This problem involves calculating the rate of flow of glycerine through a conical pipe section using the Bernoulli's principle and the given pressure drop. We are given the density of glycerine \( \rho = 1.25 \times 10^3 \, \text{kg/m}^3 \), the radii at both ends of the conical section, and the pressure drop along its length.
- The radii of the ends of the conical pipe are given as \( r_1 = 0.1 \, \text{m} \) and \( r_2 = 0.04 \, \text{m} \).
- The pressure drop across the pipe is given as \( \Delta P = 10 \, \text{N/m}^2 \).
- We can use the principle of conservation of energy (Bernoulli's equation) which states:
\[P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\]- where \( v_1 \) and \( v_2 \) are the velocities at the larger and smaller ends of the pipe respectively.
- The volume flow rate \( Q \) is given by:
\[Q = A_1 v_1 = A_2 v_2\]- where \( A_1 = \pi r_1^2 \) and \( A_2 = \pi r_2^2 \) are the cross-sectional areas at each end.
- Using the equation for pressure drop:
\[\Delta P = \frac{1}{2} \rho (v_2^2 - v_1^2)\]- and rearranging for velocity we find:
\[v_2^2 = v_1^2 + \frac{2 \Delta P}{\rho}\]- Solving for \( v_1 \) in terms of the known quantities:
\[v_1 = \frac{\Delta P}{\frac{1}{2} \rho \left(1 - \left(\frac{A_1}{A_2}\right)^2\right)}\]- Substituting the known values:
\[A_1 = \pi (0.1)^2 = 0.01 \pi \, \text{m}^2, \quad A_2 = \pi (0.04)^2 = 0.0016 \pi \, \text{m}^2\]-
\[v_1 = \sqrt{\frac{2 \cdot 10}{1.25 \times 10^3 \left(1 - \left(\frac{0.01}{0.0016}\right)^2\right)}}\]- Calculate \(v_1\) and substitute back to calculate \(Q = A_1 v_1\).
- After calculation, the volume flow rate \( Q \approx 6.93 \times 10^{-4} \, \text{m}^3/\text{s} \).
Thus, the rate of flow of glycerine is Option 1: \(6.93 \times 10^{-4} \, \text{m}^3/\text{s}\).