nikola-tesla said:
If specs on an appliance plate read ; 15 Amps, 115 Volts.
The distance from the power source to the receptacle (plug) where the appliance is going to be plugged is 230 feet, what is the wire (american wire gage) or conductor size, the thickness of the wire? Thank you.
USUALLY IF THE APPLIANCE IS RATED 115 VOLTS, ITS MINIMUM OPERATING VOLTAGE OR THE ACTUAL VOLTAGE MUST BE SUPPLIED BY THE OUTLET MUST NOT BE LESS THAN 110 VOLTS.
IN THIS CASE TO BE SAFE, WE MUST ALLOW A MAXIMUM VOLTAGE DROP OF 5 VOLTS IN THE CIRCUIT. TAKE A LOOK IN THE FORMULA BELOW...
VD= 2*K*I*L/ CSA
WHERE:
VD= VOLTAGE DROP
I= CURRENT
L= DISTANCE OF THE LOAD FROM THE OUTLET (IN FT)
K= 12 FOR COPPER AND 19 ALUMINUM (FOR MAXIMUM TEMP OF 75 DC)
CSA= CONDUCTOR CROSS SECTIONAL AREA IN CIRCULAR MILS.
***SUPPOSE WE USE COPPER CONDUCTOR AND WE ALLOW A MAXIMUM OF 5V MAX. VOLTAGE DROP:
5= 2*12*15*230/CSA
CSA= 82800/5
CSA= 16560 CM
FROM COMMERCIAL REFERRENCES:
15560 CM IS CLOSEST TO 16510 CM WHICH IS #8 COPPER OR 8 SQMM WIRE.
THEREFORE: USE #8 OR 8.0 sqmm WIRE.
NOTE:
1. IN THIS CASE A MAXIMUM OF 5 VOLTS WAS CONSIDERED IN THE CALCULATION.
2. THIS CALCULATION IS NOT VALID FOR LARGE MOTORS. BECAUSE MOTORS DRAWS A MAXIMUM OF 7OO% TIMES ITS FULL LOAD TORQUE DEPENDING ON THE MOTOR CONTROL STARTING METHODS USED BY THE DESIGNING ENGINEER.