Pre-certificati on V oltali!e Drop Calculations

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1 7 Pre-certificati on V oltali!e Drop Calculations Rule governs the maximum amount of voltage drop for an electrical installation (this includes service, feeder and branch circuit conductors). Subrule one states that the voltage drop shall be based on the calculated demand load of the installation, whether a feeder or a branch circuit. This part of the rule also states that the maximum voltage drop allowed for any installation is five percent from the consumer's service to the point of utilization. In other words the maximum voltage drop is five percent from the service entrance cap to the light or receptacle on a branch circuit. However, it is also stated that only three percent voltage drop is permitted on a feeder or branch circuit. For example, if a branch circuit that feeds receptacles has a voltage at the panel of 120 volts, then the minimum voltage allowed to be measured at the receptacle would be 120 volts less 3% or volts. Low voltage supplied to equipment causes a decrease in operating efficiency and output of electrical equipment. It is for this reason that the electrical code has established voltage drop standards. This makes voltage drop a very important consideration in any electrical installation. In order to keep the voltage drop in a circuit to the code standards, correct calculations on the size of wire feeding electrical equipment is necessary. There are at least two methods for calculating wire size for a circuit: Method one: This method uses imperial units in a formula for the calculation of voltage drop. In order to use this method, it is necessary to know the resistance constant (K) for the type of conductor used ( copper or aluminum), the load of the circuit in amps, the length of conductor, and the voltage drop you wish to achieve. This method is known as the "KILND" formula. It will provide the required size of wire in circular mils. K = resistance constant (l0.4 for copper at 20 c e) I = load of the circuit in amps L = length of conductor (total there and back) VD= % voltage drop desired Example: What size of copper conductor is required to service a load of 6000 watts, 200 ft. from the generator? The voltage at the load center is 120 volts and the desired voltage drop is 3%. Answer: Kl(L+L) VD 10.4 X 50 X ( ) 3 I =~ E I = I = 50 A AREA OF CONDUCTOR IN CIRCULAR MILS = circular mils At this point it is necessary to convert circular mils to millimeters squared. This is done by dividing the circular mils by = mm squared Turn to table D5 and look up mm squared for the conductor size. Without going lower. According to Table D5 under the nominal column a # 1 A WG copper conductor is required to keep the voltage drop to 3%.

2 8 Pre-certification Voltaee Drop Calculations The second and preferred method of calculating voltage drop is using Appendix D of the Canadian Electrical Code. Table D3 is used for Low voltage applications and Table D4 is used for Extra Low voltage applications. This method requires the load in amps, the voltage of,the circuit, the one way distance of the conductors in meters, and the desired/permitted percent voltage drop. The formula to use is listed in Table D3 of the code. Method Two Example: (using table D3 to find a wire size for a given load in amps and a given distance) What size of copper conductor is required to service a load of 6000 watts, 61 meters from the generator? The voltage at the load center is 120 volts and the desired voltage drop is 3%. First Calculate the load in amps I=E. E 1= = 50 amps Secon d Using the applicable rules, determine the conductors calculated ampacity, as well as the conductor's size, CSA insulation designation temperature and current rating. 50 X 125% = 62.5 amps From Table 2 # 6 R-90 Third Fourth Calculate the one way distance in meters of the conductor 61 meters Apply the distance correction factor from Table D3 -find the per cent load on the conductor by dividing the load in amps by the ampacity rating ofthe wire. % load = 50 amps % load =.77 X 100 = 77% 65 amps (#6 R-90 from table 2) From Table D3 correction factor for a 90 C 77 % load is 1.00 Determine the one way distance in meters to use in table D3 (1 % VD at 120 volts) Fifth circuit conductor lelll?:th in meters (one way) % VD X distance correction factor X (circuit voltage/ 120) 61 meters 3 X 1.00 (120/120) 61 3 =20.3 The one way distance in meters at 1 % voltage drop at 120 volts is 20.3 meters. Sixth Using the actual load in amps, use Table D3 to determine the figure closest to this value without going lower. Then thumb up to the top of that column and use that figure as the required wire size. The closest figure to 20.3 without going lower is 25, at the top of this column it gives the conductor size of# 1 A WG Please note the conductor size must never be smaller than that of the second step, due to rule 8-104

3 9 Pre-certification Voltal!e Drop Calculations Week # 5 Table D3 may also be used to find the maximum distance that a known size of conductor can carry a load. In this case, the table is used in a similar manner as before. The current and the voltage of the load must be known, the size and type of conductor must be known, and the desired voltage drop must also be known in order to perform this calculation. The distance correction factor is calculated the same as before. The example at the end of table D3 in the code book does not show this calculation. Example: (Using table D3 to find the max. distance that a conductor can carry a load) What is the maximum distance that a 2/0 T -90 conductor (installed in a raceway) can carry a load of 100 amps at 240 volts, keeping a minimum voltage drop of3%? First Determine the load in amps (if not given) 1= 100 amps Second Third Using the applicable rules, determine the conductors calculated ampacity, as well as the conductor's size, CSA insulation designation temperature and current rating. Don't forget that table D3 only determines voltage drop, other code rules will determine the absolute minimum size of conductor. Apply the distance correction factor from Table D3 -find the per cent load on the conductor by dividing the load in amps by the ampacity rating of the wire. % load = 100 amps (actual load on the conductor) 185 amps (rated load of2/0 T-90 from table 2) % load=54 % From the distance correction factor chart on table D3 a 90 conductor at 60% load = 1.04 (when using the DCF chart always round the % load UP to the nearest number) Fourth this Using the main table D3 find the closest number to the actual current in the left hand column, then using the correct conductor size column across the top. This will bring you to the one-way distance in meters that a 2/0 T -90 conductor can carry 100 amps at 1 % voltage drop at 120 volts, in this case 19.9 meters. But the question asked the max. distance at a 3% voltage drop and at 240 volts. This means that we must correct number by using the following formula: Table D3 value X actual voltage X distance correction X [actual volta gel (I % VD at 120 volts) drop factor meters X 3 X 1.04 X [ 240 ] meters The maximum distance that a 2/0 T-90 conductor can carry a 100 amp load and maintain a 3% voltage drop is meters.

4 0 Pre-certification Volta!!e Drop Calculations Week # 5 Voltage drop calculations are also required on extra-low voltage circuits as well. A common example of an extra-low voltage circuit is remote emergency lighting heads fed from a central battery unit. These systems may operate on 6,12,18, and 24 volts DC. When a circuit operates on a very low voltage, then voltage drop becomes even more important to consider. This should explain why in some cases #8, #10 and #12 A WG wires are used to supply these remote lighting heads, even-though they only draw a few amps. To properly calculate the size of wire needed for an extra-low voltage circuit, Table D4 needs to be used. The information given on Table D4 is used in a very similar format to that of Table D3. The main difference is that Table D4 does not have a distance correction factor to consider, but instead uses a current correction factor. Like table D3, table D4 can be used on two ways. One, to find the maximum distance of a circuit in meters for a specified voltage drop, when given the circuit load, the conductor size used and the circuit voltage. Two, to find the minimum conductor size needed to keep the voltage drop to a specified percentage, when given the circuit load, the circuit distance in meters and the circuit voltage. Example of finding the minimum size of conductor using Table D4: What size of AC-90 cable is required to supply a 12 volt DC, remote lighting circuit, if the load on the circuit is 36 watts, 30 meters from the source, and the voltage drop is the maximum permitted by code? Step 1: Calculate the load in amps. I = PIE 1=36/12 1=3 amps Step 2: Find the one-way distance of the circuit in meters. Step 3: Use rule (1) to find the maximum % voltage drop. 30 meters 5% max. Step 4: Find the circuit voltage. 12volts DC Step 5: Find the closest current value listed in Table D4 to the actual circuit current. 2.5 amps Step 6: Apply the information to the following formula: One-way distance in meters % VD allowed X circuit volts X TD4 amps 5 6 actual amps The above formula will calculate the one-way distance in meters permitted for a 5% voltage drop for a 6 volt circuit using the information given in the example question. With this number, you can then use Table D4 to find the minimum size of wire needed. Simply find the distance calculated with the formula, beside the closest current value in Table D4. Once you have found that number (or the next higher number if not exact) follow that column to the top of the table and the minimum size of wire will be listed. 30 meters 5/5 X 12/6 X 2.5/3 = 30 meters 1 X 2 X meters meters@5%vd@6 volts Usinl! the 18.0 meters on Table D4 usin!! a current of 2.5 amps = #10 A WG wire

5 1 Pre-certification Voltaee Drop Calculations Week #5 Practice Questions 1. A 5 HP, 240 volt, single phase motor is located 30 meters away from the motor control centre. If a maximum of 2.5% voltage drop is to be maintained, what is the required size of RW 90 conductor? 2. An emergency lighting circuit supplies 54 watts of remote heads at 12 volts DC. The circuit length is measured to be 18 meters. What is the minimum size of T -90 conductor installed in EMT required to maintain the maximum voltage drop permitted by code? 3. A 5 kw, 240 volt fan forced heater is to be installed 60 meters away from the load centre. Using the maximum voltage drop allowed by the code, what is the minimum size oftw-75 conductor required? 4. A2 hp, 240 volt, single-phase, submersible pump is located 100 meters away from it's source of supply. Using the maximum voltage drop allowed by the code, what is the minimum size ifnmw-u conductors required? 5. What is the maximum distance that a #10/2 AC-90 cable can carry a 3 amp load of remote emergency lighting? The circuit voltage is 12 volts DC, and the maximum voltage drop permitted is 5%.

6 2 Pre-certification Volta2:e Drop Calculations Week # 5 Practice Questions: 1. How far in meters, would a 1/0 T -90 conductor carry a load of 105 amps at 240 volts, while still maintaining a 2% voltage drop? 2. There are two 240 volt baseboard heaters connected on a residential heating branch circuit. Each of the two heaters has an output of 1500 watts. What is the maximum distance in meters that a 12/2 NMD-90 can carry this heating load according to the code? 3. A 4.8 kw, 208 volt single-phase fan forced heater is to be installed on a commercial branch circuit. How far in meters would #6 RW-90 XLPE 600v unjacketed conductors carry this load and still maintain a 3% voltage drop? 4. A 15 amp, 120 volt duplex receptacle is fed from a separate branch circuit. If the load on this receptacle is unknown, what is the maximum distance in meters that a 10/2 NMD-90 can maintain a 3% voltage drop? 5. How far in meters could a 5 Hp, 230 volt single phase pump motor circuit be ran using a 6/2 NMW -U cable, if a maximum voltage drop of 2.5% was to be maintained?

7 3 Pre-certification Service/Feeder Calculations Section 8 of the code deals with the calculation of services and feeders for various types of buildings. The first rule we need to look at is rule This rule explains how we are to calculate the area ofliving space before we can determine our service or feeder size in accordance with rules (for single dwellings) and rule (for apartments and similar buildings). Rule explains that when heating and air conditioning are interlocked so they both cannot operate simultaneously, the load with the highest demand is taken into the calculation of the service or feeder. It is also important to remember that electric space heating loads need to have demand factors applied from section 62. It is also necessary to determine if an electric range has been considered. Note that paragraph (b) has minimum service sizes based on area. Single Dwellings Example: A home with a total living area 0006 square meters has the following electrical equipment installed: W electric baseboard heating W electric range 3000 W electric water heating 4000 W electric clothes dryer 3000 VA air conditioning unit 4000 W hot tub heater Apply Rule first 90 square meters of living area remaining 216 square meters/90=2.4(3 X 1000) electric heating (15000 X 75%) range (2000 X 40%) hot 100% clothes dryer (4000 X 25%) water heater (3000 X 25%) Total 5000 watts 3000 watts watts (section 62) 6800 watts 4000 watts 1000 watts 750 watts watts From rule we need to calculate current: Amps = watts Amps = volts Rule (1) tells us to use the greater of paragraph (a) or (b) in this case (a) 174 amps is greater than (b) 100 amps Therefore our service ampacity required for this home is 174 amps. Based on the 174 amps calculated, the min. equipment size is a 200 amp panel (standard size) Using rule and , the max. olc device is 200 amps. From table 17 our common ground conductor would be # 6 A WG From table 2 the proper 90 C conductor size would be 2/0 RW-90 (rated at 185 amps) From table 6 the proper size of conduit would be 53 mm (for 3-#2/0 RW -90 XLPE)

8 4 Pre-certification Service Calculations Week # 5 Row housine Calculations When calculating the minimum size of service for buildings such as row housing, (if the building has a main service that supplies two or more dwelling units), the calculation becomes a little more complex. In order to complete this type of calculation, you must use parts of two different code rules. The first part of the calculation is the individual basic loads of each dwelling unit. This is done using the same rule as for a single family dwelling. Rule (1) Once the individual unit loads have been calculated, then another rule must be used, Rule (3a) This rule states to take the individual unit load totals in watts and subtract any heating or air conditioning loads. After subtracting the heating or air conditioning loads from each individual unit, you are to take the unit with the largest load at 100 %, the next two largest unit loads at 65%, the next 2 largest unit loads at 40%, the next 15 largest unit loads at 25%, and any remaining unit loads at 10%. This calculation will represent the total load in watts that the units will draw from the main service conductors. (non-continuous load). To complete the calculation, Rule (3b) (3c) (3d) need to be followed. Any heating loads are to be taken as demands are listed in Rule Air conditioning are always taken at 100% unless Rule (4) applies (ifboth heating and air conditioning are installed so that only one may be energized at any given time, then only the largest load ofthe two is taken into the calculation). Rule (3d) states that any lighting, heating and power loads that are not located in dwelling units shall be added at a 75% demand factor (when considering the main service conductors). Example: A row housing complex has three units each with a calculated living area of265 square meters. Each of the three units also has the following equipment: 1-4kW electric clothes dryer 1-3kW electric water heater kw electric range w electric baseboard heater w electric baseboard heater w electric baseboard heater w electric baseboard heater There is 4000 watts of outside security lighting. If the main service of the row house complex is 120/240V single phase, 3 wire, calculate the minimum ampacity of the individual unit feeders, and the minimum ampacity of the main service conductors. Also, find the maximum rating/setting of the over-current devices for both the individual unit feeders and the main service.

9 5 Answer to the example row house calculation: Pre-certification Service Calculations Week # 5 Row housin2: Calculations Individual unit loads Rule (1) 265 m 2 /90 = 3 sections of90 m 2 1 sl 90 m 2 = 5000 watts 2 nd 90 m 2 = 1000 watts 3 rd 90 m 2 = 1000 watts 4kW 25% = 1000 watts 3kW water 25% = 750 watts 12.5kW electric range: 1 sl w = 6000 watts % = 200 watts Add all unit baseboard heaters: 6750 w (3a) 1 sl 100 % = 6750 watts 75% = 0 watts Total unit demand watts/240 volts Unit feeder = 90.4 amps Table 2 = #3 RW-90 XLPE (105 amps) Take total unit demand and subtract heating & Ale largest 100% = watts watts next two largest 65% = watts less heating watts unit demand on main service watts watts/unit (non-continuous load) UNIT Heating loads 3 units X 6750 watts = watts (total) Apply (3a) pi 100% = w 75% = w Unit Heating Demand on Main = watts (non-continuous load) Loads not located inside units 4000 watts of outside 75% = 3000 watts (continuous load) Apply 8-104(5a) 3000/80% = 3750 watts Total demand on main service conductors total unit demand = watts total unit heat demand = watts total other demand = 3750 watts Total Main Service Demand watts (non-continuous load) Divided by 240volts = amps min. Therefore the minimum calculated ampacitv of the main service is amps. Using table 2, the minimum size of main service conductors would be #4/0 RW-90 (235 amps) Using rule , the max. o/c device would be 250 amps, if standard 400 amp equipment was used.

10 6 Pre-certification Service Calculations Week #5 Practice Question 1. An 8 unit non-electrically heated row housing complex is to be constructed. This building will have 3000 watts of sidewalk lighting. The service is to be 120/240 volt, single phase, 3-wire. Each unit will have a living area of 306 square meters and has the following electrical equipment: 4000 watt clothes dryer 4000V A air conditioning unit 11.5 kw range Calculate the minimum ampacity of the service conductors as well as the minimum size of conduit, overcurrent device, and ground wire.

11 7 Pre-certification Service Calculations Week #5 Practice Question 2. A four unit row-housing complex is to be constructed. This building is to have 2500 watts of exterior and sidewalk lighting. A % horse power 230 volt single phase circulating pump for the boiler heating system. The service voltage is to be 120/240 volts single phase, 3-wire. Each unit will have a living area of 365 square meters and will have the following electrical equipment: 5500 watt clothes dryer 5000 VA air conditioning unit 14 kwelectric range 3 kw electric sauna heater Calculate the minimum ampacity of the service conductors as well as the minimum size of conduit, over-current device, and ground wire.

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