Mike Dolt Enterprises
Mike Dolt Enterprises
User: FER3, Location: Tennessee
Asked on: Friday, April 15, 2011 at 4:01 pm EDT
Answered on: Tuesday, May 14, 2024 at 4:14 am EDT
Question: I have a new 200hp 460v motor, premieum efficiency, 226 fla , 1800 rpm, 1.15 SFPF= 86.7, Code F, (4.67). Distance from breaker to starter 30 feet, distance to motor from starter 5 feet, I know the OL for starter, VD using 350mcm is 0.4 or 0.1%. My question is: will 350mcm 75 degree cable be suitable for this application, taking into consideration the inrush-current? Breaker is 600amp frame with 300 amp trip.
Step 1: Calculate Motor Starting Current
Starting current (FLA) = 226 1.15 (4.67) = 1163.71 AStep 2: Determine Voltage Drop
VD = 0.4%Step 3: Calculate Allowable Voltage Drop
Allowable voltage drop = 4.67 0.4% = 0.0186 VStep 4: Calculate Circuit Impedance
Circuit impedance = Allowable voltage drop / Starting currentCircuit impedance = 0.0186 V / 1163.71 A = 0.000016 ΩStep 5: Calculate Cable Impedance
Cable impedance = (Distance from breaker to starter + Distance from starter to motor) Resistance per 1000 feetCable impedance = (30 ft + 5 ft) * 0.0000074 Ω/ft = 0.0002486 ΩStep 6: Determine if Cable Impedance is Acceptable
Cable impedance should be less than circuit impedance for acceptable voltage drop.0.0002486 Ω < 0.000016 ΩConclusion
The 350mcm 75-degree cable has an impedance of 0.0002486 Ω, which is greater than the allowable circuit impedance of 0.000016 Ω.
Therefore, 350mcm 75-degree cable is not suitable for this application, considering the inrush current. A larger cable size will be required. Flag for review