171117_SECAP of Greater Irbid Municipality_SET_rev2

For the sake of argument, let's assume the volumetric efficiency is 100%.
Let's also assume the engine is running at sea level on a standard 15C day, so the manifold air pressure is equivalent to the
air density at sea level * 25% (because it's throttled to idle)
Air consumed will be:
displacement * cycles * volumetric efficiency * manifold air pressure
2L = 0.002 m^3
Air density is 1.225 kg/m^3 * 0.25
This gives us:
0.002 m^3 * 350/min * 100% * 1.225 kg/m^3 * 0.25= 0.214 kg/min
Fuel used would be 1/15 of that (actually 1/14.7 in stoichiometric conditions), or 0.0143 kg/min
at 719.7 kg/m^3, that works out to 0.02 L/min
0.02 L/min is 1.2 L/hr (ref: https://www.quora.com/How-much-gas-does-a-4-cylinder-use-while-idling)
The number of cars car park capacity will be 2,000 with assumption od % use is 70.39%
With ideal current waiting time is one hour per car and with three times full per day
=6,000 cars per day x 1.2 L saving x 300 days in year x 9.2 KWh convert fuel in litre to KWh / 1000 to convert to MWh x
70.39% of use = 13,989 MWh x 0.246 to tCO2
146