The derivation of Equation (6.1) relies on the method of canceling units in brackets [. . .] as in the following steps. Replace each of the steps in words with its dimensionally correct mathematical equivalent, and so derive the Otto Cycle Power Output Equation.

Step 1: How much air is drawn into the cylinder? The actual volume of air drawn into the cylinder is D as the piston cycles from top dead center to the bottom dead center. The mass of air drawn into the cylinder is thus . . .?

Step 2: How much fuel? Use (F/A)Mass ratio.

Step 3: How much thermal energy is released by this amount of fuel for just one power stroke?

Needs the HV of the fuel

Step 4: Of this heat, only a portion is converted into mechanical work. The mechanical work is related to the thermal input by Z_Otto.

Step 5: How many power strokes are there? As we have seen, there is just one per every two engine revolutions, and N is the number of revolutions/minute. Multiply the mechanical work by number of power strokes/s.

Step 6: Hence, the rate of working is η_Otto × [ρ_rair × D × (A/F)_Mass] × HV × N/(2 × 60) [kJ of work per power stroke] [power strokes per second] = [kJ (work)/s] = [kW].

Step 7: Finally, cast this in the suggested form:

 

Transcribed Image Text:

Nouo X Pair x Dx N x HVfuel 120 x (A/F)Mass Ideal Power %3D