xpensive wrote:wuzak wrote:
....
Well, it has been recovered....
Like the energy is coming out of the MGU-H perhaps?
Recovered and stored for use at a different time. Which can also be said for the MGUH.
Here is a simple calculation for you.
Assume fuel energy density is 45MJ/kg, and the MGUH generates 60kW to send to the MGUK. That means that to use the full 120kW an additional 60kW has to come from the ES.
This energy has been recovered and stored from braking/MGUH over the previous lap/laps. But how much fuel flow does it represent?
If we assume no losses in conversion anywhere, the 60kW equates to 4.8kg/hr of fuel flow - ie nearly 5% of the maximum allowed fuel flow.
Since the efficiency is the output power of the PU divided by the potential power of the fuel, should not the potential power of the fuel which was used to generate and store the energy to the ES be used in the calculation for efficiency?