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.

**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
× weight exerted on the driving wheels × 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 × 20 × 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 o 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 × wt exerted on each pair
× coefficient of friction = 3 × 22 × 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

= 0.0016*W* + 0.00008*WV* + 0.0000006*WV*^{2}
Substituting the value of *V* = 80 km/h

*R *= 0.01184*W*

Assuming total resistance = hauling power, *W *×
0.01184 = 13.2 t

or

*W *=* *13.2* */ 0.01184 =* *1114.86 t* *Approx. 1115 t

On a gradient of 1 in 200, there will be an additional
resistance due to gradient equal to *W* × % of slope. Since hauling power
= total resistance,

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 o on a rising gradient of 1 in 200, curve resistance will be equal to

*R *= 0.0004 × degree of curve × wt*
*

= 0.0004
× 4 × *W* = 0.0016*W*

Hauling
power of locomotive = total resistance. Therefore,

13.2 = 0.0016*W* + 0.00008*WV* +
0.0000006*WV*^{2} + 0.005*W* + 0.0016*W* 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 × weight of wagon ×
number of wagons

= 2.5 × 20
× 20 = 1000 kg = 1 t

(b) Rolling
resistance due to locomotive

= rolling
resistance of locomotive × wt of locomotive

= 3 × 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 × 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*WV*^{2} + *W*/*g*

= 1.45 + 0.00008 × 550 × 60 + 0.0000006 × 550 × 60 ^{2}
+ (550/*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 o ,
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 × axle load × coefficient of friction

= 4 × 28.42
× 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

= 0.0016*W*
+ 0.00008*WV* + 0.0000006*WV*^{2} + *W*(1/*g*) +
0.0004*WD*

=0.0016*W*
+ 0.00008*W* × 48.3 + 0.0000006 *W* × (48.3) ^{2} + *W*
×

(1/200) ×
0.0004 × *W* × 3

(c) Hauling capacity = total resistance 22.73 =
0.01306*W*

or

*W *= 1740 t

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

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Civil : Railway Airport Harbour Engineering : Railway Engineering : Train Resistance and Tractive Power : Hauling Power of a Locomotive |

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