Question

Water is flowing with a flow rate \( Q \) in a horizontal circular pipe. Due to the low pressure created at the venturi section (Section-1 in the figure), water from a reservoir is drawn upward using a connecting pipe as shown in the figure. Take acceleration due to gravity \( g = 10 \, \text{m/s}^2 \). The flow rate \( Q = 0.1 \, \text{m}^3/\text{s} \), \( D_1 = 8 \, \text{cm} \), and \( D_2 = 20 \, \text{cm} \). The maximum height \( h \) (in meters, up to one decimal place) of the venturi from the reservoir just sufficient to raise the liquid up to Section-1 is \(\underline{\hspace{1cm}}\). \includegraphics[width=0.75\linewidth]{image22.png}

Hint:

To solve for the height difference in a venturi, use the Bernoulli equation along with the continuity equation to relate the velocities and pressures at different sections of the pipe.

Answer 1

Correct Answer: 18.5

We will use the Bernoulli equation and the continuity equation to solve the problem. The Bernoulli equation between Section-1 and Section-2 is given by:
\[ \frac{v_1^2}{2} + P_1 + \rho g h = \frac{v_2^2}{2} + P_2 \] Where:
- \( v_1 \) and \( v_2 \) are the velocities at Sections 1 and 2, respectively,
- \( P_1 \) and \( P_2 \) are the pressures at Sections 1 and 2, respectively,
- \( h \) is the height difference,
- \( \rho \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)).
Since the flow is steady, we can use the continuity equation to find the velocities at Sections 1 and 2:
\[ Q = A_1 v_1 = A_2 v_2 \] Where:
- \( A_1 = \pi D_1^2 / 4 \) and \( A_2 = \pi D_2^2 / 4 \) are the cross-sectional areas of Sections 1 and 2, respectively.
Substitute \( v_1 = \frac{Q}{A_1} \) and \( v_2 = \frac{Q}{A_2} \) into the Bernoulli equation. For the venturi, \( P_1 = P_2 \) (since both are at atmospheric pressure), so the equation simplifies to:
\[ \frac{v_2^2}{2} - \frac{v_1^2}{2} = \rho g h \] Substituting the values for \( v_1 \) and \( v_2 \), we get:
\[ \frac{\left( \frac{Q}{A_2} \right)^2}{2} - \frac{\left( \frac{Q}{A_1} \right)^2}{2} = \rho g h \] Now, substitute the values \( Q = 0.1 \, \text{m}^3/\text{s} \), \( A_1 = \pi \times (0.08)^2 / 4 \), and \( A_2 = \pi \times (0.20)^2 / 4 \) into the equation. After simplifying, we get:
\[ h \approx 19.5 \, \text{m}. \] Thus, the maximum height \( h \) of the venturi from the reservoir is approximately \( 19.5 \, \text{m} \).