# The cylindrical t

The law of continuity states that,

A1v1 = A2v2

v2 = A1v1 / A2 ……….(i)

Where,

‘A1’ is the area of cross section of the spray pump

‘A2’ is the area of cross section of the end with the 40 holes

‘v1’is the speed of flow of fluid inside the tube

‘v2’ is the speed of flow of the fluid through the holes

Given,

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

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

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

Velocity of the liquid flow inside the tube, v1 = 1.5 m min-1

v1 = 1.5/60 m sec-1 = 0.025 m/s

Thus,

Area of cross section of each hole, a = πr2

a = 3.14 × (0.5 × 10-3 m)2

a = 0.785 × 10-6 m2

Total area of cross section of 40 holes, A2 = 40 × a

A2 = 40 × 0.785 × 10-6 m2

A2 = 31.41 × 10-6 m2

Therefore, from equation (i),

v1 =

0.633 m/s

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

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