# Water flows at speed of 4.7 m/s through a horizontal pipe of diameter 3.3 cm .?

The gauge

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

### 2 Answers

- NCSLv 71 month ago
Bernoulli:

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

solves to

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)

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