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.
You can check the distance you can go with this wire gauge calculator EDIT: Sorry. This calculator is for 24 volt equipment only! Regards
Typical household wiring is AWG number 12 or 14. See if any of these of useful - www.faa.gov/certification/aircraft/av-info/dst/43-13/Ch_11-05.doc (open directly or save target as), which looks like it gives gauge 8 for 115 V, 240 feet, 15A - but check the Word document. Wire gauge calculator - http://www.powerstream.com/Wire_Size.htm Electrical conductors http://www.tpub.com/content/neets/14176/index.htm http://www.tpub.com/content/neets/14176/css/14176_20.htm - Wire gauge table
Use #8 copper or # 6 aluminum. If the appliance has a 15-amp connector (plug) you must use a 15-amp circuit breaker and 15/20-amp receptacle. Im not sure a 15-amp breaker or receptacle exists that will allow connection to these large wire sizes so youâ€™ll need to pigtail with short smaller gauge wire with approved mechanical devices. I'm not sure if doing that is allowed by the electrical code. If you use aluminum, all mechanical connection points must be approved for aluminum and must be coated with anti-oxidation paste. A # 8 copper cable may have to be special ordered in some areas and will be more expensive. Some insurance companies will not reimburse for damage if the electrical code is not followed. ...
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. 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.
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.
Hai Iam rias iam doing my project on electric trolley bus which should be taking power supply from over head grid supply and running the vehicle through drive motors. In this case iam having various Ac & Dc voltages. I need how to calculate the proper wire size in sqmm. please reply me as soon as possible with proper formulas. I have calculated all the terms as volt,amps,watt,res, ohms etc Regards A.Riaz ahamed