Question

A 1.0-cm-diameter pipe widens to 2.0 cm, then narrows to 0.5 cm. Liquid flows through the first segment at a speed of 4.0m/s. 

A: What is the speed in the second segment? 

B: What is the speed in the third segment?

C: What is the volume flow rate through the pipe?

Answer 1

Concepts and reason

The concept required to solve this problem is the equation of continuity in liquids.

Initially, Use the equation of continuity to find out the relation between the speed of liquid at first segment and speed of liquid at second segment. Rearrange the equation to calculate $\left( {{v_2}} \right)$ . Then, use continuity equation for the second and the third segment and rearrange for the speed $\left( {{v_3}} \right)$ . Finally, calculate the volume flow rate using the area and the speed $\left( {{v_1}} \right)$ .

Fundamentals

The equation of continuity states that when the fluid is in motion, then the motion of the fluid must be in such a way that mass is conserved. The equation of rate of flow of liquid is given as follows:

${A_1}{v_1} = {A_2}{v_2}$

Here, ${A_1}$ is the area of cross section at point 1, ${v_1}$ is the speed of liquid at point 1, ${A_2}$ the area of cross at point 2 and ${v_2}$ is the speed of liquid at point 2.

The expression for the area of the pipe in terms of diameter is,

$A = \pi {\left( {\frac{d}{2}} \right)^2}$

Here, A is the area and d is the diameter of the pipe.

The expression for the volume flow rate is,

$Q = Av$

Here, Q is the volume flow rate, A is the area, and v is the speed of the fluid.

(A)

The equation of rate of flow of liquid is given as follows:

${A_1}{v_1} = {A_2}{v_2}$

Rearrange the expression for $\left( {{v_2}} \right)$ .

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

The area at the first segment is,

${A_1} = \pi {\left( {\frac{{{d_1}}}{2}} \right)^2}$

Substitute 1.0 cm for ${d_1}$ .

$\begin{array}{c}\\{A_1} = \pi {\left( {\frac{{1.0{\rm{ cm}}}}{2}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)^2}\\\\{A_1} = 7.9 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The area at the second segment is,

${A_2} = \pi {\left( {\frac{{{d_2}}}{2}} \right)^2}$

Substitute 2.0 cm for ${d_2}$ .

$\begin{array}{c}\\{A_2} = \pi {\left( {\frac{{2.0{\rm{ cm}}}}{2}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)^2}\\\\{A_2} = 3.14 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The velocity for the second segment is,

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

Substitute 4.0 m/s for ${v_1}$ , $1.96 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}$ for ${A_2}$ , and $7.9 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}$ for ${A_1}$ .

$\begin{array}{c}\\{v_2} = \frac{{\left( {7.9 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}} \right)\left( {4.0{\rm{ m/s}}} \right)}}{{3.14 \times {{10}^{ - 4}}{\rm{ }}{{\rm{m}}^2}}}\\\\ = 1.0{\rm{ m/s}}\\\end{array}$

(B)

The equation of rate of flow of liquid for the second and the third segment is given as follows:

${A_2}{v_2} = {A_3}{v_3}$

Rearrange the expression for $\left( {{v_3}} \right)$ .

${v_3} = \frac{{{A_2}{v_2}}}{{{A_3}}}$

The area at the second segment is,

${A_2} = \pi {\left( {\frac{{{d_2}}}{2}} \right)^2}$

Substitute 2.0 cm for ${d_2}$ .

$\begin{array}{c}\\{A_2} = \pi {\left( {\frac{{2.0{\rm{ cm}}}}{2}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)^2}\\\\{A_2} = 3.14 \times {10^{ - 4}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The area at the third segment is,

${A_3} = \pi {\left( {\frac{{{d_3}}}{2}} \right)^2}$

Substitute 0.5 cm for ${d_3}$ .

$\begin{array}{c}\\{A_3} = \pi {\left( {\frac{{{\rm{0}}{\rm{.5 cm}}}}{2}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)^2}\\\\{A_3} = 1.96 \times {10^{ - 4}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The velocity for the third segment is,

${v_3} = \frac{{{A_2}{v_2}}}{{{A_3}}}$

Substitute 1.0 m/s for ${v_2}$ , $1.96 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}$ for ${A_3}$ , and $3.14 \times {10^{ - 4}}{\rm{ }}{{\rm{m}}^2}$ for ${A_2}$ .

$\begin{array}{c}\\{v_2} = \frac{{\left( {3.14 \times {{10}^{ - 4}}{\rm{ }}{{\rm{m}}^2}} \right)\left( {{\rm{1}}{\rm{.0 m/s}}} \right)}}{{1.96 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}}}\\\\ = 16.0{\rm{ m/s}}\\\end{array}$

(C)

The area at the first segment is,

${A_1} = \pi {\left( {\frac{{{d_1}}}{2}} \right)^2}$

Substitute 1.0 cm for ${d_1}$ .

$\begin{array}{c}\\{A_1} = \pi {\left( {\frac{{1.0{\rm{ cm}}}}{2}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)^2}\\\\{A_1} = 7.9 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}\\\end{array}$

The expression for the volume flow rate is,

$Q = {A_1}{v_1}$

Substitute 4.0 m/s for ${v_1}$ , $1.96 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}$ for ${A_2}$ , and $7.9 \times {10^{ - 5}}{\rm{ }}{{\rm{m}}^2}$ for ${A_1}$ .

$\begin{array}{c}\\Q = \left( {7.9 \times {{10}^{ - 5}}{\rm{ }}{{\rm{m}}^2}} \right)\left( {4.0{\rm{ m/s}}} \right)\\\\ = 3.16 \times {10^{ - 4}}{\rm{ }}\frac{{{{\rm{m}}^3}}}{{\rm{s}}}\left( {\frac{{1000{\rm{ L}}}}{{1{\rm{ }}{{\rm{m}}^3}}}} \right)\\\\ = 0.316{\rm{ L/s}}\\\end{array}$

Ans: Part A

The speed in the second segment is 1.0 m/s.

Part B

The speed in the third segment is 16.0 m/s.

Part C

The volume flow rate through the pipe is $0.316{\rm{ L/s}}$ .