Question

The cylindrical tube of a spray pump has a cross-section of $8.0c{m}^2$ one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is $1.5mmi{n}^{–1}$, what is the speed of ejection of the liquid through the holes?

Answer 1

Area of cross-section of the spray pump, $A_1=8c{m}^2=8\times {10}^{–4}{m}^2$
Number of holes, n = 40
Diameter of each hole, d = 1 mm = $1\times {10}^{–3}m$
Radius of each hole, $r=\frac{d}{2}=0.5\times {10}^{–3}m$
Area of cross-section of each hole, $a=\pi r^2=\pi (0.5\times {10}^{-3}{)}^2{m}^2$
Total area of 40 holes, $A_2=n\times a=40\times \pi (0.5\times {10}^{-3}{)}^2{m}^2=31.41\times {10}^{-6}{m}^2$
Speed of flow of liquid inside the tube, $V_1=1.5$ m/min = 0.025 m/s
Speed of ejection of liquid through the holes = $V_2$
According to the law of continuity, we have:
$A_1{V}_1=A_2{V}_2{V}_2=\frac{A_1{V}_1}{A_2}=\frac{8\times {10}^{-4}\times 0.025}{31.61\times {10}^{-6}}=0.633m/s$
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.