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Talja [164]
3 years ago
14

An incompressible fluid flows steadily through a pipe that has a change in diameter. The fluid speed at a location where the pip

e diameter is 8.0 cm is 1.28 m/s. What is the fluid speed at a location where the diameter has narrowed to 4.0 cm?

Physics
2 answers:
Tema [17]3 years ago
8 0

Answer:

5.10 m/s

Explanation:

The volumetric flow rate for an incompressible fluid through a pipe is constant, so we can write:

A_1 v_1 = A_2 v_2 (1)

where

A_1 is the cross-sectional area of the first part of the pipe

A_2 is the cross-sectional area of the second part of the pipe

v_1 is the speed of the fluid in the first part of the pipe

v_2 is the speed of the fluid in the second part of the pipe

Here we have:

v_1 = 1.28 m/s

r_1 = \frac{8.0 cm}{2}=4.0 cm = 0.04 m is the radius in the first part of the pipe, so the area is

A_1 = \pi r_1^2 = \pi (0.04 m)^2 =5.02\cdot 10^{-3}m^2

r_2 = \frac{4.0 cm}{2}=2.0 cm = 0.02 m is the radius in the first part of the pipe, so the area is

A_2 = \pi r_2^2 = \pi (0.02 m)^2 =1.26\cdot 10^{-3}m^2

Using eq.(1), we find the fluid speed at the second location:

v_2 = \frac{A_1 v_1}{A_2}=\frac{(5.02\cdot 10^{-3} m^2)(1.28 m/s)}{1.26\cdot 10^{-3} m^2}=5.10 m/s

DanielleElmas [232]3 years ago
7 0

The fluid speed at a location where the diameter has narrowed to 4.0 cm is 5.12 m/s

\texttt{ }

<h3>Further explanation</h3>

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

\large {\boxed {P = F \div A} }

<em>P = Pressure (Pa)</em>

<em>F = Force (N)</em>

<em>A = Cross-sectional Area (m²)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

diameter of pipe at location 1 = d₁ = 8.0 cm

speed of fluid at location 1 = v₁ = 1.28 m/s

diameter of pipe at location 2 = d₂ = 4.0 cm

<u>Asked:</u>

speed of fluid at location 2 = v₂ = ?

<u>Solution:</u>

We will use Continuity Equation as follows:

A_1 v_1 = A_2 v_2

\frac{1}{4}\pi (d_1)^2 v_1 = \frac{1}{4} \pi (d_2)^2 v_2

(d_1)^2 v_1 = (d_2)^2 v_2

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

v_2 = (\frac{8}{4})^2 \times 1.28

v_2 = 2^2 \times 1.28

v_2 = 4 \times 1.28

v_2 = 5.12 \texttt{ m/s}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
  • The Pressure In A Sealed Plastic Container : brainly.com/question/10209135
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

\texttt{ }

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

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