Question

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

Answer 1

Correct Answer - Option 1 : 640 kW

Concept:

Tractive Effort (F_t) 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,

F_t = F_a + F_r

If gradient is involved,

F_t = F_a + F_r + F_g (For ascending gradient)

F_t = F_a + F_r - F_g (For descending gradient)

Where,
F_a is the force required to giving linear acceleration
F_g is the force required to overcome the effect of gravity
F_r is the force required to overcome the resistance of train motion

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

F_a = 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 F_g:

F_g = W sin θ = Mg sin θ

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

Power output (P₀) from Driving Axles:

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

P_o = F_t × v Watt

Calculation:

Given,

F_t = 57600 N

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

From the above concept,

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