A horizontal Venturimeter carries water. The inlet and throat cross-sectional areas are 4 cm² and 2 cm², respectively. A differential mercury manometer connected between the inlet and throat shows a mercury-level difference of 3.0 cm. Take ρ(Hg) = 13600 kg/m³, ρ(water) = 1000 kg/m³ and g = 10 m/s². What is the volume-flow rate through the Venturimeter?
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Question

A horizontal Venturimeter carries water. The inlet and throat cross-sectional areas are 4 cm² and 2 cm², respectively. A differential mercury manometer connected between the inlet and throat shows a mercury-level difference of 3.0 cm. Take ρ(Hg) = 13600 kg/m³, ρ(water) = 1000 kg/m³ and g = 10 m/s². What is the volume-flow rate through the Venturimeter?

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G1 Physics · Accepted Answer

First find the pressure difference from the differential manometer. Because both tapping points contain water and lie at the same level, P₁−P₂ = [ρ(Hg)−ρ(water)]gh. Thus, P₁−P₂ = (13600−1000)×10×0.03 = 3780 Pa. From continuity, A₁v₁ = A₂v₂. Since A₁/A₂ = 4/2 = 2, we get v₂ = 2v₁. For a horizontal Venturimeter, Bernoulli's equation gives P₁−P₂ = ½ρ(v₂²−v₁²). Therefore, 3780 = ½×1000×[(2v₁)²−v₁²] = 1500v₁². Hence, v₁² = 2.52, so v₁ ≈ 1.587 m/s. The volume-flow rate is Q = A₁v₁. Now, A₁ = 4 cm² =…