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?

Answer 1

 Total cross-sectional area of 40 holes, `a_2`

`= 40 xx 22/7 xx (1xx10^(-3))^2/4 m^2`

`= 22/7 xx 10^(-5) m^2`

Cross-sectional area of tube, `a_1 = 8 xx 10^(-4) m^2`

Speed inside the tube,. `v_1 = 1.5 m min^(-1) = 1.5/60 ms^(-1)`

Speed of ejection, `v_2 = ?`

Using `a_2v_2 = a_1v_1`

We get

` v_2 = (a_1v_1)/a_2 = (8xx10^(-4) xx (1.5/60) xx 7)/(22xx10^(-5)) ms^(-1) = 0.64 ms^(-1)`

Answer 2

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:

`A_1V_1 = A_2V_2`

`V_2 = (A_1V_1)/A_2`

`= (8xx10^(-4)xx0.025)/(31.61xx10^(-6))`

`= 0.633 m/s`

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