How much energy is recovered by regenerative braking?

The method in which no energy is drawn from the supply system during braking period and some energy is fed back to the supply system is known as regenerative braking.

Calculation of Energy Returned During Regeneration

When the train is accelerated up to a certain speed, then it acquires energy, (known as kinetic energy), corresponding to that speed and this kinetic energy is given by,

$$\mathrm{KE\mathrm{\: =\: }\frac{1}{2}\mathit{mv^{\mathrm{2}}}}$$

During coasting period, a part of this stored energy is utilized in propelling the train against frictional and other resistances to motion and hence the speed of the train falls. Under ideal conditions, i.e. no resistances to motion of the train, the speed of the train would have not decreased.

In the similar way, when the train going down the gradient or moving on a level track, the speed of the train remaining the same or reduced. Under this condition, the stored energy can be converted into electrical energy and fed back to the supply system.

The amount of electrical energy returned to the supply system depends upon the following factors −

• The initial and final speeds of the train during regenerative braking.

• The resistance to the motion of the train.

• The gradient of the track in case the train is moving down the gradient.

• Efficiency of the traction system.

Now, let

• 𝑉₁ = Initial speed of train in kmph

• 𝑉₂ = Final speed of train in kmph

• 𝑊_𝑒 = Equivalent weight of train

Then, the kinetic energy stored in the train at the initial speed (𝑉₁) is given by,

$$\mathrm{KE_{1}\mathrm{\: =\: }\frac{1}{2}\times \frac{1000\mathit{W_{e}}}{9.81}\times \left ( \frac{1000\mathit{V_{\mathrm{1}}}}{3600} \right )^{2}\; kgm}$$

$$\mathrm{\Rightarrow KE_{1}\mathrm{\: =\: }\frac{1}{2}\times 1000\mathit{W_{e}} \times \left ( \frac{1000\mathit{V_{\mathrm{1}}}}{3600} \right )^{2}\; Watt\: seconds}$$

$$\mathrm{\Rightarrow KE_{1}\mathrm{\: =\: }\frac{1}{2}\times 1000\mathit{W_{e}} \times \left ( \frac{1000\mathit{V_{\mathrm{1}}}}{3600} \right )^{2}\times \frac{1}{3600}\; Watt\: hours}$$

$$\mathrm{\therefore KE_{1}\mathrm{\: =\: }0.01072\, \mathit{W_{e}V_{\mathrm{1}}^{\mathrm{2}}}\; Wh}$$

Similarly, the kinetic energy at the final speed (𝑉₂) is given by,

$$\mathrm{ KE_{2}\mathrm{\: =\: }0.01072\, \mathit{W_{e}V_{\mathrm{2}}^{\mathrm{2}}}\; Wh}$$

Therefore, the amount of energy available during regeneration is

$$\mathrm{KE_{1}-KE_{2}\mathrm{\: =\: }0.01072\, \mathit{W_{e}\left ( V_{\mathrm{1}}^{\mathrm{2}}-V_{\mathrm{2}}^{\mathrm{2}} \right )}}$$

Also, some of the energy is lost to overcome the resistance to motion and the losses in the traction system including traction motors.

The energy wasted to overcome the resistance to motion is given by,

$$\mathrm{Energy\: lost\mathrm{\: =\: }\mathit{\frac{W\times r\times S\times \mathrm{1000}}{\mathrm{3600}}}\; Watt\: hours\mathrm{\: =\: }0.2778\mathit{W\times r\times S}\:Watt\: hours }$$

Where,

• 𝒓 is the specific resistance of the train in Newton per ton.

• 𝑾 is the weight of the train.

• 𝑺 is the distance of run.

Also, while going down the gradient in the hilly track service, the energy is supplied as tractive effort due to the gradient and energy is added up to the energy available during regeneration. The energy available due to motion down the gradient is given by,

Energy available due to motion down the gradient,

$$\mathrm{\mathrm{\: =\: }\frac{98.1\mathit{GW}\times S\times \mathrm{1000}}{\mathrm{3600}}\mathrm{\: =\: }27.25\: GSW}$$

Therefore, the total energy available during the regeneration, i.e. amount of energy returned to the supply system is,

$$\mathrm{Energy\: returned\: to\: system \mathrm{\: =\: }0.01072\, \mathit{W_{e}}\left ( \mathit{V}_{1}^{2}- \mathit{V}_{2}^{2}\right )\mathrm{\: +\: }27.25\mathit{GSW}-0.2778\mathit{WrS}\; \; \cdot \cdot \cdot \left ( 1 \right )}$$

If η is the efficiency of the system, then the energy returned to the system (in watt-hours) is,

$$\mathrm{Energy\: returned\: to\: system\mathrm{\: =\: }\left [ 0.01072\, \mathit{W_{e}}\left ( \mathit{V}_{1}^{2}- \mathit{V}_{2}^{2}\right )\mathrm{\: +\: }27.25\mathit{GSW}-0.2778\mathit{WrS} \right ]\times \eta \; \; \cdot \cdot \cdot \left ( 2 \right )}$$

Numerical Example

A train weighing 450 tons has speed reduced by regenerative braking from 50 to 20 kmph over a distance of 3 km on a down gradient of 3%. Calculate the electrical energy returned to the line. The tractive effort is 35 Newton/ton and rotational inertia of 10% and efficiency of conversion is 80%.

Solution

Given data,

• Acceleration weight of train,𝑊_𝑒 = 𝑊 + 10%𝑊 = 1.1𝑊 = 1.1×450 = 495 tons

• Distance travelled,𝑆 = 3 km

• %Gradient,𝐺 = 3%

• Efficiency of conversion,𝜂 = 80% = 0.80

Hence, the energy available due to reduction in speed is,

$$\mathrm{\mathrm{\: =\: }0.01072\: \mathit{W_{e}\left ( V\mathrm{_{1}^{2}-\mathit{V}_{2}^{2}} \right )}}$$

$$\mathrm{\mathrm{\: =\: }0.01072 \times 495 \times \left ( 50^{2}-20^{2} \right )}$$

$$\mathrm{\mathrm{\: =\: }11143.44 \: Wh\mathrm{\: =\: }11.143\: kWh}$$

The tractive effort required when train going down the gradient is,

$$\mathrm{\mathit{F_{t}\mathrm{\: =\: }Wr-\mathrm{98.1}WG}}$$

$$\mathrm{\Rightarrow\mathit{F_{t}}\mathrm{\: =\: } 450 \times 40-98.1\times 450\times 3\mathrm{\: =\: } -114435 \: Newton }$$

Here, the negative sign indicates that tractive effort of 114435 N is available. Hence, the energy available on the account of moving down the gradient a distance of 3 km is given by,

$$\mathrm{\mathrm{\: =\: }\frac{\mathit{F_{t}\times S\times }1000}{1000\times 3600}\: kWh\mathrm{\: =\: }\frac{114435\times 3\times 1000}{1000\times 3600}\mathrm{\: =\: }95.36 \: kWh}$$

Thus, the total energy available is,

$$\mathrm{\mathrm{\: =\: }11.143\mathrm{\: +\: }95.36\mathrm{\: =\: }106.5\: kWh}$$

And the energy returned to the supply system is,

$$\mathrm{\mathrm{\: =\: }0.80\times 106.5\mathrm{\: =\: }85.2\: kWh}$$