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Specific energy of 400 t train on 1% up-gradient

Short answer

Calculate powered distances and tractive efforts per tonne, then divide energy input by total distance. Specific energy consumption is about 69.4 Wh per tonne-km, or 41.05 kWh for the whole 1.478 km run.

An electric train weighing 400 tonnes runs a 1% up-gradient with the following speed-time curve: i) Uniform acceleration of 1.6 kmphps for 35 seconds ii) Constant speed for 45 seconds iii) Coasting for 30 seconds iv) Braking at 2.4 kmphps Calculate the specific energy consumption if the tractive resistance is 50N/tonne, rotational inertia effect 10%, and overall efficiency of transmission and motor 75%.

Data: W=400W = 400 t, We=1.1WW_e = 1.1W, G=1 %G = 1\ \% up, r=50r = 50 N/t, η=75 %\eta = 75\ \%. Power is ON during acceleration and constant speed; OFF during coasting and braking. Values per tonne.

Speed-time curve

  • Acceleration: V1=1.6×35=56V_1 = 1.6\times35 = 56 km/h.
  • Constant speed 45 s at 56 km/h.
  • Coasting retardation (gradient + resistance oppose):
βc=98.1×1+50277.8×1.1=148.1305.58=0.4847 km/h/s\beta_c = \frac{98.1\times1 + 50}{277.8\times1.1} = \frac{148.1}{305.58} = 0.4847\ \text{km/h/s}

V3=56−0.4847×30=41.46V_3 = 56 - 0.4847\times30 = 41.46 km/h.

  • Braking: t4=41.46/2.4=17.28t_4 = 41.46/2.4 = 17.28 s.
PeriodTime (s)Distance (m)
Acceleration35563.6×352=272.22\frac{56}{3.6}\times\frac{35}{2} = 272.22
Constant speed45563.6×45=700.00\frac{56}{3.6}\times45 = 700.00
Coasting3056+41.462×3.6×30=406.09\frac{56+41.46}{2\times3.6}\times30 = 406.09
Braking17.2841.463.6×17.282=99.48\frac{41.46}{3.6}\times\frac{17.28}{2} = 99.48
Total127.281477.78

Tractive effort per tonne

Facc=277.8×1.1×1.6+98.1×1+50=488.93+148.1=637.03 N/tFconst=98.1+50=148.1 N/t\begin{aligned} F_{acc} &= 277.8\times1.1\times1.6 + 98.1\times1 + 50 = 488.93 + 148.1 = 637.03\ \text{N/t} \\ F_{const} &= 98.1 + 50 = 148.1\ \text{N/t} \end{aligned}

Energy per tonne

Ewheel=637.03×272.22+148.1×700=173413+103670=277083 J/tEinput=2770830.75=369444 J/t=102.62 Wh/t\begin{aligned} E_{wheel} &= 637.03\times272.22 + 148.1\times700 \\ &= 173413 + 103670 = 277083\ \text{J/t} \\ E_{input} &= \frac{277083}{0.75} = 369444\ \text{J/t} = 102.62\ \text{Wh/t} \end{aligned}

Specific energy consumption

SEC=102.621.47778=69.44 Wh/t-km\text{SEC} = \frac{102.62}{1.47778} = 69.44\ \text{Wh/t-km}

Total energy for the 400 t train =102.62×400=41.05= 102.62\times400 = 41.05 kWh per run.

Answer: Specific energy consumption ≈ 69.4 Wh per tonne-km (41.05 kWh for the whole run of 1.478 km).

✓ Answer checked by Claude Opus 5.5 (high effort)

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