Question

Water at position 1 in a pipe, driven by a pressure P1 = 950 kPa flows at vi = 12.00 m/s. Position 1 is 6.00 m above the ground; i.e., yı = 6.00 m. Position 2 is15.0 m above the ground; i.e., y2 = 15.0 m. At position 2, the pressure is P2 = 167 kPa. Calculate the position 2 flow rate v2 in m/s. The density of water is given as p = 997 kg/m²

Answer 1

P₁ = pressure at position 1 = 950 kPa

v₁ = speed at position 1 = 12 m/s

h₁ = height at position 1 = y₁ = 6 m

P₂ = pressure at position 2 = 167 kPa

v₂ = speed at position 2 = ?

h₂ = height at position 2 = y₂ = 15 m

Using bernoulli's equation

P₁ + $\rho$ gh₁ + (0.5) $\rho$ v₁² = P₂ + $\rho$ gh₂ + (0.5) $\rho$ v₂²

950,000 + (1000) (9.8) (6) + (0.5) (1000) (12)² = 167,000 + (1000) (9.8) (15) + (0.5) (1000) v₂²

v₂ = 39.2 m/s