Diversity Factor

codeissues code issues By James Stallcup Sr., NEC/OSHA Consultant

APPLYING DEMAND FACTOR AND DIVERSITY FACTOR PER THE NEC  Don’t be confused by these terms! There are two terms used in calculating loads in electrical systems that cause designers to get confused. These terms are “demand factor” and “diversity factor.” To better understand how these terms are applied when calculating loads, you must understand their meaning. Demand factor is the ratio of the maximum demand of  a system, or part of a system, to the total connected load on the system, or part of the system under consideration. Demand factor is always less than one. Diversity factor is the ratio of the sum of the individual maximum demands of the various subdivisions of a system, or part of a system, to the maximum demand of the whole system, or part of the system, under consideration. Diversity factor is usually more than one. For example, these terms, when used in an electrical design, desi gn, should be ap plie plied d a s follows: follows: The sum of the connected loads supp li lied ed by a fee feederder-ci cirrcuit can be multiplied by the demand factor to determine the load used to size the components of the system. The sum of the maximum demand loads for two or more feeders is divided by the diversity factor for the feeders to derive the maximum demand load. Given: Con sider four individual feeder-c feeder-circuits ircuits with connected loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and demand dema nd factors factors of 90%, 80% , 75% and 85% res respec pec-tively. Use a diversity factor of 1.5. Solution: Calculating deman d for • 250 kV kVA A x 90% = • 200 kV kVA A x 80% = • 150 kV kVA A x 75% = • 400 kV kVA A x 85% =

feeder-circui feedercircuits ts 225 kV kVA A 160 kV kVA A 112.5 kV kVA A 340 kV kVA A 837.5 kVA The sum of the individual demands is equal to 837.5 kVA 20

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If the main feeder-circuit were sized at unity diversity: kVA kV A = 837 .5 kVA kVA ÷ 1.00 = 837 .5 kVA kVA The main feeder-circuit would have to be supplied by an 850 kVA transformer. However, using the diversity factor of 1.5, the kVA = 837.5 kV kVA A ÷ 1.5 = 558 kV kVA A for the main feeder feeder.. For diversity factor of 1.5, a 600 kVA transformer could be used.  N ot otee th at a 600 k VA transfo rm rmer er can be b e used instead of  an 850 kVA when applying the 1.5 diversity factor. DEMAND FACTOR Although feeder-circuit conductors should have an ampacity sufficient to carry the load, the ampacity of  the feeder-circuit need not always be equal to the total of all loads on all branch-circuits connected to it. A study of the following sections will show that, in some cases, a “demand factor” may be applied to the total load. Remember, the demand factor permits a feeder-c fee der-circuit ircuit ampa city to be less less than 1 00% of the sum of all bra nch-ci nch-circuit rcuit load s connected to the feeder. feeder.

APPLYING DEMAND FACTOR FOR GENERAL LIGHTING 220.3(A); TABLE 220.3(A) Section 220.3(A) of the  N EC ® governs the ru les for for calculating calcul ating t he lighting load on servi services ces and feedercircuits. circ uits. Th e difference difference between calculating bran chcircuit loads and feeder-circuit loads is that a demand factor is not usually applied for a branch-circuit, but may be applied in t he case of a fee feeder-c der-circuit. ircuit. The load on a service or feeder is the sum of all of the branch loads subject to their demand factors as permitted by the rules of this Section. “Demand factor” is a percentage by which the total connected load on a service or feeder is multiplied to determine the greatest probable load that the feeder will be called upon to carry. In hospitals, hotels, apartment complexes, and dwelling units, it is not likely that all of  the lights and rece receptacles ptacles connected to eve every ry br anch-circ anch-circuit uit served by a service or feeder would be “on” at the same time. Therefore, instead of siz sizing ing the feeder to carry all of  the load on all of the branches, a percentage can be

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codeissues code issues applied to this total load, and the components sized accordingly. EC,, it can be seen Referring to Table 220.11 of the  N EC that a demand factor for lighting may be applied only for dwelling units, hospitals, hotels, motels, and warehouses. All other occupancies are calculated on a basis of total computed lighting wattage, and no demand factor is permitted. (See Figure 1)

General-purpose receptacle outlets used to cord-and-plug connect loads are considered to have noncontinuous operation and are calculated per 220.3(B)(9) and Table 220.13 . Noncontinuously operated receptacles with a VA rating of  10,000 VA or less shall be computed at 100 percent. If the VA rating of the receptacle load exceeds 10,000 VA, a demand factor of 50 percent should be applied to all VA exceeding 10,000 VA per Table 220.13. 220.13 . (See Figure 2)

GENERAL LIGHTING LOADS COLUMN 1; DEMAND LOAD 1 GENERAL LIGHTING LOAD GENERAL 3 VA PER SQ. FT. • TABLE 220.3(A) DEMAND FACTORS • TABLE 220.1 220.111

GENERAL PURPOSE RECEPTACLE  AND LIGHTING OUTLETS

SMALL APPLIANCE LOAD 1500 VA PER CIRCUIT • 220.16(A); (B) DEMAND FACTORS • TABLE 220.11

GEC GES

LAUNDRY SMALL APPLIANCE RECEPTACLE OUTLETS RECEPTACLE OUTLET

What is the demand load for the general lighti ng load of a 3500 sq. ft. dwelling unit? Step 1:

Step St ep 2: 2: Step 3:

Calculating the VA Table 220.3(A) 3500 sq. ft. x 3 VA

= 10,500 VA

220.16 220. 16(A) (A);(B ;(B)) 1500 VA x 3

= 4,500 VA

Total loads General lighting load Small appliance load Total

= 10,500 VA = 4,500 VA = 15,000 VA

Step 4:  Applying demand factors Table 220.1 220.111 Firs Fi rstt 30 3000 00 VA x 10 100% 0% = 3, 3,00 0000 VA Next 12,000 VA x 35% = 4,200 VA Total = 7,200 VA Solutio So lution: n:

Deman De mandd load load 1 is 720 72000 volt volt amps. amps.

Figure 1. The above illustration shows the calculation of a demand factor load f or a dwelling un it usin g Table 220.1 220.111

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LOAD #2 RECEPTACLES • 150

MBJ

GEC GES RECEPTACLE LOADS • 220.3(B)(9) • TABLE 220.13

CALCULATING RECEPTACLE RECEPTACLE LOAD AND APPLYING DEMAND FACTORS Step 1: Calculating VA 220.3(B)(9);; 230.42(A)(1 220.3(B)(9) 230.42(A)(1)) 150 x 180 VA = 27,000 VA Step 2:  Applying demand factors Table 220.13 First 10,000 VA x 100% = 10,000 VA Next 17,000 VA VA x 50% = 8,500 VA VA 18,500 VA Solution: The dema demand nd load load for the the receptacles is 18,500 VA.

Figure 2. The demand load for general-purpose receptacles for other than dwelling units is computed by using Table 220.13.

Figure 2

APPLYING D EMAND FACTOR FOR APPLYING COMM ERCIA ERCIAL L COOKING EQUIPMENT EQUIPMENT 220.20; TABLE 220.20 Section 220.20 in the  N EC  permits Table 220.20 to be used for load comp utation for commercial elec electrical trical cooking equipment, such as dishwashers, booster heaters, water heaters and other kitchen equipment. The demand factors shown in Table 220.20 are applicable to all equipment that is thermostatically controlled or is only intermittently used as part of the kitchen equipment. In no way do the demand factors apply to the electric heating, ventilating or air-conditioning equipment. In computing the demand, the demand load should not be less than the sum of the two largest kitchen equipment loads. (See Figure 3)

APPLYING DEMAND FACTOR FOR RECEPTACLES 220.13; TABLE 220.13

APPLYING DEM AND FACTOR FOR THE N EUTRAL APPLYING 220.22

Sec ection tion 220 .13 o f the  N EC  makes it clear that in dwelling units, general-purpose receptacles are not counted as a load. In other than dwelling units, a minimum of 180 VA is computed for each general-purpose receptacle. For hospitals, hotels, motels, and warehouses, this receptacle load can be lumped with the lighting lighting load, and the demand factors of Table Table 220.11 may be applied to th e total.

Section 220.22 of the  N EC  states that for a service or feeder, the maximum unbalanced load controls the ampacity selecte selected d for th e ground ed (neutral) condu ctor ctor.. The grou nded (neutra l) conductor service service or fe feeder eder load should be considered wherever wherever a grounded (neutra l) conductor is used in conjunction with one or more ungroun ded (phase) conductors. O n a si single ngle-phase -phase feeder feeder necdigest.org

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Figure 4. The The above calcul calculation ation show s the procedur e for computing the neutral for a service or feeder and applying demand factors per 220.22 220.22..

Figure 4

APPLYING DEM AND FACTOR FOR CONNECTIN G APPLYING ADDITIONAL LOADS TO EXISTING INSTALLATIONS 220.35 When additional loads are connected to existing facilities having feeders feeders an d service as originally compu compu ted, the maximum kVA computations in determining the load on the existing feeders and service should be used if the following conditions are met: • If the maximum data for the demand in kVA, such as demand meter ratings, is available for a minimum of one year. • If 125 percent of the demand ratings for the period of one year added to the new load does not exceed the rating of the service. Where demand meters are used, in most cases the load as computed will probably be less than the demand meter indications. (See Figure 5)

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codeissues code issues SERVICE CONDUCTORS • 400 KCMIL KCMIL THWN THWN cu. OCPD • 230.90(A) • 230.42(A)(1); (A)(2)

OCPD • 215.3 • 215.2(A) 15.1 kVA LOAD TO BE ADDED FEEDER-CIRCUIT CONDUCTORS • 215.2(A)(1)

MBJ GEC GES

EXISTING SERVICE PANEL • 78.4 kVA LOAD (EXISTING DEMAND) • 208 V, 3Ø • USE 360 V

ACTUAL AMPS RECORDED FOR A PERIOD OF 30 DAYS • 500 KCMIL • 220.35, Ex.

SUBFEEDER DIRECTORY 1 . M ot o r1

2.

3. 5. 7. 9. 11.

4. 6. 8. 10. 12.

SUBFEED DISTRIBUTION PANELBOARD

RECORDED AMPS WITH AMMETER OR USE POWER METER • 218 A • LARGEST PHASE READING

NO. 2000 KCMIL IN PARALLEL

BRANCHCIRCUITS DIRECTORY

1.Motor1 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

MOTOR BRANCHCIRCUIT

 ADDED PANELBOARD LOAD • 102 A

NOTE: ADDED LO NOTE: LOAD AD FORA SE SERVI RVICE CE CANBE COM COMPUT PUTED, ED, USING USI NG THESAME PRO PROCED CEDUREPER  UREPER 220.35, Ex .

CALCULATING LOAD IN AMPS Step 1: Finding demand 220.35 Maximum demand = 78.4 kVA

Step 1: Finding recorded demand 220.35 Maximum demand = 218 A

Step 2: Calculating existing demand 78.4 kVA x 125% = 98 kVA

Step 2: Calculating existing demand 218 A x 125% = 273 A

Step 3: Calculating total kVA 230.42(A)(1) 98 kVA + 15.1 kVA = 113.1 kVA

Step 3: Calculating existing and added load 273 A + 102 A = 375 A

Step 4: Calculating amperage Table 310.16 No. 400 KCMIL THWN copper = 335 A 113.1 kVA x 1000 = 113,100 VA 113,100 VA ÷ (208 x 1.732) = 314 A 314 A is less than 335 A

Step 4: Finding amperage for  feeder conductors Table 310.16 500 KCMIL THWN copper = 380 A

Solution: The 15. 15.11 kVA kVA demand demand load can be applied to the existing service without upgrading the elements.

Figure 5. The above illustration is the calculation for adding a load to an existing service or feedercircuit using 220.35. F igure 5

APPLYING DEM AND FACTOR TO APPLYING THE EXCEPTION TO 220.3 220.35 5 220.35, Ex.  .     C    N    I  ,    Y     O    B    Y    A    R     G    F     O    Y     S    E    T    R    U     O     C     S    N     O    I    T    A    R    T     S    U    L    L    I

CALCULATING LOAD IN AMPS

If the maximum demand data for a on e year year period is not available, the calculated load is permitted to be based on the maximum demand (measure of average power demand over a 15-minute period) continuously recorded over a minimum 30 day period using a recording ammeter or power meter connected to the highest loaded ungrounded (phase) of the feeder or service, based on the initial loading at the start of  the recording. (See Figure 6)

Step 5: Determining if load can be added 375 A is less than 380 A Solution: The 375 375 amp amp load can be applied to the existing feeder-circuit conductors.

Figure 6. The above illustration shows the optional calculation being applied for adding a load to an existing feeder-circuit using 220.35.

Fig ure 6

APPLYING DEM AND FACTOR APPLYING FACTOR FOR M OTORS 430.26 There are, in some cases, motor installations where there may be a special situation in which a number of motors are con nected to a feeder-ci feeder-circuit. rcuit. Because Because of th e par ticular application, application, certain motors do not operate together and the feeder-circuit conductors are permitted to be sized based on a historical demand factor. For example, the authority having jurisdiction may grant permission to allow a demand factor of less than 100 percent if operation procedures, production demands, or the nature of the work is such such that no t all the motors are running at one time. An engineering study or evaluation of  necdigest.org

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Equipment Load Factors, Use Factors and Diversity Factors As Well as a General Discussion of Energy Audit Procedures Barney L. Capehart

To do a good job on an energy audit, the energy auditor must understand the areas of equipment load factor, use factor and diversity factor. Definitions: First, let's define these terms.

a)

Load factor - the ratio of the load that a piece of equipment actually draws when it is in operation to the load it could draw (which we call full load).

For example, an oversized motor - 20 hp - drives a constant 15 hp load whenever it is on. The motor load factor is then 15/20 = 75%. b)

Use (or utilization) factor - the ratio of the time that th at a piece of equipment is in use to the t he total time that it could be in use.

For example, the motor above may only be used for eight hours a day, 50 weeks a year. The hours of operation would then be 2000 hours, and the motor use factor for a base of 8760 hours per year would be 2000/8760 = 22.83%. With a base of 2000 hours per year, the motor use factor would be 100%. The bottom line is that the use factor is applied to get the correct number of hours that the motor is in use. c)

Diversity factor - the probability that a particular piece of equipment will come on at the time of the facility's peak load.

The diversity factor is the most complicated of these factors. For example, we might have ten air conditioning units that are 20 tons each at a facility. In Florida we typically assume that the average full load equivalent operating hours for the units are 2000 hours per year. However, since the units are each thermostatically controlled, we do not know exactly when each unit turns on. If the ten units are substantially bigger than the facility's actual peak A/C load, then fewer than all ten units will likely come on at once. Thus, even though each unit runs a total of  2000 hours a year, they do not all come on at the same time to affect the facility's peak load. The diversity factor gives us a correction factor to use, which results in a lower total kW load for the ten A/C units. If the energy balance we do for this facility comes out within reason, but the demand balance shows far too many kW for the peak load, then we can use the diversity factor to bring the kW into line with the facility's true peak load. The diversity factor does not affect the kWh; it only affects the kW. Motor load, use and diversity factors: Sometimes motor load factors that are too low are chosen because the auditor has not properly determined the hours of use of the motors - i.e. the auditor uses an incorrect use factor. For example, just because a facility has a production shift that is 2000 hours per year it does not mean that all of the production-related production-rel ated motors in that facility are operated oper ated for 2000 hours per year. Some motors - or machines - might only be used one day a week rather than every day. Other motors might be used every day, but for only half the day, i.e. 4 hours per day. Other motors might be in use throughout the day so that their use really is 2000 hours per year .

The auditor must collect data on the use factor - or hours of use - for every motor in the facility during site visit. For each machine, line, process or operation, ask "How many hours a day does this machine (line, process or operation) operate?" This data then needs to be entered into the

energy balance. Motor load factors in many facilities are more in the range of 40% - 50%, than in the range of 80% that had been a standard assumption for many years of doing audits. Rarely do you find a motor running at 100% load factor. However, not all motors at a facility are running at the same load factors. Ventilating fans that come from a supplier as a packaged unit with a fan and a motor are most often assumed to be operating at near full load. You should probably use a load factor of 80% here, since the manufacturer of the ventilating fans should have reasonably matched these loads. Other motors may also be in this category - some engineering judgment and common sense are required to determine which other motors these are. Motors with variable loads are going to have the lowest load factors factor s in general. A dust collector fan motor will normally have quite a variable load, and would often be expected to have a low load factor. Other examples are saws, presses, milling machines, sanders and grinders, waste grinders, water pumps, hydraulic pumps, etc. If a group of motors do not all operate together all of the time, then using a diversity factor is appropriate. This is the case with a number of separate air conditioning units (considering the motors for the compressors) that are individually thermostatically thermostatical ly controlled. It could also be the case for a group of production motors if some of the motors are not in use all of the time. You should use a diversity factor in your motor calculations, since it is not often the case that a facility has all of the motors on at the same time. th e motor load Reconciling the energy balance: When you perform an energy balance, do not use the factor as the first and only adjustment made to reconcile the estimated energy use (energy balance) with the energy bills. Making this adjustment too quickly results in failure to pick up other things that have been overlooked. For example, if the energy use does not balance with the energy bills, the first step is to check to see that all of the equipment and uses have been accounted for. Do the items on the energy balance spreadsheet match your recollection of the equipment you saw in the facility? Does anything appear to be missing? Are the utility bills for total energy use and peak kW recorded correctly? The next step is to check the hours of use for lights and other equipment to see if it matches your knowledge of the facility's operation. Remember that each motor - as well as each other piece of  equipment - does not necessarily operate the same number of hours each day or year. Finally, if  some of the equipment does not come on at the same time as the facility peaks in kW use, then utilize the diversity factor to account for this. Adjusting the motor load factors should probably be the last thing you do to reconcile the energy and demand balances. Now, if all other information and all other factors are correct to the best of  your knowledge, then adjust the load factors. While motor load factors are not often in the range of  80-100%, you should be equally suspect of very low motor load factors. If you get motor load factors in the range of 20-30%, it is more likely that you have the hours of use wrong than that you have a facility which is using motors that are an average of four times too big for the job they are doing. Lumber mills and wood products facilities using lots of saws may have these low load factors. Most other places should have motors with a higher load factor.

Basic motor load measurements should be taken at the plant visit. The electrical person at the facility is generally willing to measure the current being drawn by a motor of interest. Air compressors are ones that are usually easy to do, and you should ask the plant personnel to do this for you. Let them open the motor controller or switch box and connect a clamp-on ammeter to see what the current for the motor is. You then need to know the full load current from the nameplate of the motor. The ratio of the actual current to the full load current is the approximate load factor on the motor at that time. This procedure works as long as the current is greater than or equal to about 50% of the full load current. Try to take this measurement for each of the large motors in the facility - i.e. motors of 50 hp and above; or even 20 hp or above if the facility does not have a lot of  big motors. If you have not received formal electrical safety training, you should not make these electrical measurements yourself. If the facility electrician does not want to make these for you, then let it go at that. Air handlers—use factor: Air handlers use motors and are subject to all of the comments co mments made in the motor section. In addition, you may be able to get a better handle on the hours of use for the air handlers by knowing how the A/C system works. Ask if the air handlers run constantly when the facility is occupied. They might if the facility wants the ventilation, even though the compressors might not come on except to periodically provide some temperature reduction or moisture removal. If this is the case, then the use factor for these air handler motors should reflect an hours-of -use that matches the offices or other area that the air handlers supply. In addition, the hours-of-use must also consider the compressor run hours. Thus the total hours for the air handlers must be at least the same as the compressor hours, and may be higher if the A/C unit is left on during periods that the facility is not occupied, or if ventilation is provided.

If the air handlers only come on when the thermostat thermosta t orders cooling, then the hours-of hours -of -use must be the same as the hours-of-use of the compressors. It is important to get adequate information on the operation of the air conditioning system. To get complete data on the air handler motors for an air-conditioned facility, you will need all of the standard information - size, maker, single or three phase, etc - together with the operating basis for the air handlers discussed above. You should also collect data on the drive belt system for air handlers. Record the number of belts, the lengths, and the types of belts. Ask about motor and drive lubrication and cleaning. Also check  the A/C filters to see if they are reasonably clean. Sometimes a visual inspection will show some real problems. Ask the maintenance person to open up one of the air handlers - or just look into it (SAFELY) if it is accessible - and see if the belt is tight, slack, or really loose. Do not stick your hand into an air handler that is off at the moment, and may come back on when the thermostat kicks in. Have the maintenance person turn the air handler motor off with the circuit breaker or control box. Do not put your finger on a moving drive belt. Is the belt in good shape? Is it frayed, cracked or coming apart? Does it look like the pulleys for the motor and the fan are lined up? Ask the electrician to measure the current that the air handler motor is drawing to see what its load factor is while driving the fan. It should be very near full load - but you never know. Maybe the original motor burned out and was replaced with a bigger one to "make sure it did not burn out again." Remember to take the full load current off the nameplate to find the load factor.