Question

A sump pump (used to drain water from the basement of houses built below the water table) is draining a flooded basement at the rate of 0.850 L/s, with an output pressure of 4.00 ✕ 105 N/m2. You may neglect frictional losses in both parts of this problem. (a) The water enters a hose with a 3.00 cm inside diameter and rises 2.90 m above the pump. What is its pressure (in N/m2) at this point? (b) The hose goes over the foundation wall, losing 1.20 m in height, and widens to 4.00 cm diameter. What is the pressure now (in N/m2)?

Answer 1

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a)

P = P₁ - P₂

= (4.00 * 10⁵) - (1000 * 9.8 * 2.9)

= 3.72 * 10⁵ N/m²

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b)

V = v₁ A₁

v₁ = 0.850 * 10^-3 / pi * 0.015² = 1.20 m/s

v₂ = 0.850 * 10^-3 / pi * 0.02² = 0.676 m/s

P₂ = P₁ + 1/2 p (v₁² - v₂²) + p g (h₁ - h₂)

= (4.00 * 10⁵) + 1/2 * 1000 (1.20² - 0.676²) + 1000 * 9.8 (0 - (2.9 - 1.2))

= 3.84 * 10⁵ N/m²

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