Chapter 12 Branch Circuits and Feeders 157
Copyright Goodheart-Willcox Co., Inc.
(Continued)
Sample Problem 12-12 Continued
All loads are added to determine the total load:
Total load = 29,500 VA + 10,800 VA + 50,000 VA + 1470 VA
= 91,770 VA
The total current can be calculated:
I =
P
E
=
=
91,770 VA
382 A
240 V
Using Table 310.15(B)(16), 500 kcmil conductors
will be sufficient.
The neutral feeder conductor is sized for the
120-volt loads (lighting and receptacles) only.
Neutral feeder load = Lighting load + Receptacle load
= 29,500 VA + 10,800 VA
= 40,300 VA
The current can be calculated:
I =
P
E
=
=
40,300 VA
168 A
240 V
A 2/0 AWG copper conductor should be used for the
neutral feeder conductor.
Sample Problem 12-13
Problem: A two-story office building (80′ × 80′) is
supplied with 120/208-volt service. Several loads
are supplied:
75-kVA air-conditioning unit
85-kVA heating unit
200 duplex receptacles
50 linear feet of show window
30 exterior light fixtures (175 VA each)
3 blower motors (3/4-hp, 208-volt)
Copper THW conductors are used as feeders. What
size should the current-carrying conductors be?
Solution: First, all loads must be computed. The
heating unit and air-conditioning unit will not be
used at the same time, so only the larger load is
Sample Problem 12-13 Continued
needed. The 3/4-hp, 208-volt motors draw 3.5 amps,
as shown in Table 430.250. The actual motor voltage
is 208 V × 1.732 (or 360 volts) because the service
is three-phase.
General lighting load = 2 × 80′ × 80′ × 3 1/2 VA/ft2
= 44,800 VA
Show-window lighting load = 50′ × 200 VA
= 10,000 VA
Exterior lighting load = 30 × 175 VA
= 5250 VA
Receptacle load = 200 × 180 VA
= 36,000 VA
Heating/air-conditioning load = 85,000 VA
Motor load = 3 × 3.5 A × 360 V
= 3780 VA
The total lighting load is the combination of the
three separate lighting loads.
Total lighting load = 44,800 VA + 10,000 VA + 5250 VA
= 60,050 VA
The motor load and lighting loads are continuous,
so the conductors must supply 125% of the full-load
current:
Total lighting load = 60,050 VA × 1.25
= 75,063 VA
Motor load = 3780 VA × 1.25
= 4725 VA
All loads are added to determine the total load:
Total load = 75,063 VA + 36,000 VA + 85,000 VA + 4725 VA
= 200,788 VA
The total current can be calculated:
I =
P
E
=
=
200,788 VA
557 A
360 V
Using Table 310.15(B)(16), 1250 kcmil conductors
will be sufficient.