The fuel orifice has a diameter of 2.3 mm and the coefficient of fuel flow is 0.66. The petrol surface is 5 mm below the throat. Find:

i. The air fuel ratio for a pressure drop of 0.07 bar when the nozzle lip is neglected

ii. The air fuel ratio when the nozzle lip is taken into account

iii. The minimum velocity of air flow required to start the fuel flow when nozzle lip is provided.

Take density of air and fuel as 1.2 and $750 kg/m^3.$

Given data:

$d_v=0.035 m, \hspace{0.5cm} ∴A_v = 9.62 ×10^{-4} m^2, C_{dv} = 0.85$

$d_n=2.3×10^{-3} m, \hspace{0.5cm} ∴A_n=4.15× 10^{-6} m^2, C_{dn} = 0.66$

$x=0.005 mm, ρ_a=1.2 kg/m3, ρ_f=750 kg/m3, P1-P2=0.07× 10^5 Pa$

1. Calculation of A/F if nozzle lip is neglected

$$Using \ \ \frac{A}{F} = \frac{C_{dv} A_v}{C_{dn} A_n} \sqrt{\frac{ρ_a}{ρ_f}}$$

Substituting the values we get, $$ \frac{A}{F} = 11.928:1$$

1. Calculation of A/F if nozzle lip is considered

$$Using \frac{A}{F} = \frac{C_{dv}A_v}{C_{dn}A_n} \sqrt{\frac{ρ_a (P1-P2)}{ρ_f (P1-P2-ρ_f.g.x)}}$$

Take g=9.81 $m/s^2$

Substituting the values we get, $\frac{A}{F} = 11.959:1 3. Calculation of minimum velocity of air required to start fuel flow Conditions with 1 are at entry to the carburetor and with 2 are at the venturi. We know fuel flow begins if $$(P1-P2) |start=ρ_f.g.x$$

Substituting the values,$$(P1-P2) | start = 36.79 Pa $$Using \ \ \ C_2= \sqrt{\frac{2× (P1-P2) |start}{ρ_a}} =7.83 m/sec$$