Elevation gains/losses

I made a 61 mi trip recently, 30 mi each way, first descending 3400', then returning and climbing back up. Both legs were under the same conditions, light rain and slow highway speeds (ave: ~55mph).

I averaged 16.3 mi/kWh downhill and 2.9 mi/kWh returning along the same route uphill. The whole trip average was 5.0 mi/kWh, which is about the same as if the entire trip was on level ground (in light rain, slow highway speeds, occasional heating/venting).

Being able to regen on the downhill portion means EVs are very efficient at changing elevations. Just like ICE vehicles, they take lots of fuel or battery energy to climb a hill, but EVs using regen can "collect" energy back going downhill instead of wasting it in friction brakes.

Because my trip was out-and-back driven under similar conditions, I can estimate the gains or losses due to the elevation change alone.

Math follows:

Downhill energy used: 1.9 kWh
Uphill energy used: 10.4 kWh
Delta energy over one leg: (10.4-1.9)/2 = 4.25 kW or 1.25 kW per 1000 ft elevation

A longer way to calculate this:
Average mi/kWh: 61.3/(1.9+10.4) = 4.98 mi/kWh
One way: 30.65 mi / 4.98 mi/kWh = 6.15 kW
Delta energy over one leg: 6.15 - 1.9 = 4.25 kw or 1.25 kW per 1000 ft

Physics follows:

If you look at just the potential energy gain/loss, we can predict the same:

Potential Energy = mass * g * height

where:
Mass = 2866 + 200 + 170 = 3236 lbm -> 1468 kg
g = 9.807 m/s/s
height = 3,400 ft -> 1036 m

PE = 1468 kg * 9.807 m/s/s * 1036 m = 14,914,956 J or 14,914,956 Ws

Converting Joules to kWh by dividing by 3600 sec/hr and dividing by 1000 W per kW = 4.14 kWh

and 4.14 kWh / 3.4 kft = 1.22 kWh per 1000ft

So this is pretty damn close! Yeah physics!

Neat, with a few assumptions, we can use these numbers to estimate how efficient going downhill is.

Assuming that the drive-train is 80% efficient going uphill, the actual energy required just to get the car up the hill would be

4.14 kWh / 0.8 = 5.175 kWh

The actual energy used going uphill was 10.4 kWh, so
10.4 kWh - 5.175 kWh = 5.225 kWh energy used for wind resistance, heating, etc.

As a sanity check,
30 mi / 5.225 kWh = 5.74 mi / kWh, which sounds reasonable if the trip were over flat ground.

Going downhill, we assume the energy needed for wind resistance, heating, etc. is the same, so the energy recovered would be
5.225 kWh - 1.9 kWh = 3.325 kWh

The theoretical maximum energy available is 4.14 kWh, so the downhill efficiency is
3.325 kWh / 4.14 kWh = 0.8
So the estimate is that going downhill is 80% efficient.

Assuming instead that the drivetrain going uphill is 85% efficient, then the downhill efficiency comes out to:
4.14 kWh / 0.85 = 4.87 kWh
10.4 kWh - 4.87 kWh = 5.53 kWh
30 mi / 5.53 kWh = 5.4 mi / kWh
5.53 kWh - 1.9 kWh = 3.63 kWh
3.63 kWh / 4.14 kWh = 0.88 or 88%

There are a lot of assumptions here, so please interpret these numbers as informed guesses, not the actual efficiency.

Edit 1: The only info I have found about Spark EV efficiency is this quote from http://insideevs.com/gm-general-says-spark-evs-400lb-ft-of-torque-no-misprint/

The [2013] Spark EV efficiency from DC current to delivered Wheel torque is 85% averaged over the city driving schedule and 92% when averaged over the highway schedule

Click to expand...

Edit 2: I changed "regenerative braking" to "going downhill" since if the hill is at a shallow angle regenerative braking might not be used.

Very interesting. I happen to be working on a blog post on regeneration efficiency, and I came up with 78%. Driving on flat road power at 50 MPH is about 9 kW (bit more). Going down a hill in L mode with only using the regen to keep 50 MPH was 9 kW to 10 kW regen. Going up hill on same road and speed was about 32 kW.

-9 kW - 9 kW = -18 kW (regen power going down minus flat road power)
32 kW - 9 kW = 23 kW (power going up minus flat road power)

In effect, you added 18 kW * t of energy while expending 23 kW * t.

18/23=78%

This is only for short piece of hill, so values are probably not as accurate. But it gives you a sense of efficiency for regen, at least at 50 MPH: about 80%.

SparkevBlogspot said:

Very interesting. I happen to be working on a blog post on regeneration efficiency, and I came up with 78%. Driving on flat road power at 50 MPH is about 9 kW (bit more). Going down a hill in L mode with only using the regen to keep 50 MPH was 9 kW to 10 kW regen. Going up hill on same road and speed was about 32 kW.

-9 kW - 9 kW = -18 kW (regen power going down minus flat road power)
32 kW - 9 kW = 23 kW (power going up minus flat road power)

In effect, you added 18 kW * t of energy while expending 23 kW * t.

18/23=78%

This is only for short piece of hill, so values are probably not as accurate. But it gives you a sense of efficiency for regen, at least at 50 MPH: about 80%.

Click to expand...

A slight clarification: 78% is not the actual physical efficiency. Since going uphill is not 100% efficient, the increase in potential energy going up the hill is less than the electrical energy lost from the battery. To calculate the actual regeneration efficiency, one would need to know the slope of the hill.

Edit: Changed "height" to "slope"

..but he's using instantaneous energy (aka power), where the time and distances are immaterial.

In the method I'm using, you don't need to know the slope, or anything other than power readings and speed as long as same piece of road was used and only regen (ie, L without using brake pedal).

I'm assuming power readings are at the battery terminal, energy expended vs energy that was put back in. Battery itself is not 100% efficient, but DCFC was found to be about 93% efficient. If battery is assumed 95%, regen is about 75% efficient.

However, all the values are from car's display, and accuracy suffers. Since only integer values are displayed, 10kW could be between 9.5kW and 10.49kW, 10kW and 10.9kW, 9.1kW to 10.0kW. But in all cases, it's about 10% delta (+/- 5%). Then the regen efficiency could be 70% or 80% or thereabouts. It's to get rough estimate to show that regen isn't awful.

Updated my blog with more details.

http://sparkev.blogspot.com/2016/05/regenerative-braking-efficiency.html

Here's someone's web page using a Leaf: Predicting Energy Use

They report:
Driving uphill requires about 1.5 kWh per 1,000 feet elevation gain. You'll get back about 1 kWh on the downhill side.

Here's mine based on SparkEV. I like graphs, because it gives quick view of how things could behave under different scenarios. I use % grade since that's what's commonly posted on road, but average over long distance over different grades by finding % grade manually (rise over run) should work just as well.

http://sparkev.blogspot.com/2016/03/range-polynomial-climbing-hill.html

A snippet.

Hmmm... That graph says that at 10 mph, I'm going to get fewer miles out of the battery than at 25 mph when climbing a grade.

Why would that be the case?

Static power at 0 MPH is 1.25 kW as found by bicycle guy. Then the efficiency with respect to moving is 0% at 0 MPH, close to 0% at 1 MPH and so on. That gets better as you go faster, after certain point the aerodynamics take far bigger share. 25 MPH seem to be the best case with available data, and you can see where i got the data and how the plots are generated in my blog.

As for hill, that is the same, except steeper hill would take bigger share of power compared to 1.25 kW static power. That's why as the hills get steeper, range is roughly the same at various speeds.

1.25Kw is a lot of energy to use while "idling". Have you been able to confirm this level Sevbs? Your blog is incredibly informative, btw. I also haven't been able to find out how much energy it takes to keep the battery warm when plugged in AFTER full charge, overnight at temps below freezing. I pay 19c/Kwh. 12 hours at (?) 1.5Kw = 18Kwh, or $3.42. X 30 days = $100+ a month. That's a lot of money for me. Any further info would be appreciated.

I didn't measure static power, so you'll have to ask bicycleguy in this forum. He used custom OBD setup.

As for money, while $100/mo may sound like much, it may not be if it's better than gas car. To see MPGe$ (money equivalent MPG), see this blog post. This assumes 4 mi/kWh (about 4.8 mi/kWh displayed on dash).

http://sparkev.blogspot.com/2015/05/spark-ev-miles-per-gallon-this-table.html

To figure out static power after full charge and/or power to keep it warm, use kill-a-watt on 120 EVSE that comes with the car. That's the easy part. Try it at various temperatures, and let us know. It could be that lower temp will use more power with between "too cold" and "too hot" taking almost no power.