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 1.5 m min–1, what is the speed of ejection of the liquid through the holes?
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Answer 1

Explanation:

Area of cross-section of the spray pump, A1 = 8 cm2 = 8 × 10–4 m2

Number of holes, n = 40

Diameter of each hole, d = 1 mm = 1 × 10–3 m

Radius of each hole, r = d/2 = 0.5 × 10–3 m

Area of cross-section of each hole, a = πr2 = π (0.5 × 10–3)2 m2

Total area of 40 holes, A2 = n × a

= 40 × π (0.5 × 10–3)2 m2

= 31.41 × 10–6 m2

Speed of flow of liquid inside the tube, V1 = 1.5 m/min = 0.025 m/s

Speed of ejection of liquid through the holes = V2

According to the law of continuity, we have:

A1V1 = A2V2

V2 = A1V1 / A2

= 8 × 10-4 × 0.025 / (31.61 × 10-6)

= 0.633 m/s

Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.