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Based on the energy per kg stored in a battery and gasoline; the efficiency of the electric and thermal engine and other factors, after some calculations I reached the conclusion that 60 liters of gasoline gave to a car the same autonomy as 2000 kg of batteries. I assumed that the automobile never carried more than 45 kg of batteries and that it was resupplied each time the energy in the pack run out.

Did I make some big mistakes?

I can not provide you with my method of calculation because I do not want to influence you. I need some independent confirmation or infirmation of the result I obtained.

Robert Werner
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    You should show your calculations - it may be easier to spot an error than try to repeat the research and find different numbers... – Solar Mike Mar 03 '19 at 20:48
  • You battery size is too small. The Tesla model S battery weighs about 540kg The Nissan Leaf is about half that, but only 1/3 of the capacity. Also, don't forget that electric and IC engines have very different efficiencies, when you make the comparison. – alephzero Mar 03 '19 at 21:35
  • I took the efficiency of the electric and thermal engine into account. Regarding my calculations they were done like this: 0.75g/cm^360liters47.5MJ/kg*25%=148.43kWh where the numbers are the density of gasoline, the capacity of the fuel tank, energy per kg stored in gasoline, the average efficiency of the thermal engine, the total useful energy. – Robert Werner Mar 03 '19 at 22:22
  • Have you accounted for the efficiency of electric motors as opposed to piston engines? I don't see it in those calculations. I think you should edit your question and put your full calculations in it. – GdD Mar 04 '19 at 09:26
  • Of course I accounted for the efficiency of the electric engine and battery discharge but I do not want to answer my own question. – Robert Werner Mar 04 '19 at 10:27
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    The reason people want that is SE has a tradition that askers are expected to show their own work "so far". Partly to avoid wasting our time doubling work you have already done, and partly because it is easier to spot improvements in your approach (or structural flaws) than it is explaining everything de novo. It also helps us avoid adding errors of our own. In other words your "don't influence us" doctrine is Off-topic for this answers platform. – Harper - Reinstate Monica Mar 04 '19 at 21:00

2 Answers2

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Let's do the math!

Gasoline is 32 MJ / litre, so 60 litres is 1920 MJ, or 533 kWh. However, there's only 25% efficiency in most engines, so this translates to useful 133 kWh.

Electric motors are nearly 100% efficient. Lithium ion batteries are 0.1 - 0.265 kWh / kg, so this is 502 kg - 1330 kg of batteries, depending on the exact battery chemistry.

I don't know which battery chemistry you used in calculations, but Tesla uses this: http://blog.evandmore.com/lets-talk-about-the-panasonic-ncr18650b/ which has 12 watt-hours of energy, and weight is 48.5 grams, so it's 0.247 kWh / kg, giving 538 kg of batteries needed. Tesla batteries are widely considered the best.

Note that of the 538 kg, you save at least 38 kg or even more, due to not needing a heavy gasoline engine and needing just a lightweight electric motor. Therefore, I would say the equivalence is about 60 liters of gasoline = 500 kg of batteries.

Note also, that electric cars can do regenerative braking. Thus, it might be even the case that 60 liters of gasoline in a regular non-hybrid car is equivalent to even as little as 400 kg of batteries. Of course, this depends on the driving (city vs highway) and the quality of the regenerative braking systems.

Edit: I didn't account for the structural parts holding the battery together in a nearly but not quite fatal accident. They need to be substantial. Thus, battery module weight is more than the sum of the weights of its cells.

juhist
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  • Tesla S (https://en.wikipedia.org/wiki/Tesla_Model_S) has a total weight of 1961 kg with a 40 kWh battery and 2250 kg for a 100 kWh pack which means that the density of energy per kg can be calculated and in consequence 750 kg of batteries are required to provide ~133 kWh [ 750kg((100kWh-40kWh)/(2250kg-1961kg))85%=132.35kWh ]. The total efficiency, charge discharge plus that of the electric engine, is unlikely to be over 85% because the max charge discharge efficiency reaches 90%. – Robert Werner Mar 05 '19 at 16:16
  • @RobertWerner That's probably the module weight. I agree, I had a slight calculation error: I didn't account for the structural parts holding the battery together in a collision. They need to be substantial, as the car needs to be safe even in a nearly fatal accident. – juhist Mar 05 '19 at 18:54
  • @RobertWerner And oh, charge-discharge efficiency probably means efficiency in times efficiency out. If efficiency in is 95%, efficiency out is 95%, charge-discharge efficiency should be about 90% (or more accurately, 90.25%). So, you can use 95% as the efficiency instead of 100%. – juhist Mar 05 '19 at 18:57
  • ...but if you take into account structural weight in an electric car, you should take into account gas tank system weight, fuel line weight, gas pump weight, etc. in a gasoline car... Perhaps even exhaust weight, emission treatment device weight, extra weight of gasoline engine, etc... – juhist Mar 05 '19 at 18:59
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The first 3 terms work out to be 45kg. Does that seem reasonable? A liter of fuel weigh 3/4 of a kilogram or 1.6 lbs - I'm ok with that. Multiply that by 47.5MJ/kg = 2137.5MJ The author supplied that J/kg conversion. Using 1Joule = 2.778 x 10^ I get 491.6 kWh

The author wanted that multiplied by 25% I get 123 kWh which is reasonable close to the authors 148 kWh. So at least the math is OK with me.

Bob S.
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