I'm not sure how to convert or correctly compare gasoline/alcohol to batteries in terms of weight and volume. For example in the following link Specific Energy and Energy Density of Fuels I can find the following information:

Specific Energy

$$E_{8} = \frac{E}{m}$$

Energy Density

$$E_{d} = \frac{E}{V} $$

                   Specific  Energy
Fuel      Density   Energy   Density
           kg/m3    MJ/kg     MJ/m3
Gasoline    716     -47.3    -33,867
Ethanol     784     -29.7    -23,278

And the information I've found on Li-ion batery from What's the Best Battery:

Li-ion - Gravimetric Energy Density (\$\frac{Wh}{kg}\$): 110-160

I have no idea how to compare/convert \$\frac{wh}{kg}\$ from/to \$\frac{kg}{m3}\$ nor \$\frac{MJ}{kg}\$.

(Most likely it's more complicated then even this, as a combustion engine isn't that efficient in changing potential energy into kinetic energy. I'm not concerned with that at the moment, just trying to figure from a storage point of view how to compare these energy sources).

  • \$\begingroup\$ Gasoline yields 50 MegaJoules at the axle, or 150MJ inside the engine, per gallon. \$\endgroup\$ Jul 6 '17 at 17:42

I have no idea how to compare/convert wh/kg from/to kg/m3 nor MJ/kg.

First, \$\rm\frac{[kg]}{[m]^3}\$ are the units of density and not related to energy storage. Your source was just giving the density figures to let you double-check the energy density and specific energy values for self-consistency. Converting from watt-hours per kilogram to kilograms per cubic meter wouldn't make any sense.

Second, one hour is 3600 seconds. And one watt is one joule per second. So you can very easily do the math to convert watt-hours to megajoules.

  • \$\begingroup\$ How did you get an inline formula? It doesn't seem that single dollar signs ($) allow inline formula's. \$\endgroup\$ Jul 6 '17 at 16:11
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    \$\begingroup\$ @Erik, EE.SE uses \$ to start and end inline formulas because we often want to use $ to talk about how much things cost. \$\endgroup\$
    – The Photon
    Jul 6 '17 at 16:12
  • \$\begingroup\$ so 100 wh/kg or 396,000 ws/kg or 396 MJ/kg but why do the liquid fuels have a negative value? (-47.3 MJ/kg) It seems that looking at weight, batteries are much more efficient. \$\endgroup\$ Jul 6 '17 at 16:20
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    \$\begingroup\$ @ErikPhilips nope, you are doing it wrong. 100 Wh/kg computes to 0.36 MJ/kg. 1 Ws = 1 J, and 1 MJ = 1000000 J not 1000 J. The negative value is probably because you get energy out by burning it - it's a standpoint thing. Now you realize just how bad batteries are. But you should compare values after efficiency (electric things can be very efficient, while fuel powered things often have much less efficiency). \$\endgroup\$
    – Arsenal
    Jul 6 '17 at 16:27
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    \$\begingroup\$ @Arsenal: to elaborate on that thought, burning fuels incurs the wrath of the Carnot efficiency, which limits most practical heat engines to about 30-40% efficient in a big commercial station. Internal combustion engines do worse, around 20%. \$\endgroup\$ Jul 6 '17 at 16:42

Comparing energy density of liquid fuels and batteries is a little unfair. When you've burnt your liquid fuel it's gone. When your (rechargeable) battery is flat you still have your battery. To pursue this argument to its illogical conclusion:

$$ Relative\;energy\;density= \frac {liquid\;fuel\;weight\;loss}{battery\;weight\;loss} = \frac {100\%}{0} = \infty$$

Batteries' energy density is infinitely higher than liquid fuel.

I find it easy to remember that oil / diesel / petrol gives a heat output of about 10 kWh/kg. Timber (such as wood pellet) is about 4.5 kWh/kg. I've used these numbers to mentally estimate fuel consumption rates of diesel generators and wood-fired pellet boilers, etc. with reasonable accuracy when factoring in likely efficiency.

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    \$\begingroup\$ Until we have wireless recharging on a road, both products have to be "refilled". Either a tank with liquid, or a batter with energy (electrons whatever..) \$\endgroup\$ Jul 6 '17 at 21:44

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