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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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Steam Turbine Testing
Location and Type of Test Instrumentation
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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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 Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
Sample Calculation of Test kW Load (1) -Measurement
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Steam Turbine Testing
Sample Calculation of Test kW Load (2) - Adjustment
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Steam Turbine Testing
Sample Calculation of Test Heat Rate (1) - Unaccounted for Flow
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Steam Turbine Testing
Sample Calculation of Test Heat Rate (2) - Throttle Flow and Heat Rate
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Steam Turbine Testing
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
STEAM TURBINE PERFORMANCE CALCULATION
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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
Feedwater Pump Turbine Extraction Correction CurveSample Calculation
Measured % Throttle Flow : 1.29 %
Design % Throttle Flow : 1.26 %
@ Test TFR of this curve
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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Steam Turbine Testing
Feedwater Pump Enthalpy Rise Correction CurveSample Calculation
Measured Enthalpy Rise : 2.69 Btu/lb
Design Enthalpy Rise : 3.36 Btu/lb
@ Test TFR of this curve
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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Steam Turbine Testing
Condenser Sub-cooling Correction CurveSample Calculation
Measured Sub-Cooled Temp : 0.24
Design Sub-Cooled Temp : 0
@ Test TFR of this curve
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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Steam Turbine Testing
Condenser Make-up Correction CurveSample Calculation
Measured % Make-up : 0.00 %
Design % Make-up : 0.20 %
@ Test TFR of this curve
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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%
From of this curve @ 4.04%
HR % Correction = - 0.30%
HR = 1 + % Correction
= 1 + ( - 0.30 / 100) =0.9970
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Steam Turbine Testing
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%
From of this curve @ 0.21%
kW % Correction = 0.06%kW = 1 + % Correction
= 1 + ( 0.06 / 100 ) =1.0006
+100
%Corr1kW
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Steam Turbine Testing
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%
From of this curve @ 0.21%
HR % Correction = - 0.06%HR = 1 + % Correction
= 1 + ( - 0.06 / 100 ) =0.9994
+100
%Corr1HR
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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
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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Steam Turbine Testing
Raw Data Evaluation
Test Cycle CalculationContract Cycle Calculation
Verification of Performance Test Results Test Result Calculation Contract Cycle Heat Rate
STEAM TURBINE PERFORMANCE CALCULATION
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Steam Turbine Testing
Test Result Calculation
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Steam Turbine Testing
Contract Cycle Heat Rate
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END OF DOCUMENT