Electric Drive System

The duty of a motor is the pattern of load, rest, starting and braking it experiences with time. Since motor heating depends on this pattern, the rating must be chosen for the duty class; a motor sized for continuous duty would be oversized for short-time duty, and a short-time-rated motor would overheat on continuous duty. IS 4722 / IEC 60034-1 define eight duty classes (S1 to S8).

Classes of duty

Load-time sketches

 S1 continuous           S2 short-time
 P |______________       P |____
   |                       |    |__________ rest
   +-------------> t       +-------------> t

 S3 intermittent          S6 continuous, intermittent load
 P |__   __   __          P |__   __   __
   |  |_|  |_|  |_ rest     |  |_|  |_|  |_ no-load
   +-------------> t        +-------------> t

Notes on each class

• Continuous duty (S1): the motor reaches its final steady temperature; rating = continuous rating (CMR). Load may be constant (fan) or variable continuous; for variable load, the equivalent (rms) current/torque/power method is used to size the motor.
• Short-time duty (S2): the motor can deliver more than its continuous rating because it never reaches final temperature; standard periods 10, 30, 60, 90 min.
• Intermittent periodic duties (S3, S4, S5): specified by cyclic duration factor (CDF) = (load time)/(cycle time), e.g. 15, 25, 40, 60 %; cycle time usually 10 min. Starting (S4) and braking (S5) losses add heating and must be included.
• Continuous with intermittent loading (S6): motor never stops; no-load periods give partial cooling.
• S7 and S8: continuous running with frequent electric braking/reversal or speed changes; heavy rotor heating, so inertia and number of operations per hour matter.

Applying to motor selection

1. Identify the load cycle of the driven machine.
2. Select the matching duty class.
3. Find the equivalent power (rms) over the cycle and check maximum torque against motor pull-out torque.
4. Choose a standard rating with that duty class marked on the nameplate (e.g. "S3 40 %").

Correct duty matching avoids both overheating (insulation damage) and over-sizing (poor efficiency and power factor).

The duty class of a motor describes how its load varies with time: continuous running, short running, periodic on/off, starting, braking and speed changes. Because temperature rise depends on the duty, the motor rating is specified for a duty class (IEC 60034-1 / IS 4722, classes S1 to S8).

Classes with waveforms

In the sketches, P = load, $P_v$ = losses, $\theta$ = temperature.

S1 Continuous duty: operation at constant load long enough to reach thermal equilibrium.

 P  |_________________
 th |   ___-----------  (reaches steady temp)
    +-----------------> t

Applications: pumps, fans, compressors, conveyors.

S2 Short-time duty: constant load for a short time $t_L$ (10, 30, 60, 90 min), then rest until the motor cools to ambient.

 P  |_____
    |     |__________________ rest
 th |  /\
    | /   \______ (never reaches final temp)
    +-----------------> t

Applications: sluice gates, swing bridges, valve actuators, kitchen mixers.

S3 Intermittent periodic duty: identical cycles of constant load and rest; starting losses negligible; temperature fluctuates around an average.

 P  |__    __    __
    |  |__|  |__|  |__
 th |  /\/\/\/\/\/\  (saw-tooth)
    +-----------------> t
     on off

Specified by cyclic duration factor $\text{CDF} = \dfrac{t_{on}}{t_{on}+t_{off}} \times 100\,\%$ (15, 25, 40, 60 %). Applications: cranes, hoists, presses.

S4 Intermittent periodic with starting: as S3, but each cycle has a significant starting period. Applications: lifts, machine tools.

S5 Intermittent periodic with starting and electric braking: start, load, electrical braking, rest. Applications: rolling-mill auxiliaries, centrifuges.

 S5:  start load brake rest
 P  |  /----\
    | /      \_______
    +-----------------> t

S6 Continuous duty with intermittent periodic loading: load and no-load periods, no rest.

 P  |__    __    __
    |  |~~|  |~~|  |~~ (no-load, motor running)
    +-----------------> t

Applications: pressing, drilling, punching machines.

S7 Continuous duty with starting and braking: start, load, electric braking repeatedly, no rest period. Applications: reversing rolling mills.

S8 Continuous duty with periodic speed changes: load at one speed then at another speed, no rest. Applications: multi-speed machine tools, pole-changing motor drives.

Summary

For variable duty, the motor is sized by the rms (equivalent) torque or power over a cycle and checked for maximum torque.

Equilibrium (steady-state) speed is where the motor torque equals the load torque: $T = T_l$.

Data

• Characteristic 1 (quadrants I, II, IV): $T = 200 - 0.2N$
• Characteristic 2 (quadrants II, III, IV): $T = -200 - 0.2N$
• Loaded hoist: $T_l = 100$ N-m; unloaded hoist: $T_l = -80$ N-m (the counterweight is heavier than the empty cage, so the load torque is negative).

A hoist load is an active (gravity) load, so $T_l$ keeps its sign when the speed reverses.

Quadrant I: forward motoring (hoisting the loaded cage), characteristic 1, $T_l = 100$:

$$\begin{aligned} 200 - 0.2N &= 100 \\ N &= \frac{200-100}{0.2} = 500\ \text{rpm} \end{aligned}$$

$T = +100$, $N = +500$: both positive, quadrant I.

Quadrant II: forward braking (hoisting the empty cage; counterweight pulls), characteristic 1, $T_l = -80$:

$$\begin{aligned} 200 - 0.2N &= -80 \\ N &= \frac{200+80}{0.2} = 1400\ \text{rpm} \end{aligned}$$

$T = -80$, $N = +1400$: torque negative, speed positive, quadrant II (regenerative braking; speed above no-load speed of 1000 rpm).

Quadrant III: reverse motoring (lowering the empty cage), characteristic 2, $T_l = -80$:

$$\begin{aligned} -200 - 0.2N &= -80 \\ N &= \frac{-200+80}{0.2} = -600\ \text{rpm} \end{aligned}$$

$T = -80$, $N = -600$: both negative, quadrant III.

Quadrant IV: reverse braking (lowering the loaded cage), characteristic 2, $T_l = 100$:

$$\begin{aligned} -200 - 0.2N &= 100 \\ N &= \frac{-200-100}{0.2} = -1500\ \text{rpm} \end{aligned}$$

$T = +100$, $N = -1500$: torque positive, speed negative, quadrant IV (regenerative braking during lowering).

            N (rpm)
   II: 1400 |  I: 500
  (T=-80)   |  (T=+100)
 -----------+-----------> T
  III: -600 |  IV: -1500
  (T=-80)   |  (T=+100)

Answer: equilibrium speeds are 500 rpm (I), 1400 rpm (II), -600 rpm (III) and -1500 rpm (IV).

Selecting a motor means choosing the type, rating and enclosure so that it drives the load correctly, economically and safely over its life. The main factors are:

1. Nature of supply

• AC or DC, single- or three-phase, voltage and frequency available; supply capacity for starting current.

2. Load (mechanical) characteristics

• Speed-torque characteristic of the load: constant torque (hoist, conveyor), torque proportional to speed squared (fan, pump), constant power (lathe, winder).
• Starting torque needed (high for cranes, crushers, traction; low for fans).
• Speed range and speed control; constant or variable speed.

3. Electrical characteristics of the motor

• Starting characteristics: starting torque and starting current (DOL, star-delta, soft starter, VFD).
• Running characteristics: speed regulation, efficiency, power factor.
• Speed control method and range.
• Braking requirements: plugging, dynamic or regenerative braking.

4. Rating and duty cycle

• Continuous, short-time or intermittent duty (S1 to S8).
• Overload capacity and peak torque; rms (equivalent) power for variable loads.

5. Mechanical considerations

• Type of enclosure (open, drip-proof, totally enclosed fan-cooled, flameproof) to suit the environment.
• Mounting (foot/flange, horizontal/vertical), shaft and bearings.
• Transmission: direct coupling, belt, gear; noise and vibration limits.

6. Environmental conditions

• Temperature, altitude, humidity, dust, explosive gases, corrosive atmosphere; insulation class.

7. Cost and economics

• Capital cost, running cost (efficiency, power factor), maintenance cost, life; availability of spares.

8. Size, weight and reliability where space is limited.

Assumptions: the belt's own mass and friction are not given, so only the mass of material on the belt is considered; the belt is fully loaded along its 200 m length.

Mass of material on the belt

Material rate $= \dfrac{100 \times 1000}{3600} = 27.78$ kg/s. Time for material to travel the belt $= \dfrac{200}{1.2} = 166.67$ s.

$$M = 27.78 \times 166.67 = 4629.6\ \text{kg}$$

Motor speed

$$\omega_m = \frac{2\pi \times 1200}{60} = 125.66\ \text{rad/s}$$

(i) Load inertia referred to motor shaft

Equating kinetic energies, $\tfrac12 J_l\omega_m^2 = \tfrac12 Mv^2$:

$$\begin{aligned} J_l &= M\left(\frac{v}{\omega_m}\right)^2 = 4629.6 \times \left(\frac{1.2}{125.66}\right)^2 \\ &= 4629.6 \times 9.119\times10^{-5} = 0.422\ \text{kg-m}^2 \end{aligned}$$

(ii) Accelerating torque

Total inertia referred to the motor shaft:

$$J = J_m + J_l = 0.1 + 0.422 = 0.522\ \text{kg-m}^2$$

Uniform acceleration from rest to full speed in 8 s:

$$\frac{d\omega_m}{dt} = \frac{125.66}{8} = 15.71\ \text{rad/s}^2$$ $$T = J\frac{d\omega_m}{dt} = 0.522 \times 15.71 = 8.20\ \text{N-m}$$

(As the conveyor is horizontal and friction is neglected, there is no steady load torque; with friction, its torque would be added.)

Answer: (i) load inertia referred to motor shaft $= 0.422$ kg-m²; (ii) motor torque to accelerate in 8 s $\approx 8.20$ N-m.

In a multi-motor drive, one machine has several motors, each driving a separate part (section) of it, e.g. paper machines, textile machines, rolling mills, cranes (hoist, trolley, travel motors). The sections must run at precise speed ratios (or tension), so closed-loop feedback control is used.

Block diagram

                 Master speed reference w*
                         |
         +---------------+----------------+
         |               |                |
      [ratio k1]      [ratio k2]       [ratio kn]
         |               |                |
  w1* -->(+)-->[SC]-->(+)-->[CC]-->[Conv]-->M1--> section 1
          ^-  speed      ^-  current            |
          |   ctrl       |   ctrl               |
          |              +-- current sensor <---+
          +------------- tacho/encoder <--------+
         ...
  (same loop for sections 2 ... n)
         |
   [Tension/load sensor between sections]
         +---> trims speed refs of next section

Working

1. Master reference: an operator sets the line speed $\omega^*$. Each section's speed reference is obtained by multiplying by a ratio $k_i$ (draw ratio), so all sections change speed together.
2. Outer speed loop: the speed sensor (tachogenerator or encoder) measures the actual speed $\omega_i$. The error $\omega_i^* - \omega_i$ goes to the speed controller (PI), whose output is the current (torque) reference.
3. Inner current loop: the current controller compares the reference with the measured armature (or stator) current and sets the converter firing angle/duty cycle (for DC drives) or the inverter PWM (for AC drives). It limits current during starting and overload and gives fast torque response.
4. Power converter (controlled rectifier, chopper or inverter) supplies the motor.
5. Tension/position (dancer) feedback between sections adds a trim signal so the material is neither slack nor over-stretched.
6. Load sharing: when several motors drive the same shaft, their current loops are coordinated so each carries its share.

Advantages

• Accurate speed ratios and tension; good product quality.
• Fast response to disturbances (current loop corrects supply and load changes).
• Protection against overcurrent; smooth starting of the whole line.
• Easy changing of line speed with one master control; suitable for PLC/computer control.

Applications

Paper and textile mills, steel rolling mills, printing presses, cranes, conveyor lines.

An electric drive is a system that uses an electric motor, with its control, to drive a machine. Compared with conventional drives (steam engines, diesel/IC engines, hydraulic and pneumatic drives, water wheels), electric drives have these advantages:

1. Wide range of power and speed: from milliwatts to tens of MW and from very low to very high speeds.
2. Easy control: starting, stopping, reversing, speed control and braking are simple and can be automated or done remotely.
3. Available in many torque-speed characteristics (series, shunt, induction, synchronous) to match almost any load; electronic control can shape the characteristic further.
4. High efficiency and no standby losses; no fuel consumed when idle.
5. Instant starting and full-load operation without warm-up.
6. Electric braking, including regenerative braking that returns energy to the supply.
7. Operation in all four quadrants (forward/reverse motoring and braking).
8. Clean and quiet: no exhaust, smoke or fuel storage; suits mines, hospitals, cities.
9. Low maintenance and long life; compact, reliable.
10. Short-time overload capacity.
11. Power from the grid: no fuel transport; can use renewable electricity.

Main limitation: needs a continuous electric supply; failure of supply stops the drive, and mobile use needs batteries or overhead lines.

(i) Group drive vs individual drive

In a group drive one motor drives several machines through a line shaft, belts and pulleys. In an individual drive each machine has its own motor.

 Group drive                  Individual drive
   M==line shaft==            M1-[m/c 1]
     |    |    |              M2-[m/c 2]
   [m1] [m2] [m3]             M3-[m/c 3]
   (belts, pulleys)

(ii) Constant torque drive vs constant power drive

In a DC separately excited motor, $T = k\phi I_a$ and $P = T\omega$. Below base speed, speed is raised by armature voltage control with full field; above base speed, by field weakening at rated voltage.

 T, P
   |  T const    |
   |-------------+.
   |           / |  ' .  T ~ 1/w
   |    P    /   |-------------- P const
   |       /     |
   |     /       |
   +---/---------+--------------> speed
     armature    base   field
     voltage     speed  weakening

In a group drive, one motor drives several machines through a line shaft, belts and pulleys. In a multi-motor drive, one machine has several motors, each driving a separate part of it (e.g. a crane with hoist, trolley and travel motors).

Group drive

• Merits: low initial cost (one large motor); motor can be rated for the diversity of loads, so total rating is less.
• Demerits: low efficiency; failure of the motor stops all machines; no individual speed control; belts and pulleys are unsafe, noisy and untidy; machine layout fixed along the shaft.

Multi-motor drive

• Merits: each operation gets the most suitable motor and speed; precise, automatic coordination of sections; high efficiency; compact, safe and flexible.
• Demerits: high initial cost; complex control and coordination system; needs skilled maintenance.

An electric drive is a system that controls the motion of a machine using an electric motor. Its main parts are the power source, power modulator, motor, load, and control unit with sensors.

Block diagram

          +-----------+     +-------+     +------+
 Source ->|   Power   |---->| Motor |====>| Load |
 (AC/DC)  | modulator |     +-------+     +------+
          +-----------+        |  ^ speed/current/
                ^              |  | position sensing
                |   +----------+--+
                +---| Control unit|<--- Input command
                    +-------------+     (reference)

1. Source: single- or three-phase AC (50 Hz), or DC (battery, DC mains).
2. Power modulator: controls the flow of power from source to motor so the motor gets the voltage, current and frequency needed. Examples: controlled rectifiers, choppers, inverters, cycloconverters, AC voltage controllers; also resistors and contactors in older drives. Functions: limit starting current, convert source energy to suit the motor, select motoring or braking mode.
3. Electric motor: DC (series, shunt, separately excited), induction, synchronous, BLDC, stepper; converts electrical energy to mechanical.
4. Load: the machine (pump, fan, crane, conveyor, machine tool) with its own torque-speed requirement.
5. Control unit: compares the command with feedback from sensors (speed, current, position) and generates control signals for the modulator. Built with microcontrollers, DSPs, PLCs, op-amps.
6. Sensing unit: tachometer/encoder (speed, position), CTs/Hall sensors (current), voltage sensors; used for closed-loop control and protection.

Why a series motor is started on load

• Series motor speed is roughly $N \propto \dfrac{E_b}{\phi}$ and $\phi \propto I_a$.
• At no load, armature current is very small, so the flux is very small and the speed rises to a dangerously high value ("runaway"). Centrifugal forces can damage the armature winding and commutator.
• Also, $T \propto \phi I_a \propto I_a^2$, so it develops high starting torque and is able to start heavy loads.
• Hence a series motor is always started on load and connected directly (gear, not belt) to the load, so the load can never come off.

 N
 |\              series motor
 | \             (N very high
 |  \            at light load)
 |   '--.__
 |         ''---___
 +-------------------> T or Ia

Why a shunt motor is started at no load

• The field is across the supply, so the flux is nearly constant and the speed is nearly constant from no load to full load; there is no danger of overspeed at no load.
• Starting current is limited only by a starter resistance; at starting $E_b = 0$, so $I_a = V/(R_a+R_{st})$. Starting at no (light) load keeps the starting current and the accelerating time small, so the starter and armature are not overheated.
• Torque $T \propto \phi I_a$ (only proportional to $I_a$), so its starting torque per ampere is lower than a series motor's; starting under heavy load would need a large current for a long time.
• Hence shunt motors are started at no load or light load, and the load is applied after the motor reaches speed.

(Important: the field of a shunt motor must never open while running, otherwise the speed rises dangerously.)

Assumptions: the belt's own mass and friction are neglected (not given); the whole 100 m length carries material.

Mass of material on the belt

$$\begin{aligned} \text{rate} &= \frac{200\times1000}{3600} = 55.56\ \text{kg/s} \\ t_{belt} &= \frac{100}{2.2} = 45.45\ \text{s} \\ M &= 55.56 \times 45.45 = 2525.3\ \text{kg} \end{aligned}$$

Motor speed

$$\omega_m = \frac{2\pi\times1500}{60} = 157.08\ \text{rad/s}$$

i) Load inertia referred to motor shaft

Equating kinetic energies, $\tfrac12 J_l\omega_m^2 = \tfrac12Mv^2$:

$$\begin{aligned} J_l &= M\left(\frac{v}{\omega_m}\right)^2 = 2525.3\times\left(\frac{2.2}{157.08}\right)^2 \\ &= 2525.3\times1.962\times10^{-4} = 0.495\ \text{kg-m}^2 \end{aligned}$$

ii) Torque to accelerate in 8 s

$$J = J_m + J_l = 0.1 + 0.495 = 0.595\ \text{kg-m}^2$$ $$\frac{d\omega_m}{dt} = \frac{157.08}{8} = 19.63\ \text{rad/s}^2$$ $$T = J\frac{d\omega_m}{dt} = 0.595\times19.63 = 11.69\ \text{N-m}$$

The conveyor is horizontal and friction is neglected, so no steady load torque is added.

Answer: (i) $J_l = 0.495$ kg-m²; (ii) accelerating torque $\approx 11.7$ N-m.

Initial steady state. Motor torque equals load torque:

$$100 - 0.1N = 0.05N \;\Rightarrow\; N_0 = \frac{100}{0.15} = 666.67\ \text{rpm}$$

After the change. Motor characteristic $T = -100 - 0.1N$ (valid for positive and negative speed). The passive load torque $0.05N$ always opposes motion, so the same expression holds for negative speed. New steady state:

$$-100 - 0.1N = 0.05N \;\Rightarrow\; N_f = -666.67\ \text{rpm}$$

Dynamic equation ($\omega = \frac{2\pi}{60}N$):

$$J\frac{2\pi}{60}\frac{dN}{dt} = T - T_l = -100 - 0.15N$$ $$\frac{dN}{dt} = -\frac{0.15}{J(2\pi/60)}\left(N + 666.67\right)$$

Time constant:

$$\tau = \frac{J(2\pi/60)}{0.15} = \frac{10\times0.10472}{0.15} = 6.981\ \text{s}$$

Solution:

$$N(t) = N_f + (N_0 - N_f)e^{-t/\tau} = -666.67 + 1333.33\,e^{-t/6.981}$$

Since the speed approaches $-666.67$ rpm only asymptotically, reversal is taken as complete when the speed reaches 95 % of the final value, i.e. $N = -633.33$ rpm (usual textbook assumption, Dubey).

$$\begin{aligned} t_r &= \tau\ln\frac{N_0 - N_f}{N - N_f} = 6.981\ln\frac{666.67+666.67}{-633.33+666.67} \\ &= 6.981\ln\frac{1333.33}{33.33} = 6.981\ln 40 = 25.75\ \text{s} \end{aligned}$$

(Braking to zero speed alone takes $6.981\ln 2 = 4.84$ s.)

Answer: reversal time $\approx 25.75$ s (to 95 % of the reverse steady-state speed of $-666.67$ rpm).

A hoist (lift/mine winder) has a cage, a counterweight and a rope over a drum driven by a motor. The load torque from gravity is an active, constant torque: its direction does not change when the motor reverses. Taking upward motion of the cage as positive speed and motor torque in the hoisting direction as positive, the motor works in all four quadrants of the speed-torque plane.

                 +N (cage up)
                  |
   II: Forward    |   I: Forward
   braking        |   motoring
   (empty cage up)|   (loaded cage up)
   T<0, N>0       |   T>0, N>0
 -T --------------+--------------- +T
   III: Reverse   |   IV: Reverse
   motoring       |   braking
   (empty cage    |   (loaded cage
    down)         |    down)
   T<0, N<0       |   T>0, N<0
                  |
                 -N (cage down)

 Load torque lines (constant):
   loaded cage : Tl = +Tl1 (vertical line, right)
   empty cage  : Tl = -Tl2 (vertical line, left;
                 counterweight heavier than empty cage)

Assumption: counterweight is heavier than the empty cage but lighter than the loaded cage, so the load torque is positive for a loaded cage and negative for an empty cage.

Quadrant I: forward motoring

• Loaded cage is raised. Motor torque and speed are both positive (upward).
• Power flows from supply to motor to load: motor supplies the energy to raise the load ($P = T\omega > 0$).

Quadrant II: forward braking

• Empty cage is moving up, but the heavier counterweight would pull it up faster. The motor torque is reversed (negative) to hold the speed.
• Speed positive, torque negative: $P < 0$, the motor works as a generator; energy can be returned to the supply (regenerative braking) or dissipated (dynamic braking).

Quadrant III: reverse motoring

• Empty cage is lowered. Since the counterweight is heavier, the motor must drive the cage downward against it.
• Torque and speed both negative: $P > 0$, motoring in reverse direction.

Quadrant IV: reverse braking

• Loaded cage is lowered. Gravity tends to accelerate it downward, so the motor produces positive (upward) torque to limit the speed.
• Speed negative, torque positive: $P < 0$, generating (regenerative braking, or dynamic/plugging).

With a constant load torque, steady operation in each quadrant occurs where the motor characteristic crosses the vertical load-torque line. Drives such as DC motors with dual converters or induction motors with regenerative VFDs provide all four quadrants and save energy by regeneration in II and IV.

Characteristics considered in motor selection

Electrical characteristics

1. Starting characteristics: starting torque (high for cranes, traction, crushers; low for fans) and starting current (limited by supply capacity; starter type).
2. Running (speed-torque) characteristics: constant speed (shunt, synchronous, induction), variable speed (series), match with load characteristic for stable operation.
3. Speed control: range, smoothness, method (armature voltage, field, V/f, slip control) and efficiency over the range.
4. Braking characteristics: availability of plugging, dynamic and regenerative braking.
5. Efficiency and power factor at the expected load.
6. Supply: AC/DC, voltage, phases, frequency.
7. Overload capacity and duty cycle (S1 to S8).

Mechanical characteristics

1. Type of enclosure: open, drip-proof, TEFC, flameproof, according to environment.
2. Bearings: ball/roller/sleeve; axial and radial load.
3. Transmission of drive: direct coupling, belt, gear, chain.
4. Mounting: foot, flange, vertical/horizontal.
5. Noise and vibration levels.
6. Cooling method; size and weight; cost and maintenance.

Rheostatic (dynamic) braking

The motor is disconnected from the supply and connected across a braking resistor. Driven by the stored kinetic energy, it acts as a generator; the generated energy is dissipated as heat in the resistor, producing a braking torque.

• DC shunt/separately excited motor: armature disconnected and connected to resistor $R_B$; field kept excited.

$$I_a = -\frac{E_b}{R_a + R_B}, \qquad T = K\phi I_a \;(\text{negative})$$

• DC series motor: field connection reversed (or kept in self-excited form) so the generated current builds up the flux.
• Induction motor: stator disconnected from AC and fed with DC (DC dynamic braking); the rotor cuts the stationary field and the energy is dissipated in rotor resistance.

   +----[Ra]---(Eb)----+
   |                   |
   +-------[ RB ]------+   (field excited separately)

Braking torque falls as speed falls (since $E_b \propto N$), so it cannot hold a load at standstill; mechanical brakes are used to stop finally. Simple and cheap, but energy is wasted.

Regenerative braking

The motor runs faster than its no-load speed (driven by the load, e.g. a lowering hoist or a train going downhill), so the back emf exceeds the supply voltage ($E_b > V$ for DC, or rotor speed above synchronous speed for an induction motor, slip negative). Current reverses, the machine acts as a generator and returns energy to the supply.

$$I_a = \frac{V - E_b}{R_a} < 0 \quad \text{when } E_b > V$$

• Used in electric traction, hoists, lifts, cranes, EVs.
• Advantages: energy saving, less brake wear.
• Requirements: supply must be able to accept energy (receptive line, or inverter/converter capable of reverse power flow); cannot bring the motor to rest by itself.

Set-up. Motor speed $\omega_m$ (rad/s, $\omega_m = 2\pi N/60$). Load $L_1$ directly coupled, so it runs at $\omega_m$. Load $L_2$ runs through a gear of ratio $K$, so $\omega_2 = \omega_m/K$. Gear assumed lossless (efficiency $\eta$ given in a remark).

         J1, T1           gear 1:K    J2, T2
 [Motor]======[L1]======[G]=========[L2]
  Jm, Te   w_m                         w_m / K

Equivalent inertia referred to motor shaft

The kinetic energy must be the same:

$$\frac12 J\omega_m^2 = \frac12J_m\omega_m^2 + \frac12J_1\omega_m^2 + \frac12J_2\left(\frac{\omega_m}{K}\right)^2$$ $$J = J_m + J_1 + \frac{J_2}{K^2}$$

Equivalent load torque referred to motor shaft

The power must be the same (lossless gear):

$$T_l\,\omega_m = T_1\omega_m + T_2\frac{\omega_m}{K} \;\Rightarrow\; T_l = T_1 + \frac{T_2}{K}$$

With gear efficiency $\eta$, $T_l = T_1 + \dfrac{T_2}{K\eta}$.

So both inertia and torque of the low-speed load appear smaller at the motor shaft (by $K^2$ and $K$ respectively), which is why a geared high-speed motor is used for slow, heavy loads.

Dynamic equation

The net torque accelerates the total equivalent inertia:

$$T_e - T_l = J\frac{d\omega_m}{dt}$$ $$T_e = \left(T_1 + \frac{T_2}{K}\right) + \left(J_m + J_1 + \frac{J_2}{K^2}\right)\frac{d\omega_m}{dt}$$

In terms of speed in rpm, $\dfrac{d\omega_m}{dt} = \dfrac{2\pi}{60}\dfrac{dN}{dt}$.

• If $T_e > T_l$: the drive accelerates.
• If $T_e < T_l$: the drive decelerates.
• If $T_e = T_l$: steady speed (equilibrium).

An electric drive is a system in which an electric motor, with its control equipment, is used to produce and control the motion of a machine (for example, a pump, conveyor, crane or machine tool).

Major parts

 Source -->[Power modulator]-->[Motor]==>[Load]
                ^                 |
                |   [Control]<----+ sensors
                +-----  unit  <--- command

1. Power source: AC (single- or three-phase) or DC supply.
2. Power modulator: regulates power from the source to the motor: controlled rectifiers, choppers, inverters, cycloconverters, AC voltage controllers, or starters and resistors. It limits starting current, gives speed control and allows braking.
3. Electric motor: DC motor, induction motor, synchronous motor, BLDC, stepper or servo motor.
4. Load: the driven machine; its torque-speed characteristic decides the motor and control.
5. Control unit: generates firing/switching signals using the command and feedback; analog circuits, microcontroller, DSP or PLC.
6. Sensing unit: speed, current, voltage and position sensors for closed-loop control and protection.

Advantages

• Wide range of power, speed and torque available.
• Easy starting, stopping, reversing, speed control and braking; easily automated or remote controlled.
• Operation in all four quadrants; regenerative braking saves energy.
• High efficiency; no fuel storage or exhaust; clean and quiet.
• Ready for full load immediately; short-time overload capacity.
• Low maintenance, long life, compact.

Disadvantages

• Depends on continuous electric supply; supply failure stops the drive.
• Not suitable for mobile use unless batteries or overhead supply are available.
• Power electronic converters inject harmonics and may reduce power factor.
• Initial cost of the drive with control can be high.

In a group drive a single motor drives several machines through a line shaft, belts and pulleys. In an individual drive each machine has its own separate motor.

 Group:  M ==== line shaft ====
              |      |      |
            [m/c1] [m/c2] [m/c3]

 Individual: M1-[m/c1]  M2-[m/c2]  M3-[m/c3]

Assumptions: the drive starts from rest at $t = 0$; the load torque $T_L = 5$ N-m is referred to the motor shaft and stays constant (as given) in every section.

Equivalent inertia

Load 2 runs at $\omega_{m2} = \dfrac{n_1}{n_2}\omega_m = 2\omega_m$. Equating kinetic energies:

$$\begin{aligned} J_{eq} &= J_1 + \left(\frac{n_1}{n_2}\right)^2J_2 = 0.08 + (2)^2(0.01) \\ &= 0.08 + 0.04 = 0.12\ \text{kg-m}^2 \end{aligned}$$

Speed profile

From $T_e - T_L = J_{eq}\dfrac{d\omega_m}{dt}$, in each section the acceleration is constant:

$$\alpha = \frac{T_e - 5}{0.12}$$

Sample working, section A: $\omega = 0 + 300\times0.1 = 30$ rad/s; section C: $\omega = 30 - 600\times0.05 = 0$.

 Te (N-m)
  41 |___                    ___
   5 |   |____          ____|   |_____
   0 +---+----+----+----+---+---+-----> t (s)
 -67 |        |____|____|
     0  .1   .2  .25  .3   .4  .5
          A    B   C  D   E   F

 w (rad/s)
  30 |    /------\
     |   /        \
   0 +--/----------\-----------/------> t
     |              \         /
 -30 |               \-------/
     0  .1   .2  .25  .3   .4  .5

Quadrant of operation

Answer: $J_{eq} = 0.12$ kg-m²; speed rises to 30 rad/s (A), holds (B), reverses to -30 rad/s (C, D), holds (E) and returns to zero at 0.5 s (F); quadrants I, I, II, III, IV, IV for A to F.

For matching a motor to a load, the load torque $T_l$ that the motor must overcome is split into components. With inertia, the motor torque equation is

$$T = T_l + J\frac{d\omega_m}{dt}, \qquad T_l = T_F + T_W + T_L$$

Components of load torque

1. Friction torque $T_F$ (present in all machines, at bearings, gears, couplings, brushes). Its parts:

• Static friction (stiction) $T_s$: present only at standstill; must be overcome to start. Large for some loads (e.g. loaded conveyor, rolling mill).
• Coulomb friction $T_C$: constant, independent of speed.
• Viscous friction $T_v = B\omega_m$: proportional to speed, due to lubricant.

2. Windage torque $T_W$: due to air (or fluid) resistance of moving parts, approximately proportional to speed squared: $T_W = C\omega_m^2$.

3. Torque for useful mechanical work $T_L$: depends on the nature of the load:

 TL
  |\  1/w (constant power)
  | \          w^2 (fan)
  |  \        /
  |___\______/________ constant (hoist)
  |    '.  /
  |      /'-..___
  |    /  w (linear)
  +-------------------> speed

Total friction characteristic

 TF
  | Ts (stiction)
  |*.
  |   '.  ___----- Tv + Tc (rises with speed)
  |     '--
  |  Tc
  +--------------> w

At very low speed the total friction is high (stiction) and then dips before rising with speed. Usually the friction and windage are combined with the useful load torque in one load characteristic $T_l(\omega)$.

Active and passive load torques

• Active torques: produced by gravity, tension, compression or torsion; keep their sign when the drive reverses (hoist, lift, locomotive on a gradient).
• Passive torques: always oppose motion, so change sign with speed (friction, cutting, windage).

Knowing these components and their dependence on speed lets the designer check starting (stiction), steady-state stability and the motor rating.

Data: $m = 500$ kg, $v = 1.5$ m/s, $N = 1000$ rpm, $J_m = 0.5$ kg-m², $J_{winch} = 0.3$ kg-m² (taken at motor speed), no-load (friction) torque $= 100$ N-m at 1000 rpm.

$$\omega_m = \frac{2\pi\times1000}{60} = 104.72\ \text{rad/s}$$

Motor torque

Torque to lift the weight, from power balance $T\omega_m = Fv$:

$$\begin{aligned} T_w &= \frac{mgv}{\omega_m} = \frac{500\times9.81\times1.5}{104.72} \\ &= \frac{7357.5}{104.72} = 70.26\ \text{N-m} \end{aligned}$$

Without the weight the motor must still supply 100 N-m (friction and other losses at 1000 rpm). At uniform speed there is no accelerating torque, so

$$T = 100 + 70.26 = 170.26\ \text{N-m}$$

Equivalent moment of inertia at motor shaft

The weight moving at $v$ has kinetic energy $\tfrac12mv^2$, referred to the motor shaft as $J_w = m(v/\omega_m)^2$:

$$\begin{aligned} J_w &= 500\times\left(\frac{1.5}{104.72}\right)^2 = 0.1026\ \text{kg-m}^2 \\ J &= J_m + J_{winch} + J_w = 0.5 + 0.3 + 0.1026 = 0.9026\ \text{kg-m}^2 \end{aligned}$$

Answer: motor torque $\approx 170.3$ N-m; equivalent moment of inertia $\approx 0.903$ kg-m².

Electric drives are classified by how motors are arranged with respect to machines.

• Group drive: one motor drives several machines through a line shaft, belts and pulleys.
• Individual drive: each machine has its own motor (one motor may drive all mechanisms of that machine through gears).
• Multi-motor drive: one machine has several motors, each driving a separate mechanism (e.g. an overhead crane with hoist, trolley and bridge-travel motors).

 Group          Individual        Multi-motor
 M=shaft=       M1-[m/c1]         +--- machine ---+
  | | |         M2-[m/c2]         | M1  M2  M3    |
 m1 m2 m3       M3-[m/c3]         | hoist trolley |
                                  |       travel  |
                                  +---------------+

Group drive is cheap but wasteful and inflexible. Individual drive is the normal choice in modern industry. Multi-motor drive gives the best performance and automatic control for complex machines, at higher cost.

An electric drive is a system that uses an electric motor, together with its control and power-modulating equipment, to control the motion (speed, torque, position) of a machine.

Types of electric drives

A. By arrangement of motors

1. Group drive: one motor runs several machines via a line shaft.
   • Merits: low initial cost; smaller total motor rating due to diversity.
   • Demerits: low efficiency; motor fault stops everything; no individual speed control; belts unsafe and noisy; rigid layout.
2. Individual drive: each machine has its own motor.
   • Merits: high efficiency, independent control, flexible layout, fault affects one machine, clean and safe.
   • Demerits: higher initial cost.
3. Multi-motor drive: several motors on one machine, each for one mechanism (cranes, rolling mills).
   • Merits: best control and automation, each motor sized to its job, high productivity.
   • Demerits: costly, complex coordination.

B. By type of motor/supply

C. By control

• Constant speed drives vs variable speed drives.
• Open-loop vs closed-loop (feedback) drives: closed-loop gives accurate speed and current limiting.
• Reversible / non-reversible, single-quadrant / four-quadrant drives.

D. By power modulator: rectifier-fed DC drives, chopper-fed DC drives, inverter-fed AC drives, cycloconverter-fed AC drives.

General merits of electric drives

• Wide range of torque, speed and power; easily controlled and automated.
• Four-quadrant operation and regenerative braking.
• No exhaust or fuel storage; clean, quiet; high efficiency; immediately available.

General demerits

• Need continuous power supply; supply failure stops the drive.
• Converters cause harmonics; initial cost of modern drives is high.

Equilibrium points: where motor torque equals load torque, $T_m = T_L$:

$$\begin{aligned} -(1 + 2\omega_m) &= -3\sqrt{\omega_m} \\ 2\omega_m - 3\sqrt{\omega_m} + 1 &= 0 \end{aligned}$$

Let $x = \sqrt{\omega_m}$: $2x^2 - 3x + 1 = 0 \Rightarrow (2x-1)(x-1) = 0$, so $x = 1$ or $x = 0.5$.

$$\omega_m = 1\ \text{rad/s} \quad \text{and} \quad \omega_m = 0.25\ \text{rad/s}$$

Stability criterion. An equilibrium point is steady-state stable if a small speed increase produces a net decelerating torque:

$$\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} > 0$$

Derivatives:

$$\frac{dT_m}{d\omega_m} = -2, \qquad \frac{dT_L}{d\omega_m} = -\frac{3}{2\sqrt{\omega_m}}$$

At $\omega_m = 1$:

$$\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2} + 2 = 0.5 > 0 \;\Rightarrow\; \text{stable}$$

At $\omega_m = 0.25$:

$$\frac{dT_L}{d\omega_m} - \frac{dT_m}{d\omega_m} = -\frac{3}{2\times0.5} + 2 = -3 + 2 = -1 < 0 \;\Rightarrow\; \text{unstable}$$

Check at $\omega_m = 0.25$: torque $T_m = T_L = -1.5$ N-m; at $\omega_m = 1$: $T_m = T_L = -3$ N-m.

Answer: equilibrium points at $\omega_m = 1$ rad/s (stable) and $\omega_m = 0.25$ rad/s (unstable).