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?
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?
Solution and 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 - 3m
Radius of each hole, r = \(\frac{d}{2}\) = 0.5 × 10 - 3m
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
\(V_2 =\frac{ A_1V_1 }{ A_2}\)
\(= \frac{8 × 10 ^{- 4} × 0.025 }{ 31.61 × 10 ^{- 6} }\)
= 0.633 m / s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m / s.
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Concepts Used:
Bernauli Theorem
In fluid dynamics, Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. The principle is named after Daniel Bernoulli who published it in his book Hydrodynamica in 1738.
