A liquid of density 750 kg m - 3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A 1 = 1 . 2 × 10 - 2 m 2 to A 2 = A 1 2 . The pressure difference between the wide and narrow sections of the pipe is 4500 Pa . The rate of flow of liquid is _____ × 10 - 3 m 3 s - 1 .

Step-by-step solution

From the equation of continuity, v 2 = 2 v 1 . Applying the Bernoulli's Theorem, P 1 + ρ v 1 2 2 = P 2 + ρ 4 v 1 2 2 ⇒ P 1 - P 2 = 3 ρ v 1 2 2 ⇒ 4500 = 3 × 750 × v 1 2 2    ⇒ v 1 = 2   m   s - 1 Now volume flow rate, Q = A v 1 = 1 . 2 × 10 - 2 × 2 = 24 × 10 - 3   m 3   s - 1