Branch circuit cables are rated for 75?C and feeder cables are rated for90°C. This will be a 3-phase, 575-volt system with four induction motors, specified asfollows:Motor 1: 60 hp 0.90 p.f. squirrel cage motorMotor 2: 60 hp 0.90 p.f. squirrel cage motorMotor 3: 40 hp 0.85 p.f. wound rotor motorMotor 4: 7-1/2 hp 0.80 p.f. wound rotor motorA) Assuming no line losses, find the capacity of the transformer required to supplypower to this system.B) Assume that it is desired to improve the overall power factor for this system to 0.95lagging. Determine, in kVAR, the required capacitance for this power factorcorrection.C) Assume these motors have their windings connected in a delta configuration. Whatwould be the line voltage if they were connected in wye?
I would look in the NEC for the full load amperes of these motor sizes. These are: 7.5 HP9 amps 40 HP.41 amps 60 HP.62 amps The 7.5 HP motor KVA will be KVA 575 * 9 * 1.732 8.9631 The 7.5 HP motor KW will be KW 575 * 9 * 1.732 * 0.8 7.17048 The KVAR of this motor is Sqrt(8.9631^2 - 7.1705^2) 5.3778 The 40 HP motor KVA will be KVA 575 * 41 * 1.732 40.8319 The 40 HP motor KW will be KW 575 *41 * 1.732 * 0.8 5 34.707115 The KVAR for this motor is Sqrt(40.8319^2 - 34.7071^2) 21.5095 The 60 HP motor KVA will be KVA 575 * 62 * 1.732 61.7458 The 60 HP motor KW will be KW 575 *62 * 1.732 * 0.8 5 55.5712 The KVAR for this motor is Sqrt(61.7458^2 - 55.5712^2) 26.9141 The total KVA requirement for all motors running at once is 8.9631 + 40.8319 + 61.7458 + 61.7458, which is 173.2866 KVA (note this is the requirement from a transformer, not the size of the transformer The total KW of all the motors is 7.1701 + 34.7071 + 55.5712 + 55.5712 153.0196 KW The total KVAR of all the motors is 5.3778 + 21,5095 + 26,9141 + 26.9141 80.7145 KVAR The power factor for all the motor is KW / KVA 153.0196 / 173.2866 0.883 I took a short cut at this point and used an application i wrote to calculate the required capacitor KVAR. The results follows: At 0.95 power factor, the motors' KVA will be 161.065 The motors' KW will be the same. The motors' KVAR will be 50.2926 The required capacitor's KVAR will be 53.988 The reactance of the capacitor will be 10.65 ohms The capacitance will be 3.49059248 E -4 Farads You can calculate these values as I did above, if you want to. EDIT Forgot the last answer. The line voltage is the same for both delta and wye. The leg voltage for the wye motor will be 575 / 1.732 331.98 volts TexMav
