Electrical Drives & Control

Chapter 2
Mechanics, Load Dynamics, and Thermal Ratings

Bachelors Program in Electrical Engineering
Electrical Drives & Control   •   Bachelor

Mechanics, Load Dynamics, and Thermal Ratings

Introduction and Context

An electrical drive is an electromechanical energy-conversion system in which a motor, power converter, controller, transmission, and mechanical load operate as one dynamic system. Electrical control cannot be designed correctly unless the mechanical system is represented accurately. The mechanical model determines acceleration time, speed regulation, transient torque, energy consumption, braking requirements, and motor thermal stress.

The central mechanical question is: how does the torque produced by the motor change the speed and position of the load? For a rotating system, the answer is expressed through Newton’s law for rotation. For a translating system, it is expressed through Newton’s law for linear motion. Transmission elements such as gears, belts, screws, and couplings connect these two domains and transform force, torque, speed, and inertia.

A second question is thermal: can the motor deliver the required torque repeatedly without exceeding its permissible temperature? A motor may be capable of a high short-term torque but unable to sustain it continuously. Therefore, drive selection requires both dynamic analysis and thermal-duty analysis.

Learning Objectives

After studying this chapter, the reader should be able to:

  1. Model translational and rotational mechanical motion.
  2. Distinguish inertia, friction, gravitational force, and externally applied load torque.
  3. Derive the basic equation of an electric drive.
  4. Convert load parameters through gears and other transmissions.
  5. Calculate acceleration torque, braking torque, and acceleration time.
  6. Explain motor heating, cooling, thermal time constants, and temperature rise.
  7. Evaluate continuous, short-time, and intermittent-periodic duties.
  8. Apply equivalent torque and equivalent current methods appropriately.
  9. Recognize modelling assumptions, practical limitations, and safe engineering practices.

1. Mechanical Variables and Units

The principal translational variables are displacement x in metres (m), velocity v in metres per second (m/s), acceleration a in metres per second squared (m/s²), force F in newtons (N), and mass m in kilograms (kg). Their relationships are

v = dx/dt
a = dv/dt = d²x/dt²

For a particle or rigid body undergoing translation, Newton’s second law is

∑F = ma

where ∑F is the algebraic sum of forces in N.

The corresponding rotational variables are angular displacement θ in radians, angular velocity ω in radians per second (rad/s), angular acceleration α in rad/s², torque T in newton-metres (N·m), and mass moment of inertia J in kg·m²:

ω = dθ/dt
α = dω/dt = d²θ/dt²
∑T = Jα

The radian is dimensionless in strict SI dimensional analysis, but it is retained in engineering notation to identify rotational quantities. Mechanical power is

P = Fv = Tω

where P is in watts. If speed is specified in revolutions per minute, n, then

ω = 2πn/60

Thus, torque and speed must not be multiplied directly unless speed has first been converted to rad/s.

2. Translational Mechanical Systems

A translating load may include a vehicle, elevator car, conveyor, hoist, or linear actuator. A free-body model identifies all forces acting in the direction of motion. Typical forces include motor-generated traction force, rolling or sliding friction, aerodynamic resistance, gravity, spring force, and process resistance.

For a body moving along a straight path, choose positive motion as the positive direction. Then

Fm - Ff - Fg - Fr = ma

where F_m is applied driving force, F_f is friction force, F_g is the component of weight opposing motion, and F_r represents other resistance forces. The signs must be changed when the direction of motion or force changes.

For a mass lifted vertically upward,

Fm - mg - Ff = ma

where g ≈ 9.81 m/s². During downward motion, gravitational force assists motion and the dynamic equation must be written with a consistent sign convention. In hoisting systems, gravity is not merely a loss; it may cause regenerative braking or an overhauling load.

Friction models

The simplest viscous-friction model is

Fv = Bv v

where B_v is the translational viscous-friction coefficient in N·s/m. Coulomb or dry friction is often approximated by

Fc = μN

where μ is the coefficient of friction and N is the normal force in N. The direction of Coulomb friction opposes actual or impending motion. A more realistic model may include both effects:

Ff = Fc sgn(v) + Bv v

where sgn(v) is +1 for positive velocity and −1 for negative velocity. At v = 0, static friction requires a separate inequality because the friction force takes whatever value is necessary, up to its limiting value, to prevent motion.

These models are approximations. Bearing friction, seal friction, windage, material deformation, and speed-dependent process forces can make the actual load nonlinear.

3. Rotational Mechanical Systems

Most electric motors produce rotary motion. The motor torque must overcome load torque, friction torque, and the torque needed to accelerate rotating inertias. The general equation is

Tm - TL - Tf = J dω/dt

where T_m is electromagnetic motor torque, T_L is externally useful load torque, T_f is friction and windage torque, and J is the total inertia referred to the selected shaft.

Equivalently, define the resisting torque T_res = T_L + T_f:

Tm - Tres = Jα

The acceleration torque is

Tacc = Jα = Tm - Tres

Acceleration occurs when T_m > T_res; constant speed occurs when T_m = T_res; deceleration occurs when T_m < T_res. If the motor is commanded to decelerate while the load continues to drive the motor, energy may flow from the mechanical system back to the electrical supply or braking resistor.

Rotational inertia

For a point mass m at radius r,

J = mr²

For common rigid bodies:

BodyAxisMoment of inertia
Solid cylinder or diskCentral axisJ = ½mr²
Thin ringCentral axisJ = mr²
Slender rodThrough centre, perpendicular to lengthJ = 1/12 mL²
Slender rodThrough one end, perpendicular to lengthJ = 1/3 mL²

The parallel-axis theorem is

JO = JC + md²

where J_C is inertia about a parallel axis through the centre of mass and d is the distance between axes. Inertia is a property of mass distribution, not simply mass magnitude.

4. Gear Systems and Reflected Parameters

Consider an ideal gear pair with motor-side speed ω_m, load-side speed ω_L, motor-side torque T_m, load torque T_L, and gear ratio

N = ωm/ωL

For a reduction drive, N > 1. The load rotates more slowly and receives greater torque, neglecting losses. Conservation of power gives

Tmωm = TLωL

Therefore, in the ideal case,

Tm = TL/N

If the transmission efficiency is η_g, defined as output power divided by input power, then the motor torque required for a resisting load torque is

Tm = TL/(ηg N)

This expression assumes η_g is applicable in the stated direction of power flow. Efficiency can differ during motoring and regenerative operation.

The load inertia reflected to the motor shaft is

JL' = JL/N²

A motor-side inertia J_m therefore sees total inertia

Jeq = Jm + JL/N²

For multiple rotating components, each inertia is referred to the motor shaft using the square of the relevant speed ratio. The square is important: a high-speed motor driving a low-speed load may see a substantially reduced reflected load inertia.

For a translating mass connected through a wheel of radius r to the motor shaft, the equivalent motor-side inertia is

Jtrans' = mr²

because the rim speed is v = rω. If a gearbox lies between motor and wheel, the appropriate wheel-side inertia is then divided by N² when referred to the motor shaft.

Gear-train example

A motor drives a conveyor through a gearbox with N = 5, η_g = 0.90. The conveyor shaft requires 150 N·m at steady speed. The motor and gearbox input shaft have combined inertia 0.20 kg·m², while the conveyor-side rotating inertia is 4.0 kg·m². The required motor torque at steady speed is

Tm = 150/(0.90 × 5) = 33.3 N·m

The reflected conveyor inertia is

JL' = 4.0/5² = 0.16 kg·m²

Thus, the total motor-side inertia is 0.36 kg·m². If the motor accelerates at 100 rad/s², its acceleration torque is

Tacc = 0.36 × 100 = 36 N·m

The approximate motor torque during acceleration is 33.3 + 36 = 69.3 N·m, assuming the load torque and efficiency remain constant.

5. Drive Equation and Operating Quadrants

Combining motor torque, load torque, friction, and inertia gives the fundamental drive equation:

Tm = TL + Tf + Jeq dω/dt

This equation applies to a lumped rotational model with parameters referred to one shaft. It assumes rigid coupling, or that shaft elasticity is negligible over the time scale of interest. If shaft compliance, backlash, or torsional resonance is important, a multi-mass model is required.

The signs of torque and speed define four operating quadrants:

QuadrantSpeedTorqueTypical operation
IPositivePositiveForward motoring
IIPositiveNegativeForward braking or regeneration
IIINegativeNegativeReverse motoring
IVNegativePositiveReverse braking or regeneration

Mechanical power P = Tω is positive for motoring and negative for regenerative operation, according to the usual convention that positive power flows from the electrical system to the mechanical load. Actual converter and motor conventions must be stated explicitly in a control system.

Acceleration time

If the net accelerating torque is constant,

tacc = Jeq(ω2 − ω1)/(Tm − Tres)

where ω_1 and ω_2 are initial and final speeds. The formula is valid only when T_m − T_res is nonzero and approximately constant. For a speed-dependent torque difference, use

tacc = ∫[ω1 to ω2] Jeq dω/(Tm(ω) − Tres(ω))

provided the denominator remains positive throughout the interval. If it approaches zero, acceleration time becomes very large, indicating that the operating point is near the available-torque limit.

Worked acceleration example

A drive has total inertia J_eq = 0.80 kg·m². The motor torque is constant at 80 N·m, while resisting torque is 20 N·m. The speed increases from 0 to 1200 r/min. First convert speed:

ω2 = 2π(1200)/60 = 125.66 rad/s

The net torque is 60 N·m, so

tacc = 0.80 × 125.66/60 = 1.68 s

The result is physically plausible because the net torque is positive and the assumed torque values are constant over the speed range.

6. Load-Torque Characteristics

The torque required by a load may be classified by its relationship with speed:

  1. Constant torque: T_L is approximately independent of speed. Conveyors, positive-displacement pumps, and hoists often approximate this class.
  2. Torque proportional to speed: T_L ∝ ω. This occurs in viscous drag and some fluid systems.
  3. Fan or pump load: T_L ∝ ω² and power P ∝ ω³ for geometrically similar operation. Centrifugal fans and pumps often follow this approximation over a useful range.
  4. Constant power: P is approximately constant, so T_L ∝ 1/ω. Machine tools and traction drives may have constant-power regions above base speed.

These are models rather than universal laws. Valve position, fluid density, process conditions, blade geometry, and control mode can alter the characteristic. The motor must provide torque exceeding the load torque over the complete acceleration path, not merely at rated speed.

7. Motor Heating and Thermal Models

Motor losses become heat. Important losses include stator copper loss, rotor copper loss, core loss, mechanical friction and windage loss, converter-related loss, and stray-load loss. The temperature rise above ambient is denoted by θ in °C or K; a temperature difference has the same numerical value in °C and K.

A first-order lumped thermal model is

Cth dθ/dt + θ/Rth = Ploss

where C_th is thermal capacitance in J/K, R_th is thermal resistance in K/W, and P_loss is heat generation in W. For constant loss P_loss and initial temperature rise θ_0,

θ(t) = Ploss Rth + [θ0 − Ploss Rth]e^(−t/τth)

with thermal time constant

τth = Rth Cth

For a cold motor starting at ambient, θ_0 = 0 and

θ(t) = θ_∞[1 − e^(−t/τth)]

where θ_∞ = P_loss R_th is the final steady-state temperature rise under constant loss.

During cooling with negligible internal loss, the temperature rise approximately follows

θ(t) = θ0 e^(−t/τth)

Real motors can have multiple thermal time constants because windings, iron, frame, bearings, and cooling air do not heat identically. The first-order model is useful for preliminary sizing and control protection, but manufacturer data should govern final decisions.

Thermal limits and insulation

The permissible winding temperature depends on insulation thermal class, ambient temperature, altitude, cooling method, enclosure, and service factor. Exceeding the permitted temperature accelerates insulation ageing and can cause irreversible damage. A drive protection system should therefore use measured or estimated temperature, overload models, current limits, and appropriate fault handling rather than relying only on instantaneous current.

8. Torque, Current, and Thermal Equivalence

For many motors, electromagnetic torque is approximately proportional to current over a specified operating region:

T = kT I

where k_T is the torque constant in N·m/A and I is the relevant torque-producing current. This relation is exact only for the stated motor model and operating conditions. Saturation, field weakening, commutation limits, and AC-machine control interactions can invalidate a simple proportionality.

Copper loss is approximately proportional to current squared:

PCu = I²R

Consequently, the thermal equivalent current for a duty cycle with constant resistance and approximately uniform cooling is

Ieq = √[(∑ Ii²ti)/(∑ti)]

For torque-proportional current, the equivalent torque is

Teq = √[(∑ Ti²ti)/(∑ti)]

These expressions apply when copper loss dominates and the thermal time constant is long compared with individual pulses, or when the duty cycle is evaluated using an appropriate averaged-loss approximation. They do not automatically account for iron loss, speed-dependent cooling, transient thermal gradients, or changing winding resistance.

Example of equivalent torque

A motor operates for 20 s at 40 N·m, then for 10 s at 80 N·m, repeating periodically. Neglecting other losses,

Teq = √[(40²×20 + 80²×10)/30]
Teq = √[(32000 + 64000)/30] = √3200 = 56.6 N·m

A motor with a continuous thermal rating of at least 56.6 N·m may be thermally adequate under the stated simplified assumptions, but it must also satisfy the peak torque requirement of 80 N·m and the acceleration requirement.

9. Duty Classes and Motor Ratings

Motor selection distinguishes continuous capability from short-term overload capability. Common duty descriptions include:

  • Continuous duty: operation at essentially constant load long enough to reach thermal equilibrium.
  • Short-time duty: operation at a specified load for a limited time, followed by sufficient rest for cooling.
  • Intermittent periodic duty: repeated cycles of load and rest, often with starting, braking, or electric separation periods.
  • Continuous operation with periodic load changes: operation without stopping but with varying load.
  • Duty with non-periodic load or speed: an irregular sequence specified by a load-time-speed profile.

The exact duty classification and test requirements depend on the applicable motor standard and manufacturer documentation. A nameplate rated power generally describes a defined thermal and mechanical condition, not every possible operating point.

Continuous rating

For a constant-speed motor, the continuous torque rating is constrained by temperature rise, insulation system, bearing limits, commutation or inverter limits, and mechanical strength. Continuous power is

Pcont = Tcontω

At lower speed, a self-cooled motor may have reduced cooling, so constant torque operation may be thermally restricted even though the power is lower.

Overload capacity

Overload capacity is the ratio of permitted maximum torque to rated torque:

KOL = Tmax/Trated

It is not solely a thermal quantity. Peak current, voltage, magnetic saturation, demagnetization risk in permanent-magnet machines, commutation, shaft stress, coupling strength, and converter current limits may be decisive. The duration and repetition rate of overloads must be specified.

Short-time rating

For a short-time duty, the motor starts at a known thermal state and operates for duration t_s. The temperature must satisfy

θ(ts) ≤ θpermissible

Using the first-order model with constant loss gives a direct check. If the motor begins warm, θ_0 must be included; assuming every cycle starts from ambient can produce unsafe results.

10. Applied Case: Conveyor Drive Selection

Background

Consider an illustrative conveyor that moves a 600 kg material load horizontally at 1.5 m/s. The effective rolling and process resistance is 900 N. A drum of radius 0.20 m is driven through a gearbox with ratio N = 4, efficiency η_g = 0.92. The drum and other conveyor rotating parts have inertia 8 kg·m² at the drum shaft. The motor-side inertia, including rotor and input coupling, is 0.12 kg·m². The conveyor must reach operating speed in 2 s.

Problem analysis

The steady conveyor force is 900 N, so drum torque is

Tdrum = Fr = 900 × 0.20 = 180 N·m

The drum speed is

ωdrum = v/r = 1.5/0.20 = 7.5 rad/s

The motor speed is

ωm = Nωdrum = 4 × 7.5 = 30 rad/s

The motor-side steady torque, including gearbox loss, is

Tsteady = 180/(0.92 × 4) = 48.9 N·m

The reflected rotating inertia is

Jrot' = 8/4² = 0.50 kg·m²

The translating mass can be represented at the drum as J_trans,drum = mr²:

Jtrans,drum = 600 × 0.20² = 24 kg·m²

Referred to the motor shaft,

Jtrans' = 24/4² = 1.50 kg·m²

Hence,

Jeq = 0.12 + 0.50 + 1.50 = 2.12 kg·m²

The required motor acceleration is

αm = 30/2 = 15 rad/s²

The acceleration torque is

Tacc = 2.12 × 15 = 31.8 N·m

The approximate peak motor torque during acceleration is therefore

Tpeak = 48.9 + 31.8 = 80.7 N·m

Approach and outcome

A suitable motor-drive combination must provide at least 48.9 N·m continuously at 30 rad/s, approximately 80.7 N·m for 2 s, and additional margin for starting friction, uncertainty, voltage variation, and temperature. Its converter must permit the required current without violating motor or semiconductor limits. If frequent starts occur, the starting losses must be included in the thermal duty calculation; a motor that passes the single-start torque test may still overheat under repeated cycles.

Learning points

The example demonstrates four important practices: convert force to drum torque; reflect all inertias through the square of the speed ratio; separate steady torque from acceleration torque; and check peak and thermal ratings independently. Selecting a motor solely from steady-state power would be inadequate.

11. Limitations, Safety, and Professional Practice

Mechanical models should be validated against measured speed, current, acceleration time, and temperature. Important effects omitted by a rigid-body model may include backlash, elastic shafts, belt slip, gearbox compliance, resonance, windup, changing friction, load impact, and sensor delay. Emergency stops and braking systems require separate safety analysis because stopping distance, stored energy, brake torque, and failure modes may dominate.

Engineers should document sign conventions, reference shaft, gear efficiency, load profile, ambient conditions, cooling arrangement, overload duration, and applicable standards. Protective settings must not be increased merely to prevent nuisance trips; they must remain consistent with insulation, conductor, converter, and mechanical limits. Regenerative energy must have a controlled path through the supply converter, braking resistor, storage system, or mechanical brake.

Modern drives increasingly use thermal observers that estimate rotor and stator temperature from current, speed, ambient temperature, and a thermal model. These observers can improve utilization, but they are not infallible: incorrect parameters, blocked cooling paths, sensor faults, and unmodelled duty changes can create unsafe estimates. AI-based predictive maintenance may identify abnormal vibration or temperature trends, but its alarms should complement—not replace—verified protection functions and engineering judgement.

Chapter Summary

Mechanical drive analysis begins with Newton’s laws: ∑F = ma for translation and ∑T = Jα for rotation. Motor torque is divided between useful load torque, friction and windage, and acceleration torque. Correct sign conventions are essential, particularly during braking and regenerative operation.

Gear systems change torque and speed and reflect load inertia by the square of the speed ratio. Translating masses can be converted to equivalent rotational inertia using J = mr². The drive equation permits calculation of acceleration torque and time, but speed-dependent torque requires integration or numerical simulation.

Thermal rating is governed by losses, thermal resistance, thermal capacitance, cooling, insulation, and duty cycle. A first-order thermal model provides useful temperature-rise estimates. Equivalent torque or current methods are convenient when copper loss dominates, but they do not replace detailed manufacturer data for complex duties. Motor selection must satisfy continuous, peak, acceleration, braking, converter, mechanical, and thermal requirements simultaneously.

Key Terms

Motor torque: Electromagnetic torque produced by the motor.

Load torque: Torque required by the driven process.

Acceleration torque: Torque available to change rotational speed, Jα.

Inertia: Resistance of a body to angular acceleration, measured in kg·m².

Reflected inertia: Load inertia expressed at another shaft through a transmission.

Gear ratio: Ratio of input speed to output speed.

Transmission efficiency: Output mechanical power divided by input mechanical power.

Friction torque: Resistive torque caused by contact, seals, bearings, or viscous effects.

Overhauling load: A load that can drive the motor during part of operation.

Regenerative braking: Braking in which mechanical energy is converted back into electrical energy.

Thermal resistance: Temperature rise per unit heat flow, in K/W.

Thermal capacitance: Heat required to produce a unit temperature rise, in J/K.

Thermal time constant: Product R_thC_th describing first-order heating or cooling response.

Continuous duty: Operation at a load sufficient to approach thermal equilibrium.

Short-time duty: Operation for a specified limited time followed by cooling.

Intermittent-periodic duty: Repeated operating and rest intervals.

Equivalent torque: RMS-like torque used to estimate copper-loss heating.

Overload capacity: Permitted maximum torque relative to rated torque.

Field weakening: Operation above base speed with reduced flux and limited torque.

Rigid-body model: A model neglecting elastic deformation between connected masses.

Review Questions

Short Answer Questions

  1. State the rotational form of Newton’s second law and define each quantity.
  2. Why is load inertia reflected through the square of the gear ratio?
  3. Distinguish steady load torque from acceleration torque.
  4. Define thermal resistance and thermal time constant.
  5. Why can a motor’s peak torque rating not be used as its continuous rating?

Long Answer Questions

  1. Derive the motor-side drive equation for a motor connected to a load through an ideal gearbox, including reflected inertia.
  2. Explain the four operating quadrants of a motor drive and identify when regeneration can occur.
  3. Develop a first-order motor heating model and explain the physical meaning of its parameters.
  4. Compare continuous, short-time, and intermittent-periodic motor duties, including the implications for rating selection.
  5. Explain the assumptions and limitations of equivalent torque and equivalent current methods.

Analytical / Application Questions

  1. A motor drives a 5 kg·m² load through a reduction ratio of 6. The motor-side inertia is 0.10 kg·m², and the net motor torque available for acceleration is 25 N·m. Calculate the equivalent inertia and angular acceleration.
  2. A conveyor requires 120 N·m at its drum. The gearbox ratio is 3 and efficiency is 0.90. Calculate the steady motor torque and explain how the answer changes during acceleration.
  3. A drive operates at 30 N·m for 15 s and 60 N·m for 5 s in a repeating cycle. Calculate the equivalent torque under the assumption that copper loss dominates, then identify one additional check required before selecting the motor.

Multiple Choice Questions with Answers

Question 1

For a rotating drive, which equation represents the fundamental mechanical balance?

A. T_m + T_L = Jα

B. T_m − T_L − T_f = Jα

C. T_m − Jα = 0 for all loads

D. T_mω = Jα

Correct Answer: B. T_m − T_L − T_f = Jα

Question 2

A load inertia of 9 kg·m² is connected through an ideal speed reduction ratio N = 3, defined as motor speed divided by load speed. What inertia is reflected to the motor shaft?

A. 27 kg·m²

B. 3 kg·m²

C. 1 kg·m²

D. 81 kg·m²

Correct Answer: C. 1 kg·m²

Question 3

A motor rotates at 1500 r/min and produces 20 N·m. What is its approximate mechanical power?

A. 314 W

B. 1500 W

C. 3142 W

D. 30,000 W

Correct Answer: C. 3142 W

Question 4

Which condition produces acceleration in the positive direction?

A. Motor torque is less than resisting torque

B. Motor torque equals resisting torque

C. Motor torque is greater than resisting torque

D. Motor speed is zero, regardless of torque

Correct Answer: C. Motor torque is greater than resisting torque

Question 5

For a gearbox with ratio N = 4 and efficiency η_g = 0.8, a load requires 160 N·m. What motor torque is required at steady speed?

A. 32 N·m

B. 40 N·m

C. 50 N·m

D. 512 N·m

Correct Answer: C. 50 N·m

Question 6

In a first-order thermal model, the product R_thC_th has units of:

A. W

B. K/W

C. J/K

D. s

Correct Answer: D. s

Question 7

If copper loss dominates and torque is proportional to current, which quantity is most appropriate for estimating thermal loading over a varying torque cycle?

A. Arithmetic mean torque

B. Peak speed only

C. RMS or equivalent torque

D. Minimum torque

Correct Answer: C. RMS or equivalent torque

Question 8

A motor operates at 40 N·m for 10 s and 80 N·m for 10 s. Under the simplified copper-loss assumption, its equivalent torque is:

A. 40 N·m

B. 56.6 N·m

C. 60 N·m

D. 80 N·m

Correct Answer: B. 56.6 N·m

Question 9

Which operating condition is normally associated with regenerative braking?

A. Torque and speed have the same sign during motoring

B. Torque is zero and speed is constant

C. Torque opposes the direction of motion while mechanical energy flows toward the electrical system

D. The motor is disconnected and stationary

Correct Answer: C. Torque opposes the direction of motion while mechanical energy flows toward the electrical system

Question 10

For a translating mass m connected to a rotating shaft through a wheel of radius r, the equivalent rotational inertia at that wheel is:

A. m/r²

B. mr

C. mr²

D. m²r

Correct Answer: C. mr²

Electrical Drives & Control
Chapter 2 of 14