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27 April, 19:06

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 2.02 inch. The lower end of the hose (15 m below the nozzle) is connected to the pump outlet of diameter 3.37 inch. The gauge pressure of the water at the pump is P (gauge) pump = P (abs) pump - Patm = 65.2 PSI = 449.538 kPa. Calculate the speed of the water jet emerging from the nozzle. Assume that water is an incompressible liquid of density 1000 kg/m3 and negligible viscosity. The acceleration of gravity is 9.8 m/s 2. Answer in units of m/s.

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  1. 27 April, 19:32
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    v₁ = 1,606 10⁴ m / s

    Explanation:

    For this exercise we must use Bernoulli's equation, let's use index 1 for the nozzle on the stairs and index 2 the pump on the street

    P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

    The pressure when the water comes out is the atmospheric pressure

    P₁ = P_atm

    The difference in height between the street and the nozzle on the stairs is

    y₂-y₁ = 15 m

    Now let's use the continuity equation

    v₁ A₁ = v₂ A₁

    The area of a circle is

    A = π r² = π (d/2) ²

    v₁ π d₁² / 4 = v₂ π d₂² / 4

    v₂ = v₁ d₁² / d₂²

    Let's replace

    P₂-P_atm + ½ ρ [ (v₁ d₁² / d₂²) ² - v₁² ] + ρ g (y2-y1) = 0

    P₂ - P_Atm + ρ g (y₂-y₁) = ½ ρ v₁² [1 - (d₁/d₂) ⁴]

    v₁² [1 - (d₁/d₂) ⁴] = (P₂-P_atm) ρ / 2 + g (y₂-y₁) / 2

    Let's reduce the magnitudes to the SI system

    d₂ = 3.37 in (2.54 10⁻² m / 1 in) = 8.56 10⁻² m

    d₁ = 2.02 in = 5.13 10⁻² m

    Let's calculate

    v₁² [1 - (5.13 / 8.56) 4] = 449.538 10³ 10³/2 - 9.8 15/2

    v₁² [0.8710] = 2.2477 10⁸ - 73.5 = 2.2477 10⁸

    v₁ = √ 2.2477 10⁸ / 0.8710

    v₁ = 1,606 10⁴ m / s
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