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Applied Sciences Education Research Group (ASERG)

Faculty of Applied SciencesUniversiti Teknologi MARA

Dynamic Energy Dynamic Energy Transfer – Heat, Transfer – Heat, Work and MassWork and Mass

Thermodynamics Lecture Series

CHAPTER

2

Properties of Pure Substances- A Review

Pure substance

QuotesQuotes

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

- Anonymous

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

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

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

SteamTurbine

Mechanical Energyto Generator

Heat Exchanger

Cooling Water

Pump

Fuel

Air

CombustionProducts

System Boundaryfor ThermodynamicAnalysis


Steam Power Plant

Phase Change of WaterPhase 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:SuperVapor

H2O:SuperVapor

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@

100 kPa

T, C

30, m3/kg

1

4 =

g@

100 kPa

3

5 = @100 kPa, 150°C

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

1 = f@T1

150

100

kPa

5

Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = y = yf@T f@T where where

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

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

2-2

FIGURE 2-16T-v diagram of constant-pressurephase-change processes of a puresubstance at various pressures(numerical values are for water).

99.6

45.8

179.9

T –v diagram: Multiple P


2-3

FIGURE 2-18T-v diagram of a pure substance.

T –v diagram: Multiple P

Phase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure ChangePhase Change of Water - Pressure Change

H2O:Sat. Vapor

H2O:Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:SuperVapor

H2O:SuperVapor

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

TTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPaTTsat@100 kPasat@100 kPa = 99.63 = 99.63 CCPPsat@30 sat@30 CC = 4.246 kPa = 4.246 kPa

Phase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure ChangePhase Change of Water- Pressure Change

Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = y = yf@T f@T where where

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

2 =

f@

30 °C

4.246

3

2

5

4 =

g@

30 °C

, m3/kg

1

100

P, kPa

30 C 3 = [f + x f g]@ 30 °C

4 = g@ 30 °C

1 = f@ 30 °C

2 = f@ 30 °C

5= @2kPa, 30 °C



P, C

, m3/kg

101.35

g@

100 C

1,553.8

1.2276

200 C

10 C

100 C

f@

100 C

22,090

P- diagram with respect to the saturation

lines

P- diagram with respect to the saturation

lines

Saturated Liquid-Vapor Mixture

H2O:Sat. Liq.


t

g

m

mx

Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg

Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff

Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality

Divide by Divide by total mass, total mass,

mmtt

Divide by Divide by total mass, total mass,

mmtt

gft

gavg x

m

m

1

ggffavgt mmm

fgfg wherewherewherewhere

fgf xx fgf x

ggfgt mmm

Given the pressure, P, then T = Tsat, yf < y <yg


Saturated Liquid-Vapor Mixture

H2O:Sat. Liq.


t

g

m

mx

Vapor Phase:, VVapor Phase:, Vgg, m, mgg, , gg, u, ugg, h, hgg

Liquid Phase:, VLiquid Phase:, Vff, m, mff, , ff, u, uff, h, hff

Mixture:, V, m, Mixture:, V, m, , u, h, x, u, h, xMixture’s qualityMixture’s qualityMixture’s qualityMixture’s quality



fgfg wherewherewherewherefgf x

fgfg yyy wherewherewherewherefgf xyyy

If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If

either either , u, h are known, use it to find quality, x., u, h are known, use it to find quality, x.

If x is known or has been determined, use If x is known or has been determined, use above relations to find other properties. If above relations to find other properties. If

either either , u, h are known, use it to find quality, x., u, h are known, use it to find quality, x.

y can be y can be , u, h, u, hy can be y can be , u, h, u, h

CHAPTER

3

Energy Transfer byHeat, Work, and MassGoal: Identify forms of energy interactions and ways of representing it in thermodynamics processes

IntroductionIntroduction

1. Identify the types of dynamic energies interacting with a system.

2. Distinguish the difference and relate between heat transfer and thermal energy.

3. Write the different symbols and the conventions used to represent heat transfer.

4. Differentiate between heat transfer and work done.

Objectives:Objectives:


5. Write the symbols and convention used for work done.

6. Obtain a mathematical relation representing mechanical work done for any system.

7. Obtain the amount of work done from a P – V or P - graph.

8. Write down the relationship between mass and volume flow rate.



9. Obtain a mathematical relation representing mass flow rate in terms of the mass velocities and the system’s inlet or exit area.

10.Write the specific energy carried by a flowing mass.

11.Use all mathematical relations and graphing skills to solve problems involving interaction energies.


3-1

Energy Transfer -Heat Energy Transfer -Heat TransferTransfer

Energy Transfer -Heat Energy Transfer -Heat TransferTransfer

SODA5C

SODA5C

qqininqqinin

LemLem

OvenOven

200C200C

NasiLemak 20C

NasiLemak 20C

qinqin

H2O:Sat. Liq.


P = 100 kPa

T = 99.6 C

P = 100 kPa

T = 99.6 C

Qin

qqouou

tt

qqouou

tt


3-1

FIGURE 3-9Specifying the directions ofheat and work.

Isochoric Process – Isochoric Process – Closed SystemClosed System

Isochoric Process – Isochoric Process – Closed SystemClosed System

Isobaric Process – Closed SystemIsobaric Process – Closed SystemIsobaric Process – Closed SystemIsobaric Process – Closed 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:SuperVapor

H2O:SuperVapor

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 energySource of thermal energy

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

Isobaric Process – Closed SystemIsobaric Process – Closed SystemIsobaric Process – Closed SystemIsobaric Process – Closed System

System expands, volume increasesSystem 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:SuperVapor

H2O:SuperVapor

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 energySource of thermal energy

QQinin, , kJkJ

QQinin, , kJkJ

kg

kJ

m

Qq in

in , ,t

QQ in

in

kW

s

kJor

Symbols and Convention – Heat transferSymbols and Convention – Heat transferSymbols and Convention – Heat transferSymbols and Convention – Heat transfer

Total heat entering, QTotal heat entering, Qinin = 100 kJ, m = 100 kJ, m = 5 kg= 5 kg

Total heat entering, QTotal heat entering, Qinin = 100 kJ, m = 100 kJ, m = 5 kg= 5 kg

kWs

kJ

s

kJ

t

QQ out

in 2 2 50

100

kg

kJ

kg

kJ

m

Qq in

in 20 5

100Specific heatSpecific heatSpecific heatSpecific heat

Rate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 seconds


Total heat leaving, QTotal heat leaving, Qoutout = 20 kJ, m = 20 kJ, m = 5 kg= 5 kg

Total heat leaving, QTotal heat leaving, Qoutout = 20 kJ, m = 20 kJ, m = 5 kg= 5 kg

kWs

kJ

s

kJ

t

QQ out

out 4.0 4.0 50

20

kg

kJ

kg

kJ

m

Qq out

out 4 5

20Specific heatSpecific heatSpecific heatSpecific heat

Rate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 secondsRate of heat transfer if heat is allowed to interact for 50 seconds


Net heat Net heat transfertransferNet heat Net heat transfertransfer

kWs

kJ

s

kJ

t

QQQQ outin

innet 2.1 2.1 50

60,

kg

kJ

kg

kJqqqq outininnet 12

5

)2080(,

Net specific heat transferNet specific heat transferNet specific heat transferNet specific heat transfer

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

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

J 60 20 80, kkJkJQQQQ outininnet

H2O:SuperVapor

H2O:SuperVapor

QoutQout

Qin

Qin

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

SteamTurbine

Mechanical Energyto Generator

Heat Exchanger

Cooling Water

Pump

Fuel

Air

CombustionProducts



Steam Power Plant

Qou

tQ

ou

t

Win

Win

WoutWout

QinQin

Energy Transfer – Work Done

ii

Voltage, VVoltage, V

No heat transferT increases

after some time

No heat transferT increases

after 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

ViW e

Symbols and conventions for Work – Symbols and conventions for Work – Isothermal ProcessIsothermal Process

Symbols and conventions for Work – Symbols and conventions for Work – Isothermal ProcessIsothermal Process

H2O:Sat. Vapor

H2O:Sat. Vapor

T = 30 C

P = 4.246 kPa

T = 30 C

P = 4.246 kPa

H2O:SuperVapor

H2O:SuperVapor

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


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


Boundary Work, Boundary Work, WWbb,,outout


dsFWb

Final, nFinal, nFinal, nFinal, nFIGURE 3-19A gas does a differential amount of work Wb as it forces the piston to move by a differential amount ds. InitialInitialInitialInitial

nbbb

f

ibb WWWWW ,2,1, ...

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

ibb dsFdsFdsFWW ...2211

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

11112222




Pressure 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 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

kPaA

FP ,

f

i

i

i

bb dsFWW

As ds approaches zeroAs ds approaches zeroAs ds approaches zeroAs ds approaches zero

Final, nFinal, nFinal, nFinal, n

FIGURE 3-19A gas does a differential amount of work Wb as it forces the piston to move by a differential amount ds.

InitialInitialInitialInitial11112222






InitialInitialInitialInitial11112222Pressure exerted on a surface is the ratioPressure exerted on a surface is the ratio

of force applied w.r.t. the area of surfaceof 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

i

bb PdVWW

PAF

f

i

f

i

b dsPAdsFW

ThenThenThenThen

SoSoSoSo

dVdsA ButButButBut HenceHenceHenceHence







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

if

f

i

b VVPdVPW ThenThenThenThen

iiff

f

i

b VPVPdVPW OrOrOrOr wherewherewherewhere PconstPP if


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

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




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

kg

kJPdP if

f

i

b , ThenThenThenThen

kg

kJPPdP iiff

f

i

b , OrOrOrOr wherewherewherewhere PconstPP if


3-3

FIGURE 3-20The area under the process curve on a P-V diagram represents the boundary work.

Boundary work on a P – V graph

f

i

b dsFW

f

i

b dsPAW

f

i

b PdVW

AA

PdVdAArea00


3-4

FIGURE 3-22The net work done during a cycle is thedifference between the work done by thesystem and the work done on the system.

Work done - Cyclic process

Total 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 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 powerInput power

kWt

WW outb ,,

Output powerOutput powerOutput powerOutput power


3-5

FIGURE 3-48Schematic for flow work.

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

Work done – Flow work

kJPVW f ,

kgkJPf / ,

InletInletInletInlet

ExitExitExitExit

kJPVW f ,

kgkJPf / ,

Energy Transfer – Mass Conservation

Consider mass flowing through cylinder of length and area AConsider mass flowing through cylinder of length and area AConsider mass flowing through cylinder of length and area AConsider mass flowing through cylinder of length and area A

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

kgmmm sysoutin ,

0

sysmInlet

ExitAA

Length,

orororor

For steadyFor steadyflowflow

For steadyFor steadyflowflow

s

kgmmm sysoutin ,

ThenThenThenThen

outin mm

Energy Transfer – Energy of Moving Mass


Let the mass flow through for time Let the mass flow through for time interval of interval of t, such as 10 secondst, such as 10 secondsLet the mass flow through for time Let the mass flow through for time interval of interval of t, such as 10 secondst, such as 10 seconds

Let the mass Let the mass flowflowwith a velocity with a velocity


Since volume is Since volume is Since volume is Since volume is mV Then volume Then volume flow rate is flow rate is Then volume Then volume flow rate is flow rate is t

m

t

V

s

mmV

3

,

Inlet

ExitAA

Length,

orororor

s

kgVm ,

Then massThen massFlow rateFlow rate

Then massThen massFlow rateFlow rate



Let the mass flow through for time Let the mass flow through for time interval of interval of t, such as 10 secondst, such as 10 secondsLet the mass flow through for time Let the mass flow through for time interval of interval of t, such as 10 secondst, such as 10 seconds



Since volume is Since volume is Since volume is Since volume is mAV Then volume Then volume flow rate is flow rate is Then volume Then volume flow rate is flow rate is t

m

t

lA

t

V

s

mmAV

3

,

InletExitA

A

Length,

orororor

s

kgAVm ,

Then massThen mass

Flow rateFlow rateThen massThen massFlow rateFlow rate


FIGURE 3-51The total energy consists of three parts for a nonflowing fluid and four parts for a flowing fluid.


kg

kJ,

υke

2000

2

Kinetic energy, keKinetic energy, keKinetic energy, keKinetic energy, ke

kg

kJ,

ghpe

2000

Potential energy, pePotential energy, pePotential energy, pePotential energy, pe

kg

kJu,

Internal energy, uInternal energy, uInternal energy, uInternal energy, u

Gas Mixtures – Ideal GasesGas Mixtures – Ideal GasesGas Mixtures – Ideal GasesGas Mixtures – Ideal Gases

Equation of StateEquation of State - P--T behaviour

Low density

High density

Hence, can also writeHence, can also write

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/kmol = 8.314 kJ/kmolKK

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/kmol = 8.314 kJ/kmolKK

PV =mRTPV =mRTPV =mRTPV =mRT

PV = NRPV = NRuuTTPV = NRPV = NRuuTT

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