thermodynamics i chapter 4 first law of thermodynamics open...
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![Page 1: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/1.jpg)
Thermodynamics IChapter 4
First Law of ThermodynamicsOpen Systems
Mohsin Mohd SiesFakulti Kejuruteraan Mekanikal, Universiti Teknologi Malaysia
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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.
![Page 3: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/3.jpg)
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
![Page 4: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/4.jpg)
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
![Page 5: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/5.jpg)
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;
![Page 6: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/6.jpg)
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;
![Page 7: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/7.jpg)
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
![Page 8: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/8.jpg)
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)
![Page 9: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/9.jpg)
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
![Page 10: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/10.jpg)
∑∑ ++−+++
−=∆+∆+∆
outin
CV
pekehpekeh
wqpekeu
)()(
)(
but;CVCV pekeue )( ∆+∆+∆=∆
so;
![Page 11: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/11.jpg)
Energy Balance in other forms;
∑∑ ++−+++−=∆outin
CV PEKEHPEKEHWQE )()(
Rate form;
∑∑ ++−+++−=outin
CV pekehmpekehmWQdt
dE)()(
(general form of Energy Balance for anOpen System)
![Page 12: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/12.jpg)
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
![Page 13: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/13.jpg)
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
![Page 14: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/14.jpg)
for 1 inlet/ 1 exit;
111 ,, AVmin
222 ,, AVmout
Implication of Criterion 1 (Mass) ctd.
![Page 15: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/15.jpg)
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
![Page 16: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/16.jpg)
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 )()(
![Page 17: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/17.jpg)
outouthm outV
inmininVh inW
outW
inQ
outQdt
dECV
zin
zout
∑∑ ++−++=−inout
gzV
hmgzV
hmWQ )2
()2
(22
For each exitchannel
For each inletchannel
SSSF
![Page 18: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/18.jpg)
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 )
![Page 19: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/19.jpg)
SSSF energy equation for multiple channels butneglecting ∆ke, ∆pe
∑∑ −=− ininoutout hmhmWQ
∑∑ = outin mm
![Page 20: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/20.jpg)
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
![Page 21: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/21.jpg)
Nozzle & Diffuser ( ∆KE ≠ 0 ) (To change fluid velocity)
![Page 22: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/22.jpg)
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 ∆+∆=
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Turbine ( wCV ≠ 0 )
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Turbine & Compressors ( wCV ≠ 0 )
![Page 25: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/25.jpg)
Compressors (Reciprocating) ( wCV ≠ 0 )
![Page 26: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/26.jpg)
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
![Page 27: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/27.jpg)
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 −=∆=−
![Page 28: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/28.jpg)
Throttling valve/Porous plug (const.enthalpy, h=c)
To reduce pressure without involving work
![Page 29: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/29.jpg)
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
![Page 30: Thermodynamics I Chapter 4 First Law of Thermodynamics Open …mohsin/sme1413/01.english/chap04/04.1st.law.pdf · Mohsin Mohd Sies Fakulti Kejuruteraan Mekanikal, Universiti Teknologi](https://reader034.vdocuments.us/reader034/viewer/2022042622/5f86479c0745af684f1e4f14/html5/thumbnails/30.jpg)
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
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Heat Exchangers
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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
−=−+=+
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General device (Black box analysis)
5
4
3
2
1
∑Q ∑W
Assumptions made according to situation...
∑∑ ++−++=−inout
pekehmpekehmWQ )()(
∑∑ = outin mm
also