thermodynamics i chapter 4 first law of thermodynamics open...

Thermodynamics IChapter 4

First Law of ThermodynamicsOpen Systems

Mohsin Mohd SiesFakulti Kejuruteraan Mekanikal, Universiti Teknologi Malaysia

First Law of Thermodynamics(Motivation)

A system changes due to interaction with itssurroundings.Interaction study is possible due to conservation laws.Various forms of conservation laws are studied in thischapter in the form of balance equations.

1ST LAW FOR OPEN SYSTEMS

Energy and Mass can enter/leave a systemDivided into two parts:

Conservation of Mass Principle

Conservation of Energy Principle1

Qout

Wout

Qin

Win

min

mout

CV

Mass Conservation (Mass Balance)

min

mout

mCV

Mass Balance:

∑∑ −=∆ outinCV mmm

Net changeof mass ofCV

Total of allmass entering

Total of allmass exiting

∑∑ −=dt

dm

dt

dm

dt

dm outinCV

Rate ofchange ofmass ofCV

Total of all massflow rates for allinlet channels

Total of all massflow rates for allexit channels

Mass flow rates and Speed for a channel

A

Vm

,,

s

AVm

Adt

ds

dt

dm

sAm

Vm

=

=

==

VAdt

ds

VsA

=

=

v

VV

v

AVAVm

====

For flow rates;

but

Thus;

Flow Work

For an amount of mass to cross a boundary, it needs aforce (work) to push it; called Flow Work , wflow

F

A

L

m, ρ, V

pvw

PV

LPA

LFW

dsFW

flow

flow

==

⋅=

⋅=

⋅= ∫

The force that pushes the fluid;

PAF =

Work that is involved;

Conservation of Energy for Open Systems

Energy can enter/leave an open system via

Heat, Work AND Mass Flow Entering/Leaving

Qout

Wout

Qin

Win

min

mout

ECV

outinCV EEWQE −+−=∆

Netenergychange ofCV

Net energytransfer byheat/work

Totalenergy ofenteringmass

Totalenergyofexitingmass

Energy of a Flowing Mass

An amount of mass possesses energy, E, ee = u + ke + pe

but a flowing mass also possesses flow work, so

the Energy for a flowing mass; eflow

eflow = u + ke + pe + wflow

= u + pv + ke + pe

but u + pv = h (enthalpy)

eflow = h + ke + pe (energy for a flowing mass)

Energy Balance for an Open Systemcan be written as;

outflowinflowCV EEWQE −+−=∆or;

∑∑ ++−+++−=++−+++−=

−+−=∆

outin

outin

flowflowCV

pekehpekehwq

pekehpekehwq

eewqeoutin

)()(

)()(

For eachinlet

For eachoutlet

Total of all workexcept flow work

∑∑ ++−+++

−=∆+∆+∆

outin

CV

pekehpekeh

wqpekeu

)()(

)(

but;CVCV pekeue )( ∆+∆+∆=∆

so;

Energy Balance in other forms;

∑∑ ++−+++−=∆outin

CV PEKEHPEKEHWQE )()(

Rate form;

∑∑ ++−+++−=outin

CV pekehmpekehmWQdt

dE)()(

(general form of Energy Balance for anOpen System)

Steady Flow ProcessCriteria;

All system properties (intensive & extensive)do not change with timemcv= constant; ∆mCV= 0VCV= constant; ∆VCV= 0ECV= constant; ∆ECV= 0

Fluid properties at all inlet/exit channels do notchange with time (Values might be differentfor different channels)

Heat and work interactions do not change withtime

.),,,( etcAVm

Q= constantW= constant

Implication of Criterion 1(Mass)∆mCV= 0,From Mass Conservation;

222111

11

AVAV

mm

AVAV

v

AVAVm

mm

mmm

outin

out

k

oooo

in

n

iiii

outin

CVoutin

=

=

=

==

=∴

∆=−

∑∑

∑∑∑∑

==

Recall that

so

for 1 inlet/ 1 exit;

111 ,, AVmin

222 ,, AVmout

Implication of Criterion 1 (Mass) ctd.

Implication of Criterion 1

(Volume)∆VCV= 0,

∴ WB = 0(boundary work = 0 )

∆ECV= Q – W + ΣEflow(in) - ΣEflow(out)

W = Welectrical + Wshaft + Wboundary+ Wflow+ Wetc.

0 in h

W not necessarily 0, only WB is 0

Implication of Criterion 1(Energy)

∆ECV= 0,From Energy Conservation;

∑∑ ++−+++−=outin

CV pekehmpekehmWQdt

dE)()(

= 0

Rearrange to get Steady Flow Energy Equation

(SSSF)

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

outouthm outV

inmininVh inW

outW

inQ

outQdt

dECV

zin

zout

∑∑ ++−++=−inout

gzV

hmgzV

hmWQ )2

()2

(22

For each exitchannel

For each inletchannel

SSSF

SSSF energy equation for 1 inlet / 1 exit

1 2Mass

outin mm = or 21 mm =Energy

)2

()2

( 1

21

112

22

22 gzV

hmgzV

hmWQ ++−++=−

mmm == 21

pekehwq

zzgVV

hhm

W

m

Q

zzgVV

hhmWQ

∆+∆+∆=−

−+

−+−=−

−+

−+−=−

)(2

)(

)(2

)(

12

21

22

12

12

21

22

12

( ∆ = exit – inlet )( 1 inlet / 1 exit )

SSSF energy equation for multiple channels butneglecting ∆ke, ∆pe

∑∑ −=− ininoutout hmhmWQ

∑∑ = outin mm

SSSF Applications

Nozzle & DiffuserTurbine & compressorThrottling valve @ porous plugMixing/Separation ChamberHeat exchanger(boiler, condenser, etc)

(a)

(b)

2 categories;

1 inlet / 1 outletMultiple inlet / outlet

Analysis starts from the general SSSF energyequation;

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

For devices which are open systems

Nozzle & Diffuser ( ∆KE ≠ 0 ) (To change fluid velocity)

Nozzle & Diffuser ( ∆KE ≠ 0 ) (To change fluid velocity)

1

1

1

1

AVpTm

2

2

2

2

AVpTm

1

1

1

1

AVpTm

2

2

2

2

AVpTm

Nozzle Diffuser↑V

↓V

AssumptionsConst. volume → wB = 0No other workSmall difference in height → ∆pe = 0

wCV = 0

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

pekehwq ∆+∆+∆=−1 inlet / 1 outlet

kehq ∆+∆=

Turbine ( wCV ≠ 0 )

Turbine & Compressors ( wCV ≠ 0 )

Compressors (Reciprocating) ( wCV ≠ 0 )

Turbine & Compressor ( wCV ≠ 0 )

1

1Tpm

2

2Tpm

outW

1

1Tpm

2

2Tpm

inW

Turbineproduces work (W +ve)from expansion offluid (∆P -ve)

Compressorincreases pressure (∆P+ve) using work supplied(W –ve)

AssumptionsConst. volume → wB = 0Involves other workSmall difference in height → ∆pe = 0

wCV ≠ 0

Turbine & Compressor ( wCV ≠ 0 )

1

1Tpm

2

2Tpm

outW

1

1Tpm

2

2Tpm

inW

TurbineCompressor

AssumptionsConst. volume → wB = 0Involves other workSmall difference in height → ∆pe = 0

wCV ≠ 0

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

1 inlet / 1 outlet

)( pekehmWQ ∆+∆+∆=−

Commonly insulated (q=0), and also∆ke ≈ 0 (since ∆ke << ∆h ) 12 hhh

m

W −=∆=−

Throttling valve/Porous plug (const.enthalpy, h=c)

To reduce pressure without involving work

Throttling valve/Porous plug (const.enthalpy, h=c)

To reduce pressure without involving work

1

1Tpm

2

2Tpm

1

1Tpm

2

2Tpm

AssumptionsW = 0∆ke = 0 , ∆pe = 0Q ≈ 0 (system too small)

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

1 inlet / 1 outlet pekehwq ∆+∆+∆=−

21

0

hh

h

==∆

For throttling valves

Mixing/Separation Chamber / T junction pipe

11

11,

,,hTpm

22

22,

,,hTpm

33

33

,

,,

hT

pm

33

33,

,,hTpm

22

22,

,,hTpm

11

11,

,,hTpm

AssumptionsWB = 0Same pressure at all channelsOther assumptions made accordingly

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

∑∑ = outin mm

also

Heat Exchangers

Heat Exchangers

1 2

34 B Q

AAm

Bm

Assumptions (System encompassesthe whole heat exchanger)

No work involved, W = 0Insulated, Q = 0∆ke = 0 , ∆pe = 0

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

outoutinin hmhm ∑∑ =

∑∑ = outin mm

also

)()( 3421

4231

hhmhhm

hmhmhmhm

BA

BABA

−=−+=+

General device (Black box analysis)

5

4

3

2

1

∑Q ∑W

Assumptions made according to situation...

∑∑ ++−++=−inout

pekehmpekehmWQ )()(

∑∑ = outin mm

also