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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

Water flows at speed of 4.4 m/s through a horizontal pipe of diameter 3.3 cm . The-example-1
User Atturri
by
4.1k points

2 Answers

6 votes

Final answer:

The gauge pressure in the constricted segment of a pipe is calculated using Bernoulli's equation and the principle of conservation of flow rate, converted to atmospheres.

Step-by-step explanation:

To calculate the gauge pressure in the constricted segment of the pipe, we can use Bernoulli's equation for incompressible flow, assuming negligible fluid viscosity and level horizontal flow. The equation is:

P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2,

where P is the pressure, \rho is the density of the fluid, and v is the fluid velocity.

To find the velocity in the constricted segment (v_2), we use the continuity equation which states that the flow rate must be conserved:

A_1 v_1 = A_2 v_2,

where A is the cross-sectional area. Thus:

v_2 = v_1 \frac{A_1}{A_2}.

We then solve for P_2 using the rearranged Bernoulli's equation:

P_2 = P_1 + \frac{1}{2}\rho v_1^2 - \frac{1}{2}\rho v_2^2.

Finally, we convert the pressure in pascals to atmospheres using the conversion factor (1 atm = 1.013 \u00d7 10^5 Pa). We'll calculate using the given velocities, diameters, and pressure values:

The cross-sectional areas A_1 and A_2 can be calculated using:

A = \pi \frac{d^2}{4}.

Therefore, the gauge pressure at the constricted segment in atmospheres can be determined.

User Ajmal Jamil
by
3.5k points
2 votes

Answer:

1.75 atm

Step-by-step 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

User Kokos
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4.4k points