Energy Storage Hits the Rails Out West

It’s an interesting concept. They’ve already built a test system, so I’m going to assume this was thought out and will actually work. Let’s look at the numbers.

Each car carries 230 tons and the hill is roughly 3000 feet high, so if we convert to metric and grab an envelope, mgh is about 2 billion Joules of potential energy per car. The output is designed to be 50 MW, so each car gives 40 seconds of runtime at that power, and that’s assuming 100% efficiency of the regenerative brakes. Obviously we need more than one car.

We can also look at this from a different perspective. In this case the power output is P =Fv (that’s actually a dot product, if you’re scoring at home). The incline is given as 6 to 9 degrees, so one car would have to travel at least 160 m/s for that output (at 9º), but that scales with the number of cars. 160 cars can travel at 1 m/s and give us 50 MW (or 80 cars at 2 m/s, etc.), again, modified by the efficiency of the system. Slow speeds would allow the system to “ramp up” by getting to the target speed quickly and operate safely — I doubt anyone wants these trains running down a hill at tens of meters a second.

One last bit of data is that the track is 6.5 km long. I’m assuming that’s the length in addition to the length of the train itself, i.e. it’s how far the train can go.

Let’s assume the efficiency (e) is 75% and we have n cars.

t= (30s)*n and also t = L/v where L is our effective track length. vn = L/30s = 215 m/s

We also know that en(mg)(sin9º)v = 50 MW, or vn = 200 m/s.

Not bad — the answers are within 10%. (Of course it could mean I made the same underlying error in both estimates, or two that happen to cancel)

The final factor is how long the system runs. More cars going slower extends the time, but runs into a space limit. But 50 cars at 4 m/s goes for ~25 minutes, which is not bad for a gap-stopper.