Motor- And Drive Unit Info

http://www.reliance.com/mtr/flaclcmn.htm

Basic Motor Formulas And CalculationsThe formulas and calculations which appear below should be used for estimating purposes only. It is the responsibility of the customer to specify the required motor Hp, Torque, and accelerating time for his application. The salesman may wish to check the customers specified values with the formulas in this section, however, if there is serious doubt concerning the customers application or if the customer requires guaranteed motor/application performance, the Product Department Customer Service group should be contacted.

Rules Of Thumb (Approximation)

At 1800 rpm, a motor develops a 3 lb.ft. per hpAt 1200 rpm, a motor develops a 4.5 lb.ft. per hpAt 575 volts, a 3-phase motor draws 1 amp per hpAt 460 volts, a 3-phase motor draws 1.25 amp per hpAt 230 volts a 3-phase motor draws 2.5 amp per hpAt 230 volts, a single-phase motor draws 5 amp per hpAt 115 volts, a single-phase motor draws 10 amp per hp

Mechanical Formulas

Torque in lb.ft. =HP x 5250

rpm

HP =

Torque x rpm

5250rpm =

120 x Frequency

No. of Poles

Temperature Conversion

Deg C = (Deg F - 32) x 5/9Deg F = (Deg C x 9/5) + 32

High Inertia Loads

t =WK2 x rpm

308 x T av. WK2 = inertia in lb.ft.2

t = accelerating time in sec.T = Av. accelerating torque lb.ft..

T =

WK2 x rpm

308 x t

inertia reflected to motor = Load Inertia Load rpm 2

Motor rpm

Synchronous Speed, Frequency And Number Of Poles Of AC Motors

ns =120 x f

Pf =

P x ns

120P =

120 x f

ns

Relation Between Horsepower, Torque, And Speed

HP =T x n

5250T =

5250 HP

n

n =

5250 HP

T

Motor Slip

% Slip =ns - n

ns

x 100

Code KVA/HP Code KVA/HP Code KVA/HP Code KVA/HP

A 0-3.14 F 5.0 -5.59 L 9.0-9.99 S 16.0-17.99

B 3.15-3.54 G 5.6 -6.29 M 10.0-11.19 T 18.0-19.99

C 3.55-3.99 H 6.3 -7.09 N 11.2-12.49 U 20.0-22.39

D 4.0 -4.49 I 7.1 -7.99 P 12.5-13.99 V 22.4 & Up

E 4.5 -4.99 K 8.0 -8.99 R 14.0-15.99

Symbols

I = current in amperes

E = voltage in volts

KW = power in kilowatts

KVA = apparent power in kilo-volt-amperes

HP = output power in horsepower

n = motor speed in revolutions per minute (RPM)

ns = synchronous speed in revolutions per minute (RPM)

P = number of poles

f = frequency in cycles per second (CPS)

T = torque in pound-feet

EFF = efficiency as a decimal

PF = power factor as a decimal

Equivalent Inertia

In mechanical systems, all rotating parts do not usually operate at the same speed. Thus, we need to determine the "equivalent inertia" of each moving part at a particular speed of the prime mover.

The total equivalent WK2 for a system is the sum of the WK2 of each part, referenced to prime mover speed.

The equation says:

WK2EQ = WK2

part

Npart

Nprime mover

2

This equation becomes a common denominator on which other calculations can be based. For variable-speed devices, inertia should be calculated first at low speed.

Let's look at a simple system which has a prime mover (PM), a reducer and a load.

WK2 = 100 lb.ft.2 WK2 = 900 lb.ft.2

(as seen at output shaft)WK2 = 27,000 lb.ft.2

PRIME MOVER 3:1 GEAR REDUCER LOAD

The formula states that the system WK2 equivalent is equal to the sum of WK2parts at the

prime mover's RPM, or in this case:

WK2EQ = WK2

pm + WK2Red.

Red. RPM

PM RPM2 + WK2

Load

Load RPM

PM RPM2

Note: reducer RPM = Load RPM

WK2EQ = WK2

pm + WK2Red.

1

32 + WK2

Load

1

32

The WK2 equivalent is equal to the WK2 of the prime mover, plus the WK2 of the load. This is equal to the WK2 of the prime mover, plus the WK2 of the reducer times (1/3)2, plus the WK2 of the load times (1/3)2.

This relationship of the reducer to the driven load is expressed by the formula given earlier:

WK2EQ = WK2

part

Npart

Nprime mover

2

In other words, when a part is rotating at a speed (N) different from the prime mover, the WK2

EQ is equal to the WK2 of the part's speed ratio squared.

In the example, the result can be obtained as follows:

The WK2 equivalent is equal to:

WK2EQ = 100 lb.ft.2 + 900 lb.ft.2

1

32 + 27,000 lb.ft.2

1

32

Finally:

WK2EQ = lb.ft.2

pm + 100 lb.ft.2Red + 3,000 lb.ft2

Load

WK2EQ = 3200 lb.ft.2

The total WK2 equivalent is that WK2 seen by the prime mover at its speed.

Electrical Formulas

To FindAlternating Current

Single-Phase Three-Phase

Amperes when horsepower is knownHP x 746

E x Eff x pf

HP x 746

1.73 x E x Eff x pf

Amperes when kilowatts are knownKw x 1000

E x pf

Kw x 1000

1.73 x E x pf

Amperes when kva are knownKva x 1000

E

Kva x 1000

1.73 x E

KilowattsI x E x pf

1000

1.73 x I x E x pf

1000

KvaI x E

1000

1.73 x I x E

1000

Horsepower = (Output)I x E x Eff x pf

746

1.73 x I x E x Eff x pff

746

I = Amperes; E = Volts; Eff = Efficiency; pf = Power Factor; Kva = Kilovolt-amperes; Kw = Kilowatts

Locked Rotor Current (IL) From Nameplate Data

Three Phase: IL =577 x HP x KVA/HP

ESee: KVA/HP Chart

Single Phase: IL =1000 x HP x KVA/HP

E

EXAMPLE: Motor nameplate indicates 10 HP, 3 Phase, 460 Volts, Code F.

IL =577 x 10 x (5.6 or 6.29)

460

IL = 70.25 or 78.9 Amperes (possible range)

Effect Of Line Voltage On Locked Rotor Current (IL) (Approx.)

IL @ ELINE = IL @ EN/P xELINE

EN/P

EXAMPLE: Motor has a locked rotor current (inrush of 100 Amperes (IL) at the rated nameplate voltage (EN/P) of 230 volts.

What is IL with 245 volts (ELINE) applied to this motor?

IL @ 245 V. = 100 x 254V/230V

IL @ 245V. = 107 Amperes Basic Horsepower Calculations

Horsepower is work done per unit of time. One HP equals 33,000 ft-lb of work per minute. When work is done by a source of torque (T) to produce (M) rotations about an axis, the work done is:

radius x 2 x rpm x lb. or 2 TM

When rotation is at the rate N rpm, the HP delivered is:

HP =radius x 2 x rpm x lb.

33,000=

TN

5,250

For vertical or hoisting motion:

HP =

W x S

33,000 x E

Where:

W = total weight in lbs. to be raised by motor

S = hoisting speed in feet per minute

E = overall mechanical efficiency of hoist and gearing. For purposes of estimating

E = .65 for eff. of hoist and connected gear.

For fans and blowers:

HP =

Volume (cfm) x Head (inches of water)

6356 x Mechanical Efficiency of Fan

Or

HP =Volume (cfm) x Pressure (lb. Per sq. ft.)

3300 x Mechanical Efficiency of Fan

Or

HP =

Volume (cfm) x Pressure (lb. Per sq. in.)

229 x Mechanical Efficiency of Fan

For purpose of estimating, the eff. of a fan or blower may be assumed to be 0.65.

Note: Air Capacity (cfm) varies directly with fan speed. Developed Pressure varies with square of fan speed. Hp varies with cube of fan speed.

For pumps:

HP =

GPM x Pressure in lb. Per sq. in. x Specific Grav.

1713 x Mechanical Efficiency of Pump

Or

HP =GPM x Total Dynamic Head in Feet x S.G.

3960 x Mechanical Efficiency of Pump

where Total Dynamic Head = Static Head + Friction Head

For estimating, pump efficiency may be assumed at 0.70.

Accelerating Torque

The equivalent inertia of an adjustable speed drive indicates the energy required to keep the system running. However, starting or accelerating the system requires extra energy.

The torque required to accelerate a body is equal to the WK2 of the body, times the change in RPM, divided by 308 times the interval (in seconds) in which this acceleration takes place:

ACCELERATING TORQUE =WK2N (in lb.ft.)

308t

Where:

N = Change in RPM

W = Weight in Lbs.

K = Radius of gyration

t =Time of acceleration (secs.)

WK2 = Equivalent Inertia

308 = Constant of proportionality

Or

TAcc =WK2N

308t

The constant (308) is derived by transferring linear motion to angular motion, and considering acceleration due to gravity. If, for example, we have simply a prime mover and a load with no speed adjustment:

Example 1

PRIME LOADER LOAD

WK2 = 200 lb.ft.2 WK2 = 800 lb.ft.2

The WK2EQ is determined as before:

WK2EQ = WK2

pm + WK2Load

WK2EQ = 200 + 800

WK2EQ = 1000 ft.lb.2

If we want to accelerate this load to 1800 RPM in 1 minute, enough information is available to find the amount of torque necessary to accelerate the load.

The formula states:

TAcc =WK2

EQN

308t

or

1000 x 1800

308 x 60or

1800000

18480

TAcc = 97.4 lb.ft.

In other words, 97.4 lb.ft. of torque must be applied to get this load turning at 1800 RPM, in 60 seconds.

Note that TAcc is an average value of accelerating torque during the speed change under consideration. If a more accurate calculation is desired, the following example may be helpful.

Example 2

The time that it takes to accelerate an induction motor from one speed to another may be found from the following equation:

t =

WR2 x change in rpm

308 x T

Where:

T = Average value of accelerating torque during the speed change under consideration.

t = Time the motor takes to accelerate from the initial speed to the final speed.

WR2 = Flywheel effect, or moment of inertia, for the driven machinery plus the motor rotor in lb.ft.2 (WR2 of driven machinery must be referred to the motor shaft).

The Application of the above formula will now be considered by means of an example. Figure A shows the speed-torque curves of a squirrel-cage induction motor and a blower which it drives. At any speed of the blower, the difference between the torque which the motor can deliver at its shaft and the torque required by the blower is the torque available for acceleration. Reference to Figure A shows that the accelerating torque may vary greatly with speed. When the speed-torque curves for the motor and blower intersect there is no torque available for acceleration. The motor then drives the blower at constant speed and just delivers the torque required by the load.

In order to find the total time required to accelerate the motor and blower, the area between the motor speed-torque curve and the blower speed-torque curve is divided into strips, the ends of which approximate straight lines. Each strip corresponds to a speed increment which takes place within a definite time interval. The solid horizontal lines in Figure A represent the boundaries of strips; the lengths of the broken lines the average accelerating torques for the selected speed intervals. In order to calculate the total acceleration time for the motor and the direct-coupled blower it is necessary to find the

time required to accelerate the motor from the beginning of one speed interval to the beginning of the next interval and add up the incremental times for all intervals to arrive at the total acceleration time. If the WR2 of the motor whose speed-torque curve is given in Figure A is 3.26 ft.lb.2 and the WR2 of the blower referred to the motor shaft is 15 ft.lb.2, the total WR2 is:

15 + 3.26 = 18.26 ft.lb.2,

And the total time of acceleration is:

WR2

308

rpm1

T1

+rpm2

T2

+rpm3

T3

+ - - - - - - - - - +rpm9

T9

Or

t =

18.26

308

150

46+

150

48+

300

47+

300

43.8+

200

39.8+

200

36.4+

300

32.8+

100

29.6+

40

11

t = 2.75 sec.

Figure ACurves used to determine time required to accelerate induction motor and blower

Accelerating Torques

T1 = 46 lb.ft. T4 = 43.8 lb.ft. T7 = 32.8 lb.ft.

T2 = 48 lb.ft. T5 = 39.8 lb.ft. T8 = 29.6 lb.ft.

T3 = 47 lb.ft. T6 = 36.4 lb.ft. T9 = 11 lb.ft.

Duty Cycles

Sales Orders are often entered with a note under special features such as:

"Suitable for 10 starts per hour"Or

"Suitable for 3 reverses per minute"Or

"Motor to be capable of accelerating 350 lb.ft.2"Or

"Suitable for 5 starts and stops per hour"

Orders with notes such as these can not be processed for two reasons.

1. The appropriate product group must first be consulted to see if a design is available that will perform the required duty cycle and, if not, to determine if the type of design required falls within our present product line.

2. None of the above notes contain enough information to make the necessary duty cycle calculation. In order for a duty cycle to be checked out, the duty cycle information must include the following:

a. Inertia reflected to the motor shaft. b. Torque load on the motor during all portions of the duty cycle

including starts, running time, stops or reversals. c. Accurate timing of each portion of the cycle. d. Information on how each step of the cycle is accomplished. For

example, a stop can be by coasting, mechanical braking, DC dynamic braking or plugging. A reversal can be accomplished by plugging, or the motor may be stopped by some means then re-started in the opposite direction.

e. When the motor is multi-speed, the cycle for each speed must be completely defined, including the method of changing from one speed to another.

f. Any special mechanical problems, features or limitations.

Obtaining this information and checking with the product group before the order is entered can save much time, expense and correspondence.

Duty cycle refers to the detailed description of a work cycle that repeats in a specific time period. This cycle may include frequent starts, plugging stops, reversals or stalls. These characteristics are usually involved in batch-type processes and may include tumbling barrels, certain cranes, shovels and draglines, dampers, gate- or plow-positioning drives, drawbridges, freight and personnel elevators, press-type extractors, some feeders,presses of certain types, hoists, indexers, boring machines,cinder block machines, keyseating, kneading, car-pulling, shakers (foundry or car), swaging and washing machines, and certain freight and passenger vehicles. The list is not all-inclusive. The drives for these loads must be capable of absorbing the heat generated during the duty cycles. Adequate thermal capacity would be required in slip couplings, clutches or motors to accelerate or plug-stop these drives or to withstand stalls. It is the product of the slip speed and the torque absorbed by the load per unit of time which generates heat in these drive components. All the events which occur during the duty cycle generate heat which the drive components must dissipate.

Because of the complexity of the Duty Cycle Calculations and the extensive engineering data per specific motor design and rating required for the calculations, it is necessary for the sales engineer to refer to the Product Department for motor sizing with a duty cycle application.

Last Updated September 1, 1998

http://en.wikipedia.org/wiki/Torque

Conversion to other units

A conversion factor may be necessary when using different units of power, torque, or angular speed. For example, if rotational speed (revolutions per time) is used in place of angular speed (radians per time), we multiply by a factor of 2π radians per revolution.

Adding units:

Dividing on the left by 60 seconds per minute and by 1000 watts per kilowatt gives us the following.

where rotational speed is in revolutions per minute (rpm).

Copyright ©2007, Baldor Electric Company. All Rights Reserved.

Some people (e.g. American automotive engineers) use horsepower (imperial mechanical) for power, foot-pounds (lbf·ft) for torque and rpm for rotational speed. This results in the formula changing to:

The constant below in, ft·lbf/min, changes with the definition of the horsepower; for example, using metric horsepower, it becomes ~32,550.

Use of other units (e.g. BTU/h for power) would require a different custom conversion factor.

http://www.indrives.com/calculations/simplified_calculation.html

Simplified Calculation

Simplified Calculation method of a Hercules Drive SystemNote: This is a simplified way of calculation a Hercules Drive System and is intended to give a rough estimation of the general size of a drive system. Always contact Integrated Drives for a certified calculation.

GeneralInput

PowerOutput Power

Start to define max needed output torque and max needed output speed

Max output Power in kW

T = Torque in Nmn = Speed in rpm9550 = To get result in kWOpen Calculator in new window

Continue with average needed output torque and average needed output speed

Average output Power in kW

T = Torque in Nmn = Average Motor speed in rpm9550 = To get result in kWOpen Calculator in new window

Hercules Motor and Power unit:The Hercules motor concept is designed for an L10- life of 40 000 hours corresponding to 5 years of continuous 24 hours per day operation. A normal demand from process industries like Chemical, Pulp & Paper, Mineral Processing etc. Here we assume 40 000 hours at average torque and speed. We also assume that the working pressure for the motor is 200 bar at average torque.

Motor size in cc/rev

= Displacement of the motor in cc/revT = Average torque in Nm200 = 200 bar in working modeOpen Calculator in new window

Average Input Power in kW

T = Average torque in Nmn = Average speed in rpm9550 = To get result in kW0.8 = Estimated overall efficiency of 80%Open Calculator in new window

Pump flow needed at average speed in litre/min

Q = Litre/min= Displacement of the motor in cc/rev

n = Average speed in rpmOpen Calculator in new window

Max torque and speed: Usually the max torque is expected when starting. High starting friction, starting with full load, starting at low temperature etc. Often called "Brake away conditions". In the case of a Hercules Drive system, the prime mover always starts unloaded and double the output torque from the motor do not overload the prime mover. Consequently the prime mover do not need to be oversized because of starting conditions. In the case max speed and max torque occur at the same time, please contact Integrated Drives for assistance.

UNDERSTANDING TORQUE

Edward H. Cowern, P.E.

New England District Manager, Baldor Electric Company65 South Turnpike Road • Wallingford, Connecticut 06492

In the process of applying industrial drive products, we occasionally are misled into believing that we are applying horsepower. The real driving force is not horsepower, it is TORQUE. This paper is developed to give a deeper understanding of torque, its relationship to horsepower, and the types of loads we most frequently encounter.

INTRODUCTION

Torque is the twisting force supplied by a drive to the load. In most applications, a substantial amount of torque must be applied to the driven shaft before it will even start to turn. In the

Table 2 illustrates the very dramatic changes in horsepower requirements for relatively small changes in speeds that occur with “Variable Torque” loads.

% Speed Change

% Torque Change

% of Original HP

% HP Change

–20 –36 51 –49–15 –28 61 –39–10 –19 73 –27– 5 –10 86 –140 0 100 0

+5 +10 116 +16+10 +21 133 +33

English System, the standard units of torque as used in the power transmission industry are pound inches (lb. in.), or pound feet (lb. ft.) and, in some cases, for very low levels of torque, you will encounter ounce inches (oz. in.).

TORQUE BASICS

At some time, we have all had difficulty in removing the lid from a jar. The reason we have this trouble is simply that we are unable to supply adequate torque to the lid to break it loose. The solution to our dilemma may be to: 1) grit our teeth and try harder, 2) use a rubber pad, or cloth, to increase the ability to transmit torque without slippage, or 3) use a mechanical device to help multiply our torque producing capability. Failing on all of the above, we may pass the jar to someone stronger who can produce more torque.

If we were to wrap a cord around the lid and supply a force to the end of the cord through a scale, as shown in Figure 1, we could get the exact measurement of the torque it takes to loosen the lid.

The torque required would be the force as indicated on the scale., multiplied by the radius of the lid.

For example, if the indicated force on the scale at the time of “breakaway” was 25 lbs. and the radius of the lid was 1.5 inches, the torque required would have been:

T = 25 lbs. x 1.5 in. = 37.5 lb. inches

Although this example does give a reasonable illustration of “torque”, it does not represent a very common example of requirements on industrial equipment.

There is, however, one additional important point that can be derived from the jar and the lid example; namely “Sticksion”. “Sticksion” is a term generated to indicate the amount of torque

+15 +32 152 +52+20 +44 173 +73

Table 2Most variable speed drives are inherently capable of handling “Variable Torque” loads provided that they are adequately sized to handle the horsepower requirement at MAXIMUM speed.

HIGH INERTIA LOADS*

A discussion of load types would not be complete without including information on “High Inertia Loads”.

Inertia is the tendency of an object that is at rest to stay at rest or an object that is moving to keep moving.

In the industrial drive business, we tend to think immediately of flywheels as having high inertia; but, many other types of motor driven equipment, such as: large fans, centrifuges, extractors, hammer mills, and some types of machine tools, have inertias that have to be identified and analyzed in order to produce satisfactory applications.

*A load is generally considered to be “High Inertia” when the reflected inertia at the motor shaft is greater than five times the motor rotor inertia.

THE HIGH INERTIA PROBLEM

The high inertia aspect of a load normally has to be considered only during acceleration and deceleration. For example, if a standard motor is applied to a large high inertia blower, there is a possibility that the motor could be damaged or fail completely on its first attempt to start. This failure could occur even though the motor might have more than adequate torque and horsepower capacity to drive the load after it reaches the required running speed.

A good example of high inertia that most of us are familiar with is a jet plane taking off. In this case, the maximum output of the engines is required to accelerate the weight of the plane and contents. Only when it has reached take-off speed and is nearly ready to leave the ground do the engines start doing the useful work of moving the plane to the final destination.

Similarly, when the plane lands, the reversed thrust of the engines and the brakes are used

required to break a load loose on its way to making the first revolution.

Generally speaking, the breakaway torque requirement to start a machine will be substantially greater than that required to keep it running once it has started. The amount of “sticksion” present in a machine will be dependent on the characteristics of the machine as well as the type of bearings that are used on the moving parts.

Table I indicates typical values of breakaway torque for various general classifications of machinery.

BREAKAWAY & STARTING TORQUECHARACTERISTICS OF VARIOUS TYPES OF

LOADS

Torque% of RunningTorque

Types of Machines

BreakawayTorque

120% to 130%

General machines with ball or roller bearings

BreakawayTorque

130% to 160%

General machines with sleeve bearings

BreakawayTorque

160% to 250%

Conveyors and machines with excessive sliding friction

BreakawayTorque

250% to 600%

Machines that have "high" load spots in their cycle, such as some printing and punch presses, and machines with "cam" or "crank" operated mechanisms.Table 1

Assuming that the “sticksion”, or breakaway torque, has been overcome and the load has started, a continuing amount of torque must be supplied to handle the running torque requirements of the machine.

In a high percentage of industrial applications, the torque requirement of the load is independent of the speed at which the machine is driven. This type of load is generally called a “constant torque load”.

Constant torque loads will be used to introduce the basic concepts of horsepower. Additional load types will be introduced after the discussion of

to slow down and stop the inertia of the plane.

In the motor and drive industry, the inertia of a rotating body is referred to as the WR2 or WK2 . In the English System. “W” is the weight in pounds and “R” or “K” is the Radius of Gyration in feet. It is usually easy to obtain the weight of the body, but determining the radius of gyration can be a little more difficult. Figure 8 gives the formulas for determining the radius of gyration and WR2 of two frequently occurring cylindrical shapes.

In most cases, the WR2 of flywheels can be determined by utilizing one, or both, of these normal shapes. In the case of flywheels having spokes, the contribution made by the spokes can generally be ignored and the inertia calculation based only on the formula for a Hollow Circular Cylinder as shown in Figure 8. The weight of the spokes should be included. If exact calculations are required, formulas are available to enable the calculation of WR2 values of nearly any shape.

In most cases, equipment manufacturers will be able to provide the exact inertia values for a given application.

Motor manufacturers can be asked to supply the maximum WK2 limits for any specific application requirement. (Please note WK2 and WR2 are used interchangeably and they are the same).

The values shown in Table 3 are published in NEMA (National Electrical Manufacturers Association) standards MG-1. This table gives

horsepower.

HORSEPOWER

Many years ago, the invention of the steam engine made it necessary to establish a unit of measurement that could be used as a basis for comparison for how much work could be done by an engine. The unit that was chosen was related to the animal that was to be replaced by the new sources of power — the horse.

After a great deal of testing, it was found that the average work-horse could accomplish work at a rate equal to 33,000 ft. lbs. in one minute. This would be equal to lifting 1 ton (2,000 lbs.) 16.5 feet, or 1,000 lbs., 33 feet in one minute.

This unit, once established, has become the Western Hemisphere’s standard for measuring the rate at which motors and other drives can produce work. For example, a 1 H.P. motor can produce 33,000 ft. lbs. of work in one minute.

Torque and horsepower are related to each other by a basic formula which states that:

Horsepower = (Torque x Speed)/Constant

The value of the constant changes depending upon the units that are used for torque. The most frequently used combinations are as follows:

HP = (T X S)/5252

T = TORQUE IN LB. FT.S = SPEED IN RPM

OR

a listing of the normal maximum values of WK2 that could be safely handled by standard motors. This table can be used as a guide. If the required WK 2 exceeds these values, the motor manufacturer should be consulted. It is also important to note the details of paragraphs 1, 2, and 3 that are associated with this table. If the number of starts required or the method of starting is not “across the line”, the manufacturer should be consulted.

WHY IS HIGH INERTIA A PROBLEM?

Prior to the time that a standard induction motor reaches operating speed, it will draw line current several times the rated nameplate value. The high current does not cause any problem if it is of short duration; but, when the high currents persist for an extended period of time, the temperature within the motor can reach levels that can be damaging.

Speed, RPM

HP3600 1800 1200 900 720 600 514

Load WK 2 (Exclusive of Motor WK 2 ), Lb-Ft 2

1 — 5.8 15 31 53 82 1181.5 1.8 8.6 23 45 77 120 1742 2.4 11 30 60 102 158 2283 3.5 17 44 87 149 231 3355 5.7 27 71 142 242 375 544

7.5 8.3 39 104 208 356 551 79810 11 51 137 273 467 723 104815 16 75 200 400 685 1061 153820 21 99 262 525 898 1393 201825 26 122 324 647 1108 1719 249130 31 144 384 769 1316 2042 295940 40 189 503 1007 1725 2677 388150 49 232 620 1241 2127 3302 478860 58 275 735 1473 2524 3819 568075 71 338 904 1814 3111 4831 7010100 92 441 1181 2372 4070 6320 9180

125 113 542 1452 2919 5010 7790 11310

150 133 640 1719 3456 5940 9230 —

200 172 831 2238 4508 7750 — —

250 210 1017 2744 5540 — — —

300 246 1197 3239 — — — —

35 281 1373 3723 — — — —

HP = (T X S)/63,025

T = TORQUE IN LB. IN.S = SPEED IN RPM

OR

HP = (T X S)/1,000,000

T = TORQUE IN IN. OUNCES.S = SPEED IN RPM

Re-arranging these formulas to obtain torque, we can arrive at the equations:

T = (HP X 5252)/S

T = TORQUE IN LB. FT.S = SPEED IN RPM

OR

T = (HP X63,025)/S

T = TORQUE IN LB. FT.S = SPEED IN RPM

OR

T = (HP X 1,000,000)/S

T = TORQUE IN LB. FT.S = SPEED IN RPM

In order to save time, graphs and tables are frequently used to show values of torque, speed and horsepower.

The previous discussion applies to calculations for all single speed loads where the required torque and speed for a given operating condition are known.

VARIABLE SPEED DRIVES

When adjustable speed drives such as DC SCR units, magnetic couplings, or variable frequency drives are to be utilized, a determination of load type must be made.

As previously mentioned, the most common type of load is the “constant torque” load. The relationships of torque and horsepower to speed for a “constant” torque load is shown in Figure 3.

0400 315 1546 — — — — —

450 349 1714 — — — — —

500 381 1880 — — — — —

The table shown above lists the load WK2 which integral-horsepower polyphase squirrel-cage induction motors, having performance characteristics in accordance with Part 12*, can accelerate without injurious heating under the following conditions:

1. Applied voltage and frequency in accordance with 12.44.

2. During the accelerating period, a connected load torque equal to or less than a torque which varies as the square of the speed and is equal to 100 percent of rated-load torque at rated speed.

3. Two starts in succession (coasting to rest between starts) with the motor initially at the ambient temperature or one start with the motor initially at a temperature not exceeding its rated load operating temperature.

* Locked-rotor torque in accordance with 12.38.1, breakdown torque in accordance with 12.39.1, Class A or B insulation system with temperature rise in accordance with 12.43, and service factor in accordance with 12.51.2.

Figure 9 (a) shows typical plots of available torque from a standard motor vs. speed. Also plotted on curve (a) is the typical speed torque curve for a Variable Torque load. The values of A1, A2, A3, and A4 are the values of torque available to overcome the effect of the inertia

In the case of “constant torque” loads, the drive must be sized to handle:

1) The torque required to breakaway the load.

2) The torque required to run the load.

3) The output speed required to operate the machine at the maximum required speed.

Please note that only after the load has 1) been started and 2) adequate torque is available to run it, does speed become a factor.

Only after these three items have been determined, is it possible to calculate the required horsepower for the application.

Most variable speed drives are inherently “constant torque” devices; therefore, no special considerations are involved in handling “constant torque” loads.

CONSTANT HORSEPOWER

A load type that occurs most frequently in metal working applications, is the Constant Horsepower load.

On applications requiring constant horsepower, the torque requirement is greatest at the lowest speed and diminishes at higher speeds. In order to visualize this requirement, consider the torque requirements of a drill press, as shown in Figure 4.

and accelerate the load at different motor speeds as the motor speed increases.

Referring to Figure 9 (b), you will see that during the accelerating period this motor will draw line current that initially starts at 550% of rated current and gradually drops off as the motor approaches rated speed. A great deal of heat is generated within the motor during this high current interval. It is this heat build up that is potentially damaging to the motor if the acceleration interval is too long.

HOW LONG WILL IT TAKE?

Calculating the time to accelerate a direct coupled load can be determined quite easily by utilizing the following formula:

t = (WR2 x N)/308T

T = AVERAGE ACCELERATING TORQUE IN LB. FT.

N = REQUIRED CHANGE IN SPEED

WR2 = INERTIA IN LB. FT.2

t = TIME IN SECONDS

The same formula can be rearranged to determine the average accelerating torque required to produce full speed in a given period of time.

T = (WR2 x N)/308t

Referring back to Figure 9 (a), the accelerating torque would be the average value of the shaded area. In most cases, for standard motors through 100 HP, it is reasonable to assume that average accelerating torque available would be 150% of the motor full load running torque and that accelerating times of 8-10 seconds, or less, would not be damaging provided that starting is not repeated frequently. When load inertias exceed those shown in Table 4, the application should be referred to the motor supplier for complete analysis.

REFLECTED INERTIAS

Up to this point, the only load inertias that have been considered have been rotating inertias directly connected to the motor shaft.

When a large hole is being drilled, the drill is operated at a low speed, but it requires a very high torque to turn the large drill in the material.

When a small hole is being drilled, the drill is operated at a high speed, but it requires a very low torque to turn the small drill in the material.

A mathematical approach to this type of requirement would indicate that the HP requirement would be nearly constant regardless of the machine speed. Figure 5 shows the relationships of torque and horsepower to speed on constant horsepower loads.

As previously mentioned, this load type occurs most frequently on metal working applications, such as: drilling or boring, tapping, turning (lathes), planing, milling, grinding, wire drawing, etc. Center driven winders winding materials under constant tension also require constant horsepower. Constant horsepower can also be a requirement on some types of mixers.

On many applications, the load is connected to the motor by belts or a gear reducer. In these cases, the “Equivalent Inertia” or “Reflected Inertia” that is seen at the motor shaft is the important consideration.

In the case of belted or geared loads, the “Equivalent Inertia” is given by the following formula:

EQUIVALENT WR2 = WR2 LOAD [N/NM]2 x 1.1*

WR2 LOAD = INERTIA OF THE ROTATING PART

N = SPEED OF THE ROTATING PART

NM = SPEED OF THE DRIVING MOTOR

*Please note: the x 1.1 factor has been added as a safety factor to make an allowance for the inertia and efficiency of the pulleys (sheaves) or gears used in the speed change.

This formula will apply regardless of whether the speed of the load is greater than, or less than, the motor speed.

Once the equivalent inertia has been calculated, the equations for accelerating time, or required torque, can be solved by substituting the equivalent WR2 in the time or torque equation to be solved.

WHAT CAN BE DONE

When loads having high inertias are encountered, several approaches can be used. Some of the possibilities are:

1. Oversize the motor.

2. Use reduced voltage starting.

3. Use special motor winding design.

4. Use special slip couplings between the motor and load.

5. Oversize the frame.

LINEAR MOTION

Occasionally, applications arise where the load to be accelerated is traveling in a straight line

An example of this might be a food mixer used to mix a variety of batters and dough. In this case, dough would require low speed and high torque. Thin batters would require high speed and low torque. This is “Constant Horsepower”.

Spring coilers, fourslide machines, punch presses and eyeletting presses will frequently have torque requirements falling somewhere between the characteristics of constant horsepower and constant torque.

A general test for deciding if a machine might require “Constant Horsepower” would be to study the machine output. When a machine is designed to produce a fixed number of pounds per hour regardless of whether it is making small parts at high speed, or large parts at a lower speed, the drive requirement is apt to be “Constant Horsepower”.

Although details of selecting drives for constant horsepower loads are beyond the scope of this presentation, some possibilities are as follows.

“Constant Horsepower” loads can be handled by oversizing drives such as standard SCR units or slip couplings. This is done by matching the drive’s output torque with the machine’s requirement at the low speed. Depending upon the speed range that is required, this can result in gross oversizing at the high speed. More practical approaches involve using stepped pulleys, gearshift transmissions and metallic or rubber belt adjustable pitch pulley drives. Some additional and more sophisticated approaches are DC (SCR) drives operating with a combination of armature control at full field power up to base speed and field weakening above base speed. Some variable frequency drives can also be used at frequencies above 60 HZ., with

rather than rotating. In this case, it is necessary to calculate an equivalent WR2 for the body that is moving linearly. The equation for this conversion is as follows:

EQUIVALENT WR2 = W(V)2/39.5 (SM)2

W = WEIGHT OF LOAD IN POUNDS

V = VELOCITY OF THE LOAD IN FEET PER MINUTE

SM = SPEED OF THE MOTOR IN RPM WHEN LOAD IS MOVING AT VELOCITY V

Once the equivalent WR2 has been calculated, acceleration time, or required accelerating torque, is calculated by using the same equations for rotating loads.

SUMMARY

The turning force on machinery is torque, not horsepower.

Horsepower blends torque with speed to determine the total amount of work that must be accomplished in a span of time.

In all cases, the horsepower required for single speed application can be determined by utilizing the torque required at rated speed along with the required speed.

When variable speed drives are to be utilized, an additional determination of load type has to be made. Most applications require either Constant Torque or Variable Torque. Metal cutting and metal forming applications frequently will require Constant Horsepower.

High inertia loads need to be approached with some caution due to high currents absorbed by the motors during the starting period. If there is any question regarding safe accelerating capabilities, the application should be referred to the motor manufacturer.

An understanding of torque is essential for proper selection of any drive product.

I would like to thank U. S. Electrical Motors, Division of Emerson Electric, for allowing me to utilize this material for presentation at the 1981 National Conference on Power Transmission. This basic text was originally authored by

voltage held constant to achieve a moderate amount of constant horsepower speed range.

VARIABLE TORQUE

The final load type that is often encountered is the “Variable Torque” load. In general, variable torque loads are found only in centrifugal pumps, fans and blowers.

A cross section of a centrifugal pump is shown in Figure 6. The torque requirement for this load type can be thought of as being nearly opposite that of the “Constant Horsepower” load. For a variable torque load, the torque required at low speed is very low, but the torque required at high speed is very high. Mathematically, the torque requirement is a function of the speed squared and the horsepower is a function of the speed cubed.

The relationships of torque and horsepower to speed on “Variable Torque” loads are shown in Figure 7.

The key to drive sizing on “Variable Torque” loads is strictly related to providing adequate torque and horsepower at the MAXIMUM speed

myself while employed by U. S. Electrical Motors.

GLOSSARY OF TERMS

Torque … Twisting force measured in inch pounds or foot pounds.

Horsepower … A measurement of work done per unit of time. 33,000 foot pounds per minute = 1 H.P.

Sticksion … A word used to describe the torque required to breakaway a load.

Constant Torque Load … A load where the driving torque requirement is independent of speed.

Variable Speed Drives … A driving device whose speed is adjustable to provide for changes in speed flow or rate.

Load Type … Classifications of loads by their torque and horsepower requirements as related to speed.

Constant Horsepower … A load type where the torque requirement is greatest at low speeds and reduces at higher speeds.

Variable Torque … A load type where the torque required to drive a load increases with speed. This load type is usually associated with centrifugal pumps and blowers.

Inertia … The tendency of a load to resist increases or decreases in speed.

High Inertia Loads … Loads exhibiting a flywheel characteristic.

WR2 or WK2 … A measure of inertia related to the weight and radius of gyration of a rotating body.

Radius of Gyration … A radius at which the entire weight of a body can be assumed to exist for purposes of inertia calculations.

NEMA … National Electrical Manufacturers Association. A body charged with establishing many industry standards for electrical equipment.

that will be required. MAXIMUM must be emphasized since a 9% increase in speed over the normal maximum will produce a 30% increase in the horsepower requirement.

It is impossible to speculate on the number of motors that have been burned out because people have unknowingly changed pulley ratios to obtain “more output” from their centrifugal pumps or blowers.

Direct Connected Loads … A load coupled directly to the motor shaft where the load speed is the same as the motor speed.

Reflected Inertia … Used to relate load inertia to the motor shaft for loads driven through speed increasing or decreasing belt or gear ratios. Also called “Equivalent Inertia”.

Linear Motion … Straight line motion as encountered in cars and conveyors of various types.