Convert my CT200h to plug-in?

My Lexus CT200h is EPA rated to get 42 miles per gallon combined. In it, over the 2 1/2 years I've had it, my total net has been 51.0 miles per gallon. The CT200h has a hybrid power train with a 1.4 kWh (kilowatt hour) NiMH (nickel metal hydride) battery. This battery is good for a mile or so at very low speed in so-called "EV mode." Plug-in hybrid electric vehicles (PHEVs) will have much larger batteries, enabling them to travel further and faster in EV mode. For example, the Prius PHEV is estimated to be able to travel 11 miles and at a maximum EV mode speed of 62 m.p.h. It achieves this with a 4.4 kWh battery pack. And the Chevy Volt now sports a 16.5 kWh battery pack that is estimated to provide an EV mode range of 38 miles.

I wondered if it would be possible to install a larger battery back and charging capability to my CT200h to convert it into a PHEV. As it happens, the answer is yes. And a huge variety of installations are possible, all the way from 2 kWh to 15 kWh capacity. Note that the Nissan Leaf, a pure electric vehicle (EV), sports a 24 kWh battery pack and a claimed range of 75 miles.

Should I install any conversion and, if so, which one? This can be looked at from an economic viewpoint and from a CO₂ emission viewpoint. I'll look at both. As a baseline, my most common drive is the daily commute from my home in Anaheim Hills to my office in Long Beach. This round trip is about 62.5 miles and I do it, on average, four times each week. I'd be able to charge the vehicle at work, though I'd have to park at our laboratory facility about two blocks from our corporate offices. Such a strategy would enable me to utilize a battery pack with a 35 mile range to, on a typical day, never have to use the internal combustion engine.

What would this look like? The 7 kWh pack theoretically provides this 35 mile range, and I'd assume it would do so by discharging no more than 90%, but let's be conservative and assume that I'd need 14 kWh of charging per day. Assuming the charge system is 85% efficient, I'd draw about 16.5 kWh from the grid. At $0.16/kWh, this would cost about $2.64. I'll round down to $2.50 because I don't really quite go 70 miles on my daily commute.

Currently, with my 51 m.p.g. average, I use about 1.22 gallons of fuel in my commute. At a current price of $3.699/gallon, this costs about $4.50 per day. Thus, the conversion would save me something like $1.86 per day and, for 200 trips per year, I'd save $372/year. It's not completely clear from Plug In Supply's pricing page but it looks like I'd pay $9,875 for the 7 kWh system fully installed. Clearly, from a purely financial point of view, this makes no sense.

As to CO₂ emissions, this is a bit more problematic to compute. I'd be getting half of my electrical energy from our house, where the City of Anaheim provides our electricity, and the other half in Long Beach, where Southern California Edison is the provider. And, assuming that some combination of coal, nuclear, and natural gas provides almost all of the electricity, I won't figure in emissions resulting from extraction and transportation of these fuels. This is reasonable, I'm not figuring the emissions resulting from extracting, refining, and transporting the gasoline I burn.

The calculation for the CO₂ emissions from burning 1.22 gallons of fuel can't be exact as I don't know how much ethanol is in the fuel, and the mix of the various hydrocarbon chain lengths. I'm going to assume that the gasoline consists of n-heptane, C₇H₁₆ at that its density is 6 pounds per gallon. The chemical reaction would be C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O. A mole of heptane has a mass of 100.2 grams. This mole results in 7 moles of carbon dioxide, each with a mass of 44 grams for a total of 308 grams.

Now, I'll be a bit general and figure that I use 1/51=0.0196 gallons of gasoline per mile. This gasoline weighs 0.118 pounds or 53.4 grams. This produces (308/100.2)*53.4 or 164 grams of CO₂ per mile (this is the typical metric for vehicular carbon dioxide emissions) for a total on my commute of 62.5*164=10,250 grams or 10.25 kg of CO₂.

That was the easy part. The electrical emissions are much more problematic because I've not been able to determine the mix of sources for Long Beach electricity. I'll just speculate. In any case, I need to determine the emissions related to 7 kWh from the City of Anaheim and 7 kWh from Southern California Edison in Long Beach.

I'm not able to find the appropriate mix of generating facilities to use to calculate for Long Beach, so I'll back into it from information from California State University Long Beach on this page, where it's stated that the renewable generation of 656,000 kWh avoided the emission of 471 metric tons of carbon dioxide. CSULB is only a couple of miles from our office, so this will have to suffice. So the generation of 7 kWh would involve the emission of 8.25*(471,000,000/656000) or 5,923 grams of CO₂.

From Anaheim, I'll use the figures from my previous post on the Nissan Leaf. Using those numbers (and sparing my readers the gory details) I can figure that those 8.25 kWh involve the emission of 9,051 grams of CO₂. The total emissions are thus 9,051+5,923 or 14,974 grams of carbon dioxide. Call it 15 kg. This is half again as much as my CT200h emits burning gasoline and the main culprit, as was the case for the Nissan Leaf I looked at, is that the City of Anaheim derives a surprising amount of its electricity from the burning of coal. And this information comes straight off of our bi-monthly bill!

So, when all is said and done, it makes no economic sense, and certainly no sense with respect to CO₂ emissions, to undertake such a conversion. That's comforting because I don't have a spare $10K to throw at such a project!

And all of this probably overestimates the gains. The careful reader may have noticed that the Chevy Volt uses a 16.5 kWh pack to go 38 miles or 0.43 kWh/mile. The Nissan Leaf uses a 24 kWh pack to go 75 miles, or 0.32 kWh/mile. The Plug-In Supply site, where I got these figures, claims 35 miles on 7.33 kWh or 0.29 kWh/mile. It's true that these figures represent a Prius rather than a CT200h but I'm skeptical that the Prius is that much more efficient in terms of aerodynamics and rolling resistance than a Chevy Volt or a Nissan Leaf.

A quick note on the eGallon

Screen shot of my results from DOE eGallon site

In a previous post I mentioned the eGallon concept from a Department of Energy (DOE) web site. It purports to tell a visitor how much he or she would pay to drive as far in an electric vehicle as a gallon of gas takes them in an "average vehicle." It breaks down only as far as by state (or U.S. average). So, for example, if I use California, it tells me that a gallon of regular gasoline costs $3.99 and that my eGallon costs $1.53.

But my average mileage over the life of my vehicle is 50.86 m.p.g. At my most recent fill up I paid $4.059/gallon. I'll use the Nissan Leaf for a comparison, the vehicles are broadly similar in important ways. Each has a C_d (drag coefficient) of 0.29 and, while the frontal area of the CT200h is a bit larger, the Leaf weighs more. The Leaf is rated by the EPA to consume 29 kWh/100 miles for the 2013 model year.

So, on a gallon of fuel, I go 50.86 miles. The Leaf would need (50.86/100)*29 kWh = 14.75 kWh to go that distance. If I assume that the charging system is 85% efficient, I'd pay for 14.75/.85=17.35 kWh. On my most recent electric bill I paid $0.1611/kWh for electricity above the "basic lifeline" rate, so these 17.35 kWh would cost me $2.80 and that's the price of my eGallon. Quite a difference between that number and $1.53, the "true" number is 83% higher whereas the number for my gasoline cost is not far away from what I actually pay. The computed eGallon price would be even further from ReGallon cost ("Rob's eGallon") if DOE had used the 2013 model year numbers for the Leaf in lieu of previous years' 34kWh/100 miles. If I use the 2013 model year number for the Leaf and the EPA combined estimate (42 m.p.g.) for the Lexus CT200h that I drive, an eGallon would cost $2.31, only 51% higher than the site's number.

You can read about their methodology here. The confounding factors are the actual cost of electricity and the fuel economy utilized for the ICE (internal combustion engine) vehicle. For reference, the plot below (you can click it to enlarge and be able to read the numbers) shows an AeGallon ("actual eGallon") for a range of actual fuel economies from 12 m.p.g. (the driver currently in a vehicle getting less than that is not a likely candidate for an EV) to 70 m.p.g. (a hypermiler in a Prius). For this plot, I'll use the same electricity consumption as the DOE site uses, i.e., 35 kWh/100 miles, a blended rate from 5 top selling EVs. Electricity prices on the plot range from $0.09 to $0.20 per kWh. You can calculate your number yourself, it's as simple as 0.4118*(m.p.g.)*(electricity cost per kWh). You'll note that, for combinations of high mileage vehicles and expensive electricity, the eGallon may be more expensive than a gGallon (i.e., a gallon of gasoline).

On the plot, the "front" axis is m.p.g. for the vehicle being replaced with an EV, the rearward extending axis is the price of a kilowatt hour of electricity, and the vertical axis is the price of an eGallon in dollars. You can see that, for low mileage vehicles being replaced, the eGallon is quite inexpensive, regardless of electricity costs. But as replaced vehicle fuel economy climbs, the eGallon becomes much more expensive. The DOE site simply uses a single fleet average fuel economy (28.2 m.p.g.) and does not correct for the 85% charging efficiency I estimated.