Electric Traction: Power Output and Energy Output at Driving Axles

Electric TractionPower SystemsUtilisation of Electrical Power

Power Output from the Driving Axles


  • 𝐹𝑑 = Tractive effort in Newton

  • 𝑉 = Speed of the train in kmph

Since, the power output is defined as the work done per unit time, i.e.,

$$\mathrm{Power\: output,\mathit{P}\mathrm{\: =\: }Rate\: of \: work\: done\mathrm{\: =\: }\frac{Work \: done}{time}}$$

$$\mathrm{\Rightarrow \mathit{P}\mathrm{\: =\: }\frac{Tractive\: effort\times Distance}{Time}\mathrm{\: =\: }Tractive\: effort\times Speed}$$

$$\mathrm{\therefore \mathit{P\mathrm{\: =\: }F_{t}\times V}\: Nm/s}$$

$$\mathrm{\Rightarrow \mathit{P\mathrm{\: =\: }F_{t}\times V}\times \frac{1000}{3600}\, Watts}$$

$$\mathrm{\therefore \mathit{P\mathrm{\: =\: }\frac{F_{t}\times V}{\mathrm{3600}}}\: kW\: \cdot \cdot \cdot \left ( 1 \right )}$$

Now, if 𝜼 is the efficiency of gear transmission, then the power output of motor (or power input to driving axles) is given by,

$$\mathrm{Power\: input,\mathit{P_{i}\mathrm{\: =\: }\frac{F_{t}\times V}{\eta }}\: Watt\: \cdot \cdot \cdot \left ( 2 \right )}$$

$$\mathrm{\Rightarrow \mathit{P_{i}\mathrm{\: =\: }\frac{F_{t}\times V}{\mathrm{3600\eta }}}\: kW\: \cdot \cdot \cdot \left ( 3 \right )}$$

Energy Output from the Driving Axles

Assume the run of the electric train according to the simplified trapezoidal speed-time curve.

Then, the total energy required for the run (or energy output from driving axles) is given by

$$\mathrm{Total\: energy\mathrm{\: =\: }Energy\: required\: during\: acceleration\mathrm{\: +\: }Energy\: required \: during\: free\: run}$$

$$\mathrm{\Rightarrow Total\: energy\mathrm{\: =\: }Average\: power\: during \: acceleration\times acceleration\: period\mathrm{\: +\: }Average\: power\: during\: free \: run \times duration\: of\: free \: run}$$

From the figure, we have,

$$\mathrm{Total\: energy\mathrm{\: =\: }\frac{1}{2}\frac{\mathit{F_{t}\times V_{m}}}{\mathrm{3600}}\times \frac{\mathit{t_{\mathrm{1}}}}{3600}\mathrm{\: +\: }\frac{\mathit{F_{t}^{'}V_{m}}}{3600}\mathrm{\: +\: }\frac{\mathit{t_{\mathrm{2}}}}{3600}\: \: kWh …(4)}$$


  • π‘½π’Ž is the maximum speed of train in kmph.

  • 𝑭𝒕 is the tractive effort required during acceleration (in Newton)

  • 𝑭𝒕′ is the tractive effort required during free run in Newton

  • 𝒕1 is the time of acceleration in seconds

  • 𝒕2 is the duration of free run in seconds

In practice, the energy output from driving axles or total energy required for run is expressed in watt-hours per ton-km, instead of kWh. Therefore,

$$\mathrm{\mathrm{\: =\: }\frac{Energy\: output\: in\: watt\: hours}{Weight\: of\: train\: in\: tons\times Distance \: of\: run\: in\: km} \: \: \: \cdot \cdot \cdot \left ( 5 \right )}$$

The quantity given by equation (5) is known as the specific energy output from the driving axles.

Numerical Example (1)

An electric train weighing 100 tons is to be driven up an incline of 2% at a speed of 35 kmph. If the train resistance at this speed is 1 kg per ton, find the power output from driving axles.


Given data,

  • Weight of train,π‘Š = 100 tons

  • Gradient,𝐺 = 2%

  • Specific resistance of train,π‘Ÿ = 1 kg per ton

  • Speed,𝑉 = 35 kmph

Therefore, the tractive effort required is,

$$\mathrm{\mathit{F_{t}\mathrm{\: =\: }\mathrm{98.1}WG\mathrm{\: +\: }Wrg}} $$

$$\mathrm{\Rightarrow \mathit{F_{t}}\mathrm{\: =\: }98.1\times 100\times 2\mathrm{\: +\: }100\times 1\times 9.81\mathrm{\: =\: }20601\: Newton} $$

$$\mathrm{\therefore Power\: output\: from\: driving\: axles,\: \mathit{P\mathrm{\: =\: }\frac{F_{t}\times V}{\mathrm{3600}}}\mathrm{\: =\: }\frac{20601\times 35}{3600}} $$

$$\mathrm{\mathit{P}\mathrm{\: =\: }200.29 \: kW} $$

Numerical Example (2)

Calculate the specific energy output of train whose weight is 100 tons, if maximum speed of 46 kmph and for a given run of 1.5 km and the tractive effort during acceleration is 9810 N and acceleration period is 30 seconds. The tractive effort required during free run is 981 N and the duration of free run is 73 seconds. Assume a trapezoidal speed-time curve.


Given data,

  • Maximum speed,π‘‰π‘š = 46 kmph

  • Distance travelled,𝑆 = 1.5 km

  • Tractive effort during acceleration,𝐹𝑑 = 9810 N

  • Tractive effort during free run,𝐹𝑑′ = 981 N

  • Period of acceleration,𝑑1 = 30 sec

  • Period of free run,𝑑2 = 73 sec

Now, the energy output in watt-hours is given by,

$$\mathrm{\mathrm{\: =\: }\frac{1}{2}\frac{\mathit{F_{t}\times V_{m}}}{\mathrm{3600}}\times \frac{\mathit{t_{\mathrm{1}}}}{3600}\mathrm{\: +\: }\frac{\mathit{F_{t}^{'}V_{m}}}{3600}\mathrm{\: +\: }\frac{\mathit{t_{\mathrm{2}}}}{3600}}$$

$$\mathrm{\mathrm{\: =\: }\frac{1}{2}\times \frac{9810\times 46}{3600}\times \frac{30}{3600}\mathrm{\: +\: }\frac{981\times 46}{3600}\times \frac{73}{3600}\mathrm{\: =\: }0.5223\mathrm{\: +\: }0.2542\mathrm{\: =\: }0.7765 \: kWh}$$

$$\mathrm{\therefore Energy\: output\: in\: watt\: hours\mathrm{\: =\: }776.5 \, Wh}$$

Thus, the specific energy output is given by,

$$\mathrm{\mathrm{\: =\: }\frac{Energy\: output\: in\: watt\: hours}{Weight\: of\: train\: in\: tons \times Distance \: of \: run\: in\: km}}$$

$$\mathrm{\mathrm{\: =\: }\frac{776.5}{100\times 1.5}\mathrm{\: =\: }5.17\: Whton-km}$$

Updated on 19-May-2022 13:54:27