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