2

Train Movement & Energy Consumption

2.1. SPEEI].TIME CUBVES

The curve drawn between speed and time speed (in km/hour) on the
.taking
Y-axis and time (in seconds or minutes) on X-axis is [n6wn i spied_tiie curvi,
The speed+ime curv,e gives.complete informetion of the motion oi tt. truir. iti,
curve grves the speed at various times alter the start of run directly. Slope of
the
curve at any point gives the acceleration at the corresponding instani or spied. The
area covered !ythe curve, the tim: a\is and the ordinates" through th; instani
between which the time is taken represents the distance io ini ;;;."r-p;;_
ing time. "ou.,"a
Speed.rim-. curve-(Fig.2.l) mainly consists of (i) initial acceleration (ii)
c)nslant speed run or free run. (r'Ii) coasring (iu) retardatio, Typi;;i'rp.*i
time curve for suburbrn run with a short di-stanie between stops"ii,ho;;i;
fig. 2.1 .

. (i) Accelerario[ : It consist ol tr.r,o parts known as (a) constant accelera_

tion or acceleration while notching up and 16j sp--ed curve .roning o. acceleration
on the speed curvc.

,.(q) Con aat A celerotior or A relerqtio Duriug Notchint lLp: Durinp

n)tcnrtrl up p--flod (0 to rr) the currcnt ii mlintained approximrte'iy irnrtunt
the voltage_across the m)tor is gradually increased by'iutring -out the anE
resistance. Thus tractive effort ii constanr and rber;ro; it'"aJ*ii;ii'
startins
iiril1,lt I
constant during this period.
(b) Speed Curve Running or Acceleration ott Spee:l Curve : During speed
curve running (r1 to ,r) the
voltage actiog across tbe motor
remains ccnstant and current
starts decreasing with the increase v
in speed according to the
characte stics of the motor and o
flnally the current taken by the H
I
motor becomes constant. Durins -t
this period, though the traii I fr il
aco€lerates but acceleration I
iY
r,
decreases with the increase in I I I
speed and finally becomes zero
at the speed at which the tractive aox L2 600 oco 1.! ,000 t.{o(!
tr
effort developed by the motor rwE tN SECS
becomes extctly equal to the rfq ; ! I SPEF D- TIUE C lRy€
resistanae to motion of the train. Fig.2.1
l8 ELICTRIC TRACTION

(ii) Free Run or Constofit Speel Run: At the end of speed curve running
i.e.ztt2the train attains the mfximum speed. During this period the train runs
with constant speed attained at 12 and constant power is drawn.

_ (rir) Coasting At the end of free running period (i.e. at ,s) power supply
:
is cut ofr and the train is allowed to run under its own momentum. The speed of
train starts decreasins on acrount ol resistance to the motion of train. Therate
of decrease of speed -<luring coasting period is klown as coasting retardation.

, _ (it') Retardotiot or Brakins Period: At the e1d of coasting period (i.e. at ,a)
the.brakes are applied to bring the train to rest. During this period speed decreases
rapidly and finally reduces tolerr-r.

2.2. TYPICAL SPEED.TIME CURVES

_ l. t|fian ot City Setrice: ln urban or city service the distalce between

the two stops is compaiatively very short (say I km- or so). The time required for
thrs run is very small (few minutes). The acceleration as well as retardation is
required to be high so that high average speed and shoft time of run is obtained.
The acccleration and retardation for ur-ban iervice is betleen l'5 and 4 km. p.h.p.s.
aod between 3 and 4 k m.p.h.p.s. respectively. Free run is not present in tLis run.
The_coasting retardation is a6out 0:15 k m.p.h.p.s. The coasting period is also
small. Typical speed-time curve for urban service is shown in fig. i.2.

_ 2. Suburba Strice: In this run the distance between two stops is little
longer than urban service but smaller than main line service (say betw-een I km
and 8_km). Free rBn is still not possible. Coasting is for a comparatively longer
period. Acceleration and retardation required are a1 high as for urban service.

-T-l-
t
45
$* _=---t_
€

0
3 S*
b:
r-l- \
* 5 \
\
t
lrt \
I I
I o I
o2040@60 o ?o .b @ 80 loo ,20 ro Go t6a
1=--+ lME .tN SECS fllt'rE tt SECS
ltl,l sER c€ SPEED flta€ At?n -----t-
tu JPaA gRnc€ 9t:toEuCCllRvF
Fis.2.2 Fig.2.3

3. Mqin Line Seryice : Th: distanc: b:tween two stops in main line
service is considerably more (usually mrrethan l0 km). In thii service free run
is of longer duration. The duration ofacceleration and retardation isasmall
fraction ( [ total running tiru:. Sirce r.rte of acceleration and retardation does Dot
affect the average and schedqle spceds therefore th-.se are of little importance in
main line servic€.
- l9
TRAIN MoVEMENT & eNrr.cv @NSUMPTIoN

DEto ctPtr
EJtt//to PU4/;U0
,00
-I ].
s v I \
75
\ I I

e!.r I I t
I 50 f I
I

I i"
I I

I 25
I
I

I I

Rr& rrar/c I I I I

tCCf,I f,?Ar/Ol I I I I

I I I
I
0 ar0 40! 6M A00 t0t0
/l6ucs

Typical Speed-Curre a
Fig.2.4

are given below :

Characteristics of each type of scrvice
stance
Maximum between Special
S. No. Type of Acceleration J Retardation I
Remarks
in km P'h'P s speed in stations
Service in km.P.h.P.s.
km. p.h in km
l-
l

No tee
running
Urban l'5 to 4 00 3to4 120 I
period,
coasting
period-
small
-No-Eee-
L tsto 3to4 ttoSkm running
2. suburban +'t.,0 J 120
period,
coasting
period-
long

More Free run-

3 Main line I o'0 to o't 1.5 160
than ning and
10 km coasting
periods-
, long.
Accelera-
tion and
braking
periods-
compara-
tieely
small.

a
I
20 ELEC RIC lNACTION

2.3. CREST SPEED, AVERAGE SPEEO ANt] SCHEDULE SPEED

Crest Speed: The maximum speed attained by the vehicle during the run
is known as crest speel.

Aterage Speed :
The mean of the speed from start to stop i.e. the distance
covered between two stops divided by the actual time of run is known as alerage
sryed.

M athemat ica I I y uu.."g" rpe.d -. E!!1!91!9!I*IllgPl

Actual trme ol run
ScLcdule Speed: The ratio of distanc€ covered between two stops and total
time of run including time of stop is known as schedule speed-

Distance between stops

Mathematically schedule speed:
Actual time of ruD+ stop time

2.4. TA0TORS AFFE0TING ScHEDULE SPEED

--
The schedule speed of a given train when running on a given service (i.e.
with a given distance between stations) is affected by the tollowing factors :
(i) Acceleration and braking retardation.
(ii) Maximum or crest speed.
(iii) Durarion of stop.

. . ^ (i). Elfect of .Arceleratkn awl Broking Retoftlation : For a given run and
with fixed crest speed the increase in acceleration will result in Aeciease in actual
time of run and therefore increase in schedule speed. Similarly increase in brakins
retardation will aflect the schedure speed. variatibn in accelerition ana r"taroiiiof;
will have more eflect on schedule speed in case of shorter distance
to longer distance run. -n in "o.piiiio,
(iD_ Efect of Moximum Speed: For a constant distance run and with
^ a,cceleration
fixed and retardation the actual time oi run *iii'd""..ur" and theri_
fore schedule speed will increase with the inciiasi ln crest ipeed. fne etreci of
yar.iation in crest speed on schedule speed is considerable' i, oifong
drstance run. "rr"
.(iii.). .Efecl of Dtration of Stop: For a given average speed the schedule
speed wttr rncrease by reducing the duration of !top. The lariaiion in duration of
stop wrll atlect the schedule speed more in case of shorter djstance run as comparcd
to longer distance run.
a
2,5. SIMPLIFIED SPEED.TIME CURVES

. .The speed-time curve'of an. urban service _can be replaced by an equivalent

speed-time curve of simple quadrilateral shape The speed' time iurve oi u main
line service is best and most easily replaced by a
traiezoid. Since the area of thi
speed-tiTe curve rerresents the total distance travelled-hence tbe areas of tne two
curves should be same-

I
 \
TRAIN MoVEMENT & rxsncy coNsrrMprloN 2l

The following examples illustrate the method of calculations.

q a
3 \
N
(t
B $
t
I 1 I

----->T/uE -->7/ E
Approximate Speed-Time Curves
Fig 2.5
(i) Calculation By Trapezoidal Speed Time Curves.
Let
a: acceleration in k m.p.h.p.s. E
6
=
9:retardation in k m.p.h.p.s.
Vrn:crest speed in k m.p.h.
I
T:Total time of run in seconds.
I
Time for acceleration in seconds, lr:YlL I
trl *.r rl t,
t
Time for retardation in seconds, ,r-T-
- Y4,
r.
---.......,-flMttls'Cs
tar?E?otal t eo-ltlE cuay€
Time for free running in seconds, ,r:T-(tr+13) Fig. 2.6
:r- /Yr+!"
\z fi
\ ./
Total distance of run in km, S:Distance travelled d uring acceleration
+distance traYelled during free run
idistance travelled during braking
:*v* *m-+v-
ffi+l r- -lh
Substituting r:Y, o:Yana r,:r-(Y- +f
) "" e"t

'- ,k:0.*#-['- (?, ?)].,*r-

or ':r*&;n#o r-#;-## *-rkp
of S :
v", T Y,,2
3,om--2,r00;-?ro-0
Y,o2
p

or -l,t;. Gl+f )-ffi+s:o

or ,-. (r* ** )-v,n r+3,600 s:o

I
22 ELECTRIC TRACTION

This is a quadtatic equation for V",

tt
Substituting
; * 5| : K, we get

K vrnz- Vr), T+3,600 S:o

T+ T.-41K x 3,600 S rlr^/ nrr--T-
F 3600 S
2K 2K
The +ve sign can not be adopted, as value ofVnr obtained by using fve
sign will that is possible in practice. Hence - ve sign will be
be much h;gher than
rrsed and therefore we have

\*:*- I T' -1,600 s

J +-Rz- x
From the above expression unknown quantity can be determined by sub-
[tituting the values of known quantities.
(ii) Simplified Quadrilateral Speed Tinu Curve.
Let a:Acceleration in knr.p.h.p.s.
pc:Coasting retardation in km p.h.p.s F

0: Braking retardation in km.p.h.p.s. o

Vr:Maximum speed at the end of
acceleration in km.p.h.
Yr:Speed at the end of coasting in
I'
I

k.m.p.h. I

T:Total time of run in seconds. I

Time of acceleration in seconds t'- ! I - ,r -+,- lt

----4tt
'TtuEtlttlcs
Time of coasting in seconds, ,r:U')'
9c

Qua dr ilate ra I Speed-Time C urYe

Time of braking in seconds, tr: !p Fig. 2.7
rotar distance travelled in km' t;?r'.',?ffi,,.;ili,::t
l|,ilt"frIi:f't'
*distance travelted during retardation
:+ vrx -ritn*LjL,ffi-+lv,* {;-
'- v,r, V,r, . vrr2 , v,ra
r t,zwt T,zgo
i,iw+ t,zw
:*,* ("+")+ #tu'-t"'
- r:rL,r* (r-r,)+ ,lrroorr_,,,, since ,,+,,+rr:r
* r:#(v!+vr)- #,r*-ffi
TRAIN MOVEMENT & LNERGY CONSUMP]ION 23

or
T (V,+%)- vr v, v, ..v,
S.-7-200 x
, ,OOx O -Z-,ZOo, ;
r
(V,* vr)- v, Yc
vrYr
or S : 71-,
266 f,ffi -zf i-
or ?,200S:T(V,+V.)-V1 U, (**t)
From the above expression the value of unknown quantity can be determined
by substituting the values of known quantities.
Example 2.1. A trafu nus lt sn aYemge speed of4i km per hour between
strtiotrs 2 5 irm apart.
-its Tbe train acceler.tes rt 2 fm.p.h.p.-s. and retcrds rt 3
km.p.h.p.s, Finrl mrximum speed assuming a trapezoidal sPe€-d time curve.
Caliulaie rlso tbe distsncc traYelted-by it before the brakes are rPPli€d. DeriYe thc
formula used for calculalion of maximum
"{iii'.u. u*. Traction & utiri*riut 19781

Solutlon: Acceleratiirn a:2km.p.h.p.s.

Retardation 9: 3 km.p.h.p.s.
Distance of run, S:2'5 km

rime of run, r: flx:ooo: fjx36o0:200

seconds

Using equation u^: ,!*- Tr 3600 s

4K' K

K:l+ 2p I
-t-
l5
wrrere 4 6 12

V,,,- ,Hnr_ 4 (200f 3600x 2 5

We gel nraximum spce<i,
I x (5i l2)! 5112
:240- \/ 57 ioo-2I,600 :50'263 km p'h' Ans'

Distance travelled before the brakes are aPplied

ll:'jffifi j,
=2 5- -72o1n: z r-
:;:l',"l,TJ. -7ZcrV1 383 km A,,

Example 2.2. A trein hrs scbedule -speed of 60 km' per hour betrce[ tho
-O"iut,ioi
$oDs whicb are 6 km the crest speed over the rutr' s3suBiDg
in" t't"in accelerates at 2 km p h'p's' and retards rt
"p".t.
till"ra-"i-l*Ji-tii"" *."J.-
3 klm.p.h.p.s. Duration of stops is 60 seconds.
lpb. (Jniv. Elec. Traction Juty, 197{
Solution: Acceleration, c:2
km.p.h'p s'
Retardation,P:3 km P.h. P.s.
Distance of run' S:6 km.
Schedule speed, V,:60 km P'h'

Schedule time, T,:fi x 36oo seaoqds= o.a

x360Q:360 seconds
 ELECTRIC TRACTION

- Actual time of run, T:To-Time of stop:360-60:300 seconds.

Using the equation
'r2
u*:+-J 4K'- 3600 s
K
where K :
lt t15
2a'2O
-I-_-I- 4 ' 6 12

We get maximum speed, Vm: 300 300 3600x 6

,'.- 5 5 5
'n 12 4x (
2 ) 2
:360-\/ (3oo1r-r t 0
:360-278 8:81 2 km/br Ans.

Example 2.3. An electric traitr is to have a braking retardation of3 2 km.

p.hrp.s. If the ratio of maximum speed to ayerage speed is 1 3, the time for stop
is 26 seconds and acceleration is 0'8 km.p.h.p.s. find its schedule speed for a run 1 5
km. Assume trapezoidal spe€d time curve.
Solution :Acceleration, a:0.8 km.p.h.p.s.
Retardation, 3:3'2 km.P h.P.s
Distance of run, S: l'5 km.
Let the actual time of run be T seconds

Average speed, V.,:

3,600S 3,600x l5 5,400
km per hour
T T T

Maximum speed, V,,:l'3 vo:l'3x '#:ry km per hour

t
Since V,,2 ( + 29 v,, T+ 3,600 s:0
7't?ox
v- T-3,600 s
r-.:,ooox r's 7,020-5,400
v ri:-1
2o.
- '29
_L-
1

2x0
I
8
+ 2 x3'2 -sieE-
6'4 :45'53
1,620 x 5

o,?:rr'028 km/hour
and v.o:* :
Actual time or .*, r:
3'?9 s 3'6-WXl 5
-154 seconds.
Schedule time, Ts:Actual time of run+Time of stop
:154*26:180 seconds

$qhedule speed, V , _ _ _Tc_

S x 3,600
U#6@S:30 km.p.h. Ans.
 TRAIN MOVEMENT & ENERGY CONSUMPTION 25

.- Exampl€ 2.4. The schedule speed with I 200 tontre troin on an electdc
railway with stations 777 metres apart i. ZZ n i.,
f.. f,o", *aif," maximum speed
higter tban the ayerage running ipeed.
i:20_ Pelcent The brakiog ."t" i" 3.22
s: an.d.lhe duraaiotr of stop is 20 secon'ds.' Finrl the acceterition required.
tm_..P:h-.p_
assume a simplified speed-time curye with free runuing
at ahe maximtrm speed.'
lAgra Unir'. Troction & Utilisition of Elec. power'1976 Supp.l
Solution: Schedule speed, y,:27.3 km.p.h.
Distance of run, 5:777 metres:0.777 km
Retardation, p:3.22 km.p.h.p.s.

Schedule time of run, - rs 3.600xS 3.600v0

=ff1 '-{!fr121 -roz.: sgc.

Actual tinte of run, T:Ta_duration of stop :102.5_20

Average speed,
Maximum speed,
".:{+I!:!q#Pj;ffr:
V,,:1.2 t.2x 33.9:40.7 km/ht
Y
":
Since v-z (**#)-* r+3,600 s:o
... I I V"l T_3600
2;_28 _-aT
S

ot _ tl L_ 40 7x8Z'5-3600x0 77
?:o.rsa
2a ' 2x3'22 (40.712
ll
ot'
;-0'338 644 o 338 o l5s o'183

or a:2'73 km.p.h.p.s. Ans.

,.^ --_E!|:lpt" 2.5-

A,suburban electric traitr-has-a maximum speerl of 65 km.p.h.
:^1"^,1!99rh..p"_.qincluding a srrtiotr stop of 30 seconds is 43;5 km.pi. tftm
is 13 km.p.h.p.s, find the yalue of retardrrion when tie a"erage
i9_.."]:11t,J"1
orstalce stops is g km.
beaweeo
lGorakhpur Univ. Utilisation of Electrical power and Traction I9Z5l
Solttion: Schedule speed, v,:43.5 km.p.h.
Distance of run, S:3 km.
Acceleration, *: 1.3 km.p.h.p.s.
Maximum speed, V,r:65 km.p.h.
Duration of stop:30 seconds

schedure tim: or.un, r,:3-699 J:Sf :24a seconds

Actual time of run, T:T,-duration of stop:24g_30:2lg seconds

since v-. ({-+i)

-v", r+3.600 s-o
.,' l--l V-T-3600S 65x218-3600x3
", Z; r2p : ----T;--:ffi:o.zer
 26 ELTCTRIC TRACTION

rl
or .r;6 * 16 -0 798

or +:0'7e8-o'385:o'413

or p:ffi:
I
t zt km.P.h.P.s. Ans'

Example 2.6. A lreio is required to run between tfio stations l'6 km' apa
O t r. p3i'i""i'-- tl" run is totob-e
of made lo a simplified
"t ", "ilii-gi''"p."4iir" tt" t"*i'ot is be linited to t4 kmlhour
;;;ittt;i;;i.ffd "*"".'r ;;;;;.tin!
spe-ed-
a'na urarjng re'rrdatiotrs to 0'16
acceleretion lo 2'0 kD. p.h.p.".-
il-. ;--d; iru gi r.r.ri.t ri.s rJspectively' ie'ermine the duralion of accelerstitrg'
corstirg ad brakitrg Periods.

Soh*ion : S: l'6 km
Distance of run,
Average sPeed, V":40 km Per hour
Maximum sPeed, Vr:64 ko Per hour
Acceteration, a:2'0 km'P h'P s'
Coasting retardation, 9o:0' l6 km.p.h.p's.
Braking retardation, P--3 2 km p'h'P S'

Duration of acceleration, ,r:\:f;J,:n seconds Ans'

Actual time of run, T: 3

'!iqg-r :3'6qx I 6:144 seconds Ans'

Let the speed belore applying brakes be V,

then duration of coasting, ,,--+ -' !1Ju 'seconds

Duration of :$
brakinc, ,. - Lp'- ,."ooat

Since actual tine of run, f :/r$(* la

r+e-rz+ 6I#, +$
or Vs ( 0 16 :32*400-r44
3'2 )
or ,,: o5$|6:48'5 km per hour
Duration of coasting, u:u*u"'":6{ff-e6 '85 seconds Ats.
v! 48'5
l5'15 second s Ans.
Puration of braking, t p 32

I
TRAIN MOVEMENT & ENERGY CONSUMPTION

2.6. MECHANICS OF TRAIN MOVEMENT

uatu
' Essential driving mechanism of an electric ,ornrtupt 0f
locomotive is shown in lig. 2.8. The armature
ofthe drivins motor has a pinion of diameter r/'
attached to it The tractive effort lt the edge 6trR t/ttEt
of the pinion is transferred to the driving wheel by ,ltftl
means of a gear wheel.
(; - IDAD

0
Let the driYing motor exert a torque T
(in N-m)
Tractive effort at the edge of the pinion is
ql An 5 t 55t0 0r I tAcl l$ [ ff1Rr
given by the expression. ^t
Fig.2.8
d'
T=F a
2T
or F
a

Tractive effort transferred to the drivi ng wheel

O:rr,(A):rT ( Dd ) :,tT ( 2t) )( 7d ) \T D
where r/ is diameter o[ gear wheel in metres, D is
diameter - of driving wheel
in metres and is the efficieniy of transmission, )'' is the gear ratio and is equal
4

to7d
The maximum frictional force between the whetl and the track:pw where
p i" tni iiJnnt"nt iToitnrion b€tween the wheel and the track. and w is the weiqlt
irti"ir"ir1-i-in"frirlng i*t"t (called adhesive weight)without Slipping.will not take
,i*" ,irfii, ii".iive effoit F>r.W. For motion of tiins slipPing lractive
;tr";i F;h;il4 be less than or at the most equal to /rw but in no case Sreater
than rrw.
The magnitude of the tractive effort that can be employed for propulsion,
tuerefori aepe-ndi upon the weight coming over the driving.wheels aad the coefr-
between the d;iving wheel and the track .The coefficient of
"i."i "ilatJti.,
uJft"rion if a"to"a as the ratio of tr,ctive efTort to slip the wheels and adhesive
weight
Tractive effort to sliP the wheels
i.e. coemcient of adhesion, P: Adhes lve welg ht

The normal value of coefficient of adhesion with- clean dry rails is 0'25 and
with wet-or-gteasy rails the value may be as low as 0'08'
Electric traction has a very rmportant advantlge over steam
traction and
tr,ut i. inl""ourrt oigt*t"t adhe:iv; iveight-in a motor coach 100 perc€nt of.the
;;;hi i;';;iltiri,ig-*'tt."tt. inperient an eleciric locotnotivc 70 percent where as in a ,
;;;il' iil;;t;"i.ti trtiin so The cocfficient of adhesion in electric
due ro thc forlowing rwo reasons.
iilil;"il..g;uiiirt un tuot iir steum rraction
(il In electric traction the torque excrted is conlinuous where as io steam
traction ihe torque is pulsatitrg which causes joltitrg aod skiddrDg'
 28

*.*,.11J,*lX ;l::.*, ll':i::: l*il,

ll:..i they wheers
rractron
a re .,,,,,
are close to each other'
J::'T;' :^::i
. Since higher'va;;; ;;t't
ar erectnc can be used in electric
train can be
-ro" lT"jlli,"I9',t at a faster rate. Tl tracrion, therefore,
ti'Ie' especially w1;;ih;3i:.:-:*:rcrate
urbanareas. _."tance betweensioi,l.lijili;r,ilir,l"r"rllilr^r{,11;l
2.7. TEACTIVE EFFORT FOR
PB()PULSI()N (]F TRAIN
p*,nJff i:ll[?#;;t:','l'r_1,i1,.,",r*tr:gin;Til,.",:,:::,";:*,1,J;
"'fl
Total tractive effort required
to run a train on track

.
.r ",lfti,ir.:$fl::1{:*:ffiii,r"l#fHi,l"i':;::rii
(l) Tractie Efforr Fot
i,*q"i-J.rouj'f#l'[,:(jjii:,*:#1[.,;.1fi
:.fl 1y":,,,;xi*,31;HJ[",.'*
Consider a train of weighr W
toflnes being accelerated
The weight ol train I,000 w -o ---''''..su al
at d km_p.h.p.s.
kgl
Mass of train, m: 1,000 W kg

-A':o zzz8 a 6/5,

..";:::','.:*,rf,j;i;1"",,*,tffi

*N**ffi*'l#i*ffi
ffi***n.'ffi*'fi*l
Hence tractive effort required
for acceleration
Fa:277.9 We ,r newtons
(2) When the train is on rhe st^^- o r^-^- ,.
ffiHf ihf :{l',}:;1'iIiT'il',lJitil..':'.,,"i;'ffi ""1,f
Il''l'".'tfi il',:o:"?,To,?;
Hence force du€ to gradient:I000 \{ 5j1 Q ftg
: looo w
ft-tg stn". sin o:fi the loage
: l0 WG kg: l0x 9.81 Wc Newtons gradient
:98'I WG Newtons

r
tRArN MoVEMENT & ENERGY coNsuMprloN 29

When the train is going up .r gradient, the tractive effort will be required to
balance this force due to gradleni but while going down the gradient, this force will
add to the tractive effort.

_ (3) jResrstnn ce to Molion: When the train moves, there is a force opposing
the motion on friction and wind etc. this force is known as jrictioi
account of
&rce and, depends upon various factors such as shape, size and condition of track.
This i; expressed in newtons,'tonne of the dead weight. For a normal train the
value of specific resistance has been 40 to 70 newtons per tonne.
Tractive effort is also required to overcome the force of friction.
.'. F":Wx r newtons
where r is the specific resistance in newtons per tonne of the dead weight
Total tractiye effort required, Fr:F"tFatF :277 8 W6 a*98 1 WG*Wr
l-ve sign for the motion up the gradient and -ve sign for the motion down
the gradient.
2.8, POWER OUTPUT FH()M THE I]RIVING AXLES
Power, P:Rate of doing work

-TractiYe "nor,r. ff
:Tractive effort X speed
:Fr X y where Fr is in newtons and u is in m/s

:FL, yy
ffi *"r":ffifw where V is in km.p.h.

Example 2.7. A 150 tonne E.M.U. train has four motors each ileliy€ring a
shaft torque of 35C0 N-m dlritrg acceleration from rest. Fitrd out the time taken

for the train to attain a speed of 41 km. p-h. startitrg from r"st, on gradietrt
,OIO
with a wheel diameter of 1140 mm. and gear ratio of4 5. Assume traitr resistatrce
as 0'5 kg/tonne, rotary inertia as 10o/o and gear emcielcy as 95%.

Solution: Gear ratio, y:4 5

Gear efrciency, q:0 !J
Driving wheel diameter, D:1140 mm:l'14 metres
Torque developed by each motor:3,500 N-m
Total torque developed by 4 motors, T: 3,500 x 4: 14,000 N-m
Tractive effort, F1:17
2y _
095x 14,000x2x4 5 :1,05,000
N
D l'14
...(r)
kt the acceleration of the train be a km. p.h.p.s.
then Ft:277'8 We c-l'98 1 WG-l-Wr
:277'8x 150x l l x o.+ 98'l x I50 x ** 150 x0 5 x 9'81 ...t )
:45,837 a*8093 25
Equating expressions (i) a d (if) we get
45,837 q*8093' 25 : i,05,000
I
30 ELECTRIC TRACTION

1.05,qq0-!093 25, 2.1l3 km. p.h.p.s.

or "d - 45.817

Time taken for the train to attain a speed of 40 km. p.h

, = y.-r-k:18.e2 seconds Ans

Example 2.8. train wcighing 120 Joane is 1o be driven uP an incline-of 2

A
nelceota.asoeedof3ikm.p.h.If3hetrainresistanceetthis'speediszRgpe]
find the current required at l'fl)J V d.c. if thc efficiency -of the molors
aoo
tlnoe,
'g"liiiJi. u p".i;oi. Ii tt. "otr"ot were cut off, how long w6uld the trsio tske to
come to lest ?
Solution: Tractive effort, Fr-98't WG-l W,
:98 tx t20x2-120x2x9 81:25,900 N
SPeed, V:36 km.P.h.

power output of driving "'.'.r: fi,Yo :25'900'I 259 L*

Po\Yer output
Power input - Efficiency o f motors and geals

:i't:,,0, kw
- Power ioDut in
.%n;Ce kwx 1.000
Current requrred: of-the-line-
-
:294 1[ L{oo - 196 2 amperes Ans.

kt the ccastiug retardation be Pc

then Fr:277 8 We(-Pc)+98'l WGtWr
Assuming We:110 percent of dead weight we get
o:-27'l'8 xI l Wp"+98 1x2W+2x9'8lYv
or ,6:+#t-0.7 km. p.h.p.s.

Time taken to come to ..u: J:19,:Sl

F" \t1
4 seconds Ans'

2.9. An eleclric train is rccel€rsted uniformly from rest to a speed

';[..-;1"
Examnle
40k-. ,"iioa of acceleration being 20 seconrts' If it corsts for
"f
6b..o-ods ii"il.i iesistarce of 50 N pei tonne and is brought to rest itr
"6rii"rt
a furaher period of "l0 seconds by braking, determire
(i) the rcceleratiotr (ii) the coasting retrrdation
(iii)the braking retardation (iY) distance travelled and
(v) schedute speed with station stops of 10 seconds duratior' Allow le PeI991!
for rotatiooatinertia.- lPb. Univ' Elec' Ttoction 19731
Solution : Maximum speed, Vr-40 km.p.h. '
Time of accelerarion, t,- 20 seconds
TRAIN MOVTMENT & NNTNCY CONSUMPTION 3l

Time of coasting, lr- 60 seconds

Time of braking, la =- l0 seconds

(i) Acceleration,
v, 40 km. p.h.p.s. Ans
": t:#:,
Wr Wx50
(ii) Coasting retardation, Po: 277:8 w.
x l'lxW
277'8
:0'1636 km. P.h.P.s. Ans'
Vr:Yr-Pc'!:40-0'1636 x 60 :30 km' P'h'
(iii) Braking retardation, U:?-i+:, km. p.h.p.s. Ans.

(,'v) Distatrce travelled, t:#; * 'u+jJ(i'u *ffi n

40x20 . (40+30)x60 -30I]! v-

-aN ' - 72oo-' 7,2m '*
:0'736 km Ans.
Schedule time, Ts: rr + rr+ 13+ duration of stop
:20+60+ l0+ 10: 100 seconds

Schedule speed,
s:3+#lf9:27 5 km' p'h' Ans'
''10"!
',: time ctrwe rs
Example 2.10. An etectric train has quadrilateral sPeed
follows :

(i) Uoiform rcc€leration from rest at 2 km p h p s for 30 seconds ;

(:i) Coasting for 50 seconds ;(iii) Uniform brating to rest for 20 seconds'
If the train is moving a uoiform up grsdient of-10/1'Ofl)' train n;sistance is
lr N.'trrne, rriuional ineria efrect 10"i oT dead weight 8trd durati?'' of stop 3)
secords, find the
- schedule-- spe€il.
:- IA'.M.!.E. Sec. B' Utilisation of Elec' Potver &tmner 197\
Solution : Trme of acceleration, rr--30 seconds
Time of coasting, lg:50 seconds
Time of braking, tr:20 seconds
Acceleration, e:2 krn' P h'P's'
Maximum speed, Yr : a,t : l v lQ : 60 km/hour
Lt thc coasting rctardation be Pc
Ft:277'8 W" (-PJ+98'1 wc+wr
or O:-277'8x l'l w9,+98 l x l x W+40 W
or p"- -;8;+*+-rr|ffi :o r52 km. p.h.p.s.
Vr:Vr-p" ,r:60-0 452x 50:37 4 km per hour
37'4
Braking retardatir^ P:t - -1 8? km' p'h'p.s'
 32
ELECTAIC TXACTION

Distance trave[ed,r:
,h#* +hkD *fr$ rtrl
: uro#q* t'# *so+ 37#f l'03 km
---
Schedule time, T,: rr * ,r*/s*duration of stop
:30+50+20+30: 130 seconds
Schedure speed, i.?-:3,6q0i01.03:28
"-: 5 km per hour Aas
2..97 ENERCy 0UTPUT FR0M DR|V|NG
AXTES
\/ |r:lr,nt the run according to trapezoidal
speed_time curve
Total energy required
forit" ,un -- 'during
u*rrr-'r.orl."i accele.ari.,n
:f,}:ffi#X,T,t'#;f;
:+ ?k "lTi',?'.l"."JJ::".-,J,;::T"i.::ll'"[::;:;"-.,
.rr*-r'-+Ju#
where V- is the maximum speed
'#
k*h
in km. p.h.
rr is the time of acceleration
in seconds
,r is the time of lree run in seconds
Fr is the effort required during
accelerution in newtons
and Fr, is the effort required
during f.* i" ;;;;;;.
Instead of expressine thr ^"
1fl.1:".;"J:$,{:'"rl,;#'';:*;.,#Y:,-;','f iJlf :'"o"iflJ'Jl::lli"J,:X;
i"" w"igh, * ,nii."rt#inulo't in watt-hours
u,.".,lil'l"n
. . Th.
iilll'J":xlTilff.y;;;,,#
en€ rgy input to the m
ffiffiH,.,r comparing r he

;:x*ii*',r**illgii,lffi r,:iF,ff:fl ffiH*it!,J*i"[{

The energy consumption can
be expressed in watt_hours per
i.e, ,,, .,-.n.tg1 consumprion of train in **.0"".r
""''" tonne-km
Weight of rr.rin in tonneiz
Oir,"##.uo i,, trn:\
The above quantitv i, crlleJ th: sperifc
energy ,.],Ji,n
2.I 0. spEcrFic' rnriif 'oiiiff
"orrr^),,,"^
tr1egy,ifrrt,jlrEoF r,rrr.,r,ro
I€t the track iavea gradient of G p3rcent through
out its run.
Energy output to accelerate the tru'n
from rest to a speed v"l
I Fr \/- ,.
---2- -1106" 3;;*- run
 -

TRAIN Mo!'EMTNT & rNmcv coNsuMPTloN 33

I FiY? x t*tr since r, -Y!L

T 3,600 -E-
"3,600 0c""" d.

I V-2
w" r r98 l WG*Wrl kwh
[277 8
7 G6OO,11
since Fr:277 8 W" af 98'l WG*Wr
Energy output to run the traio at the speed V- against the gradient and
resistance to motion
Fr'V-,, kwh
3,600^ 3m0- a
:t#**n srnce
3-EOO -D', tle distance travelled
during free run

:[wr]-e8'l wc]x {- kwn

Total energy output for the run

:r;#oorrz77'8 w.a*98 l wG*wrl-| [wr*e8 l wc]x ffi wn

:rffiffi lz77'Bv'tca*e8'I wGlwrl+twrtes l wc]
D'x ,000
1
X --3600- wh

:Y.u.ffi1, x277 8W""+ Y*ffir [e8'r wGtwr]

*S9-twr-fe8'r wcl

wr]
:0.01072 v,,r, w,+[e8.l wG-f
[oJfu-*",],l#
:o.olo72 v-, w"+# [98.] wc+wr] tD,,+DI
where distance moved during accelerating period
2ffi:D"'
: O Ol072 V,.r W.+o'2778 (98 1 WG*Wr) Dr
whcrc D1 is the distance travelted during acccleration and fitc run in km
Specific energy output
I
En output for the run in watt-hours a
elg hto train in tonnes x D ls tance of run in km
0.01072 V-r W,+0.2778 (98 1 WGf Wr) Dr
WD
_0.01072 v"., ,H ,o rrr, -'; tr, , G lt)'/iatt-hours Per tonne-km
l)
I
 ELECTRIC TRACTION
34

If the track is level one, G:0

qlgie:t$'+o zzzt
and specific energy output: $'
watt-hours/tonne-km
Specific energy output
- Efrciency of the motors
Specific energY consumPtron

0 010?2 V'nr - W" . n.ll:* .D.1 x -l watt-hourj tonne-km

Dr-^ W'ru'Z"o D't"-"
2.II.O FACT()ES AFFECTING ENERGY C()NSUMPTI()II

Energy consumption in propelling the train is required for

t. having the linear and angular acceleration;
2. working against gravity while moving uP the gradient;
3. working against the resistance to motion and
4. supplying losses in motors and other electrical system'
Enersv reouired being the product of power and.time depends upon
rhe
poru.r t.qui-t.d anh the duration for which the Pot,rer ls taken'

(l) The power (in watt5) required in accelerJting the train on level track
,. the wiight. _of rhe train
W, and
A"o.nOi |ooo-in! i"te ot a."eteritiJn' elTective
v"'
;if;i.-"iIiih.'inJ- oi'u"".r.roti* v,,"the ueing cqual to.77 17 w€ d' where as
upon rate of acceleration' effective
iil.t-iiir-ii,i ir"it-ttort.io"p.nat io. *r,i.h the rrain accelerates, being equal weight
;?i.iil.'ilril; ,pJ.o i-t i ti"[.
to 0'0 t072 V,r We a ,. "*i
(2) The
'up power (in watts) requireri in overggmrjlq the force of gravity
-gr",ti.n,
t.ing lq,oi . to 21 lJ4 yG--Y" depends.upon the
'Wft]t" it.
wtrile go-iirc
goinE rUe'eiadient rhe energy i:' returned-bxck The energy
n."J,i.'nt. "iin down
'*utt-t...) 'i' given bv thi':xpre'sion 27 114 wGv'n' and
:;;;i;; rt ;;i;;;d 8'uo *iieut of troin' maximum speed and
i;Ht#.:L;'",iji.;;d;',il';ilil'i,
duration of run.
(3) The powlr (in wattsl required in overcoming- the resistance to motion'
o zn[ wr V^ ano ei,er-g'y riquired.- 0 2778 wr V'a' ,watt-seconds
The
i..o*f-tl inii""it" tt it it dcpends upon the resistance to motion,
l'_#"ti# r.. tlli,ey ,"qoir"o
miximum speed and time of run'
a given speed time curve is indeoendent of
The energy output of train for ""o'ii"'1'-ti"t upon tvpe or
tu tvpit"ri?i"3'";;"1'",4. s,;i .^iigl depends t-he

drive emPloYed. a
OF AN ELECTHIC
-"i
2.I2. FACT()RS AFFECTING SPECIFIC ENEBGY C()N.SUMPTI(]N
rnnn oprnattue 0N A GIvEN ScHEDULE SPEED
The soecifrc energy corsumptio;rof a train op:ratiili at a given schedule
sp:al d:p:ni; u2c1 ths followinl factor';
(a) distance betwcen stopi (h) acctler'rtion (c) r":t'Jrtio;r ( rI) mrrimum
spee,t (ri nature of route and (.f) the type of tr'tin equtvalent'
The specific energy output is given by the expression'
TRArN MoVEMENT & ENtRGy coNsuMPrIoN 35

output'001072V-'z W", 0'2778 r(98 I G+r) D, warr

Specific energy D:W -- D-
hours/tonne-km
Soecific enersy output is independent of locomotive oYerall emciency but the
soecific inersv consiinption being equal to specific energy output divided by loco-
tiroiir. ou"riit efliciency depends upon the overall efficiency of the locomotive'
Greater the overall effciency less will be the specific energy consumption for a given
specific energy outPut at axles.
The expression clea y shows that specific enetgy consumption depends upon
the maximum'sDeed V-. rhe distance travelled by the lrain while power is on, Dr,
til.ioiiin" ..Jiitu*.,- ., gradient G and distance between stops, Greater the
aistaici tetween stops- leis will be the specific energy consumption' For a given
iu" ut i niu.o schedirle speed, greater ttie value of acceleration and retardation,
more wili be the oeriod of coasting and rherefore less the period during which power
iio, ir, O, wiil'be small and, therefore, specif,c energy consumption will accor-
ai*i" u! -t*. Sieip eradieni will involvi more energy consumption even if
i"gEni ruiiu" biaking ii u-sed. Similarly more the train resistance, greater will be the
specific energy consumPtion.
How the sDecific energy consumption falls with the inclease value of in
accelerution 1or rriardation; a-nd distance of run is illustrated rn figs.2.9 and 2'10
respectively.
O

F E
s
!a
E= =
a:
s=
sl et
sr
&: Ei
I I
-..------.,> acc E LERtflq| I
I .* trl6
/rv Y.'{.P.,r.P.S. otslAflcE 0F NN

Fig 2.9 "a

Fis. 2.10
Examole 2.11, A 3c0 tonnes E M U train is
s-tarted with uniform I
."""r"riiiin'I'o'ai*"t*i " ip*t'"ieiirr. p.r,. io 25 seconds or-level section. Find
;i;;;;;tfi;;;;si consumpt'i'otr'i" u"i,tine a'iimplified..trapezoidal curve' with rotrtio-
;;i;#;il;6;ii. i"to.un[io, i r.,n p t ,' the diststrce betweetr two stotions as
pln
5 l(g tontre'
5 tm., efficienci-of molors as 0 9 nnd ' resistance
'"- train . Sec. E. Elec. Traction Mav, 197 2l
G.u.t.e
Solution : Maximum speed, V--=40 km' p h'

Distance between two stations, D - 3 km

Resistance of motion, r: 5 x 9' 81 - 49
05 Nltonne

!L - r.or
Duratiotr of braking ,":YU4:$:r:334 seconds
 35 ELECTRIC TRASIION

Distance travelled durine braking- km

| ].;#- *'q#,..#:o'0741
. D',- 3-0 0741-2 9259 km

: 0
! ! of,2Jn' o' zzza' wn4onnJt*
Specifi c energv outnut $" + $ \
0107? x (40)'*
-0 1.08+0.2778 x 49. osx2'9152

:6'11 *13'29- 19'46 whi tonne-km

13.\L:,,u wh/tonne-km Ans'
Specific energy consumPtion-

Exampte 2.12. Celculrte the specific energy consum-ption ifa m^eximum speed
of t2'20 m/s inrl for a given run of i525 metreJ'an acceleration of0'366 mls' are
desirerl. Train resista-nce durirg acceleration is 52 6 newtons/Ifi[ kg rnd duriog
coosting is 612 newaonsilo)o kg, 10?(, being sllowsble for rotatiotrcl he ia 'lhe
' Assme I
efficienEy of the eqlipment during the rcceleralion period is 50%.
ouedrilsleral soeed-time curve.
lA.M.I.E. Sec. B. Utilisation ol Elec. Power May, 19751
solu'ion : Effective weight of train w'l'l w
is the weiglt of train in kg
Acceleration a:0 366 m/s2
Maximum sPeed, V",:12 2 m/s
Distance of run. D: 1525 metres

Acceleration p"'ioa' rr:I'l: ffi : 33'3 seconds

Tractive eflort required during acceleration, Ft:Wu c*Wr

:l t WxO 366*0'0526 W
:0'4552 W newtons
Total energy required for the lun
=.Etrergy required during acceleration as there is no free run
I : Average power during acceleraiion x acceleration period
:I F1 V,,y rr:Ix0 4552 Wx l2'2x 33 3 joules
:92'56 W joules or watt'seconds
:36m-
92.s6 W :0
0257 W watt-hours

output:
Energy required for the run 0 0257 W
Specific energy WxD Wx I ,525
l'685x l0-5 Wh per kg-m
Specific energy output
Specific energy consumPtion: 1l

l'685 x l0-5 :- 3'37 Wh/kg-m

--T5 Ans.

!
I

TTAIN MOVEMENT & ENEtr,GY CONSUMPTION 31

Example 2.I3. The speed-time curve of an el€ctric tmin on a uniform rising

gradient of I in l0o comprises
(i) Uniform acceleratiotr from rest at 2 km p.h.p.s. for 30 secondc
(ii) Coastitrg with power off for 70 seconds
(iii) Braking at 3 km p.h.p.s. to stand still.
Th€ weight of the trrin is 250 tornes, the train resistatrce otr level track b€ing
5 kgi totrtre, atrd rllorYrnce for rotary iDertis l0?o.
C.lculate the maximum power developed by tractioD motors and aotal distence
treyelled by the traiD. Assume trsnsnission efficietrcy as 9790.
lAgra Unit. Truction & Utilisation of Elec. Pox'er 19791
Solutiotr : Weight of train, W:250 tonnes

Effcctive weisht of train, w": ( t* l3O) w:t l w

The traia resistance, r 5 kg/tonne:5x9'81-4905 N/tonne
Gradient. G --l Percent
Acceleration, a:2 km P'h.P.s.
Maximum velocity, V,,,:a tr:2x 30:60 kmrhour
rractive effort'"t""" a' r' :
)f;.1 I:,,1r'$ I Ifrt.YI r ro *, * rro
x 49'05
...1,89,5 77 newtons
Maximum power output form the driving axles
:Il-'k :'jffi"x 6o: 3' 16o kw

Maximum power deveroped tl{irry$;,

kw Ans.

l,et the coasting retardation be Pc

then Ft:277 8 We (-Pc)+98WG*Wr 1
or o:-2778x l l wxpc+98'l wx1.1.wx4905
or e"-tfi*y+i:o'48rs km p.h.p.s.

Vz:Vr-Pc r.:60-70x 0 4815:26'3 km;hour

Braking retardation, F:3 km p.h.p.s.
v- 26.3 tl //
Braking period, h- t:--T-: seconds

Total distance travelled by the train, S

V, tr . lvr- vr) r,I , v2 rr
:2,1m, - 7700--1- 7,200
: tilol *sr#l :r. km
x 7o +2i+,;;' 11 1 2 Ans.
 J6
ELECIR]C TRACTION

Examp-le 2I4. A train weighing 20g tontres eccelerates

uniformly from
-U"iogrest
to a
speed of 45 km o.h.o.s. up i g.iuiert_or r-i, sOri,iteiime
r"f."i ilri
ii th6n cut off -and
legol$j - The power train .rnil Ooi"o--" uoiform gradient of
I in ltXtO for a qeriod of {0 second-s when fraf"s are
"ppi"j fri Calcuhte
s€conds so as to bring lhe train uniformly to rest on iUirtinOi"or. ":.-iLia'r?fS
(it rhe
mLxipuJn power ourpu. from the driving ixtes
rails in kwh, assuming an efrciency ottOgo. '
tii) the eneig'i;.k;;'ir; .b"-:;;iii"il;
Assume tractive effort to be 44.rcwtotrs per tonne at all speeds, and allow logo
ror rorariotrar ineftia. lAgra univ. iiiiiria uiiiiiiii'itErii niii tc?ti)
Solution : Weight of train, W:203 tonnes
Effective weight of train, We: l.t W:l.txZO3:223.3 toanes
The ttain resistance, r:44 newtons/tonne

cradient, c: _rlo :o.zz

Maximum speed, V,.:45 km p.h.
Acceleration period, rr:30 seconds

Accelerarion,
":+:1* : r.5 km p.h.p.s.
Tractive eflort required, Fr:277.8 We or*98.1 WG*Wr
:277'8x223'3x 1.5+98.1x 203 x 0.2+203x44
: 1,05,964 newtons
(i) The maximum powef output from driving axles
FrxV- 1.05.964
3,60d' - ffi x +s: t:24 55 kw Ans'
Total energy required for the run:Energy requ,ed during aca€leration
as there
ls no lree run
Ft V- tr .Kwh
-E' J,6-00- x 3600-
1,05,964x 45
-t,, . _
so6f xr,lo6
30
s s2 kwh
(ii) The energy taken from conductor rails

: 'OB :9 2 kwh Ans'

ssd!i+t::il'1"Il_[,lf r':g.nf,:,ffi
is assumed to be teyet. Each 'srop I. ,r eril"""ia.
""J""#fi "*,:il,TlJi:,,ffi
ali;;i"r":"riJirg
"i,f,;,.Hi
time curve,- calculate the maxim-um simptifiert speert-
as"u.io!-ii-J to be 2 lm.
p.h.p.s. atrd retardatior ro be 3.2 km.p.h.p...
"peea i[;;;?
'iil""i""i""f"r"tioo is,tt)16 ronnes,
-iri "ligfitii"..,
rotatiotrat inertia is I0 Dercenr nf rhe
per.onre. If the overal efficiency i. 80g", ie;i
'?;ic;;L
""iiii, l'"ii"r"""" aewrons
'i;i^r#.r_
ftom drivitrg axles (b) the specific energy co$umptiotr powe] output
in watt_hours per tontre_km.
Solution : Schedule speed, Va:45 km.p.h.
Distance of run, D_2.g km.
Acceleration, or:2 km.p.h.p.s.
I

39
TT,AIN UOVEMENT & ENEIGY CONSUM9TION

Retardation' P:3'2 km'P'h'P'r'

Duration of stoP:30 scconds

schetrurc time of run, T,: DX1L600-21YP-224 seconds

T:224-30:194 seconds.
Actual time of run'
_- t I I , t .ta:0.40625
K:z;+N:'i+e.f g
D
Maximum rn".a, v*: |*-;/ (&) 600K
3

ro4 /, --1%--
:z;iimx-J(Tfr T t6mt 'E-
:1t-'4cr;25)--T-'ffi 25-
:58 km.P.b' Ao3'
v:l- 5-9 :29 **--
Acceleration time' ,r:; :Z -" .econds

Duiation of braking, ,": Y3:l-!':

I 5',2
ra l25 scconds

Timeoffreerun,r,:T-(,{fu):194-(29*18'125):146'8?5secosds
Wc a*Wr
Tractive effort during acc€leration , Fr 271'8
:277'8xtO x t't x Z+ tO x l0: 10'418'56 Newtons
(a) Maxirnum Power outPut
:tk- **: 44p"otr!! :r67'8s5 kw Ao"'

(r) Distance travelled during braking period

:1ffi:r* 56x 18 125 :0.146 km
3,600
Distance travelled with Power on
Dr:2 8-O 146:2 654 km

0'01072 v,l', p+o ,rrt r

Spercific energY outPut - ----5- # wh/tonne-km

_o oroJ2 x58rx r.l+0.2?78x40 x2;6#

- 28
:24 8 wh/tonne-km

wh/tonnc'km Ana'
Spccific encrgy consumption: H:"

ry;x##qq?Lb-ffi iHii["il',,.li"1*ffit14-#f:ii*y;
*,ll ;lit-*r;rr+1rum.*$a14;1+!ffi;i$ffi ;;
 I

40 ELECTRIC TRACTION

Solulion : Weight of trains, W:450 tonoes

Average speed, Vo:40 km. P.h,
Distance of run, D:3 km
' Acceleration, e:2 km.p.h'P's.
Retardation, 0:3 km. P.h.P.s.

o-rlo 0.5 percent

Gradient,
Tractive resistance, r:5 kg/tonne:5x 9 81 N/tonne
Effective weight of train, W.:l'l W:l'1x450:495 tonnes
Overall effciencY, (5
'2:0
Dj(!,600 _3x 3,600_ 270
Actual time of run, T: 40
seconds

ll I 5

":# N-T' u
612
V,r: ,[ - 3 D
Maximum sPeed,
K

-rvw- 270 | 2'70 1' _3,600

--rnfx3
\zxslz /
: 324-28t' 17 : 42' 83 kmi hour
Tractive eflort required during acceleration,
Ft:277 8 Ws a l' 98'1 WG I-Wr
:277'8x 495x2I98 1x 450x 0 5*450x 49 05
:3,19,167 newtons
Tractive eflort during free run, Fi:98' I wG + wr
--98 tx450x0 5 i'450x49'05
- 44,145 newtons

Acceleration nerioa, r,:Yl:$:21 42 seconds

Braking Period, 28 seconds

',:Y:9+:14 :
Free run pertod, t,-T -1t,1-
tr1:Z1O- (21'42+ l4'28',

Total energy requirtd for the run

"'"f;o'oo,
-,XI,X6b "*-*Effi"ffi-r*n
:il',{r'#ffi
2x 3,600x 3'600'--- -' t;ffiffi1
x2!'42t
x234.3

- lr' 291 + 34' r82 :45'5 kwh

Energy consumption fot the run:ffi
:?0 kwh Ans'

I
 47
Ti,AIN MOVEMENT & ENERGY CONSUMPTION
ADHESI()N WEIGHT
,.IS. O'IO WEIGHT, ACCELERATING WEIGHT AND
and train to be pulled by the
Deatl Weisht: The total weight of locomotive
locomotive is kn5wn as dead weigl't'
the weight of loco-
Acceleratin| Weisht Thedead weight - of. th: lra]L':e
-otiu.ino't in
"
LJionsidereo to be divided into two
parts'
""un as weight of wheels'
(1) The weight, which requires angular acceleration such
axles, gears etc.
and (2) The weight, which requires linea acceleration'
Hence the elTective weight, which is ']t greater than ,*"u9-- *tinht
-more
is called as
accelerating rr€,:gtl. Accelerat.; ;"tig"ht
"i;[Ln i to l0 percent than dead
*"'*t'uor"rrw
to be carried on the driving wheels is
lreight: Thetotal weight
known as the adhesie ueight.

.."u,",?:?18'i;;,';-"."n"13,-1",{;illi1;,'['::il#jf.f1"'"i]ib;]:**TJ-
ii not to
increase
Fn-;::*H: llil;i:m:H;;;;i;;-i;;;ffi.I"s ir the axri toaa
of Etec Pouer teTel
beyonrl 22 tonnes. l,;;;U#:\i;;;;'-i'"uiiliiii'
W1 tonne
Let the weight of the locomotive be
:
Solution
o.uO *.igUt of train and locomotive, W:(400tW1)
tonnes
Acceleration, d- 1 km. P.h'P's'
Track resistance, r:40 newtons/tonno
Gradient, G--2 Per cent
Coefficient of adhesion P:0 2
Effeclive weight o[ train and locomolive
We:l l (400+WL)-
t'u"tiu" tffott' Ft-2'77'8 We af 98'l WG+W/

:lzn'z$a+ra'r G+']w
:lZ77 8x 1 1x 1+98'1x 2+401 W
:541'78 W newtous
s41.78 w :0'05523 W tontres
9'81x I 0
:0.05523 (400+wL)
tonnes .
'. (i)

that can be possible with W1 adhesive

Maximum value of tractive effort
weight of locomotive ,..(i,)
Fr:F Wr:0'2 WL

Comparing expressions (i) and (ii) we get

0'2 wL:0'05523 (400 +wt)

or *,:'-&;!1-3+# : tr;k--152'6 tonnes Aus.

r
42 ELEC'TRIC TRACTION

Number of axles regrired:

O*;;gS--
:Ltli-' n*'
Example 2.18. An electric locomotive is required- to haul .e train of 12 coaches
we-ishini i0 toro"" otr the maitr litre seryice riquiring an initial acceleration of
o:g t . i.fr.p].. up a gradiert of 1 in lfi) Estimate the sdhesive w€igbt and hence
"rcn
th; ;;f";;i;tfing-axles the locomotive must hrYe .if the permissible axle loarling
i;20 t*ilp"4 aifa. Assnrning for rotatiotral iDertia to be 4o,( for the. coaches
;;rt I5;Z for'the iocomotive. Miximum coef8ci€trt of adhesion is 0 2 and the tract-
ii"" i".i.irr"" itg/tooo". lB.E.tt. Elec' Troction & Litilisation 19781
Solution : Dead weight of train, W: 12x 30:360 tonnes
Accelerating weight of train, W,:l'Mx 360:374 4 tonnes
Lct adhesive weight of locoinotive:WL
AcceleBting weight of locomotive: l'15 Wr
Acceleration a:0 8 km. 9'h.P.s.
Gradient' G:1%
Tractive resistance, r:5 kg per tonne
:5 x 9 81:49 05 Nitonne
Coefficient of a adhesion, F:0 2
Tractive effort, Ft:\277'8 d) (W,+l 15 WL)+(98 1 G*r) (W*Wr)
:(2?7'8x0'8)(374 4+ l'15 wt)+(98 l 1 49 05)(3601 Wa)rs

4 411'15 wL) + l47 ls 1-]9Ll]{].l to,lu.,

(37
-.222'24 \-9 8 t
. (13 882'10 041 wL) tonncs
--l0o-0 ..(i)
Maximum value of tractive efTort that can be possible with Wr adhesive
weight of locomotive:p Wr:0'2 Wr. ...(i,)
Equating expressions (i) and (ii) we get
0'2 Wr: l3'882+0'041 wL

or w':ffi:37'3 1oor",

Number of drivinS axles-ff:4'37=5 Ans.

Exampte 2.19. An electric locomotive of l0O aoanes can just &cceler&te a

train of 500 6Bnes (trailing weight) with
' ar scceleration of 1 km. p.h.p.s. on an up
grartietrt of 1/1000. Tractive resistrnce of the tmck is45 trewtors per tontre and lhe
iotetional inrirtia is t0ol.. If this locomotiYe is helped by atrother locomotiYe of 120
totrnes, find (i) the treilfug weight that can be haulerl up the saDe-gradieEt under he
same coaditiiirs anrl (ii) tfie miximum grsdient the trailing haul€d lord remaining
utrchrtrged.
Assume aithesive weight expressert ss percentsge of aotsl dead lYeight to b€
the same for both the loconotiYe.
lA.M.l.E. Sec. B. Utilisatian of Ele<'. Poter Suamu 19761

I
 43
TRAIN MOVEMENT & ENERGY CONSUMPTION

tonnes
Solutior Dead weight of train and locomotive, W:1001-500:600
Effective weight, We-l l W:l'Ix600:660tonncs
Accelerationcr I km.P'h'P s
I
Gradien' C dOO - O't';
Tractive resistance, r:45 newtons per tonne
Tractive effort required,
FFZ17'8 Wa a-|98'l WG*Wr
:277'8x 660x I +98' l'x 600x 0' I +600x 45
:zz ...(,)
:2,16,234 newton.:
ffiffi tonnes

Maximum tractive effort of filst locomotive

Adhesive weight
:P WLX Dead we ight
:Fx 0'8:80 P ...(i0
100 x
--
-2 of total dead weight'
Assuming adhesive weight of locomotive to be 809/0
Equating expression (i) and (ii) we get
8o tt-22
or Coefficient of adhesion, r:ft-ozlS
With two locomotive total adhesive weight
:(100+ 120) x 0'8:176 tonnes
Total tractive effort that can be developed
...(,,,)
: px176:0'275x 176 toones

Letthetrailingweightthatcanbehauted,upthesam€gradientunderthe
same conditions, be \Y tonnes.
No\" the total dead weight
W':(100+120+W) tonn s

Tractive effoft tequired,

Ft:277'9xW'e a*98'1 W' G*W'r
:(277'8x 1'l x I +98 1x 0'l +45) W',
:350'39 W' newtons
:9.gl360'39- - - w' tonnes:o 036737 w' tonnes '..(,v)
x l,om
Equating erpressions (iii) and (iv) we get
0'036?37 W':O'275x176

or w,:9###:r,320 tonnes

Trailing weight:W'- 100- 120= 1,320-220: I '100 tonnes Ars

I

(i,) Total hauled load

:500* l00i 120:720 tonnes
44 ELI CTRiC TRACTION

Tractive effort required,

r,- (ztt'tx S x.+et r c+r ) w

:Q77'Sx l'l x I *98'l G*45) 720 newtons
. (350 58+98'l G)x 720 tonnes :(v)
looox 9'8-l-
Equating expressions (tr,) and (I) we get
(350'58+98 1 G)x 720
O'275x 116
l, 0x 9 8t
or 350'58 +98 1G: 0 275 x 175
12
x 9810 :659 45

or G_659
4s__ ?50'58 :3.15% Ans.

EXERCISES
For Ans Refer Art.
1 What do you understand by speed-time curves
- ? What is its 2'l
use in practicr'! [B.T.E Rsjasthan Nov: 1972]
z Sketch a tvDical speed-lime curve for the main line service 2.5
'hbw
and show thd simplified speed-time curve can be used for
determining the maximum speed with the given values of
actual timtof run, acceleration, retardation and the distance
between stoPs.
[Agra Univ.'Traction & Utilisation of Elec. Power 1976 Supp.]
3. Draw a complete speed-time curve and discuss how different 2.2
Darameters of this curre change with the type of the train
ierr ice. [Pb. Univ. Elec. Traction July. 19741
4. Draw the speed-time curves for each of the following )')
traction systems
(r) Tram .Car (ii) Diesel'electric (iii) Trolley buses
(iv) Suburban electric train and (v) Urban serviccs.
5 Define 'crest speed' 'average spced'and 'schedulc speed'and 2.3 & 2.4
discuss tbe faciors which affect lhe schedule speed of a train
[B.H.U. Elec. Traction & Utilisatiotr 1978]
6 Explain the meaning o[ the schedule speed of a train. What 2.3 & 2.4
are the factors aflecting schedule spced ?
[A rhabsd UDiv. Utilisation of Elei. Energy & Traction 1975]
7 Explain ctearty "free running", "coasting" and "braking" 2.2
with relerence to electrtc traction systems.
[A.M.I.E. S€c. B. Ljtilisation of Elec. Power Dec. I974]
tRAtN MoVEMENT & ENERGY @NSt MPTIoN 45

Draw the speed-time curve for an electric train- accelerating 2.5

8
uniformly to a speed V,,, the power is cut off and.after a
iim" r *6"n it attains v', coasti'ng is allowed to continue up
to the end o[ the run, the braking being neglected' Derive a
suitable eouation to determine V",.
tA.li4.I.E. Sec. B. Utilisatior of Elec. Pofier Mly 19741
9 Exolain how an actual swed'time curve for an electric train
,(
*i'"-i* *" be replaced by a curve having simple. geometric
shaoe deduce. from first irinciples, relationshiP between
'maxiirum the speed, running time and
ac,ceteration, ietardation,
distatrce between stops 'S. assuming
a simplified lpeed--Igl:
curve. [A.M.I.E. Sec, Utilisgtiotr of Elec. Power I)ec' 19751
10. Prove that the crest sDeed of main line servicc with trapezoi- 2.5
dal speed-time curre tan be obrained by the formula
Ur,-Lr*-J ZRt Tr 3,600 s
K

where
ll
K- u- n
a:acceleration in km. p.h.p.s-
9:retardation in km. p.h.P.s.
Va:crest speed in km. P.h.
T: Total time in seconds
S:Total distance of run in km.
[Pb. Uriv Elec. Trgction July 1973, 197{]
11. Derive an exoression for the tractive effort transferred to the 2.6
driving wheeis in terms of torque developed by the motor,
gear ritio, diameter of wheel and efficiency of transmission'
12. Derive an expression for the tractive effort. 2.1

13. Explain the terms "dead weight" "effective weight" and 2.6 &
2.t3
- " in
"adhesive weight a locomotive.
IB.H.U. Elec. Traction & Utilisation 19781
t4 Explain the terms "specific €nergy output" and "specific 2. r0
energy consumption".
15 For I typical trapezoidal speed-time curve, show how you 2.10
would ciiculate ihe energy i:onsumed during two consecutive
stoDs. use either eraohical or semi-graphical method'
' IA.M.LE. Sic. '8. Utilisation of Etec. Power Dec. 19741
l6 Describe the ptocedure of catculating the specific energy 2.10
consumDtion of an electric train.
i,lii"U"tira Univ. Utilisgtion of Elec Energy & Traction 19751
17. Derive an expression for specifrc energy consumption o[ a 2.t0
suburban traiir for a typical inter slation run.
1n.H.U. Uec. Traction & Utilisation 19781
t8. Explain the factors affecting- thespecific energy consumprion' 2.12
[Pb: Uoiv Elec' Trection 1973]

I
 --

4t EI,ECTRIC TRACTION

19. Discuss the factors that govern the total weight of electric z.l3
locomotive. Differentiate tetween dead weight, accelcrating
weight and adhe'i\e weight.
20. Explain the terms "adhesive weight" and "coellicient of 2.6 &
2.13
'
adhesion.
t.lgi. Uoi". & Utilisation oI Elec Power 1976'
Traction
fsZb;i.Ml.E. B.
Sec.Utilisation of Elec. Porrer strmmer 19761
2.6
-21. Exolain theUn-iv.
sisnficance of the coefficicnt of ldhesion'
Utilisation of Elec Energy & Traction 19?51
Iefi"t.UnA
2.6
22. Comment on the following statements.
tit The adhesion coefficiettt of a locomotive employing d c
ihunt .orot will be higher thun lhe using d c :!rie' m!-tor'
"^'-'il.iil.i.g
Sec. B. Uillisation of Elec Power winter 19761
(ii\ A hisher ratio of adhesive weight to total weight-is
i"l""Uf.i6i irUurUan rritins, where the distance between the
is verY small'
'"'i.i.ni.I.n.stoPs
consecutive
S[". B. Uiilisatlon of Elec. Power'summer 1977]
23 Explain the following :
(i) What do you unders'and by speed-time curves ? 2l
(il) what difference exists in speed+ime curves in case of a 22
suburban and urban services ? [Gorakhpur Univ' 1975]
liiif w["t ii ,ne"nr by schedule speed o[ a train ?
[GorakhPur UtriY 1975]
2.4
(iv) What are those factors rryhich affect the schedule speed
of a train ?
(v) Does variation in duation of stop affect the average 2.4
speed ? If not why ?
2.4
(vi) How does increase in acceleration and braking.retarda-
;;; ;fi& itre schedule speed of a train for a given run
with fixed crest speed ?
(vii) How does
-f- increase in crest speed affect the -schedule 2.4
rp"la iir"" iun with fixed' acceleration and braking
retardation"?
in 2.4
(viii) How does increase duration of stop affect the schedule
speed ?
How acceleration o[ tbe revolving Parts o[a train is 2.7
(ix)
allowed for in rhe acceleration for train movement I
2.12
(x) What are those factors which affect the specific energy
consumPtion of a ffain ?
2.12
(xi) How is soecific energy consumption affected by variation
in dislanie between stops ?
2.r2
(xiD How does the value of acceleration and retardation afiect
it soe.ific energy consumptlon for a given run at a glven
"
schedule sPeed ?

Giil) What factors goYern the total weight of electric locomo- 2.13
tive ?

I
 r-

TRAIN MOVEMENT & ENERGY CONSUMPTION 47

24. Fill up the blanks :

(a) Variation in acceleration and retardation will have- 2.4
effect on schedule speed in case of shorter distanc€ run
in comparison to longer distance run,
(r) The area covered by the speed-time curve, the time-axis 2.1
and the orOinare through ihe instant between which the
time is taken represents the--in the corresponding
time.
(.) The effect of variation in crest speed on schedule speed 2.4
is considerable in case of---distance run.
(d) The effect of vanation in duration of stop on schedule 2.4
1s---;n iase of shorter distance run as compared to
longer distance run.
(e) The soeedtime curve of ln urban service can be replaced 2.5
Ui- rn' .quiuut.nt speed time curve of simple --shape'
(,f ) The speed-time curve ofa main line service is best and 2.5
most easily replaced bY a---.
(g) The coefficient of adhesion in electric traction is--- 2.6
than that in steam traction.
(h) Steep gradient will involve--energy consumption' 2.12
(,) More the train resistance,---will be the specific energy 2.12
consumption.
(i) Effective weight is--than dead weiglrt of the train. 2.13

PROBLEMS

l.A trairI runs lt an averlge spccJlol 50 km.p.h. betwecn stiltions situated

2 5 km aDart. The train ilccclerltes ai 2 km.p.h.p.s. ond retards at 3 km.p.h p.s.
Oiaw ttrd speed-time curve for the ruq and find its maximum speed assuming- a
irafezoidal ipeed-time curve. [Pb. Uriv. Elec.
]1TtHf.11]
2. An electric train is accelerated at 1'5 km.p.h.p.s.
-km and is braked at
3'0 km.p.h.p.s. The train has an average speed of 45 p.h. on a level track of
t.500 metrls between stalions. Determine (i) actual time of run (ii) maximum
soeed (iii) dislance travelled before applying brikes ahd (iv) sehedule speed Assume
time for itop as l5 seconds and run according to trapezoidal speed'time curve'
lAns. (i) 120 s (ii) 60 [m.p.h.-(iii) 1 3333 km (iv) 40 km.p.h.]
3. A 50 tonne motor coach train has 6 motors each delivering a shaft
toroue of 2,000 N-m during acceleration from rest. Determine the acceleration and
tim6 taken for the train to attain a speed of 40 km.p.h. starting from rest on a
sradient of 2l Dercent (Gear ratio 4 : I , gear efficiency 90o/o, wheel diameter 100 cm.
irain resistance 50 N per tonne ; rotationil inertil l0ro). Derive the formulae you
may emptoy. tAns. 4'7i km.p.h.p.s., 8'4 sl
4. A 250 tonne t ain with l0ll rotational inerti4 effect is started with
uniform acceleration and reaches a speed of 50 km.p.h. in 25 seconds on level
road. Find the specific enerpy ccnsirnpticn if the journey is to be made according
48 LEcrRIc TRACTIoN

p h p s"
to a simplified trapezoidal speed-time curve' the acceleration is 2 km
hrakins retardation 3 km p. h. p.s:";;;'i ftt?iiu'"t- utt*een the two 'tations is 2 4
[ri. lifi.i"n.v of rotors is b'9. track resistance 5 kg ronne -[Ans' 28'3 wh/tonne'km]
Et;;:i;;;ii;;p'ri'ii'iii
.

tPb. u,t".
5. The distance between two stops on the level 50 on a suburban electric
km per hour. The
railwav is 2 km and the mean,p'J"d"iroi"i".ii"',t"fp ir p s lf the tractive resls-
ffiil?c;;;G oi-i z tm.p.t.p"'i- undiituiaiut 4^kd h
rhe energv
ffiii"il" i'ilil i*^iui,o itt"
uet*tto
t*io'
t*o-
*tGt"
stops'
300 r6nn6s'
- Take average -calculate
efficiency as 70o/o and
consumption for the run- tntillj""i,
iiri,i. ii"pir"ia"l speed-time .ui*. eitl*un"" for rotational
,o o*nt

An electric train weighing 300 tonnes is-"tntiiaccelerated from rest up a gra-

6. rt t of 64 km/hour is
dient I in 200 at a unilorm t^S 'l''o:a [rn'p coitrt for one.minute
p
l*tl-::.d along level track
Power is lhen cut on u'nO ittt irai'i
^ttaircd.
The brakes are now applied iitiigut'iil" i*inio.tett .*ittr^" retardation of l 2 km
';til;;'
'"o i;;8;f
o.h.o.s. Find (i) the schedute ""[nt"uip';; -;ll;* .for
a starion stop of 20
'llowing
I d'o'o for rotational inertia
ieco'nds (ii) the specihc .nt'gv -
tut Jtt" train resistance as 5 kp ner tonne.
"oa- In;: iii i"rTt/noo' (ii) 46 s wh per tonne per km l
with
fotational inertia effect is startedtrack'
7. A 200 tonne train with 8o'o".p?.;;i'4t k .ph.in 25soo level
rmiform acceleration und ,.o.nlr'"u consumption' if
Assuming a simplified -trapez"'iii
i"J* H"a the specific energv
'uti*!in -6-km
hrakine retardatron rs r Km p ;:; t::';iilil
it"pt 2 ' efficiencv of
motors"as 0 9 and track re\i\lance "li.'*Ii;:H:. : May 19711
rnr'rr'^ iJ "--' B : Electric Tractiotr
22.3 wh, tonne-kml
[Ans.
km apart ;tt I ::Ltd'lt speed of 50
8. A trxin ru'rs hetween stations Jtiit itil" accelerates at 2 4 km p h p s'
km.o.h. wirh station stop\ "r
'zb ';;;;1;
|li ",.i'i,ii ir r r.. o n, e:,-.,lii'"t5xi. :l,t',X n:''$:::flt :I?*fil:iffi i:i
#[?H"*t:';3'i""lli'i"'.'il'31";"""rd;';'H'ril;;l;;;".$t?llJfx-ifii
time curve.
tg'1113'o an average speed
9. An electric train weighing 4o0-tonnes Fot

r:*'"x,*-t."',*l'rmid;:{{,,:lt:r'#f mJ'",f "i."i}HtL*,1'"'"T,'r#

,**tJ,Hil3.lfr:,Fft
;ti:"3j:;fl ;Tl:ffiqi#'n,ilJ'il.;h""J:"ff ,y;l:
? Calculate the
What is meant by 'adbesive weigtrt' ot^a
locomolive
t0.
ff '$',t'ri!$filliiil"lil*,,r,U*1T:;fi**HT,,,if#"*::*
LUTo'
and allowance for rotatronat lnef[la [Ans. 120 tonnes[

tonnes (includingloco) on Jiilioo uno adelerales it at the rate ot

ll.Al500vd,c'locomotiveisrequired^tostaftup,a-^trainweighing1000-
"1'Jl'"ili"i
[i#r,d'.,n*fr *t1fr s:';:,*; q55iltg,';ffi#