Basic Motor Theory (15) – Finish

Losses And Efficiency

Friction and Windage
These losses include bearing friction, brush friction, and windage. They are also known as mechanical losses. They are constant at a given speed but vary with changes in speed. Power losses due to friction increase as the square of the speed and those due to windage increase as the cube of the speed.

Armature Copper Losses
These are the I2 R losses of the armature circuit, which includes the armature winding, commutator, and brushes. They vary directly with the resistance and as the square of the currents.

Field Copper Losses
These are the I2 R losses of the field circuit which can include the shunt field winding, series field winding, interpole windings and any shunts used in connection with these windings. They vary directly with the resistance and as the square of the currents.

Core Losses
These are the hysteresis and eddy current losses in the armature. With the continual change of direction of flux in the armature iron, an expenditure of energy is required to carry the iron through a complete hysteresis loop. This is the hysteresis loss. Also since the iron is a conductor and revolving in a magnetic field, a voltage will be generated. This, in turn, will result in small circulating currents known as eddy currents. If a solid core were used for the armature, the eddy current losses would be high. They are reduced by using thin laminations, which are insulated from each other. Hysteresis and eddy current losses vary with flux density and speed.

For generations or motors, the efficiency is equal to the output divided by the input. However, in a generator, the input is mechanical while the output is electrical. In a motor the opposite is true, therefore:
Motor Efficiency = (Input – Losses) / Input
Generator Efficiency = Output / (Output + Losses)

Section 3: Horsepower Basics

In 18th century England, coal was feeding the industrial revolution and Thomas Newcomen invented a steam driven engine that was used to pump water from coal mines. It was a Scott however, by the name of James Watt, who in 1769 improved the steam engine making it truly workable and practical. In his attempt to sell his new steam engines, the first question coal mine owners asked was “can your engine out work one of my horses?” Watt didn’t know since he didn’t know how much work a horse could do. To find out, Watt and his partner bought a few average size horses and measured their work. They found that the average horse worked at the rate of 22,000 foot pounds per minute. Watt decided, for some unknown reason, to add 50% to this figure and rate the average horse at 33,000 foot pounds per minute.
What’s important is that there is now a system in place for measuring the rate of doing work. And there is a unit of power, horsepower.
If steam engines had been developed some place else in the world, where the horse was not the beast of burden, we might be rating motors in oxen power or camel power. Today, motors are also rated in Watts output.
hp = lb x fpm / 33,000
hp = ft-lb x rpm / 5,252
kW = hp x 0.7457
hpMetric = hp x 1.0138
Horsepower as defined by Watt, is the same for AC and DC motors, gasoline engines, dog sleds, etc.

Horsepower and Electric Motors
Torque = force x radius = lb x ft = T
Speed = rpm = N
Constant = 5252 = C
HP = T x N / C
Torque and DC Motors
T = k Ia

At overload, torque increases at some rate less than the increase in current due to saturation

D2 L and Torque
258AT = 324 D2 L
259AT = 378 D2 L
With the same frame diameter, the 259AT has 17% more D2 L and thus 17% more and 17% more Torque. Motor torque increases with an increase in iron and copper, combined with current. It can then be said that it takes iron and copper to produce torque and torque makes products. Or to put it another way, what you purchase to make product is TORQUE and that is IRON and COPPER. The rate of doing work is power and HORSEPOWER is a unit of power.

Speed and DC Motors

Shunt wound DC motors
With motor load, temperature and field current held constant, speed is controlled by armature voltage.
E = ((Z / a) x x P x (N / 60) x 10-8 ) + (I Ra + I Rip + I Rb )

The sum of the voltage drop in the armature circuit can be represented as IR
N = (E – IR) / K
Speed example: given motor is design G6219, frame MC3212, 50 hp, 1150 rpm, 500 volt armature, 85 amps full load, 0.432 armature circuit resistance hot, 0.206 armature circuit resistance cold
Edrop = IR = 85 amp x 0.432 = 36.72 volts

500 v arm – 36.72 v drop = 463.28 working volts
Volts per rpm = 463.28 / 1150 rpm = 0.40285
Nbase speed = 1150 rpm = (500 v – 36.72 v) / 0.40285
With 250 v on the armature, there is 213.28 working volts (250 – 36.72)
213.28 / 0.40285 = 529 rpm (not 1/2 speed, 575 rpm)
N = 529 rpm = (250 v – 36.72 v) / 0.40285
N = (E – IR) / K = (E – IR) / 0.40285
K changes with changes in load and temperature
HPMetric = HP x 1.0138
kW = HP x 0.7457



2 Responses to “Basic Motor Theory (15) – Finish”

  1. Evan silas Says:

    I really like this post.i really need your post on electrical machine for my course exam

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