Jet diameter of a simple carburettor

Given: $d_t = 0.032$ m, $p = 101.3$ kPa, $T = 300$ K, $\Delta p = 5000$ Pa, $C_{da} = 0.85$, $C_{df} = 0.65$, $\rho_f = 740$ kg/m$^3$.

Air density and throat area

$$\begin{aligned} \rho_a &= \frac{p}{RT} = \frac{101\,300}{287 \times 300} = 1.177\ \text{kg/m}^3 \\ A_t &= \frac{\pi}{4}(0.032)^2 = 8.042 \times 10^{-4}\ \text{m}^2 \end{aligned}$$

Air flow

$$\begin{aligned} \dot{m}_a &= C_{da} A_t \sqrt{2 \rho_a \Delta p} \\ &= 0.85 \times 8.042 \times 10^{-4} \times \sqrt{2 \times 1.177 \times 5000} \\ &= 0.85 \times 8.042 \times 10^{-4} \times 108.48 = 0.0742\ \text{kg/s} \end{aligned}$$

This is 267 kg/h. The throat velocity is $\sqrt{2\Delta p/\rho_a} = 92.2$ m/s, which is well below sonic, so the incompressible assumption is acceptable.

Fuel flow for A/F = 15

$$\dot{m}_f = \frac{0.0742}{15} = 4.944 \times 10^{-3}\ \text{kg/s}$$

Jet area and diameter

$$\begin{aligned} A_f &= \frac{\dot{m}_f}{C_{df}\sqrt{2\rho_f \Delta p}} = \frac{4.944 \times 10^{-3}}{0.65 \times \sqrt{2 \times 740 \times 5000}} \\ &= \frac{4.944 \times 10^{-3}}{0.65 \times 2720.3} = 2.796 \times 10^{-6}\ \text{m}^2 \\ d_f &= \sqrt{\frac{4 A_f}{\pi}} = 1.887 \times 10^{-3}\ \text{m} \end{aligned}$$

Answer: air flow $= 0.0742$ kg/s (267 kg/h); fuel flow $= 0.00494$ kg/s (17.8 kg/h); jet diameter $\approx 1.89$ mm.