Question

10. Water at a pressure of 3.60 atm at street level flows into an office building at a speed of 0.60 m/s through a pipe 4.40 cm in diameter. The pipes taper down to 1.50 cm in diameter by the top floor, 23.0 m above. Calculate the water pressure in such a pipe on the top floor.

Answer 1

At the street level,(1),

P₁ = 3.60atm = 3.60 x (1.01 x10⁵ N/m²) = 3.636 x 10⁵ N/m² = 363600 N/m²

h₁ =0

Area of the pipe(A₁) =R₁² = (2.2)² cm² =4.84 cm²

v₁ =0.60m/s

of water = 1000kg/m³

At the top floor,(2),

P₂ =P₂(let)

h₂=23m

Area of the pipe(A₂) = R₂² =(0.75) cm² = 0.5625 cm²

To find v₂ at this level, we can use A₁v₁ = A₂v₂

Putting the values, (4.84)(0.60) = (0.5625)v₂

v₂ = 5.163 m/s

Applying Bernoulli's equation at (1) and (2),

P₁ + gh₁ + 1/2v₁² = P₂ + gh₂ + 1/2v₂²

Putting the values,

363600 N/m²​​​​​​​ + 0 + 1/2(1000kg/m³)(0.60m/s)² = P₂ + (1000kgm³)(9.8m/s²)(23m) + 1/2(1000kg/m³)(5.163m/s)²

P₂ = 125051.72 N/m² or 1.238 atm.