Respuesta :
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:
[tex]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} }[/tex]
Where
v₁ = 0.65 m/s
d₁ = 3 cm = 0.03 m
d₂ = 0.3 cm = 0.003 m
[tex]v_{2} =\frac{0.03^{2}*0.65 }{0.003^{2} } =65m/s[/tex]
b) The pressure at the pump outlet is:
[tex]\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[/tex]
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
