Electric Traction

Data: $D = 2$ km, $V_{avg} = 40$ km/h, $V_{max} \le 60$ km/h, $\alpha = 2$, $\beta_c = 0.15$, $\beta = 3$ km/h/s.

Running time:

$$T = \frac{D}{V_{avg}} = \frac{2}{40}\times3600 = 180\ \text{s}$$

Check with 60 km/h reached

If the train accelerates to 60 km/h ($t_1 = 30$ s), coasts to $V_2$ and brakes, the time condition $30 + \frac{60-V_2}{0.15} + \frac{V_2}{3} = 180$ gives $V_2 = 39.47$ km/h, but the distance covered would be 2.21 km, more than 2 km. So the train does not need to reach 60 km/h; the peak speed $V_1$ is found from both conditions (and checked $\le 60$).

 V  V1 /\
      /  \___  coasting (-0.15)
     /       ----___ V2
    / accel 2       |\ brake 3
   /________________|_\____> t
   0   t1      t2      t3

Equations (speed in km/h, time in s; distance in km = area/3600)

$$\begin{aligned} \text{Time:}\quad & \frac{V_1}{2} + \frac{V_1 - V_2}{0.15} + \frac{V_2}{3} = 180 \\ \text{Distance:}\quad & \frac{V_1^2}{2\times2} + \frac{V_1^2 - V_2^2}{2\times0.15} + \frac{V_2^2}{2\times3} = 2\times3600 = 7200 \end{aligned}$$

From the time equation: $V_2 = \dfrac{7.1667\,V_1 - 180}{6.3333}$. Substituting in the distance equation and solving (numerically):

$$V_1 = 54.88\ \text{km/h}, \quad V_2 = 33.68\ \text{km/h}$$

$V_1 < 60$ km/h, so the speed limit is respected.

Periods

$$\begin{aligned} t_1 &= \frac{V_1}{\alpha} = \frac{54.88}{2} = 27.44\ \text{s} \\ t_2 &= \frac{V_1 - V_2}{\beta_c} = \frac{54.88 - 33.68}{0.15} = 141.33\ \text{s} \\ t_3 &= \frac{V_2}{\beta} = \frac{33.68}{3} = 11.23\ \text{s} \end{aligned}$$

Check: $27.44 + 141.33 + 11.23 = 180$ s.

Distances: acceleration $= \frac{54.88\times27.44}{2\times3600} = 0.209$ km; coasting $= \frac{(54.88+33.68)\times141.33}{2\times3600} = 1.738$ km; braking $= \frac{33.68\times11.23}{2\times3600} = 0.053$ km; total = 2.0 km.

Answer: acceleration 27.4 s, coasting 141.3 s, braking 11.2 s (peak speed 54.9 km/h, braking starts at 33.7 km/h).

Data: $D = 0.777$ km, $V_s = 27.2$ km/h, stop $= 20$ s, $\beta = 3.22$ km/h/s, $V_m = 1.2\,V_a$. Simplified trapezoidal curve (acceleration, free running, braking). The train mass does not affect the kinematics.

Running time and speeds

$$\begin{aligned} T_s &= \frac{0.777}{27.2}\times3600 = 102.84\ \text{s} \\ T &= 102.84 - 20 = 82.84\ \text{s} \\ V_a &= \frac{0.777\times3600}{82.84} = 33.77\ \text{km/h} \\ V_m &= 1.2\times33.77 = 40.52\ \text{km/h} \end{aligned}$$

Trapezoidal relation

$$D\times3600 = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$\begin{aligned} 2797.2 &= 40.52\times82.84 - \frac{40.52^2}{2}\left(\frac1\alpha + \frac{1}{3.22}\right) \\ \frac1\alpha + \frac{1}{3.22} &= \frac{2(3356.7 - 2797.2)}{1641.9} = 0.6815 \\ \frac1\alpha &= 0.6815 - 0.3106 = 0.3709 \\ \alpha &= 2.70\ \text{km/h/s} \end{aligned}$$

Check: $t_1 = 40.52/2.696 = 15.03$ s, $t_3 = 40.52/3.22 = 12.58$ s, free run $= 82.84 - 15.03 - 12.58 = 55.23$ s; distance $= 40.52(55.23 + (15.03+12.58)/2)/3600 = 0.777$ km.

Answer: Required acceleration ≈ 2.70 km/h/s.

A self-contained electric vehicle (self-contained locomotive) carries its own source of electrical energy on board, so it needs no overhead wire or third rail. The energy is produced or stored on the vehicle and supplied to electric traction motors that drive the wheels.

Types

• Diesel-electric locomotive: diesel engine drives a generator/alternator; output feeds traction motors.
• Battery electric vehicle: lead-acid or lithium-ion batteries feed dc motors (shunting locos, mine locos, e-cars, e-rickshaws).
• Petrol-electric / hybrid vehicles: engine plus generator and battery.
• Steam/gas-turbine electric locomotives (rare).

Transmission systems

The transmission connects the prime mover to the driving axles. In electric vehicles it is mainly electrical, followed by a gear drive.

1. DC–DC transmission (diesel-electric)

 Diesel -> DC generator -> DC series motors -> gear -> axle
 engine    (sep. excited)   (axle hung)

• Generator voltage varied by its field; engine runs at its best speed.
• Simple and robust, but generator commutator limits power.

2. AC–DC transmission

 Diesel -> 3-ph alternator -> rectifier -> DC series motors

• Alternator is lighter and cheaper, no commutator on the generator; widely used.

3. AC–DC–AC (AC–AC) transmission

 Diesel -> alternator -> rectifier -> inverter (VVVF) -> 3-ph IM

• Uses rugged cage induction motors, high adhesion, regenerative/rheostatic braking; modern locomotives.

4. Battery vehicles: battery → chopper (dc motor) or inverter (ac motor) → motor → reduction gear/differential.

Mechanical part: a single-reduction spur gear (gear ratio about 3–5) between motor pinion and axle, with axle-hung nose-suspended or frame-mounted motors.

Advantages

• Independent of track electrification; can run on any route; low initial cost of line.
• Electric transmission gives smooth control and high starting torque.

Disadvantages

• Heavy and costly per kW; low overall efficiency (diesel 25–30 %).
• Limited overload capacity; maintenance of engine; fuel cost and pollution; battery vehicles have limited range.

Electric traction is the system of driving vehicles (trains, trams, trolley buses) by electric motors, with energy taken from an overhead line, third rail or on-board source.

Advantages of electric traction

1. Clean: no smoke or exhaust; ideal for cities and tunnels.
2. High starting torque and fast acceleration, so higher schedule speed and more trains per track.
3. Regenerative braking saves 15–30 % energy and reduces brake wear.
4. Low maintenance: about half the maintenance cost of steam or diesel locomotives.
5. High overall efficiency when power comes from central (e.g. hydro) stations.
6. Higher coefficient of adhesion and no fuel storage on board; lower centre of gravity.
7. Locomotive is ready to start at once; no idle running.
8. In Nepal, uses domestic hydropower instead of imported fuel.

Disadvantages: high capital cost of overhead lines and substations, interference with telecommunication lines, and dependence on supply.

Comparison

Example: Kathmandu's trolley bus (Tripureshwor–Suryabinayak, 1975–2009) used two overhead wires, as the rubber tyres cannot return current through rails.

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 2$ km, stop $= 26$ s.

Trapezoidal relation

$$D = \frac{1}{3600}\left[V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)\right]$$ $$K = \frac1\alpha + \frac1\beta = \frac1{0.8} + \frac1{3.2} = 1.25 + 0.3125 = 1.5625$$

Average speed $V_a = \dfrac{3600D}{T}$, so $V_m T = 1.3\times3600D = 1.3\times7200 = 9360$.

$$\begin{aligned} 7200 &= 9360 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times2160}{1.5625} = 2764.8 \\ V_m &= 52.58\ \text{km/h} \end{aligned}$$

Times and speeds

$$\begin{aligned} T &= \frac{9360}{52.58} = 178.01\ \text{s} \\ V_a &= \frac{7200}{178.01} = 40.45\ \text{km/h} \\ T_s &= 178.01 + 26 = 204.01\ \text{s} \\ V_s &= \frac{2\times3600}{204.01} = 35.29\ \text{km/h} \end{aligned}$$

Check: $t_1 = 52.58/0.8 = 65.7$ s, $t_3 = 52.58/3.2 = 16.4$ s, free run $= 95.9$ s.

Answer: Schedule speed = 35.3 km/h (maximum speed 52.6 km/h).

Tractive effort $F_t$ is the force developed by the locomotive at the rim of the driving wheels. It must supply three components:

Components

1. Force for acceleration (including rotating parts). With effective mass $W_e = 1.1W$ (10 % allowance for rotational inertia), $W$ in tonnes and $\alpha$ in km/h/s:

$$F_a = 1000\,W_e \times \frac{1000}{3600}\alpha = 277.8\,W_e\,\alpha\ \ \text{N}$$

2. Force to overcome gradient $G$ (% = rise in metres per 100 m of track):

$$F_g = 1000\,W\,g\,\sin\theta \approx 1000\,W\times9.81\times\frac{G}{100} = 98.1\,W\,G\ \ \text{N}$$

3. Force to overcome train resistance $r$ (N per tonne: friction, air resistance, track):

$$F_r = W r\ \ \text{N}$$

Up and down gradient

$$F_t = 277.8\,W_e\,\alpha \pm 98.1\,W G + W r\ \ \text{N}$$

• Up gradient (+): the weight component opposes motion, so more tractive effort is needed.
• Down gradient (−): the weight component helps motion, so less effort is needed. If $98.1G > r$, the train accelerates even with power off, and braking is needed to hold speed.

      F_t -->  [train]
            /   | W g
          /     v   (W g sin theta acts down slope)
        /  theta

Example

A 400 t train ($W_e = 440$ t) accelerating at 1.5 km/h/s with $r = 50$ N/t on a 1 % gradient:

• Up: $F_t = 277.8\times440\times1.5 + 98.1\times400\times1 + 400\times50 = 183348 + 39240 + 20000 = 242588$ N ≈ 242.6 kN
• Down: $F_t = 183348 - 39240 + 20000 = 164108$ N ≈ 164.1 kN

Power at the wheels: $P = F_t v$ (W, with $v$ in m/s).

Importance of electric traction in Nepal

Electric traction uses electric motors to drive vehicles, with energy from overhead lines, rails or batteries. It is very relevant to Nepal because:

1. Hydropower surplus: Nepal has large hydro potential (about 83,000 MW theoretical, ~42,000 MW feasible) and now spills energy in the wet season. Traction gives a domestic market for this clean energy.
2. Reduced fuel import: petroleum products are a large part of the import bill and trade deficit; electric vehicles cut this.
3. Pollution: Kathmandu valley has severe air pollution from diesel vehicles; electric traction has zero local emission.
4. Efficiency and cost: electric motors are 85–95 % efficient; running cost per km is much lower than diesel.
5. Hilly terrain: high starting torque and regenerative braking suit steep gradients.
6. Projects: Janakpur–Jaynagar railway, the planned Mechi–Mahakali electric railway, Kerung–Kathmandu railway study, metro/monorail studies for Kathmandu, electric buses (Sajha Yatayat), e-rickshaws and the earlier Kathmandu trolley bus (1975–2009). Government tax incentives promote EVs.
7. Energy security and employment in the electricity sector.

Challenges: high capital cost, reliability of supply, charging/feeder infrastructure, and difficult terrain for railways.

Self-contained vehicles

A self-contained vehicle produces or stores its electrical energy on board, so no overhead line is needed. Examples: diesel-electric locomotives, battery electric vehicles (e-buses, e-cars, mine and shunting locos), and hybrid vehicles. They can run on any route but are heavier, costlier per kW and less efficient than line-fed vehicles.

Transmission systems

 DC-DC: Diesel -> DC generator -> DC series motors -> gear -> axle
 AC-DC: Diesel -> alternator -> rectifier -> DC series motors
 AC-AC: Diesel -> alternator -> rectifier -> VVVF inverter -> IM
 Battery: Battery -> chopper/inverter -> motor -> gear/differential

• DC–DC: generator field controls voltage; simple but commutator of generator limits rating.
• AC–DC: lighter alternator with silicon rectifier; most common in diesel locos.
• AC–DC–AC: rugged three-phase induction motors, better adhesion, regenerative or rheostatic braking.
• Battery vehicles: chopper control (dc motor) or inverter with PMSM/induction motor; regenerative braking recharges the battery.

The final mechanical drive is a single-reduction gear from motor pinion to axle (axle-hung, nose-suspended motor) or a differential in road vehicles.

Assumptions: $W_e = 1.1W$; 1 tonne train taken (answers per tonne); power is ON only during acceleration; during coasting the net force is gradient minus resistance.

Speeds and distances

Acceleration: $V_1 = 2\times10 = 20$ km/h.

Coasting on a 1 % down gradient: gradient force $= 98.1\times1 = 98.1$ N/t, resistance $= 40$ N/t, so the net force helps motion. Coasting acceleration:

$$\alpha_c = \frac{98.1 - 40}{277.8\times1.1} = 0.1901\ \text{km/h/s}$$

Speed at end of coasting:

$$V_2 = 20 + 0.1901\times50 = 29.51\ \text{km/h}$$

Braking in 15 s: $\beta = 29.51/15 = 1.967$ km/h/s.

Schedule speed

$$\begin{aligned} T_s &= 75 + 15 = 90\ \text{s} \\ V_s &= \frac{433.04}{90}\times3.6 = 17.32\ \text{km/h} \end{aligned}$$

Specific energy consumption

Tractive effort per tonne during acceleration (down gradient, so $G$ is negative):

$$\begin{aligned} F_t &= 277.8\,(1.1)\,\alpha - 98.1G + r \\ &= 277.8\times1.1\times2 - 98.1 + 40 = 553.06\ \text{N/t} \end{aligned}$$

Energy at wheels per tonne $= F_t \times d_1 = 553.06\times27.78 = 15362.8$ J/t.

Energy input (efficiency 75 %) $= 15362.8/0.75 = 20483.7$ J/t $= 5.690$ Wh/t.

$$\text{SEC} = \frac{5.690\ \text{Wh/t}}{0.43304\ \text{km}} = 13.14\ \text{Wh/t-km}$$

Answer: Schedule speed = 17.3 km/h; specific energy consumption = 13.14 Wh per tonne-km.

An electric traction system converts electrical energy from the grid (or an on-board source) into mechanical motion of a train. Its main components and their roles:

 Grid -> [Traction substation] -> [Feeder/OHE or 3rd rail]
        -> [Current collector] -> [Locomotive: transformer,
           converter, control] -> [Traction motor] -> [Gear]
        -> [Wheels/axle] -> Rail return -> substation

1. Traction substation

• Steps down grid voltage (132/66 kV) to traction voltage (25 kV ac, or rectified 600–3000 V dc).
• Spacing 40–60 km for 25 kV ac, 3–5 km for dc; reliability here sets service continuity.

2. Overhead equipment (OHE) / third rail and feeders

• Catenary wire with contact wire, droppers, masts and section insulators; third rail for metros.
• Keeps contact wire at a uniform height and tension so collection is smooth at high speed; voltage drop affects motor performance.

3. Current collector

• Pantograph (high speed), bow or trolley pole (trams), shoe for third rail.
• Must keep continuous contact without sparking; poor collection causes arcing and wear.

4. Locomotive transformer and power converter

• On ac locos: tap-changer transformer, rectifier, chopper or VVVF inverter; on dc: chopper or resistance control.
• Controls voltage/frequency to the motors, giving smooth acceleration, speed control and regenerative braking.

5. Traction motors

• DC series motor (high starting torque, speed adjusts to load) or three-phase induction motor (rugged, light, high adhesion).
• Determine tractive effort, speed range and efficiency.

6. Mechanical transmission

• Pinion and gear wheel (single reduction), axle-hung nose-suspended or frame-mounted motors.
• Gear ratio matches motor speed to wheel speed and sets tractive effort.

7. Braking system

• Regenerative, rheostatic and mechanical/air brakes.
• Safety, energy saving and reduced wear.

8. Control, protection and auxiliaries

• Driver controls, microprocessor control, circuit breakers, lightning arrestors, relays, compressors, blowers, lighting.
• Protects equipment and keeps the train running safely.

9. Track and return circuit

• Running rails carry the return current; bonded rails and earthing limit touch voltage and stray currents (corrosion).

10. Signalling and communication

• Track circuits and signals for safe headway; must be immune to traction current interference.

Overall performance (schedule speed, energy use, reliability) depends on all these working together: e.g. good adhesion needs smooth motor control; regenerative braking needs a receptive supply.

Assumption: braking at 3.3 km/h/s from the speed at end of coasting (the braking time then comes out as 11.6 s; the stated 10 s does not match this rate, see the note at the end).

(i) Acceleration

$$\alpha = \frac{48}{24} = 2\ \text{km/h/s}$$

(ii) Coasting retardation

Resistance 50 N/t, effective mass $1.1W$:

$$\begin{aligned} 277.8\times1.1\,\beta_c &= 50 \\ \beta_c &= \frac{50}{305.58} = 0.1636\ \text{km/h/s} \end{aligned}$$

(iii) Schedule speed (20 s stops)

Speed at end of coasting:

$$V_2 = 48 - 0.1636\times60 = 38.18\ \text{km/h}$$

Braking time: $t_3 = 38.18/3.3 = 11.57$ s.

$$\begin{aligned} T_s &= 95.57 + 20 = 115.57\ \text{s} \\ V_s &= \frac{939.55}{115.57}\times3.6 = 29.27\ \text{km/h} \end{aligned}$$

Effect of 15 s stops

$$\begin{aligned} T_s &= 95.57 + 15 = 110.57\ \text{s} \\ V_s &= \frac{939.55}{110.57}\times3.6 = 30.59\ \text{km/h} \end{aligned}$$

Increase $= 30.59 - 29.27 = 1.32$ km/h (about 4.5 %).

Answer: (i) 2 km/h/s, (ii) 0.164 km/h/s, (iii) 29.27 km/h; with 15 s stops the schedule speed rises to 30.59 km/h.

Note: if the braking period is taken as exactly 10 s from 38.18 km/h, the distance is 931.2 m and the schedule speeds are 29.41 km/h (20 s stops) and 30.76 km/h (15 s stops); the conclusion is the same.

Electric traction systems are classed by the supply to the vehicle: DC systems (600–750 V for trams and metros, 1500 V or 3000 V for main lines) and AC systems (single-phase 25 kV, 50 Hz industrial-frequency, older 15 kV 16⅔ Hz, and three-phase 3.3–3.6 kV). Today 25 kV, 50 Hz single-phase is standard for main lines, and 750 V dc for metros.

Comparison

DC system

Advantages: dc series motor gives high starting torque and natural speed–load matching; lighter and cheaper vehicle; less insulation and clearance (suits metros, tunnels); no telecom interference; good for frequent stop urban service. Disadvantages: costly rectifier substations close together; heavy overhead conductor or third rail; high losses; energy wasted in starting resistance (older); stray currents corrode underground pipes.

AC system

Advantages: fewer and simpler substations fed directly from the grid; light overhead line; low losses, longer feed distance; modern power electronics give efficient control and regenerative braking; lower overall cost for long-distance and heavy traffic. Disadvantages: heavier and costlier locomotive; unbalance and harmonics on the 3-phase grid; inductive interference in communication lines; higher insulation requirement.

Application

• DC: tramways, trolley buses, metro and suburban lines with frequent stops (e.g. 750 V dc third rail).
• AC 25 kV: main-line, long-distance and heavy freight (e.g. Indian Railways; proposed Nepal railways).
• Composite (ac line, dc motors via rectifier; or VVVF induction motors) combines the advantages and is now the usual choice.

Data: $W = 400$ t, $W_e = 1.1W$, $G = 1\ \%$ up, $r = 50$ N/t, $\eta = 75\ \%$. Power is ON during acceleration and constant speed; OFF during coasting and braking. Values per tonne.

Speed-time curve

• Acceleration: $V_1 = 1.6\times35 = 56$ km/h.
• Constant speed 45 s at 56 km/h.
• Coasting retardation (gradient + resistance oppose):

$$\beta_c = \frac{98.1\times1 + 50}{277.8\times1.1} = \frac{148.1}{305.58} = 0.4847\ \text{km/h/s}$$

$V_3 = 56 - 0.4847\times30 = 41.46$ km/h.

• Braking: $t_4 = 41.46/2.4 = 17.28$ s.

Tractive effort per tonne

$$\begin{aligned} F_{acc} &= 277.8\times1.1\times1.6 + 98.1\times1 + 50 = 488.93 + 148.1 = 637.03\ \text{N/t} \\ F_{const} &= 98.1 + 50 = 148.1\ \text{N/t} \end{aligned}$$

Energy per tonne

$$\begin{aligned} E_{wheel} &= 637.03\times272.22 + 148.1\times700 \\ &= 173413 + 103670 = 277083\ \text{J/t} \\ E_{input} &= \frac{277083}{0.75} = 369444\ \text{J/t} = 102.62\ \text{Wh/t} \end{aligned}$$

Specific energy consumption

$$\text{SEC} = \frac{102.62}{1.47778} = 69.44\ \text{Wh/t-km}$$

Total energy for the 400 t train $= 102.62\times400 = 41.05$ kWh per run.

Answer: Specific energy consumption ≈ 69.4 Wh per tonne-km (41.05 kWh for the whole run of 1.478 km).

Advantages of electric traction

1. No pollution at the point of use; suits cities, tunnels and underground railways.
2. High starting torque and acceleration, giving higher schedule speed and line capacity.
3. Regenerative braking returns energy to supply and reduces brake wear.
4. Lower maintenance and running cost; electric locomotives have long life and high availability.
5. High overall efficiency when energy comes from large power stations or hydropower.
6. Better adhesion and smooth, jerk-free control; can haul heavier trains on gradients.
7. Ready for service at once; no fuel or water stops.
8. Uses domestic energy (hydropower in Nepal) instead of imported fuel.

Traction fed from a separate distribution line

In this system the vehicle carries no energy source; it collects energy continuously from a distribution network laid along the route (overhead contact line or third rail), fed from traction substations. Electric trains, tramways and trolley buses use it.

 Grid (132/66 kV) -> traction substation -> feeders
   -> overhead contact wire / third rail
   -> collector (pantograph, trolley, shoe) -> vehicle
   -> traction motors -> running rails (return) -> substation

Supply systems used:

• DC system: 600–750 V (trams, metros), 1500–3000 V (suburban/main line). Rectifier substations every few km; dc series motors.
• Single-phase AC system: 25 kV, 50 Hz (or 15 kV, 16⅔ Hz). Substations every 40–60 km fed directly from the grid; on-board transformer with rectifier/inverter.
• Three-phase AC system: 3.3–3.6 kV, using two overhead wires plus rails; induction motors; now obsolete because of complex overhead wiring.
• Composite system: single-phase ac line with on-board conversion to dc (rectifier) or to variable-frequency three-phase ac (VVVF inverter) for traction motors; the modern standard.

Merits: vehicle is lighter and cheaper, high power available, high efficiency, regeneration possible. Demerits: high capital cost of track electrification; vehicles tied to the electrified route; supply failure stops all traffic; interference with communication lines.

Electric traction is the system of moving vehicles (trains, trams, trolley buses, electric locomotives) using electric motors, with energy supplied from a fixed distribution system along the track or from an on-board source.

For vehicles fed from a distribution line, a current collector keeps a sliding contact with the supply conductor while the vehicle moves.

1. Third (conductor) rail system

• An insulated steel rail laid beside or between the running rails; a collector shoe slides on it (top, side or bottom contact).
• Used for 600–750 V dc metros and suburban lines (e.g. London Underground, many older metros).
• Collection current high; low voltage only (safety).

2. Overhead system

Overhead wire hung above the track; collector on the roof presses against it.

a) Trolley collector (trolley pole)

• A pole with a grooved wheel or slider at its end, pressed upward by springs.
• Used in tramways and trolley buses (two poles, since return is not through rails).
• Suitable only for low speed (about 30 km/h); can leave the wire at junctions; must be reversed at route ends.

b) Bow collector

• A light metal strip (bow) on a frame, 1 m wide, sliding on the wire.
• Used in tramways; speeds up to about 30–35 km/h; needs reversing with direction (or reversible type).

c) Pantograph collector

• A hinged, diamond- or half-diamond (Z) frame raised by springs or air pressure, with a carbon/copper contact strip.
• Keeps uniform pressure over a wide range of wire height; works in both directions and at high speeds (over 300 km/h); can collect large currents.
• Used on all modern ac and dc main-line locomotives and EMUs.

   ===== contact wire ==========
        \____/    pantograph head
         /  \
        /    \   frame (springs/air)
   ____/______\____ vehicle roof

Comparison

The pantograph is preferred today for its reliability, high-speed performance and ability to handle high voltage; the third rail remains common for underground metros where tunnel height is small.

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 1.5$ km, stop $= 26$ s.

Trapezoidal relation (speeds in km/h, time in s)

$$3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$K = \frac1{0.8} + \frac1{3.2} = 1.5625$$

Since $V_a = 3600D/T$ and $V_m = 1.3V_a$: $V_m T = 1.3\times3600\times1.5 = 7020$.

$$\begin{aligned} 5400 &= 7020 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times1620}{1.5625} = 2073.6 \\ V_m &= 45.54\ \text{km/h} \end{aligned}$$

Running time and schedule speed

$$\begin{aligned} T &= \frac{7020}{45.54} = 154.16\ \text{s} \\ V_a &= \frac{5400}{154.16} = 35.03\ \text{km/h} \\ T_s &= 154.16 + 26 = 180.16\ \text{s} \\ V_s &= \frac{1.5\times3600}{180.16} = 29.97\ \text{km/h} \end{aligned}$$

Check: $t_1 = 45.54/0.8 = 56.9$ s, $t_3 = 45.54/3.2 = 14.2$ s, free running $= 154.16 - 71.1 = 83.0$ s.

Answer: Schedule speed ≈ 29.97 km/h (≈ 30 km/h); maximum speed 45.5 km/h.

a) Power drawn by a dc series motor varies as $\sqrt{T}$

In a series motor the field current is the armature current, so (below saturation) $\phi \propto I$:

$$T \propto \phi I \propto I^2 \Rightarrow I \propto \sqrt{T}$$

The supply voltage $V$ is constant, so the power drawn is

$$P = VI \propto I \propto \sqrt{T}$$

Example: if the load torque rises four times (steep gradient), the current and power drawn only double; speed falls automatically since $N \propto E/\phi \propto 1/I$.

For a shunt motor, $\phi$ is constant, so $T \propto I$ and $P \propto T$: a four-fold torque needs four-fold power. Hence a series motor puts much less strain on the supply and substation during heavy loads such as starting and climbing. This is a main reason dc series motors are used for traction.

b) Shunt motor is not suitable for traction

1. Power demand ∝ torque: since $T \propto I$ at constant flux and speed is nearly constant, heavy starting and gradient torques draw very large current and power from the supply (series motor: only $\sqrt T$).
2. Constant speed characteristic: traction needs a speed that falls on up-gradients and rises on light loads (falling characteristic). A shunt motor tries to keep the same speed, so power peaks are high.
3. Unequal load sharing: locomotives run several motors in parallel. Wheel diameters differ slightly by wear; with flat speed–torque curves, a small speed difference makes one shunt motor take a very large share of load, overloading it. Series motors with drooping curves share load nearly equally.
4. Supply voltage fluctuations: traction line voltage varies widely. In a shunt motor flux follows voltage; a sudden voltage rise or dip causes large current surges, since back emf cannot change instantly with speed. In a series motor the field current is the armature current, so surges are small.
5. Low starting torque per ampere: series motor torque rises with $I^2$, giving high starting torque with moderate current; a shunt motor gives only $T \propto I$.
6. Interruption of supply (e.g. pantograph bounce) is dangerous with shunt motors due to heavy current on reconnection.

(The shunt motor's only merit, easy regenerative braking, is outweighed by these drawbacks.)

Data: $\alpha = 1.2$ km/h/s, $\beta = 3.8$ km/h/s, $V_m/V_a = 1.6$, $D = 2.5$ km, stop $= 45$ s.

Trapezoidal relation (speeds in km/h, time in s)

$$3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$K = \frac1{1.2} + \frac1{3.8} = 0.8333 + 0.2632 = 1.0965$$

With $V_a = 3600D/T$ and $V_m = 1.6V_a$: $V_m T = 1.6\times3600\times2.5 = 14400$.

$$\begin{aligned} 9000 &= 14400 - \frac{V_m^2}{2}\times1.0965 \\ V_m^2 &= \frac{2\times5400}{1.0965} = 9849.6 \\ V_m &= 99.25\ \text{km/h} \end{aligned}$$

Running time and schedule speed

$$\begin{aligned} T &= \frac{14400}{99.25} = 145.10\ \text{s} \\ V_a &= \frac{9000}{145.10} = 62.03\ \text{km/h} \\ T_s &= 145.10 + 45 = 190.10\ \text{s} \\ V_s &= \frac{2.5\times3600}{190.10} = 47.34\ \text{km/h} \end{aligned}$$

Check: $t_1 = 99.25/1.2 = 82.7$ s, $t_3 = 99.25/3.8 = 26.1$ s, free running $= 145.10 - 108.8 = 36.3$ s (positive, so the curve is valid).

Answer: Schedule speed ≈ 47.3 km/h (maximum speed 99.2 km/h).

In the dc system of track electrification, the contact line or third rail carries dc at 600–750 V (tramways, metros) or 1500–3000 V (suburban and main lines). Substations take ac from the grid and convert it to dc with rectifiers; dc series motors (or chopper-controlled motors) drive the vehicle.

Merits

1. Ideal motor characteristic: dc series motor gives high starting torque, falling speed with load, power $\propto \sqrt{T}$, and good load sharing.
2. Simple, light and cheap locomotive: no heavy transformer or rectifier on board.
3. Lower energy per train-km in frequent-stop urban service due to good acceleration.
4. Small clearance and insulation at low voltage: suits tunnels, underground metros and the third-rail system.
5. No interference with telecommunication and signalling lines (no alternating magnetic field).
6. Single-phase unbalance does not arise: the substation rectifiers draw balanced three-phase load from the grid.
7. Speed control is simple (series-parallel, chopper), and regenerative braking is possible with series-excited control.

Demerits

1. High line current at low voltage: heavy, costly overhead conductor or third rail and supports.
2. Large voltage drop and $I^2R$ loss; substations must be close (3–5 km at 750 V, 10–20 km at 3 kV).
3. Costly substations: rectifier equipment needed at every substation, many of them; high capital cost.
4. Stray currents returning through earth cause electrolytic corrosion of underground pipes and cables.
5. Energy wasted in starting resistances (with resistance control).
6. Regenerated energy can be used only if another train is drawing power nearby (rectifier substations cannot feed back to the grid without inverters).
7. Third rail at ground level is a safety hazard.

A trapezoidal speed-time curve (acceleration, free run, braking) represents main-line (long-distance) service, while a quadrilateral curve (acceleration, speed curve, coasting, braking) represents urban and suburban service.

Trapezoidal curve: main-line service

• Stations are far apart (often more than 10 km), so the train runs most of the time at constant (crest) speed.
• The acceleration and braking periods are short compared with the free-run period.
• So the actual curve is closely approximated by three straight lines: constant acceleration, constant speed, constant retardation. This shape is a trapezium.
• It is used for main-line and long-distance express trains.

Quadrilateral curve: urban and suburban service

• Stops are close together (about 1 km for urban and 1–8 km for suburban), so the train never runs long at constant speed.
• After acceleration the speed rises slowly along the motor curve, then power is switched off and the train coasts, then brakes.
• Approximating this with straight lines gives four sides: acceleration, speed-curve running, coasting and braking. This shape is a quadrilateral.
• Used for city and suburban trains, metro and tram services, where high acceleration and braking matter more than top speed.

 Speed                       Speed
  |   ________                |    /\
  |  /        \               |   /  ``--._
  | /          \              |  /         \
  |/            \             | /           \
  +--------------\-- t        +/-------------\-- t
  Trapezoidal (main line)     Quadrilateral (urban)

For a simplified trapezoidal curve, the distance between stops equals the area under the speed-time curve.

Given: $V_a = 50$ km/h, $D = 2.5$ km, $\alpha = 2$ km/h/s, $\beta = 3$ km/h/s.

Running time

$$\begin{aligned} T &= \frac{3600\,D}{V_a} = \frac{3600 \times 2.5}{50} = 180\ \text{s} \end{aligned}$$

Maximum (crest) speed

For a trapezoidal curve (speeds in km/h, time in s, $D$ in km):

$$D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$$

Let $K = \frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167$. Then

$$\begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600\,D}{K}} \\ &= \frac{180}{0.8333} - \sqrt{\frac{180^2}{0.6944} - \frac{3600 \times 2.5}{0.4167}} \\ &= 216 - \sqrt{46656 - 21600} = 216 - 158.29 \\ &= 57.71\ \text{km/h} \end{aligned}$$

Times of each period

$$\begin{aligned} t_1 &= \frac{V_m}{\alpha} = \frac{57.71}{2} = 28.85\ \text{s} \\ t_3 &= \frac{V_m}{\beta} = \frac{57.71}{3} = 19.24\ \text{s} \\ t_2 &= T - t_1 - t_3 = 180 - 28.85 - 19.24 = 131.91\ \text{s} \end{aligned}$$

Distance before brakes are applied

Distance covered during braking:

$$D_3 = \frac{V_m t_3}{2 \times 3600} = \frac{57.71 \times 19.24}{7200} = 0.154\ \text{km}$$ $$D_1 + D_2 = D - D_3 = 2.5 - 0.154 = 2.346\ \text{km}$$

Speed-time curve

 V (km/h)
 57.7 |     ______________________
      |    /                      \
      |   /                        \
      |  /                          \
      | /                            \
    0 +/------------------------------\---- t (s)
      0   28.85                  160.76  180

Answer: Maximum speed $V_m = 57.71$ km/h; distance travelled before brakes are applied $= 2.346$ km (accelerating 28.85 s, free run 131.91 s, braking 19.24 s).

Crest, average and schedule speed

• Crest speed ($V_m$): the maximum speed reached by the train during a run.
• Average speed ($V_a$): distance between two stops divided by the actual running time (stop time not included). $V_a = \dfrac{\text{distance}}{\text{running time}}$
• Schedule speed ($V_s$): distance between two stops divided by running time plus stop time. $V_s = \dfrac{\text{distance}}{\text{running time} + \text{stop time}}$, so $V_s < V_a < V_m$.

Numerical: speed-time curve for the run

Given: $V_a = 42$ km/h, $D = 1400$ m $= 1.4$ km, $\alpha = 1.7$ km/h/s, $\beta = 3.3$ km/h/s. A simplified trapezoidal curve is assumed.

Running time:

$$T = \frac{3600 \times 1.4}{42} = 120\ \text{s}$$

With $K = \frac{1}{2}\left(\frac{1}{1.7}+\frac{1}{3.3}\right) = 0.4456$:

$$\begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600D}{K}} \\ &= \frac{120}{0.8913} - \sqrt{\left(\frac{120}{0.8913}\right)^2 - \frac{3600\times1.4}{0.4456}} \\ &= 134.64 - \sqrt{18128.5 - 11309.7} = 134.64 - 82.58 \\ &= 52.07\ \text{km/h} \end{aligned}$$

Times and distances:

 V (km/h)
 52.07 |      _________________
       |     /                 \
       |    /                   \
       |   /                     \
       |  /                       \
     0 +-/-------------------------\--- t (s)
       0  30.63             104.22  120

Answer: Crest speed $= 52.07$ km/h; acceleration 30.63 s, free run 73.59 s, braking 15.78 s.

Electric braking stops or slows a traction motor by making it act as a generator, so the kinetic energy of the train is turned into electrical energy that is either wasted in resistors or returned to the supply. It reduces wear of brake shoes and gives smooth, controllable braking; mechanical brakes are still needed to hold the train at rest.

1. Plugging (reverse current braking)

• The connections of the armature (dc motor) or two supply phases (induction motor) are reversed, so the motor produces torque opposite to motion.
• A resistor is inserted to limit the very large current.
• Braking is quick but wasteful: energy from both the supply and the train is lost as heat. Used only rarely in traction (for example, emergency stopping).

2. Rheostatic (dynamic) braking

• The motor is disconnected from the supply and its armature is connected across a braking resistor.
• DC series motor: the field connections are reversed relative to the armature so the machine self-excites as a generator; two motors may be cross-connected (each excites the other's field) to share load.
• Induction motor: the stator is disconnected from ac and fed with dc (dc injection), creating a stationary field; the rotor currents produce braking torque and energy is lost in the rotor resistance.
• Simple and does not depend on the line, but energy is wasted as heat.

3. Regenerative braking

• The motor runs as a generator and returns energy to the supply line. This happens when the back emf exceeds the supply voltage (e.g. going downhill).
• DC series motors cannot regenerate directly because their field becomes weak, so they are switched to separate (shunt) excitation during braking. With chopper drives, regeneration is done by the chopper.
• Induction motor: regeneration happens naturally when the rotor runs above synchronous speed; with VVVF inverter drives the frequency is lowered so the motor runs super-synchronously.
• Most efficient (about 20–40% energy saving in hilly or frequent-stop routes), but needs a receptive supply or other trains to absorb the energy.

 Supply ===+=== line
           |
       [Motor as generator] --> energy back to line
           |                    (regenerative)
           +--[ R ]--           (rheostatic: heat)

A speed-time curve shows how the speed of a train varies with time between two stops; its slope gives acceleration and its area gives distance travelled.

Speed-time curves for different services

(a) Urban: stops ~1 km, no free run
 V |   /\__
   |  /    ``--.
   | /          \
   |/            \
   +--------------\--- t

(b) Suburban: stops 1-8 km, short free run, long coast
 V |   /--.___
   |  /       ``--.
   | /             \
   |/               \
   +-----------------\--- t

(c) Main line: stops > 10 km, long free run
 V |   ______________
   |  /              \
   | /                \
   |/                  \
   +--------------------\--- t

• Urban: high acceleration (1.5–4 km/h/s) and braking (3–4 km/h/s), no free-running period, short coasting.
• Suburban: similar to urban but with a short free run and a longer coasting period.
• Main line: long free run at high speed; acceleration and braking periods are small in comparison.

Terms (refer to a complete curve)

 V |      B  C
   |     ____
   |    /    ``--. D
   |   /          ``-- E
   |  /A              \
   | /                 \
   |/                   \
   +---------------------\F--- t
   O t1   t2   t3    t4   t5

1. Notching period (O–A): the train starts and speed rises at nearly constant acceleration while starting resistance is cut out step by step (notch by notch) and the motor current is kept nearly constant.
2. Accelerating period (O–B): the total time from start to end of acceleration; it includes the notching period (constant acceleration) and the speed-curve running period (A–B), where the motor runs on its natural characteristic and acceleration falls as speed rises.
3. Free run period (B–C): the motor stays connected and the train runs at constant (crest) speed; motor output just balances train resistance.
4. Coasting period (C–D): power is switched off; the train moves on its own kinetic energy and speed falls slowly due to friction and air resistance. Coasting saves energy.
5. Retardation (braking) period (D–F): brakes are applied and the train comes to rest at the next stop.

A simplified trapezoidal speed-time curve is assumed.

Given: $V_s = 25$ km/h, $D = 800$ m $= 0.8$ km, stop time $= 20$ s, $V_m = 1.2\,V_a$, $\beta = 3$ km/h/s.

Schedule time and running time

$$\begin{aligned} T_s &= \frac{3600 \times 0.8}{25} = 115.2\ \text{s} \\ T &= T_s - \text{stop time} = 115.2 - 20 = 95.2\ \text{s} \end{aligned}$$

Average and crest speed

$$\begin{aligned} V_a &= \frac{3600 \times 0.8}{95.2} = 30.25\ \text{km/h} \\ V_m &= 1.2 \times 30.25 = 36.30\ \text{km/h} \end{aligned}$$

Acceleration

For the trapezoidal curve:

$$D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$$ $$\begin{aligned} 0.8 &= \frac{36.30 \times 95.2}{3600} - \frac{36.30^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{3}\right) \\ 0.8 &= 0.96 - 0.18303\left(\frac{1}{\alpha}+0.3333\right) \\ \frac{1}{\alpha}+0.3333 &= \frac{0.16}{0.18303} = 0.8741 \\ \frac{1}{\alpha} &= 0.5408 \quad\Rightarrow\quad \alpha = 1.849\ \text{km/h/s} \end{aligned}$$

Check: $t_1 = 36.30/1.849 = 19.63$ s, $t_3 = 36.30/3 = 12.10$ s, free run $= 95.2 - 19.63 - 12.10 = 63.47$ s (positive, so the trapezoidal curve is valid).

Answer: Required acceleration $\alpha \approx 1.85$ km/h/s.

A traction motor must give high starting torque, series-type speed-torque characteristic (speed falls as load rises), simple speed control, easy braking, rugged construction and good load sharing when several motors work in parallel. The main types used are listed below.

1. DC series motor

• Characteristics: torque $\propto I^2$ before saturation, so starting torque is very high; speed falls sharply with load (self-regulating); power demand on the supply varies roughly as $\sqrt{T}$, so heavy loads do not overload the line; motors share load well in parallel.
• Speed control: series-parallel control, field weakening, chopper control.
• Applications: the classic traction motor for dc railways, tramways, metro and suburban trains, and in diesel-electric locomotives.
• Drawback: commutator and brushes need maintenance; regenerative braking needs change to separate excitation.

2. DC compound motor

• Characteristics: cumulative compound gives high starting torque with a definite no-load speed, so it does not race on light load; regenerative braking is easier than with series motors.
• Applications: trolley buses and some hilly-route dc vehicles where regeneration is useful.

3. AC single-phase series (commutator) motor

• Characteristics: series-type speed-torque curve similar to dc series motor; works on low-frequency ac (16⅔ or 25 Hz) to limit commutation trouble; speed controlled by a tap-changing transformer.
• Applications: older single-phase low-frequency ac main-line railways (Europe).

4. Three-phase induction motor

• Characteristics: nearly constant speed on fixed frequency, simple and rugged, automatic regenerative braking above synchronous speed. With VVVF inverters it gets high starting torque and smooth speed control.
• Applications: modern electric locomotives, metro trains, EMUs and electric buses; earlier used on 3-phase hill railways.

5. Synchronous and other modern motors

• Linear induction motors (some metros, maglev), permanent-magnet synchronous motors (high efficiency in new trains and electric vehicles), and brushless dc motors in e-rickshaws and electric buses.

A speed-time curve is a graph of the speed of a train against time between two stops. Its slope gives acceleration or retardation, and the area under it gives distance travelled. It is used to find schedule speed, energy consumption and motor rating.

Parts of a typical speed-time curve

 V |       B   C
   |      .------.
   |     /        ``--. D
   |    / A            \
   |   /                \
   |  /                  \
   | /                    \
   +------------------------\E--- t
   O  t1   t2   t3     t4    t5

1. Constant acceleration / notching (O–A): starting resistance is cut out notch by notch; current and torque stay almost constant, so acceleration is constant.
2. Speed-curve running (A–B): all resistance is out; the motor works on its natural characteristic, so acceleration falls as speed rises.
3. Free running (B–C): constant (crest) speed; tractive effort equals train resistance.
4. Coasting (C–D): supply is switched off; the train runs on stored kinetic energy and speed drops slowly. Saves energy.
5. Braking (D–E): brakes applied to stop the train at the next station.

Example: in a metro run of 1 km, the train accelerates at 2 km/h/s for about 20 s to about 40 km/h, coasts for about 50 s, and brakes at 3 km/h/s for about 12 s.

Urban service

• Stations are close (about 1 km); high acceleration (1.5–4 km/h/s) and high braking (3–4 km/h/s) are needed to keep a good schedule speed.
• No free-run period; acceleration is followed by coasting, then braking. The curve is approximately quadrilateral.

Suburban service

• Stations are 1–8 km apart; acceleration 1.5–4 km/h/s, braking 3–4 km/h/s.
• A short free run may follow acceleration, and the coasting period is longer than in urban service.

 Urban                    Suburban
 V |  /\                  V |  /--.__
   | /  ``-.                | /      ``--.
   |/      \                |/           \
   +--------\-- t           +-------------\-- t

Schedule speed is the ratio of the distance between two stops to the total time of the run including the stop time:

$$V_s = \frac{D}{T_{run} + T_{stop}}$$

It is lower than average speed and is the speed that matters to passengers and to timetable planning. A high schedule speed means fewer trains are needed for the same service.

Factors affecting schedule speed

1. Acceleration and retardation: higher acceleration and braking shorten the run time, especially for short distances (urban service), so schedule speed rises. They are limited by passenger comfort and adhesion.
2. Maximum (crest) speed: for a given distance and rates, a higher crest speed reduces running time. Its effect is larger on long runs (main line) than on short urban runs.
3. Duration of stops: schedule time includes stop time, so longer stops reduce schedule speed. The effect is large in urban service where stops are frequent.
4. Distance between stops: for the same acceleration, braking and crest speed, schedule speed increases with distance because acceleration and braking take a smaller share of the time.
5. Coasting: longer coasting saves energy but lowers average and schedule speed.
6. Gradient and curves of track: up-gradients and sharp curves force lower speeds.
7. Train resistance and load: heavier trains accelerate more slowly for the same motor power.
8. Motor characteristic and supply voltage: a drop in line voltage reduces motor speed and acceleration.

 V | constant D, Vm
   |  /------\       higher alpha, beta  --> shorter T
   | /        \      longer stop time    --> lower Vs
   |/          \
   +------------\-- t

Summary: for urban service, acceleration, braking and stop time are the main factors; for main-line service, crest speed is the main factor.

A simplified trapezoidal speed-time curve is used.

Given: $V_s = 60$ km/h, $D = 6$ km, stop time $= 60$ s, $\alpha = 2$ km/h/s, $\beta = 3$ km/h/s.

Schedule time and running time

$$\begin{aligned} T_s &= \frac{3600 \times 6}{60} = 360\ \text{s} \\ T &= 360 - 60 = 300\ \text{s} \end{aligned}$$

(Average speed $V_a = 3600 \times 6 / 300 = 72$ km/h.)

Crest speed

$$D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)$$

With $K = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167$:

$$\begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600D}{K}} \\ &= \frac{300}{0.8333} - \sqrt{360^2 - \frac{3600 \times 6}{0.4167}} \\ &= 360 - \sqrt{129600 - 51840} = 360 - 278.85 \\ &= 81.15\ \text{km/h} \end{aligned}$$

Check: $t_1 = 81.15/2 = 40.57$ s, $t_3 = 81.15/3 = 27.05$ s, free run $t_2 = 300 - 40.57 - 27.05 = 232.38$ s.

Answer: Crest speed $V_m \approx 81.15$ km/h.

Systems of track electrification are classified by the type of supply fed to the train: DC system, single-phase ac system, three-phase ac system and composite systems (single-phase to dc, or single-phase to three-phase on the locomotive).

1. DC system (600–750 V, 1500 V, 3000 V)

• DC series motors fed through third rail or overhead line; substations with rectifiers every 3–5 km (low voltage) or 15–30 km (3 kV).
• High starting torque, simple and light motors, good for frequent stops.
• Used for tramways, metro and suburban services.

2. Single-phase low-frequency ac (15–16 kV, 16⅔ or 25 Hz)

• AC series motors with tap-changing transformer on the locomotive.
• Needs special low-frequency generation or frequency converters.
• Used on older main lines in Europe.

3. Three-phase ac (3.3–3.6 kV, 16⅔ Hz)

• Three-phase induction motors; two overhead wires plus rail.
• Simple robust motors, automatic regeneration, but complex overhead and nearly constant speed.
• Used on some hill railways; now obsolete.

4. Composite systems

• Single-phase to dc (25 kV, 50 Hz): a transformer and rectifier on the locomotive feed dc series motors. Cheap single overhead wire at high voltage, substations 40–50 km apart, supply from the national grid at 50 Hz. The most widely used main-line system today (India, many countries).
• Single-phase to three-phase: phase/frequency converter or VVVF inverter on board feeding induction motors. Modern locomotives use this.

Comparison

Electric traction is the propulsion of vehicles (trains, trams, trolley buses, electric buses) using electric motors, with the electrical energy taken from a supply line or produced or stored on the vehicle.

Types of electric traction based on supply source

A. Self-contained (non-electrified track) systems — the vehicle carries its own source.

1. Diesel-electric: a diesel engine drives a generator that feeds dc or ac traction motors.
   • Advantages: no overhead line, can run on any route, low line cost, high availability.
   • Disadvantages: low overall efficiency (about 25%), heavy locomotive, high running and maintenance cost, pollution, limited overload capacity.
2. Battery-electric: batteries feed dc motors.
   • Advantages: no pollution, simple, quiet; good for short distances, shunting, mines and e-rickshaws/e-buses.
   • Disadvantages: limited range, heavy batteries, long charging time, high battery replacement cost.

B. Electrified track (supply from distribution network) — power is taken from an overhead wire or third rail.

1. DC system: 600–750 V (tramways, metro), 1500–3000 V (suburban, main line), using dc series motors.
2. AC system: single-phase 25 kV, 50 Hz (most common), single-phase low frequency, or three-phase.
3. Composite system: ac in the line and rectifier or inverter on board.

Advantages of electric traction from line supply

• Clean, no smoke; ideal for cities and tunnels.
• High starting torque and acceleration, so higher schedule speed.
• Low maintenance and running cost; long life of locomotives.
• Regenerative braking returns energy to the line.
• Higher overall efficiency when fed from hydro power (important for Nepal).

Disadvantages

• Very high initial cost of overhead line, substations and electrification.
• Failure of supply stops all trains on the section.
• Trains can run only on electrified routes.
• Interference with telecommunication lines (ac systems).
• Additional cost of signalling changes.

A speed-time curve of a traction system is the graph of train speed against time for a run between two stations. Its slope is acceleration, the area under it is distance travelled, and it is used to find schedule speed, energy use and motor rating.

1. Urban (city) service

• Stops are about 1 km apart; trains must start and stop often.
• High acceleration (1.5–4 km/h/s) and braking (3–4 km/h/s) are needed to keep a reasonable schedule speed.
• There is no free-run period; acceleration is followed by a short coasting and braking.
• Curve shape is approximately quadrilateral.

 V |   /\
   |  /  ``-._
   | /        \
   |/          \
   +------------\--- t
   accel  coast  brake

2. Suburban service

• Stops 1–8 km apart; acceleration and braking similar to urban service.
• A short free-run period may appear after acceleration, followed by a long coasting period.

 V |   /---.___
   |  /        ``--.
   | /              \
   |/                \
   +------------------\--- t
   accel free  coast  brake

3. Main-line service

• Stops are more than 10 km apart, so the train runs mostly at crest speed.
• Acceleration and braking are low (0.6–0.8 and about 1.5 km/h/s) and occupy a small part of the time.
• Long free run, short coasting before braking; curve approximately trapezoidal.

 V |   ____________________
   |  /                    ``-.
   | /                         \
   |/                           \
   +-----------------------------\--- t
   accel     free run     coast brake