Question 10.11: The next problem concerns conductors used in a three-phase s......

The solution to this problem is very similar to the solution in the previous example. First, find the maximum voltage drop that can be permitted at the load by multiplying the source voltage by 5%:

\mathrm{E}=560 \mathrm{~V} \times 0.05

\mathrm{E}=28 \mathrm{~V}

The second step is to determine the maximum amount of resistance of the conductors. To compute this value, the maximum voltage drop is divided by the starting current of the motor.

\mathrm{R}=\frac{\mathrm{E}}{\mathrm{I}}

R=\frac{28 ~V}{168 ~A}

R=0.167 ~\Omega

The next step is to compute the length of the conductors. In the previous example, the lengths of the two conductors were added to find the total amount of wire resistance. In a single-phase system, each conductor must carry the same amount of current. During any period of time, one conductor is supplying current from the source to the load, and the other conductor completes the circuit by permitting the same amount of current to flow from the load to the source.

In a balanced three-phase circuit, three currents are 120° out of phase with each other (Figure 10 –13). These three conductors share the flow of current between source and load. In Figure 10 –13, two lines labeled A and B have been drawn through the three current waveforms. Notice that at position A the current flow in phase 1 is maximum and in a positive direction. The current flow in phases 2 and 3 is less than maximum and in a negative direction. This condition corresponds to the example shown in Figure 10 –14. Notice that maximum current is flowing in only one conductor. Less than maximum current is flowing in the other two conductors.

Observe the line marking position B in Figure 10 –13. The current flow in phase 1 is zero, and the currents flowing in phases 2 and 3 are in opposite directions and less than maximum. This condition of current flow is illustrated in Figure 10 –15. Notice that only two of the three phase lines are conducting current and that the current in each line is less than maximum.

Since the currents flowing in a three-phase system are never maximum at the same time, and at other times the current is divided between two phases, the total conductor resistance will not be the sum of two conductors. To compute the resistance of conductors in a three-phase system, a demand factor of 0.866 is used.

In this problem, the motor is located 2500 ft from the source. The conductor length is computed by doubling the length of one conductor and then multiplying by 0.866:

\mathrm{L}=2500 ~\mathrm{ft} \times 2 \times 0.866

\mathrm{L}=4330~ \mathrm{ft}

Now that all the factors are known, the size of the conductor can be computed using the formula:

\mathrm{CM}=\frac{\mathrm{K} \times \mathrm{L}}{\mathrm{R}}

where

\mathrm{K}=\frac{17~ \Omega-\mathrm{CM}}{\mathrm{ft}}

\mathrm{I}=4300~ \mathrm{ft}

R=0.167~ \Omega

\mathrm{CM}=\frac{17~ \Omega-\mathrm{CM} / \mathrm{ft} \times 4330 ~\mathrm{ft}}{0.167 ~\Omega}

\mathrm{CM}=\frac{73,601~ \Omega-\mathrm{CM}}{0.167 ~\Omega}

\mathrm{CM}=440,724.551

Three 500-kcmil conductors will be used.