1. **Problem Statement:** A Venturimeter has an entrance diameter $d_1 = 0.3$ m and a throat diameter $d_2 = 0.2$ m. The discharge coefficient $C_d = 0.96$. The specific weight of the gas $\gamma = 19.62$ N/m$^3$. The pressure difference between the entrance and throat is measured as $h = 0.06$ m of water. Calculate the volume flow rate $Q$ of the gas. 2. **Formula and Concepts:** The Venturimeter measures flow using Bernoulli's equation and continuity equation. The volume flow rate $Q$ is given by: $$Q = C_d A_2 \sqrt{\frac{2 \Delta P}{\rho (1 - (A_2/A_1)^2)}}$$ where: - $A_1$ and $A_2$ are cross-sectional areas at entrance and throat, - $\Delta P$ is the pressure difference, - $\rho$ is the density of the gas. The pressure difference $\Delta P$ can be found from the manometer reading: $$\Delta P = \rho_{water} g h$$ Since specific weight $\gamma = \rho g$, for water $\gamma_{water} = 9800$ N/m$^3$ (standard), so: $$\Delta P = \gamma_{water} h = 9800 \times 0.06 = 588 \text{ Pa}$$ 3. **Calculate Areas:** $$A_1 = \pi \frac{d_1^2}{4} = \pi \frac{0.3^2}{4} = \pi \times 0.0225 = 0.0707 \text{ m}^2$$ $$A_2 = \pi \frac{d_2^2}{4} = \pi \frac{0.2^2}{4} = \pi \times 0.01 = 0.0314 \text{ m}^2$$ 4. **Calculate Density of Gas:** Given specific weight $\gamma = 19.62$ N/m$^3$, and $\gamma = \rho g$, with $g = 9.81$ m/s$^2$: $$\rho = \frac{\gamma}{g} = \frac{19.62}{9.81} = 2 \text{ kg/m}^3$$ 5. **Calculate Volume Flow Rate $Q$:** Calculate the area ratio squared: $$\left(\frac{A_2}{A_1}\right)^2 = \left(\frac{0.0314}{0.0707}\right)^2 = (0.444)^2 = 0.197$$ Calculate the denominator term: $$1 - 0.197 = 0.803$$ Calculate the term under the square root: $$\frac{2 \Delta P}{\rho (1 - (A_2/A_1)^2)} = \frac{2 \times 588}{2 \times 0.803} = \frac{1176}{1.606} = 732.3$$ Square root: $$\sqrt{732.3} = 27.07 \text{ m/s}$$ Finally, calculate $Q$: $$Q = C_d A_2 \times 27.07 = 0.96 \times 0.0314 \times 27.07 = 0.816 \text{ m}^3/\text{s}$$ **Answer:** The volume flow rate of the gas is approximately **0.816 m$^3$/s**.