answer to this calculation would be appreciated

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hidroela

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Location
Texas
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.
 

hidroela

Member
Location
Texas
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 ]
 

hidroela

Member
Location
Texas
Re: answer to this calculation would be appreciated

Hey charlie my home work is now there

[ February 28, 2003, 11:31 AM: Message edited by: hidroela ]
 
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