Storing the Energy
Some quick calculations show: a car needs around 4kWh to travel 100km at a speed of 50 km/h and 8kWh to travel the same distance at 100 km/h (the linearity is coincidential!).
A typical 12V 50Ah lead acid battery weighs in at around 20kg and stores 600 Wh of electric energy. Neglecting all of the losses that come in, we would need around 14 of those
to do the trip mentioned above. Thus, the batteries alone weigh 280 kg. Juicy! Prices vary around 1000-1500€ for such a pack.
There is the theoretical alternative of using the famous LiIon batteries. They store the same amount of energy at around 1/10 of the weight but also at 10 times the price.
A 5-figure amount just for the batteries? I don't think so. I'd rather wait for either prices to drop or new storage technologies that are cheaper in the first place.
So for now let me conclude, that home-brew car converters are stuck with lead acid (gel) batteries.
Losses, Losses...
As you might have guessed, not all of the 600Wh stored in a battery actually make it to the wheels of the car. There is some friction down the path:
- Discharge losses
- Discharge Reserves
- Inverter losses
- Motor losses
- Mechanical losses
Lets look into them a bit closer.
Discharge Losses
The slower a battery is discharged, the more of the power stored inside it will actually make it out of the battery. When a battery is discharged very quickly, some of the power is lost to internal
heating of the cells. A while ago I calculated some actual values for a set of cells using the manufacturers data sheet.
At a 60 min. discharge time the figure was 60% capacity (53% at 30 Min. discharge).
That means that from the 8 kWh mentioned above, we only get 4.8 kWh. 3.2 kWh gone if we drive fast.
Discharge Reserves
A battery musn't be drained all the way down to 0V. Discharging it to around 30% is recommended. This is considered in the data sheets mentioned above.
Inverter Losses
The inverter consists of Semiconductor switches. Even when these switches are turned on, some of the energy is dissipated as heat. During the process of switching
(which happens around 8000 times per second)
the so called
switching losses occur. Over all, about 90-95% of the power that goes into the inverter comes out the other end. Sticking with the example above, thats another 400Wh gone.
Motor Losses
The motor consists of a set of coils which generate a magnetic field. The coils have an ohmic resistance which dissipates power. Also, some of the magnetic field isn't causing any movement of the shaft.
This is called
stray inductance. Over all, the effiency of an inverter driven electric motor is around 90-95%. Another 350Wh gone in the worst case.
Mechanical losses
All the bearings and gears account for mechanical losses. An effiency figure I've come across is 97% effiency. That would be around 70Wh gone and 3.8 kWh left of the original 8 kWh.
Conclusion
Neglecting the losses obviously wasn't a good idea. Now the calculations above are worst case. As a rule of thumb it looks like we can only use half of the nominal capacity for propulsion.
Thus, the set of batteries mentioned above will allow us to cover 50km at 100 km/h or 100 km at 50 km/h.
Voltage Considerations
Electric power is the product of voltage and current. Thus, drawing 20kW of power from a 100V battery pack causes a current of 200A. If we use 200V the same power causes a current of 100A etc.
There are a couple of reasons, why we don't want high currents:
- Thicker wires needed
- Higher currents through the batteries cause more heat to be generated inside them
- Higher currents through the inverter cause more heat to be generated (for the motor, this doesn't really matter, because it needs more windings as the voltage increases. Thus the ohmic
losses are pretty much independent of the voltage range used)
Basically, lower currents are easier to handle and make the whole system more efficient because of lower ohmic losses. But then, they require higher voltages. And higher voltages means more
battery cells. This either means a lot of mass, or a lot of small batteries. But small batteries have a higher ohmic resistance. So, if the batteries become too small, the higher voltage
offers no advantage. Thus, the choice of the batteries is a trade-off between feasibility, weight, range and cost. Right now, I'm aiming towards 15-20 batteries in the 50Ah class.
