Dear Electrical Specialists,
For 3 phase induction motors, whether the motor was operating on Star-Delta or DOL starter, I was blissfully using the formula I = w/v/0.93/(3^0.5)/0.89 to derive the current and accordingly size the cables, where
I => Current in Amps
W => Power in Watts
V => Voltage in Volts.
However, I saw a table where the current drawn by the Star-Delta Startered Motor is shown very low.
The typical values indicated are as follows:
a. Motor output - 75kW ie 100 HP.
b. Full Load Current Particulars
1. Line = 131 Amps. (What is the formula for this?)
2. Phase = 75.6 Amps (What is the formula for this?)
c. Direct On-Line Starter
1. Overload Relay range = 90-135 Amps (what is the formula for this?).
2. Maximum backup fuse = 200 Amps (what is the formula for this?).
3. Recommended cable for Al. = 95 sq.mm (Based on which current?)
4. Recommended cable for Cu. = 70 sq.mm (Based on which current?)
d. Star-Delta Starter
1. Overload Relay range = 50-90 Amps (what is the formula for this?).
2. Maximum backup fuse = 160 Amps (what is the formula for this?).
3. Recommended cable for supply side
3.1. For Al. conductor = 95 sq.mm (Based on which current?)
3.2. For Cu conductor = 70 sq.mm (Based on which current?)
4. Recommended cable for Motor side
4.1. For Al. conductor = 50 sq.mm (Based on which current?)
4.2. For Cu conductor = 35 sq.mm (Based on which current?)
Will be greatful if the queries solicited above are clarified.