What am I calculating in each case? The cost per distance of an electric car and the same quantity for a gasoline powered car. The ratio of the two tells us how much farther we can go, per dollar, with an electric car.

Assumptions

Both vehicles are traveling at the same speed, and require the same brake power. The brake power is the power delivered to the wheels.

Variables

P = Power required at wheels of vehicle to drive at speed V (hp or Watts)

V = Vehicle speed (mile/hr or m/s)

η

_{e}= Overall efficiency of electric engine

η

_{g}= Overall efficiency of gasoline engine

C

_{e}= Cost of electric energy ($/kW-hr or $/J)

C

_{g}= Cost of gasoline ($/gal or $/liter)

ρ = Density of gasoline (kg/liter)

Δh = Specific energy of combustion of gasoline (J/kg)

DD

_{e}= Distance per Dollar of electric car (miles/$ or km/$)

DD

_{g}= Distance per Dollar of gasoline car (miles/$ or km/$)

E

_{e}= Fuel economy of electric car (miles/gal or km/liter)

E

_{g}= Fuel economy of electric car (miles/gal or km/liter)

Q

_{g}= Volumetric flow rate of gasoline (gal/hr or liter/hr)

Analysis

Step 1. Analysis of the Gasoline Car.

The Distance per dollar, DD

_{g}is the ratio of the fuel economy to the cost of the gasoline fuel consumed by the engine, DD

_{g}= E

_{g}/C

_{g}. The fuel economy is the ratio of the vehicle speed to the volumetric flow rate of fuel consumed, E

_{g}= V / Q

_{g}.

The thermal power released by the combustion of gasoline is ρΔHQ

_{g}. (We multiply by flow rate by the density because the thermal energy is usually listed on a mass-basis, but the flow rate of fuel is a volume-basis.) Due to thermodynamic inefficiencies inherent in the combustion process (and transmission losses), the amount of mechanical power delivered to the wheels is less than this. The ratio of the power at the wheels to the thermal power by combustion is the overall gasoline engine efficiency, η

_{g}. The mechanical power delivered to the wheels, P = ρΔhQ

_{g}η

_{g}. Rearranging this equation, we have Q

_{g}= P / (ρΔhη

_{g}).

Substituting this expression into the first equation in this section, we have DD

_{g}= ρV&eta

_{g}Δh / (PC

_{g}).

Step 2. Analysis of the Electric Car

The distance per dollar for the electric car is given as the ratio of the "fuel" economy to the price of electricity needed to charge the batteries, DD

_{e}= E

_{e}/ C

_{e}.

The electric car economy is the ratio of the speed of the vehicle to the rate of electrical power supplied by the batteries. The ratio of the power at the wheels to the electrical power supplied by the batteries is the overall efficiency η

_{e}. Thus, E

_{e}= Vη

_{e}/ P, where P is the power at the wheels. Thus, DD

_{e}= Vη

_{e}/ (PC

_{e}).

Step 3. Compute Ratio

The ratio of the electric vehicle distance per dollar to the gasoline vehicle distance per dollar is given by dividing the two equations at the end of each previous section. Note that the P and V cancel out since they represent the power at the wheels and the vehicle speed. We assumed these things were the same for both cars. Thus,

DD

_{e}/DD

_{g}= η

_{e}C

_{g}/ (η

_{g}ρΔh

_{c}C

_{e})

Sample Calculation

For an electric car, the overall efficiency is 75%. For the gasoline car, the efficiency is about 20%. The cost of gasoline is $4.25/gallon. The cost of electrical energy is about $0.106/kW-hr (according to my ComEd bill). Finally, the density of gasoline is about 740 kg/m

^{3}and the specific combustion energy (actually enthalpy is the correct term here) is about 47,000,000 J/kg.

Inserting these terms into the our equation for the DD ratio yields.

DD

_{e}/DD

_{g}= (0.75/0.2) x (2.734) x (4.25 / 10.6) = 4.1 !!!

The middle term accounts for the density, combustion term, and unit conversion factors.

Conclusion

For a typical electric car, the distance driven per dollar is 4 times that for a gasoline-powered car. This analysis does not imply that the overall costs of electric vehicles is lower, since a real economic analysis has to include many other factors, not the least of which is initial cost.

## 4 comments:

Not mention new batteries every 3 years

Not to mention new batteries every three years.

This makes sense, with the ICE at 20%, I've heard a modern electric plant can run to from 75 to 80

% eficiency (?). I suppose cost of ownership is a factor (i.e. replacing batteries), but todays ICE simply cannot convert fuel to power as effectively as a coal or gas fired power plant.

Does 75% electric efficiency include the losses from your house's electrical socket to your car battery? In other words, you may not be charging that battery at 10.5 cents per Kw-hr.

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