Basic Motor Theory (8)

DC Machines, Principles Of Operation Generator

In a generator, moving a conductor through a stationary magnetic field generates voltage. If a coil is rotated through a magnetic field as shown in Figure 4, an alternating voltage will be produced. To make this voltage available to a stationary external circuit, two slip rings and brushes must be provided. For the external circuit to produce DC voltage, it is necessary to reverse the polarity of the external leads at the same time the voltage in the coil is reversed. This is accomplished by segmenting a slip ring to form what is called a commutator. An elementary two segment commutator is illustrated in Figure 5. This single coil, two piece commutator will yield an unidirectional but pulsating voltage as shown in Figure 6. However, when a large number of commutator segments or bars is used, the resulting voltage will be more uniform as shown in Figure 7.

Figure 4. Brushes and slip rings provide AC voltage


Figure 5. Brushes and Commutator provides DC voltage

Figure 6. Unidirectional, Pulsating Voltage

Figure 7. Uniform DC Voltage

As stated above, the generated voltage in a single conductor is:
E = N B v x 10-8
where:
B = flux density in lines per square inch
= length of the conductor in inches
v = velocity in inches per second
This equation can be developed to the following equation for DC machines:
E = (Z / paths) x x poles x (rpm / 60) x 10-8
where:
Z = total number of conductors
= flux per pole in lines
This equation represents the average voltage. For a given machine, it can be reduced to:
E = K1 S
where:
= flux per pole
S = speed in rpm
K1 = all other factors Motor
As stated previously, if current is supplied to a conductor in a magnetic field, a force will be produced. The force developed in a single conductor is:
F = (B I) / 10
where:
F = force in dynes
B = flux density in lines per square centimeter
= length of the conductor in centimeters
I = current in amperes
This equation can be developed to the following for DC motors:
T = 11.73 x (Z / paths) x x poles x IA x 10-10
where:
T = torque in ft-lb
Z = total number of conductors
= flux per pole in lines
I = current in amperes
For a given machine, this can be reduced to:
T = K2 IA
where:
= flux per pole in lines
IA = current in amperes
K2 = all other factors
K2 is not the same as the K1 for voltage. The above torque is not the output torque of the shaft, but rather the total torque developed by the armature. Part of this total torque is needed to overcome the inertia of the armature itself.
The horsepower output of any motor can be expressed as:
HP = T x N / C
where:
T = output torque in ft-lb
N = speed in rpm
C = the constant 5252