for Thermodynamic

Analysis

System Boundary

for Thermodynamic

Analysis

Steam Power Plant

Phase Change of WaterPhase Change of WaterPhase Change of Water

Water interacts with thermal energyWater interacts with thermal energy

H2O:

Sat. Liq.

Sat. VaporSat. Vapor

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:Sat. Vapor

H2O:Sat. Vapor

Qin

P = 100 kPa

T = 150 C

P = 100 kPa

T = 150 C

H2O:Super

Vapor

H2O:Super

Vapor

Qin

P = 100 kPa

T = 30 C

P = 100 kPa

T = 30 C

H2O:

C. liquid

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2OSat.

liquid

Qin


99.6

2=f@100kP

a

T, C

30, m3/kg

1

4=g@100kPa

3

5 =@100 kPa, 150C

3 = [f + x f g]@100 kPa

1 =f@T1

150100

kP

a

5

Compressed liquidCompressed liquid: Good: Good

estimation for propertiesestimation for propertiesby taking yby taking y == yyf@Tf@T wherewhere

y can be eithery can be either , u, h or, u, h ors.s.


2/72

Copyright The McGraw-Hill Companies, Inc. Permission required for reproduc tion or display.

2-2

FIGURE 2-16

T-vdiagram ofconstant-

pressurephase-change

processes of a

puresubstance at

variouspressures

(numerical values

are for water).

99.6

45.8

179.9

T v diagram: Multiple P

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

2-3

FIGURE 2-18

T-vdiagram of a

pure substance.

T v diagram: Multiple P

Phase Change of Water - Pressure ChangePhase Change of WaterPhase Change of Water -- Pressure ChangePressure Change

H2O:

Sat. Vapor

H2O:

Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Super

Vapor

H2O:Super

Vapor

T = 30 C

P = 2 kPa

T = 30 C

P = 2 kPa

H2O:

Sat. Liq.


T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

T = 30 C

P = 100 kPa

T = 30 C

P = 100 kPa

H2O:

C. liquid

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:

Sat. liquid

Water when pressure is reducedWater when pressure is reduced

Tsat@100 kPa = 99.63 C

Psat@30 C = 4.246 kPa

TTsat@100sat@100 kPakPa = 99.63= 99.63 CC

PPsat@30sat@30 CC = 4.246= 4.246 kPakPa

Phase Change of Water-Pressure ChangePhase Change of WaterPhase Change of Water-- Pressure ChangePressure Change

Compressed liquidCompressed liquid: Good: Good

estimation for propertiesestimation for propertiesby taking yby taking y == yyf@Tf@T wherewhere

y can be eithery can be either , u, h or, u, h ors.s.

2=f@30C

4.246

3

2

5

4=g@30C

, m3/kg

1

100

P, kPa

30 C 3 = [f + x f g]@ 30 C

4 = g@ 30 C

1 = f@ 30C

2 = f@ 30 C

5= @2kPa, 30 C

Tsat@100 kPa = 99.63 C

Psat@30 C = 4.246 kPa




P, C

, m3/kg

101.35

g@100

C

1,553.8

1.2276

200C

10 C

100C

f@100C

22,090

P- diagramwith respect tothe saturation

lines

P- diagramwith respect tothe saturation

lines

Saturated Liquid-Vapor Mixture

H2O:

Sat. Liq.


t

g

m

mx =

Vapor Phase:, VVapor Phase:, Vgg, m, mgg,, gg,, uugg,, hhgg

Liquid Phase:,Liquid Phase:,VVff, m, mff,, ff,, uuff,, hhff

Mixture:, V, m,Mixture:, V, m,, u,, u, h, xh, xMixtures qualityMixtures qualityMixtures quality

Divide bytotal mass, mt

Divide byDivide bytotal mass,total mass, mmtt

gf

t

g

avg xm

m +

= 1

ggffavgt mmm +=

fgfg =wherewherewhere

fgfxx +=

fgf x +=

) ggfgt mmm +=

Given the pressure, P, thenT = Tsat, yf < y


3/73

Saturated Liquid-Vapor Mixture

H2O:

Sat. Liq.


t

g

m

mx =

Vapor Phase:, VVapor Phase:, Vgg, m, mgg,, gg,, uugg,, hhgg

Liquid Phase:,Liquid Phase:, VVff, m, mff,, ff,, uuff,, hhff

Mixture:, V, m,Mixture:, V, m,, u,, u, h, xh, xMixtures qualityMixtures qualityMixtures quality

Given the pressure, P, thenT = Tsat, yf < y


4/74


3-1

FIGURE 3-9

Specifying the

directions of

heat and work.

Isochoric Process Closed System

Isochoric ProcessIsochoric ProcessClosed SystemClosed System

Isobaric Process Closed SystemIsobaric ProcessIsobaric Process Closed SystemClosed System

H2O:

Sat. Liq.


P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:

Sat. Vapor

H2O:

Sat. Vapor

Qin

P = 100 kPa

T = 150 C

P = 100 kPa

T = 150 C

H2O:Super

Vapor

H2O:Super

Vapor

Qin

P = 100 kPa

T = 30 C

P = 100 kPa

T = 30 C

H2O:C. liquid

Qin

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2OSat.

liquid

Qin

Source of thermal energySource of thermal energySource of thermal energy

System expands, volume increasesSystem expands, volume increasesSystem expands, volume increases

Isobaric Process Closed SystemIsobaric ProcessIsobaric Process Closed SystemClosed System

System expands, volume increasesSystem expands, volume increasesSystem expands, volume increases

H2O:

Sat. Liq.


P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2O:

Sat. Vapor

H2O:

Sat. Vapor

P = 100 kPa

T = 150 C

P = 100 kPa

T = 150 C

H2O:Super

Vapor

H2O:Super

Vapor

P = 100 kPa

T = 30 C

P = 100 kPa

T = 30 C

H2O:C. liquid

P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

H2OSat.

liquid

Source of thermal energySource of thermal energySource of thermal energy

Qin, kJQQinin, kJ, kJkg

kJ

m

Qq in

in,= ,

t

QQ in

in =

kW

s

kJor

Symbols and Convention Heat transferSymbols and ConventionSymbols and Convention Heat transferHeat transfer

Total heat entering, Qin = 100 kJ, m = 5 kgTotal heat entering,Total heat entering, QQinin = 100 kJ, m = 5 kg= 100 kJ, m = 5 kg

kWs

kJ

s

kJ

t

QQ ou t

in 2250

100 ===

=

kg

kJ

kg

kJ

m

Qq inin 20

5

100 ===Specific heatSpecific heatSpecific heat

Rate of heat transfer if heat is allowed to interact for 50 seconRate of heat transfer if heat is allowed to interact for 50 secoRate of heat transfer if heat is allowed to interact for 50 seconn


Total heat leaving, Qout = 20 kJ, m = 5 kgTotal heat leaving,Total heat leaving, QQoutout = 20 kJ, m = 5 kg= 20 kJ, m = 5 kg

kWs

kJ

s

kJ

t

QQ ou tou t 4.04.0

50

20 ===

=

kg

kJ

kg

kJ

m

Qq ou t

ou t 45

20===Specific heatSpecific heatSpecific heat

Rate of heat transfer if heat is allowed to interact for 50 seconate of heat transfer if heat is allowed to interact for 50 secoate of heat transfer if heat is allowed to interact for 50 seconn


Net heat transferNet heat transferNet heat transfer

kWs

kJ

s

kJ

t

QQ

QQou tin

innet 2.12.150

60, ===

==

kg

kJ

kg

kJ

m

Qqq innet

innet 125

60,,

====

Net specific heat transferNet specific heat transferNet specific heat transfer

Net rate of heat transfer if heat is allowedto interact for 50 seconds

Net rate of heat transfer if heat is allowedNet rate of heat transfer if heat is allowedto interact for 50 secondsto interact for 50 seconds

J602080, kkJkJQQQQ ou tininnet ====

H2O:

SuperVapor

H2O:

SuperVapor

QoutQout

Qin

Qin


5/75

Example: A steam power cycle.Example: A steam power cycle.

SteamTurbine

Mechanical Energy

to Generator

Heat

Exchanger

Cooling Water

Pump

Fuel

Air

Combustion

Products

System Boundary

for Thermodynamic

Analysis

System Boundary

for Thermodynamic

Analysis

Steam Power Plant

Qout

Qout

Win

Win

WoutWout

QinQin

Energy Transfer Work Done

ii

Voltage, VVoltage, V

No heat transfer

T increasesafter some time

No heat transfer

T increasesafter some time

H2O:

SuperVapor

H2O:

SuperVapor

Mechanical work:Piston moves up

Boundary work isdone by system

Mechanical work:Piston moves up

Boundary work isdone by system

Electrical work is done on systemElectrical work is done on system

H2O:

Sat. liquid

Wpw,kJWpw,kJ

We = Vit, kJWe = Vit, kJ

ViWe =

Symbols and conventions for Work IsothermalProcess

Symbols and conventions for WorkSymbols and conventions for Work IsothermalIsothermalProcessProcess

H2O:

Sat. Vapor

H2O:

Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:

Super

Vapor

H2O:

Super

Vapor

T = 30 C

P = 2 kPa

T = 30 C

P = 2 kPa

H2O:

Sat. Liq.


T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

T = 30 C

P = 100 kPa

T = 30 C

P = 100 kPa

H2O:

C. liquid

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:Sat. liquid

Water when pressure is reduced no thermal energy or heatWater when pressure is reduced no thermal energy or heatno thermal energy or heat

Tsat@100 kPa = 99.63 C

Psat@30 C = 4.246 kPa



System expands, volume increases-boundary work is doneSystem expands, volume increasesSystem expands, volume increases--boundary work is doneboundary work is done


Boundary Work,Wb,out

Boundary Work,Boundary Work,WWbb,,outout

dsFWb =

Final, nFinal, nFinal, nFIGURE 3-19A gas does a differentialamount of work Wbas it forces

the piston to move by adifferential amount ds.

InitialInitialInitial

nbbb

f

i

bb WWWWW ,2,1, ... +++==

Total work done by system toexpand from initial to final stateTotal work done by system toTotal work done by system toexpand from initial to final stateexpand from initial to final state

nn

f

i

bbdsFdsFdsFWW +++== ...2211

Work done involves force moving an objectin the direction of movement

WorkWork done involvesdone involves force moving an objectforce moving an objectin the direction of movementin the direction of movement

111222




Pressure exerted on a surface is the ratioof force applied w.r.t. the area of surfacePressure exerted on a surface is the ratioPressure exerted on a surface is the ratioof force applied w.r.t. the area of surfaceof force applied w.r.t. the area of surface

kPa

A

FP ,=

==f

i

i

i

bb dsFWW

As ds approaches zeroAsAs dsds approaches zeroapproaches zero

Final, nFinal, nFinal, n

FIGURE 3-19A gas does a differential

amount of work Wbas it forces



111222





FIGURE 3-19A gas does a differential




111222

Pressure exerted on a surface is the ratioof force applied w.r.t. the area of surfacePressure exerted on a surface is the ratioPressure exerted on a surface is the ratioof force applied w.r.t. the area of surfaceof force applied w.r.t. the area of surface

==f

i

f

ibb PdVWW

PAF=

==f

i

f

i

b dsPAdsFW

ThenThenThen

SoSoSo

dVdsA

=ButButBut HenceHenceHence


6/76





FIGURE 3-19

A gas does a differential




111222

When the pressure is kept constant,Isobaric process,

When the pressure is kept constant,When the pressure is kept constant,Isobaric process,Isobaric process,

( )iff

i

b VVPdVPW == ThenThenThen

iiff

f

i

b VPVPdVPW == OrOrOr wherewherewhere PconstPP if ===


Specific Boundary Work,b,out

Specific Boundary Work,Specific Boundary Work,bb,,outout


FIGURE 3-19

A gas does a differential




111222

When the pressure is kept constant,Isobaric process,

When the pressure is kept constant,When the pressure is kept constant,Isobaric process,Isobaric process,

( )kg

kJPdP if

f

i

b , == ThenThenThen

kg

kJPPdP iiff

f

i

b , == OrOrOr wherewherewhere PconstPP if ===


3-3

FIGURE 3-20

The area under

the processcurve on a P-V

diagramrepresents the

boundary work.

Boundary work on a P V graph

=f

i

bdsFW

=f

i

b dsPAW

=f

i

b PdVW

==AA

PdVdAArea00


3-4

FIGURE 3-22

The net work

done during acycle is the

difference

between thework done by

thesystem and the

work done on

the system.

Work done -Cyclic process

Total work is area of A minus areaof B. Total work is shaded areaTotal work is area of A minus areaTotal work is area of A minus areaof B.of B. Total work is shaded areaTotal work is shaded area

=f

i

b PdVW

=f

i

b Pd

kWt

WW inb ,,

=

Input powerInput powerInput power

kWt

WW ou tb ,,

=

Output powerOutput powerOutput power


3-5

FIGURE 3-48

Schematic for

flow work.

Flow work is energy required to overcome resistance at theBoundary both while entering and leaving the systemFlow work is energy required to overcome resistance at theFlow work is energy required to overcome resistance at theBoundary both while entering and leaving the systemBoundary both while entering and leaving the system

Work done Flow work

kJPVWf ,=

kgkJPf /, =

InletInletInlet

ExitExitExit

kJPVWf ,=

kgkJPf /, =

Energy Transfer Mass Conservation

Consider mass flowing through cylinder of length and areaConsider mass flowing through cylinder of length and areaConsider mass flowing through cylinder of length and areal

Mass must be conserved in any process.Mass must be conserved in any process.Mass must be conserved in any process.

kgmmm sysoutin ,=

0=

sysm

InletExitA

A

Length,l

r

r

ororor

For steadyflow

For steadyFor steadyflowflow

s

kgmmm sysou tin ,

=

ThenThenThen

= ou tin mm


7/77

Energy Transfer Energy of Moving Mass


Let the mass flow through for timeinterval of t, such as 10 secondsLet the mass flow through for timeLet the mass flow through for timeinterval ofinterval of t, such as 10 secondst, such as 10 seconds

Let the mass flowwith a velocityLet the mass flowLet the mass flowwith a velocitywith a velocity

r

Since volume isSince volume isSince volume is mV= Then volumeflow rate is

Then volumeThen volumeflow rate isflow rate is t

m

t

V

=

s

mmV

3

,

=Inlet

ExitAA

Length,l

r

r

ororor

s

kgVm ,

=Then massFlow rate

Then massThen massFlow rateFlow rate



Let the mass flow through for timeinterval of t, such as 10 secondsLet the mass flow through for timeLet the mass flow through for timeinterval ofinterval of t, such as 10 secondst, such as 10 seconds

Let the mass flowwith a velocityLet the mass flowLet the mass flowwith a velocitywith a velocity

r

Since volume isSince volume isSince volume is mAV == l Then volumeflow rate is

Then volumeThen volumeflow rate isflow rate is t

m

t

lA

t

V

=

=

s

mmAV

3

,

== rInletExitA

A

Length,l

r

r

ororor

s

kgAVm ,

r

==

Then massFlow rate

Then massThen massFlow rateFlow rate


FIGURE 3-51

The total energy consists of three parts for anonflowingfluid and four parts for a flowing

fluid.


kg

kJ,

?ke

2000

2r

=

Kinetic energy, keKinetic energy,Kinetic energy,keke

kg

kJ,

ghpe

2000=

Potential energy, pePotential energy,Potential energy, pepe

kg

kJu,

Internal energy, uInternal energy, uInternal energy, uGas Mixtures Ideal GasesGas MixturesGas Mixtures Ideal GasesIdeal Gases

Equation of StateEquation of State - P--T behaviour

Low density

High density

Hence, can also writeHence, can also write

where

N is no of kilomoles, kmol,

M is molar mass in kg/kmole and

Ru is universal gas constant; Ru=MR.

Ru = 8.314 kJ/kmolK

where

NN is no of kilomoles, kmol,

MM is molar mass in kg/kmole and

RRuu is universal gas constant; RRuu=MR=MR.

RRuu = 8.314 kJ/= 8.314 kJ/kmolkmolKK

PV =mRTPV =PV =mRTmRT

PV = NRuTPV =PV = NRNRuuTT

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