Respuesta :
Answer:
The velocity at section is approximately 42.2 m/s
Explanation:
For the water flowing through the pipe, we have;
The pressure at section (1), P₁ = 300 kPa
The pressure at section (2), P₂ = 100 kPa
The diameter at section (1), D₁ = 0.1 m
The height of section (1) above section (2), D₂ = 50 m
The velocity at section (1), v₁ = 20 m/s
Let 'v₂' represent the velocity at section (2)
According to Bernoulli's equation, we have;
[tex]z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}[/tex]
Where;
ρ = The density of water = 997 kg/m³
g = The acceleration due to gravity = 9.8 m/s²
z₁ = 50 m
z₂ = The reference = 0 m
By plugging in the values, we have;
[tex]50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}[/tex]50 m + 30.704358 m + 20.4081633 m = 10.234786 m + [tex]\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}[/tex]
50 m + 30.704358 m + 20.4081633 m - 10.234786 m = [tex]\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}[/tex]
90.8777353 m = [tex]\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}[/tex]
v₂² = 2 × 9.8 m/s² × 90.8777353 m
v₂² = 1,781.20361 m²/s²
v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s
The velocity at section (2), v₂ ≈ 42.2 m/s
