1. **State the problem:** We have a venturimeter installed in a pipeline of diameter $D_1 = 100$ mm carrying oil with specific gravity $SG = 0.87$. The maximum differential gauge reading is $h_{Hg} = 50$ cm of mercury. The flow rate is $Q = 20$ L/s, and the discharge coefficient is $C_d = 0.984$. We need to find the maximum throat diameter $D_2$ that causes full gauge deflection. 2. **Convert units and known values:** - Diameter of pipe: $D_1 = 100$ mm = 0.1 m - Flow rate: $Q = 20$ L/s = 0.02 m³/s - Mercury column height: $h_{Hg} = 50$ cm = 0.5 m - Specific gravity of oil: $SG = 0.87$ (density of oil $\rho_o = SG \times 1000 = 870$ kg/m³) - Density of mercury $\rho_{Hg} = 13600$ kg/m³ 3. **Calculate pressure difference $\Delta P$ from mercury column:** $$\Delta P = \rho_{Hg} g h_{Hg}$$ where $g = 9.81$ m/s². 4. **Calculate velocity in the pipe $V_1$ using flow rate and pipe area:** $$A_1 = \frac{\pi}{4} D_1^2 = \frac{\pi}{4} (0.1)^2 = 0.007854 \text{ m}^2$$ $$V_1 = \frac{Q}{A_1} = \frac{0.02}{0.007854} = 2.547 \text{ m/s}$$ 5. **Apply Bernoulli and continuity equations for venturimeter:** Continuity: $$A_1 V_1 = A_2 V_2$$ Pressure difference related to velocity difference: $$\Delta P = \frac{\rho_o}{2} (V_2^2 - V_1^2)$$ 6. **Express $V_2$ in terms of $A_2$ and $V_1$:** $$V_2 = \frac{A_1}{A_2} V_1$$ 7. **Substitute $V_2$ into pressure difference equation:** $$\Delta P = \frac{\rho_o}{2} \left( \left(\frac{A_1}{A_2} V_1\right)^2 - V_1^2 \right) = \frac{\rho_o V_1^2}{2} \left( \frac{A_1^2}{A_2^2} - 1 \right)$$ 8. **Solve for $\frac{A_1}{A_2}$:** $$\frac{A_1^2}{A_2^2} = 1 + \frac{2 \Delta P}{\rho_o V_1^2}$$ Calculate $\Delta P$: $$\Delta P = 13600 \times 9.81 \times 0.5 = 66708 \text{ Pa}$$ Calculate denominator: $$\rho_o V_1^2 = 870 \times (2.547)^2 = 870 \times 6.487 = 5641.7$$ Calculate ratio: $$1 + \frac{2 \times 66708}{5641.7} = 1 + \frac{133416}{5641.7} = 1 + 23.64 = 24.64$$ 9. **Calculate $\frac{A_1}{A_2}$:** $$\frac{A_1}{A_2} = \sqrt{24.64} = 4.96$$ 10. **Calculate $A_2$ and then $D_2$:** $$A_2 = \frac{A_1}{4.96} = \frac{0.007854}{4.96} = 0.001583 \text{ m}^2$$ $$D_2 = \sqrt{\frac{4 A_2}{\pi}} = \sqrt{\frac{4 \times 0.001583}{3.1416}} = \sqrt{0.002016} = 0.0449 \text{ m} = 44.9 \text{ mm}$$ 11. **Account for discharge coefficient $C_d$:** The actual flow rate is related by: $$Q = C_d A_2 V_2$$ Since $V_2 = \frac{A_1}{A_2} V_1$, the effective area is reduced by $C_d$, so adjust $A_2$: $$A_2 = \frac{A_1}{4.96 C_d} = \frac{0.007854}{4.96 \times 0.984} = 0.001609 \text{ m}^2$$ $$D_2 = \sqrt{\frac{4 \times 0.001609}{3.1416}} = \sqrt{0.002048} = 0.0453 \text{ m} = 45.3 \text{ mm}$$ **Final answer:** The maximum throat diameter $D_2$ is approximately **45.3 mm** to show full gauge deflection at 20 L/s flow rate.