GTG by Khosrawi

8/10/2019 GTG by Khosrawi

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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


Combustion ChamberCombustion Chamber


Turbine Hot gases converted to work




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Compressor


Combustion Chamber


TurbineTurbine Hot gases converted to work




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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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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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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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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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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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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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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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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


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


Re= Reynolds number



46/60



47/60



48/60



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Pipe Line Hydraulic Calculation

By:

M. Khosravy



50/60

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



51/60

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



52/60

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.



53/60

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.



54/60

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.



55/60

RPA Hydraulic Module Calculate

All Of Them !!!

It can be also help you to find thermo

physical properties of natural gas.



56/60

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


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


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


Re= Reynolds number



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Thank You for

Your Attention


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GTG by Khosrawi