Question

Water flows at 0.65 m/s through a 3 cm diameter hose that terminates in a 0.3 cm diameter nozzle. At what speed does the water pass through the nozzle? If the pump at one end of the hose and the nozzle at the other end are at the same height, and if the pressure at the nozzle is 1 atm, what is the pressure at the pump outlet?

Answer 1

Answer:

P2 = 5.18 10⁴ Pa

Explanation:

This is a fluid mechanics exercise where underneath use Bernoulli's equation

Let's use subscript 1 for the thick part of the hose and subscript 2 for the thin part

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

Since the system is at the same height y₁ = y₂

Now we must use the continuity equation to find the muzzle velocity (v₂)

          v₁ A₁ = v₂ A₂

          v₂ = v₁ A₁ / A₂

The nozzles are circular, so the area of ​​a circle is

         A₁ = π r₁² = π d₁² / 4

         A₂ = π d₂² / 4

         v₂ = v₁ d₁² / d₂²

We substitute in the first equation

          P₁ + ½ ρ  v₁² = P₂ + ½ ρ  v₁ d₁² / d₂²²

We clear the pressure P₂

         P₂ = P₁ + ½ ρ  v₁² (1 - d₁² / d₂²)

Let's reduce the magnitudes to the SI system

      d₁ = 3 cm = 0.03 m

      d₂ = 0.3 cm = 0.003 m

      P₁ = 1 atm = 1,013 10⁵ Pa

Let's calculate

        P2 = 1,013 10⁵ + ½ 1 10³ 0.65 (1 - (3 / 0.3)²)

        P2 = 1,013 10⁵ - 4.95 10⁴

        P2 = 5.18 10⁴ Pa

Answer 2

Answer:

a) The speed is 65 m/s

b) The pressure at the pump outlet is 21.85 atm

Explanation:

a) The speed the water pass through the nozzle is:

$A_{1} v_{1} =A_{2} v_{2}\\A_{1} =2\pi d_{1} ^{2}/2 \\A_{2} =2\pi d_{2} ^{2} \\Clearing v_{2} \\v_{2} =\frac{d_{1}^{2}v_{1} }{d_{2}^{2} }$

Where

v₁ = 0.65 m/s

d₁ = 3 cm = 0.03 m

d₂ = 0.3 cm = 0.003 m

$v_{2} =\frac{0.03^{2}*0.65 }{0.003^{2} } =65m/s$

b) The pressure at the pump outlet is:

$\frac{1}{2}\rho v^{2} +P=\frac{1}{2} \rho u^{2} +P_{2} \\ClearingP_{2} \\P_{2} =2.214x10^{6} Pa=21.85atm$