Question

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

Answer 1

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:

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

Answer 2

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

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Further explanation

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

Pressure = Force / Cross-sectional Area

or symbolized:

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

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

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Given:

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

Asked:

speed of fluid at location 2 = v₂ = ?

Solution:

We will use Continuity Equation as follows:

$A_1 v_1 = A_2 v_2$

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

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

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

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

$v_2 = 2^2 \times 1.28$

$v_2 = 4 \times 1.28$

$v_2 = 5.12 \texttt{ m/s}$

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Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

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