A horizontal Venturimeter carries water. The inlet and throat cross-sectional areas are 6 cm² and 3 cm², respectively. The inlet pressure exceeds the throat pressure by 6 kPa. Neglect losses. If the volume-flow rate is N × 10⁻⁴ m³/s, find N.
Physics · Fluid Mechanics · equation of continuity and bornullie theorem
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Question

A horizontal Venturimeter carries water. The inlet and throat cross-sectional areas are 6 cm² and 3 cm², respectively. The inlet pressure exceeds the throat pressure by 6 kPa. Neglect losses. If the volume-flow rate is N × 10⁻⁴ m³/s, find N.

Physics · Fluid Mechanics · equation of continuity and bornullie theorem

G1 Physics · Accepted Answer

From continuity, A₁v₁ = A₂v₂. Since A₁/A₂ = 6/3 = 2, we have v₂ = 2v₁. For a horizontal Venturimeter, P₁−P₂ = ½ρ(v₂²−v₁²). For water, ρ = 1000 kg/m³. Thus, 6000 = ½×1000×(4v₁²−v₁²) = 1500v₁². Therefore, v₁² = 4, so v₁ = 2 m/s. Now, Q = A₁v₁ = 6×10⁻⁴×2 = 12×10⁻⁴ m³/s. Hence, N = 12.