2
$\begingroup$

Is there a database showing how many grams of a particular fuel generates how many Kilowatt hours of electricity?

$\endgroup$
2
  • $\begingroup$ Google "levelized cost of electricity". The fuel cost and energy conversion efficiency are two of the primary drivers of LCoE. The size and engineering of the plant have a lot to do with final number, so you can expect a range of about a factor of three for many types of fuel. Lazard has published very detailed breakdowns of LCoE for many years. These include the fuel cost percentage of levelized cost $\endgroup$
    – Phil Sweet
    Aug 17 '18 at 14:20
  • $\begingroup$ This is the Lazard document I was thinking of. I was about to walk out the door when I wrote the above comment. Lazard’s Levelized Cost of Energy Analysis, version 11.0. Some of the price and conversion numbers are on page 19. But look at the entire presentation. This is how it's done. $\endgroup$
    – Phil Sweet
    Aug 17 '18 at 19:20
1
$\begingroup$

The burning of the fuel generates thermal energy, i.e. hot gases.

The energy what it generates is the heat of combustion. Its ideal value is well known, depends mainly in the binding energy of the of the fuel and the combustion products.

Here comes the first problem: the burning is typically not perfect and it depends significantly on the circumstances (mainly temperature, pressure).

All the motors, generators, power plants convert this energy to electrical energy. It happens typically by the conversion to mechanical energy (typically rotating parts), and then the mechanical is converted to electrical energy.

All energy conversion using a different path have typically a very bad efficiency, typically below $20\%$.

The conversion of the thermal energy to mechanical or electrical energy is subject of the Second law of thermodynamics. It is the same Law which says that

  • heat moves always from the warmer body to the colder, never back
  • you can't build a ship which would cool the water of the sea and using its energy to move
  • more technically, the maximal efficiency is $\frac{T_2-T_1}{T_2}$, for example if the $1000K$ burning products are cooled down to $300K$, then the maximal efficiency is $\approx 70\%$.

The conversion of the mechanical energy to electrical energy is much easier, because (ideally) there is no heat exchange, it happens by coils in generators, the efficiency is typically over $90\%$.

Practical generators have an efficiency between $30\%$ and $70\%$, depending on the circumstances.

Find the burning heat of the fuel (anywhere, it is public data), and find the efficiency of the device which burns it. After that, you need only some multiplications.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.