(alternative test) · 3 steam turbine testing variation in measured test data steady-state : -...

(Alternative Test)

STEAM TURBINE THERMAL PERFORMANCE TESTING

2008. 04. 10증기터빈설계1팀

1

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 CalculationVerification of Performance Test Results

2


Location and Type of Test Instrumentation

3


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 7Time (hours)

Tem

pera

ture

(F)

Diffe

renc

e fro

m st

eady

-st

ate

Deviation from ReferenceInstability

Two-hour Performance Test

4


Permissible Deviation of Variables

Temp +/- 30 O

F from design temp +/- 7 O

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

Variable Permissible Deviationfrom Design Condition Permissible Fluctuation

Main Steam

5


Effect of Measurement Error

1%

1ºF

1%

1ºF

1%

1ºF

1%

1ºF

-0.76%

+0.27%

+0.69%

-0.33%

-

-

-

-

-0.60%

+0.21%

+0.59%

-0.26%

Error HP Efficiency % IP Efficiency %

Throttle Pressure

Throttle Temperature

Cold Reheat Pressure

Cold Reheat Temperature

Hot Reheat Pressure

Hot Reheat Temperature

IPCrossover Pressure

IP Crossover Temperature

FOSSIL UNIT1% h HP = 0.16% HR & 0.30% KW 1% h IP = 0.12% HR & 0.12% KW1% h LP = 0.50% HR & 0.50% KW

NUCLEAR UNIT1% h HP = 0.26% ~ 0.41% HR & KW 1% h LP = 0.59% ~ 0.74% HR & KW

6


STEAM TURBINE PERFORMANCE CALCULATION

Raw Data EvaluationTest Cycle Calculation

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

Contract Cycle CalculationVerification of Performance Test Results

7


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 *Δ P / ρ )Where g : Gravitational constant

ΔP: Differential pressureρ : 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-sectional area(A) of the pipe:

Q = A * sqrt( 2g *Δ P / ρ )And mass flow rate(m) is calculated by multiplying Q by the density of the fluid at the operatingpressure and temperature:

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

8


ASME PTC 6.0 Flow SectionQ = 1890.07 * d2 * Fa * Cq * ΔP/ v * [ 1 / 1-β4 ]

Where d : Nozzle Diameter (inches)D : Pipe Diameter (inches)β : d / Dv : Specific VolumeΔP: Differential Pressure Fa : Thermal Expansion FactorCq : 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 = ρ * V * d / μ = 48 * Q / ( π * d * μ)

Where ρ : densityV : Velocity (ft/sec)μ : Dynamic viscocityQ : Flow (lb/hr)

9


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 * ΔP / v * [ 1 / 1-β4 ] with predetermined Cq(4) Determine Rd = 48 * Q / ( π * d * μ) with predetermined Q (5) Iterate until Assumed Rd = Determined Rd

Cq = 1.0054 - 0.185 * Rd-0.2 * (1 - 361239 / Rd) 0.8

10


Sample Calculation for Final Feedwater Flow

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Secondary Flow Elements

V-CONEORIFICE

PITOT TUBE Forward-Reverse Tube

12


Sample Calculation for Secondary Flow

13


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 * (H throttle-H FFW) + Q HRT * ( H HRT – H CRH)

KW LoadWhere

K : Watt-hour meter constant, 0.003125WHMCF : Watt-hour meter correction factorCTRCF : Current tranformer ratio correction factorPTRCF : Potential tranformer ratio correction factorCTR : Current tranformer ratioPTR : Potential tranformer ratio

Q throttle: Turbine throttle steam flow, lbm/hrQ HRT : Turbine hot reheat steam flow, lbm/hrH 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

14


Sample Calculation of Test kW Load (1) -Measurement

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Sample Calculation of Test kW Load (2) - Adjustment

16


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

17


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

18


Raw Data EvaluationTest Cycle CalculationContract 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


19


Group 1 Correction for ASME PTC 6 Alternative Test Group 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 Drop Correction 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

20


Top Heater Terminal Temperature Difference Correction Curve for Load

Sample CalculationMeasured TTD : 3.64 ℉Design TTD : 5.00 ℉

From this curvekW % Correction @ 3.64 ℉ = 0.032%

kW = 1 + %Correction / 100= 1 + 0.032 / 100 = 1.0003

표준원전: -315kW @518,859kW

⎠

⎞⎜⎜⎝

⎛+=100

%Corr1kW ⎜

21


Top Heater Terminal Temperature Difference Correction Curve for Heat Rate

Sample CalculationMeasured TTD : 3.64 ℉Design TTD : 5.00 ℉

From this curveHR % Correction @ 3.64 ℉= -0.032%

kW = 1 + %Correction / 100= 1 - 0.032 / 100 = 0.9997

표준원전: -0.7kCal @2,312kCal/kWh

⎠

⎞⎜⎜⎝

⎛+=100

%Corr1HR ⎜

22


Top Heater Extraction Pressure Drop Correction

Sample CalculationMeasured ΔP (from TBN to HRT) = 10.13 psiDesign ΔP (from TBN to HRT) = 13.80 psi

Measured Pressure @ TBN = 478.80 psiaExpected Pressure @ HTR with Design ΔP = Pressure @ TBN - ΔP_Design = 478.8 - 13.8 = 465.0 psiaSaturation Temperature @ 465.0 psia = 459.59 ℉

Measured Pressure @ HTR = 468.67 psiaSaturation Temperature @ 468.67 psia = 460.39 ℉

Change of TTD caused by test ΔP = 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

23


Top Heater Drain Cooler Approach Temperature Correction Curve

Sample CalculationMeasured DC : 12.26 ℉Design DC : 10.00 ℉

@ Test TFR of this curvekW % 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

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−×−=

−×+=

F deg

F deg

10)DC(DC

100%Corr1kW

10)DC(DC

100%Corr1 HR

dt

dt

24


Feedwater Pump Turbine Extraction Correction Curve

Sample CalculationMeasured % Throttle Flow : 1.29 %Design % Throttle Flow : 1.26 %

@ Test TFR of this curvekW % 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%Corr1 HR

dt

dt

25


Feedwater Pump Enthalpy Rise Correction Curve

Sample CalculationMeasured Enthalpy Rise : 2.69 Btu/lbDesign Enthalpy Rise : 3.36 Btu/lb

@ Test TFR of this curvekW % 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

26


Condenser Sub-cooling Correction Curve

Sample CalculationMeasured Sub-Cooled Temp : 0.24 ℉Design Sub-Cooled Temp : 0 ℉

@ Test TFR of this curvekW & 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

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−×−=

−×+=

F deg

F deg

5)SCT(SCT

100%Corr1kW

5)SCT(SCT

100%Corr1HR

dt

dt

27


Condenser Make-up Correction Curve

Sample CalculationMeasured % Make-up : 0.00 %Design % Make-up : 0.20 %

@ Test TFR of this curvekW % 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(%MU

100%Corr1kW

1%)%MU(%MU

100%Corr1HR

dt

dt

28


Sample Calculation for Group1 Correction with Correction Curves

TTD – Top Feedwater HeaterELPD - Top Feedwater HeaterDC – Top Feedwater Heater Feedwater Pump Turbine ExtractionFeedwater Pump Enthalpy RiseCondensate Sub-cooling Condenser Make-up FlowCombined Factor

3.64 deg F2.12% (10.13 psi)12.26 deg F158391 lbm/h2.69 Btu/lm-0.24 deg F0 %

5.00 deg F3.0% (13.0 psi)10.00 deg F152482 lbm/h3.36 Btu/lm0.00 deg F0.2 %

1.00031.00021.00000.99970.99951.00001.00051.0001

0.99970.99981.00001.00031.00051.00000.99950.9999

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

29


Group 2 Correction

Turbine Initial Steam Pressure CorrectionTurbine Initial Steam Temperature CorrectionTurbine 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 design conditions, as follows;

Pd x υt

Pt x υdQc = Q t x

Where Qc : 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 Load

Sample CalculationMeasured 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

31


Initial Pressure Correction Curve for Heat rate

⎟⎟⎠

⎞⎜⎜⎝

⎛+= 100%Corr1HR

Sample CalculationMeasured Pressure : 1076.82 psiaDesign Pressure : 1035 psia

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

From of this curve @ 4.04%HR % Correction = - 0.30%

HR = 1 + % Correction= 1 + ( - 0.30 / 100) = 0.9970

32


Initial Moisture Correction Curve for LoadSample Calculation@ Steam Generator Outlet Pressure : 1109.98 psiaMoisture : 0.003%Enthalpy : 1188.64 Btu/lb from steam table

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

Design Moisture : 0.45%

From of this curve @ 0.21%kW % Correction = 0.06%

kW = 1 + % Correction= 1 + ( 0.06 / 100 ) = 1.0006

⎟⎟⎠

⎞⎜⎜⎝

⎛+= 100%Corr1kW

33


Initial Moisture Correction Curve for Heat RateSample Calculation@ Steam Generator Outlet Pressure : 1109.98 psiaMoisture : 0.003%Enthalpy : 1188.64 Btu/lb from steam table

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

Design Moisture : 0.45%

From of this curve @ 0.21%HR % Correction = - 0.06%

HR = 1 + % Correction= 1 + ( - 0.06 / 100 ) = 0.9994

⎟⎟⎠

⎞⎜⎜⎝

⎛+= 100%Corr1HR

34


Exhaust Pressure Correction

Sample CalculationMeasured Exhaust Pressure : 32.1 psiaDesign Exhaust Pressure : 38.0 psia

From of this curve @ 33.1 psiakW & HR % Correction = 0.12%

kW = 1 + % Correction= 1 + ( 0.12 / 100) = 1.0012

HR = 1 - % Correction= 1 - ( 0.12 / 100 ) = 0.9988

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

−=

+=

100%Corr1HR

100%Corr1kW

35


Sample Calculation for Group2 Correction with Correction Curves

Initial Pressure, psiaInitial Moisture, %Exhaust Pressure, mm HgCombined Factor

1076.8 psia0.21%33.1 mm Hg

1035 psia0.45%38.0 mm Hg

1.00301.00061.00121.0048

0.99700.99940.99880.9952

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 / kWh kW LOAD after Group1 Correction = 1,053,385 kW

36


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

% HR Correction = x x 100 x KW mo P tW n ΔP mo - ΔP vwo

where subscript mo indicates highest load test, and

Wn / W mo = 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)

(ΔP mo - ΔP 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

37


Sample CalculationMeasured Initial Steam Pressure : 1076.82 psiaMeasured Steam Chest Pressure : 1073.73 psiaMeasured CV#4 Discharge Pressure : 906.79 psiaMeasured CV#4 Position : 28.97%

1) By interpolation between 23.43% and 30.08% in the %Lift column of this table Total Flow (W mo) = 12,185,173 lbm/hrCV#4 Flow (W m ) = 2,152,956 lbm/hr

∴ W m / W mo = 0.1767

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

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

3) ΔP 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%Corr1 HR

100%Corr1kW

38


Correction for MW Thermal Output

MW Thermal Output = Net heat to the cycle x 3,412.14= Q SG Outlet x ( H SG Outlet - H SG inlet ) / 3,412.14

Where Q 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 H SG Outlet and H SG Inlet using Steam Table H SG Outlet = STMPTH (1,204.10 psia, 456.6 ℉) = 438.03 Btu/lb H 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 = 0.136 kW = 1 + % Correction = 1 + ( 0.136 / 100) = 1.0014 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+= 100%Corr1kW

39


Raw Data EvaluationTest Cycle CalculationContract Cycle Calculation

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


40


Test Result Calculation

41


Contract Cycle Heat Rate

42

END OF DOCUMENT

(alternative test) · 3 steam turbine testing variation in measured test data steady-state : -...

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