158 Modern Commercial Wiring
Copyright Goodheart-Willcox Co., Inc.
Sample Problem 12-15
Problem: A 200-unit hotel is supplied with three-
phase, 120/208-volt service. Each unit is 300 ft2, has a
30-amp air conditioner (208-volt) and has a heating
load of 8 kilowatts. The hotel also has a 30-kilowatt
continuous load for general equipment. What is the
total current that must be supplied to the hotel?
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 needed. Section
220.14 explains that the receptacle loads do not need
to be included.
General lighting load = 200 × 300 ft2 × 2 VA/ft2
= 120,000 VA
This load can be reduced in accordance with Table
220.42:
50% of 0–20 kW = 10,000 VA
40% of 20–100 kW = 32,000 VA
30% of 100–120 kW = 8000 VA
Sample Problem 12-14
Problem: A restaurant is supplied with three-wire,
120/240-volt service. The restaurant is 50′ × 80′ and
has several loads:
(1) 12-kW electric range (120-volt)
(1) 10-kW water heater (240-volt)
(2) 8-kW fryers (240-volt)
(2) 1.5-kW coffeemakers (120-volt)
(1) 2.5-kW steam table (120-volt)
(2) 3.0-kW toasters (120-volt)
(1) 2.5-kW disposal unit (120-volt)
(1) 1.5-kW outdoor sign (120-volt)
(1) 40-kW heating unit
(26) duplex receptacles
Copper THW conductors are used as feeders. What
size should the current-carrying feeders be?
Solution: First, the loads must be computed.
General lighting load = 50′ × 80′ × 2 VA/ft2
= 8000 VA
Sign lighting load = 1500 VA
Receptacle load = 26 × 180 VA
= 4680 VA
Heating load = 40,000 VA
The cooking equipment will be combined as the
cooking equipment load:
Electric range: 12,000 VA
Water heater: 10,000 VA
Fryers: 16,000 VA
Coffeemakers: 3000 VA
Steam table: 2500 VA
Toasters: 6000 VA
Disposal unit: 2500 VA
Cooking equipment load = 52,000 VA
Table 220.56 contains load-reduction factors for
commercial kitchen equipment. For six or more pieces
of equipment, the feeder conductors can be designed
for 65% of the load.
Reduced cooking equipment load = 52,000 VA × 0.65
= 33,800 VA
The total lighting load is the combination of the two
separate lighting loads.
Total lighting load = 8000 VA + 1500 VA
= 9500 VA
The lighting loads are continuous, so the conductors
must supply 125% of the full-load current:
Total lighting load = 9500 VA × 1.25
= 11,875 VA
All loads are added to determine the total load:
Load = 11,875 VA + 4680 VA + 40,000 VA + 33,800 VA
= 90,355 VA
The total current can be calculated:
I =
P
E
=
=
90,355 VA
377 A
240 V
Using Table 310.15(B)(16), 500 kcmil conductors will
be sufficient.
(Continued on the following page.)