Water flows at speed of 4.7 m/s through a horizontal pipe of diameter 3.3 cm .?
pressure P1 of the water in the pipe is 1.3 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.1 cm .
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 105 Pa . The
density of water is 1000 kg/m3
. The viscosity
of water is negligible.
Answer in units of atm
- NCSLv 71 month ago
p₁ + ρgh₁ + ½ρv₁² = p₂ + ρgh₂ + ½ρv₂²
and for flow continuity,
v₁*A₁ = v₂*A₂
let ₁ be the left side
4.7m/s*(3.3cm)² = v₂*(2.1cm)²
v₂ = 11.6 m/s
The ρgh terms cancel for a horizontal pipe; plug in v₂:
1.3*1.013e5Pa * ½*1000kg/m³*(4.7m/s)² =
p*1.013e5Pa * ½*1000kg/m³*(11.6m/s)²
p = 0.74 atm
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- Anonymous1 month ago
p1 = 1.3*1.013*10^5 Pa
V1 = 4.7 m/sec
V2 = V1*(A/a)^2 = 4.7*(3.3/2.1)^2 = 11.61 m/sec
p2 = ?
h1 = h2, therefore the 3rd terms in Bernoulli's equation can be omitted
p1+ρ/2*V1^2 = p2+ρ/2*V2^2
p1-p2 = 500(11.61^1-4.7^2) = 0.563*10^5
p2 = (1.3*1.013-0.563)*10^5 = 0.754*10^5 Pa (0.73 atm)