Question

Water at a gauge pressure of 3.8 atm at street level flows into an office building at a speed of 0.62 m/s through a pipe 5.3 cm in diameter. The pipe tapers down to 3.0 cm in diameter by the top floor, 18 m above, where the faucet has been left open. Calculate the flow velocity and the gauge pressure in such a pipe on the top floor. Assume no branch pipes and ignore viscosity.

Answer 1

Answer:

  P₂ = 206945 Pa

Explanation:

given,

gauge pressure = 3.8 atm

1 atm = 101325 Pa

speed of flow,v₁ = 0.62 m/s

diameter of pipe,d₁ = 5.3 cm

diameter of pipe after tapering, d₂ = 3 cm

height above the ground = 18 m

Using equation of continuity

  A₁v₁ = A₂ v₂

$\dfrac{\pi}{4}d_1^2 v_1 = \dfrac{\pi}{4}d_2^2 v_2$

$v_2 = v_1(\dfrac{d_1^2}{d_2^2})$

$v_2 =0.62\times (\dfrac{5.3^2}{3^2})$

    v₂ = 1.94 m/s

using Bernoulli's equation

$P_1 + \dfrac{\rho v_1^2}{2} = P_2 + \dfrac{\rho v_2^2}{2} +\rho g h$

$P_2= P_1 + \dfrac{\rho}{2}(v_1^2 - v_2^2) -\rho g h$

$P_2=3.8\times 101325 + \dfrac{1000}{2}(0.62^2 - 1.94^2) -1000\times 9.8\times 18$

  P₂ = 206945 Pa

hence, pressure of the pipe at top floor is equal to   P₂ = 206945 Pa