Diversity Factor

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I I L L L L U U S S T T R R A A T T I I O O N N S S C C O

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R R T T E E S S Y Y O O F F G G R R A A Y Y B B O

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, , I I N N C C . . By James Stallcup Sr., NEC/OSHA Consultant

By James Stallcup Sr., NEC/OSHA Consultant

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APPLYING DEMAND

FACTOR AND DIVERSITY

FACTOR PER THE

NEC

Don’t be confused by these terms! Don’t be confused by these terms!

There are two terms used in calculating loads in There are two terms used in calculating loads in electrical systems that cause designers to get confused. electrical systems that cause designers to get confused. These terms are “demand factor” and “diversity factor.” These terms are “demand factor” and “diversity factor.” To better understand how these terms are applied when To better understand how these terms are applied when calculating loads, you must understand their meaning. calculating loads, you must understand their meaning.

Demand factor

Demand factor is the ratio of the maximum demand of is the ratio of the maximum demand of a system, or part of a system, to the total connected load a system, or part of a system, to the total connected load on the system, or part of the system under consideration. on the system, or part of the system under consideration. Demand factor is always less than one.

Demand factor is always less than one. Diversity factor

Diversity factor is the ratio of the sum of the individualis the ratio of the sum of the individual maximum demands of the various subdivisions of a maximum demands of the various subdivisions of a sys-tem, or part of a syssys-tem, to the maximum demand of the tem, or part of a system, to the maximum demand of the whole system, or part of the system, under consideration. whole system, or part of the system, under consideration. Diversity factor is usually more than one.

Diversity factor is usually more than one.

For example, these terms, when used in an electrical For example, these terms, when used in an electrical desi

design, should gn, should be apbe applieplied ad as folls follows:ows: The sum

The sum of the of the connected loads suppconnected loads supplilied by a ed by a feefeeder-der-cicir- r-cuit can be multiplied by the demand factor to determine cuit can be multiplied by the demand factor to determine the load used to size the components of the system. The the load used to size the components of the system. The sum of the maximum demand loads for two or more sum of the maximum demand loads for two or more feed-ers is divided by the divfeed-ersity factor for the feedfeed-ers to ers is divided by the diversity factor for the feeders to derive the maximum demand load.

derive the maximum demand load. Given:

Given: Con

Consider four sider four individual feeindividual feeder-cder-circuits with ircuits with connectedconnected loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and loads of 250 kVA, 200 kVA, 150 kVA and 400 kVA and dema

demand facnd factors of 90%, 80%tors of 90%, 80%, 75% , 75% and 85% and 85% resrespecpec- -tively. Use a diversity factor of 1.5.

tively. Use a diversity factor of 1.5. Solution:

Solution:

Calculating deman

Calculating demand d for feeder-for feeder-circuicircuitsts • • 250 250 kVkVA A x x 90% 90% = = 225 225 kVkVAA • • 200 200 kVkVA A x x 80% 80% = = 160 160 kVkVAA • 150 • 150 kVkVA A x x 75% = 75% = 112.5 112.5 kVkVAA • • 400 400 kVkVA A x x 85% 85% = = 340 340 kVkVAA 837.5 kVA 837.5 kVA

The sum of the individual demands is equal to 837.5 kVA The sum of the individual demands is equal to 837.5 kVA

If the main feeder-circuit were sized at unity diversity: If the main feeder-circuit were sized at unity diversity:

kV

kVA = A = 837837.5 kV.5 kVA ÷A ÷1.00 = 8371.00 = 837.5 kV.5 kVAA

The main feeder-circuit would have to be supplied by an The main feeder-circuit would have to be supplied by an 850 kVA transformer.

850 kVA transformer.

However, using the diversity factor of 1.5, the kVA = However, using the diversity factor of 1.5, the kVA = 837.5

837.5 kVkVA ÷A ÷1.5 = 558 1.5 = 558 kVkVA for the main feeA for the main feederder..

For diversity factor of 1.5, a 600 kVA transformer could For diversity factor of 1.5, a 600 kVA transformer could be used.

be used. N

Notote the that a at a 600 600 kkVA VA transfotransformrmer can ber can be used e used instead instead of of an 850 kVA when applying the 1.5 diversity factor.

an 850 kVA when applying the 1.5 diversity factor. DEMAND FACTOR

DEMAND FACTOR

Although feeder-circuit conductors should have an Although feeder-circuit conductors should have an ampacity sufficient to carry the load, the ampacity of ampacity sufficient to carry the load, the ampacity of the feeder-circuit need not always be equal to the total the feeder-circuit need not always be equal to the total of all loads on all branch-circuits connected to it. of all loads on all branch-circuits connected to it. A study of the following sections will show that, in A study of the following sections will show that, in some cases, a “demand factor” may be applied to the some cases, a “demand factor” may be applied to the total load. Remember, the demand factor permits a total load. Remember, the demand factor permits a fee

feeder-cder-circuit ampaircuit ampacity to be lecity to be less than 1ss than 100% 00% of the sumof the sum of all bra

of all branch-cinch-circuit loadrcuit loads connected to s connected to the feederthe feeder..

APPLYING DEMAND FACTOR

FOR GENERAL LIGHTING

220.3(A); TABLE 220.3(A)

Section 220.3(A) of the

Section 220.3(A) of the N NEC EC ®®

governs the ru

governs the rules fles foror

calcul

calculating tating the lighting load he lighting load on on serviservices and feeder-ces and feeder-circ

circuits. Thuits. The diffee difference between calculating branrence between calculating branch- ch-circuit loads and feeder-ch-circuit loads is that a demand circuit loads and feeder-circuit loads is that a demand factor is not usually applied for a branch-circuit, but factor is not usually applied for a branch-circuit, but may be

may be applied in tapplied in the case of a he case of a feefeeder-cder-circuit.ircuit. The loadThe load on a service or feeder is the sum of all of the branch on a service or feeder is the sum of all of the branch loads subject to their demand factors as permitted by loads subject to their demand factors as permitted by the rules of this Section.

the rules of this Section.

“Demand factor” is a percentage by which the total “Demand factor” is a percentage by which the total connected load on a service or feeder is multiplied to connected load on a service or feeder is multiplied to determine the greatest probable load that the feeder will determine the greatest probable load that the feeder will be called upon to carry. In hospitals, hotels, apartment be called upon to carry. In hospitals, hotels, apartment complexes, and dwelling units, it is not likely that all of complexes, and dwelling units, it is not likely that all of the lights and

the lights and recereceptacles connected to ptacles connected to eveevery brry branch-circanch-circuituit served by a service or feeder would be “on” at the same served by a service or feeder would be “on” at the same time. Therefore, instead of

time. Therefore, instead of sizsizing the feeder to ing the feeder to carry all of carry all of the load on all of the branches, a percentage can be

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applied to this total load, and the components sized applied to this total load, and the components sized accordingly.

accordingly.

Referring to Table 220.11 of the

Referring to Table 220.11 of the N NECEC,,it can be seenit can be seen

that a demand factor for lighting may be applied only for that a demand factor for lighting may be applied only for dwelling units, hospitals, hotels, motels, and warehouses. dwelling units, hospitals, hotels, motels, and warehouses. All other occupancies are calculated on a basis of total All other occupancies are calculated on a basis of total computed lighting wattage, and no demand factor is computed lighting wattage, and no demand factor is permitted.

permitted. (See Figure 1)(See Figure 1)

APPLYING DEMAND FACTOR FOR RECEPTACLES

220.13; TABLE 220.13

S

Secection 220tion 220.13 o.13 of thef the N NEC EC makes it clear that in dwellingmakes it clear that in dwelling units, general-purpose receptacles are not counted as a units, general-purpose receptacles are not counted as a load. In other than dwelling units, a minimum of 180 VA load. In other than dwelling units, a minimum of 180 VA is computed for each general-purpose receptacle. For is computed for each general-purpose receptacle. For hos-pitals, hotels, motels, and warehouses, this receptacle load pitals, hotels, motels, and warehouses, this receptacle load can be lumped with

can be lumped with the ligthe lighting load, and hting load, and the demandthe demand factors of T

factors of Table 220.11 able 220.11 may be applied to thmay be applied to the total.e total.

General-purpose receptacle outlets used to cord-and-plug General-purpose receptacle outlets used to cord-and-plug connect loads are considered to have noncontinuous connect loads are considered to have noncontinuous opera-tion and are calculated per

tion and are calculated per 220.3(B)(9)220.3(B)(9)andand Table 220.13Table 220.13..

Noncontinuously operated receptacles with a VA rating of Noncontinuously operated receptacles with a VA rating of 10,000 VA or less shall be computed at 100 percent. If the 10,000 VA or less shall be computed at 100 percent. If the VA rating of the receptacle load exceeds 10,000 VA, a VA rating of the receptacle load exceeds 10,000 VA, a demand factor of 50 percent should be applied to all VA demand factor of 50 percent should be applied to all VA exceeding 10,000 VA per

exceeding 10,000 VA per Table 220.13Table 220.13..(See Figure 2)(See Figure 2)

APPL

APPLYING DYING DEMAND EMAND FFACTOR FORACTOR FOR

COMM

COMMERCIAERCIAL COOKING EQUIL COOKING EQUIPMENTPMENT

220.20; TABLE 220.20

Section 220.20 in the

Section 220.20 in the N NEC EC _{permits Table 220.20}_{permits Table 220.20} _to_to

be used for

be used for load compload computation utation for commercial elfor commercial elecectricaltrical cooking equipment, such as dishwashers, booster heaters, cooking equipment, such as dishwashers, booster heaters, water heaters and other kitchen equipment. The demand water heaters and other kitchen equipment. The demand factors shown in Table 220.20 are applicable to all factors shown in Table 220.20 are applicable to all equipment that is thermostatically controlled or is only equipment that is thermostatically controlled or is only intermittently used as part of the kitchen equipment. In intermittently used as part of the kitchen equipment. In no way do the demand factors apply to the electric no way do the demand factors apply to the electric heat-ing, ventilating or air-conditioning equipment. In ing, ventilating or air-conditioning equipment. In com-puting the demand, the demand load should not be less puting the demand, the demand load should not be less than the sum of the two largest kitchen equipment loads. than the sum of the two largest kitchen equipment loads. (See Figure 3)

(See Figure 3)

APPL

APPLYING DEMYING DEMAND FACTOR FOR THE NAND FACTOR FOR THE NEUTRALEUTRAL

220.22

Section 220.22 of the

Section 220.22 of the N NEC EC states that for a service orstates that for a service or feeder, the maximum unbalanced load controls the feeder, the maximum unbalanced load controls the ampacity sel

ampacity selecteected d for thfor the grounde grounded (neutral) condued (neutral) conductorctor.. The grou

The grounded (neutranded (neutral) conductor servicl) conductor service or e or fefeeder loadeder load should be considered whereve

should be considered wherever a r a grounded (neutragrounded (neutral)l)

conductor is used in conjunction with one or more conductor is used in conjunction with one or more ungroun

ungrounded (phase) conductors. Oded (phase) conductors. On a n a sisinglengle-phase fe-phase feedereder

Step 1:

Step 1: Calculating the VACalculating the VA

Table 220.3(A) Table 220.3(A)

3500

3500 sq. sq. ft. ft. x x 3 3 VA VA = = 10,500 10,500 VAVA

St

Step 2ep 2:: 22220.0.1616(A)(A);(B;(B))

1500

1500 VA VA x x 3 3 = = 4,500 4,500 VAVA

Step 3:

Step 3: Total loadsTotal loads

General

General lighting lighting load load = = 10,500 10,500 VAVA

Small

Small appliance appliance load load = 4,500 = 4,500 VAVA

T

Toottaall = = 1155,,00000 0 VVAA

Step 4:

Step 4: Applying demand factors Applying demand factors

Table Table 220.1220.111 Fi Firsrst t 303000 00 VVA A x x 10100%0% = = 3,3,00000 0 VVAA Next Next 12,000 12,000 VA VA x x 35% 35% = = 4,200 4,200 VAVA T Toottaall = = 77,,22000 0 VVAA So

Solutiolution:n: DeDemanmand load load 1 is 7d 1 is 720200 vol0 volt ampt amps.s. What is the demand load for the general lighti What is the demand load for the general lightingng load of a 3500 sq. ft. dwelling unit?

load of a 3500 sq. ft. dwelling unit? SMALL APPLIANCE LOAD SMALL APPLIANCE LOAD 1500 VA PER CIRCUIT 1500 VA PER CIRCUIT • 220.16(A); (B) • 220.16(A); (B) DEMAND FACTORS DEMAND FACTORS • TABLE 220.11 • TABLE 220.11 GENER

GENERAL AL LIGHTING LOADLIGHTING LOAD 3 VA PER 3 VA PERSQ. FT.SQ. FT. • TABLE 220.3(A) • TABLE 220.3(A) DEMAND FACTORS DEMAND FACTORS • TABLE • TABLE 220.1220.111

GENERAL PURPOSE RECEPTACLE

AND LIGHTING OUTLETS

GES

GEC

GEC _{SMALL APPLIANCE}_{SMALL APPLIANCE}

RECEPTACLE OUTLETS RECEPTACLE OUTLETS LAUNDRY LAUNDRY RECEPTACLE RECEPTACLE OUTLET OUTLET

GENERAL LIGHTING LOADS GENERAL LIGHTING LOADS COLUMN 1; DEMAND LOAD 1 COLUMN 1; DEMAND LOAD 1

Figure 1. The above illustration shows the calculation of a

demand factor load f

demand factor load for a dwelling unor a dwelling unit usinit using Table 22g Table 220.10.111

RECEPTACLE LOADS RECEPTACLE LOADS • 220.3(B)(9) • 220.3(B)(9) • TABLE 220.13 • TABLE 220.13 MBJ MBJ GECGEC GES GES LOAD #2

LOAD #2 _{CALCULATING RECEP}_{CALCULATING RECEP}_{TACLE LOAD}_{TACLE LOAD}_AND_AND

APPLYING DEMAND FACTORS

Step 1:

Step 1: Calculating VACalculating VA

220.3(B)(9)

220.3(B)(9); ; 230.42(A)(1230.42(A)(1))

150 x 180

150 x 180 VA = 27,000 VAVA = 27,000 VA

Step 2:

Table 220.13 Table 220.13 First 10,000 VA x 100% = First 10,000 VA x 100% = 10,000 VA10,000 VA Next Next 17,000 V17,000 VA A x x 50% 50% = = 8,500 V8,500 VAA 18,500 VA 18,500 VA Solution:

Solution: The The demademand loand load for thd for thee

receptacles is 18,500 VA. receptacles is 18,500 VA. RECEPTACLES RECEPTACLES • 150 • 150 Figure 2.

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

Figure 1 Figure 1 Figure 2 Figure 2 I I L L L L U U S S T T R R A A T T I I O O N N S S C C O

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using one ungrounded (phase) conductor and a grounded using one ungrounded (phase) conductor and a grounded (neutral) conductor, the grounded (neutral) conductor (neutral) conductor, the grounded (neutral) conductor will carry the same amount of current as the will carry the same amount of current as the unground-ed (pha

ed (phase) cse) condonductor. A twouctor. A two-wire fee-wire feeder is seldom used,der is seldom used, so in considering the grounded (neutral) feeder current, so in considering the grounded (neutral) feeder current, always assume that there is a grounded (neutral) always assume that there is a grounded (neutral) con-ductor and two or more ungrounded (phase) concon-ductors. ductor and two or more ungrounded (phase) conductors. If the

If there are two re are two ungrounded (phase) cungrounded (phase) conductors that onductors that areare connec

connected to ted to the same phase, and a the same phase, and a grounded (neutral)grounded (neutral) conductor, the grounded (neutral) conductor would conductor, the grounded (neutral) conductor would be require

be required to d to carry the total current from bothcarry the total current from both ungrounded (phase) c

ungrounded (phase) conductors, which would noonductors, which would not bet be

an accepted practice. an accepted practice.

For three-wire DC or single-phase AC, four-wire For three-wire DC or single-phase AC, four-wire three-phase, three-wire phase, and five-wire three-phase, three-wire phase, and five-wire two-phase systems, a further demand factor of 70 percent phase systems, a further demand factor of 70 percent should be applie

should be applied to d to that pthat portion of the unortion of the unbalancebalancedd load in excess of 200 amperes. There should be no load in excess of 200 amperes. There should be no reduction of the grounded (neutral) conductor ampacity reduction of the grounded (neutral) conductor ampacity for that portion of the load that consists of for that portion of the load that consists of electric-discharge li

discharge lighting, eleghting, electronic compctronic computer/data uter/data processiprocessingng or similar equipment, when supplied by four-wire, or similar equipment, when supplied by four-wire, wye-c

wye-connected, onnected, three-phase three-phase syssystems.tems. (See Figure 4)

(See Figure 4)

APPL

APPLYING DEMYING DEMAND FACTOR FOR CONNECTINAND FACTOR FOR CONNECTINGG

ADDITIONAL LOADS TO EXISTING INSTALLATIONS

220.35

When additional loads are connected to existing When additional loads are connected to existing facili-ties having fee

ties having feeders anders and service as originally cd service as originally compuomputed,ted, the maximum kVA computations in determining the the maximum kVA computations in determining the load on the existing feeders and service should be used load on the existing feeders and service should be used if the following conditions are met:

if the following conditions are met: •

• If the maximum data for the demand in kVA,If the maximum data for the demand in kVA, such as demand meter ratings, is available for such as demand meter ratings, is available for a minimum of one year.

a minimum of one year. •

• If 125 percent of the demand ratings for the periodIf 125 percent of the demand ratings for the period of one year added to the new load does not exceed of one year added to the new load does not exceed the rating of the service. Where demand meters are the rating of the service. Where demand meters are used, in most cases the load as computed will used, in most cases the load as computed will probably be less than the demand meter indications. probably be less than the demand meter indications. (See Figure 5)

(See Figure 5)

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LOAD #3

TWO LARGEST LOADS

• 10 kW

• 8 kW

• 8 kW FOUR PIECES OFFOUR PIECES OF

COOKING EQUIPMENT COOKING EQUIPMENT TWO KETTLES TWO KETTLES TWO MICROWAVES TWO MICROWAVES THREE THREE BOILERS BOILERS TWO STEAMERS TWO STEAMERS THREE FRYERS THREE FRYERS

TOTAL KITCHEN LOAD IS 82 kW

CALCULATING LOAD

FOR COOKING EQUIPMENT

Step 1:

Step 1: Calculating percentageCalculating percentage

Table 220.20

16 pieces allowed 65%

Step 2:

Table 220.20;

Table 220.20; 220.20220.20

82 kW

82 kW x 65% = 53.3 kVAx 65% = 53.3 kVA

Solution:

Solution: The deThe demand load of 53mand load of 53.3 kV.3 kVAA

is greater than the sum of the

two largest l

two largest loads of 18 kVA.oads of 18 kVA.

Figure 3. The demand load for commercial cooking Figure 3. The demand load for commercial cooking equipment is computed based upon the

equipment is computed based upon the number andnumber and

the perc

the percentage per Table 220.20entage per Table 220.20

ASSUME THE FOLLOWING LOADS AND

COMPUTE THE NEUTRAL LOAD

Step 1:

Step 1: Calculating total neutral loadCalculating total neutral load

220.22

Di

Discschahargrge lie lighghtiting lng loaoadsds = 30= 300 am0 amp pep per phr phasasee

Inca

Incandesndescent cent lightlighting ling loadsoads = = 50 a50 amp pmp per per phasehase

O

Othther er rresesiiststivive le looaadsds = = 29295 5 amamp p pper er pphahasese

T

Toottaal l nneeuuttrraal l llooaaddss = = 66445 5 aammp p ppeer r pphhaassee

Step 2:

Step 2: Applying dema Applying demand factors for nd factors for inductive andinductive and

resistive loads resistive loads 220.22 220.22 F Fiirrsst t 22000 0 aammp p x x 110000%% = = 22000 0 aammppss N Neexxt 1t 1445 a5 ammpps x s x 7700%% = = 110011..5 a5 ammppss R Reessiissttiivve e llooaaddss = = 330011..5 5 aammppss Disch

Discharge arge lighlighting ting at at 100%100% = = 300 300 ampampss

T

Toottaal l nneeuuttrraal l llooaadd = = 660011..5 5 aammppss

Solution:

Solution: The The neutraneutral load al load after defter demanmand factors thad factors that havet have

been applied is 601.5 amps.

SUBPANEL SUBPANEL • 408.20 • 408.20 • 408.16(A) • 408.16(A) SERVICE CONDUCTORS SERVICE CONDUCTORS • ARTICLE 100 • ARTICLE 100 MBJ MBJ GEC GEC GES GES SERVICE EQUIPMENT SERVICE EQUIPMENT • ARTICLE 100 • ARTICLE 100 FEEDER-CIRCUIT FEEDER-CIRCUIT • ARTICLE 100 • ARTICLE 100 BRANCH-CIRCUIT BRANCH-CIRCUIT • ARTICLE 100 • ARTICLE 100 Figure 4. T

Figure 4. The above cahe above calcullculation showation shows the procedurs the proceduree for computing the neutral for a service or feeder and for computing the neutral for a service or feeder and applying

applying demand factors demand factors per 220per 220.22.22..

Figure 3

Figure 4

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APPL

APPLYING DEMYING DEMAND FACTOR TOAND FACTOR TO

THE EXCEPTION TO

THE EXCEPTION TO 220.3220.355

220.35, Ex.

If the maximum demand data

If the maximum demand data for a onfor a one yee year period isar period is

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

the recording. (See Figure 6) (See Figure 6)

APPL

APPLYING DEMYING DEMAND FAAND FACTOR FOR MCTOR FOR MOTORSOTORS

430.26

There are, in some cases, motor installations where there There are, in some cases, motor installations where there may be a special situation in which a number of motors may be a special situation in which a number of motors are con

are connected to nected to a feedera feeder-ci-circuit. Becrcuit. Because of thause of the pare particu- ticu-lar applic

lar application, certain motors do ation, certain motors do not not operate togetheroperate together and the feeder-circuit conductors are permitted to be and the feeder-circuit conductors are permitted to be sized based on a historical demand factor.

sized based on a historical demand factor.

For example, the authority having jurisdiction may grant For example, the authority having jurisdiction may grant permission to allow a demand factor of less than 100 permission to allow a demand factor of less than 100 per-cent if operation procedures, production demands, or the cent if operation procedures, production demands, or the nature of the work

nature of the work is sis such that nouch that not all the motors are run-t all the motors are run-ning at one time. An engineering study or evaluation of ning at one time. An engineering study or evaluation of

CALCULATING LOAD IN AMPS CALCULATING LOAD IN AMPS Step 1:

Step 1: Finding demandFinding demand 220.35 220.35

Maximum demand = 78.4 kVA Maximum demand = 78.4 kVA Step 2:

Step 2: Calculating existing demandCalculating existing demand 78.4 kVA x 125% = 98 kVA 78.4 kVA x 125% = 98 kVA Step 3:

Step 3: Calculating total kVACalculating total kVA 230.42(A)(1) 230.42(A)(1)

98 kVA + 15.1 kVA = 113.1 kVA 98 kVA + 15.1 kVA = 113.1 kVA Step 4:

Step 4: Calculating amperageCalculating amperage Table 310.16 Table 310.16 No. 400 KCMIL

No. 400 KCMIL THWN copper = 335 ATHWN copper = 335 A 113.1 kVA x 1000 = 113,100 VA 113.1 kVA x 1000 = 113,100 VA 113,100 VA ÷ (208 x 1.732) = 314 A 113,100 VA ÷ (208 x 1.732) = 314 A 314 A is less than 335 A 314 A is less than 335 A Solution:

Solution: The 1The 15.5.1 kV1 kVA demanA demand load can bed load can be applied to the existing service applied to the existing service without upgrading the elements. without upgrading the elements. MBJ MBJ OCPD OCPD • 230.90(A) • 230.90(A) • 230.42(A)(1); (A)(2) • 230.42(A)(1); (A)(2) GEC GEC GES GES 15.1 kVA LOAD 15.1 kVA LOAD TO BE ADDED TO BE ADDED FEEDER-CIRCUIT FEEDER-CIRCUIT CONDUCTORS CONDUCTORS • 215.2(A)(1) • 215.2(A)(1)

EXISTING SERVICE PANEL EXISTING SERVICE PANEL

• 78.4 kVA LOAD (EXISTING DEMAND) • 78.4 kVA LOAD (EXISTING DEMAND) • 208 V, 3Ø • 208 V, 3Ø • USE 360 V • USE 360 V OCPD OCPD • 215.3 • 215.3 • 215.2(A) • 215.2(A) SERVICE CONDUCTORS SERVICE CONDUCTORS • 400 KC

• 400 KCMILMIL THWTHWNN cu.cu.

Figure 5. The above illustration is the calculation Figure 5. The above illustration is the calculation for adding a load to an existing service or for adding a load to an existing service or feeder-circuit using 220.35.

circuit using 220.35.

CALCULATING LOAD IN AMPS

Step 1:

Step 1: Finding recorded demandFinding recorded demand

220.35

Maximum demand = 218 A

Step 2:

Step 2: Calculating existing demandCalculating existing demand

218 A x 125% = 273 A

Step 3:

Step 3: Calculating existing andCalculating existing and

added load

273 A + 102 A = 375 A

Step 4:

Step 4: Finding amperage for Finding amperage for

feeder conductors

Table 310.16

500 KCMIL

500 KCMILTHWN copper = 380 THWN copper = 380 AA

Step 5:

Step 5: Determining if load can be addedDetermining if load can be added

375 A is less than

375 A is less than 380 A380 A

Solution:

Solution: The 3The 375 a75 amp load can be mp load can be appliedapplied

to the existing feeder-circuit

conductors. conductors. ACTUAL AMPS ACTUAL AMPS RECORDED RECORDED FOR A PERIOD FOR A PERIOD OF 30 DAYS OF 30 DAYS • 500 • 500KCMILKCMIL • 220.35, Ex. • 220.35, Ex. RECORDED AMPS RECORDED AMPS WITH AMMETER OR WITH AMMETER OR

USE POWER METER

• 218 A • 218 A • LARGEST PHASE • LARGEST PHASE READING READING MOTOR MOTOR BRANCH-CIRCUIT CIRCUIT NO. 2000 KCMIL NO. 2000 KCMIL IN PARALLEL IN PARALLEL ADDED ADDED PANELBOARD PANELBOARD LOAD LOAD • 102 A • 102 A BRANCH-CIRCUITS CIRCUITS SUBFEEDER SUBFEEDER SUBFEED SUBFEED DISTRIBUTION DISTRIBUTION PANELBOARD PANELBOARD NO

NOTE: ADDEDTE: ADDEDLOLOADADFORAFORASESERVIRVICECECANBECANBECOMCOMPUTPUTED,ED,

USI

USINGNGTHESAMETHESAMEPROPROCEDCEDUREPER UREPER 220.35, Ex 220.35, Ex .. DIRECTORY DIRECTORY 1.Motor 1.Motor 1 1 2.2. 3 3.. 44.. 5 5.. 66.. 7 7.. 88.. 9 9.. 1100.. 1 111.. 1122.. DIRECTORY DIRECTORY 1 . M o 1 . M ot o r t o r 1 1 2 .2 . 3 3.. 44.. 5 5.. 66.. 7 7.. 88.. 9 9.. 1100.. 1 111.. 1122..

Figure 6. The above illustration shows the optional calculation Figure 6. The above illustration shows the optional calculation being applied for adding a load to

being applied for adding a load toan existing feeder-circuitan existing feeder-circuit using 220.35. using 220.35. F Fiigguurree 55 FFiigguurree 66 I I L L L L U U S S T T R R A A T T I I O O N N S S C C O

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motor operation may provide information that will allow motor operation may provide information that will allow a demand factor of less than 100 percent.

a demand factor of less than 100 percent. (See Figure 7)(See Figure 7)

CONCLUSION

For engineers and contractors, the above demand factors For engineers and contractors, the above demand factors are the most widely used on a regular basis because of are the most widely used on a regular basis because of their un

their uniqueness to eleciqueness to electrical desigtrical design. Wn. With thith the appe appliclica- a-tion of demand factors, smaller components can be tion of demand factors, smaller components can be uti-liz

lized in ted in the eleche electrical systrical system antem and greater d greater savingsavings can bs can bee passed on to the consumer. Due to the high cost of passed on to the consumer. Due to the high cost of wiring, look

wiring, look for dfor desiesigners to ugners to utilitilize these tecze these techniqueshniques more than ever before.

more than ever before. A

Adddditionitional infal informormation ation on on this tothis topic mpic may bay be foe found und inin chapters 22 and

chapters 22 and 23 23 of of the bthe book ook ‘‘StallStallcup’cup’s Electris Electricalcal D

Desigesign n BoBookok, 2, 2002 002 EdEditionition,’ available from ,’ available from ththe Ne NFFPPAA..

NOTE 1: DUE TO HISTORICAL DATA, A NOTE 1: DUE TO HISTORICAL DATA, A DEMAND FACTOR OF 75 PERCENT DEMAND FACTOR OF 75 PERCENT SHALL BE PERMITTED TO BE APPLIED SHALL BE PERMITTED TO BE APPLIED PER AHJ AS PERMITTED BY 430.26 PER AHJ AS PERMITTED BY 430.26 TO

TOOCPD IN SERVICEOCPD IN SERVICE EQUIPMENT EQUIPMENT • 430.62(A) • 430.62(A) • 430.63 • 430.63 4 CONDUCTORS 4 CONDUCTORS PARALLELED ON PARALLELED ON PHASE A PHASE A FEEDER-CIRCUIT CONDUCTORS FEEDER-CIRCUIT CONDUCTORS (SHOWING PHASE A ONLY) (SHOWING PHASE A ONLY) • 430.26 • 430.26 CB’S CB’S • 3-POLE • 3-POLE GROUP 1 GROUP 1 5 - 20 HP 5 - 20 HP MOTORS MOTORS GROUP 2 GROUP 2 5-25 HP 5-25 HP MOTORS MOTORS FEEDER SIZED FEEDER SIZED AT 75% D AT 75% DEMANDEMAND

CONDUCTORS CONDUCTORS • THWN • THWN • cu. • cu. GROUP 4 GROUP 4 5 - 30 HP 5 - 30 HP MOTORS MOTORS GROUP 3 GROUP 3 5- 15 HP 5- 15 HP MOTORS MOTORS NOTE 2: FOR SIMPLICITY, ONLY ONE

NOTE 2: FOR SIMPLICITY, ONLY ONE CONDUIT AND 4-CONDUCTORS PER CONDUIT AND 4-CONDUCTORS PER PHASE ARE SHOWN.

PHASE ARE SHOWN. NOTE 3:

NOTE 3: FOR SIMPLICITYFOR SIMPLICITY, THE , THE DISCONNECTS AND

DISCONNECTS AND CONTROLLERS CONTROLLERS ARE NOT SHOWN.

ARE NOT SHOWN.

SUPPLY SUPPLY • 3Ø • 3Ø • 208 V • 208 V

Figure 7. The above illustration shows

Figure 7. The above illustration shows the sizing of the sizing of a feeder-a feeder-circuit circuit using 430.26using 430.26

.

Sizing Largest Motor Load Sizing Largest Motor Load Step 1:

Step 1: Calculating amps of motorsCalculating amps of motors Table 430.150 Table 430.150 15 HP = 46.2 15 HP = 46.2 A x A x 5 = 5 = 231 A231 A 20 HP = 59.5 20 HP = 59.5 A x A x 5 = 5 = 297 A297 A 25 HP = 75.8 25 HP = 75.8 A x A x 5 = 5 = 374 A374 A 30 H 30 HP = P = 88 88 A A x 5x 5 = = 44440 0 AA T Toottaal l LLooaadd = = 1133442 2 AA Step 2:

Step 2: Applying demand factors Applying demand factors 430.26

430.26 1342

1342 A A x x 75% 75% = = 1007 1007 AA Solution:

Solution: The deThe demand load mand load is 100is 1007 a7 amps.mps.

Sizing Conductors Sizing Conductors Step 1:

Step 1:Paralleling 4 lines per phaseParalleling 4 lines per phase 310.4

310.4

Amps per conductor = 1007 Amps per conductor = 1007 A ÷ 4A ÷ 4 Amps per conductor = Amps per conductor = 252 A252 A Step 2:

Step 2:Sizing conductor for feeder Sizing conductor for feeder Table 310.16

Table 310.16 252 A

252 A requires 250 KCMILTHWNrequires 250 KCMILTHWNcu. conductorscu. conductors 250 KCM

250 KCMILILTHWNcu. = THWNcu. = 255 A255 A 255 A x 4 = 1020 A 255 A x 4 = 1020 A 1020 A supplies 1007 A 1020 A supplies 1007 A Solution:

Solution: The The size THsize THWN WN copper conductors arecopper conductors are 4 - 250 KCMIL per phase.

4 - 250 KCMIL per phase.

Figure 7 Figure 7 I I L L L L U U S S T T R R A A T T I I O O N N S S C C O

O U U

O Y Y

, , I I N N C C . .

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Equipment Load Factors, Use Factors and Diversity Factors As Well as a General Equipment Load Factors, Use Factors and Diversity Factors As Well as a General

Discussion of Energy Audit Procedures Discussion of Energy Audit Procedures

Barney L. Capehart Barney L. Capehart

To do a good job on an energy audit, the energy auditor must understand the areas of equipment 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.

load factor, use factor and diversity factor.

Definitions:

Definitions:First, let's define these terms.First, let's define these terms.

a)

a) Load factorLoad factor- the ratio of the load that a piece of equipment actually draws when it is in- 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). 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 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%.

motor load factor is then 15/20 = 75%. b)

b) Use (or utilization) factorUse (or utilization) factor- the ratio of the time th- the ratio of the time that a piece of equipment is in use to tat a piece of equipment is in use to the totalhe total

time that it could be in use. 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 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 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 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 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.

number of hours that the motor is in use. c)

c) Diversity factorDiversity factor- the probability that a particular piece of equipment will come on at the time- the probability that a particular piece of equipment will come on at the time

of the facility's peak load. of the facility's peak load.

The diversity factor is the most complicated of these factors. For example, we might have ten 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 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, 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 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 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 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. 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 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 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 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 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.

affect the kWh; it only affects the kW.

Motor load, use and diversity factors:

Motor load, use and diversity factors:Sometimes motor load factors that are too low are chosenSometimes 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 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 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-rel

hours per year it does not mean that all of the production-related motors in that facility are operated motors in that facility are operatedated

for 2000 hours per year. Some motors - or machines - might only be used one day a week rather 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 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 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 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 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 this machine (line, process or operation) operate?" This data then needs to be entered into the

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energy balance. Motor load factors in many facilities are more in the range of 40% - 50%, than in 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 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.

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 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 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 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 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 may also be in this category - some engineering judgment and common sense are required to determine which other motors these are.

determine which other motors these are.

Motors with variable loads are going to have the lowest load factor

Motors with variable loads are going to have the lowest load factors in general. A dust collector fans in general. A dust collector fan

motor will normally have quite a variable load, and would often be expected to have a low load 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, factor. Other examples are saws, presses, milling machines, sanders and grinders, waste grinders, water pumps, hydraulic pumps, etc.

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 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 appropriate. This is the case with a number of separate air conditioning units (considering the motors for

motors for the compressors) the compressors) that are that are individually thermostaticalindividually thermostatically controlled. ly controlled. It could It could also be also be thethe

case for a group of production motors if some of the motors are not in use all of the time. You 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 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.

has all of the motors on at the same time.

Reconciling the energy balance:

Reconciling the energy balance:When you perform an energy balance, do not use thWhen you perform an energy balance, do not use the motor loade motor load

factor as the first and only adjustment made to reconcile the estimated energy use (energy balance) 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 with the energy bills. Making this adjustment too quickly results in failure to pick up other things that have been overlooked.

that have been overlooked.

For example, if the energy use does not balance with the energy bills, the first step is to check to 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.

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 Do the items on the energy balance spreadsheet match your recollection of the equipment you saw in the facility?

you saw in the facility?

Does anything appear to be missing? Does anything appear to be missing?

Are the utility bills for total energy use and peak kW recorded correctly? 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 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 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 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 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.

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 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 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 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 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 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 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 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.

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Basic motor load measurements should be taken at the plant visit. The electrical person at the 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 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 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 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 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 of the motor. The ratio of the actual current to the full load current is the approximate load factor on the motor

on the motor at that timeat that time. This procedure works as long as the current is greater than or equal to. 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 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 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.

big motors. If you have not received formal electrical safety training, you should not makeIf 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 electrical measurements yourself. If the facility electrician does not want to make thesethese

for you, then let it go at that. for you, then let it go at that. Air handlers—use factor:

Air handlers—use factor:Air handlers use motors and are subject to all of the coAir handlers use motors and are subject to all of the comments made inmments made in

the motor section. In addition, you may be able to get a better handle on the hours of use for the air 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 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 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. 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 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 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 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 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.

that the facility is not occupied, or if ventilation is provided. If the air handlers only come on when the thermosta

If the air handlers only come on when the thermostat orders cooling, then the hourst orders cooling, then the hours-of -use-of -use must bemust be

the same as the hours-of-use of the compressors. 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 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 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 standard information - size, maker, single or three phase, etc - together with the operating basis for the air handlers discussed above.

the air handlers discussed above.

You should also collect data on the drive belt system for air handlers. Record the number of belts, 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.

the lengths, and the types of belts. Ask about motor and drive lubrication and cleaning. Also check Ask about motor and drive lubrication and cleaning. Also check

the A/C filters to see if they are reasonably clean. 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 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

up one of the air handlers - or just look into it (SAFELY)(SAFELY)if it is accessible - and see if the belt isif it is accessible - and see if the belt is

tight, slack, or really loose.

tight, slack, or really loose. Do not stick your hand Do not stick your hand into an air handler that is into an air handler that is off at theoff at the

moment, and may come back on when the thermostat kicks in.

moment, and may come back on when the thermostat kicks in.Have the maintenance personHave the maintenance person

turn the air handler motor off with the circuit breaker or control box. Do not put your finger on a turn the air handler motor off with the circuit breaker or control box. Do not put your finger on a moving drive belt.

moving drive belt.

Is the belt in good shape? Is the belt in good shape?

Is it frayed, cracked or coming apart? Is it frayed, cracked or coming apart?

Does it look like the pulleys for the motor and the fan are lined up? 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 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 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 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.