Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), thereis no change in height. Use the Bernoulli equation to calculate the velocity of thewater exiting the nozzle. (Hint: The density of water is 1000 kg/m3 and gravity g is9.8 m/s2. Pay attention to units!)]a.12 m/sb.13 m/sc.7m/sd. 14m/s
Question
Water is flowing in a fire hose with a velocity of 1.0 m/s and a pressure of 200000Pa. At the nozzle the pressure decreases to atmospheric pressure (101300 Pa), thereis no change in height. Use the Bernoulli equation to calculate the velocity of thewater exiting the nozzle. (Hint: The density of water is 1000 kg/m3 and gravity g is9.8 m/s2. Pay attention to units!)]a.12 m/sb.13 m/sc.7m/sd. 14m/s
Solution
The Bernoulli equation is a statement of the conservation of energy principle for flowing fluids. It accounts for gravitational potential energy, kinetic energy, and fluid pressure. The equation is:
P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂
where: P₁ and P₂ are the pressures at the two points, ρ is the fluid density, v₁ and v₂ are the velocities at the two points, h₁ and h₂ are the heights above some reference level at the two points, and g is the acceleration due to gravity.
Given in the problem: P₁ = 200000 Pa, P₂ = 101300 Pa, ρ = 1000 kg/m³, v₁ = 1.0 m/s, h₁ = h₂ (so ρgh₁ and ρgh₂ will cancel out), and g = 9.8 m/s².
We are asked to find v₂.
Substituting the given values into the Bernoulli equation and solving for v₂, we get:
200000 Pa + ½(1000 kg/m³)(1.0 m/s)² = 101300 Pa + ½(1000 kg/m³)v₂²
Solving the above equation for v₂ gives:
v₂ = sqrt[(2/ρ) * (P₁ - P₂ + ½ρv₁²)] = sqrt[(2/1000 kg/m³) * (200000 Pa - 101300 Pa + ½(1000 kg/m³)(1.0 m/s)²)] = sqrt[(2/1000 kg/m³) * (98600 Pa)] = sqrt[(2/1000 kg/m³) * (
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