On a 1 percent down gradient with resistance 40 N/t, what happens during coasting?

Gradient force 98.1 N/t exceeds resistance, so the net force helps and the train gains speed (about 0.19 km/h/s)

For which period is energy taken from the supply?

Only the acceleration period, since power is on only then

Which expression gives the tractive effort per tonne during acceleration?

277.8 x 1.1 x 2 - 98.1 + 40 = 553.06 N/t

2081 Baisakh · 8 marks

Quadrilateral speed-time curve on down gradient

Short answer

Compute acceleration, coasting, and braking distances with effective mass 1.1W; divide total distance by run plus stop time. Schedule speed is 17.3 km/h. Specific energy consumption is 13.14 Wh per tonne-km.

An electric train has quadrilateral speed-time curve as follows: i) Uniform acceleration from rest at 2 kmphps for 10 seconds. ii) Coasting for 50 seconds. iii) Braking period of 15 seconds. The train is moving a uniform down gradient of 1% tractive resistance 40 newtons per tonne, rotational inertia effect 10% of dead weight, duration of stop 15 seconds and overall efficiency of transmission gear and motor as 75%. Calculate its schedule speed and specific energy consumption of run.

Assumptions: $W_e = 1.1W$ ; a 1 tonne train is taken (answers per tonne); power is ON only during acceleration; during coasting the net force is the gradient force minus the resistance.

Speeds and distances

Acceleration: $V_1 = 2\times10 = 20$ km/h.

Coasting on a 1 % down gradient: gradient force $= 98.1\times1 = 98.1$ N/t, resistance $= 40$ N/t, so the net force helps motion. Coasting acceleration:

\alpha_c = \frac{98.1 - 40}{277.8\times1.1} = 0.1901\ \text{km/h/s}

Speed at end of coasting:

V_2 = 20 + 0.1901\times50 = 29.51\ \text{km/h}

Braking in 15 s: $\beta = 29.51/15 = 1.967$ km/h/s.

Period	Time (s)	Speeds (km/h)	Distance (m)
Acceleration	10	0 → 20	$\frac{20}{3.6}\times\frac{10}{2} = 27.78$
Coasting	50	20 → 29.51	$\frac{20+29.51}{2\times3.6}\times50 = 343.80$
Braking	15	29.51 → 0	$\frac{29.51}{3.6}\times\frac{15}{2} = 61.47$
Total	75		433.04

Schedule speed

\begin{aligned} T_s &= 75 + 15 = 90\ \text{s} \\ V_s &= \frac{433.04}{90}\times3.6 = 17.32\ \text{km/h} \end{aligned}

Specific energy consumption

Tractive effort per tonne during acceleration (down gradient, so the gradient force $98.1G$ with $G = 1$ is subtracted):

\begin{aligned} F_t &= 277.8\,(1.1)\,\alpha - 98.1G + r \\ &= 277.8\times1.1\times2 - 98.1\times1 + 40 = 553.06\ \text{N/t} \end{aligned}

Energy at wheels per tonne $= F_t \times d_1 = 553.06\times27.78 = 15362.8$ J/t.

Energy input (efficiency 75 %) $= 15362.8/0.75 = 20483.7$ J/t $= 5.690$ Wh/t.

\text{SEC} = \frac{5.690\ \text{Wh/t}}{0.43304\ \text{km}} = 13.14\ \text{Wh/t-km}

Answer: Schedule speed = 17.3 km/h; specific energy consumption = 13.14 Wh per tonne-km.

✓ Answer checked by Claude Opus 5.5 (high effort)

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Quadrilateral speed-time curve on down gradient

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