Question

Water is flowing in the pipe shown in the figure below, with the 7.70-cm diameter at point 1 tapering to 3.45 cm at point 2, located y = 11.5 cm below point 1. The water pressure at point 1 is 3.20 ✕ 104 Pa and decreases by 50% at point 2. Assume steady, ideal flow. What is the speed of the water at the following points? Point 1? Point 2?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of water = = 1000 kg/m³

Pressure of water at point 1 = P₁ = 3.2 x 10⁴ Pa

Diameter of the pipe at point 1 = D₁ = 7.7 cm = 0.077 m

Cross-sectional area of the pipe at point 1 = A₁ = D₁²/4

Speed of flow of water at point 1 = V₁

Height of point 1 compared to point 2 = H = 11.5 cm = 0.115 m

Pressure of water at point 2 = P₂ = P₁/2 = (3.2x10⁴)/2 Pa = 1.6 x 10⁴ Pa

Diameter of the pipe at point 2 = D₂ = 3.45 cm = 0.0345 m

Cross-sectional area of the pipe at point 2 = A₂ = D₂²/4

Speed of flow of water at point 2 = V₂

By continuity equation,

A₁V₁ = A₂V₂

(D₁²/4)V₁ = (D₂²/4)V₂

D₁²V₁ = D₂²V₂

(0.077)²V₁ = (0.0345)²V₂

V₂ = 4.981V₁

By bernoulli's equation,

11905.18V₁² = 17128.15

V₁ = 1.2 m/s

V₂ = 4.981V₁

V₂ = (4.981)(1.2)

V₂ = 5.98 m/s

A) Speed of flow of water at point 1 = 1.2 m/s

B) Speed of flow of water at point 2 = 5.98 m/s