Question

Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 100 kPa respectively. Determine the diameter of the pipe at section (2), D, if the velocity at section (1) is 20 m/sec and viscous effects are negligible.

Answer 1

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;

$z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}$

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 + \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}$50 m + 30.704358 m + 20.4081633 m = 10.234786 m + $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

90.8777353 m = $\dfrac{v_2^2}{2 \times 9.8 \, m/s^2}$

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