Question

The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5m min -1. What is the speed of ejection of the liquid through the holes?

Answer 1

Given

• Cross-sectional area of tube: \[ A_1 = 8.0 \,\text{cm}^2 = 8.0 \times 10^{-4} \,\text{m}^2 \]
• Number of holes: \[ N = 40 \]
• Diameter of each hole: \[ d = 1.0 \,\text{mm} = 1.0 \times 10^{-3} \,\text{m} \]
• Radius of each hole: \[ r = \frac{d}{2} = 0.5 \times 10^{-3} \,\text{m} \]
• Speed of liquid in tube: \[ v_1 = 1.5 \,\text{m min}^{-1} = \frac{1.5}{60} \,\text{m s}^{-1} = 0.025 \,\text{m s}^{-1} \]

1. Total Area of All Holes

Area of one hole: \[ a = \pi r^{2} = \pi (0.5 \times 10^{-3})^{2} = \pi \times 0.25 \times 10^{-6} = 0.25\pi \times 10^{-6} \,\text{m}^2 \] Total area of 40 holes: \[ A_2 = N a = 40 \times 0.25\pi \times 10^{-6} = 10\pi \times 10^{-6} \approx 3.14 \times 10^{-5} \,\text{m}^2 \]

2. Use Continuity Equation (Incompressible Flow)

\[ A_1 v_1 = A_2 v_2 \] where \( v_2 \) = speed of ejection through the holes. Hence: \[ v_2 = \frac{A_1 v_1}{A_2} = \frac{8.0 \times 10^{-4} \times 0.025}{3.14 \times 10^{-5}} \] Calculate numerator: \[ 8.0 \times 10^{-4} \times 0.025 = 2.0 \times 10^{-5} \] So: \[ v_2 = \frac{2.0 \times 10^{-5}}{3.14 \times 10^{-5}} \approx 0.64 \,\text{m s}^{-1} \]

Final Answer

The speed of ejection of the liquid through the holes is approximately \[ \boxed{v_2 \approx 0.63 \,\text{m s}^{-1}}. \]