turbine test

8/7/2019 turbine test

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(Alternative Test)

STEAM TURBINE THERMAL PERFORMANCE TESTING

2008. 04. 101


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Steam Turbine Testing

STEAM TURBINE PERFORMANCE TEST (ALTERNATIVE)

Raw Data Evaluation Location and Type of Test Instrumentation Variation in Measured Test Data Permissible Variation of Variables

Effect of Measurement Error Test Cycle CalculationContract Cycle Calculation

Verification of Performance Test Results


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Location and Type of Test Instrumentation


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Variation in Measured Test DataSteady-state :- Average Value Over Short Intervals Does Not ChangeStable :

- Value Does Not Fluctuate Over TimeDifference From Reference :- Difference Between Average Value During Test and Value in Reference Case

300

400

500

600

700

800

900

1000

1100

0 1 2 3 4 5 6 7

Time (hours)

T e m p e r a

t u r e

( F )

D i f f e r e n c e

f r o m

s t e a

d y -

s t a t e

Deviation from ReferenceInstability

Two-hour PerformanceTest


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Permissible Deviation of Variables

Temp +/- 30O

F from design temp +/- 7O

F from average value

Press +/- 3.0 % of the absolute press +/- 0.25% of the absolute press

RTH Steam Temp +/- 30 O F +/- 7 O F

+/- 0.05 psi +/- 0.02 psi

Not Specified +/- 1.0%

+/- 5.0%

+/- 10.0%

Exhaust Pressure

Power Factor

Voltage

Speed

VariablePermissible Deviation

from Design ConditionPermissible Fluctuation

Main Steam


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Effect of Measurement Error

1%1F

1%

1F

1%

1F

1%

1F

-0.76%+0.27%

+0.69%

-0.33%

-

-

-

-

-0.60%

+0.21%

+0.59%

-0.26%

Error HP Efficiency% IP Efficiency %

Throttle PressureThrottle Temperature

Cold Reheat Pressure

Cold Reheat Temperature

Hot Reheat Pressure

Hot Reheat Temperature

IPCrossover Pressure

IP Crossover TemperatureFOSSIL UNIT1% h HP = 0.16% HR & 0.30% KW1% h IP = 0.12% HR & 0.12% KW

1% h LP = 0.50% HR & 0.50% KW

NUCLEAR UNIT1% h HP = 0.26% ~ 0.41% HR & KW1% h LP = 0.59% ~ 0.74% HR & KW


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STEAM TURBINE PERFORMANCE CALCULATION

Raw Data Evaluation

Test Cycle Calculation Primary Flow & Secondary Flow Test Cycle Heat Rate & kW Load

Contract Cycle CalculationVerification of Performance Test Results


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Primary Flow & Secondary Flows Calculation

Calculation of Flow Rate from Differential PressureThe average velocity(V) of the flow in the pipe can be calculated by the following formula:

V = sqrt ( 2g * P / )

Where g : Gravitational constantP: Differential pressure : Density of fluid

However velocity is rarely calculated in practice, but volumetric and mass flow rate are.Volumetric flow rate(Q) is calculated by multiplying the average velocity by the cross-sectionalarea(A) of the pipe:

Q = A * sqrt( 2g * P / )

And mass flow rate(m) is calculated by multiplying Q by the density of the fluid at the operatingpressure and temperature:

M = Q = A Cq * sqrt ( 2g * P / )Where Cq : Flow coefficient determined by calibration


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ASME PTC 6.0 Flow SectionQ = 1890.07 * d2 * Fa * Cq * P/ v * [ 1 / 1-4 ]

Where d : Nozzle Diameter (inches)

D : Pipe Diameter (inches) : d / Dv : Specific VolumeP: Differential PressureFa : Thermal Expansion Factor Cq : Flow coefficient from calibration data

Cq = Cx - 0.185 * Rd-0.2 * (1 - 361239 / Rd)0.8

Where Cx : Coefficient determined by calibrationRd : Reynolds Number

Rd = * V * d / = 48 * Q / ( * d *)

Where : densityV : Veloci ty (ft/sec) : Dynamic viscoci tyQ : Flow (lb/hr)


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Primary Flow Calculation Method(1) Assume Reynolds Number

(2) Determine Cq = Cx - 0.185 * Rd-0.2 * (1 - 361239 / Rd)0.8 with assumed Rd

(3) Determine Q = 1890.07 * d2 * Fa * Cq * P / v * [ 1 / 1-4 ] with predetermined Cq(4) Determine Rd = 48 * Q / ( * d *) with predetermined Q(5) Iterate until Assumed Rd = Determined Rd

Cq = 1.0054 - 0.185 * Rd-0.2

* (1 - 361239 / Rd)0.8


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Test Cycle Heat Rate and KW Load

KW Load = * K * WHMCF * CTRCF * PTR CF * CTR * PTR# Counts

Test Time(hrs)

Heat Rate = ( Heat Supplied Heat Returned ) / KW Load

=

Q throttle* (Hthrottle-HFFW) + QHRT * ( HHRT HCRH)

KW Load

Where

K : Watt-hour meter constant, 0.003125

WHMCF : Watt-hour meter correction factor CTRCF : Current tranformer ratio correction factor PTRCF : Potential tranformer ratio correction factor CTR : Current tranformer ratioPTR : Potential tranformer ratio

Q thrott le: Turbine throttle steam flow, lbm/hr

Q HRT : Turbine hot reheat steam flow, lbm/hr H throtle : Turbine throttle steam enthalpy , Btu/lbmH HRT : Turbien hot reheat steam enthalpy, Btu/lbmH CRH : Turbien cold reheat steam enthalpy, Btu/lbmH FFW : Final feedwater enthalpy, Btu/lbm


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Sample Calculation of Test kW Load (1) -Measurement


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Sample Calculation of Test kW Load (2) - Adjustment


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Sample Calculation of Test Heat Rate (1) - Unaccounted for Flow


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Sample Calculation of Test Heat Rate (2) - Throttle Flow and Heat Rate


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Raw Data Evaluation

Test Cycle Calculation

Contract Cycle Calculation- Group 1 Correction for ASME PTC 6.0 Alternative Test- Group 2 Correction- Correction for Control Valves Throttling and MW Thermal Output

Verification of Performance Test Results



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Group 1 Correction for ASME PTC 6 Alternative TestGroup 1 Correction Curves for The Alternative Procedure

Final Feed Water Temperature Correction- Top Heater Terminal Temperature Difference- Top Heater Drain Cooler Approach Temperature Difference- Top Heater Extraction Line Pressure DropCorrection for Auxiliary Extraction Steam Flow

- Extraction Steam to Station Heating- Extraction Steam to Air Pre-heater (Fossil Only)- etc.Corrections for Main Steam De-superheating Flow (Fossil Only)

Corrections for Reheat Steam De-superheating Flow (Fossil Only) Auxiliary Turbine Extraction CorrectionCondensate Sub-cooling CorrectionCondenser Make-up Correction


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Top Heater Terminal Temperature Difference Correction Curve for LoadSample Calculation

Measured TTD : 3.64

Design TTD : 5.00

From this curve

kW % Correction @ 3.64 = 0.032%

kW = 1 + %Correction / 100

= 1 + 0.032 / 100 =1.0003

: -315kW @518,859kW

+100

%Corr1kW


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Top Heater Terminal Temperature Difference Correction Curve for Heat RateSample Calculation

Measured TTD : 3.64

Design TTD : 5.00

From this curve

HR % Correction @ 3.64= -0.032%

kW = 1 + %Correction / 100

= 1 - 0.032 / 100 =0.9997

: -0.7kCal @2,312kCal/kWh

+100

%Corr1HR


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Top Heater Extraction Pressure Drop CorrectionSample Calculation

Measured P (from TBN to HRT) = 10.13 psiDesign P (from TBN to HRT) = 13.80 psi

Measured Pressure @ TBN = 478.80 psiaExpected Pressure @ HTR with Design P = Pressure @ TBN - P_Design = 478.8 - 13.8 = 465.0 psiaSaturation Temperature @ 465.0 psia =459.59

Measured Pressure @ HTR = 468.67 psia

Saturation Temperature @ 468.67 psia =460.39

Change of TTD caused by test P = 460.39 - 459.59 = 0.8

From Top Heater Terminal Temperature Difference Correction CurveskW % correction @ 4.2= 0.019%HR % correction @ 4.2= - 0.019%

kW = 1 + %Correction / 100 = 1 + 0.019 / 100 =1.0002HR = 1 + %Correction / 100 = 1 - 0.019 / 100 =0.9998


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Top Heater Drain Cooler Approach Temperature Correction CurveSample Calculation

Measured DC : 12.26

Design DC : 10.00

@ Test TFR of this curve

kW % Correction = 0.018%HR % Correction = 0.022%

kW = 1-[ (0.018/100) x (12.26-10) / 10 ]= 1.0000

HR = 1+[ (0.022/100) x (12.26-10) /10 ]= 1.0000

=

+=

Fdeg

Fdeg

10)DC(DC

100%Corr1kW

10)DC(DC

100%Corr1 HR

dt

dt


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Feedwater Pump Turbine Extraction Correction CurveSample Calculation

Measured % Throttle Flow : 1.29 %

Design % Throttle Flow : 1.26 %


kW % Correction = -1.015%HR % Correction = 1.019%

kW = 1 + [ (-1.019 / 100) x (1.29-1.26) ]= 0.9997

HR = 1 + [ (1.015 / 100) x (1.29-1.26) ]= 1.0003

+=

+=

1%)%TF(%TF

100%Corr1kW

1%)%TF(%TF

100%Corr1HR

dt

dt


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Feedwater Pump Enthalpy Rise Correction CurveSample Calculation

Measured Enthalpy Rise : 2.69 Btu/lb

Design Enthalpy Rise : 3.36 Btu/lb


kW % Correction = 0.079%HR % Correction = - 0.078%

kW = 1 + [ (0.079 / 100) x (2.69 - 3.36) ]= 0.9995

HR = 1 + [ (-0.078 / 100) x (2.69 - 3.36) ]= 1.0005

Btu/lb

Btu/lb

1)H_rise(H_rise

100%Corr1kW

1)H_rise(H_rise

100%Corr1HR

dt

dt


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Condenser Sub-cooling Correction CurveSample Calculation

Measured Sub-Cooled Temp : 0.24

Design Sub-Cooled Temp : 0


kW & HR % Correction = 0.027%

kW = 1 - [ (0.027 / 100) x (0.24 - 0.00) / 5 ]= 1.0000

HR = 1 + [ (0.027 / 100) x (0.24 - 0.00) / 5 ]= 1.0000

Fdeg

Fdeg

5)SCT(SCT

100%Corr1kW

5)SCT(SCT

100%Corr1HR

dt

dt


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Condenser Make-up Correction CurveSample Calculation

Measured % Make-up : 0.00 %

Design % Make-up : 0.20 %


kW % Correction = - 0.255%HR % Correction = 0.259%

kW = 1 + [ (-0.255 / 100) x (0.0-0.2) ]= 1.0005

HR = 1 + [ (0.259 / 100) x (0.0-0.2) ]= 0.9995

1%

)%MU(%MU100

%Corr1kW

1%)%MU(%MU

100%Corr1HR

dt

dt


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Sample Calculation for Group1 Correction with Correction Curves

TTD Top Feedwater Heater

ELPD - Top Feedwater Heater DC Top Feedwater Heater Feedwater Pump Turbine ExtractionFeedwater Pump Enthalpy RiseCondensate Sub-coolingCondenser Make-up Flow

Combined Factor

3.64 deg F

2.12% (10.13 psi)12.26 deg F158391 lbm/h2.69 Btu/lm-0.24 deg F0 %

5.00 deg F

3.0% (13.0 psi)10.00 deg F152482 lbm/h3.36 Btu/lm0.00 deg F0.2 %

1.0003

1.00021.00000.99970.99951.00001.0005

1.0001

0.9997

0.99981.00001.00031.00051.00000.9995

0.9999

Test Cycle Desing Cycle HR Factor kW Factor

Heat Rate after Group 1 Correction = Test Heat Rate / Combined HR Factor = 2308.87 / 0.9999 = 2308.23 kcal / kWh

KW Load after Group 1 Correction = Test kW Load / Combined KW Factor = 1,053,485 / 1.0001 = 1,053,385 KW

TEST HEAT RATE = 2,308.87 kcal/ kWhTEST kW LOAD = 1,053,485 kW


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Group 2 CorrectionTurbine Initial Steam Pressure CorrectionTurbine Initial Steam Temperature Correction

Turbine Hot Reheat Steam Temperature Correction (Fossil Only)Turbine Initial Steam Moisture (Nuclear only)Pressure Drop of Steam through the reheater systems (Fossil Only)Turbine Exhauster Pressure

Use of Group 2 Correction Curves requires correction of the test throttle flow to designconditions, as follows;

Pd x tPt x d

Qc = Q t x

WhereQc : Corrected Throttle FlowQt : Test Throttle FlowPd : Design Turbine Initial Steam PressureUd : Design Turbine Initial Steam Specific VolumePt : Test Turbine Initial Steam PressureUt : Test Turbine Initial Steam Specific VolumeFor nuclear unit, use Steam Quality

instead of Steam Specific Volume...


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Initial Pressure Correction Curve for LoadSample Calculation

Measured Pressure : 1076.82 psiaDesign Pressure : 1025 psia

%Change in pressure= (1076.82-1035) / 1035 = 4.04%

From of this curve @ 4.04%

kW % Correction = 0.30%

kW = 1 + % Correction

= 1 + ( 0.30 / 100) =1.0030

+100

%Corr1kW


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Initial Pressure Correction Curve for Heat rate

+100

%Corr1HR

Sample Calculation

Measured Pressure : 1076.82 psiaDesign Pressure : 1035 psia

%Change in pressure= (1076.82-1035) / 1035 = 4.04%


HR % Correction = - 0.30%

HR = 1 + % Correction

= 1 + ( - 0.30 / 100) =0.9970


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Initial Moisture Correction Curve for LoadSample Calculation@ Steam Generator OutletPressure : 1109.98 psia

Moisture : 0.003%Enthalpy : 1188.64 Btu/lbfrom steam table

@ Turbine inletPressure : 1076.82 psiaMoisture : 0. 21%from steam tableEnthalpy : 1188.64 Btu/lb= H @ SG out let

Design Moisture : 0.45%


kW % Correction = 0.06%kW = 1 + % Correction

= 1 + ( 0.06 / 100 ) =1.0006

+100

%Corr1kW


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Initial Moisture Correction Curve for Heat RateSample Calculation@ Steam Generator OutletPressure : 1109.98 psia

Moisture : 0.003%Enthalpy : 1188.64 Btu/lbfrom steam table

@ Turbine inletPressure : 1076.82 psiaMoisture : 0. 21%from steam tableEnthalpy : 1188.64 Btu/lb= H @ SG out let

Design Moisture : 0.45%


HR % Correction = - 0.06%HR = 1 + % Correction

= 1 + ( - 0.06 / 100 ) =0.9994

+100

%Corr1HR


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Exhaust Pressure CorrectionSample Calculation

Measured Exhaust Pressure : 32.1 psiaDesign Exhaust Pressure : 38.0 psia

From of this curve @ 33.1 psia

kW & HR % Correction = 0.12%

kW = 1 + % Correction

= 1 + ( 0.12 / 100) =1.0012HR = 1 - % Correction

= 1 - ( 0.12 / 100 ) =0.9988

+

100%Corr1HR

100%Corr1kW


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Sample Calculation for Group2 Correction with Correction Curves

Initial Pressure, psia

Initial Moisture, %Exhaust Pressure, mm HgCombined Factor

1076.8 psia

0.21%33.1 mm Hg

1035 psia

0.45%38.0 mm Hg

1.0030

1.00061.00121.0048

0.9970

0.99940.99880.9952

Test Cycle Design Cycle HR Factor kW Factor

Contract Cycle Heat Rate = Heat Rate after Group1 Correction / Combined HR Factor = 2,308.23 / 0.9952 = 2,311.27 kcal / kWh

Contract Cycle KW Load = kW Load after Group1 Correction / Combined kW Factor = 1,053,385 / 1.0048 = 1,050,996 kW

HEAT RATE after Group1 Correction = 2,308.23 kcal / kWhkW LOAD after Group1 Correction = 1,053,385 kW


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Correction for Control Valves Throttling and MW Thermal OutputCorrection for Control Valves Throttling

% HR Correction = x x 100 x KWmo P tWn

P mo -

P vwo

where subscript mo indicates highest load test, and

Wn / Wmo = the ratio of flow through the final valve(s) to total flow during the highest load test(decimal fraction of total flow being subject to extra throttling)

( P mo - P vwo) / P t = the ratio of (1) the difference between pressure drop across the final valve(s)during the highest load test and the pressure drop across the same valve(s) atVWO conditions to (2) throttle pressure (extra pressure drop due to final valvesnot being wide open)

P t = absolute throttle pressure

K = 0.15 for turbines with nuclear steam supply operating predominantly in the moisture region= 0.10 for turbines operating predominantly in the superheated region


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Sample CalculationMeasured Init ial Steam Pressure : 1076.82 psiaMeasured Steam Chest Pressure : 1073.73 psiaMeasured CV#4 Discharge Pressure : 906.79 psiaMeasured CV#4 Posit ion : 28.97%

1) By interpolation between 23.43% and 30.08%in the %Lift column of this tableTotal Flow (Wmo) = 12,185,173 lbm/hr CV#4 Flow (Wm ) = 2,152,956 lbm/hr

Wm / Wmo = 0.1767

2) Assuming 1% pressure drop at Main Stop Valves P vwo @ Design= P initial@ Design x 0.99 - PCV#4@ Design= 1035 x 0.99 - 1014.97 = 9.68 psia

P vwo = P initial@ Measured x ( P vwo @ Design / P initial@ Design ) = 1076.82 x (9.68 / 1035) = 10.07 psia

3) P mo = P steam chest @ Measured - P CV#4 Discharge@ Measured = 1073.73 - 906.79 = 166.94 psia

% kW & HR Correction = 0.1767 x [ (166.94-10.07) / 1076.82 ] x 100 x 1.5 = 0.386

kW = 1 + % Correction = 1 - ( 0.386 / 100) =1.0039HR = 1 - % Correction = 1 + ( 0.386 / 100 ) =

0.9961

=

+=

100

%Corr1HR

100%Corr1kW


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Correction for MW Thermal Output

MW Thermal Output= Net heat to the cycle x 3,412.14

= QSG Outlet

x ( HSG Outlet

- HSG inlet

) / 3,412.14

WhereQ SG Outlet : Steam Generator Outlet Flow (lbm/hr)H SG Outlet : Steam Generator Outlet Enthalpy (Btu/lbm)H SG Inlet : Steam Generator Inlet Enthalpy (Btu/lbm)

3412. 14 : Conversion factor from BTU to kW

Sample Calculation1) Measured final feedwater flow : 12,879,666 lbm/hr

Measured unaccounted for flow : 20,222 lbm/hr

Q SG Outlet = 12,879,666 - 20,222 = 12, 859,444 lbm/hr

2) Calculate HSG Outlet and H SG Inlet using Steam TableH SG Outlet = STMPTH (1,204.10 psia, 456.6 ) = 438.03 Btu/lbH SG Intlet = STMPMH (1,109.98 psia, 0.003%) = 1,188.64 Btu/lb

3) MW Thermal Output = 12,859,444 x ( 1188.64 - 438.03) / 3,412.14 = 2828.84 MW

From Design MW Thermal Output of 2825 MW

% kW Correction = (2828.84 / 2825) - 1 = 0.136kW = 1 + % Correction = 1 + ( 0.136 / 100) =1.0014

+100

%Corr1kW


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Raw Data Evaluation

Test Cycle CalculationContract Cycle Calculation

Verification of Performance Test Results Test Result Calculation Contract Cycle Heat Rate



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Test Result Calculation


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Contract Cycle Heat Rate


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