Motor Calculation Steps:
 Branch Circuit Conductor Size_Part II_430.22
 Motor Overloads_Part III_430.32
 Motor Branch Circuit ShortCircuit & GroundFault Protection_Part IV_430.52
 Motor Feeder Conductors_Part II_430.24
 Motor Feeder ShortCircuit & GroundFault Protection_Part V_430.62
MikeHolt_Section04_Motors, Transformers, & HVAC
Unit 04: Motors & Transformers
NEC Article 430
Motors
 Motor windings are connected
 parallel for low voltage operation
 series for high voltage operation
 At a motors shaft, 1 HP of mechanical energy = 746W of electrical energy
 HP is based on output watts only, input isn’t considered
 1 HP = 746 W
 HP = ^{Output W} / _{746}
 FullLoad Ampere Rating (FLA)
 Equals Nameplate Amps
 Starting FLA is minimum 68 times FLA rating
 Locked Rotor Current (LRC) is 6 times FLA rating
 FLA formulas:
 Single Ø
 FLA = ^{746W x HP} / _{E x Eff x PF}
 Three Ø
 FLA = ^{746W x HP} / _{E x Eff x PF X 1.732}
 Motor Running Current
 I = E / Z
 Z = √R_{2} + X_{L}
 Single Ø


 W = watts
 HP = horsepower
 E = voltage
 Eff = efficiency factor
 PF = power factor
 Z =

 Overload Device Trip Sizing
 125% FLA
 SF 1.15 & up
 Temp rise of 40°C and below
 115% FLA
 All other motors
 125% FLA
 Motor Types_Direct Current
 Reverse rotation by switching lead polarity on either the armature or field, and leaving the other (armature or field) unchanged
 ShuntWound
 Constant speed under load
 Armature and field winding are in parallel
 Design has a builtin system for regulating its own speed
 Disk drives, recording equipment
 SeriesWound
 Good torque characteristics
 Armature and field winding are in series
 Poor speed regulation
 slows down when load applied
 unloaded run at high RPM, possibly to point of motor ‘runningaway’
 Starter motor
 Motor Types_Alternating Current
 Reverse rotation
 3 Ø
 Swapping 2 of the 3 power leads
 1 Ø
 Change the start winding polarity in relation to the run winding
 Check the Nameplate for details
 3 Ø
 SquirrelCage Induction
 Most common type
 Stator produces rotating magnetic field
 Rotor is a closed loop series of coils (windings)
 Stator magnetic field has faster RPM then Rotor RPM
 No physical connection between rotor and stator
 rotor is bars connected at ends by shorting rings
 all major industrial applications
 Synchronous
 Rotor locked in step and synchronous with stator rotating magnetic field
 Maintain speed with high accuracy
 Able to operate unloaded
 acts like capacitor
 used for power factor correction
 clock motors, large industry driving loads, compressors, crushers, large pumps
 WoundRotor
 Used in special applications due to complexity
 only operate on 3 Ø AC power
 Resistors used to limit starting inrush current
 Applications requiring speed control
 Mostly replaced by VFDs and induction motors
 Universal
 Fractional HP motors
 Operate well on AC or DC
 has disadvantage to DC motors, communications
 parts wear out
 vacuum cleaners, drills, mixers, light household appliances
 Reverse rotation

Motor Efficiency
 η = 0.7457 x hp x Load / P_{i}
 Load = P_{i} x η / hp x 0.7457
 η = Efficiency as operated in %
 P_{or} = Nameplate rated horsepower
 Load = Output power as a % of rated power
 P_{i }= 3Ø power in kW
 hp = Nameplate rated horsepower
 Troubleshooting
 Current increases
 running above HP rating
 supply voltage is below nameplate rating
 Current increases
Transformers

 Transformer is a machine used to transfer electrical energy from one system to another by induction with no physical connection between the 2 systems (except for autotransformers)
 Used for isolation and to raise or lower voltage
 Efficiency typically is 9597%
 Transformer Losses
 conductor resistance
 Directly proportional to conductor length
 Inversely proportional to conductor crosssectional area
 P = I^{2} * R used often for problems
 flux leakage
 Electromagnetic flux line leakage not absorbed by secondary winding
 eddy currents
 Expanding/collapsing ACinduced electromagnetic field causes circulating eddy currents and heat in iron core
 Separating iron cores with (lacquer) insulation reduces eddy current losses
 hysteresis losses
 energy required for iron core molecules to realign with changing polarity due to iron magnetism created by AC current expanding/collapsing electromagnetic field
 Directly proportional to AC frequncy
 higher the Hz, the more times per second the molecules must realign, the greater the hysteresis losses
 conductor resistance
 Transformers are rated in kVA (kilovoltamperes)
 Primary current increases in direct proportion to the secondary current (good txfr current explanation pg 160)
 Autotransformer
 use a single winding for the primary and secondary windings
 inexpensive
 lack of electrical isolation
 used to move voltage up/down, i.e. between 240v and 208v
 Transformer Formulas
 Efficiency
 Eff = ^{Output W} / _{Input W}
 Input W = ^{Output W} / _{Eff}
 Output W = Eff * Input W
 Turns Ratio = Primary : Secondary
 Current
 1 Ø
 I = ^{VA}/_{E}
 3 Ø
 I = ^{VA}/_{(E * 1.732)}
 1 Ø
 Efficiency
 Note: See pg 157Fig 421 for an interesting note on secondary voltages to neutral (480/240 Txfr has 240V across 2 hots and 208V from 1 hot to neutral)Unit 06: Conductor Sizing and Protection Calculations
NEC Article 240, 310, 400, 402
 Conductor heat is directly proportional to wire resistance and the square of current flow
 Conductors 8 AWG & larger must be stranded when installed in a raceway
 310.106(C)

Conductor sizing determining factors:
 current flow & resulting heat (ampacity adjustment)
 number of conductors in same raceway (bundle amp adjustment)
 ambient temperature
 terminal temperature ratings
 continuous load factor
 overcurrent protection
 conductor insulation temperature rating

Table 310.104(A)
 Conductor insulation properties
 letter type
 operational temperature
 application
 insulation
 outer cover properties
 In general, only conductors in this table can be used unless permitted elsewhere in the NEC
 Conductor insulation properties

Conductor Insulation Letter Ratings

Letter Designation Description No H 60° H 75° HH 90°, Dry 2 90°, Wet N Nylon Outer Cover T Thermoplastic U Underground W Wet or Damp X Crosslinked Polyethylene R (rubber – replaced by:) Thermoset  ?? Why is H & W identified in THWN2 cable? Isn’t this redundant??


Article 400
 Tables 400.4 / 400.5(A) / 400.5(B)
 Flexible cords and cables
 Article 402
 Tables 402.3 / 402.5
 Fixture wires

Conductor sizes
 110.14(C) (7045
 Temperature Limitations
 AWG
 18 to 4/0 (4ought)
 larger than 4/0 is expressed in kcmil
 250 kcmil & up
 kcmil
 kcmil = 1000’s (K) of (C)ircular (M)ils
 Mil means 1/1000 inch
 Circular Mil means the area of a circle 1/1000 inch in diameter
 20 AWG = 1 kcmil
 Table 310.15(B)(16)(A & B)
 **Conductor Insulation Temperature Ratings
 Equipment terminations
 Conductor sizing
 Equipment Ratings
 ≤ 100 amps
 60°C
 110.14(C)(1)(A)(1)
 Must be sized to table 310.15(B)(16)
 75°C
 110.14(C)(1)(A)(3)
 Must be sized to table 310.15(B)(16)
 Motor terminals
 must be design letters B, C, D (on nameplate)
 110.14(C)(1)(A)(4)
 Can be used within 75° column of table 310.15(B)(16)
 60°C
 ≥ 100 amps
 75°C
 110.14(C)(1)(B)(1)
 Larger than 1 AWG
 Must be sized to table 310.15(B)(16)
 75°C
 ≤ 100 amps
 110.14(C) (7045

Branch Circuits Conductor sizing
 Minimum 14 AWG
 Smaller than 14 AWG permitted for:
 class 1 remote controls
 725.43
 Fixture wire
 402.6
 Motor Control Circuits
 Table 430.72(B)
 class 1 remote controls

Overcurrent
 Article 240
 Transformers secondary circuit can use the primary circuit for overcurrent protection
 Conductors must be protected at the point receiving the supply
 310.15(B)(16)
 Except for 240.4(A) to (G)
 310.15
 Conductor amperage correction and adjustments
 Must be ontheline side of terminals
 110.9
 Must have short circuit protection
 110.10 note
 Causes by
 Overload
 short circuit
 ground fault

Overcurrent ≤ 800 amps
 240.4(B)
 Next higher rating permitted providing:
 240.6(A)
 not supplying more than 1 receptacle cordandplug
 conductor sizing requires
 ambient temperature correction
 310.15(B)(2)(A)
 conductor bundling adjustment
 310.15(B)(3)(A)
 last two points don’t correspond with fuse nor circuit breaker ratings
 ambient temperature correction
 not over 800 amps

Overcurrent≥ 800 amps
 240.4(C)
 must have a rating not less than the overcurrent device

Overcurrent Small Conductors
 240.4(D)
 Exceptions
 240.4(E)
 240.4(G)
 Must not exceed:

Wire Size, AWG Amperage 18Cu 7 16Cu 10 14Cu 15 12Al 15 12Cu 20 10Al 25 10Cu 30

 Exception 240.4(G)
 Must comply with table 240.4(G)
 Motors, Article 430
 Overcurrent protection, Article 430
 A/C equipment, Article 440
 Overcurrent protection, Article 440.22

Conductor Ampacity

Conductor Ampacity Calculation
 Table 310.15(B)(16)
 Table 310.15(B)(2)(a)_ambient temp above 86°F/30°C
 Table 310.15(B)(3)(a)_4+ hot conductors

Table 310.15(B)(16)
 allowable ampacity of insulated conductors
 Provided:
 No more than 3 hot/ungrounded/currentcarrying conductors
 if 4 or more, correct using Table 310.15(B)(3)(a)
 ambient temperature of 86°F/30°C
 if not correct using Table 310.15(B)(2)(a)
 No more than 3 hot/ungrounded/currentcarrying conductors
 **When using this table, if the ambient temperature where installed is outside 78°86°C, the ampacity must be adjusted using table 310.15(B)(2)(a)
 **When using this table, if there are more than 3 bundled currentcarrying conductors, the ampacity must be adjusted using table 310.15(B)(3)(a)

Table 310.15(B)(2)(a)
 Ampacity Correction
 Ambient temperature correction formula
 Corrected Amps = Table 310.15(B)(16) amps * Ambient Temp. Correction Factor table 310.15(B)(2)(a)

Table 310.15(B)(3)(a)
 Conductor Ampacity Adjustment after table 310.15(B)(16)
 Used when more than 3 bundled currentcarrying conductors installed in any crosssectional wireway
 Conductor Bundling Ampacity adjustment formula
 Adjusted Amps = Table 310.15(B)(16) amps * Bundled amp Adjustment Factor table 310.15(B)(3)(a)
 Reduce conductor ampacity when:
 4 or more hot conductors
 raceway or cable is longer than 24″
 adjust according to table 310.15(B)(3)(a)
 Conductor Ampacity Adjustment after table 310.15(B)(16)

Wet locations
 300.5(B)
 300.9
 310.10(C)
 300.6(D)

Rooftops
 310.15(B)(3)(6), pg 70148
 When exposed to sunlight and less than 7/8″ above a roof, add 60°F/33°C to table 310.15(B)(2)(a)
 Exception: Cable XHHW2 not subject to this rule

Cable Types AC and MC
 Special amperage, adjustment rules do not apply
 No outer jacket
 No more than 3 currentcarrying conductors
 Conductors are 12 AWG copper
 No more than 20 conductors installed without spacing, with length ≥ 24″

Conductor Ampacity Notes:
 Neutral conductor
 Only considered a currentcarrying conductor according to 310.15(B)(5)
 Equipment conductors
 Never considered currentcarrying conductors according to 310.15(B)(6)
 Corrections based on conductors insulation temperature rating
 determined by Table 310.15(B)(16)
 not by terminal rating, 110.14(C)

Correct for multiple conditions when:
 more than 3 hot conductors, and
 ambient temperature is not between 78°86°F
 Combine adjustments of ambient temperature and conductor buldling
 Adjusted & Corrected Ampacity = Table 310.15(B)(16) amps * Temperature factor * Bundled adjustment factor

Lower Ampacity Rule
 When more then 1 ampacity applies, use the lower amperage rating

Higher ampacity Exception
 310.15(A)(2)(Ex)
 If different amperage’s apply to different areas of one circuit, the higher amperage rating can be used if the lower amperage cable area :
 length is not over 10′ or
 ≥ 10% of total circuit length

Neutral, Bonding, and Grounding Conductors
 refer to 310.15(B)(5) and (6)
 Does not apply to conductors in cable trays
 refer to 392.80

Neutral Conductor CurrentCarrying Determination

Circuit Type Neutral Carrying Current Notes 2wire Yes 310.15()()() 3wire, 1 Ø, 120/240V, Wyeconnect No 310.15(B)(5)(a) 4wire, 3 Ø, 120/208V or 277/480V, Wyeconnect No 310.15(B)(5)(a) 3wirefrom4wire, 3 Ø, Wyeconnect Yes 310.15(B)(5)(b) 4wire, 3 Ø, 120/208V or 277/480V, Wyeconnect
(if more than 50% of neutral is nonlinear load)
Yes 310.15(B)(5)(c)  NonLinear load is when the current wave form does not follow the voltage wave form
 i.e.: Triplen Harmonic Currents
 Unbalanced 3wire Wye Secondary Neutral Current Formula
 (I_{Neutral} = √(I_{Line1}^{2} + I_{Line2}^{2}) – (I_{Line1} * I_{Line2})

 Neutral conductor


Miscellaneous Overcurrent Protection and Conductor Sizing

Branch Circuits
 Overcurrent protection
 210.20(a)
 Must have a rating of not less than 125% of continuous loads, plus 100% of noncontinuous loads
 Conductor sizing
 Must have an ampacity of not less than the maximum load served
 Must be the larger of these 2 options:
 conductors must be sized with a rating of not less than 125% of continuous loads, plus 100% of noncontinuous loads, based on terminals
 Temperature rating ampacity of table 310.15(B)(16)
 Minimum size allowable amperage not less than maximum load to be served
 Overcurrent protection

Appliance OverCurrent Protection
 422.11(E)
 Not to exceed marked rating
 If rating is unmarked
 Rated ≤ 13.30 amps
 not to exceed 20 amps
 Rated ≥ 13.30 amps
 422.11(E)(3)
 Not to exceed 150% of rated current
 Note: Next higher rating permitted when amperage falls inbetween (2406(A)pg7095)
 Rated ≤ 13.30 amps

Water Heater Branch Circuit OverCurrent Protection
 422.13
 ≤ 120 gallon is a continuous load
 not less than 125% continuous loads, refer to 210.19(A)(1)
 Branch circuit conductor
 422.10
 125% of amperage of continuous load
 Overcurrent protection device
 sized to 422.11(E)(3)
 Electric Space Heating Branch Circuit
 424.3(B)
 considered a continuous load
 Branch circuit and overcurrent protection
 210.19(A)(1)
 210.20(A)
 Must be sized not smaller that 125% of the total load (heater and motor)
 REVIEW MHPG 269QUESTIONSFIG. 6643 STEPS


Feeders

Feeder OverCurrent Protection
 215.3
 Must have a rating of not less than 125% of continuous loads, plus 100% of noncontinuous loads

Feeder Conductor Sizing
 215.2(A)(1)
 Must be sized to larger of:
 215.2(A)(1)(A) (has exceptions)
 minimum feeder conductor ampacity is not less than 125% of continuous loads, plus 100% of noncontinuous loads
 215.2(A)(1)(A)(Ex2)
 Feeder conductors with a junction box at both ends into 90°C terminals can have an ampacity of not less than 100% of continuous loads and 100% of noncontinuous loads based on the 90°C column of table 310.15(B)(16)
 110.14(C)(2), pg 7045
 ADD PHONE PICTURE
 or
 215.2(A)(1)(B)
 determined after ampacity adjustment & correction & not less than the load to be served
 215.2(A)(1)(A) (has exceptions)

Feeder Circuits
 75°C
 125% of continuous loads for circuits ≥ 100 amps
 215.2(A)(1)(A)
 Conductors sized to 75°C on table 315(B)(16) / 110.14(C)(1)(b)
 Rated ≤ 800 amps
 overcurrent protection to 240.4(B)
 Rated ≥ 800 amps
 overcurrent protection to 240.4(C)
 Continue from pg 12
 ADDED FROM PHONE
 Neutral sized to 100% of continuous load to 75°C column of table 310.15(B)(16), 110.14(C)(1)(B), based on maximum unbalanced load according to 220.61 and not smaller then required for equipment grounding conductors by 220.122
 215.2(A)(1)(Ex3)
 125% of continuous loads for circuits ≥ 100 amps
 90°C
 Sized to 100% of continuous load according to 90C column on table 310.15(B)(16), 110.14(C)(2)
 215.2(A)(1)(a)(Ex1)
 Rated ≤ 800 amps
 overcurrent protection to 240.4(B)
 Rated ≥ 800 amps
 overcurrent protection to 240.4(C)
 Neutral conductor
 Sized to 100% of continuous load to 75C column on table 310.15(B)(16) based on maximum unbalanced load according to 220.61 and not smaller than 250.122 for equipment grounding conductors
 215.2(A)(1)(Ex3)
 110.14(C)(1)(b)
 Sized to 100% of continuous load to 75C column on table 310.15(B)(16) based on maximum unbalanced load according to 220.61 and not smaller than 250.122 for equipment grounding conductors
 Exception 3
 Neutral conductors must have an ampacity of not less than 100% of continuous and noncontinuous loads
 Feeder Overcurrent Protection
 215.3
 Device sizing requirements for continuous and noncontinuous loads
 215.3
 Feeder Taps
 Tap is a conductor (not a service conductor) that has overcurrent protection rated higher than normally allowed in 240.2
 Overcurrent protection must be placed where branch circuits or feeder conductors receive power
 Except as permitted by (a)to(h)
 10′ feeder tap conductors
 No overcurrent protection needed provided:
 Ampacity is greater than the:
 Calculated load according to article 220
 rating of overcurrent protection device equipment and termination containing overcurrent protection supplied by tap conductors
 must not extend beyond the equipment they supply
 are installed within a raceway leaving an enclosure
 must have an ampacity of not less than a 10% rating of the overcurrent protection device protecting the feeder
 Ampacity is greater than the:
 No overcurrent protection needed provided:
 NOTE: 408.36 = overcurrent protection for panelboards
 25′ feeder tap conductors
 No overcurrent protection needed provided:
 Ampacity is not less than 1/3rd the rating of the overcurrent protection device protecting the feeder
 terminate in an overcurrent protection device rated not more than the tap conductor ampacity from 310.5, table 310.15(B)(16)
 No overcurrent protection needed provided:
 Feeder Tap Conductors with unlimited length outside
 No overcurrent protection needed for unlimited length provided:
 protected from physical damage within a raceway
 Terminates at a single overcurrent protection device limiting load to the ampacity of the outside feeder tap conductors
 overcurrent protection device for outside feeder tap conductors is part of building feeder disconnect
 Disconnecting means is readily accessible
 No overcurrent protection needed for unlimited length provided:
 Sized to 100% of continuous load according to 90C column on table 310.15(B)(16), 110.14(C)(2)
 75°C


Unit 6 Conclusion
 NEC tables are based on 86°F/30°C
 Conductor amperages must be corrected for other temperatures
 NEC requires ampacity adjustments when bundling more than 3 currentcarrying conductors
 Section 110.14(C) is a confusing part of code
 Do not size conductors to a higher rating then the terminals
 With requirement of a rating of not less that 125% of continuous loads, plus 100% of noncontinuous loads
 Noncontinuous load is not added when not mentioned
 NEC tables are based on 86°F/30°C
Unit 07: Motor and Air Conditioning Calculations
NEC Article 430, 440, Tables 310.15(B)(16), 430.247 to 430.250, 430.22(E)
 A motor generally draws 6 times more current at startup then when running
 Consider motors continuous duty unless noted noncontinuous duty
 Use table 430.22(E) for noncontinuous duty motors
 FLC is to Full Load Code as FLA is to Full Load Actual
 FLC from NEC tables 430.247250
 FLA from motors nameplate
Motor Protection_2 parts

Short Circuits and Ground Faults
 Protection is Fuses and Circuit Breakers
 Full Load Current (FLC)
 Conductor amperage, branch circuit short & ground faults overcurrent protection
 430.6(A)(1)
 Tables:
 430.247
 DCV
 430.248
 1 Ø
 430.249
 2 Ø
 430.250
 3 Ø
 430.247
 430.22
 Conductor Ampacity
 Single motor, cintinuous duty
 Ampacity no less than 125% of the motors FLC listed in table 430.247430.250
 430.52 & 430.62
 Short Circuits and Ground Faults Overcurrent Protection Device sizing
 430.110
 Disconnect switches Ampacity rating

Overload Protection
 Closetomotorrunningcurrent but with sufficient time delay to start
 Usually ‘heaters’, overload sensing devices in magnetic starters
 Full Load Amps (FLA)
 430.6(A)(2)
 Use motor nameplate data to size overload protection
 Motor FLA = ^{746 x HP} / _{E x Eff x PF}
 Motor FLA = ^{746 x HP} / _{E x 1.732 x Eff x PF}
 ?? When to use 1 formula over the other??
 746 = watts per hour standard
 HP / E / Eff / PF all read from nameplate
 1.732 = square root of 3

Load Voltage Current ↑ & – ↑ – ↓ & ↑
Motor Conductor Sizing Procedures
Motor Conductor Sizing Procedures 


Motor Type  Motor Conductor Sizing Rule  NEC Tables  Final Conductor Size 
1 Ø, Continuous Duty  not less than 125% of the motors FLC  430.247 to 430.250  310.15(B)(16) 
3 Ø, Continuous Duty  not less than 125% of the motors FLC  430.247 to 430.250  310.15(B)(16) 
NonContinuous DutyCycle  not less than 85% of the motors FLA  430.22(E)  310.15(B)(16) 
Motor Feeder Conductors Supplying Multiple Motors  not less than 125% of the largest motors FLC plus the sum of the additional motor FLCs fed from the same feeder  430.24(1)  310.15(B)(16) 
Motor Feeder Conductors Supplying Multiple Motors
 ?? DISCREPANCY: code says “sum of FLC ratings of all other motors in the group” but the MH examples only add in the 2nd highest motor??
Motor Overload Protection
Motor Overload Protection Sizing 


Overload Rating 
Motor Characteristic  Notes 
125%  SF 1.15 & ↑  Normal overload protection sizing 
40°C Rise & ↓  
140%  SF 1.15 & ↑  Used when insufficient for motor starting or to carry the load 
40°C Rise & ↓  
115%  OL rating when motor is not rated to SF 1.15 & ↑ nor a Temperature Rise of 40° & ↓  
130%  OL rating when motor is not rated to SF 1.15 & ↑ nor a Temperature Rise of 40° & ↓ and when insufficient for motor starting or to carry the load  
Branch shortcircuit & ╧ fault protection sufficient  NonContinuous Duty Cycle  shorttime, intermittent, periodic, or varying duty motors 
Note: OL values can not be exceeded, round down. 430.32(C) 
Motor ShortCircuit & GroundFault Protection
 430.52(B) & (C) compliance
 Shortcircuit & ╧fault protection requires:
 fast current rise
 short duration
 fast response time

Motor BranchCircuit Shortcircuit & ╧fault Protection Table 430.52
Motor Type
NonTime Delay
FastActing
DualElement Fuse
TimeDelay
Inverse Time Breaker
Instantaneous Trip
Direct Current 150% 150% 150% Wound Rotor 150% 150% 150% All Other Motors 300% 175% 250%
 Shortcircuit & ╧fault protectionNext Size Up Rule
 When protecting device values derived from table 430.52 don’t correspond with the ratings listed in 240.6(A), the next higher OC device rating can be used
 430.52(C)(1)(Ex1)
 Does not apply to feeder shortcircuit & ╧fault protection
 DO THESE EXAMPLES, MHpg 317319
Motor Feeder Shortcircuit & ╧fault Protection
 430.62
 Sized not more than the largest rating of the branchcircuit, plus the sum of the FLC of the other motors in the group
 BranchCircuit Next Size Up Rule does not apply, so round down
 DO THESE EXAMPLES, MHpg 321322
Motor VA Calculations
 Motor output
 1 HP = 746 watts
 1 Ø Motor Input VA Calculation
 Input VA = motor volts * motor FLC amps
 1 Ø VA = E*FLC
 P=IE
 1 Ø VA = E*FLC
 Input VA = motor volts * motor FLC amps
 3 Ø Motor Input VA Calculations
 3 Ø Input VA = Motor Volt Rating * Motor ampere Rating * 1.732
 1.732 = √3
 3 Ø VA = E*FLC*1.732
 3 Ø Input VA = Motor Volt Rating * Motor ampere Rating * 1.732
Variable Speed Drive Calculations
Variable Speed Drive Conductor Sizing
 430.122(A)
 not less than 125% of the rated input current
Variable Speed Drive BranchCircuit Protection
 430.130(A)(2)
 must be in accordance with the manufacturers instructions
Variable Speed Drive OverCurrent Protection
 must be in accordance with the manufacturers instructions
Fire Pump Calculations
Fire Pump Conductor Size
 695.6(B)(2)
 not less than 125% of the motor FLC as listed in tables 430.248 or 430.250
 must comply with voltagedrop requirements of 695.7
Fire Pump ShortCircuit and ╧ fault Protection
 695.4(B)(2)(a)
 must be sized to carry the motor lockedrotor current indefinitely
Fire Pump OverCurrent Protection
 695.4(B)(2)(a)(1)
 must be sized to carry the motor lockedrotor current indefinitely
 Nextsize Up Rule in use if the lockedrotor current does not correspond to a standard overcurrent device, round up
 240.6(A)
Air Conditioning Calculations
Mulitmotor Equipment ShortCircuit and ╧ fault Protection
 440.4(B)
 Must have a nameplate identifying branchcircuit shortcircuit and ╧fault protective device
 RLA here, not FLA ??
Mulitmotor Equipment Conductor Size
 440.4(B)
 Must have a nameplate identifying minimum conductor ampacity
 RLA here, not FLA ??
MotorCompressor & Other Motors ShortCircuit and ╧ fault Protection
 440.22(B)(1)
 Must not be greater than 175% of the largest motorcompressor nameplate load current rating, plus the current of other motors
 If unable to start at 175%, 225% is the maximum
MotorCompressor & Other Motors Conductor Size
 440.33
 sized to 125% of the largest motorcompressor nameplate load current rating, plus the sum of the other ratedload current of other motors