Chapter 8, Problem 8.103P

To determine

(a)

The point on the front surface where the fluid acceleration is maximum.

Answer to Problem 8.103P

The required value of the placement of the point is $\frac{9}{8}\frac{U_{\infty }{}^2}{a}.$.

Explanation of Solution

Given Information:

Formula used:

$a_{max}={\left(V\frac{dV}{dx}\right)}_{\max }$

Calculation:

We know the formula for the maximum acceleration, i.e.: $a_{max}={\left(V\frac{dV}{dx}\right)}_{\max }$

Put the values in the above equation;

${\left(V\frac{dV}{dx}\right)}_{\max }=\frac{9}{8}\frac{U_{\infty }{}^2}{a}$.

To determine

(b)

The value of maximum acceleration.

Answer to Problem 8.103P

The required value of the acceleration is $\frac{\pi }{4}$ or a ${45}^o.$

Explanation of Solution

Given information:

Formula used:

$V_{alongsurface}=1.5U_{\infty }\sin \left(\frac{x}{a}\right)$

Calculation:

To solve this problem, along the sphere surface, the flow is purely tangential, so $v_r=0$.

Therefore, $V_{alongsurface}=1.5U_{\infty }\sin \left(\frac{x}{a}\right)$ ;

X is along the surface in the above equation;

$\frac{dV}{dx}=\frac{1.5U_{\infty }}{a}\cos \left(\frac{x}{a}\right)$ ;

So, the acceleration $=V\frac{dV}{dx}=\frac{9U_{\infty }{}^2}{4a}\sin \left(\frac{x}{a}\right)\cos \left(\frac{x}{a}\right)=$ maximum when $\frac{x}{a}=\frac{\pi }{4}$ or ${45}^o$.

To determine

(c)

Sphere diameter for given parameter and comment.

Answer to Problem 8.103P

The required value of the diameter is $0.0115m.$

The diameter shows that the sphere is large $23mm$ diameter and it shows that fluid flow past small bodies can cause large acceleration, even thousands of times of g’s.

Explanation of Solution

Given Information:

$v=1m/s$

Formula used:

$\frac{9U_{\infty }{}^2}{8a}$

Calculation:

Since from the first part we have $U_{\infty }=1m/s$ ;

And the maximum acceleration is ten times the acceleration due to gravity;

$a_{max}=10\left(9.81m/s^2\right)$ ;

We know that, $\frac{9U_{\infty }{}^2}{8a}$ ;

Put the values in the above equation;

$\Rightarrow \frac{9{\left(1m/s\right)}^2}{8a}$

$a=0.0115m$

The final value resembles a large sphere $23mm$ diameter and it shows that fluid flow past small bodies can cause large acceleration, even thousands times of g’s.