---
title: "Specific energy of 400 t train on 1% up-gradient"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:train-400t-up-gradient-specific-energy"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/train-400t-up-gradient-specific-energy
site: BCT Engineering
---

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

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2080 Bhadra, 8 marks
- Video: [ELECTRIC TRACTION - PART - 09 - SPECIFIC ENERGY OUTPUT AND ENERGY CONSUMPTION  & TWO PROBLEMS](https://www.youtube.com/watch?v=T8Bavig_uLw) — zenmurali Sachithanandam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/train-400t-up-gradient-specific-energy
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

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%.

## 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.

## Answer

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

#### Speed-time curve
- Acceleration: $V_1 = 1.6\times35 = 56$ km/h.
- Constant speed 45 s at 56 km/h.
- Coasting retardation (gradient + resistance oppose):
$$
\beta_c = \frac{98.1\times1 + 50}{277.8\times1.1} = \frac{148.1}{305.58} = 0.4847\ \text{km/h/s}
$$
$V_3 = 56 - 0.4847\times30 = 41.46$ km/h.
- Braking: $t_4 = 41.46/2.4 = 17.28$ s.

| Period | Time (s) | Distance (m) |
|---|---|---|
| Acceleration | 35 | $\frac{56}{3.6}\times\frac{35}{2} = 272.22$ |
| Constant speed | 45 | $\frac{56}{3.6}\times45 = 700.00$ |
| Coasting | 30 | $\frac{56+41.46}{2\times3.6}\times30 = 406.09$ |
| Braking | 17.28 | $\frac{41.46}{3.6}\times\frac{17.28}{2} = 99.48$ |
| Total | 127.28 | 1477.78 |

#### Tractive effort per tonne
$$
\begin{aligned} F_{acc} &= 277.8\times1.1\times1.6 + 98.1\times1 + 50 = 488.93 + 148.1 = 637.03\ \text{N/t} \\ F_{const} &= 98.1 + 50 = 148.1\ \text{N/t} \end{aligned}
$$

#### Energy per tonne
$$
\begin{aligned} E_{wheel} &= 637.03\times272.22 + 148.1\times700 \\ &= 173413 + 103670 = 277083\ \text{J/t} \\ E_{input} &= \frac{277083}{0.75} = 369444\ \text{J/t} = 102.62\ \text{Wh/t} \end{aligned}
$$

#### Specific energy consumption
$$
\text{SEC} = \frac{102.62}{1.47778} = 69.44\ \text{Wh/t-km}
$$
Total energy for the 400 t train $= 102.62\times400 = 41.05$ kWh per run.

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In calculating energy input from the supply, which periods are counted?
   A. Acceleration and constant speed only
   B. All four periods
   C. Acceleration and braking only
   D. Coasting and braking only
   Answer: A. Power is ON during acceleration and constant speed; it is OFF during coasting and braking. (Key point: Power ON periods)

2. Which expression gives tractive effort per tonne during acceleration?
   A. 277.8 x We x alpha only
   B. 98.1 G + r
   C. 277.8 x alpha - 98.1 G
   D. 277.8 x 1.1 x alpha + 98.1 x G + r
   Answer: D. Effort = accelerating force (with 10% rotational inertia) + gradient force 98.1G + resistance r = 637.03 N/t. (Key point: Tractive effort in acceleration)

3. What is the coasting retardation and why is it used to find the speed at end of coasting?
   A. 0.4847 km/h/s, because the brakes are applied
   B. 2.4 km/h/s, set by braking
   C. 0.4847 km/h/s, because gradient and resistance both oppose motion once power is off
   D. Zero, since the train coasts freely
   Answer: C. Gradient plus resistance act alone: (98.1+50)/(277.8x1.1) = 0.4847, giving V3 = 41.46 km/h. (Key point: Coasting retardation)

4. How is specific energy consumption obtained?
   A. Wheel energy x efficiency divided by time
   B. Total energy of all periods divided by 400
   C. Wheel energy / efficiency, converted to Wh/t, divided by total distance in km = 69.4 Wh/t-km
   D. Energy during braking divided by distance
   Answer: C. Input = 277083/0.75 = 369444 J/t = 102.62 Wh/t; divided by 1.478 km gives 69.44 Wh/t-km. (Key point: Specific energy)

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