First, consider the case with negligible auxiliary loads (no air conditioning).
For a Civic-sized engine (1.8 liters), this US DOE worksheet estimates about 0.3 US gallon/hour fuel consumption at idle.
Here is a conservative starter calculation:
- The Civic starter is rated at 1.0 kW (83A$\times$12V). A 3 second start therefore produces 3 kJ. Assume an additional 25% in battery internal dissipation that must be replaced.
- As you note, this energy must be replenished by the ICE (internal combustion engine). Max ICE efficiency is only 30%. The incremental efficiency, which is what matters for this small additional load, is no doubt higher, but I’ll use 25% as a conservative estimate.
Alternator efficiency is not great either; I’ll use a conservative 50%.
With these values, it requires 3.0 kJ$\times$(1.25 / 0.25 / 0.5) = 30 kJ worth of fuel to recharge the battery (Note the overall charging efficiency is only 10%!).
Now, the energy density of gasoline is 120 MJ per US gallon (42.4 MJ/kg), so the amount of fuel required to recharge the battery, including all the inefficiencies, is 30 kJ $\div$ 120 MJ/gal = 0.00025 US gallon.
So, the “crossover” idle time in this case, above which it is more efficient to stop and restart, is 0.00025 gal $\div$ 0.3 gal/hour $\approxeq$ 8.3 $\times10^{-4}$ hours, or about 3 seconds.
Now suppose an air conditioner (PDF) is consuming 1 kW of electrical power.
- With the engine running, the A/C requires (via the alternator) an additional engine fuel consumption equivalent to 1 kW / 0.5 / 0.25 = 8 kW, or 29 MJ/hour, or 0.24 gal/hour of gasoline. For a duration $t$, the total fuel consumption with the engine running is (0.3 + 0.24)$t$ = 0.54$t$ (with $t$ in hours).
- With the engine stopped, the A/C still consumes 1 kW, or 3.6 MJ per hour. With that low 10% charging efficiency, it requires 36 MJ worth of fuel (or 0.3 gal) to recharge an hour’s worth of A/C operation. Adding in the starter contribution, the total fuel requirement is 0.00025 + 0.3$t$ (with $t$ again in hours).
Equating these two new fuel requirements, the crossover time with the A/C on increases, but only to about 4 seconds.
Although the battery charging efficiency is low, the waste of the idling fuel consumption dominates the calculation.
Note that I don’t have a reference for the 25% battery re-charge inefficiency.
Unfortunately, that’s an important number when running an A/C, since it reduces the advantage of shutting off the engine. At some high load level (in the neighborhood or 4 kW) that disadvantage outweighs the advantage of turning off the engine.
Further (experimental) data to confirm the above estimates can be found here: http://www.iwilltry.org/b/projects/how-many-seconds-of-idling-is-equivalent-to-starting-your-engine/
In my case it consumes about the same amount of fuel as 7 seconds of idling. However, the additional fuel consumption observed seems almost entirely due to a faster idle speed setting for the first 20 seconds after starting. Any good driver would start moving within 1-2 seconds after starting, which would effectively eliminate the fast idle losses. If you can begin extracting useful work from your engine within 1 second after starting the engine then it appears starting the engine consumes fuel equivalent to about 0.2 seconds of idling.
Best Answer
The fuel injector system does not always inject a constant amount of fuel into the engine.
If you are driving down a hill and remove your foot from the gas pedal (a little) the engine will stay the same rpm but with less fuel passing thou the system.
And if you are driving up a hill you need to press down the gas pedal to maintain the same rpm and speed, more or less you need to push more fuel into the engine.
This is why you will not have a constant fuel consumption at a given rpm.
Then when you remove some load, let's say you are on the top of the hill, you will have a small surplus of "energy released" inside the engine that will push the rpm higher until you remove your foot from the gas pedal and reducing the amount of fuel sent into the engine.
The faster you travel the more wind resistance you will have, therefore you need more power to overcome this resistance. And to get more power out from a engine you need to send in more fuel (or use the amount have in there in a more efficient way).
But why do we need to switch gears up and down then, why can't we just inject more fuel into the engine?
This is where we go into chemistry since you can only make gas burn within a specified rate between fuel and air, and the volume inside the engine has a fix size and therefore has a maximum volume of gas that you can burn per rotation. Send in more fuel than this and the engine will drop in efficiency.
And when you then switch down a gear you will increase the rpm at this speed and that way increase the amount of fuel you can burn in the engine within this timeframe and you will in most of the cases more power out from the engine.
But at the end of the day there is a lot of other constraints in a combustion engine that will limit how it works beside this simplified version I just describe here.