# GENERATOR SELECTION. a. Three phase - 120/208V, 3 phase, 4W wye; 277/408, 3 phase, 4W wye; * 120/240V 3 phase, 4W Delta

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2 2. List Loads (for each load, list the following) a. Voltage b. Phase c. Horsepower if motor, STARTING CODE LETTER d. Running KW, Running KW= starting KW if PF = 1 e. Starting KW f. Running KVA, Running Power Factor g. Starting KVA, Starting Power Starting h. Motor full load amps from nameplate See Table 2 Onan book for motor data 3. Resistive Loads Resistive loads consist of incandescent lights, water heaters, electric heaters, stoves, and electric furnaces. Power factor = 1, KW = KVA. Running and starting KW are the same. 4. Motors and Inductive Loads Motors are inductive loads with KVA always larger than KW. The power factor running usually is between 0.6 and P.F. = KW KVA KW = (KVA) (PF) KVA = KW PF Determine locked rotor code letter from Table 2, A-R, for a given horsepower motor. If the motor is existing, use nameplate data. If this is a new installation, use shaded blocks in Table 2. Find starting KVA from Table 3. Find starting power factor from Table 2. Calculate KW starting = (KVA start) (PF start). To find starting KVA if only locked rotor amps are known; 1 phase KVA start = (locked rotor amps) (voltage) 3 phase KVA start = 3 ( locked rotor amps) ( voltage) For submersible water pumps only, use page 24-17,18 for motor data. 6/17/02 General Selection 2/20

3 To find running KW and KVA if only full load current is known; 1 phase KVA = ( I full load current) ( voltage) Kwrun = (KVArun ) ( power factor running ) See Table 2 for given HP, PF = KW KVA 5. Voltage Dip Determine maximum percentage voltage dip. This is basically what sizes the generator. If voltage dip is too large, the motor will over heat or not start because there is not enough torque to accelerate the motor. Motor torque falls off approximately as the product of reduced squared. For example, with a 30% voltage dip resultant torque, it will be: T= (.70 volts) 2 =.49 or 49% available torque. Voltage dip is usually expressed as a percent dip of nominal voltage as measured by light beam oscillograph or pen recorder on the first cycle. This is the voltage dip due to the generator winding resistance before the regulator tries to make the voltage recover. This will last anywhere from 1 cycle to 30 cycles depending on the response of the generator. This voltage dip is usually called the initial or transient voltage dip. After the regulator makes the generator recover, there is a recovery voltage dip from 5 cycles until the motor reaches 70%-90% of synchronous speed. The sustained voltage dip must be kept low enough to insure that relays and motor starter do not drop out. 6. Voltage Dropout Problems on Motor Control Systems Recovery dip should be limited to 5% if there are control relays and motor starters unless special precautions are taken in the control system power supply. If the voltage dips low enough to drop out the starters, the motor starters will chatter, burning out the motor. There are three ways to prevent system oscillation. a. Size generator for 5% initial voltage dip ( RMS). This will result in a larger generator size. b. On small systems where the generator is the only power source, utilize D.C. voltage from the generator battery as the control power. c. If the generator is a standby unit, utilize a constant voltage transformer in the control system power supply, such as a Sola CVS. This will keep the voltage to the control relays to within 95% of nominal voltage with up to 30% dip on the primary. 6/17/02 General Selection 3/20

4 7. Recovery Voltage Dip This voltage dip occurs from 5-30 cycles after load is applied. This dip should be no more than 5%. This dip is not given in the manufacturers standard literature. Each application must be checked with the generator manufacturer. 8. Total Voltage Dip The total voltage dip is the sum of the generator voltage dip and the voltage drop to the load on the conductors. This total should not exceed 10% on starting. 9. Resistive Voltage Dip For resistive loads, the starting and running voltage dip in the conductors is the same. The formula for voltage drop is: 1 phase Voltage Drop Vd= 2(I) (Z) (L) I = Current Amps Z = Total impedance ohms per 1,000 feet of conductor L = One way length of conductors kilo feet P.F. = Power Factor = 1.0 (use this when looking up Z) Refer to Electrical Design/Operation Maintenance Manual R (pages 8-1, 8-6 for voltage drop tables). Size feeders for resistive loads at 2% voltage drop. 3 phase Voltage Drop Vd = ( 3 ) ( I) ( Z) ( L) P.F. = 1.0 % Vd = Vd (100) 2 volts (100) Base volts Example: 240 volts =.8% Use same tables for 1 phase, or 3 phase voltage drop. Reactive Dip Example Loads Voltage: 120/240V, 1 phase Resistive load: 2 KW, 240V; 1 KW, 120V; 1.5 KW, 120 volt. Total resistive load: 6/17/02 General Selection 4/20

5 I = 2 KW.23KV = 8.7 A on L1, and L2 I = I = 1 KW.115 KV = 8.7 A on L1 1.5 KW.115 KV = on L2 Total Resistance Load: L1 L2 2 KW 8.7A 8.7A 1 KW 8.7A KW Use the 21 amps for sizing the generator. A voltage drop calculation normally does not need to be done if the generator is close to the distribution panel where the loads receive their power, say 25 feet or less. Conductors for loads must be sized for their ampacity, use NEC first, then check for voltage drop. For the 2KW load, 2 #12 Cu wire would be used with a 2 pole 15 amp breaker. For the 1 and 1.5 KW loads 2 #12 Cu with a 1 pole 20 amp breaker would be used. Refer to O&M Manual, Section 3 for sizing conductors. These are the wire sizes for ampacity. Now check voltage drop. NEC ampacity tables do not take into account voltage drop. Conductor Voltage Drop for Resistive Loads and Wire Sizes I = 8.7A Length = 250 feet #12 CU in conduit PF = 1 Vd = 2 I Z L from page #12, Z = 1.62 / Kft. Vd= (2) (8.7A) (1.62 /Kft.) (.25Kft.) = 7.04 volts %Vd = Vd 100 = (7.04V) (100) = 1.92% Base volts 230V So #10 is o.k. for voltage drop. 6/17/02 General Selection 5/20

6 Motor Voltage Drop In Feeder Conductors Motor 2HP, 230V, 1 Phase, code letter J, capacitor start, conductor length = 300 ft. Full load current from NEC table , I = 12 amp. Onan Book, Table 2, 2 HP, Code J Starting KVA = 16.4 KVA Run KVA = 2.6 KVA Start KW = ( KVA start) ( PF start) = (16.4) (.9) = KW Run KW = 1.8 KW Istart = 16.4 KVA = amp.23kv Irun= 12 amp Inrush = amp start 12 amp run = 5.9 times running current So if we size the wire for 1% voltage drop on running, we will have about a 6% voltage drop on starting. If this 6% voltage drop is added to the generator recovery voltage drop of 5%, the total overall voltage drop will be about 11%. NEMA says motors need 90% of voltage to start. Size Conductors Vd = 2 I Z L Z = Vd 2 I L L =.3Kft. Vd = 1% = (.01) (230V) = 2.3V Z = 2.3 =.319/ Kft. 2(12 amp) (.3 P.F. running, page 8-4 O&M, CU wire #4, Z =.2365/Kft. Voltage Drop for #4 CU Vd = 2 I Z L = (2) (12) (.2365) (.3 Kft.) = 1.70V % Vd = (1.70V) (100) =.74% 230V 6/17/02 General Selection 6/20

8 5. Look at the largest numbers in columns 10, 11, 12 and 13. Generator output must be greater than these numbers. WARNING THIS DOES NOT TAKE INTO ACCOUNT VOLTAGE DROP ON GENERATOR. CONSULT MANUFACTURER AT THIS POINT. Look at Example pages Max. KVA Max. KW Cont. KVA Cont. KW Onan generator 20ES This applies to Onan only, other generators by other manufacturers have different characteristics. In this case, a 20 KW generator would handle the continuous KVA and KW, and maximum KVA and KW. Refer to pages 3-6, Onan T-009 Generator Selection Guide. Sizing for Submersible Water Pump Motors Submersible water pump motors have a higher than normal locked rotor code letter and running current than Onan tables shown for average motors. The running current is higher than NEC when run at their service factor amps. Use the following attached tables: 6/17/02 General Selection 8/20

9 6/17/02 General Selection 9/20

10 6/17/02 General Selection 10/20

11 Voltage Drop Equations Voltage Drop Equations Voltage Drop = 3 I ( R Cos θ + X Sin θ ) L 3 Voltage Drop = 21 (R Cos θ + Sin θ ) L 1 Voltage Drop = in volts (V) I = Current in Amperes (A) R = Conductor Resistance in ohms/1000 ft. X = Conductor inductive reactance in ohms/ 1000 ft. L = One way length of circuit (source to load) in thousands of feet (K ft.) Z = Complex impedance ohms/ 1000 ft. Obtain from Tables. θ = Phase angle of load. Cos θ = Power Factor: Motors see 6-5,6-6,.6-.8 is usual see 5-1 t0 5-8 for power factor calculations, also 8-2 Given Voltage Drop, Find wire size: Voltage Drop 3 = 3 I ( Z ) L Z = Voltage Drop 3 I L = Vd 3 IL Voltage Drop 1 = 2 I ( Z ) L Z = Voltage Drop 2 I L = Vd 2 IL Procedure (Example) 1. Assume a voltage drop, say 2%, Base voltage 230V, 1 Vd =.02(230V) = 4.8 volts drop. Direct burial Copper. 2. Current and Distance must be known I = 30A, L =.56K Ft. =.56K Ft. Power Factor must be known, PF = Solve for Z: Z = Vd/2IL = 4.8V/2 (30A) (.5K Ft.) =.16Ω/ K Ft. 6/17/02 General Selection 11/20

12 4. Look up Z in tables at.16ω/ K Ft, 85 P.F. Copper Direct burial in nonmagnetic conduit. #1 CU, Z =.16Ω/ K Ft smaller impedance so use #1 Always use wire with the next smaller impedance Z per 1,000 feet than that calculated. Example: Three phase, Direct Burial Copper 1. Vd = 3 ( I) (Z) (L) Three Phase Voltage Drop 2. Given: Voltage 230 V, 3 phase, load 5 KW P.F. = 1 heater, L = 480 ft., find wire size. 3. Assume a voltage drop, say 2%, base voltage, 230 volts. Voltage drop maximum = (.02) (230 volts) = 4.6 Volts. I = 5 KW = 21.7A.23 3 Solve for Z: Z = Voltage Drop = 4.6 Volts =.0254 ohms/k FT. 3 IL 3 (21.7A) (.48 K FT) Look Z up in Table on voltage drop P.F. = 1.0 Copper Direct Burial Table 75 degrees C. Use next smaller Z for wire size. Nonmagnetic conduit. 500 MCM =.0270 Ω/ KFT, 600 MCM =.023Ω/ KFT Use 600 MCM because its impedance is less than that calculated. Power Factor PF = Cos θ = KW KVA Given 10 KW, 12 KVA Load, find PF: PF = 10KW = KVA 6/17/02 General Selection 12/20

13 6/17/02 General Selection 13/20

14 6/17/02 General Selection 14/20

15 6/17/02 General Selection 15/20

16 6/17/02 General Selection 16/20

17 Submersible Motors Engineering Manual 4-inch Three Wire Submersible Water Well Motors 60 Hertz Representative Loading and Performance Data 6/17/02 General Selection 17/20

18 Submersible Motors Engineering Manual 6-Inch Submersible Water Well Motors 60 Hertz Representative Loading and Performance Data 6/17/02 General Selection 18/20

19 For full load currents of 208-volt and 220-volt motors, increase the corresponding 230- volt motor full-load current by 10 and 15 percent, respectively. * These values of full-load current are for motors running at speeds usual for belted motors and motors with normal torque characteristics. Motors built for especially low speeds or high torques may require more running current, and multispeed motors will have full-load current varying with speed, in which case the nameplate current rating shall be used. For 90 and 80 percent power factor the above figures shall be multiplied by 1.1 and 1.25 respectively. The voltages listed are rated motor voltages. The currents listed shall be permitted for system voltage ranges of 110 to 120, 220 to 240, 440 to 480, and 550 to 600 volts. 6/17/02 General Selection 19/20

20 ARTICLE 430 MOTOR CIRCUITS, CONTROLLERS Table Full load Currents in Amperes Single-Phase Alternating- Current Motors The following values of full-load currents are for motors running at usual speeds and motors with normal torque characteristics. Motors built for especially low speeds or high torques may have higher full-load currents and multispeed motors will have full-load current varying with speed, in which case the nameplate current ratings shall be used. To obtain full load currents of 208-volt and 200-volt motors, increase corresponding 230-volt motor full-load currents by 10 and 15 percent, respectively. The voltages listed are rated motor voltages. The currents listed shall be permitted for system voltage ranges of 110 to 120 and 220 to /17/02 General Selection 20/20

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