Answer:
3.14 × 10⁻⁴ m³ /s
Explanation:
The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.
Q = Area x velocity
Given:
Diameters of 3 sections of the pipe are given as
d1 = 1.0 cm, d2 = 2.0 cm and d3 = 0.5 cm.
Speed in the first segment of the pipe is
v1 = 4 m/s.
From the equation of continuity the flow rate through different cross-sections remains the same.
Flow rate = Q = A1 v1 = A2 v2 = A3 v3.
Q = A1v1
=π/4 d²1 v1 = π/4 * 0.01² ×4.0 m³/s = 3.14 × 10⁻⁴ m³ /s
