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;
![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}](https://tex.z-dn.net/?f=z_1%20%2B%20%5Cdfrac%7BP_1%7D%7B%5Crho%20%5Ccdot%20g%7D%20%2B%20%5Cdfrac%7Bv%5E2_1%7D%7B2%20%5Ccdot%20g%7D%20%3D%20z_2%20%2B%20%5Cdfrac%7BP_2%7D%7B%5Crho%20%5Ccdot%20g%7D%20%2B%20%5Cdfrac%7Bv%5E2_2%7D%7B2%20%5Ccdot%20g%7D)
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;
50 m + 30.704358 m + 20.4081633 m = 10.234786 m + ![\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_2%5E2%7D%7B2%20%5Ctimes%209.8%20%5C%2C%20m%2Fs%5E2%7D)
50 m + 30.704358 m + 20.4081633 m - 10.234786 m = ![\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_2%5E2%7D%7B2%20%5Ctimes%209.8%20%5C%2C%20m%2Fs%5E2%7D)
90.8777353 m = ![\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}](https://tex.z-dn.net/?f=%5Cdfrac%7Bv_2%5E2%7D%7B2%20%5Ctimes%209.8%20%5C%2C%20m%2Fs%5E2%7D)
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