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Chapter 7: Problem 1045

An incompressible fluid flows steadily through a cylindrical pipe which has radius \(2 \mathrm{r}\) at point \(\mathrm{A}\) and radius \(\mathrm{r}\) at \(\mathrm{B}\) further along the flow direction. It the velocity at point \(\mathrm{A}\) is \(\mathrm{V}\), its velocity at point \(\mathrm{B}\). (A) \(2 \mathrm{~V}\) (B) V (C) \((\mathrm{V} / 2)\) (D) \(4 \mathrm{~V}\)

Short Answer

Expert verified
The velocity of the fluid at point B is four times the velocity at point A, due to the conservation of mass in the incompressible fluid flow. Therefore, the solution is (D) \(4V\).

Step by step solution

01

Understanding the Principle of Conservation of Mass

In fluid dynamics, the principle of conservation of mass is often expressed as the Continuity Equation for steady flow, which states that the product of the cross-sectional area of the pipe and the velocity of incompressible fluid remains constant at all points along the pipeline.
02

Define the Continuity Equation

The Continuity Equation is expressed as \(A_1V_1=A_2V_2\), where \(A_1\) and \(V_1\) are the cross-sectional area and velocity at point A, and \(A_2\) and \(V_2\) are the cross-sectional area and velocity at point B.
03

Express the Areas in Terms of Radii

The cross-sectional area of a pipe with radius \(r\) is given by the formula \(A=\pi r^2\). Therefore, the areas at points A and B are \(A_1=\pi(2r)^2\), and \(A_2=\pi r^2\), respectively.
04

Substitute Areas and Velocity at A into the Continuity Equation

Substituting the expressions for \(A_1\), \(A_2\), and \(V_1\) into the Continuity Equation, we get \[4\pi r^2 V=\pi r^2 V_2\].
05

Solve for Velocity at B

Dividing both sides of the equation by \(\pi r^2\), and simplifying, we get \(V_2=4V\). This means the velocity of the fluid at point B is four times the velocity at point A. Therefore, the solution to the problem is (D) \(4V\).

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