Question

Water flows at speed of 4.4 m/s through a
horizontal pipe of diameter 3.3 cm . The gauge
pressure P1 of the water in the pipe is 2 atm .
A short segment of the pipe is constricted to
a smaller diameter of 2.4 cm
(IMAGE)
What is the gauge pressure of the water
flowing through the constricted segment? Atmospheric pressure is 1.013 × 10^5 Pa . The density of water is 1000 kg/m^3
. The viscosity
of water is negligible.
Answer in units of atm

Answer 1

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

Answer 2

Following are the calculation to the pressure:

$\to A_1\ v_1 = A_2\ v_2$

$\to 4.4 \times 3.3^2 = 2.4^2 \times v_2\\\\\to v_2= \frac{4.4 \times 3.3^2}{2.4^2}$

         $= \frac{4.4 \times 10.89}{5.76}\\\\= \frac{47.916}{5.76}\\\\=8.31875\\\\= 8.31\ \frac{m}{s}$  

Using the bernoulli principle:

$\to P_1 + 0.5\pho v_1^2 = P_2 + 0.5\rho v_2^2$

$\to 2 \times 1.01 \times 10^5 + 0.5 \times 1000 \times 4.4^2 = P_2 + 0.5 \times 1000 \times 8.31^2\\\\\to 2.02 \times 10^5 + 0.5 \times 10^3 \times 19.36 = P_2 + 0.5 \times 1000 \times 69.0561\\\\\to 211680 = P_2 + 34528.05\\\\\to P_2= 211680 -34528.05\\\\\to P_2 = -177151.95\\\\\to P_2 = 177151.95\ atm$

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