# Electric Traction: Power Output and Energy Output at Driving Axles

## Power Output from the Driving Axles

Let,

• 𝐹𝑡 = 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,

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

Where,

• 𝑽𝒎 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.

### Solution

Given data,

• Weight of train,𝑊 = 100 tons

• Specific resistance of train,𝑟 = 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.

### Solution

Given data,

• Maximum speed,𝑉𝑚 = 46 kmph

• Distance travelled,𝑆 = 1.5 km

• Tractive effort during acceleration,𝐹𝑡 = 9810 N

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

• Period of acceleration,𝑡1 = 30 sec

• Period of free run,𝑡2 = 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,

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

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