Transcript
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M. Khosravy
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Performance Evaluation Targets
1. Enhance Turbine Efficiency
2. Increase Output
3. Improve Heat Rate4. Recover Lost Performance
5. Improve Exhaust Temperature
6. Maximize Compressor Efficiency an Airflow
7. Reduce Fuel Costs
8. Avoid Forced Outages
9. Improve Maintenance Planning10. Improve Relative Firing Temperature
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Standard Methods ofPerformance Evaluation
ASME PTC 22: Gas turbine power plants performance test
code ISO 2314 : Gas turbines - Acceptance tests
JIS B8401 : Gas turbine Open cycle performance test
BS 3135 : Specification for gas turbine acceptance test
DIN 4341 : Acceptance rules for gas turbines ..
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ASME PTC 22
This Code provides for the testing of gas turbinessupplied with gaseous or liquid fuels (or solid fuels
converted to liquid or gas prior to entrance to the gasturbine).
Tests of gas turbines with emission control and/orpower augmentation devices, such as injection fluidsand inlet air treatment, are included.
It may be applied to gas turbines in combined cycleplants or with other heat recovery systems.
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ASME PTC 22
This Code provides for comparative (back to
back) tests designed to verify performancedifferentials of the gas turbine, primarily for
testing before and after modifications,
uprates, or overhauls.
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ASME PTC 22
The Code does not apply to the following:
1. Gas turbines where useful output is other than power to drive a generatoror other load device
2. Environmental compliance testing for gas turbines for stack emissionsand sound levels. Procedures developed by regulatory agencies, ANSI, orother PTC Committees are available to govern the conduct of suchtesting.
3. Overall plant power output and thermal efficiency of gas turbinecombined cycle and cogeneration facilities. Refer to PTC 46.
4. Performance of specific components of the gas turbine
5. Performance of auxiliary systems of the gas turbine power plant, such asinlet cooling devices, fuel gas booster compressors, etc.
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Required Instruments to PerformPeriodic Performance Tests
Pressure measurement (PTC 19.2)
Temperature measurement (PTC 19.3) Fluid meters (PTC 19.5)
Flue gas analyses (PTC 19.10)
Measurement of shaft power (PTC 19.7)
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Brayton Cycle Unlike diesels, operate on STEADY-FLOW
cycle
Open cycle, unheated engine1-2: Compression
2-3: Combustion
3-4: Expansion through
Turbine and Exhaust
Nozzle
(4-1: Atmospheric
Pressure)
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Basic
Compon
ents
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Basic
Components
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CompressorCompressor
Draws in air & compresses it
Combustion Chamber
Fuel pumped in and ignited to burn with compressedair
Turbine Hot gases converted to work
Can drive compressor & external load
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Compressor
Draws in air & compresses it
Combustion ChamberCombustion Chamber
Fuel pumped in and ignited to burn with compressedair
Turbine Hot gases converted to work
Can drive compressor & external load
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Compressor
Draws in air & compresses it
Combustion Chamber
Fuel pumped in and ignited to burn with compressedair
TurbineTurbine Hot gases converted to work
Can drive compressor & external load
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Companies Data Sheets
Output
(MW)
El. efficiency
(%)
Pr. ratio
(-)
Exh. flow
(kg/s)
Exh.temp.
deg C
SGT-100 4.35-5,25 30.0 13 17.7 527
SGT-200 6.75 31.5 12.3 29.3 466
SGT-300 7.70 30.7 13.9 29.8 545
SGT-400 12.9 34.0 16.9 39.7 570
SGT-500 17.0 32.1 12 92 375
SGT-600 24.8 34.2 14 80 543
SGT-700 29.1 36.0 18 91 518
SGT-800 45.0 37.0 20 122 546
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Ideal Brayton Cycle In reality, gas turbines operate on an open cycle
Fresh air is continuously drawn into the compressor and exhaust gases are
thrown out
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Ideal Brayton Cycle (cont.)
The open gas-turbine cycle can be modeled as a closed cycle
The combustion process is replaced by a constant-pressure heat-addition
process and the exhaust process is replaced by a constant-pressure heat-
rejection process
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Ideal Brayton Cycle (cont.)
The idealized closed loop cycle is the Brayton cycle, which consists of the
following four internally reversible processes
1!2 Isentropic compression
2!3 Constant-pressure heat addition
3!4 Isentropic expansion4!1 Constant-pressure heat rejection
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Thermodynamic Analysis
The four processes of the Brayton cycle are executed in steady-
flow devices
When changes in kinetic and potential energies are neglected,
the energy balance for one of the processes can be expressed as
Therefore, heat transfers to and from the working fluid are
( ) ( ) inletexitoutinoutin hhwwqq =+
( )
( )
in 3 2 3 2
out 4 1 4 1
p
p
q h h c T T
q h h c T T
= =
= =
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Thermal Efficiency
The thermal efficiency of the ideal Brayton cycle under the cold-
air-standard assumptions becomes
Processes 1!2 and 3!4 are isentropic, and P2 = P3 and P4 = P1
( )( )
( )
( )
net outth, Brayton
in in
4 1
3 2
1 4 1
2 3 2
1
1
11
1
p
p
w q
q q
c T T
c T T
T T T
T T T
= =
=
=
( ) ( )
4
3
1
4
3
1
1
2
1
2
T
T
P
P
P
P
T
Tkkkk
=
=
=
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Thermal Efficiency (cont.)
Substituting these expressions into the thermal efficiency
relation yields
Where rP is the pressure ratio
( ) kkPr
1Braytonth, 11 =
1
2
PrP=
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Thermal Efficiency (cont.)
The thermal efficiency
increases with both the
pressure ratio (rP) and the
specific heat ratio (k)
The plot to the right shows
the thermal efficiency as a
function of the compression
ratio
The two major application
areas of gas-turbine enginesare aircraft propulsion and
electric power generation
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Brayton Cycle: The Ideal Cycle for Gas Turbine Engines
Deviation of Actual Gas-Turbine Cycles from Idealized Ones
The deviation of actual compressor and turbine behavior can be
accurately accounted for by utilizing the isentropic efficiencies
of the turbine and compressor
The actual and isentropic statesof a gas-turbine cycle are:
12
12
hh
hh
w
w
a
s
a
sC
=
s
a
s
aT
hh
hh
w
w
43
43
=
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A Problem
A simple Brayton cycle using air as the working fluid has
a pressure ratio of 8. The minimum and maximum
temperatures in the cycle are 310 and 1160 K. Assuming
an isentropic efficiency of 75 percent for the compressor
and 82 percent for the turbine, determine
(a) The air temperature at the turbine exit.
(b)The net work output.
(c) The thermal efficiency.
Use constant specific heats at room temperature.
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Actual GT Evaluation
Our study of gas power cycles will involve the study of
those heat engines in which the working fluid remains in
the gaseous state throughout the cycle. We often study
the ideal cycle in which internal irreversibilities and
complexities (the actual intake of air and fuel, the actual
combustion process, and the exhaust of products of
combustion among others) are removed.
We will be concerned with how the major parameters of
the cycle affect the performance of heat engines. The
performance is often measured in terms of the cycle
efficiency.
thnet
in
W
Q=
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SIMULATION BY COMPUTER
(thermodynamical and analytical models)
REAL TIME THERMAL BALANCE
(AS A DYNAMIC REFERENCE STATE)
Gas Turbine
DATA ADQ.
SYSTEM
NETWORK
EnviromentalData and Load
(by a control
system demand)
(optionally
Control SetPoint)
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ACCEPTABLE PERFORMANCE TEST
STATE
Data collection at
determinate operation
condition
Codes ASME PTC
Data validation and/or
Filtration
MEASURE POINTSMEASURE POINTS DATADATA
EnvironmentalEnvironmental conditioncondition andand compressorcompressor admitionadmition P, T,P, T, TTww
IGVIGV %%
Compressor exitCompressor exit andand airair extractionsextractions P, TP, T
Fuel admission in the combustorFuel admission in the combustor P, T,P, T, HHV,LHVHHV,LHV,,, m, m
CombustorCombustor (TG(TG admissionadmission)) PP
TGTG combustioncombustion gasesgases exitexit P, TP, T
Gases in HRSG exitGases in HRSG exit P, T, xP, T, xii
Steam generatedSteam generated by HRSGby HRSG P, T, mP, T, m
Steam admission at APSteam admission at AP--TV Y BPTV Y BP--TVTV P, TP, T
Steam exit at APSteam exit at AP--TV and steam admission at BPTV and steam admission at BP--TV.TV. P, TP, T
Exhaust steamExhaust steam PP
Condenser exit conditions and feet water.Condenser exit conditions and feet water. P, T, mP, T, m
Suction and exit condenser and feet pumps.Suction and exit condenser and feet pumps. P, TP, T
Electric generator, pumps, fans and auxiliary.Electric generator, pumps, fans and auxiliary. VoltVolt.. andand AmpAmp..
Measure variables (directly or indirectly) in plant
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EQUALIZED RECONCILIATION METHOD
Anomalies Classification
Extern
Intern
Environmental Condition: variation in envioronmental
conditions (P, T, Humidity)
Fuel Quality: variation in fuel quality (HHV, LHV, density, viscosity)
Intrinsic: anomalies (erosion, roughness)
Induced: anomalies generated by other components
Loop Control: intervention by control system
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RECONCILIATION METHOD
Module Declaration
1, 2, 3, 4, 5,
1, 2, 3, 4, 5,
( , , , , : , )
( , , , , : , )
c c i i i
c c i i i
R HR M M M M M HR W
W W M M M M M HR W
=
=
, ,, , ,test test BASE COMP test COMB test TIT TSH P
nn ThermodynamicThermodynamic ModelModel
nn InputInput datadata fromfrom acceptableacceptableperformanceperformance testtest
ToutcompressorToutcompressor(HHVpGT, Tinturb, wfuelGT, yCO2dry, yO2dry, yCOdry, tamb,RHamb, pamb, Wwaterfuel, Wsteaminjection, tsteamincc, psteamincc)
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GAS TURBINE
Key Description p[bar] T[C] h[kJ/kg] [kg/s]
. Optimal Actual Optimal Actual Optimal Actual Optimal Actual
00 Environmental Conditions 0.9687 0.9687 19.9531 19.9531 20.2385 20.2385 274.1395 268.5795
01 Compressor admition 0.9687 0.9502 19.9531 21.6406 20.2385 21.9727 274.1395 268.5795
02 Compresor exit 20.3244 19.8250 479.2276 517.8233 500.6305 543.1487 107.6584 109.5478
03 Combustor EV inlet --- --- --- --- --- --- 2.9027 3.3380
04 Expansor HP Admission 19.8244 19.1159 1190.1877 1162.1420 1385.4273 1349.7632 110.5612 112.8859
05 Expansor HP exit 8.9673 11.7578 968.0208 968.0208 1105.1069 1105.1069 112.8511 138.8969
06 Combustor SEV inlet --- --- --- --- --- --- 3.4643 3.0720
07 Expansor LP Admission 7.9673 10.7109 1129.0320 1221.7390 1307.7281 1425.5865 147.3154 141.9710
08 Expansor LP exit 0.9837 0.9799 634.8411 634.8411 699.7636 699.7636 280.5065 274.9895
09 Gas exhaust 0.9687 0.9687 94.6566 94.6566 98.8792 98.8792 280.5065 274.9895
010 Air inlet ! cooler high
pressure
20.3244 20.3228 479.2276 517.8233 500.6136 543.1304 33.2900 26.0131
011 Air exit ! cooler high
pressure
20.2562 20.2562 333.6875 333.6875 343.5631 343.5631 33.2900 26.0131
012 Air inlet ! cooler low
pressure
13.5479 12.9809 394.3492 394.3492 408.3841 408.3841 133.1911 133.0186
013 Air exit ! cooler low
pressure
11.7078 11.7078 349.2500 349.2500 360.1055 360.1055 133.1911 133.0186
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Pipe Line Hydraulic Calculation
By:
M. Khosravy
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An Example !
A pipeline has 4 segments as follow:Segment
No. Length (Km) Elevation (+m) Temp ("C)
1 15 20 502 20 40 30
3 50 10 40
4 15 0 20
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Question #1
If the pipe diameter and inlet/outlet pressure are
known, what is the flow velocity of natural gas in it?
Outlet temperatureOutlet pressure
Inlet temperature
Inlet pressure
Inside diameter
Wall thickness
Outside diameter
"C20T2Psi883P2
"C50T1
Psi1152P1
inches51.5Di
Inches0.25t
Inches52Do
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The Solution is :
ft/s27.34velocity
MMSCM/Day68.41=2,415,957,168Q(SCF/D)
If the pipe efficiency consider 92% and effective
roughness of 0.0018#.
The average erosion velocity is 59.2 ft/s socalculated velocity is OK.
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Question #2
If we want to branch and use gas along pipeline as
follow, what is the suit pipe diameter according to
desire pressure drop?
891.53883900044.109834430154
950.8890010001044.503969840503
1050.8100011004044.313135968.41202
1126.2110011522047.955650268.41151
Pav
Psi
Outlet Press.
(Psi)
Inlet Press.
(Psi)
Elevation
H (+m)
Inside Diameter
(inches)
Flow Rate
(MMSCM/D)
Distance
(Km)
Seg.
No.
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Question #3
If the flow rate and inlet pressure with pipe diameter
are known for each segments, how we can calculate
pressure drop in each segments?
20893.8917887.755490004730154
40963.2841925.61031000104740503
301063.4071025.9541100404768.41202
501123.3481094.2011152204768.41151
Ave.
Temp.
(!C)
Pav
Psi
Outlet
Press.
(Psi)
Inlet
Press
. (Psi)
Elevation
H (+m)
Inside
Diameter
(inches)
Flow Rate
(MMSCM/D)
Distance
(Km)
Seg.
No.
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RPA Hydraulic Module Calculate
All Of Them !!!
It can be also help you to find thermo
physical properties of natural gas.
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You can use any friction factorcorrelation in RPA!
1
f
.4 log
k
D
3.7065
1.2613
.Re f
k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
A 2 .2.457 ln
7
Re
0.9
.0.27 k
D
16
A 337530
Re
16
k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
f .8
8
Re
121
A 2 A 3
3
2
1
12
A 4
k
D
1.1098
2.8257
7.149
Re
0.8981
f .1
.4 log
k
D3.7065
.5.0452Re
log A 4
2
4 k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
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Pipe Line Hydraulic Calculation
By:
M. Khosravy
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An Example !
A pipeline has 4 segments as follow:Segment
No. Length (Km) Elevation (+m) Temp ("C)
1 15 20 502 20 40 30
3 50 10 40
4 15 0 20
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Question #1
If the pipe diameter and inlet/outlet pressure are
known, what is the flow velocity of natural gas in it?
Outlet temperatureOutlet pressure
Inlet temperature
Inlet pressure
Inside diameter
Wall thickness
Outside diameter
"C20T2Psi883P2
"C50T1
Psi1152P1
inches51.5Di
Inches0.25t
Inches52Do
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The Solution is :
ft/s27.34velocity
MMSCM/Day68.41=2,415,957,168Q(SCF/D)
If the pipe efficiency consider 92% and effective
roughness of 0.0018#.
The average erosion velocity is 59.2 ft/s socalculated velocity is OK.
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Question #2
If we want to branch and use gas along pipeline as
follow, what is the suit pipe diameter according to
desire pressure drop?
891.53883900044.109834430154
950.8890010001044.503969840503
1050.8100011004044.313135968.41202
1126.2110011522047.955650268.41151
Pav
Psi
Outlet Press.
(Psi)
Inlet Press.
(Psi)
Elevation
H (+m)
Inside Diameter
(inches)
Flow Rate
(MMSCM/D)
Distance
(Km)
Seg.
No.
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Question #3
If the flow rate and inlet pressure with pipe diameter
are known for each segments, how we can calculate
pressure drop in each segments?
20893.8917887.755490004730154
40963.2841925.61031000104740503
301063.4071025.9541100404768.41202
501123.3481094.2011152204768.41151
Ave.
Temp.
(!C)
Pav
Psi
Outlet
Press.
(Psi)
Inlet
Press
. (Psi)
Elevation
H (+m)
Inside
Diameter
(inches)
Flow Rate
(MMSCM/D)
Distance
(Km)
Seg.
No.
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RPA Hydraulic Module Calculate
All Of Them !!!
It can be also help you to find thermo
physical properties of natural gas.
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You can use any friction factor
correlation in RPA!
1
f
.4 log
k
D
3.7065
1.2613
.Re f
k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
A 2 .2.457 ln
7
Re
0.9
.0.27 k
D
16
A 337530
Re
16
k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
f .8 8
Re
121
A 2 A 3
3
2
1
12
A 4
k
D
1.1098
2.8257
7.149
Re
0.8981
f .1
.4 log
k
D
3.7065
.5.0452
Re log A 4
2
4 k=Roughness in feet
D=inside diameter in feet
Re= Reynolds number
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Thank You for
Your Attention
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