A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

1 answer
Question:

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:0.0072 m³/sExplanation:Using Bernoulli's law P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal 1/2ρv₂² - 1/2ρv₁² = P₁ - P₂ flow rate is constant A₁v₁ = A₂v₂A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²A₂ = πr₂² = π (0.0225)² = 0.00159 m²v₁  = (A₂ / A₁)v₂v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂ substitute v₁  into the Bernoulli's equation1/2ρv₂² - 1/2ρv₁² = P₁ - P₂ 500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa341.924 v₂² = 7000v₂² = 20.472v₂ = √ 20.472 = 4.525 m/svolume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

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