Planning The Electrical System  Safe means for the entrance of electrical supply into a building  Means of disconnecting all electrical power in a building  Means of limiting to a safe level the maximum amount of electrical current that can enter a building  Common point for grounding the electrical equipment  Safe means of subdividing the electrical supply to serve individual loads or group of loads May 2007

Types of Service Entrance Panels  Circuit Breaker Type  Fuse Type

Circuit Breaker Type Service Entrance Panel  When breaker trips, it may be reset after the problem has been corrected  Circuit breakers are easy to switch off  Circuit breakers are available in sizes to protect any circuit  Circuit breakers will take a short periods of overload without tripping  Circuit breakers cannot be tampered with  A larger size circuit breaker cannot be installed accidentally

Fuse Type Service Entrance Panel  Cost less to purchase  Less convenient to use  When fuses “blows”, It must be replaced unless equipped with a special reset- type fuses

Determining the size Service Entrance Switch  Size of dwelling  Appliances  Large Appliances (Ranges, Dryer, Water Heater, ect.)  2 Small Appliance loads Circuits  Laundry Load Circuit  General Lighting load Circuit

One Family Dwelling  2000 square feet  Appliances  Range – 240 Volts – 12 kilowatts  Dryer – 240 Volts – 5.5 kW  Water Heater –240 Volts – 5 kW  2 Small Appliance Load Circuits – 120V  Laundry Load Circuits – 120V  General Lighting Load Circuit – 120V

Computed Load for Dwelling  General Lighting load:  2000 sq. ft. X 3 volt-Amperes = 6000VA  6000VA divided by 120 Volts = 50 Amps

Laundry Circuits  At least one laundry circuit is required by NEC  Must be 20 amp  The circuit may not serve any lighting outlets or outlets in any other room in the house  3 circuits X 1,500 Watts = 4500 Watts  4,500 Watts divide by 120 Volts = 37.5 A

Major Appliances  Refrigerator350 Watts  Freezer350 Watts  Dryer 7,000 Watts  Dish Washer700 Watts  Range/Oven 1,150 Watts  Water Heater 4,500 Watts Total 24,050 Watts Well Pump 2,000 Watts

Heating and Cooling  100% of the nameplate rating of Central Heating and Air Unit  65% of the nameplate rating of central electric space heating, including supplemental heating elements  65% of the name plate rating of electrical space heating if less than four separately controlled units  40% of the nameplate ratings of electric space heating of four or more separately controlled units 9,600 Watt Central AC/Heat Pump

Loads except Heating and Cooling  Small Appliance Load 4,500 Watts  General Purpose Load 6,000 Watts  Fixed Appliance Load 24,050 Watts  Other Items (Well Pump) 2,000 Watts Total Load 36,550 Watts

Calculate Total Load  Heating/cooling 9,000 Watts  First 10 KW of other 10,000 Watts  Remainder of other 10,620 Watts 36,500 – 10,000 = 26,550 Watts 26,550 X 40% = 10,620 Watts Total Load 30,220 Watts

Total Amperage Requirement Total Load divided by 240 volts 30,220 Watts divided by 240 Volts = Amps

SERVICE ENTRANCE SWITCH SIZE Farm Shop

Computing Farm Shop Load  1200 sq. ft. Farm Shop  Volt, 50 amp Welder50 A  1 – 230 Volt, 5 hp Air Compressor28 A  1 – 230 Volt, 2 hp Bench Grinder12 A  1 – 230 Volt, 3 hp Radial Arm Saw17 A  1 – 230 Volt, 3 hp Tilting Arbor Saw17 A  1 – 115 Volt, ¾ hp Drill Press(13.8)  1 – 115 Volt, 1/3 hp Exhaust Fan (7.2)  6 – Additional 120 Volt Outlet (6 X 1.5 A = 9 A)  3 – Lighting Circuits with 15 outlets = 22.5 A

Converting 115 Volt items to 240 Volts  Drill Press13.8 Amps  Exhaust Fan 7.2 Amps  Additional Outlets 9.0 Amps  Lighting Outlet22.5 Amps Total Amperage52.5 Amps

To Convert 115 Volts to 240 Volts  52.5 Amps X120 Volts = 6300 VA  6,300 VA divided by 240 Volts or ½ of 52.5 amps= A

Total Service Load  Welder50 Amp  Air Compressor28 Amp  Bench Grinder12 Amp  Radial Arm Saw17 amp  Tilting Arbor Saw17 amp  Converted 115 Volt items26.25 Amp Total Load Amp

Load without diversity  Loads most likely to operate at one time to produce the heaviest loads are:  Welder50 Amps  Exhaust Fan 3.60 Amps  Half of Lighting Amps Total64.85 Amps Largest Motor is 5 hp; 28 amps X 125% = 35 Amps

Computing Total Demand  Not less than the first 60 amps of all loads  100% of largest demand load Amps  50% of the next 60 amps 30 Amps  25% of remainder of other load Amps – Amps= % of 25.4 = 6.36 Amps Total Demand Amps