Question #9
A retail store has 3000 square feet and 30 feet of show window. The service is a 120/240 volt single phase 3-wire service, and there is an actual connected lighting load of 8500 VA.
There are a total of 80 duplex receptacles. Given these facts, the total calculated load is which of the following:
A. 9000 VA
B. 12200 VA
C. 16200 VA
D. 28400 VA
Question #10
A project requires the installation of twelve 1.4 ampere, 120 volt, fluorescent lights fixtures on two 20 amp branch circuits, as well as three 120 volt, 5.6 ampere electric fans on individual circuits in a building with a 120/240 single phase three-wire electric service.
The minimum neutral current allowed for these loads is which of the following:
A. 0 amps
B. 5.6 amps
C. 16.8 amps
D. 33.6 amps
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C – Reference Section [250.32(B)(1)]
A – Reference Table [250.122]
A – Reference Section [280.21]
A – Reference Sections [210.11(A)] and [220.423(A)] for minimum load requirements. Divide the load by 120 volts, then divide this value into the circuits to determine the number of circuits required.
3 VA ÷ 2680 feet 2 = 8040 VA;
120 volts × 15 amps (per circuit) = 1800;
8040 ÷ 1800 4.5 (rounded up to 5)
C – Reference Table [220.3(A)] to see that 3 VA is required for every square feet of living area.
2400 square feet × 3 VA = 7200 VA.
The small appliance load in Section [220.11(C)(2)] is 1500 VA and comes to 2 small appliance loads.
1500 VA × 3 = 3000;
Section [210.11(C)(2)] requires 1500 VA for a dwelling laundry circuit.
7200 + 3000 + 1500 = 11700 VA.
B – Use formula Csecondary = 75kVA × 1000 / 1.73 × 208v = 208 amps
C – The first step is to calculate how much current will be going to the capacitor bank once it is energized, which is the same formula you would use to calculate the full load current of a transformer, except you need to use kVAR’s instead of kVA.
Ccapacitor = 92kVAR × 1000 / 1.73(3 phase) × 480v = 110.8 ampsNext you need to find the minimum ampere rating of the conductors in the capacitor bank using the requirement in Section [460.8(A)], and you will find that the ampacity cannot be more than 135% (which is 1.35).
Multiply the capacitor current of 110.8 × 1.35 = 149.58 rounded up to 150 amps.
B – Reference [Section 220.17] lists a demand factor of 0.75 for four or more appliances.
8 ranges × 3.5 kW each = 28 × 0.75 demand factor = 21 kVA
D – Reference Chapter 9, Annex D, Example D3
A – Balance the load by placing 4 fixtures and 2 fans on one circuit and 8 fixtures and one fan on the other as shown below:
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