Question

Constants Periodic Table Suppose the top surface of the vessel in the figure (Figure 1) is subjected to an external gauge pressure P2. Derive a formula for the speed, vi, at which the liquid flows from the opening at the bottom into atmospheric pressure, Po. Assume the velocity of the liquid surface, V2, is approximately zero. Express your answer in terms of the variables P2, 41, 42,P, and appropriate constants. v1 = v2/?? + g(12 â y1)] Submit Previous Answers â Correct Figure < 1 of 1 > Part B If P2 = 0.86 atm and Y2 - Y1 = 2.6 m , determine V1 for water. Express your answer using two significant figures. y2-yi mo ÎΣΦ ?

Answer 1

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