1 MM 416E ENERGY ENGINEERING COMPLEMENTARY NOTES Prof. Dr. Şenol BAŞKAYA PART-1 GENERAL CONCEPTS Utilization Factors: Demand for electricity varies from hour to hour and from season to season. Depending on the energy demand and the economical conditions a power plant may be loaded at its full capacity or at partial load or shut down. Time utilization factor: 8760 year a in time production Actual F T = Capacity Factor: capacity on Installati plant the of output Power F C = Load utilization factor: production imum Annual production Actual F L max = Reserve Calculations: () t e E t E μ . 0 = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ a t ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = 1 ln 1 0 0 E R t D μ μ , tD: depletion time Heating Value, Energy Released in Combustion: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = F kg kJ LHV s F kg m Q F F & . [ ] t kW Heat Transferred to Working Fluid in Boiler: B F B Q Q η . . . = where hB: Boiler efficiency TPP Efficiency: F el TPP Q P . = η Annual Electricity Generation: ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ a kWh el [ ] [] ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = = a h F kW P AEG E L el IC 8760 . .
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
MM 416E ENERGY ENGINEERING COMPLEMENTARY NOTES
Prof. Dr. Şenol BAŞKAYA
PART-1 GENERAL CONCEPTS
Utilization Factors: Demand for electricity varies from hour to hour and from season to season. Depending on the energy demand and the economical conditions a power plant may be loaded at its full capacity or at partial load or shut down.
Time utilization factor: 8760
yearaintimeproductionActualFT =
Capacity Factor: capacityonInstallati
planttheofoutputPowerFC =
Load utilization factor: productionimumAnnual
productionActualFL max=
Reserve Calculations:
( ) teEtE μ.0= ⎥⎦⎤
⎢⎣⎡at ⎟⎟
⎠
⎞⎜⎜⎝
⎛+= 1ln1
0
0
ERtDμ
μ , tD: depletion time
Heating Value, Energy Released in Combustion:
⎥⎦
⎤⎢⎣
⎡−⎥⎦
⎤⎢⎣⎡ −
=Fkg
kJLHVs
FkgmQ FF &.
[ ]tkW
Heat Transferred to Working Fluid in Boiler:
BFB QQ η...
= where hB: Boiler efficiency TPP Efficiency:
F
elTPP
Q
P.=η
Annual Electricity Generation: ⎥⎦
⎤⎢⎣
⎡a
kWhel
[ ] [ ] ⎥⎦
⎤⎢⎣⎡−==ahFkWPAEGE LelIC 8760..
2
Annual Fuel Consumption:
[ ] ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡−⎥⎦
⎤⎢⎣⎡=
at
ahF
htmAFC Lfuel 8760..&
⎥⎦⎤
⎢⎣⎡=
⎥⎦
⎤⎢⎣
⎡⋅⎥
⎦
⎤⎢⎣
⎡−
⎥⎦
⎤⎢⎣
⎡
=akg
kWhkWh
FkgkWh
H
akWh
AEGAFC
t
elTPP
tU
el
η
PART-2 COST ANALYSIS
General energy conversion system:
3
General energy production and profit analysis:
4
Profit spoon:
Total Production Cost: OthPersAmrFT CCCCC +++= ⎥⎦
⎤⎢⎣
⎡
elkWhTL
Fuel Cost: ⎥⎦
⎤⎢⎣
⎡⋅⎥⎦
⎤⎢⎣
⎡−
⎥⎦
⎤⎢⎣
⎡−−
=−
−
t
elTPP
tU
F
kWhkWh
FkgkWh
H
FkgFTLg
Cη
⎥⎦
⎤⎢⎣
⎡ −
elkWhFTL
Amortization Cost: ⎥⎦
⎤⎢⎣
⎡
⎥⎦⎤
⎢⎣⎡ −
=
akWh
E
aAmrTLYA
Cel
Amr ⎥⎦
⎤⎢⎣
⎡ −
elkWhAmrTL
Yearly Amortization: [ ] ⎥⎦
⎤⎢⎣⎡−=a
ARAmrTLTICYA 1. ⎥⎦⎤
⎢⎣⎡ −
aAmrTL
Total Investment Cost: [ ] ⎥⎦
⎤⎢⎣
⎡=
elel kW
TLSICkWICTIC . [ ]TL
Amortization Ratio: ( )( ) 11
1−+
+=
A
A
n
n
FFFAR ⎥⎦
⎤⎢⎣⎡a1
Example 1. For a TPP the following data are given:
PIC=300 MWe, SIC=1,5x109 [TL/kWe], ηTPP=0,30 [kWhe/kWht], Hu=4305 [kcal/kg], gF=100x106 [TL/t], FL=0,75 [-], F=15 [%], nAmr=10 years, nEC=30 years, nphs=40 years, use linear amortization
a) Calculate annual fuel consumption [t/a]. b) Calculate annual electricity generation [kWh/a]. c) Calculate CF [TL- F/kWhel], CAm [TL-Am/kWhel], Cother=0, CT [TL/kWhel]. d) Calculate total profit [TL] in economical life time (Csell=140x103 [TL/kWhe]=constant, CF
increases after nAmr linearly to 115x103 [TL/kWhel] at nEC). e) How can you utilize TPP between nEC and nphy.
5
Answer:
a) [ ] ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡−⎥⎦
⎤⎢⎣⎡=
at
ahF
htMAFC LF 8760..
.
[ ][ ]−⎥
⎦
⎤⎢⎣
⎡−
=
TPPt
U
tICF
FkgkWhH
kWPMη.
.
= [ ]
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡− t
elt
e
kWhkWh
FkgkWh
kW
30,0861
4305000.300 = 200.000
hFkg − = 200
hFt −
[ ] ⎥⎦⎤
⎢⎣⎡ −
=⎥⎦⎤
⎢⎣⎡−⎥⎦
⎤⎢⎣⎡ −
=aFtx
ah
hFtAFC 6
.10314,18760.75,0.200
b) [ ] [ ] ⎥⎦⎤
⎢⎣⎡−=ahFkWPE LelIC 8760..
[ ] [ ] ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡−=
akWhx
ahkWE e
e910971,18760.75,0.000.300
c) ⎥⎦
⎤⎢⎣
⎡ −=
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
⎥⎦
⎤⎢⎣
⎡−
=
⎥⎦
⎤⎢⎣
⎡⋅⎥⎦
⎤⎢⎣
⎡−
⎥⎦
⎤⎢⎣
⎡−−
=
−
e
t
et
t
elTPP
t
F
F kWhFTLx
kWhkWhx
FkgkWh
Fkgt
tTLx
kWhkWh
FkgkWhLHV
FkgFTLg
C 3
6
107,663,0
8614305
100010100
η
⎥⎦⎤
⎢⎣⎡
⎥⎦⎤
⎢⎣⎡ −
=
akWh
AEG
aAmrTLYA
Cel
Amr , TYxAOYA =
[ ]TLxkWTLxxkWxSICPTY
eeIC
129 10450105,1000.300 =⎥⎦
⎤⎢⎣
⎡==
( )( ) ⎥⎦
⎤⎢⎣⎡≅
−++
=−+
+=
aFFFAO
A
A
n
n 12,01)115,0(
)115,0(15,011
110
10
[ ] ⎥⎦⎤
⎢⎣⎡=⎥⎦
⎤⎢⎣⎡=
aTLx
axTLxYA 1212 109012,010450
⎥⎦
⎤⎢⎣
⎡ −=
⎥⎦⎤
⎢⎣⎡
⎥⎦⎤
⎢⎣⎡ −
=ee
Amr kWhAmrTLx
akWhx
aAmrTLx
C 3
9
12
107,4510971,1
1090
⎥⎦
⎤⎢⎣
⎡=+++=
eOthPersAmrFT kWh
TLxCCCCC 3104,112
6
d)
[ ] [ ] [ ] [ ] [ ]TLxxxxxxaxna
kWhxEkWhTLCCC Amr
e
eTsellnprofit Amr
12933 105441010971,1104,11210140 =−=⎥⎦⎤
⎢⎣⎡
⎥⎦
⎤⎢⎣
⎡−=−
( ) ( )AmrECFF
Fsellnnofit nnxExCC
CCC nAmrnec
necECAmr−
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ −+−=−− 2Pr
( ) ( ) [ ]TLxxxxxxxxCECAmr nnofit
15933
33Pr 1094,1103010971,1
2107,66101151011510140 =−⎥
⎦
⎤⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛ −+−=−−
[ ]TLxCCCECAmrAmr nnofitnofittotalofit
15PrPrPr 10484,2=+= −−−−
e) ......Peak Load .....
PART-3
COMBUSTION AND EMISSION ANALYSIS
Combustion analysis:
Fuel Demand: ⎥⎦⎤
⎢⎣⎡
hkg
[ ]
[ ]−⎥⎦
⎤⎢⎣
⎡−
=
Bt
tfuel
FkgkWh
LHV
kWQm
η.
.
&
TL/kWhe
n [a]
CF
CT CAmr
CProfit-nAmr CF=CT
Csell
CPro.-nAmr-nEC
7
[ ]
⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡−
=
t
elPP
t
elelfuel
kWhkWh
FkgkWh
LHV
kWPm
η.&
Combustion Air Demand:
⎥⎦
⎤⎢⎣
⎡−−
∀⎥⎦⎤
⎢⎣⎡ −
=∀FkgANm
hFkgm afuela
3
.&& [Nm3-A/ h] ⎥⎦
⎤⎢⎣
⎡−−
∀=∀FkgANmn ata
3
. 221
21O
n−
=
Combustion Gas Flow:
⎥⎦
⎤⎢⎣
⎡−−
∀⎥⎦⎤
⎢⎣⎡ −
=∀ −−− FkgGNm
hFkgm wetDryfgfuelwetDryfg
3
.&& ⎥⎥⎦
⎤
⎢⎢⎣
⎡ −h
GNm3
( )
tawetdryfgtwetdryfg n ∀−+∀=∀ −−−− .1 , OHdryfgfg 2∀+∀=∀ −
Example 2. Make the following emission calculations for the TPP given in Example 1. The following data is given: SO2=2000 [ppm], O2=7 [%], VHth=VGth-dry=5 [Nm3/kg-F].
a) Calculate εSOel [kg-SO2/kWhel]. b) Calculate annual SO2 emission ASO2 [t-SO2/a].
Answer: a)
[ ][ ] ⎥
⎦
⎤⎢⎣
⎡ −=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−∀
⎥⎦
⎤⎢⎣
⎡
⎥⎦
⎤⎢⎣
⎡
−−
= −−ee
t
TPPtu
Drydryfg
kmolDry
elSO kWhSOkg
kWhkWh
kWhFkg
HFkgfgNm
SOkmolSONmV
SOkmolSOkgM
fgNmSONmppm 2
3
2
23
2
2
362
3 11
..
..
102 ηε
5,1721
21=
−=n
( ) ⎥⎦
⎤⎢⎣
⎡−
=−+=∀−+∀=∀ −−−−−− Fkg
Nmn wetdryfgatwetdryfgtwetdryfg
3
5,7)15,1(5.1
[ ][ ] ⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡ −
⎥⎥⎦
⎤
⎢⎢⎣
⎡
−
−
⎥⎦
⎤⎢⎣
⎡
⎥⎦
⎤⎢⎣
⎡
−−
=−e
t
t
Dry
DryelSO kWh
kWhkWh
FkgxFkgfgNm
SOkmolSONm
SOkmolSOkg
fgNmSONmppm
3,01
8614305
15,7
.
.4,22
..64
10200 3
2
23
2
2
362
3
2ε
⎥⎦
⎤⎢⎣
⎡ −=−
eelSO kWh
SOkg 202857,02
ε
b)
⎥⎦
⎤⎢⎣
⎡−−
⎥⎦⎤
⎢⎣⎡
⎥⎦
⎤⎢⎣
⎡ −=⎥⎦
⎤⎢⎣⎡
⎥⎦
⎤⎢⎣
⎡ −=− −
2
29222 1000
110971,102857,02 SOkg
SOta
kWhxxkWh
SOkga
kWhxEkWh
SOkgSOA e
e
e
eelSOε
⎥⎦⎤
⎢⎣⎡ −
=−aSOtSOA 2
2 5,56311
10
PART-4 THERMAL POWER SYSTEMS, SYSTEM STRUCTURE,
ENERGY CONVERSION AND APPLICATIONS Rankine PC (ST-TPP):
TrICGMSPBRPCnetTPP ηηηηηηηη ......= GMTel PP ηη ..= B
Example 3. The flow diagram of a TPP- steam power cycle is given below. Answer: a)
a) Sketch the steam power cycle on h-s diagram. b) Calculate the extraction pressures Pex4 - Pex6 (bar) and extraction steam mass flow rates m4 – m6 [kg/s] ( ΔTAPP = 5 ° C) c) Calculate PelGR [MWe] (ηM= 1, ηG= 0.98, x7 = 0.95 ) d) Calculate mCW [t/h] and mCW / m3
a) Calculate steam pressures P4, P31, P32 [bar] and condensate outlet temperatures t311, t321 [oC] (For all
pipes and HE’s: ΔP~0, ΔQ~0). b) Sketch the steam power cycle on h-s diagram and determine h3, h31, h32, h4 [kJ/kg]. c) Calculate extraction mass flow rates m31, m32 and m4 [kg/s]. d) Calculate power generated Pel [MWe] (ηm=1, ηG=0,98). e) Calculate coal consumption MF [t/h] (ηB=0,85, Hu=4305 kcal/kg).
Example 5. The flow diagram of a simple GT-Power plant is given below. 3
a) Skech the GT-Power cycle on h – s diagram b) Calculate Pel [MWe] c) Calculate emission factor ε el [kg-NO / kWhel] and M NO [kg –NO / h]
(VGth = 10 Nm3 / kg, VHth = 9.5 Nm3 / kg, Hu = 10 kWh /kg, MN= 14 kg/kmol, MO = 16 kg/kmol) d) Calculate CT [TL / kWhel]) (Coth = 1 [Ykr / kWhel], gF =1 [YTL/kg], F = 10 [%] , n Amr = 10 years, SIC= 700 [YTL / kWe, FL = 0.4 ) e) Selling price of electricity for the time being is approximately 10 Ykr / kWhel . Discuss the results. What can you do to this plant so that it can be operated economically?