# Audel electrical course for apprentices and journeymen - Rosenberg P.

ISBN: 0-764-54200-1

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Refer to the elevator problem (p. 000) and convert it into electrical power

Problem

What horsepower of electric motor will it take to lift 1000 lbs a height of 100 ft in 30 seconds? Use an 80 percent efficiency factor. How many kilowatts will be required to supply this horsepower?

™ u • i l L X W 100 X 1000 Mechanical hp = 55^X330 = 16,500 = 6 06 hp

el=if=76 hp Ans- (1)

1 hp = 0.746 kW so 7.56 X 0.746 = 5.64 kW

Ans. (2)

The following are useful formulas to be used with efficiency: Output Output Input — Losses

Efficiency

Input

Input Output + Losses Input

Output

Efficiency Output = Input X Efficiency

The following formulas are listed for reference:

Watts

hp

746

Work, Power, Energy,Torque, and Efficiency 155

Watts = hp X 746 kW

hp = 0746 kW = hp X 0.746

Problem

How many horsepower are 2460 watts?

Watts 2460 „ „ ,

hp = -74T = ~746 = 3'3 hp

Problem

A motor draws 30 kW. How many horsepower is this?

kW 30 hp = 0.746 = 0/746 = 402 hp

hp = kW X 1.34 = 30 X 1.34 = 40.2 hp

Torque

Applied torque is a measure of a body’s tendency to produce rotation. Resisting torque is the tendency of a body to resist rotation.

A definition of torque is a twisting or turning force that tends to produce rotation, as of a motor.

Torque is measured by the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force:

T = F X L where

T = torque turning or twisting efforts in pounds-feet F = unbalanced force exerted to produce rotation L = lever arm length

Torque is usually expressed in pound-feet. Be careful to observe that this is not foot-pounds.

An example of torque is illustrated in Figure 13-2. The torque tending to turn the cylinder in the brick wall would be T = F X L or T = 100 lb X 12 ft X 1200 pound-feet.

156 Chapter 13

Prony Brake

The Prony brake is used for finding the brake horsepower of all types of engines and motors. In running the test, the brake lever is arranged as in Figure 13-3 to bear on a scale S. The pressure F on the scale is regulated by bolt R. The revolutions per minute of the wheel are counted. Here, L is the length of the lever arm. The brake horsepower is calculated as follows:

2 X L X 3.1416 X F X rpm

Brake Horsepower =---------------- ------------

J J ^ \J\J\J

Notice that the product FL in the numerator is the torque of the system.

In electrical units, a watt-hour represents the electrical energy expended if work is done for 1 hour at the rate of 1 watt. By the same token, a kilowatt-hour represents the electrical energy expended if work is done for 1 hour at the rate of 1 kilowatt.

1 joules = 1 watt-second 3600 joules = 1 watt-hour

3,600,000 joules = 1 kilowatt-hour

Electrical energy is purchased and metered on the kilowatt-hour basis. The rates are usually based on a sliding scale. That is, the first so many kilowatt-hours are at a higher rate than the ensuing

steps will be. Also, on commercial and industrial power customer

rates, the maximum 15-minute demand in continuous kilowatts will set the steps used in calculating the power bill. The reason for this is that if a commercial customer used a continuous demand of

Work, Power, Energy,Torque, and Efficiency 157

FRICTION BAND ON PULLEY OR FLYWHEEL

Figure 13-3 Example of a Prony brake for measuring brake horsepower.

10,000 kW, the load on the system would be steady, but if the customer drew 10,000 kW for 15 minutes or an hour, and then dropped to 5000 kW the rest of the time, the power company would have to have a system that could supply the 10,000-kW demand, while the customer used only 5000 kW the majority of the time.

The power company is entitled to a ready-to-serve charge, because we want them to be ready to supply whatever the required demand at any time. The same is true if a company has its own power plant. The investment to serve must be there regardless of whether power is utilized continuously or for only a short time.

Heat may be obtained from electrical energy, both as a loss and as work done. Heat energy is expressed in two ways: calories and Btu (British thermal units).

The calorie is the amount of heat required to raise the temperature of 1 gram of water by 1 Celsius degree.

The Btu is the amount of heat required to raise the temperature of 1 pound of water by 1 Fahrenheit degree.

Going back to Table 13-1, we find that 1 kW will produce

0.9480 Btu per second.

A current of 1 ampere maintained for 1 second in a 1-ohm resistor produces 0.239 calorie of heat. A current of 2 amperes in the same circuit produces four times the calories (four times the heat).

158 Chapter 13

Multiply 0.239 by the square of the current (in amperes), by the resistance (in ohms), and by the time (in seconds): The result is heat in calories (H):

H = 0.239 I2Rt Calories = 0.239 X amperes2 X ohms X seconds Btu = 0.000948 X amperes2 X ohms X seconds

Formulas

Work

hp

Efficiency

Input

Output

hp

W X L

L X W

33,000 X t (in minutes) Output Output

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