Electrical Systems - Manufacturer Requirements

EXAMPLES

1. A circuit in a boat consists only of a 12 volt bilge pump, rated at 12 amps. The bilge pump and its wiring are in the engine compartment and the conductors will run in a bundle of 8 conductors except where they break out of the bundle to go to the pump, a distance of 40 inches from the bundle.

FACTS

Circuit Load	12 amps
Location	Engine Space
Bundling	8 conductors
Voltage	Less than 50 volts

CONDUCTOR

The boat builder decides to use 75° C temperature rated conductors for the bilge pump.

ENGINE SPACE CORRECTION

The factor for 75° C conductors is 0.75.

CALCULATION

In Table 5, under 75° C, a 16 AWG conductor will carry 15 amps; however, when corrected for use in an engine space, it may only carry:

0.75 x 15 = 11.25 amps therefore 16 AWG conductors are too small.

Try 14 AWG:

0. 75 x 20 = 15 amps 14 AWG, 75° C conductors may be used.

Tables 5A through 5E may be used in lieu of these calculations.

NOTE:
Bundling corrections DO NOT apply to conductors used in circuits less than 50 volts.

2. A lighting and receptacle circuit is planned for the 120 volt AC system on a boat. The maximum permanent lighting load is known to be 5 amps. The receptacle load is unknown but the receptacles are rated at 15 amps. The conductors go through the engine space in a bundle of 8 conductors, 3 of which are DC and 5 of which are current carrying AC conductors.

FACTS

CONDUCTOR

The boat builder decides to use 75° C temperature rated conductors for this lighting and receptacle circuit.

ENGINE SPACE CORRECTION

The factor for 75° C conductors is 0.75. The correction factor applies to the length of the conductors used in the engine space. The lengths outside the engine space need no correction. If a conductor runs inside and outside an engine space, it must be sized for the engine space.

CIRCUIT LOAD

Since the circuit load is unknown, the size of the circuit protection device will determine the anticipated load. In this case, 20 amps was selected.

CALCULATION

In Table 5, under 75° C, a 12 AWG conductor will carry 25 amps. Corrected for engine spaces it may only carry:

0.75 x 25 = 18.75 amps 12 AWG, 75° C conductors are too small.

Try 10 AWG

0. 75 x 40 = 30 amps

Bundling Correction - Even though there are 8 conductors in the bundle, only 5 are current carrying in circuits of 50 volts or more. Therefore, the correction factor from Note 2 used is for bundles of 4 to 6 conductors, or 0.60.

Try 10 AWG, corrected for engine spaces:

0.60 x 30 = 18 amps 10 AWG 75° C conductors are too small.

Try 8 AWG

Correct for engine spaces: 0.75 x 65 = 48.75
Correct for bundling: 0.60 x 48.75 = 29.25

8 AWG, 75° C conductors may be used in the engine spaces. Please note that outside the engine spaces, 10 AWG conductors may be used, as the following bundling calculation shows:
0.60 x 40 = 24 amps

Conductors with a higher insulation temperature rating could be used to reduce the size of the conductor. (Table 5C may be used in lieu of these calculations).

3. For the same example used in (c)(2) above, there is another method for determining proper conductor size. Again, use a 20 ampere load and a 75° C rated conductor. This method avoids the iterative process used in examples (1) and (2). Simply divide the load, 20 amperes, by the correction factor for engine space:

CALCULATION

Then divide the result (26.67 amperes) by the correction factor for bundling:

Enter the 75° C column of Table 5 and select the first conductor size whose allowable amperage exceeds 44.44 amperes. In this case, a 10 AWG conductor is allowed to carry 40 amperes, which is not enough. An 8 AWG conductor can carry 65 amperes. The selection will have to be an 8 AWG conductor or larger. If an 80° C rating is used, then a 10 AWG conductor would be adequate. (Table 5C may be used in lieu of these calculations.)

TABLE 6 - Flexible Cords and Cables (from Article 400, NEC)