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A fireman standing on a 15 m high ladder
operates a water hose with a round nozzle of
diameter 1.99 inch. The lower end of the hose
(15 m below the nozzle) is connected to the
pump outlet of diameter 3.28 inch. The gauge
pressure of the water at the pump is (IMAGE)
Calculate the speed of the water jet emerging from the nozzle. Assume that water is an
incompressible liquid of density 1000 kg/m^3
and negligible viscosity. The acceleration of
gravity is 9.8 m/s^2
Answer in units of m/s.

A fireman standing on a 15 m high ladder operates a water hose with a round nozzle of diameter 199 inch The lower end of the hose 15 m below the nozzle is conne class=

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MathPhys

Answer:

31.6 m/s

Explanation:

Mass is conserved, so the mass flow at the outlet of the pump equals the mass flow at the nozzle.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

Writing v₁ in terms of v₂:

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

Plugging in values:

579,160 Pa − (1000 kg/m³)(9.8 m/s²)(15 m) = ½ (1000 kg/m³) (1 − (1.99 in / 3.28 in)⁴) v₂²

v₂ = 31.6 m/s

The speed of the water jet emerging from the nozzle is 31.6 m/s

Equation of continuity:

Since the mass of fluid coming in must be equal to the mass of fluid going out. The concept of the equation of continuity is introduced which is as  below:

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Bernoulli's principle:

Now using the Bernoulli's equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

here, h₁ = 0 and P₂ = 0:

P₁ + ½ ρ v₁² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (v₂ d₂²/d₁²)² = ½ ρ v₂² + ρgh₂

P₁ + ½ ρ (d₂/d₁)⁴ v₂² = ½ ρ v₂² + ρgh₂

P₁ − ρgh₂ = ½ ρ (1 − (d₂/d₁)⁴) v₂²

579,160  − (1000)(9.8)(15 ) = ½ (1000) (1 − (1.99/3.28)⁴) v₂²

v₂ = 31.6 m/s will be the of water jet.

Learn more about  Bernoulli's principle:

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