Venturi Diameter 61E8C5

1. **Problem Statement:** Water flows through a Venturi meter with flow rate $Q = 2$ m$^3$/s. The inlet diameter is $D_1 = 0.8$ m. Mercury manometer shows a height difference $h = 5$ cm. Find the throat diameter $D_2$. 2. **Relevant Formulas and Principles:** - Continuity equation for incompressible flow: $$A_1 V_1 = A_2 V_2$$ where $A = \frac{\pi D^2}{4}$. - Bernoulli's equation between points 1 and 2: $$P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2$$ - Pressure difference from manometer: $$P_1 - P_2 = \rho_m g h$$ where $\rho_m$ is mercury density. 3. **Known values:** - $Q = 2$ m$^3$/s - $D_1 = 0.8$ m - $h = 5$ cm = 0.05 m - Density of water $\rho = 1000$ kg/m$^3$ - Density of mercury $\rho_m = 13600$ kg/m$^3$ - Gravity $g = 9.81$ m/s$^2$ 4. **Calculate areas:** $$A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.8)^2 = 0.5027 \text{ m}^2$$ 5. **Calculate velocity at inlet:** $$V_1 = \frac{Q}{A_1} = \frac{2}{0.5027} = 3.978 \text{ m/s}$$ 6. **Express velocity at throat using continuity:** $$A_2 V_2 = A_1 V_1 \Rightarrow V_2 = \frac{A_1}{A_2} V_1 = \frac{0.5027}{A_2} \times 3.978$$ 7. **Use Bernoulli and pressure difference:** $$P_1 - P_2 = \rho_m g h = 13600 \times 9.81 \times 0.05 = 6664.8 \text{ Pa}$$ Bernoulli gives: $$P_1 - P_2 = \frac{1}{2} \rho (V_2^2 - V_1^2)$$ So: $$6664.8 = \frac{1}{2} \times 1000 \times (V_2^2 - 3.978^2)$$ $$V_2^2 - 15.82 = \frac{2 \times 6664.8}{1000} = 13.33$$ $$V_2^2 = 15.82 + 13.33 = 29.15$$ $$V_2 = \sqrt{29.15} = 5.4 \text{ m/s}$$ 8. **Find $A_2$ from $V_2$:** $$A_2 = \frac{Q}{V_2} = \frac{2}{5.4} = 0.3704 \text{ m}^2$$ 9. **Calculate $D_2$ from $A_2$:** $$D_2 = \sqrt{\frac{4 A_2}{\pi}} = \sqrt{\frac{4 \times 0.3704}{3.1416}} = \sqrt{0.471} = 0.686 \text{ m}$$ **Final answer:** $$\boxed{D_2 = 0.686 \text{ m}}$$