430 Electric Traction

schedule speed. Variation in aceleration and retardation will have more effect on schedula
speed in case of shorter distance run in comparison to longer distance run.

2. Effect of Maximum Speed. For a constant distance run and with fixed acceleration
and retardation the actual time of run will decrease, and therefore, schedule speed will
increase with the increase in crest speed. The effect of variation in crest speed on schedule
speed is considerable in case of long distance run.

3. Effect of Duration of Stop. For a given average speed the schedule speed will increase
by reducing the duration of stop. The variation in duration of stop will affect the schedule speed
more in case of shorter distance run as compared to longer distance run. It is because stops in
case of urban and suburban services are kept very small (say 15 to 20 seconds). When the
distance between stations is much larger and duration of actual run is long as in kase of main
line service, station stop has less conspicuous effect on the schedule speed.

11.5. SIMPLIFIED SPEED-TIME CURVES

In order to study the performance
of a service at different schedule
speeds the speed-time curves are
replaced by simple geometric

T
shaped curves. From these
simplified curves the relationships s
between acceleration, retardation,
average speed and distance can be
easily worked out. These can have
either quadrilateral or trapezoidal TIME,t TIME, t -

shape.
The speed-time curve of an (a) (b)
urban service can be replaced by Approximate Speed-Time Curves
A.
an equivalent speed-time curve of Fig. 11.4
simple quadrilateral shape. The speed-time curve of a main line service is best and most easily
replaced by a trapezoid./Since the area of the speed-time curve represents the total distance
travelled hence the areas of the two curves should be same. The values of acceleration and
retardation are also kept the same as those of the
original speed-time curve/In
case of simplhiied
trapezoidal speed-time curve, speed curve running and coasting periods aré replaced by constant
speed period, as illust"ated in fig. 11.4 (a). While in case of quadrilateral speed-time curve,
initial acceleration and coasting periods ore
extended, as illustrated in fig. 11.4 (6).
The following examples illustrate the methods of
caleulations.
1. Calculations By Trapezoidal Speed-Time Curve
g
Let = Acceleration in kmphps
B Retardation in kmphps
V=Crest speed in kmph Vm
T Total time of run in seconds.
Time for acceleration in seconds, t, =

Time for retardation in seconds, t =

TIME IN SECONDS

Trapezoidal Speed-1Time Curte

Fig. 11.5
Train Movement and Energy Consumption 431

Time for free

running in seconds, T
t, = -

(t, + t,) =
T-| +
Total distance of run in km,S =
Distance travelled during acceleration +distance travelled
during free run + distance travelled during braking

V.00+ V 3,600,600
Substituting t =, ta and t, "T-| we get
=
=

S
7,200 a
7,200
2
or S =
7,200a
T
3,600
. m
3,600a 3,600 7,200 B
VT V V
3,600 7,200a 7,200B .(11.1)
V 1. 1. S =00
3.6002
3,600 2a 23.600S
3,600
=

or
Va a VT+3,600S=
2p
0

This is a quadratic equation for V, Substituting + K, we get

KV-VmT +3,600 S = 0

or V = TVT-4K x3,600S T
2K T2 3,600 S
2K K
The +ve sign cannot be adopted, as value of V obtained by using
much higher than that is possible in
+ve sign will be
practice. Hence ve sign will be used
-

and,
therefore, we have
T T2 3,600 S
K .(11.2)
From the above equation unknown quantity can be
determined by substituting the value of known quantities.
2. Calculation By GQuadrilateral Speed-Time
Curve. Let
2
a Acceleration in kmphps
Pe= Coasting retardation in kmphps
Braking retardationin kmphps in kmph
, Maximum speed at the endofacceleration
in kmph
2 Speed at the end of coasting
T Total time of run in seconds

Time of acceleration in seconds, t TIME IN SECONDS

Quadrilateral Speed-Time Curve

Fig. 11.6
432 Electric Traction

Time of coasting in seconds, t2 =

V.
Time of braking in seconds t =

Total distance travelled in km,

S =Distance travelled during acceleration+ distance travelled during coasting
+distance travelled during retardation

=v,x 3,600 +Vgxx-a. V * 3,600

3,60022 3,600
VgaV 4+)+ +
*g)*7206+)
7,200 7,200 7,200 7,200 7,200
or S
7,200 V(T-4)
(T-)+1,200 Since t+
t2 + t3T
T
or S = .7,200+ Va)-a V
7,200 7,200
or S =

7,200
(V+V,)-7,200
B (V,+
7,2007,200 Va)7,200B
C 7,200 B
VV
7,200
7,200 a
aa
or 7,200 S =T(V, +V2) -V,V2|+ .(11.3)

We have V2 =V, -

Bh =
V, -

P.(T -

4 -

t) =
V, -p.T-

or V2
V= V-PT+Pev
(11.4)
1-
Solving equations (11.3) and (11.4) values of S, V1, V, etc. can be obtained.
Example 11.1. The speed-time curve of a train consists of:
() Uniform acceleration of 6 km/h/s for 25 s;
(ti) Free running for 10 minutes;
(iii) Uniform deceleration of
6 km/h/s to stop the train;
(iv) A stop of 5 minutes.
Find the distance between the stations, the
average and schedule speeds
A.M.I.E. Sec B. Electric Drives and Their Control Winter 1996|
Solution: Acceleration, a = 6 kmphps

Accelerating period, t 25 seconds

Maximum speed, V, a t = 6 x 25 150kmph
Time for free running, t, = 10 x 60 = 600 seconds

Retardation, B=6 kmphps

Timefor retardation,i, = 2526 seconds
6

1 150x 25 25 km
Distance travelled during acceleration period, S,
3.600 =x 3,600 48
 Train Movement and
Energy Consumption 433

Distance travelled 150x 6000

during free running, S, = =
25 km
3,600 3,600
Distance travelled
during braking period, S, =
n =
150X Z =
248 km
7,200 7,200
Total distance between 25
stations, S S, +S, + Sg = 48 25+ 25
48
26.042 km Ans.
Average speed, V. =9X3,60026.042 x 3,600
T 25+600+25
=
144.23 kmph Ans.
Schedule speed, Sx3.600 26.042x 3,600
VsT+stop line 650+5x 60
Example 11.2. An electric train has an
98.68 kmph Ans.
average speed of 42 kmph on a level track between stops
1,400 m apart. It is accelerated at 1.7 kmphps and is
3.3 braked at
kmphps. Draw the speed-time curve for the run.
Solution Acceleration, a = 1.7 kmphps
Retardation, ß = 3.3 kmphps
Distance of run, S =1,400 m 1.4 km
Average speed, Va 42 kmph
V 52 KMWHOUR
Time of run, =x3,600
V
T x3600
42
120 seconds
Using equation V
T2 3,600 S
2K 4K K
-t - =73.6s
30.6 s 15.8 s
where K = 1.
K= 2B 3.4 0.4456 TIME IN SECONDS

Fig. 11.7

120 (120)2 3600 1.4

Maximum speed, m2x 0.4456 (2x0.4456)3 0.4456 52 kmph
30.6 seconds
Acceleration period, t, =
L =

1.7
=

V,
Braking period, t = 3.3
15.8seconds
120- (30.6 + 15.8) = 73.6 seconds.
Free running period, t,
= T - (, + t) =

11.7.
The is shown in fig.
desired speed-time curve
two stops 1.6 km apar
45 km per hour 1s requlrea between
mple 11.3. A schedule speed of
the r u n if the stop 18 or
20
second duration. The values of
1nd the maximum speed over and 3.2 kmphps respectively. Assume a simpliñe
cceleration retardation are 2.4 kmphps
and [B.TE. U.P. Electric Traction 2003]
Fapezoidal speed-time curve.
Solution: celeration, a = 2.4 kmphps

Retardation, ß = 3.2 kmphps

Distance of run, S =1.5 km

45 kmph
Schedule speed. V, =
434 Electric Traction

Sx 3,6000 1.5
Schedule time, T, X8,b00= 1 20 seconds
V 45
Actual time for run, T = T, - stop duration = 120 - 20 100 seconds

Using equation (11.2) we have

T T 3,600 S 1 1
Maximum speed, Vn 2K 4K? K where K 2a 22x2.4 * 2x3.2
100 1002
3,600X1.5 = 74 kmph Ans.
2x 0.3646 4x (0.3646) 0.3646
Example 14. An electric train is to have acceleration and braking retardation of 0.8 km/h/s and
3.2 km/h/s respectively. If the ratio of nmaximum to average speed is 1.3 and time for stops 26
seconds, find schedule speed for a run of 1.5 km. Assume simplified trapezoidal speed-time curve.
Gorakhpur Univ. Utilisation of Electrical Power & Traction Sept. 1984]
Solution: Let the actual time of run be T seconds

Average speed, V. 3,600 S 3,600x1.55,400 km per hour

T T T
Maximum speed, V, = 1.3 V. =
1.3 x = 020 km per hour
T T

Since V22-VT+3,600 S = 0
2
V= 3,600S 7,020xT-3,600 x1.5 7,020-5,400
1 +
1 5/6.4
2a 2B 2x0.8 2x3.2
or
V =
1,620 x 5
=
45.58 kmph
and V = n = 45.53 35.028 kmph
1.3
Actual time of run T = 3,600S 3,600x1.5 154 seconds
V 35.028
Schedule time, T, = Actual time of run + time of stop 154 + 26 180 seconds

Schedule speed, V, = * Sx3,6001.5x3,600 30 kmph = Ans.

T 180
Bxample 11.6. The distance between two stations is 1 km and the schedule speed is 30 kmph.
station stopping time 20 seconds. Assuming braking retardation 3 kmphps and maximum spead
L.26 times the average speed. Determine the acceleration required to run the service if the speed-
time curve is approximated [B.T.E. U.P. Electric Traction 2001
by a
trapezoidal curve.

Solution: Schedule time of run, T, = Sx,600 1X3,600

120 seconds
V 30
Actual time of run, T = T, - duration of stop = 120 20 = 100 seconds

Average speed, Va = S*3,600= 1x3,600 36 kmph

T 100
Maximum speed, V, = 1.25 V. = 1.25 x 36 45 kmph
Using equation (11.1) we have

VT +3,600 S = 0
Train Movement and 435
Energy Consumption

or
1 VT- 3,600 S 45x 100 -3,600 x1 _ 4
2x 2 V (45)
or
= =0.2777
or a 1.8 kmphps Ans.
2x 0.2777
Example 11.6. A
suburban electric train has a maximum
speed of 70 kmph. The schedule speed
including station stop of 30 seconds is 45
a
kmph. If the acceleration is 1.5 kmphps, find the value
of retardation when the
average distance between is 4 km. stops
[A.M.I.E. Sec B. Utilisation of Elec. Power Winter 1984]

Solution: Schedule time of run, T, = 3,600xS3,600X4 =320 seconds

V. 45
Actual time of run, T duration of stop
=
T, -
=
320 -

30 290 seconds
ace
V 20-T+ 3,6003
or = ,500 S 70x290-3,600x 4
(70)2
or
23
.204
2a
204- 1.5= 0.87
r
2x 0.87
0.575 kmphps Ans.
Example 11.7. A train is required to run between two stations 1.6 km
apart at an average speed
of 40
kmph. The run is to be made to a simplified quadrilateral speed-time curve. If the maximun
speed is to be limited to 64 kmph, acceleration to 2.0 kmphps and coasting and braking retardation
to 0.16kmphps and 3.2 kmphps respectively, determine the duration of acceleration, coasting and
braking periods. [Gorakhpur Univ. Utilisation of Electrical Power
with Traction 1980:
Pb. Uni. Electric and Utilization
January 1991]
Solution: Distance of run, S =
1.6 km
Average speed, V, = 40 km per hour
Maximum speed, V, = 64 km per hour
Acceleration, a = 2.0 kmphps

Coasting retardation, B. = 0.16 kmphps

Braking retardation, ß = 3.2 kmphps

Duration of acceleration, t, = =

2.0
3 2 seconds Ans.

Actual time of run, T = S, 600S 3,600 x 1.6

= 144 seconds

Va 40
Let the speed before applying brakes be Vg

then duration of coasting, t, = 2= 64-V, seconds

0.16

Duration of braking, tg
=
seconds
Since actual time of run, T = t, + tg + lg

144 32+ D4-V V

0.16 3.2
438
Electric Traction
G
Substituting sin 0 =

100
in expression (11.10) we get,
F. =
1,00O Wx =
100
10 WG kg= 10 WG x 9.81 98.1 WG
When the train is going up a gradient, the tractive effort
newtons...(11.11)
will be required to
force due to gradient but while going down the balance this
effort.
gradient, the force will add to
the tractive
3. Tractive Effort For
Overcoming Train Resistance. Train resistance
the forces resisting the motion of a train
when it is running at uniform consists of all
and level track. Under these
circumstances the whole of the energy speed on a straight
axles is expended against train resistance.
Train resistance is due to
output from the driving
various parts of the rolling stock (iü) friction at the (i) the friction at the
track and (iüi) air
mechanical resistance component of resistance. The first
two components constitute the
train resistance depends train resistance. The
upon various factors, such
and is expressed in as shape, size and condition of track etc.
newtons per tonne of the dead weight. For a normal
specific resistance has been 40 to 70 the train value of
The general equation for train newtons/tonne.
resistance is given as
Rk + kV + k,v2
where k, k, and .11.12)
ks are constants
resistance in newtons and V is the depending upon the train and the track, R is the
mechanical resistance and the last termspeed in kmph. The first two terms
represents air resistance. represent the
Tractive effort required to overcome the
train resistance,
F. =W Xr newtons
where r is the specific resistance (11.13)
in newtons per tonne of the
Total tractive effort dead weight.
required,
F, =Fa t F, +
F, 277.8 Wa 98.1 WG+
=
Wr
+ve sign for the motion up the gradient and (11.14)
-

ve sign for the motion down the gradient.

11.8. POWER OUTPUT FROM THE DRIVING AXLES
Power, P Rate of doing work
=

= Tractive effort xsCe Tractive effort =

time
x
speed =
F, x v

where F, is in newtons and v is in m/s.

E x Vx ,000 watts=
or P =
= x VYkW
3,600 Watts 3,600 (11.15) where V is in kmph.
Example 11.8. An electric train
each wheel is 90 cm weighing 200 tonne has eight motors geared to
diameter.
train to a speed of 48 Determine the torque driving wheels.
kmph in 30 developed by each motor to accelerate
seconds up a gradient of 1 in 200. The the
of 50 newtons
per tonne, the effect of tractive resistance 1s
is 4 to 1 and rotational inertia is 10% of the train weight, the
gearing efficiency is 80 percent.
gear ratio

Solution Agra Univ. Traction & Utilisation of Elec. Power 1983)

Weight of train, W 200 tonnes =

Diameter of driving wheels, D 90 cm

= 0.9 m =

Percentage gradient, G =
x100 =
0.5%
200
Train Movement and Energy 439
Consumption
Tractive resistance, 50
r =
newtons/tonne
Gear ratió, y = 4
Gearing efficiency, n = 0.80
Equivalent accelerating weight of the train
W 1.10W 1.1 x 200 220 tonnes
Maximum speed, V 48 kmph
Time taken in accelerating the train to a speed of 48 kmph, t, = 30 seconds

Acceleration, a =
n 1.6 1.6 kmphps
Tractive effort required, F, = 277.8 W, a + 98.1 WG + Wr
277.8 x 220 x 1.6 + 98.1 x 200 x 0.5 +200 x 50 = 1,17,596 N

Total torque developed, T _FXD1,17,596

nx 2y
x 0.9 16,537 N-m
0.8x 2x 4
=

Torque developed by each motor = 0 ; 0 = 2,067 N-m Ans.

8
Example 11.9. A 200 tonne motor coach having 4 motors each developing 6,000 Nm torque during
acceleration, starts from rest. If the gradient is 30 in 1,000; gear ratio 4, gear transmission
efficiency 90%, wheel radius 45 cm, train resistance 50 N/tonne; addition of rotational inertia 10%,
calculate the time taken to attain a speed of 50 kmph. If the line voltage is 3,000 V de and
efficiency of motors 85%, find the current during notching period.
Pb. Univ. Electric Drives and Utilization Dec. 1989)
Weight of train, W = 200 tonnes
Solution:
Diameter of driving wheels, D = 2 x 45 90 cm = 0.9 m
30
x 100 3%
Percentage gradient, G
=.

1,000
Tractive resistance, r = 50 N/tonne
Gear ratio, y = 4

= 90% or 0.9
Gear transmission efficiency, n
of the train,
Equivalent accelerating weight
W 1.10 x 200 = 220 tonnes

T = 4 x 6,000 =
24,000 Nm
Total torque developed,

Tractive effort, F, =
nT2Y = 0.9 x24,000X2X4 =
1,92,000 N
0.9
acceleration of the train be a kmphps
Let the
Then F, 277.8 W, a+98.1 WG + Wr
=

x 200 x 3 200 50
1,92,000 277.8 x 220 a + 98.1
= x
or
1,23,140 2.015 kmphps
a
277.8 x 220

train to attain a speed of 50 kmph

Time taken for the
t= 50= 24.82 seconds Ans.
2.015
1,92,000 x 50
=x 2,666.7 kW
the driving axles =
. 3,600 3,600
Power output from
Power input =
Power output= 2,6bb. =
3,137 kW
m 0.85
Power input in kW *
Total
current
drawn =

V
1,0003,187X1,000=
3,000
1.046 A
440 Electric Traction

Current drawn per motor = 1,04 = 261.5SA Ans.

4
Example 11.10. A train weighing 120 tonnes is to be driven up an incline of 2 percent at a speed
o 36 kmph. If the train resistance at this speed is 2 kg per tonne, find the current required at
1,600 V de if the efficiency of the motors and gearing is 88 percent,. If the current were cut off.
how long would the train take to come to rest
Solution: Tractive effort, F, = 98.1 WG + Wr = 98.1 x 120 x 2 + 120 x 2 x 9.81 =25,90O N

Speed, V = 36 kmph
Power output of driving axles = - FXV 25,900x 36 =
259 kW
3,600 3,600

Power input Power output Z259 = 294.3 kWN

fficiency of motors and gears 0.88
294.3x 1,000o
Current required = Ower input in kW x1,000
Voltage of the line 1,500
=
196.2 amperes Ans.
Let the coasting retardation be B,
then F, = 277.8 W-B) +98.1 wG+ Wr
Assuming W. = 110 per cent of dead weight, we have

0 -277.8 x 1.1 WP, + 98.1 x 2W +2 x

9.81 W
98.1x 2.2
or Be271.8x 1.1 =
0.7 kmphps

Time taken to come to rest = =

Ans. 51 seconds
Be 0.7
Example 11.11. An electric train accelerates uniformly from rest to a speed of 48 km/hour in 24
seconds. It then coasts for 69 seconds against a constant resistance of 58 N/tonne and is braked
to rest at 3.3 km/hour/second in 11 seconds.
Calculate (i) the acceleration (ii) coasting retardation
and (ili) the schedule speed, if the station
stops are of 20 second duration. What would be the
effect on schedule speed of reducing the station
stops to 15 second duration. other conditions
remaining same? Allow 10% for rotational inertia.
Solution: Time of acceleration, t, = 24 seconds
Time of coasting, t2 = 69 seconds
Time of braking, t = 11 seconds
Maximum speed, V, = 48 kmph
Acceleration, a = = 2 kmphps Ans.
Let the coasting retardation be Pe
F 277.8 W-B) +98.1 WG+ Wr
F-277.8 x
1.1 WP.+98.1 Wx0 + 58 w
58
or 2 7 7 . 8 x 1 . 1 = 0.19 kmphps Ans.

Speed at the end of coasting period V, =

V, B. x
t 48 0.19x 69
-
= - =
34.9 kmph
Distance travelled, S =
7,200
,+V,)tz,_Vta
7,200 7.200
km
48x 24 (48+34.9) x
6934.9*11_ 1.0 km
7,200 7,200 7,200
Schedule time, T, t , + t +l + duration of stop = 24+69+ 11+20 124 seconds

(ii) Schedule speed, V. =

S*3,600 =
L03,b00 = 29 kmph Ans.
T 124
Train Movement and EnerEy Consumption 441

When the duration of stop is 15 seconds

Schedule time, T 24 69 + 11+ 15 119 seconds
and schedule speed, V; =
08,D00
119
=
30.25 kmph Ans.
Example 11.12. An electric train has quadrilateral
speed-time curve as
follows
(i) Uniform acceleration from rest at 2 kmphps for 30 seconds;
(ii) Coasting for 60 seconds; (iil) Uniform braking to rest for 20 seconds.
If the train 1 moving a uniform up gradient of 10/1,00o, train resistance is 40 N/tonne,
rotational inertia etfect 10% of dead weight and duration of stop 30 seconds, find the schedule

Solution Time of acceleration, t, 30 seconds

=

Time of coasting, t2 50 seconds

Time of braking, ta 20 seconds
Acceleration, a = 2kmphps
Maximum speed, V, = a t, = 2 x 30 = 60 km/hour

10
Gradient, G = 100 1%
1,000
Let the coasting retardation be P.
F = 277.8 W,-B) + 98.1 WG + Wr
or 0-277.8 x 1.1 W P. +98.1 x 1x W + 40 W

98.1+40 138.1 = 0.452 kmphps

or Pe 271.8 x1.1 277.8x 1.1

V2 V1B.t2= 60 -
0.452 x 50 =
37.4 km per hour
Braking retardation, B 2= 20
1.87 kmphps

Distance travelled, S = yh.(V+V,)+V km

7,200 7,200 7,200
60x3030 60+37.4 x 50+ 37.4X20 =1.03
7.200 7,200 7,200
km
1.03 kmn
7,200
30 + 50+ 20 30 = 130 seconds.
t + t + duration of stop
=
+
Schedule time, T, t ,
xS=3,600x
3.600xS 1.03 = 28.52 km per hour Ans.
3,600XL.00
Schedule speed, V.
= 3,600 130
an electric train on a uniform rising gradient of 1 in 100
Example 11.13. The speed-time
curve of
comprises rest at 2 kmphps for 30 seconds
9Uniform acceleration from
tor 70 seconds
) Coasting with power off
at 3 kmphps
to stand stil
) Braking the traln resistance on level track being 5 nne,
The weight of the train
is 260 tonnes,
inertla 10%
Allowance for rotary motors and total distance t
by
developed 97%. by traction
Calculate the maximum power as
transmission efficiency
r a i n . Assume Agra Univ. Traction & Utilisation of Elec. Power 1979]
tonnes
W = 250
Solution: Weight of train,
W
=
|100W 1 . 1
Efective weight oftrain, W, 49.05 Nt nne
9.81 =

r =5 kg/tonne
=
56 x
Train resistance,
442 Electric Traction
Gradient, G = 1 percent
Maximum velocity, Vm = a t, = 2 x 30 = 60 km/hour
Tractive effort requirod, F, = 277.8 W, a + 98.1 WG + Wr
=
277.8 x 1.1 x 250 x
2 +98.1 x 250 x 1 + 250 x
49.05
= 1,89,577 newtons

Maximum power output from the driving axlos

V
Fm 1,89,57x
3,600 3,600
60 = 3,160 kW
Maximum power developed by traction motors

3,160 =
3,258 kW Ans.
0.97
Let the coasting retardation be B. then
F. = 277.8 W-B) + 98.1 WG + Wr

or 0 =-277.8x 1.1 W x B, +98.1 W x 1+ W x 49.05

or P. 98.1+49.05
0.4815
277.8 x 1.1 kmphps
V2 V -

B.t2 60 =
70 x 0.4815 26.3 km/hour
Braking period, ' = 2 = =8.77 seconds
Total distance travelled by the train,

S = Vi +V,)4,Vsh
7,200 7,200 7,200
60x30 60 +26.3 26.3 x 8.77
7,200 7,200
-
1.12 km Ans.
7,200
11.9. ENERGY OUTPUT FROM DRIVING
AXLES
Assuming the run
according to trapezoidal speed-time curve
Total energy required for the run =
Energy required during acceleration + energy required
during free run
Average power during acceleration X acceleration period
+average power during free run x duration of free run

1 FVm x ,nx_kWh
.(11.16)
2 3,6003,600 3,600 3,600
where Vm is the maximum speed in
time of firee-run in seconds,
kmph, t, is the time of acceleration in seconds, t is the
and F is the tractive effort
F, is the tractive effort
required during acceleration in newtons
required during free run in newtons.
Instead of expressing the energy in kWh, it is more
convenient for the purp0se ot
comparison to introduce the weight of the train and the distance of run and to express the
energy in watt-hours per tonne-km
i.e. Energy output in watt- hours
Weight of the train in tonnes x distance of run .(11.17)
in km
This quantity is called specific energy outputy/and is used for comparing the
performances of trains operating to different schedules. dynamical
The energy input to the motors is
called the energy consumption of the train, since 1u
is the energy consumed for
propelling the train. The total energy drawn from the distribution
 w-3NNOLHM NI
NOLLdNNSNO ADNINI AHOIAS
X
w-NNOLHM NI
NOILdWnsNOO ADAN HI3dS
446 Electric Traction

Acceleration a = 0.366 m/s

Maximum speed, V, = 12.2 m/s
Distance of run, S = 1,525 metres

Acceleration period ,
= 12.2 =
33.3 seconds
0.366
Train resistance during acceleration,r = 52.6 N/1,000 k g = 0.0526 N/kg

Tractive effort required during acceleration,

F, = W,a+ Wr = 1.1 Wx 0.366 +0.0526 W = 0.4552 W newtons
Total energy required for the run = Energy required during acceleration as there is no free run
Average power during acceleration x acceleration period

FVh =x0.4552 Wx12.2x 33.3 joules

92.56 Wjoules or watt-seconds
92.56 W =0.0257 W watt-hours
3,600
nergy required for the r u n 0.0257 W
Specific energy output
=

WxS Wx1,525
= 1.685x 10> Wh/kg-m

1.685 x 10-5
Specific energy consumption
= pecitic enerEy output
0.5
3.37 x 10- Wh/kg-m Ans.
Example 11.16. An electric train has quadrilateral speed-time curve as follows:
(i) Uniform acceleration from rest at 2 kmphps for 30 seconds
(ii) Coasting for 50 seconds
(iii) Braking period of 20 secondis.
The train is moving a uniform up gradient of 1%, tractive resistance is 40 newtons per tonne
rotational inertia effect 10% of dead weight, duration of station stop 16 seconds and overall
efficiency of transmission gear and motor as 75%. Calculate the value of its schedule speed and
specific energy consumption of run. [Pb. Univ. Electric Drives and Utilization Nou 1988]
Solution: Maximum speed, V, =at, = 2 x 30 = 6 0 kmph
Refer fig. 11.6 and also solution to example 11.12.
Coasting retardation, B. = 0.452 kmphps

V 37.4 kmph
Braking retardation, B 1.87 kmphps
Total distance travelled, S = 1.03 km
Schedule time, T, = t, + t2 + t t stop duration = 30 + 50 + 20 + 15 = 115 seconds

Schedule speed, V,=3,600x S 3,600x1.03 = 32.24 kmph Ans.

T 115
Distance travelled when power is on,
S =
Distance travelled during acceleration period
V = 60*30 = 0.25 km
7,200 7,200
0.01072 V,
Specific energy output = -
S
+0.2778 (98.1 G+r)
0.01072 x 60x
X1.1 +0.2778 (98.1x1 40) x1.03 +
03
50.5 Wh/tonne-km
Train Movement and 447
Energy Consumption

Specific energy consumption = . = 67.34 Wh/tonne-km Ans.

. .75
Example 11.17. An electric train has quadrilateral speed-time curve as follows:
() Uniform acceleration from rest at 2 kmphps for 30 seconds.
(i) Coasting for 50 seconds.
(iii) Braking period of 20 seconds.
The train is moving a uniform down gradient of 1%, tractive resistance 40 newtons per
tonne, rotational inertia effect 10% of dead weight, duration of stop 15 seconds and overall
efficiency of transmission gear and motor as 75%. Calculate its schedule speed and specific
energy consumption of run. [Pb. Univ. Electric Drives and Utilization May 1989)
Solution: Maximum speed, V, = a t, = 2 x 30 =
60 kmph
Let the coasting retardation be Be
Tractive effort, F, = 277.8 W, (-B) - 98.1 WG + Wr
or 0 =277.8 x 1.1 W B. - 98.1 x W x 1 + 40 W

- 58.1
or Pe = -
0.19 kmphps
277.8x 1.1
V2 Vi- Bt = 60 -

(-0.19) x 50 =
69.5 kmph
Distance travelled, S =_ 1 + * ' 2 x t +a3 km
7,200 7,200 7,200
60x30 60+69.5 5 0 + . * 0 = 0.25 + 0.899 +0.193
7,200 7,200 7,200
1.342 km
Schedule time, T, = t, + tz + tg + stop duration = 30 + 50 + 20+ 15 115 seconds

Schedule speed, V, = 3,600x S _ 3,600 X1.342 = 42 kmph Ans.

115

Specific energy output 0.01072

S
Vx W 0.2778 (98.1 G+r)
0.25
O.01072 X60x 1.1 + 0.2778 [98.1(-1) 40J342 +
1.342
28.6 Wh/tonne-km
= = 38.14 Wh/tonne-km Ans.
Specific energy consumption 0.75
100 has a rotational inertia of 10%. This train
Example 11.18. An electric train weighing tonnes
stations which are 2.5 km apart has an average speed of 50 km/hour.
while running between two
The acceleration and retardation during Draklng are respectively 1 km/hour/second and 2 km p e r
two stations is 1% and the train is to
hour per second. The percentage
graalen Detween these
move up the incline. The track r e s i s t a n c e 1s
*NtOnne. l t the combined efficiency of the electric
power at the driving axle (i) total energy consumption and
train is 60%, determine (i) maximum Assume t h a t Journey estimation is being made on simplified
(ii) specific energy consumption.
trapezoidal speed-time curve.

of train, W,
=
W = 1.1 W =
1.1 x 100 =
110 tonnes
Solution: Effective weight

Distance of run, S =2.5 km

= 50 kmph
Average speed, Va
3,600 x 2.5
of r u n
=
3,
S,600xS 180 seconds
Duration
V 50

1 = 0.75
K24 28 2x1 2x2
448 Electric Traction

T
Maximum speed, Vm2K TY3,600 S
2K K

180
2x 0.75 Vl2x0.75)
180 3,600x2.5 = 71 kmph
0.75

Acceleration period, t, = n == 71 seconds

Braking period, tg = 7 35.5 seconds

Time of free run, t = 180 - (71 + 35.5) = 73.5 seconds
Tractive effort required, F, = 277.8 Wa + 98.1 wG + Wr
277.8 x 110 x 1 + 98.1 x 100 x 1+ 100 x 40 44,368 N
=

) Maximum power at the driving axle

FXVn 44,368XTL 875 kW- =

Ans.
3,600 3,600
Total energy output =
Fx3.600 3,600 VX2
3,600 3,600
where F is effort required during free and is
run equal to (98.1 WG + Wr)
i.e. 98.1 100 1 100
x x + x 40 13,810 N
Total energy output
.
x *44,368x e x 13,810x71
3,600
73.5
3,600
14.19 kWh

ii) Total energy consumption =***=23.65 kWh Ans,

0.6
(ii) Specific energy consunption = Total energy consumption in watt- hours
Weight of train in tonnes x distance of run in km
23.65 x1,000 = 94.6 Wh/tonne-km Ans.
100x 2.5
Example 11.19. A train weighing 203 tonnes accelerates uniformly from rest to a speed of 45 kmph
up a gradient ofl in 500, the time taken
coasts down a uniform gradient of 1 in
being 30 seconds. The power is then cut off and train
1,000 for a period of 40 seconds when brakes are applied
for a period of 16 seconds so as to
)
bring the train
uniformly
the maximum power output from the driving axles
to rest on this
gradient.
(i) the energy taken from
Calculate
the conductor
rails in kWh, assuming an efficiency of 60%.
Assume train resistance to be 44 newtons per tonne at all speeds, and allow 10% for
rotational inertia. [Agra Univ. Traction & Utilisation of Elec. Power 1978
Solution: Weight of train, W = 203 tonnes
Effective weight of train, We 1.1 W 1.1
The train
x 203 223.3 tonnes
resistance, r = 44 newtons/tonne
Gradient, G =x100 = 0.2%
500
Maximum speed, Vm 45 kmph
Acceleration period, t,
30seconds
Acceleration, a =
=30 1.5 kmphps
=

Tractive effort required, F, =

277.8 W,.a + 98.1 WG + Wr
277.8 x 223.3 x 1.5+98.1 x 203 x 0,2+203 x 44 1,05,964newtons
Train Movement and Energy Consumption 449

(i) The maximum power output from driving axles

X n 1,05,64 x 45 1,324.55 kW
= =
Ans.
3,600 3,600
Total energy required for the run =
Energy required during acceleration as there is no free run

1Fx_kWh=*3.6004530
2 3,600 600 3,600
= 5.52 kWh

(i) The 52
energy taken from conductor rails =
9.2 kWh Ans.
0.6
Example 11.20. It is
proposed to put an electric trolley service in a city. The schedule speed is
to be 45 kmph.The distance between stops is 2.8 km. The track
is of 30 seconds duration.
is assumed to be level. Each stop
Using simplified speed-time curve, calculate the maximum speed
assuming the acceleration to be 2 kmphps and retardation to be 3.2 kmphps. The dead
of the car is 16 tonnes, rotational inertia weight
is 10 percent of the dead weight and track resistance
is 40 newtons per
tonne. the overall efficiency is 80%, calculate («) maximum power output
If
from driving axles
(b) the specífic energy consumption in watt-hours per tonne-km.
Solution: Schedule speed, V, =
45 kmph
Schedule time of run, T. x 3,600 2.8x3,600o 224
=

V 45 seconds. =

Actual time of run, T = 224 30 = 194 seconds

0.40625
2a 23 6.4 32

Maximum speed, Vm
T 3,600 S
K

194 194
3,600x 2.8 58 kmph Ans.
2x0.40625 V2x0.40625 0.40625
58
Acceleration time, t, = =
= 29 seconds
Duration of braking, tg = = = 18.125 seconds
B
Time of free run, t2 = T- ( +t = 194 (29 + 18.125) = 146.875 seconds
Tractive effort during acceleration,
F 277.8 W,a + Wr
277.8.x 16 x 1.1 x 2 + 16 x 40 =
10,418.56 Newtons
(a) Maximum power output =
n kW =10,418.56 x58 167.855 kW Ans.
3,600 3,600
(6) Distance travelled during braking period
= a =1 58x 18.125 = 0.146 km
2 3,600 3,600
Distance travelled with power on S 2.8 0.146 = 2.654 km

Specific energy output = o.01072 VLx+0.2778r Wh/tonne-km

S
0.01072x 582
x 1.1 + 0.2778 x 40 x 24.7 Wh/tonne-kmn
2.8 2.8

Specific energy consumption =

0.8
=
30.9 Wh/tonne-km Ans.
450 Electric Traction

Example 11.21. An electric train weighing 200 tonnes runs a uniform upgradient of 1% with the
following speed-time curve:
() Uniform acceleration of 2 kmphps for 30 seconds
(ii) Constant speed for 40 seconds
(ii) Coasting for 30 seconds
(iv) Braking at 2.5 kmphps to rest
() Stop at station 15 seconds.
If the tractive resistance is 40 N/tonne, rotational inertia effect 10% of dead weight and
overall e fficiency of transmission and motor is 75% determine
() schedule speed (li) specific energy consumption (ii) total energy consumption (iv) distance
between two stations. [Pb. Univ. Electric Traction June 1993
Solution: Speed at the end of accelerating period,
V = at, = 2 x 30 60 kmph
Let the coasting retardation be B, then
F = 277.8 W,(-B) + 98.1 WG + Wr

or 0 =-277.8 x 1.1 W x B. +98.1 x 1 W + 40 W

138.1
or P. 277.8x 1.1
0.452 kmphps
Speed at the end of coasting period,
V V1- Bex tg 60 0.452 x 30 46.44 kmph

Braking duration, t, = 2 = 46.44 = 18.6 seconds

2.5

Total distance travelled, S = L , Y ' 2 4 * 2 x t . V2t4

7,200 3,600 7,200 7,200
60x 30 60x40 60+ 46.44x 30+46.44 XL6.0 =1.48 km Ans.
7,200 3,600 7,200 7,200
Distance travelled when power is on, .

S,=50X
30 60 x 40=
7,200 3,600
0.9167 km

60

45 -

30

V2
15-

30 60 90 120

-
TIME IN SECONDS
Fig 11.1
1.48 x 3,600
Schedule speed, V, =. Sx3,600 30+ 40 + 30+18.6+15
+ +
tg +4 +stop duration 39.88 kmph Ans.
Train Movement and Energy Consumption 451

Specific energy output = 0.01072 Vx +0.2778 (98.1G+r)

S S
0.01072 x 602 X1.1+0.2778 (98.1 x1+ 40) x O.9167
.48 1.48
52.45 Wh per toonne-km

Specific energy consumption = 52.45= 69.93 Wh/tonne-km Ans.

0.75
Total energy consumption =
Specific energEy consumption x W xS
69.93 x 200 x 1.48 20,700 Wh or 20.7 kWh Ans.

11.12. DEAD WEIGHT, ACCELERATING WEIGHT AND ADHESIVE WEIGHT

Dead Weight. The total weight of locomotive and train to be pulled by the locomotive is
known as dead weight.
of locomotive and
Accelerating Weight. The dead weight of the train i.e. the weight
train can be considered to be divided into two parts.
1. The weight, which requires angular acceleration such as weight of wheels, axles, gears
etc. and
The weight, which requires linear acceleration.
2 Hence the effective weight, which is greater than dead weight is called the accelerating

weight. Accelerating weight is taken 5 to 10 percent more than dead weight. the
wheels is known
Adhesive Weight. The total weight to be carried on the driving
as

adhesive weight.
is to be hauled by a locomotive up a gradient of 2% with
Example 11.22. 400 tonne goods train adhesion is
acceleration of 1 kmphps. Coefficient of
20%, track resistance 40 N/tonne and effective
Find the weight of the locomotive and number of axles
rotating m a s s e s 10% of the dead weight.
increase beyond 22 tonnes.
if the axle load is not to
Elec. Power 1979; B.T.E. U.P. Electric Traction 2002]
[Agra Uniu. Traction & Utilisation of
locomotive be W tonnes
Let the weight
Solution: of the
Dead weight of train and locomotive, . W
=
(400+ W) tonnes
Gradient, G = 2 percent
Coefficient of adhesion, a = 0.2

Effective weight of train and locomotive, W, = 1.1 (400 +

WN)
+ 98.1 WG + WVr
Tractive effort, F, = 277.8 W, a
277.8a +98.1 G+r|W [277.8 x 1.1 x 1 + 98.1 x 2 + 40] w

541.78 W newtons
541.78 WN = 0.05523 W tonnes = 0.05523 (400 + W) tonnes
9.81 x 1,000
effort that can be possible with W; adhesive weight of locomotive
Maximum value of tractive
F W L = 0.2 W, (0)
and (ii) get
Comparing expressions (i)
we

0.2 W 0.05523 (400+ W)

0.05523x 400 22.092 = 152.6 tonnes Ans.
or W 0.2-0.05523 0.14477

= W 152. = 7 Ans.
Number of axles required Allowable load 22
452 Electric raction

Example 11.23. An electric locomotive is required to haul a train of 12 coaches each weighin.
30 tonnes on the main line service requirinK an initial acceleration or 0.8 kmphps up a gradien
of 1 in 100. Estimate the adhesive weight and henece the number of driving axles the locomotive
e
must have if the permissible axle loading is 20 tonnes per axle assuming rotational inertia
De 4 for the coaches and 16% for the locomotive. Maximum coefficient of adhesion is 0.2 to
and
the tractive resistance 6 IB.H.U. Elec. Traction && Utilisation 19781
kg/tonne.
Solution: Dead weight of train,
W 12 x 30 = 360 tonnes

Accelerating weight of train,

W. =
1.04 x 360 = 374.4 tonnes
Let adhesive weight of locomotive =
W
Accelerating weight of locomotive = 1.15 WL
Tractive resistance, r = 5 kg per tonne = 5 x 9.81 = 49.05 N/tonne

Coefficient of adhesion, = 0.2

Tractive efort, F, = (277.8 a)(W, + 1.15 W) + (98.1 G + n(W + W)
= (277.8 x 0.8)(374.4 + 1.15 W) + (98.1 + 49.05)(360 + W) newtons

222.24 (374.4 +1.15 W)+147.15(360+W)

tonnes
1,000x 9.81
= (13.882 + 0.041 W,) tonnes ...)
Maximum value of tractive effort that can be possible with Wi, adhesive weight of locomotive
W = 0.2 W
Equating expressions () and (i) we get
0.2 W, 13.882 + 0.041 W,

or WL 13.882
0.159
87.3 tonnes=

Number of driving axles =

87.3
4.37 5 Ans.
20
Example 11.24. A locomotive of 100 tonnes can just accelerate a train of 400 tonnes with an
acceleration of 1.5 kmphps up a gradient of 1 in 100. Assume an adhesive weight of locomotive
70% of total dead weight. The tractive resistance is 40 N/tonne and rotational inertia increases
the dead weight by 10%. If the above locomotive is aided by another locomotive of 160 tonnes
with 80% adhesive weight, determine:
(i) The additional trailing weight that can be hauled up the same gradient under the same
conditions;
() The maximum gradient, trailing hauled load remaining unchanged;
(iin The acceleration, if both gradient and hauled load remain unchanged.
Solution: Dead weight of train and locomotive, W = 100 + 400 500tonnes
Effective weight, W. = 1.1 W = 1.1 x 500 = 550 tonnes

Tractive effort required, F, = 277.8 x W,a +98.1 WG + Wr

= 277.8 X 550 x 1.5 + 98.1 x 500 x 1 500 x 40

2,98,235 newtons = 2,98,235 ..

9.81x 1,000
30.4 tonnes
Maximum tractive effort for the front locomotive

W X Adhesive weight x 100 x 0.7 = 70

Dead weight
Equating expressions i) and (i) we get
70 30.4
or 0.434
Adhesive weight of second locomotive = 0.8 x 150 = 120 tonnes
Total adhesive weight with both iocouiotives= 70 + 120 190tonnes
Train Movement and Energy
Consumption 453

Let the trailing weight that can be hauled, up the same gradient and under the same conditions
be W, tonnes
Now the total dead weight, W' (100 150
= +
W) tonnes
Tractive effort required, F (277.8 x
=
1.1 x 1.5 +98.1 x 1 + 40) W'
= 596.47 W' newtons = 596.47 WW'
= 0.0608 W' tonnes ..iii)
9.81 x 1,000
Total tractive effort that can be developed,
X adhesive weight of both locomotives 0.434 x 190 tonnes w)
Equating expressions (ii) and (iv) we have
0.0608 W' = 0.434 x 190

or W 0.434 x 190 = 1,356 tonnes

0.0608
Trailing weight, W = W 100 150 1,356 250 1,106tonnes Ans.
(ii) Total hauled load, W" = 400 + 100 + 150 = 650 tonnes

W
Tractive eftort required. =
|271.8xxa+
W" 98.1 G+r|W
= [277.8 x 1.1 x 1.5 + 98.1 G + 40] x 650 newtons

= (498.37 + 98.1 G) x 650 newtons

(498.37 + 98.1 G)x 650

tonnes
9.81 x 1,000

Equating expressions (iv) and (v) we get

(498.37+98.1 G) x 650 = 0.434 x 190
9.81 x 1,000

0.434 x190 x 9.81x 1,000

or G 650
-

198.37 x 98.1 = 7.6%

Ans
(ii) Tractive effort required F"
=
|277.8xxa'
W
+98.1 G+r W
= {277.8 x.1.1 x a' +98.1 x 1 + 40] x 650

= [305.58a' + 138.1] x 650 newtons

- 1305.58a'+138.1]x650 tonnes .(U)

9.81x 1,000
and (vi) get
Equating expressions (iv)
we

305.58 a' +138.1 660 = 0.434 x 190

9,810

or
[o.434x 190 x9,810_188.1 =3.62 kmphps Ans.
650 305.58
Examole 11.25. A train with a 1ocomotIve having 4motors has a total mass of 260 tonnes.
40 in 20 seconds on a 1% up gradient, The
Startinx from rest the train attains a speed or kmph
gear ratio is 3, the gear eliciency o ,
he wheel diameter 95 cm, train resistance (average) is
and rotational
18
inertla the
10. F'ind torque developed by each of the motors
40 N per tonne
the adhesive coefficient is 0.25.
and the minimum weight of the locomotive, given
[Pb. Univ. Electric Drives and Utilisation June 1991]
Solution: Total mass of train and locomotive, W 250tonnes
Diameter of driving wheels, D 95 0.95
= cm =
m
 Electric Traction
454

Percentage gradient, G = 1%
Train resistance,r =40 N/tonne
Gear ratio, y = 3
Gear efficiency, n = 95% or 0.95

Equivalent accelerating weight of train, W, 100+x250

=

100
275 tonnes
Time taken in accelerating the train to a speed of 40 kmph,
t 2 0 seconds

Acceleration, a = = 2 kmphps
4 20
Tractive effort required, F, = 277.8W, a +(98.1 G + r)W
277.8 x 275 x 2 + (98.1 x 1 + 40) x 250

1,87,315 N=1,87,315
9.81x 1,000
19.0943 tonnes
Total torque developed, T =*D 1,87,315 x 0.95
31,220 Nm
nx 2y 0.95x 2x3

Torque developed by each motor = =7,805 Nm Ans.

4
Coefficient of adhesion, =0.25
Adhesive weight, W, = = = 76.4 tonnes
0.25
Assuming an adhesive weight of locomotive to be 75% of its dead weight, we have
Dead weight of locomotive = = = 102 tonnes Ans.
0.75 0.75

EXERCISES
1. What do you understand by speed-time curves ? What is its use in practice?
IGorakhpur Univ. Utilisation of Electrical Power & Traction 19831
2. Draw the speed-time curves for each of the following traction systems:
) Tramcar (i) Diesel electric (i) Trolley buses (iv) Suburban electric train and (iv) Urban services.
3. What is a speed-time curve? Discuss briefly the distinct periods in a typical speed-time curve of a train
running on main line. IPb. Univ. Electric Traction December 1991]
4 Draw a typical speed-time curve for train movement, and explain (a) acceleration (b) free running (c)
coasting (d) retardation.
Draw the speed-time curve for urban service and suburban service train movement and simplity these
curves by approximation [B.T.E. U.P. Electric Traction 2001]
5. Draw and explain a typical speed-time curve for an electric train and explain what do you understand by
erest speed, average speed and schedule speed. [Agra Univ. Traction & Utilisation of Electric Power 1982
6. Discuss the factors which affeet the schedule speed of a train. What diíference exists in speed-time curves
in case of
urban and suburban services?
IGorakhpur Univ, Utilisation of Electrioal Power &Traetion 1981; Sept. 19851
7. Define erest speed' and 'schedule spoed' and discus8 the factors which affect the schedule speed of a train
(B.H.U. Elec. Traction & Utilisation 1978|
rOm the simplified speed-time curve, determine the maximum speed, when actual time of run, values ot
acceleration, retardation and the distance between stops are given. (B.T.E. U.P. Electric Traction 2003]