Chapter 4, Problem 4.7P

To determine

(a)

The position of the maximum fluid accelerating along $AB$.

Answer to Problem 4.7P

The position of the maximum fluid accelerating along $AB$ is $-1.205R$.

Explanation of Solution

Given information:

The fluid velocity along the streamline $AB$ is $V=u\text{i}=U_0\left(1+\frac{R^3}{x^3}\right)\text{i}$.

Compare the component of velocity.

$\begin{array}{c}u\text{i}=U_0\left(1+\frac{R^3}{x^3}\right)\text{i}\\ u=U_0\left(1+\frac{R^3}{x^3}\right)\end{array}$

Write the expression for the acceleration vector in the $x$ direction.

$a_x=u\frac{\partial u}{\partial x}$ ...(I)

Here, the acceleration is $a_x$.

Substitute $U_0\left(1+\frac{R^3}{x^3}\right)$ for $u$ in Equation (I).

$\begin{array}{c}{a}_x=U_0\left(1+\frac{R^3}{x^3}\right)\frac{\partial }{\partial x}\left[U_0\left(1+\frac{R^3}{x^3}\right)\right]\\ =U_0\left(1+\frac{R^3}{x^3}\right)U_0\frac{\partial }{\partial x}\left(1+\frac{R^3}{x^3}\right)\\ =U_0^2\left(1+\frac{R^3}{x^3}\right)\left(-3\frac{R^3}{x^4}\right)\\ =-3U_0^2\left(1+\frac{R^3}{x^3}\right)\left(\frac{R^3}{x^4}\right)\end{array}$

Differentiate the above expression with respect to $x$.

$\begin{array}{c}\frac{d{a}_x}{dx}=\frac{d}{dx}\left[-3U_0^2\left(1+\frac{R^3}{x^3}\right)\left(\frac{R^3}{x^4}\right)\right]\\ =-3U_0^2\left(\frac{R^3}{x^4}\right)\left(\frac{-3R^3}{x^4}\right)-3U_0^2\left(1+\frac{R^3}{x^3}\right)\left(\frac{-4R^3}{x^5}\right)\\ =-3U_0^2\left(\frac{R^3}{x^4}\right)\left[\left(\frac{-3R^3}{x^4}\right)\left(1+\frac{R^3}{x^3}\right)\left(\frac{-4}{x}\right)\right]\end{array}$

Calculation:

Equate $\frac{d{a}_x}{dx}$ to zero for the position of maximum acceleration.

$\frac{d{a}_x}{dx}=0$ ...(II)

Substitute $-3U_0^2\left(\frac{R^3}{x^4}\right)\left[\left(\frac{-3R^3}{x^4}\right)\left(1+\frac{R^3}{x^3}\right)\left(\frac{-4}{x}\right)\right]$ for $\frac{d{a}_x}{dx}$ in Equation (II).

$\begin{array}{l}-3U_0^2\left(\frac{R^3}{x^4}\right)\left[\left(\frac{-3R^3}{x^4}\right)+\left(1+\frac{R^3}{x^3}\right)\left(\frac{-4}{x}\right)\right]=0\\ \left(\frac{-3R^3}{x^4}\right)-\left(\frac{4}{x}+\frac{4R^3}{x^4}\right)=0\\ \left(\frac{-3R^3}{x^4}\right)=\left(\frac{4}{x}+\frac{4R^3}{x^4}\right)\\ \frac{4}{x}=-\frac{7R^3}{x^4}\end{array}$

Further simplify the above expression.

$\begin{array}{c}\frac{4}{x}=-\frac{7R^3}{x^4}\\ x^3=-\frac{7}{4}{R}^3\\ x=\sqrt[3]{-\frac{7}{4}{R}^3}\\ x=-1.205R\end{array}$

Conclusion:

The position of the maximum fluid accelerating along $AB$ is $-1.205R$.

To determine

(b)

The time required to travel for a fluid particle from $A$ to $B$.

Answer to Problem 4.7P

The time required to travel for a fluid particle from $A$ to $B$ is $\infty$.

Explanation of Solution

Given information:

The fluid velocity along the streamline $AB$ is $V=u\text{i}=U_0\left(1+\frac{R^3}{x^3}\right)\text{i}$.

Write the expression for the velocity in terms of displacement.

$u=\frac{dx}{dt}$ ...(III)

Here, displacement is $x$ and the time is $t$.

Substitute $U_0\left(1+\frac{R^3}{x^3}\right)$ for $u$ in Equation (III).

$\begin{array}{l}{U}_0\left(1+\frac{R^3}{x^3}\right)=\frac{dx}{dt}\\ U_{0}dt=\frac{dx}{\left(1+\frac{R^3}{x^3}\right)}\\ U_{0}dt=\left(\frac{x^3}{x^3+R^3}\right)dx\\ U_{0}dt=\left(\frac{x^3+R^3-R^3}{x^3+R^3}\right)dx\end{array}$

Further solve the above expression.

$\begin{array}{c}{U}_{0}dt=\left(1-\frac{R^3}{x^3+R^3}\right)dx\\ U_{0}dt=dx-\frac{R^3}{x^3+R^3}dx\end{array}$

Integrate the above expression.

$U_0{\displaystyle \underset{t_i}{\overset{t_f}{\int }}dt}={\displaystyle \underset{x_i}{\overset{x_f}{\int }}dx}-{\displaystyle \underset{x_i}{\overset{x_f}{\int }}\frac{R^3}{x^3+R^3}dx}$ ...(IV)

Here, the initial limit for time is $t_i$, the final limit for time is $t_f$, the initial limit for the position $x_i$ and the initial limit for the position $x_f$.

Substitute $0$ for $t_i$, $t$ for $t_f$, $-4R$ for $x_i$ and $-R$ for $x_f$ in Equation (IV).

Further solve the above expression.

$\begin{array}{c}{U}_{0}t=3R-\frac{R}{6}\ln \left(0\right)+\frac{R}{6}\ln \left(\frac{9}{21}\right)+\frac{R}{\sqrt{3}}{\tan }^{-1}\left(-\sqrt{3}\right)-\frac{R}{\sqrt{3}}{\tan }^{-1}\left(-3\sqrt{3}\right)\\ U_{0}t=\infty \end{array}$

Thus, for a fixed velocity $U_0$ the time taken to reach the stagnation point is infinite.

Conclusion:

The time required to travel for a fluid particle from $A$ to $B$ is $\infty$.