Question:

A Venturi meter with a throat diameter \( d = 2 \, \text{cm} \) measures the flow rate in a pipe of diameter \( D = 6 \, \text{cm} \), as shown in the figure. A U-tube manometer is connected to measure the pressure drop. Assume the discharge coefficient is independent of the Reynolds number and geometric ratios. If the volumetric flow rate through the pipe is doubled \( Q_2 = 2Q_1 \), the corresponding ratio of the manometer readings \( \Delta h_2 / \Delta h_1 \), rounded off to the nearest integer, is: \includegraphics[width=0.5\linewidth]{q60 CE.PNG}

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In Venturi meter problems, remember that the pressure difference is proportional to the square of the flow rate. Use this relationship to calculate changes in pressure or manometer readings.
Updated On: Jan 24, 2025
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Solution and Explanation

Step 1: Relation between flow rate and pressure difference. The flow rate through a Venturi meter is given by: \[ Q = C_d A \sqrt{\frac{2 \Delta P}{\rho}}, \] where: - \( C_d \) is the discharge coefficient, - \( A \) is the cross-sectional area, - \( \Delta P \) is the pressure drop, - \( \rho \) is the fluid density. The pressure drop \( \Delta P \) is proportional to \( Q^2 \): \[ \Delta P \propto Q^2 \quad \Rightarrow \quad \Delta h \propto Q^2, \] where \( \Delta h \) is the manometer reading. Step 2: Ratio of manometer readings. If the flow rate is doubled \( Q_2 = 2Q_1 \): \[ \frac{\Delta h_2}{\Delta h_1} = \left( \frac{Q_2}{Q_1} \right)^2 = \left( \frac{2Q_1}{Q_1} \right)^2 = 2^2 = 4. \] Step 3: Conclusion. The ratio of manometer readings \( \Delta h_2 / \Delta h_1 \) is \( 4 \).
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