Re: answer to this calculation would be appreciated

A school and book storage warehouse serviced by one entrance. The school contains the following:

85,000 sq. ft. of school area

550 ea. fluorescent fixtures 180W 277v

400 ea. 120v duplex outlets

50 ea. power pole outlets rated 250W 120v

400 ft. multi-outlet assembly that will have a number of appliances likely to be used simultaneously.

4 ea. 15KW range outlets

3 ea. 5KW dishwashers

2 ea. 10KW hot water heater booster

4 ea. 2KW disposals

4 ea. 1KW toasters

4 ea. 3KW food warmers

1 ea. 125HP 480v 3 ? Code F squirrel cage A/C motor @ .9 pf

10 ea. 7.5HP 480v 3 ? blower motors operating @ 80% pf

The book storage contains the following:

80,000 sq. ft. of area?

100 ea. 120v duplex outlets

12 ea. 1/2HP 120v blower motors for gas space heaters

The service entrance is 480/277v 3 ? 4-wire. The lighting will be treated as continuous where Code will allow. Work problem optional where possible.

a. Service entrance conductors in amps. Ungrounded conductor.

b. Feeder size in amps for ungrounded conductor to warehouse.

General lighting

Scholl

85000 x 3VA = 255,000 x 1.25 = 318,750VA

3VA from table Table 220.3(A)

1.25 for continuous use of lighting system of three hours o more

we compare to actual load 550 x 180W = 99,000VA so we take the largest

Book storage

80,000 x 0.25VA = 20,000 x 1.25 = 25,000

0.25VA from table Table 220.3(A)

Table 220.11 Lighting Load Demand Factors (Warehouses (storage))

20,000VA - 12,500VA = 7500 x 0.5 = 3,750 + 12,500 = 16520VA

Scholl Receptacles

400 x 180VA = 72,000 220.3

Article 220.13 Receptacle Loads ? Nondwelling Units (180VA)

50 x 250W = 12,500

400 x 180VA = 72,000

Article 220.13 Receptacle Loads ? Nondwelling Units (180VA)

156,500VA apply Table 220.13 Demand Factors for Nondwelling Receptacle Loads

156,000 - 10,000 = 146,500 x .5 = 73250 + 10,000 = 83250VA

Book storage

100 x 180VA = 18,000 apply Table 220.13 Demand Factors for Nondwelling Receptacle Loads

18,000 - 10,000 = 8,000 x 0 .5 = 4,000 + 10,000 = 14,000VA

Scholl Cooking Equipments

ranges (Commercials)

4x 15,000 = 60,000VA

dishwashers

3 x 5,000 = 15,000VA

water heater

2 x 10,000 = 20,000VA

disposal

4 x 2,000 = 8,000VA

toaster

4 x 1,000 = 4,000VA

food warmer

4 x 3,000 = 12,000VA

total load 119,000VA apply demand factor from

Table 220.20 Demand Factors for Kitchen Equipment ? Other Than Dwelling Unit(s)

119,000 x 0.65 = 77,350VA

A/C

?we go to Table 430.150 Full-Load Current, Three-Phase Alternating-Current Motors

and after interpolation 480 volt 150.60 Amp

150.60 x 480 x √ 3 = 125,206.48 AV

largest motor

150.6 x 480 x √ 3 x 0.25 = 31,301.62 VA

blower motors

?we go to Table 430.150 Full-Load Current, Three-Phase Alternating-Current Motors

10.62 x 480 x √ 3 x 10 = 88,560.51VA

Book storage

blower motors

?we go to Table 430.148 Full-Load Currents in Amperes, Single-Phase Alternating-Current Motors

9.06 x 120 x 12 = 13,060.01VA

Compute load

standard optional book storage

318,750 255,000 16,520

83,250 156,500 14,000

77,350 119,000 -----

125,206 125,206 -----

31,301 ------ ------

88,560 ?88,560 13,060

724,417 744,648 43,580

standard

?Table 220.34 Optional Method ? Demand Factors for Feeders and Service-Entrance Conductors for Schools

724,417 we divide by 85,000(sqf) = 8.522

8.522 - 3 = 5.522

3 x 85,000 = 255,000 at 100%

5.522 x 85,000 = 469,370 (at 75%) x 0.75 = 352,027.5

255,000 + 352,027.5 = 607,027.5VA (school side )

607,027.5VA + 43,580VA (book storage side) = 650,607.5VA

a) 650,607.5 / 480 * √ 3 = 782.52 Amp

b) 43,580 / 480 * √ 3 = 52.41 Amp

i already did the problem just curious is there are not error

thanks for you help

[ February 27, 2003, 09:45 PM: Message edited by: hidroela ]