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Underground water is to be pumped by a 78 percent efficient 5-kW submerged pump to a pool whose free surface is 30 m above the underground water level. The diameter of the pipe is 7 cm on the intake side and 5 cm on the discharge side. Take the density of water to be 1 kg/L = 1000 kg/m3. Assume the elevation difference between the pump inlet and the outlet and the effect of the kinetic energy correction factors to be negligible

Respuesta :

Answer:

a. Q = 10.15 x 10⁻³ m³/s

b. Δp = 202.76 kPa

Explanation:

For solve the pipe system can use Bernoulli equations:

Patm / ρ + V₁² / 2 + g Z₁ + e pump / m = P atm / ρ + V₂² / 2 + g Z₂

e pump / ρ * Q = V₂² / 2  + g Z₂

a.

Q = V₂ * A₂

e pump / ρ * V₂ * A₂ = V₂² / 2  + g Z₂

Given: e pump = 0.78 * 5 kW = 3.9 kW, ρ = 1000 kg / m³ , A₂ = π * 5 ²cm /4 , Z₂ = 30 m

A₂ = 1.96 x 10 ⁻³ m²

V₂ = 5.18 m/s

Q = 10.15 x 10⁻³ m³/s

b.

A₁ = π * 7² cm / 4 = 3.85 x 10⁻³ m²

Δp = ρ * [ ( Q / A₁ )² / 2  -  (Q / A₂)² / 2 + e pump / ρ *Q]

Δp = 202.76 kPa