Venturimeter with CO2 and meriam oil

Given: CO₂ density $\rho = 3$ kg/m³, $p_1 = 170$ kPa, manometer fluid S.G. = 0.827 ($\rho_m = 827$ kg/m³), manometer reading $h = 8$ cm $= 0.08$ m, $d_1 = 10$ cm, $d_2 = 6$ cm. The gas is treated as incompressible (small pressure change), horizontal meter, no losses.

(a) Pressure at the throat

The U-tube manometer measures the pressure difference between sections 1 and 2:

$$p_1 - p_2 = (\rho_m - \rho)\,g\,h = (827 - 3)(9.81)(0.08) = 646.7\ \text{Pa}$$ $$p_2 = 170\,000 - 646.7 = 169\,353\ \text{Pa} \approx 169.35\ \text{kPa}$$

(b) Gas rate

$$\beta = \frac{d_2}{d_1} = 0.6,\qquad \beta^4 = 0.1296$$

Bernoulli with continuity: $p_1 - p_2 = \dfrac{\rho}{2}V_2^2\left(1 - \beta^4\right)$.

$$V_2 = \sqrt{\frac{2(646.7)}{3\times 0.8704}} = \sqrt{495.3} = 22.26\ \text{m/s}$$ $$\begin{aligned} A_2 &= \frac{\pi}{4}(0.06)^2 = 2.827\times 10^{-3}\ \text{m}^2\\ Q &= A_2V_2 = 0.06293\ \text{m}^3/\text{s} = 226.5\ \text{m}^3/\text{h} \end{aligned}$$

Answer: (a) $p_2 \approx 169.35$ kPa; (b) $Q \approx 0.0629$ m³/s $\approx 226.5$ m³/h.