Question

A high velocity water jet of cross section area = 0.01 m² and velocity = 35 m/s enters a pipe filled with stagnant water. The diameter of the pipe is 0.32 m. This high velocity water jet entrains additional water from the pipe and the total water leaves the pipe with a velocity 6 m/s as shown in the figure. 

The flow rate of entrained water is \(\underline{\hspace{1cm}}\) litres/s (round off to two decimal places).

Hint:

The flow rate of entrained water can be determined by subtracting the flow rate of the jet entering the pipe from the total flow rate leaving the pipe.

Answer 1

Correct Answer: 130 - 134

The total flow rate of water leaving the pipe is the product of the total velocity and the cross-sectional area of the pipe: \[ Q_{\text{total}} = A \cdot V_{\text{out}}, \] where:
- \( A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.32}{2}\right)^2 = 0.0804 \, \text{m}^2 \) is the cross-sectional area of the pipe,
- \( V_{\text{out}} = 6 \, \text{m/s} \) is the velocity of water leaving the pipe.
Thus, the total flow rate is: \[ Q_{\text{total}} = 0.0804 \times 6 = 0.4824 \, \text{m}^3/\text{s} = 482.4 \, \text{L/s}. \] The flow rate of the jet entering the pipe is: \[ Q_{\text{in}} = A_{\text{jet}} \cdot V_{\text{in}} = 0.01 \cdot 35 = 0.35 \, \text{m}^3/\text{s} = 350 \, \text{L/s}. \] The flow rate of entrained water is: \[ Q_{\text{entrained}} = Q_{\text{total}} - Q_{\text{in}} = 482.4 - 350 = 132.4 \, \text{L/s}. \] Thus, the flow rate of entrained water is: \[ \boxed{130.00 \, \text{to} \, 134.00 \, \text{L/s}}. \]