Chapter 8, Problem 8.29P

To determine

(a) Is this point outside the body?

Answer to Problem 8.29P

(a) Yes this point is outside the body.

Explanation of Solution

Given information:

$U_{\infty }=20m/s,\rho =998\frac{kg}{m^3}$ , uniform water stream combines with a source at the origin to form a half-body.

at $(x,y)=(0,1.2),p=12.5kpa\langle p_{\infty }$

Calculation:

We know that,

The point on the half body surface just above the m is at $y=\frac{\pi a}{2}$ where $a=\frac{m}{U_{\infty }},U_{\infty }=\frac{m}{a}$

Apply Bernoulli equation,

$\begin{array}{l}{p}_{\infty }+\frac{\rho }{2}{U}_{\infty }{}^2=(p_{\infty }-12500)+\frac{\rho }{2}\left(U_{\infty }{}^2+{\left(\frac{m}{1.2}\right)}^2\right)\\ m=12500\times 2\times {(}^{1.2}/998\\ m=6\frac{m^2}{s}\end{array}$

As we know, $a=\frac{m}{U_{\infty }}=\frac{6}{20}=0.33m$

And $y=\frac{\pi a}{2}$ put the value of $a$ and we get,

$y=\frac{\pi (0.33)}{2}=0.47m$

Given point is located at

$y=1.2m$ which above then $y=0.47m$

So this point is outside the body,

$m=6\frac{m^2}{s}$.

To determine

(b) The appropriate strength $m$?

Answer to Problem 8.29P

(b) $m=6\frac{m^2}{s}$

Explanation of Solution

Given information:

$U_{\infty }=20m/s,\rho =998\frac{kg}{m^3}$ , uniform water stream combines with a source at the origin to form a half-body.

at $(x,y)=(0,1.2),p=12.5kpa\langle p_{\infty }$

Calculation:

We know that,

The point on the half body surface just above the m is at $y=\frac{\pi a}{2}$ where $a=\frac{m}{U_{\infty }},U_{\infty }=\frac{m}{a}$

Apply Bernoulli equation,

$\begin{array}{l}{p}_{\infty }+\frac{\rho }{2}{U}_{\infty }{}^2=(p_{\infty }-12500)+\frac{\rho }{2}\left(U_{\infty }{}^2+{\left(\frac{m}{1.2}\right)}^2\right)\\ m=12500\times 2\times {(}^{1.2}/998\\ m=6\frac{m^2}{s}\end{array}$

As we know, $a=\frac{m}{U_{\infty }}=\frac{6}{20}=0.33m$

And $y=\frac{\pi a}{2}$ put the value of $a$ and we get,

$y=\frac{\pi (0.33)}{2}=0.47m$

Given point is located at

$y=1.2m$ which above then $y=0.47m$

So this point is outside the body,

$m=6\frac{m^2}{s}$.

To determine

(c) The pressure at the nose of the body?

Answer to Problem 8.29P

(c) $p=(p_{\infty }+199.6)kpa$

Explanation of Solution

Given information:

$U_{\infty }=20m/s,\rho =998\frac{kg}{m^3}$ , uniform water stream combines with a source at the origin to form a half-body.

Calculation:

Let pressure at nose is $p$

And we know that velocity at nose is zero,

Apply Bernoulli equation,

$\begin{array}{l}{p}_{\infty }+\frac{\rho }{2}{U}_{\infty }{}^2=(p)+\frac{\rho }{2}{\left(0\right)}^2\\ p=p_{\infty }+\frac{\rho }{2}{20}^2\\ p=(p_{\infty }+199.6)kpa\end{array}$.