A significant cost for any tramway operator is the one introduced by the vehicles themselves. This is not only important in the annual costs of supplying electricity for traction power and auxiliaries, but also in the initial infrastructure construction as well as its ongoing maintenance. In this article I will present the factors that determine the current demand of a modern tram vehicle, consider some of the losses that need to be overcome, and some ways of achieving these.
Defining the demand
In order to design an efficient overhead and power distribution system, the first thing we need to establish is the current demand.
The current is proportional to the power demand, and forms two parts: that required to accelerate or decelerate, and that to keep the tram running at a constant speed. With the rate of acceleration normally required, the former is by far the highest, except in locations with steep gradients.
Rail vehicles running at moderate speeds have very low rolling resistance, much less than that of road vehicles and in the order of 5kg per Tonne weight. So, for a laden weight of 60t, we need a tractive effort of 300kg (or approximately 3kN).
To determine the power demand, we must multiply tractive effort by speed in metres per second. So at 50km/h (14m/s), we have a mechanical power demand of 42kW. To this we need to add electrical losses, which in this case should be less than 10%, so we get a traction power demand of approximately 46kW, which at 750V is 61 amps (61A). This seems like a very low amount, but then we have to account for reaching this maximum operational speed in the first place.
Modern trams have rapid acceleration – 1.3 m/s2 is often used – so to reach 50km/h on a level surface will take 11 seconds. To calculate the tractive effort required to attain this, we multiply the acceleration rate by the vehicle’s mass in tonnes. In our example, this gives 78kN, to which we add the rolling resistance to give 81kN – far greater than the 3kN required at a constant speed.
To give the mechanical power, we multiply this by the speed; so at standstill this is zero, and ramps up to 1.13MW at 50km/h. The energy is the average power multiplied by time, which gives 6.23MJ. The rolling energy loss during this period is 231kJ, so 6MJ is left as kinetic energy, which is recovered during braking, either as re-useable energy or heat.
To achieve 81kN, the high current in the motors gives rise to what is known as ‘copper loss’, and this can total around 100kW. Added to this will be inverter losses, which can add another 20kW. This loss is there for the entire acceleration period, and gives an energy
loss in the acceleration period of 1.32MJ, or a total electrical energy input of 7.55MJ. The electrical power required also ramps up from 100KW to 1.25MW; the latter gives a current of 1643A at 750V.
So at the start of the braking period we have 6MJ of kinetic energy; from this we must subtract the losses: 231KJ for rolling resistance and 1.32MJ for copper and inverter loss, giving us a remainder of 4.45MJ. This is an average power of 405kW over the deceleration of 11 seconds, with a peak of 810kW available to be returned to the line. This represents 1080A at 750V.
Considering an average distance of 500m between tramstops, we have acceleration and braking times of 11 seconds (which will each cover 77m), giving a remaining distance of 346m at a constant 14m/s, taking 25 seconds. The power during this period is 46kW, giving an energy of 1.15MJ. The total input energy is therefore 8.7MJ, so the ideal recovery would be 51%. To attain this, the overhead line must be receptive and itself be loss free, which of course it is not.
We also have to deal with very high peak currents, here up to 1643A. A 150mm2 wire will have a typical resistance of 0.06 ohms for a 500m stretch. At 1643A, this has a loss of 162kW, with a drop of 99V. During regeneration, with a peak current of 1080A, the loss is 70kW. During constant speed running, the loss is only 225W (5.6kJ). These figures of course are for a distance of 500m from source, but of course the substations and feeders will also have losses. They are also peaks, but the average loss will be about 69.5KW, 764kJ for acceleration; and 35kW, 385kJ for braking. This increases the total input energy to 9.46MJ, and reduces returned energy to 4.07MJ. The recovery is now only 43%, and this assumes the line is fully receptive. The rms (heating effect) value of the line current is 860A.
It can be seen that there would be a great advantage if we could keep the high peaks out of the overhead. If we had onboard energy storage that could store the regenerated 4.45MJ – and use it to assist in providing the 7.55 MJ required for acceleration – we would greatly reduce the acceleration peak current.
It would also help if we could reduce the losses in the motors in this period. A figure of 100kW is used here for induction (asynchronous) motors that are mainly used at present. These are mechanically simple and robust machines, but are electrically complex, being a cross between a motor and a transformer.
All electric motors work on the reaction between a magnetic field and current in a conductor, which produces a force proportional to the product of both. In an induction motor, the magnetic field is produced by current in the stator, known as magnetising current. This is out of phase with the voltage, and itself represents no power, but it does heat the windings, which produces a loss. The rotor consists of short-circuited windings and takes the form of uninsulated copper or aluminium bars in slots, with short-circuiting rings at both ends.
In the case of aluminium, they are cast in situ into the laminated steel core. The rotor runs inside the stator (except in the case of some hub-motors where the situation is reversed), with an airgap to allow for mechanical clearance; this is a compromise between electro-magnetic and mechanical conditions. The alternating rotating field created by the three-phase stator induces currents in the rotor, and produces a torque. This causes the rotor to rotate in the direction of the magnetic field, but can never catch up as then there would be no net field cutting the rotor conductors. The rotor then runs with a ‘slip’ from synchronous speed, this being proportional to rotor loss. To operate efficiently, the slip should be as low as possible and hence the use of variable frequency inverters to match the supply frequency to rotor speed.
The motor has three principle losses, magnetising and load currents in stator and rotor resistance. A permanent magnet motor only has one of these losses, stator load. Until about 25 years ago, permanent magnets that could provide sufficient flux to compare with electromagnets as used in the traditional dc motor were not economically available, but they now use neodymium-boron technology. Not only do they provide the required flux, but they are also small and light.
We are now able to make motors that are highly efficient but also lighter. They can be operated either as synchronous or brushless dc machines and are also capable of being incorporated into a wheel, removing the need for gears and cardan shafts. The losses can be about a third of those of an induction motor, and as this is in both accelerating and braking modes, it allows a much better recovery of regenerated energy.
Reducing to 30kW the loss of 100kW in the previous example, reduces input energy to 7.93MJ and increases recovered energy to 5.22MJ, giving a recovery rate of 66% if all the energy can be accepted – all the more reason to have onboard storage.
Supercapacitors are the best way of achieving this; they have a small loss themselves, along with their control circuits, but an overall recovery of over 60% should be possible. If we just store the regenerated energy, we will reduce the peak power demand to about 260kW, giving a line current of only 347A – a lot less than 1643A – and an rms line current of 156A. Incorporating extra energy storage would further smooth out the current to a near constant value of 110A.
Creating six-fold savings
What we have considered here is only a vehicle’s traction energy requirement. We must also consider auxiliaries such as heating, ventilation and cooling. By using modern LED lighting, 1kW should be sufficient to cover lighting and a fully-laden tram should not require much heating in normal conditions. A full tram will produce about 20kW of body heat, so we are therefore left with cooling.
We might have to remove around 40kW, but this should require less than 20kW with an efficient HVAC system. We should assume around 30kW to cover all vehicle auxiliaries, which is 40A; giving a total of 150A. This is a lot better than the rms value in the original case of 900A including auxiliaries, therefore the OLE and distribution system could in theory be reduced in capacity six-fold.
The resulting current is smoother, allowing a much more favourable maximum demand tariff to be attainable. We would attain second-generation performance with first-generation power demand.
The automotive sector is doing much of this development on permanent magnet motors and energy storage – it is high time that our industry caught up.
Of course, reducing vehicle weight is another way of significantly reducing demand, as well as the associated wear and tear on infrastructure elements. Steerable wheel-motors and composite materials are ways forward in this respect – but that is a subject for another article…
Article originally appeared in March 2018 TAUT (963).