Chapter 12 Branch Circuits and Feeders 183 Copyright Goodheart-Willcox Co., Inc. The e total lighting load is the combination om of the h two separate lighting loads. d Total lighting load = 800 00 VA + 1500 VA = 9500 95 VA The lighting loads are co c on tinuous, so the con- ductors must supply 125% of the full-load current: Total lighting load d = 9500 VA × 1.25 = 11,875 VA All loads are added e to determine the total load d: Load = 11,87 8 75 VA + 4680 VA + 40,000 VA + 33,800 3 VA = 9 90 ,355 VA The total c cu rrent can be calculated: I = P _ E E = 90,355 VA _____ 240 V V = 377 A U Us ing Table 310.15(B)(16), 500 k kc mil conductors will be sufficient. ffi SAMPLE PROBLEM 12-15 Problem: A 200-un ni t hotel is supplied wit th three-phase, 120/20 08 -volt service. Each unit t is 300 ft2, has a 30-am mp air conditioner (208-volt), o and has a heating g load of 8 kilowatts. The hotel h also has a 30-kilowatt o continuous load for gen- eral equipment. . What is the total current t that must be supplie ed to the hotel? Solution: First, st all loads must be com mp uted. The heating u un it and air-conditioning u un it will not be used a at the same time, so only the th larger load is needed. d Section 220.14 explains n s that the receptacle loads l do not need to be in nc luded. (Continued) (C General = 200 20 × 300 ft2 × 2 VA/ft2 lighting load = = 120,000 VA This load can b be reduced in accordance with Table 220.42: : 50% of f 0–20 0 kW = 10,000 VA 40% % of 20–100 kW = 32,000 VA 3 30 % of 100–120 kW = 8000 VA The Th total reduced lighting load is dete er mined: Lighting load = 10,000 VA + 32,0 2, 00 0 VA + 8000 VA = 50,000 VA A Calculate the air-conditioning ni load. The volt- age is 208 V × 1.732 (360 v vo lts) due to the three- phase supply: Air-conditioning g load l = 360 V × 30 A = 10,800 VA This is larger th t ha n the heating load, so the air- conditioning g load is used in the total load. The load per ro r oo m is 10,800 VA, so it must be multi- plied by y 200 2 to determine the total load: Tot o ta l air- = 10,800 VA × 200 c co nditioning load = 2,160,000 VA General = 30,000 VA A × × 1.25 equipment load = 37,500 0 VA All loads are added to determ mi ne the total load: Total load = 50,000 VA VA + 2,160,000 VA + 37,500 0 VA = 2,24 2 47 ,500 VA The total current c ca n be calculated: I = _P E E = ___,____AV0502472, 360 V V = 6243 A The h e total current is 6243 A.