ELECTRIC TRACTION - PART - 11 - TYPES OF TRACTION SYSTEMS

ADVANTAGES OF ELECTRIC TRACTION SYSTEM

1. It is a clean traction system.

2. Well suited for underground railways.

3. No need for separate generator is required for lights and

    fans, power can be drawn directly from the lines.

4. Speed control is possible.

5. It has the advantages of rapid acceleration and retardation.

6. Less maintenance is required.

7. It can be started instantly.

8. It can take large overloads.

9. Pollution free.

10. Very economical system.

DISADVANTAGES OF ELECTRIC TRACTION

1. Its initial cost is very high.

2. Failure of supply paralyses the whole system.

3. It produces electro-magnetic interference with the

    neighboring telecommunication lines.

Electric traction is suitable for suburban and urban raiway where frequent starting and stopping and high schedule speeds are required.

It is capable of handling greater volume of traffic.

The system can be subdivide into and run in sections during the periods of light traffic.

Frequent service of trains can be maintained leading to increased traffic during these periods.

SYSTEMS OF TRACTION

1. DIRECT STEAM ENGINE SYSTEM

Steam engine drive is the most widely used traction system in almost all the UNDERDEVELOPED countries.

ADVANTAGES

1. Simplicity in design.

2. No telecommunication interferences

3. Low capital cost, because track electrification is not

    required.

4. Seep control is simple

DISADVANTAGES

1. Poor thermal efficiency.

2. Noisy in operation.

3. Due to unbalanced reciprocating part there is considerable wear on the track.

4. Corrosion of steel structure due to smoke emitted by the engine.

5. It pollutes the atmosphere.

6. Low efficiency.

7. It takes own time for starting.

8. No suitable for underground system.

9. High maintenance cost.

10. Speed of locomotive is very low.

2. DIRECT INTERNAL COMBUSTION SYSTEM  

It is suitable for road and light railway work, it is unsatisfactory work on railways. [buses, cars and trucks]

ADVANTAGES

1. It has low initial cost.

2. Very compact and self-contained unit.

3. Speed control with gear and arrangement is quite simple.

4. It has a efficiency of 25%.

5. Its braking arrangements are simple.

DISADVANTAGES

1. The life of the equipment (vehicle) is short.

2. Overload capacity is low.

3. Manintenace and running costs are fairly high.

4. It produces air pollution.

5. It requires gear arrangement for speed control.

3. DIESEL ELECTRIC DRIVE

This system is used in Indian Railways.

In this system gear system is eliminated. The diesel engine is coupled to a dc generator which supplies d.c traction motors.

ADVANTAGES

1. For conversion from steam engine to diesel traction, no modification of existing track is required.

2. No overhead transmission system is required hence capital is low.

3. The efficiency of the system is comparatively higher, about 25%.

4. The haulage capacity is larger as compared with steam locomotive.

5. Simple starting method.

Speed control with gear and arrangement is quite simple.

6. It has a efficiency of 25%.

7. Its braking arrangements are simple.

DISADVANTAGES

1. The diesel engine has a shorter life span.

2. Overload capacity is very much limited.

3. Maintenance and maintenance costs are fairly high.

4. Cooling systems are needed for diesel engine as well as for the motor generator set.

4. BATTERY – ELECTRIC SYSTEM

In this system vehicle carries batteries which run d.c. motors used for driving the vehicle. This system is not suitable for railways.

It is mainly used in mines, ports and large factories.

The batteries are connected in parallel for starting and running half of the maximum speed.

The batteries are connected in series for running at maximum speed.

ADVANTAGES

1. Low maintenance cost.

2. convenient to use.

3. Pollution free.

DISADVANTAGES

1. The major disadvantage is limited capacities of the batteries and the problem of charging them frequently.

ELECTRIC TRACTION - PART – 10 - COEFFICIENT OF ADHESION & THREE PROBLEMS

COEFFICIENT OF ADHESION

This is the ratio of the tractive effort force just necessary slip the wheels on the track to the adhesive weight.

It reduces with increase in the speed.

It is represented as μ.

μ = tractive effort to slip wheels/adhesive weight

It is usually less than one.

The normal value of μ with dry rails is 0.25 and maximum value is 0.3 when the track has been cleaned well.

If the rails are wet or greasy the value lies between 0.5 to 0.2

IMPORTANCE OF COEFFICIENT OF ADHESION

There is a minimum value of tractive effort at which driving wheels will not slip and the maximum value depends upon the dead weight over the driving axles. [Axle means a shaft on which a wheel rotates]

F is directly proportional to W

F = μ W [if ‘F’ is Newton and ‘W’ in ton]

F = μ 98.1 W in Newton

This means to haul (draw slowly or heavily) a train, there is a certain minimum weight of locomotive. Again, the maximum allowable weight on each driving axle is limited by the strength of track bridges etc., to between 15 and 30 ton.

TRAILING WEIGHT

That part of the weight of the locomotive engine which rests upon the rear pair of driving-wheels.

PROBLEM – 01

A locomotive accelerates a 400 ton train up a gradient of 1m in 100 at 0.8 kmphps. Assuming the coefficient of adhesion to be 0.25, determine the minimum adhesive weight of the locomotive. Assume train resistance 40 N per ton and allow 10 percent for the effect of rational inertia.

PROBLEM – 02

A train weighing 250 ton is accelerated up a one percent gradient with an acceleration of 1 kmphps. Determine the minimum adhesive weight of a locomotive for this purpose if the coefficient of adhesion is 0.2. Assume train resistance as 50 N/t and allowance for rotational inertia 10 percent.

PROBLEM – 03

A goods train weighing 300 ton is to be hauled by a locomotive up a gradient of 1% with an acceleration of 1 kmphps coefficient of adhesion 20% track resistance 45 N/t and effective masses 10% of dead weight. If axle load is not to exceed 20 ton determine the weight of locomotive and number of axles. 

ELECTRIC TRACTION – PART – 09 - SPECIFIC ENERGY OUTPUT AND ENERGY CONSUMPTION & TWO PROBLEMS

PROBLEM -01

An electric train has an average speed of 42 km per hour on a level track between stops 1500 m apart. It is accelerated at 1.7 kmphps and braked at 3.3 kmphps. Draw the speed-time curve and estimate the specific energy consumption.

Assume tractive resistance as 50 N/ton and allow 10% for rotational inertia.

PROBLEM -02

A 400 ton electric train runs up an ascending gradient of 1% with the following curve.

1. Uniform acceleration of 1.6 km/hr./sec. for 25 sec.

2. Constant speed for 50 sec.

3. Coasting for 30 sec.

4. Braking at 2.56 km/hr./sec.

Calculate the specific energy consumption if train resistance is 50 N/ton, effect of rotational inertia 10%, overall efficiency of transmission gear and motor 75%. 

ELECTRIC TRACTION – PART – 08 – QUADRILATERAL SPEED TIME CURVE & TWO PROBLEMS

PROBLEM -01

The following data relate to a 200 ton electric train running according to the following quadrilateral speed-time curve.

(1) Uniform acceleration from rest to the at 2kmphps for 30 sec.

(2) Coasting retardation for 50 sec.

(3) Duration of braking 15 sec.

(4) Up gradient – 1%

(5) Train resistance – 40 T/ton

(6) Overall efficiency of gear and motor – 75%

Find the schedule speed.

PROBLEM -02

An electric train accelerates uniformly from rest to a speed of 50 km/hr. It the coasts for 70 seconds against a constant resistance of 60 N/t and is then braked to rest with uniform retardation of 3 kmphps in 15 sec. Calculate 
(1) Uniform acceleration

(2) Coasting retardation

(3) Schedule speed if station stops are of 25 sec. duration. 
Allow 10% for rotational inertia. 

How will the schedule speed the affected if duration of stops is reduced to 20 seconds, other factors remaining the same?

ELECTRIC TRACTION - PART - 07 - TRAPEZOIDAL SPEED TIME CURVE & FOUR PROBLEMS

PROBLEM - 01

An electric train has an average speed of 50 km per hour on  a level track between stops 1500 m apart. It is accelerated at 1.7 kmphps and is braked at 3.3 kmphps. Draw the speed-time for the run.

PROBLEM - 02

An electric train is to have acceleration and braking, retardation of 0.8 kmphps and 3.2 kmphps respectively. if the ratio of maximum to average speed is 1.3 for stops of 30 sec.Find schedule speed for a run of 1.5 km. Assume simplified trapezoidal speed-time curve.

PROBLEM - 03

An electric train has a schedule speed of 25 km per hour between station 800 m apart. The duration of stop is 25 seconds, the maximum is speed is 20 percent higher than the average running speed and braking retardation is 3 kmphps. calculate the rate of acceleration required to operate this service.

PROBLEM - 04

A suburban electric train has a maximum speed 70 km per hour. The schedule speed including a station stop of 20 seconds is 45 kmph. If the acceleration is 3.5 kmphps, find the value of retardation when the average distance between stops is 4 km.

ELECTRIC TRACTION – PART – 06 - IMPORTANT TERMS IN TRACTION AND TWO PROBLEMS

DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT

DISTANCE is a scalar quantity that refers to "how much ground an object has covered" during its motion.

DISPLACEMENT is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

DIFFERENCE BETWEEN SPEED AND VELOCITY

SPEED – The rate of change in distance with respect to time.   Since speed is built from distance, a scalar quantity, then speed is also a scalar quantity.  This means it carries no direction information with it.

VELOCITY – The rate of change in displacement with respect to time.  Since displacement is a vector quantity, then velocity is also a vector quantity.  It has both magnitude and direction.

Both speed and velocity are typically measured in units of miles per hour, kilometers per hour (Km/hr), or meters per second (m/s).

INSTANTANEOUS SPEED

The speed at any given instant in time.

CREST SPEED OR PEAK SPEED

This is the maximum speed of the train during a run.

AVERAGE SPEED

This is the average speed of a train during a run. It is equal to the total distance divided by the total time.

Average speed = total distance covered / total time taken

SCHEDULE SPEED

This is the ratio of the distance between two destinations and the total time to cover the distance, including time wasted in stoppages.

Schedule speed = Distance / [time for run + time for stop]

FACTORS AFFECTING SCHEDULES SPEED

(a) Peak speed                 (c) Braking retardation

(b) Acceleration               (d) Duration of stoppages.

PROBLME - 01

A train has a speed of 60 km/hr. between two stops, 5 km apart. The duration of stop is one minute. The acceleration and retardation are 2 km/hr./sec. and 3 km/hr./sec. respectively. Calculate the maximum speed of the train.

PROBLME - 02

A train is required to run between two stations 1.5 km apart at the average speed of 40 km/hr. The run is to be made to a simplified quadrilateral speed-time curve. If the maximum speed is to be limited to 60 km/hr. acceleration to 2 km/hr. /sec. and coasting and braking retardation of 0.16 km/hr./sec. and 3.2 km/hr./sec. respectively. Calculate the duration of acceleration, coasting and braking periods.

ELECTRIC TRACTION - PART - 05 - SPEED TIME CURVE FOR TRAIN MOVEMENT & CATEGORIES OF RAILWAY SERVICES

DIFFERENCE BETWEEN ACCELERATION AND RETARDATION

ACCELERATION refers to the rate in change of velocity of a moving body. If a body is moving at a constant velocity, there is no change and hence it has no acceleration.

RETARDATION is the application of a force that produces negative acceleration. Synonyms would be braking, deceleration, damping, etc. Gravitational force operates downward (in a negative direction) so, in most frames of reference, gravity is a retarding force.

SPEED – TIME CURVE

The curve drawn between speed (km per hour) and the time (in minutes or seconds) is called ‘speed time curve’.

Speed-Time curve is a graph showing the variation of speed as function of time.

The speed time curve consists of

1. NOTCHING OR STARTING PERIOD [OA]

During this period, the traction motor accelerates from rest.

2. ACCELERATING PERIOD [AB]

During this period, according to the torque-speed characteristic of the motor, the torque gradually decreases and the speed increases, but the train still continues to accelerate.

3. FREE RUNNING PERIOD [BC]

During this period, the train runs at constant speed.

4. COASTING PERIOD [CD]

(Running with power switched off and therefore, there is retardation due to frictional and windage forces)

During this period, the power is cut off and the train moves due its kinetic energy.

The train slows down due to retardation production.

5. BRAKING PERIOD [DE]

During this period, brakes are applied and the train is brought is stop.

INFORMATION OBTAINED FROM SPEED – TIME CURVE

1. The curve can give speed of the train at any instant.

2. The slope of the curve, at the instant, gives acceleration or 
    retardation (a decrease in rate of change).

3. If the slope is positive it is an acceleration, if negative, it is 
    retardation and if zero, the speed is constant at that instant.

4. The area under the curve give the total distance covered.

ELECTRIC TRACTION – PART - 04 – SPECIFIC ENERGY OUTPUT FROM DRIVING AXLES AND SPECIFIC ENERGY CONSUMPTION

DIFFERENCE BETWEEN ENERGY CONSUMPTION AND SPECIFIC ENERGY CONSUMPTION OF A TRAIN

ENERGY CONSUMPTION

The energy input to the motor is called the energy consumption of the train as it is this energy which is utilized for movement of the train.

SPECIFIC ENERGY CONSUMPTION

It is defined as the energy consumed per Ton-metre. 

ENERGY OUTPUT

Total energy consumption Et= Ea + Eg + Er

Ea = Energy required for acceleration

Eg = Energy required to overcome gradient

Er = Energy required to overcome train resistance

Ea = 0.01072WeVm² watt-hour

Eg = 27.25WGD’ watt-hour

Er = 0.2778WrD’ watt-hour

Where 

We  is the accelerating weight

Vm is the maximum speed in km per hour

G is the Gradient

W is the dead weight of the train

D’ is the distance for which force due to resistance exists and

r is the train resistance.

SPECIFIC ENERGY OUTPUT

Espec- Outpu = Et / WD in watt-hour/ton-km

Espec-output = = [(0.01072WeVm²)/WD] + [(27.25GD’)/D] + [(0.2778rD’)/WD] watt-hour per ton-km
Where D is the total run length

FACTORS AFFECTING SPECIFIC ENERGY CONSUMPTION

1. DISTANCE COVERED

The greater distance covered between stops, the lesser will be the specific energy consumption.

Suburban service it is higher compared to the main line service.

Typical values for the two services are 60 and 25 per ton-km.

2. ACCELERATION AND RETARDATION

For a given run and a give schedule speed, the specific energy consumption is lower and for higher the acceleration and retardation since with a longer coasting period can be obtained and for a smaller period the supply is switched on.

3. TRAIN RESISTANCE

The specific energy consumption depends upon the train resistance which depends upon the nature of track, speed of the train and shape of the rolling stock.

The train resistance also depends upon the front and rear portion of the train.

4. GRADIENT

While going up steep gradients, more energy is needed through the specific energy consumption may be modified by regenerative braking.

5. MAXIMUM SPEED

The specific energy consumption increases with increase in maximum speed.

6. TRAIN EQUIPMENT

By using more efficient train equipment the specific energy consumption may be reduced.

SKIN FRICTION

The resistance of the air comes into play on front end of the locomotive known as head resistance and on the sides of the top and under sides of the coaches.

Head resistance may be streamlined the shape of the engine.

ELECTRIC TRACTION - PART - 03 - POWER OUTPUT FROM DRIVING AXLES

Ft = tractive force and V is the train velocity

Power output = Ft x V watts, 
where velocity V is metre per second.

Power output = Ft x (1000/3600) x V in watts
                    = Ft x V / 3600 in kW.

Where velocity V is in km per hour.

If η is the efficiency of transmission gear, then power output of motors is Ft = [Ft X V]/η in watts.
where velocity V is in metre per second.

Ft = [Ft x V] / [3600 x η] in kW.
where velocity V is in km per hour.

PROBLEM - 01

A 200 ton motor coach driven by 4 motors takes 20 seconds to attain a speed of 42 km per hour, starting from rest on an ascending  gradient of 1 in 80. The gear ratio is 3.5, gear efficiency 92%, wheel diametre 92 cm, train resistance 40 N/t and rotational inertia 10 percent of the dead weight.

Find the torque developed by each motor.

PROBLEM - 02

A 200 ton motor coach having 4 motors, each developing a torque of 8000 N-m during acceleration, starts from rest. if up-gradient is 30 in 1000, gear ratio 3.5, gear transmission efficiency 90 percent, wheel diametre 90 cm, train resistance 50 N/t, rotational inertia 10% of the dead weight, calculate the time taken by the coach to attain a speed of 80 km per hour.

If the supply voltage is 3000 V and motor efficiency 80%, calculate the current taken during the accelerating period.

PROBLEM - 03

A train weighting 500 tons is moving down a gradient of 20 in 1000. The speed of train is to be maintained at 50 km per hour through the use of regenerative braking. If the tractive resistance 40 N/ton, and the efficiency of conversion is 75%. Calculate the power fed to the line.

ELECTRIC TRACTION PART – 02 - TRACTIVE EFFORT REQUIRED FOR PROPULSION OF A TRAIN

DIFFERENCE BETWEEN MASS AND WEIGHT MASS

MASS

It is the quantity of matter contained in a body.
Mass is denoted using m or M. Mass is a scalar quantity.

WEIGHT 
It is the force with which earth pulls a body downwards.
Weight usually is denoted by W. Weight is mass multiplied by the acceleration of gravity. 
[W = m x g] Weight is a vector quantity.

DEAD WEIGHT

It is the gross weight of the train, including the locomotive moving the track.

ACCELERATION
It is the rate of change of velocity of an object.

ACCELERATING WEIGHT
It is a weight due to rotational inertia because of angular acceleration, the total effective weight of the train will be more than its dead weight. This is known as accelerating weight.
The accelerating weight is about 10% more than the dead weight.

INERTIA
It is the resistance of any physical object to any change in its state of motion, including changes to its speed and direction. It is the tendency of objects to keep moving in a straight line at constant velocity.

ROTATIONAL INERTIA
It is a scalar, not a vector and is dependent upon the radius of rotation according to the formula
Rotational inertia = mass x radius^2.
Rotational inertia, rotational inertia is the measure of an object's resistance to change in its rotation.

ADHESIVE WEIGHT
It is the weight on the driving wheels of a locomotive, which determines the frictional grip between wheels and rail.
[OR]
The total weight to be carried on the driving wheels is called the adhesive weight.

GRIP
The friction between a body and the surface on which it moves (as between an automobile tire and the road).

FRACTIONAL GRIP
The adhesion between the wheels of a locomotive and the rails of the railroad track.

TOTAL TRACTIVE EFFORT = Ft = Fa + Fg + Fr

Fa = force required for giving linear acceleration to the train

Fg = force required to overcome the effect of gravity

Fr = force required to overcome resistance to train motion

Ft = 277.8 We x a +or- 98.1 W x G + W x r

We - accelerating weight, a - acceleration, W - Dead weight, G - percentage gradient and r - Resistance to train motion