Example 10.1: The distance between two stops is 1.2 km.

A schedule speed of 40
kmph is required to cover that distance. The stop is of 18-s duration. The values of
the acceleration and retardation are 2 kmphp and 3 kmphp, respectively. Then,
determine the maximum speed over the run. Assume a simplified trapezoidal
speed time curve.

Solution:

Acceleration = 2.0 kmphp.

Schedule speed Vs = 40 kmph.

Distance of run, D = 1.2 km.

Actual run time, T = Ts stop duration

= 108 18

= 90 s.

where
Example 10.2: The speed time curve of train carries of the following parameters:

1. Free running for 12 min.

2. Uniform acceleration of 6.5 kmphp for 20 s.
3. Uniform deceleration of 6.5 kmphp to stop the train.
4. A stop of 7 min.

Then, determine the distance between two stations, the average, and the schedule
speeds.

Solution:

Acceleration ( ) = 6.5 kmphps.

Acceleration period t1 = 20 s.

Maximum speed Vm = t1

= 6.5 × 20 = 130 kmph.

Free-running time (t2) = 12 × 60

= 720 s.
 The distance travelled during the acceleration period:

The distance travelled during the free-running period:

The distance travelled during the braking period

The distance between the two stations:

D = D1 + D2 + D3

= 0.36 + 26 + 0.362

= 26.724 km.
Example 10.3: An electric train is to have the acceleration and braking
retardation of 0.6 km/hr/sec and 3 km/hr/sec, respectively. If the ratio of the
maximum speed to the average speed is 1.3 and time for stop is 25 s. Then
determine the schedule speed for a run of 1.6 km. Assume the simplified
trapezoidal speed time curve.

Solution:

Acceleration = 0.6 km/hr/s.

Retardation = 3 km/hr/s.

Distance of run D = 1.6 km.

T
Example 10.4: The distance between two stops is 5 km. A train has schedule
speed of 50 kmph. The train accelerates at 2.5 kmphps and retards 3.5 kmphps and
the duration of stop is 55 s. Determine the crest speed over the run assuming
trapezoidal speed time curve.
Solution:

Acceleration ( ) = 2.5 kmphps.

Retardation ( ) = 3.5 kmphps.

By using the equation:

Example 10.5: A train is required to run between two stations 1.5 km apart at an
average speed of 42 kmph. The run is to be made to a simplified quadrilateral
speed time curve. If the maximum speed is limited to 65 kmph, the acceleration to
2.5, kmphps, and the casting and braking retardation to 0.15 kmphs and 3 kmphs,
respectively. Determine the duration of acceleration, costing, and braking periods.

Solution:

Distance between two stations D = 1.5 km.

Average speed Va = 42 kmph.

Maximum speed Vm = 65 kmph.

Acceleration ( ) = 2.5 kmphps.

Coasting retardation c = 0.15 kmphps.

Barking retardation = 3 kmphps.

Before applying brakes; let the speed be V2.

 The actual time of run, T = t1 + t2 + t3

Example 10.6: A train has schedule speed of 32 kmph over a level track distance
between two stations being 2 km. The duration of stop is 25 s. Assuming the
braking retardation of 3.2 kmphps and the maximum speed is 20% grater than the
average speed. Determine the acceleration required to run the service.

Solution:

Schedule speed Vs = 32 kmph.

Distance D = 2 km.

Duration of stop = 25 s.

Braking retardation = 3.2 kmphps.

Example 10.7: A suburban electric train has a maximum speed of 75 kmph. The
schedule speed including a station stop of 25 s is 48 kmph. If the acceleration is 2
kmphps, the average distance between two stops is 4 km. Determine the value of
retardation.

Solution:

Maximum speed Vm = 75 kmph.

The distance of run (D) = 4 km.

 Schedule speed (Vs) = 48 kmph.

Acceleration ( ) = 2 kmphps.

The duration of stop = 25 s.

Example 10.8: An electric train is accelerated at 2 kmphps and is braked at 3

kmphps. The train has an average speed of 50 kmph on a level track of 2,000 min
between the two stations. Determine the following:

1. Actual time of run.

2. Maximum speed.
3. The distance travelled before applying brakes
4. Schedule speed.

Assume time for stop as 12 s. And, run according to trapezoidal.

Solution:

Acceleration ( ) = 2 kmphps.

Retardation ( ) = 3 kmphps.
Average speed (Va) = 50 kmph.

Distance D = 2,000 min = 2 km.

The duration of stop = 12 s.

The distance travelled before applying brakes

D1 + D2 = D - D3
 =2 0.17 = 1.83 km.

Example 10.9: An electric train has an average speed of 40 kmph on a level track
between stops 1,500 m apart. It is accelerated at 2 kmphps and is braked at 3
kmphps. Draw the speed time curve for the run.

Solution:

Average speed Va = 40 kmph.

The distance of run (D) = 1,500 m = 1.5 km.

Acceleration ( ) = 2 kmphps.

Retroaction ( ) = 3 kmphps.

Using the equation (Fig. P.10.1):

where
Fig. P.10.1
Example 10.10: An electric train has quadrilateral speed time curve as follows:

1. Uniform acceleration from rest at 1.5 kmphps for 25 s.

2. Coasting for 45 s.
3. The duration of braking 20 s.

If the train is moving a uniform up gradient of 1.5%, the reactive resistance is 45

N/ton, the rotational inertia effect is 10% of dead weight, the duration of stop is 15
s, and the overall efficiency of transmission gear and motor is 80%. Find schedule
speed.

Solution:

Time for acceleration t1 = 25 s.

Time for coasting t2 = 45 s.

Time for braking t3 = 20 s.

Acceleration ( ) = 1.5 kmphps.

Maximum speed Vm = 1

= 1.5 × 25 = 37.5 kmph.

According to the equation:

 TRACTIVE EEFFORT (FT)

It is the effective force acting on the wheel of locomotive, necessary to propel the
tractive effort Ft. The tractive
effort is a vector quantity always acting tangential to the wheel of a locomotive. It
is measured in newton.

The net effective force or the total tractive effort (Ft) on the wheel of a
locomotive or a train to run on the track is equals to the sum of tractive effort:
condition of track, etc. for the wet and greasy track conditions. The value of the
coefficient of adhesion is much higher compared to dry and sandy conditions.

IMPORTANT DEFINITIONS

1 Dead weight

It is the total weight of train to be propelled by the locomotive. It is denoted by

W

2 Accelerating weight

It is the effective weight of train that has angular acceleration due to the rotational
We

This effective train is also known as accelerating weight. The effective weight of
the train will be more than the dead weight. Normally, it is taken as 5 10% of
more than the dead weight.

3 Adhesive weight

The total weight to be carried out on the drive in wheels of a locomotive is known
as adhesive weight.

4 Coefficient of adhesion

It is defined as the ratio of the tractive effort required to propel the wheel of a
locomotive to its adhesive weight.

Ft W

= ,

where Ft is the tractive effort and W is the adhesive weight.

The energy returned to the supply system:

where is the efficiency of the system.

Advantages of regenerative breaking

1. In regenerative breaking, a part of the energy stored by the rotating parts is converted
into the electrical energy and is fed back to the supply. This will lead to the minimum
consumption of energy, thereby saving the operating cost.
2. High breaking retardation can be obtained during regenerative breaking.
3. Time taken to bring the vehicle to rest is less compared to the other breakings; so that,
the running time of the vehicle is considerably reduced.
4. The wear on the brake shoes and tyre is reduced, which increases the life of brake shoe
and tyre.

Disadvantages

In addition to the above advantages, this method suffers from the following
disadvantages.

1. In addition to the regenerative breaking, to bring the vehicle to standstill, mechanical

breaking is to be employed.
2. In case of DC traction, additional equipment is to be employed for regenerative
breaking, which increases the cost and sometimes, substation are equipped with
mercury arc rectifiers to convert AC to DC supply.
3. The electrical energy returned to the supply will cause the operation of substations
complicated.

Example 10.26: A 450-ton train travels down gradient of 1 in 75 for 110 s during
which its speed is reduced from 70 to 55 kmph. By the regenerative braking,
determine the energy returned to the lines if the reactive resistance is 4.5 kg/ton
and the allowance for the rotational inertia is 7% and the overall efficiency of the
motor is 80%.
Solution:

Accelerating weight Wa = 1.075 × 450

= 483.75 ton.

Track resistance r = 9.81 × 4.5

= 44.145 N-m/ton

= 9,904.782 W-hr

= 9.904 kW-hr.

Factiva effort required during retardation:

= Wr 98.1 WG

= 450 × 44.45 98.1 × 450 × 4/3

= 19,865.25 58,860

= 38,994.75 N.

The distance travelled during the retardation period:

As the train is moving in downward gradient, so that the tractive effort will provide
additional energy to the system. The energy available when the train moves over a
gradient is given as:
 The period of regeneration = 110 s.

Overall efficiency ( ) = 80%.

The kinetic energy of the train at a speed of 70 kmph is:

The kinetic energy of the train at the speed of 55 kmph is:

The energy available due to the retardation by the regenerative braking:

= 2,588.495 15,979.713

= 20.68 kW-hr.

The energy returned to the supply system:

= 0.8 × total energy available

= 0.8 × (20.68+9.904)

= 24.467 kW-hr.