Question

The schedule speed of a electric train is 40 kmph. The distance
between two stations is 3 km with each stop is of 30 s duration. Assuming the
acceleration and the retardation to be 2 and 3 kmphps, respectively. The dead
weight of the train is 20 ton. Assume the rotational inertia is 10% to the dead
weight and the track resistance is 40 N/ton. Calculate:
1. The maximum speed.
2. The maximum power output from driving axles.
3. The specific energy consumption is watt-hours per ton-km. The overall
efficiency is 80%, assume simplified speed–time curve.

Answer 1

Explanation:

if the explanation is correct mark me as the brainliest

Answer 2

Answer:

The maximum speed = 49.193Kmph

The maximum output = 177.958 kW.

Explanation:

Given,

Schedule Speed Vs = 40kmph.

The distance between two stations (D) = 3km.

The duration of stop = 30s

Acceleration ($\alpha$) = 2kmphps

Retardation ($\beta$) = 3 kmphps

The dead weight of the train (w) = 20 ton.

The track resistance (r) = 40 N/ton.

The overall efficiency (n) = 80%

The schedule time for run Ts = 3600 X D / Vs

                                                 = 3600 X 3/ 40 = 270s.

The actual time of run T = 270 – 30 = 240s.

       (i)

The maximum speed, Vm = $\frac{T}{4X^{2} } - \sqrt{\frac{T^{2} }{4X^{2} } - \frac{3600D}{X} }$

        Where X:

         $=\frac{1}{2\alpha } + \frac{1}{2\beta } \\ = \frac{1}{2X2 } + \frac{1}{3X3 } = 0.416$

        Vm :

       $=\frac{240}{2X0.416} - \sqrt{\frac{(240)^{2} }{4X(0.416)^{2} } - \frac{3600 X 3}{0.416} }\\= 288.46 - \sqrt{83,210.059-25,961.538} \\= 49.193 kmph$

   

       (ii) The acceleration time T' = $\frac{Vm}{\alpha } = \frac{49.193}{2} = 24.59s.$

               The duration of breaking, T''' = $\frac{Vm}{\beta } = \frac{49.193}{3} = 16.397s.$

                The free running time T'' = T - (T' + T''')

                                                          = 240 - (24.59 + 16.397) = 199.012 s.

         The transactive effort during Acceleration:

         Ft = 277.8We X $\alpha$ + Wr = 13,023.2 N

           The maximum power output = $\frac{F1Vm}{3600} = \frac{13,023.2 X 49.193}{3600} = 177.958 kW$

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