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When a train is going down a gradient of 2 in 100 against a path (track) resistance of 40 N/Tonne, the available traction effort is 57600 N. If the speed is to be maintained at 40 km/hr, how much electric power will be available to reinvest the source by regenerative braking?
1. 640 kW
2. 1280 kW
3. 320 kW
4. 2560 kW

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Correct Answer - Option 1 : 640 kW

Concept:

Tractive Effort (Ft) for Propulsion of Train:

The tractive effort is a force developed by the traction unit at the rim of driving wheels for moving the unit itself and its train.

The tractive effort required to train propulsion on a level track is given as,

Ft = Fa + Fr

If gradient is involved,

Ft = Fa + Fr + Fg (For ascending gradient)

Ft = Fa + Fr - Fg (For descending gradient)

Where,
Fa is the force required to giving linear acceleration
Fg is the force required to overcome the effect of gravity
Fr is the force required to overcome the resistance of train motion

1. Value of Fa: If M is the stationary mass of the train and a is linear acceleration.

Fa = Ma

Apart from the stationary mass train has a rotating part like the wheel, axles, motor-generator set, etc and it is about (8% to 15 %) of stationary mass.

2. Value of Fg:

Fg = W sin θ = Mg sin θ

3. Value of Fr: It consists of mechanical resistance and wind resistance.

Power output (P0) from Driving Axles:

If Ft is tractive effort and v is the train velocity, then

Po = Ft × v Watt

Calculation:

Given,

Ft = 57600 N

\(v=40 km/hr=40× \frac{5}{18}=11.11\ m/sec\)

From the above concept,

P0 = Ft × v = 57600 × 11.11 ≈ 640 × 103 W = 640 kW

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