Water of density 1000 kg/m ^3 flows through a horizontal pipe having a wide section of cross-sectional area 40 cm ^2 and a narrow constriction of cross-sectional area 20 cm ^2 . If the pressure difference between the wide section and the narrow constriction is 6000 Pa , the volume flow rate of water in the pipe is:

Step-by-step solution

Let the velocity of water in the wide section be v₁ and in the narrow constriction be v₂ . Using the equation of continuity: A₁ v₁ = A₂ v₂ (40 10⁻⁴) v₁ = (20 10⁻⁴) v₂ v₂ = 2 v₁ Applying Bernoulli's theorem for horizontal flow: P₁ + 1 2 v₁^2 = P₂ + 1 2 v₂^2 P₁ - P₂ = 1 2 (v₂^2 - v₁^2) Substitute the given values P = 6000 Pa , = 1000 kg/m ^3 , and v₂ = 2 v₁ : 6000 = 1 2 1000 ((2v₁)^2 - v₁^2) 6000 = 500 (4v₁^2 - v₁^2) 6000 = 500 3v₁^2 1500 v₁^2 = 6000 v₁^2 = 4 v₁ = 2 m/s The volume flow rate Q is given by: Q = A₁ v₁ Q