Answer:
a) v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]
b) Water does not flow,
Explanation:
a) For this exercise we will use Bernoulli's equation, where index 1 is at the exit and index 2 on the surface of the water
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
This case does not indicate at the surface pressure is P₂, the pressure at the outlet is P₁ = P₀, the surface velocity is zero v₂ = 0
P₀ + ½ ρ v₁² + ρ g y₁ = P₂ + 0 + ρ g y₂
½ ρ v₁² = P₂-P₀ + ρ (y₂ -y₁)
v₁² = 2 (P₂-P₀) /ρ + 2 (y₂ -y₁)
v₁ = √ [2 (P₂-P₀) /ρ + 2 (y₂ -y₁)]
b) Reduce the pressure to SI units
P₂ = 0.86 atm (1.01 10⁵ Pa / 1 atm) = 0.8686 10⁵ Pa
P₀ = 1.01 10⁵ Pa
ρ = 1 10³ kg / m³
Let's calculate
v₁ = √ [2 (0.8686 -1.01) 10⁵/10³ + 2 (2.6)]
v₁ = √ [-0.2828 10² + 5.2] = Ra [-23]
Water does not flow, this is because the pressure of the inner part is less than atmospheric, so that the water flows the pressure P₂> = P₀
For example if the pressure P₂ = P₀
v₁ = √ 5.2
v₁ = 2.28 m / s
