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?m/s

B: What is the speed in the third segment?m/s

C: What is the volume flow rate through the pipe?L/s

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 determine the relation between the speed of liquid at the first segment and the speed of liquid at the 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 fluid's motion must be in such a way that mass is conserved. The equation of the rate of flow of liquid is given as follows:

\(A_{1} v_{1}=A_{2} v_{2}\)

Here, \(A_{1}\) is the area of the 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, \(\mathrm{A}\) is the area and \(\mathrm{d}\) is the diameter of the pipe. The expression for the volume flow rate is, \(Q=A v\)

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 \(1.0 \mathrm{~cm}\) for \(d_{1}\)

\(A_{1}=\pi\left(\frac{1.0 \mathrm{~cm}}{2}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)\right)^{2}\)

$$ A_{1}=7.9 \times 10^{-5} \mathrm{~m}^{2} $$

The area at the second segment is, \(A_{2}=\pi\left(\frac{d 2}{2}\right)^{2}\)

Substitute \(2.0 \mathrm{~cm}\) for \(d_{2}\)

\(A_{2}=\pi\left(\frac{2.0 \mathrm{~cm}}{2}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)\right)^{2}\)

$$ A_{2}=3.14 \times 10^{-5} \mathrm{~m}^{2} $$

The velocity for the second segment is, \(v_{2}=\frac{A_{1} v_{1}}{A_{2}}\)

Substitute \(4.0 \mathrm{~m} / \mathrm{s}\) for \(v_{1}, 1.96 \times 10^{-5} \mathrm{~m}^{2}\) for \(A_{2},\) and \(7.9 \times 10^{-5} \mathrm{~m}^{2}\) for \(A_{1}\)

\(v_{2}=\frac{\left(7.9 \times 10^{-5} \mathrm{~m}^{2}\right)(4.0 \mathrm{~m} / \mathrm{s})}{3.14 \times 10^{-4} \mathrm{~m}^{2}}\)

\(=1.0 \mathrm{~m} / \mathrm{s}\)

Part A The speed in the second segment is \(1.0 \mathrm{~m} / \mathrm{s}\).

The equation of continuity states that the product of the area of the liquid's cross-section and speed is the same at both points. since, \(A_{1}>A_{2}\) implies that \(v_{1}<v_{2}\). Therefore, the equation of continuity remains true.

 

(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 \(2.0 \mathrm{~cm}\) for \(d_{2}\)

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

The area at the third segment is, \(A_{3}=\pi\left(\frac{d_{3}}{2}\right)^{2}\)

Substitute \(0.5 \mathrm{~cm}\) for \(d_{3}\)

\(A_{3}=\pi\left(\frac{0.5 \mathrm{~cm}}{2}\left(\frac{10^{-2} \mathrm{~m}}{1 \mathrm{~cm}}\right)\right)^{2}\)

$$ A_{3}=1.96 \times 10^{-4} \mathrm{~m}^{2} $$

The velocity for the third segment is, \(v_{3}=\frac{A 2 v 2}{A 3}\)

Substitute \(1.0 \mathrm{~m} / \mathrm{s}\) for \(v_{2}, 1.96 \times 10^{-5} \mathrm{~m}^{2}\) for \(A_{3},\) and \(3.14 \times 10^{-4} \mathrm{~m}^{2}\) for \(A_{2}\)

\(v_{2}=\frac{\left(3.14 \times 10^{-4} \mathrm{~m}^{2}\right)(1.0 \mathrm{~m} / \mathrm{s})}{1.96 \times 10^{-5} \mathrm{~m}^{2}}\)

\(=16.0 \mathrm{~m} / \mathrm{s}\)

The equation of continuity states that the product of the area of the liquid's cross-section and speed is the same at both segments.

 

(C) The area at the first segment is, \(A_{1}=\pi\left(\frac{d_{1}}{2}\right)^{2}\)

Substitute \(1.0 \mathrm{~cm}\) for \(d_{1}\).

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

The expression for the volume flow rate is, \(Q=A_{1} v_{1}\)

Substitute \(4.0 \mathrm{~m} / \mathrm{s}\) for \(v_{1}, 1.96 \times 10^{-5} \mathrm{~m}^{2}\) for \(A_{2}\), and \(7.9 \times 10^{-5} \mathrm{~m}^{2}\) for \(A_{1}\)

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

Part C The volume flow rate through the pipe is \(0.316 \mathrm{~L} / \mathrm{s}\).

The volume flow rate can be defined as the product of the pipe area and the speed of the liquid. The volume flow rate can also be defined as the volume passing through an area per unit of time.

 

Part A

The speed in the second segment is \(1.0 \mathrm{~m} / \mathrm{s}\).

Part B

The speed in the third segment is \(16.0 \mathrm{~m} / \mathrm{s}\).

Part C

The volume flow rate through the pipe is \(0.316 \mathrm{~L} / \mathrm{s}\).