# Hauling Power of a Locomotive

The hauling power of a locomotive depends upon the weight exerted on the driving wheels and the friction between the driving wheel and the rail. The coefficient of friction depends upon the speed of the locomotive and the condition of the rail surface. The higher the speed of the locomotive, the lower the coefficient of friction, which is about 0.1 for high speeds and 0.2 for low speeds. The condition of the rail surface, whether wet or dry, smooth or rough, etc., also plays an important role in deciding the value of the coefficient of function. If the surface is very smooth, the coefficient of friction will be very low.

Hauling power = number of pairs of driving wheels x weight exerted on the driving wheels x coefficient of friction Thus, for a locomotive with three pairs of driving wheels, an axle load of 20 t, and a coefficient of friction equal to 0.2, the hauling power will be equal to 3 x 20 x 0.2 t, i.e., 12 t.

Example 25.2 Calculate the maximum permissible load that a BG locomotive with three pairs of driving wheels bearing an axle load of 22 t each can pull on a straight level track at a speed of 80 km/h. Also calculate the reduction in speed if the train has to run on a rising gradient of 1 in 200. What would be the further reduction in speed if the train has to negotiate a 4° curve on the rising gradient? Assume the coefficient of friction to be 0.2.

Solution

(a) Hauling power of the locomotive = number of pairs of driving wheels x wt exerted on each pair x coefficient of friction = 3 x 22 x 0.2 = 13.2 t

(b) The total resistance negotiated by the train on a straight level track at a speed of 80 km/h:

R = Resistance due to friction + resistance due to wave action and track irregularities + resistance due to wind = 0.0016W + 0.00008 WV + 0.0000006 WV2 Substituting the value of V = 80 km/h R = 0.01184W

Assuming total resistance = hauling power,

W x 0.01184 = 13.2 t or

W = 13 2 = 1114.861 Approx. 1115 t

0.01184

(c) On a gradient of1 in 200, there will be an additional resistance due to gradient equal to W x % of slope. Since hauling power = total resistance,

2 0 5

13.2 = 0.0016W + 0.00008 WV + 0.0000006 WV2 + W -

100

= W (0.0016 + 0.00008V + 0.0000006V2 + 0.005)

Since W = 1114.8 t,

13.2 = 1114.8 (0.0016 + 0.000008 V + 0.0000006 V2)

On solving the equation further,

V = 48.13 km/h

Reduction in speed = 80 - 48.13 = 31.87 km/h = 32 km/h

(d) On a curve of 4° on a rising gradient of 1 in 200, curve resistance will be equal to

R = 0.0004 x degree of curve x wt = 0.0004 x 4 x W = 0.0016W Hauling power of locomotive = total resistance. Therefore,

13.2 = 0.0016W + 0.00008 WV + 0.0000006 WV2 + 0.005 W + 0.0016W By substituting the value of W = 1114.8 t in the equation and solving further,

V = 43.68 km/h

Further reduction in speed = 48.13 - 43.68 = 4.45 km/h. Therefore, Maximum permissible train load = 1115 t Reduction in speed due to rising gradient = 31.87 t Further reduction in speed due to curvature = 4.45 km/h

Example 25.3 Compute the steepest gradient that a train of 20 wagons and a locomotive can negotiate given the following data: weight of each wagon = 20 t, weight of locomotive = 150 t, tractive effort of locomotive = 15 t, rolling resistance of locomotive = 3 kg/t, rolling resistance of wagon = 2.5 kg/t, speed of the train = 60 km/h.

Solution

(a) Rolling resistance due to wagons = rolling resistance of wagon x weight of wagon x number of wagons

= 2.5 x 20 x 20 = 1000 kg = 1 t

(b) Rolling resistance due to locomotive

= rolling resistance of locomotive x wt of locomotive = 3 x 150 = 450 kg = 0.45 t

(c) Total rolling resistance = rolling resistance due to wagons + rolling resistance due to locomotive = 1.00 + 0.45 t = 1.45 t

(d) Total weight of train = weight of all wagons + wt of locomotive

= 20 x 20 + 150 = 550 t

(e) Total train resistance = rolling resistance + resistance dependent on speed + resistance due to wind + resistance due to gradient

= 1.45 + 0.00008 WV + 0.0000006 WV2 + W/g = 1.45 + 0.00 0 08 x 5 5 0 x 60 + 0.00 0 0 0 06 x 55 0 x 602 + (5 5 0/g)

= 1.45 + 2.64 + 1.19 + (550/g) = 5.28 + (550/g) where g is the gradient.

(f) Tractive effort of locomotive = Total train resistance

15 = 5.28 + (550/g) or

g = 56.5 = 1/56 = 1 in 56

Therefore, the steepest gradient that the train will be able to negotiate is 1 in 56.

Example 25.4 Calculate the maximum permissible train load that can be pulled by a locomotive with four pairs of driving wheels with an axle load of 28.42 t each on a BG track with a ruling gradient of 1 in 200 and a maximum curvature of 3°, travelling at a speed of 48.3 km/h. Take the coefficient of friction to be 0.2.

Solution

(a) Hauling capacity of locomotive

= no. of pairs of driving wheels x axle load * coefficient of friction = 4 x 28.42 x 0.2 = 22.736 t

(b) Total resistance of train = resistance due to friction + resistance due to speed + resistance due to wind + resistance due to gradient + resistance due to curve

= 0.0016W + 0.00008WV + 0.0000006 WV2 + W(1/g) + 0.0004 WD = 0.0016W + 0.00008 Wx 48.3 + 0.0000006W x (48.3)2 + W x (1/200) x 0.0004 x W x 3

(c) Hauling capacity = total resistance

22.73 = 0.01306W or

W= 1740t

Therefore, the maximum weight of the train is 1740 t.

#### Summary

There are various types of resistances or forces that oppose the movement of a train on a track. These resistances may be due to the atmosphere, track condition, gradient, curvature, or any other factor. The tractive effort of a locomotive should be sufficient to overcome these resistances so that the desired speed can be maintained. Steam locomotives have gradually been replaced with diesel and electric locomotives on Indian Railways. There are numerous advantages of electric traction over steam and diesel traction.

#### Review Questions

1. Differentiate between the hauling capacity and the tractive effort of a locomotive.

2. (a) A BG locomotive has three pairs of driving wheels with an axle load of 20 t. If this locomotive is running at a speed of 120 km/h, what is the train weight in t that the locomotive can pull on a straight level track?

(b) What is the train weight that the same locomotive will be able to haul on a 2° curve and a 1 in 100 gradient?

3. (a) List and explain the various resistances that a locomotive in motion has to overcome.

(b) Determine the maximum permissible train load that a locomotive with four pairs of driving wheels of a 22.86 t axle load each can pull on a level broad gauge track at a speed of 90 km/h. Also determine the reduced speed of the train if it has to ascend a gradient of 1 in 200 with the same train load.

(Assume the hauling capacity of the locomotive to be one-sixth of the load on the driving wheels).

4. What are the requirements of a locomotive? Briefly describe the merits of the different types of traction commonly used in India.

5. A train with 20 wagons, each weighing 18 t, is supposed to run at a speed of 50 km/h. The tractive effort of a 2-8-2 locomotive with a 22.5 t load on each driving axle is 15 t. The weight of the locomotive is 120 t. The rolling resistance of the wagons and locomotive are 2.5 kg/t and 3.5 kg/t, respectively. The resistance, which depends upon the speed, is computed to be 2.65 t. Find out the steepest gradient for these conditions.

6. Discuss how the hauling capacity of a locomotive is worked out. Compute the steepest gradient that a train of 20 wagons and a locomotive can traverse. Use the following data: weight of each wagon = 20 t, weight of locomotive (with tender) = 150 t, tractive effort of locomotive = 15 t, rolling resistance of wagons = 2.5 kg/t, speed of the train = 60 km/h.

7. What will be the gradient for a BG track when the gradient resistance together with curve resistance due to a 3° curve is equal to the resistance due to ruling gradient of1 in 200? What would be the resistance when an 8° curve is provided on an MG line and a train with a total weight of 914.85 t is passing over it.

8. What resistances does a locomotive have to overcome for hauling a train in hilly terrains? A goods train with 80 wagons weighing 30 t each is to run at a speed of 50 km/h, while ascending a 0.25% gradient with 2° curves. The train is hauled by a 2-8-4 locomotive with 18.5 t load on each driving axle. Find out whether the locomotive will be able to haul the load at the desired speed. Assume the coefficient of rail-wheel friction to be 0.2.

9. Name the different train resistances that a locomotive has to overcome in hauling a train under adverse circumstances. Explain the factors that would affect speed-dependent resistances. What do you understand by gradient compensation on curved alignment?

10. Compare the various characteristics of steam, diesel and electric traction.