Dimensioning the Traction System of a Rail Vehicle

The traction system consists of a vehicle body that carries the pantographs, power converters, and auxiliary converters. It also has bogies that carry wheelsets, traction motors, and gear systems. The performance requirement is checked in the first 5 steps, the electrical calculation is done in the 6th step, followed by thermal calculation in the 7th step and the final Evaluation is carried out in the 8th step. Inputs like operational conditions, required tractive power, driving resistance and a few electrical parameters are used. In this article, I will guide you through each step.

1.?????Calculating driving resistance

For developing the traction system we need to know what resistances the train will face and how to overcome those. Mainly train resistance consist of vehicle resistance, track resistance, and acceleration resistance(Racc). Vehicle resistance is further divided into rolling resistance(Rro), Starting resistance(Rst), and Aerodynamic resistance (Rae). Track resistance is further divided into Grade resistance (Rgr) and Curve resistance (Rcu). ?

R [kN] = Rro + Rst + Rae + Rgr + Rcu + Racc

So, these resistances are calculated depending upon the operating conditions given by the customer like speed, slope, wind, curve, etc.

Where f is the resistance coefficient of particular resistance.

?????????????Q is Vertical wheel force.

?????????????Cd is the Aerodynamic drag coefficient.

?????????????ρa is air density.

?????????????v is driving speed.

?????????????α is the angle with the horizontal plane.

?????????????λ is the rotational inertia coefficient.

Also, a general definition of Running resistance can be used. Which states R = co + c1*v + c2*v2. Where c0,c1 & c2 are factors that can be estimated using empirical formulas from the standards or past calculations.?

2.?????Setting the Tractive effort value

The force at coupler or value of tractive effort is required by the customer. Depending upon the driving condition and weight of the train this value can be calculated. The value of the Tractive force(Z) generated by the traction unit must overcome the total train resistance(R). So, The condition Z ≥ R must be fulfilled.

3.?????Check for Adhesion coefficient at Wheel-Rail contact

Now, we need to check whether our vehicle will be able to deliver the required tractive effort or not. This can be done by checking the required adhesion coefficient at wheel-Rail contact and particular vehicle mass. Tractive effort is always limited by adhesion(friction between wheel and rail). The following factors influence the achievable adhesion at wheel-rail contact.

·???????Longitudinal sliding velocity

·???????Drive slip/ Brake slip

·???????Driving speed

·???????Wet or dry contact patch

·???????Rail roughness

·???????Dynamic disturbances

(If you want to know more about Wheel-Rail contact, please read my previous article.)

4.?????Calculate the required Tractive power (P)

Once we have the Tractive force(Z) and Maximum speed(Vmax) Tractive power can be calculated as P = Z * Vmax

5.?????Draw the tractive effort diagram

From all the above-calculated values we can now draw the tractive effort diagram which consists of five characteristic points.

·???????The 1st characteristic point is calculated from the general definition of the running resistance discussed earlier depending upon the slope and operational profile.

·???????The 2nd point is for maximum speed which can be known by the customer.

·???????Now from the running resistance and maximum speed we have the value of tractive effort needed that is to be delivered at maximum speed. After adding the safety margin of 5 or 10 percent we get 3rd point. This third point also indicates the maximum power. After knowing the maximum power a parabolic line can be drawn indicating constant power.

· Point number 4 can be known from the data provided by the customer like the weight of the train and starting acceleration. That leads to the maximum tractive force that our vehicle can deliver.

·???????At the intersection of constant power and constant tractive force line, we get the 5th characteristic point. Sometimes the tractive force is not constant that is because of the change in switching frequencies of the converters and due to that high peak currents are observed in motor currents. That ultimately leads to no constant tractive force. ?

6.?????Electrical calculation

Once we get to know the required tractive power, we should now calculate the required current that we need to take from the catenary to generate this tractive power. In this step electrical parameters like line currents, motor currents, and peak currents are calculated. These parameters are required for comparing it with the ratings of particular equipment and converter specifications. Simulation tools can also be used for this.

7.?????Thermal calculation

In this step, we need to calculate the losses and temperatures of the equipment not only in the static condition but also in dynamic conditions by running a speed profile of the locomotive.

8.?????Check for the compatibility of the traction system with the infrastructure

In the infrastructure, we have signaling systems or a track circuit that sits between the two rails and supplies voltage. When our locomotive passes through this track, the wheelset short circuits the track circuit, and a short circuit current is generated. So, here we make sure that if the traction system is DC then the track circuit must be AC. By doing so traction system will never interfere with the track circuit. The problem is every country has some different specifications of the track circuit. So, we have to always check the interference currents and frequencies are always under threshold values of the signaling equipment and the infrastructure. This is an important step for the homologation of the traction system.

Now, If your customer says "I want a locomotive that can pull 500 tons of load." you can provide them with a solution. Thank you for reading.

Jinay Bhagat