SRR-3 12-87 • Instantaneous response to short circuits. • Up to 7.2 A field boost power. • Actuated when the generator voltage drops too low to maintain field current. • Compatible with Frequency Compensation. Boost is inhibited by low frequency. • For 50 or 60 Hertz brushless generators. • Each model covers a wide range of generator application. • Totally encapsulated to eliminate the effects of vibration, dust, or moisture. • CSA approved. The CBS 305 and CBS 320 Current Boost Systems are electronic devices designed to provide the Basler APR series of static voltage regulators with power to sustain exciter current when severe overload- ing of brushless generators occurs such as during motor starting or during short circuit conditions. Power for the current boost systems is derived from a power CT installed in two phases of the generator output. Class 300 Equipment CBS 305, 320 Current Boost Systems P. O. BOX 269 HIGHLAND, ILLINOIS, U.S.A. 62249 PHONE 618-654-2341 FAX 618-654-2351 DESCRIPTION Page 2 SPECIFICATIONS Pages 2 through 4 ORDERING Page 4 OUTLINE DRAWINGS Pages 7 and 8 FEATURES CBS 320 CBS 305 Current Transformer
Transcript
CBS 305, 320 Current Boost Systems
SRR-312-87
• Instantaneous response to short circuits.
• Up to 7.2 A field boost power.
• Actuated when the generator voltage drops too low to maintain field current.
• Compatible with Frequency Compensation. Boost is inhibited by low frequency.
• For 50 or 60 Hertz brushless generators.
• Each model covers a wide range of generator application.
• Totally encapsulated to eliminate the effects of vibration, dust, or moisture.
• CSA approved.
The CBS 305 and CBS 320 Current Boost Systems are electronicdevices designed to provide the Basler APR series of static voltageregulators with power to sustain exciter current when severe overload-ing of brushless generators occurs such as during motor starting orduring short circuit conditions. Power for the current boost systems isderived from a power CT installed in two phases of the generatoroutput.
Class 300 EquipmentCBS 305, 320
Current Boost Systems
P. O. BOX 269 HIGHLAND, ILLINOIS, U.S.A. 62249 PHONE 618-654-2341 FAX 618-654-2351
DESCRIPTIONPage 2
SPECIFICATIONSPages 2 through 4
ORDERINGPage 4
OUTLINE DRAWINGSPages 7 and 8
FEATURES
CBS 320
CBS 305
Current Transformer
CBS 305, 320 Current Boost Systems
CT SECONDARY TURNS:BE21331-001: Terminals 1 and 2 - 209 turnsBE21433-001: Terminals 1 and 2 - 209 turnsBE21432-001: Terminals 1 and 2 - 308 turns
Terminals 1 and 3 - 616 turns
CONTROL SIGNAL FROM APR REGULATOR:10mA, 1.5Vdc
SHOCK: Withstands 15 Gs in each of three mutuallyperpendicular planes.
VIBRATION:Withstands the following vibration spectrum:
CT SECONDARY INSULATION RATING:2500 Vac HI-Pot
STORAGE TEMPERATURE:-65°C (-85°F) to +85°C (+185°F)
OPERATING TEMPERATURE RANGE:-40°C (-40°F) to +60°C (+140°F)
HUMIDITY: Totally protected from humidity and con-densation by encapsulation.
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DESCRIPTION
OUTPUT POWER
MAXIMUM INPUT CURRENT FROM CT:CBS 305 and CBS 320: 7A continuousCBS 305: 33A for 30 secondsCBS 320: 45A for 30 seconds
POWER DISSIPATION:CBS 305 and CBS 320: 24 watts in non-boost modeCT-BE21331-001: 25 watts maximumCT-BE21433-001: 35 watts maximumCT-BE21432-001: 70 watts maximum
TRANSIENT RESPONSE: Less than 2 cycles fromregulator’s boost signaluntil boost outputreaches 90Vdc.
SPECIFICATIONS
The Current Boost System consists of a current boostmodule auto-and a current transformer (CT). Thesystem provides supplemental excitation source withthe regulator which allows the field to receive full forcingduring generator overloading and short circuits.
The Current Boost System allows three wire and fourwire generators to provide current during sustainedsingle phase and multiple phase line-to line faults. TheCBS also permits four wire generators to support phaseA and C line to neutral faults. Phase B line-to-neutralfaults will also have fault sustaining capabilities sincethe regulator and CBS input power is taken from theother unshorted phases.
The Current Boost Module uses the CB+ and CB-outputs of the APR regulator as a “turn-on” signal.Boost power to the module is supplied from a CTinstalled in phase A and C of the generator output.During short circuits, the CT is the source of all excita-tion current for the generator - see figure 2.
Because the APR regulator sensing is used to turn theboost on and off, the Current Boost System operating
level is automatically adjusted to the nominal generatorvoltage. The Current Boost System also follows theregulator’s underfrequency curve to allow the primemover to pick up large loads in one step.
To verify that the available CTs will accommodate thegenerator’s output cable, refer to “HOW TO ORDER -Current Transformer.”
Three CTs are available:a.BE21331-001 for use with the CBS 305 only and on
generators with rated line currents up to 800A.b.BE21433-001 for use with the CBS 320 on genera-
tors with rated line currents up to 800A.c.BE21432-001 for use with either the CBS 305 or
CBS 320 on generators with rated line currents upto 2400A.
The CBS 305 and CBS 320 are totally encapsulated toeliminate all effects of moisture, dust, vibration, andshock as hazards to operation.
DC Output Power Field ResistanceForUse DC Volts DC Amps Ohms Ohms
SAMPLE SPECIFICATION:A current boost system shall be provided with theBasler APR 63-5 voltage regulator to sustain exciterfield current during severe generator output overloadsor short circuit. The operation of the current boostsystem shall be controlled by the voltage regulator.
Power for the current boost system shall be provided
by a CT installed in two phases of the generator output.
The current boost system shall be a Basler ElectricCompany Mode CBS 305 current boost system, with acurrent transformer selected based on Basler recom-mendation and generator performance data.
HOW TO ORDER - Current Boost System:
HOW TO ORDER - Current Transformer (CT):Follow the step-by-step procedure below to select theproper turns-ratio for your application. Based on theresulting turns-ratio, wire or bus size, and ratedgenerator output current, determine the part numberof the CT having the proper rating and window size.
Step 1. Calculate the exciter field current duringgenerator short circuit:
IField
= E/Rf
Where:IField
= Exciter field Current at short Circuit.R
f = Exciter field resistance from generator data.
E = CBS field forcing voltage (90 Vdc forCBS 305,180 Vdc for CBS 320)
Step 2. From the generator manufacturer’s shortcircuit saturation data (plot of exciter field currentversus line amps with generator output short cir-cuited), determine the generator three phase linecurrent during a short circuit that would result from theexciter field current calculated in step 1.
When the CBS is Orderto be used with
APR63-5 CBS 305APR125-5 CBS 320
If for your generator system: Then proceed to:
This results in excessivegenerator linecurrent. Step 3A
This results in acceptablegenerator linecurrent. Step 38
Step 3A.a. Determine the acceptable generator three phaseline current at short circuit (typically 300% of nominal).Refer to Table 1 and locate the range that covers thisvalue in column 1.
b. Using the generator manufacturer’s short circuitdata (plot of exciter field current versus line amps withthe generator output short circuited), determine theexciter short circuit field current required to generatethe acceptable generator line current (1
2). Refer to
Table 1 and locate the current value equal to this valuein column 3. If the desired field current is in betweenthe column 3 values, go to the next higher value.
c. Check that the rated generator full load line currentdoesn’t exceed the maximum continuous line currentin column 2. If the generator full load line current doesexceed the maximum continuous line current rating,consult with the factory.
d. To obtain reduced field current, place a currentlimiting resistor in series with the exciter field. Calcu-late the value of this resistor as follows:
RS = E/I
2 - R
f
Where:R
S= Value of series field resistance to be added (in
ohms). This resistance must not be so great as torestrict normal forcing.
E = CBS field forcing voltage (90 Vdc for CBS 305,and 180 Vdc for CBS 320).
I2= Field current required to produce acceptable
generator line current at short circuit from b. above.R
f= Exciter field resistance from generator data.
e. Proceed to step 4.
Step 3B.
CBS 305, 320 Current Boost Systems
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HOW TO ORDER, continued
a. In Table 1. column 1, locate the range of thegenerator three phase short circuit line current thatcovers the value determined in step 2.
b. Check that rated generator line current does notexceed the maximum continuous line current incolumn 2.
c. In Table 1, column 3, locate the exciter field currentthat is closest to the value determined in step 1. If thevalue is between the values in column 3, go to the nexthigher value.
d. Proceed to step 4.
Step 4.
a. Find the turns-ratio at the point of intersection of thevalues found in step 3.
b. The first numeral of the turns-ratio indicates thenumber of turns of each of two (2) generator lines thatpass through the CT “window”. The second numberindicates secondary turns provided by the CT termi-nals.
Step 5.Verify that the CT “window” size is sufficient for thegenerator conductors by comparing the conductorcross-section (including insulation) with the windowarea shown in figure 5, 6, or 7.
Table 1 - Turns-Ratio Selection
Column 1 Column 2 Column 3
3 Phase Line Maximum Exciter Field Current at Rated CBSCurrent @ Continuous Output Voltage (90 Vdc for CBS 305 orShort Circuit Line Current 180 Vdc for CBS 320)
1.8Adc 3.6Adc 7.2Adc35-70 25 8:209 16:209 32:209
70 - 140 50 4:209 8:209 16:209BE21331-031
140- 275 100 2:209 4:209 8:209for CBS 305
or275 - 550 200 1:209 2:209 4:209
BE21433-001for CBS 320.
550 - 1100 400 1:209 1:209 2:209
1100 - 2200 800 1:209 1:209 1:209
1600 - 3200 1200 1:616 1:616 1:308 BE21432-001for CBS 305
3200 - 6400 2400 1:616 1:616 1:616 and CBS 320
CBS 305, 320 Current Boost Systems
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EXAMPLE
EXCESSIVE
(Step 3A) You determine that 450A would constitute anacceptable generator line current at short circuit. Fromthe generator manufacturer’s data, you determine thatan exciter field current of 5.14A is required for thegenerator system to deliver an acceptable 450A atshort circuit.
In Table 1, column 1, the value of 450A is between thevalues of 275 to 550A. Draw a horizontal line from thispoint across the chart.
In Table 1, column 3, the value of 7.2A is the closesthigher value to the required value of 5.14A (from step3A). Draw a vertical line from this point down the chart.In Table 1, column 2, the maximum continuous linecurrent of 200A is greater than the full load line currentof 15OA. therefore, the limits of the CT are not ex-ceeded. Determine the necessary resistance to beadded to the exciter field to achieve the 5.14A exciterfield current:
RS= - R
f = - 15 ohms = 17.5 -15 ohms
= 2.5 ohms
Note - The series resistance must not be so great as torestrict normal forcing. (Step 4) From Table 1, theintersection of the two lines (from step 3A) is the turns-ratio, 4:209.
4 turns on the primary: 209 turns on the secondary.
of the HOW TO ORDER - (CT) procedure on page 4.
Sample Generator Data:100KW, 125kVA, 60Hz, 3-phase, 480V, 150A at full rated load, 15 ohm field.IF = 5.14Adc @ 300% short circuit (450A line)IF = 6.0 Adc @ 350% short circuit (525A line)
(Step 1) Determine the field current that will be provided by a Basler CBS 305 during short circuit:
IF= = = 6.0A
Where: IF= Field Current
E = APR63-5/CBS305 Maximum dc output voltage.R = Exciter field resistance.
(Step 2) From generator manufacturer’s data, you determine that the exciter field current of 6.0A from theCBS 305 would result in a short circuit line current of 525A- for your generator system you consider this to be:
ER
900 Vdc15 ohms
EI2
90 Vdc5.14
(Step 5) The first numeral of the turns-ratio (4), thenumber of conductors per phase (1) and the numberof phase (2) multiplied together results in: 8 conduc-tors (through the CT window). From the generatormanufacturer’s data, 0.537 is the conductor diameter;therefore, for 8 conductors the CT window must be atleast 0.537 in X 4.30 inches. Basler CT P/N BE21331-001 is 0.9 in. X 4.375 in. and meets the requirement.
ACCEPTABLE
(Step 38) In table 1, column 1, the value of 525A (fromstep 2) is between the values of 275-550A. Draw ahorizontal line from this point across the chart.In Table 1, column 2, the maximum continuous linecurrent of 200A is greater than the full load line currentof 150A, therefore, the limits of the CT are not ex-ceeded.
n Table 1, column 3, the value of 7.2A is the nexthigher value after the required value of 6.OA (from step1). Draw a vertical line from this point down the chart.
(Step 4) From Table 1, the intersection of the two lines(from step 3A) is the turns-ratio, 4:209,
4 turns on the primary: 209 turns on the secondary
(Step 5) The first numeral of the turns-ratio (4) thenumber of conductors per phase (1) and the numberof phases (2). multiplied together results in: 8 conduc-tors (through the CT window). From the generatormanufacturer’s data, 0.537 is the conductor diameter;therefore, for 8 conductors, the CT window must be atleast 0.537 in. x 4.30 in. Basler CT P/N BE21331-001 is0.9 in. x 4.375 in. and meets the requirement.
CBS 305, 320 Current Boost Systems
OUTLINE DRAWINGS
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Figure 3 - Outline Drawing, Current Boost Module (CBS 305)
Figure 4 - Outline Drawing, Current Boost Module (CBS 320)
