# Traction Mechanics – Numerical Problems

on May 12, 2013

Q 1: A train weighing 5000T   starts by a locomotive weighing 125T. The starting resistance of locomotive and train are 6 & 4kg/T respectively. What is the total train resistance at the start?

5000*4+125*6=20750Kg

Ans:       =20.75T

Q 2 : A train weighting 1500T is moving on a flat and straight territory at a speed of 100kmph. What is its train resistance?  The proportionality constant for the journal, flange and wind resistance are 1.43, 0.0054 and 0.000253 respectively. Locomotive weighs 125T and resistance is 1.5kg/T

Train resistance =[(1.43+.0054*100+.000253*100^100)*1500]+1.5*125/1000

(1.43+.54+2.53)1500 + 0.1875 = 6.94T

Q 3:  A train and locomotive weighing 4135T running up a 1/100 grade. What is grade resistance experience by the train?

Ans:       4135*(1/100)=41.35T

Q 4: A train weighing 3000T is moving on a flat territory and on a curve of 20 whose length is more than the length of the train. What is the maximum curve resistance of the train? Calculate the equivalent grade for this resistance.

Curve Resistance: (0.4*2*3000)/1000=2.4T;

Curve Resistance/Weight of the Train;

2.4/3000 = 1/1250

Q 4:  Express 1 in 100 grade having 20 curve as equivalent grade.

20 Curve is equivalent 1/1250 grade as calculated in q no. 3.

Ans:1 in 92.59

Q 5:  A train is going down the gradient of 1/150. Calculate the acceleration?

Acceleration will be provided by gravitational effect=WgG

W*fa=WgG  or fa=gG = 9.81/150 =0.065m/sec2

Q 6:  A BOX N loaded train weighing 5200T is hauled by electric locomotive weighing 125T  on a gradient of 1/150. Calculate the train resistance when full train is passing on a 20 curve at a speed of 40kmph. Take the proportionality constant for the journal, flange and wind resistance as 0.64388, 0.010472, and 0.00007323 respectively and loco resistance as 1.4kg/T.

Train Resistance: (0.64388+0.010472*40+0.00007323*40*40)=1.179928 Kg/T

1.350555*5200/1000=6.135T

Loco Resistance: 1.4*125/1000=0.175T

Grade resistance: 5200/150= 34.67T

Curve Resistance: 0.4*2*5200/1000=4.16T

Total: 6.135+0.175+34.67+4.16=45.14T

Q 7: In question 6, what is the minimum adhesion required for the train not to slip while running.

Minimum % Adhesion= TE*100/W

(45.14/125)*100=36.11%

Q 8:  In question 6, the train comes to stop at this critical location, calculate the starting tractive effort required to start the train. Assume starting resistance for the train and locomotive as 4 and 6 Kg/T respectively.

Train Resistance: (5200*4+125*6)/1000=21.55T

Curve Resistance=4.16T

Total Resistance = 60.38T

It is important to note that the train resistance of the running  train was only 6.13T which has increased to 21.55T if it comes to stop and required to start again.

Q 9 : A 5000T train is hauled by a locomotive weighing 125T on a 1/150 grade of 5km of the straight stretch. Due to inclement weather, adhesion has dropped to 15%. Assume uniform train and locomotive resistance of 1.25kg/T, calculate the minimum velocity head required for the train to negotiate the grade with an exit speed of 15kmph. Neglect the length of the train for the purpose of calculation.

Train Resistance=(5125*1.25)/1000=6.41T

Grade Resistance = 5000/150=33.34T

Total resistance = 39.75T

Permissible TE without slipping = 125*15/100=18.75T

Velocity Head should provide compensation for train resistance of 39.75-18.75=21T

Now velocity head shall work for a train resistance of 21T=21000*9.8Newton=205.8KNewton for a distance of 5000 Meter

Let the velocity head is V and it loses to 15kmph for compensation of train resistance of 21T

½*W*(V1*V1-V2*V2)=(Tr-Ta)*g*D (Where V1 and V2 are entry and exit speed; W is weight of train; Tr and Ta are tractive effort required and available; D is the distance of grade)

½*5000*1000(V*V-15*15)(10/36)(10/36)=205.8*1000*5000

192.9(V*V-225)=205.8*1000*5

V=74.6kmph

Q 10 A 4875T train hauled by a locomotive weighing 125T running at a speed of 60kmph attack a continuous gradient of 1/150 for a distance of 5km of which 1km distance on a curve of 20. The train is allowed to roll on no traction. Calculate the approximate distance to it will travel on the gradient. Take the proportionality constant for the journal, flange and wind resistance as 0.64388, 0.010472, and 0.00007323 respectively for a total weight of the loco and train.

Let the train travels a distance of D km from the bottom of the gradient to the top, and the curve of 1km is in this stretch. The KE of the train will be lost in balancing the train resistance, then.

1/2WV*V=Tr*g*D+Tg*g*D+Tc*g*1000

Tr= a + (1/2)*bV + (1/3)*c*V2

0.64388+1/2*0.010472*60+1/3*.00007323*60*60 = 1.04kg/T= 5.2T

Tg=5000/150 =33.34T

Tc=0.4*2*5000=4T

½*5000*1000*60*60*5/18*5/18=(5.2+33.34)*1000*9.81*D*1000+4*1000*1000*9.81

½*5000*60*60*5/18*5/18=39.54* 9.81*D*1000+4*1000*9.81

D=1.732km

Q 11: A train weighing 5000T hauled by a locomotive running at 60kmph on a down gradient of 1/150 is required to be braked. Calculate the braking force required to stop the train within a distance of 1000m when there is no curve during the braking distance. Calculate the rate of de-acceleration and time taken to stop the train. In case the train is not required to be controlled and speed not to exceed 60kmph, how much energy can be regenerated per km. assuming no limit of regeneration up-to nil speed. In another case, say the making braking effort can be 20T and uniform up-to a speed of 10kmph

KE = ½*5000*1000*60*60*5/18*5/18=694*106N-m

Train Resistance from previous question; 6.2*1000*9.81*1000=60.8*106 N-m

Total : 694*106+327*106– 60.8*106 =960.2*106 N-m

Tb= (960.2*106 N-m)/ (1000*9.81) =97.89T

Tb=W*fa/g or fa= Tb*g/W = 97.89*9.8/5000= 0.19m/sec2

t=2*D/ (V1+V2) =2*1000/ (60*5/18*5/18) =120sec

Regeneration shall neutralize the net effect of grade and train i.e. regeneration of  (327-60.8) 106 = 266.2*106N-m of energy

266.2*106N -m=266.2*106Watt-sec

266.2*1000*1000/1000*3600=74kwh or units of energy

Now suppose the maximum braking effort at 60 kmph to 10 kmph is 20T, and distance travelled for the speed to reduce from 60 to 10 kmph is 900m, then the possible energy regeneration will be

20000*9.81*900 =49.05kwh

Q 12: Calculate the rail horse power required to run the train of question no. 6. What will be the balance speed if the train is hauled by the locomotive of 6000HP which can deliver maximum TE of 50T.  What will be the efficiency of locomotive if  the OHE current drawn is 275A at pf of 0.9.

Rail Horse Power (HP)=Tr (kgf)*V (kmph)/270

=46.03*1000*40/270 = 6819 HP

The train will not be able to achieve speed of 40 kmph due to HP limitation of 6000HP at rail level.

Balancing speed =6000*270/ 46.03*1000 = 35kmph

Iohe =  HP*735.5/OHE Voltage * pf * Effy. of loco

Effy of Loco = HP*735.5/OHE Voltage * pf * Iohe

= [6000*735.5/25000*0.8*275]*100         =80%

Q 13: A 2000T train is running at a speed of 100kmph with a tractive effort of 12T. Calculate the OHE current at 25kV, assuming an efficiency of locomotive as 80% and off of 0.9.

Rail Horse power= 12*1000*100/270=4444HP

OHE current = 4444*735.5/ (25000*0.8*0.9) = 181.6A

Q 14: What is the approximate OHE current at 25kV when the locomotive is drawing 900A current at 4th notch? Assuming the maximum TM voltage of 750V with equal voltage at different notch. Assume locomotive efficiency of 0.8, pf of 0.9. Auxiliary load is 135kW with an efficiency of 0.9 and of of 0.9.

Ohe current for traction= [{(4/32)*750*900}/0.8*0.9]/25000 = 4.68A

Ohe current for Auxiliary load= 135*1000/0.9*0.9*25000 = 6.66A

Total Current = 11.34A

Q 15: An EMU train weighing 1000T accelerate to a speed of 90kmph at uniform acceleration of 0.50m/sec2, runs at a constant speed of 90 kmph for 60 seconds, coasts upto speed of 72kmph. From here brakes are applied and train stops within 20 seconds. Draw the trapezoidal curve of the train run and the energy consumed during the run. Calculate the energy regenerated if the same is effective upto a speed of 15kmph. Assume average train resistance of 5T during acceleration and 10T during constant speed and coasting; negligible during braking.

Ans: Convert all parameters into mks system i.e. 90, 72 and 18 kmph are 25, 20 and 5 m/sec.

BC(time): Train resistance=10T, W=1000T; De-acceleration Resistance= W.fr/g;

Fr= 10*9.81/1000=.098m/sec2

Energy Consumed during OA= W*fa*D+Travg*D*9.8

[{(1000*1000*0.5*1250)}+5*1000*1250*9.8]/1000*3600=189.23

Energy Consumed during AB= 10*1000*1500*9.8/1000*3600= 40.8

Energy Regenerated during CD= 1000*1000*1*[{20*20-5*5}/2*1]/1000*3600 = 52.1

Total Energy consumed: 189.23+40.8-52.1=177.9Kwh

GTKM = 1000*(1.250+1.500+1.148+0.200)=4.098(1000GTKM)

SEC = 177.9/4.098 = 43.4

 Parameter OA AB BC CD Distance 25*50=1250m 25*60=1500m (25+20)51/2*=1148m 20*20/2=200m Time 25/0.50=50secs 60secs (25-20)/.098=51sec 20secs Acceleration 0.50m/sec2 0 .098m/sec2 20/20=1m/sec2 Speed 25m/sec 25m/sec 25m/sec to 20m/sec 20m/sec to stop Tr avg 5T 10T 10T Nil Energy W*fa*D+Tr*D*g Tr*D Nil Tb*D20to5m/sec

### There Are 4 Comments

1. umar says:

Zabrdas

2. ganthimathinathan says:

Is there any workout for express train worked by AC Loco with 24 coaches in a gradient of 1 in 60 for a distance of 5000 meter with a exit speed of 15 kmph to calculate the entry speed in inclement weather

• Mahesh Kumar Jain says:

Question No. 8 is on similar lines and velocity head requirement has been worked you. For the parameters stated by you, it can be calculated without any issue. Still, you have the problem, please address the quarry again.

3. pawan says:

A 400 tonne train travels down a gradient 1 in 70 for 100 sec during which
period its speed is reduced from 80 kmph to 40 kmph by regenerative braking.
Find the energy returned to the line, if the tractive resistance is 5 Kg/tonnes,
allowance for rotational inertia of 5% and the overall efficiency of motors is 75%.

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