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Motor Drive Calculations Guide

This document provides formulas for calculating horsepower, torque, and inertia for AC and DC motor applications. It discusses: 1) Formulas for calculating horsepower based on torque and speed, and torque based on horsepower and speed. 2) A formula for calculating accelerating torque based on inertia, change in speed, and time to accelerate. 3) Methods for calculating the equivalent inertia of multiple rotating or linearly moving parts reflected to the motor shaft speed. 4) Equations for calculating inertia of solid cylinders and shafts based on weight and radius of gyration. 5) How inertia values are adjusted for different shaft materials or when using speed reducers.

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Hoang Truong
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163 views1 page

Motor Drive Calculations Guide

This document provides formulas for calculating horsepower, torque, and inertia for AC and DC motor applications. It discusses: 1) Formulas for calculating horsepower based on torque and speed, and torque based on horsepower and speed. 2) A formula for calculating accelerating torque based on inertia, change in speed, and time to accelerate. 3) Methods for calculating the equivalent inertia of multiple rotating or linearly moving parts reflected to the motor shaft speed. 4) Equations for calculating inertia of solid cylinders and shafts based on weight and radius of gyration. 5) How inertia values are adjusted for different shaft materials or when using speed reducers.

Uploaded by

Hoang Truong
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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AC and DC Drives

Applications

MOTOR APPLICATION FORMULAS


Horsepower Required The inertia of complex, concentric rotating parts may be
calculated by breaking the part up into simple rotating
HP = Torque (Ib-ft) x Speed (RPM) cylinders, calculating their inertial and summing their
5250 values, as shown in the following diagram.

HP = Torque (Ib-in) x Speed (RPM)


63,000

Torque (Ib-ft) = HP x 5250


Speed (RPM)

2 2
Accelerating Torque (Ib-ft) = WK (Ib-ft ) x∆ RPM
308 x t (sec)
WK2 of Rotating Elements
Where In practical mechanical systems, all the rotating parts do
WK2 = Inertia (Ib-ft2) reflected to motor shaft not operate at the same speed. The WK2 of all moving
∆ RPM = Change in Speed parts operating at each speed must be reduced to an
t = Time (seconds) to accelerate equivalent WK2 at the motor shaft, so that they can all be
added together and treated as a unit, as follows:

t = WK (Ib-ft ) x ∆ RPM = Time to accelerate (sec)


2 2
N 2
308 x T (Ib-ft) Equivalent WK2 = WK2 ( Nm )
RPM = FPM
.262 x diameter (inches) Where,
WK2 = inertia of the moving part
2 N = speed of the moving part (RPM)
Inertia reflected to motor = Load inertia Load RPM
( )
Motor RPM Nm = speed of the driving motor (RPM)

When using speed reducers, and the machine inertia is


Inertia (WK2)
reflected back to the motor shaft, the equivalent inertia is
The factor WK2 is the weight (lb) of an object multiplied by equal to the machine inertia divided by the square of the
the square of the radius of gyration (K). The unit measure- drive reduction ratio.
ment of the radius of gyration is expressed in feet.
For solid or hollow cylinders, inertia may be calculated by
using the equations given here. WK2 of Linear Motion
The inertia of solid steel shafting per inch of shaft length is Not all driven systems involve rotating motion. The equiva-
given in Table 1. To calculate hollow shafts, take the differ- lent WK2 of linearly moving parts can also be reduced to
ence between the inertia values for the OD and ID. For the motor shaft speed as follows:
shafts of materials other than steel, multiply the value for
steel by the appropriate factor given in Table 2.
Equivalent WK2 = W (V) 2
39. 5 (Nm)2

Where,
W = weight of load (lb)
V = linear velocity of rack and load or
conveyor and load (FPM)
Nm = speed of the driving motor (RPM)

This equation can only be used where the linear speed


bears a continuous fixed relationship to the motor speed,
such as a conveyor.

18

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