Venturi throat diameter is 20 mm with coefficient if discharge of 0.85, Fuel orifice diameter is 1.25 mm with coefficient of fuel flow of 0.66; the fuel surface is 5 mm below the throat. Compute:

1) Air fuel ratio for depression of 0.07 bar when nozzle lip is neglected

2) Air fuel ratio when nozzle lip is considered.

Given data: ${d_v} = 0.022m$, $Av = \pi \times 0.0222/4 = 3.80 \times 10 - 04$, ${C_{dv}} = 0.85$, ${d_n} = 0.0125m$, ${A_n} = \pi \times 0.001252/4 = 1.2272 \times 10 - 06$, ${C_{dn}} = 0.66$, $X = 0.005mm$,

Assuming ${\rho _a} = 1.2\frac{{kg}}{{{m^3}}}$ ,${\rho _f} = 750\frac{{kg}}{{{m^3}}}$ ,${P_1} - {P_2} = 0.07 \times {10^5}Pa$

1) Calculation of A/F if nozzle lip is neglected:

using

$$\frac{A}{F} = \frac{{{C_{dv}}{A_v}}}{{{C_{dn}}{A_n}}}\sqrt {\frac{{{\rho _a}}}{{{\rho _f}}}} = \frac{{0.85 \times 3.8 \times {{10}^{ - 04}}}}{{0.66 \times 1.2272 \times {{10}^{ - 6}}}}\sqrt {\frac{{1.2}}{{750}}} = 15.95:1$$ Substituting the valves we get,A/F=15.95:1

$$\frac{A}{F} = 15.95:1$$

2) Calculation of A/F if nozzle lip is considered:

Using

$$\frac{A}{F} = \frac{{{C_{dv}}{A_v}}}{{{C_{dn}}{A_n}}}\sqrt {\frac{{{\rho _a}({P_1} - {P_2})}}{{{\rho _f}({P_1} - {P_2} - {\rho _f}.g.x)}}} $$

$$ = \frac{{0.85 \times 3.8 \times {{10}^{ - 04}}}}{{0.66 \times 1.2272 \times {{10}^{ - 6}}}}\sqrt {\frac{{1.2 \times 0.7 \times {{10}^5}}}{{750(0.7 \times {{10}^5} - 750 \times 9.81 \times 0.005)}}} $$ $$\frac{A}{F} = 15.95:1$$