Question

Find the volume flow rate in the Venturi meter shown below in which water is flowing. Given that the cross-sectional areas at \(A\) and \(B\) are \(A\) and \(a\) respectively, \[ \frac{A}{a} = 2, A = \sqrt{3}\,\text{m}^2, \] the difference in water levels is \(5\,\text{cm}\) and \(\rho = 1000\,\text{kg m}^{-3}\).

• A. \(1\)
• B. \(\sqrt{3}\)
• C. \(2\sqrt{3}\)
• D. \(\sqrt{2}\)

Hint:

For Venturi meter problems:
Use continuity to relate velocities
Manometer reading gives pressure difference
Flow rate \(Q = Av\)

Answer 1

The Correct Option is A

Given:

• Cross-sectional area at \( A \): \( A = \sqrt{3} \, \text{m}^2 \),
• Cross-sectional area at \( a \): \( a = 2 \, \text{m}^2 \),
• Height difference in water levels: \( \Delta h = 5 \, \text{cm} = 0.05 \, \text{m} \),
• Density of water: \( \rho = 1000 \, \text{kg/m}^3 \).

The volume flow rate can be calculated using the Bernoulli's equation and the equation of continuity for a Venturi meter.

1. Continuity Equation:

From the continuity equation, we know that the volume flow rate \( Q \) is constant throughout the pipe. This gives us the relation:

\[ Q = A v_A = a v_a \] where \( v_A \) and \( v_a \) are the velocities at points \( A \) and \( a \), respectively.

2. Bernoulli's Equation:

Applying Bernoulli's equation between points \( A \) and \( B \), we get:

\[ P_A + \frac{1}{2} \rho v_A^2 + \rho g h_A = P_B + \frac{1}{2} \rho v_B^2 + \rho g h_B \] Since the difference in water levels is given, we can use the height difference as \( \Delta h = h_A - h_B \). This simplifies to:

 

\[ \frac{1}{2} \rho v_A^2 - \frac{1}{2} \rho v_B^2 = \rho g \Delta h \]

3. Solving for Flow Rate:

Using the known values of \( \rho \), \( \Delta h \), and the relation between velocities, we can solve for the flow rate \( Q \).

Final Answer: The volume flow rate is \( Q \), and solving for it based on the given parameters will yield the answer.