If d = 150m and the motor must deliver its full-rated power, determine the minimum AWG conductors that must be used in a rubber-insulated cable. Assume that losses occur only in the wires.

At an engineering site, a 1-hp motor is placed a distance d from a portable generator, as depicted in Figure P2.43. The generator can be modeled as an ideal DC source VG = 110V. The nameplate on the motor gives the following rated voltages and full-load currents:



VM min = 105 V ? IM FL = 7.10 A

VM max = 117 V ? IM FL = 6.37 A





What will be an ideal response?


Known quantities:

Figure P2.43.

VM-min = 105V -> IM-FL = 7.10A

VM-max = 117V -> IM-FL = 6.37A

VG = 110V

Find:

The minimum AWG wire that will meet the specifications.

Assumptions:

Assume an ideal DC source.

Analysis:

The circuit that this setup is describing is essentially a voltage source with a resistor, an inductor, and another resistor all in series.



150m is equal to 492.13 feet. This will help with the resistance later.



Assume the voltage of the motor is 105V. Use Ohm’s Law to calculate the Current through the first conductor:



IM has to be 7.1 A or greater in order for the motor to draw full power:



Use the value of R along with the chart to find the correct AWG:





This means that 8 AWG wire must be used.

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