Chapter 8, Problem 8.25P

To determine

(a)

The appropriate sink strength: m.

Answer to Problem 8.25P

The appropriate sink strength: $m\approx 1980\frac{m^2}{s}$

Explanation of Solution

Given:

$\Gamma =8500m^2/s$

The given circulation yields the circumferential velocity at $r=40m$ :

$v_{\theta }=\frac{\Gamma }{2\pi r}=\frac{8500}{2\pi \left(40m\right)}$

$v_{\theta }\approx 33.8\frac{m}{s}$

Assuming sea − level density, $\rho =1.225kg/m^3$, to find radial velocity Bernoulli is to be used:

$p_{\infty }+\frac{\rho }{2}{\left(0\right)}^2=\left(p_{\infty }-\Delta p\right)+\frac{\rho }{2}\left(v_{\theta }^2+v_r^2\right)=p_{\infty }-2200+\frac{1.225}{2}\left({\left(33.8\right)}^2+v_r^2\right)$

So,

$v_r\approx 49.5\frac{m}{s}=\frac{m}{r}=\frac{m}{40}$

Therefore,

$m\approx 1980\frac{m^2}{s}$.

To determine

(b)

Pressure at $r=15m$.

Answer to Problem 8.25P

Pressure, $p_{absolute}=85kPa$

Explanation of Solution

Given:

$r=15m$

$m\approx 1980\frac{m^2}{s}$ (From part (a))

At $r=15m$, compute,

$v_r=\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$r$}\right.=\raisebox{1ex}{$1980$}\!\left/ \!\raisebox{-1ex}{$15$}\right.$

$v_r\approx 132m/s$

And

$v_{\theta }=\raisebox{1ex}{$\Gamma $}\!\left/ \!\raisebox{-1ex}{$2\pi r$}\right.=\raisebox{1ex}{$8500$}\!\left/ \!\raisebox{-1ex}{$2\pi \left(15\right)$}\right.$

$v_{\theta }\approx 90m/s$

Then we use Bernoulli again to compute the pressure at $r=15m$ :

$p+\frac{1.225}{2}\left[{\left(132\right)}^2+{\left(90\right)}^2\right]=p_{\infty }$,

Or

$p=p_{\infty }-15700Pa$

If we assume sea − level pressure of $101kPa$ at 8, then:

$p=101kPa-16kPa$

$p\approx 85kPa$.

To determine

(c)

The angle ß at which streamlines cross the circle at $r=40m$.

Answer to Problem 8.25P

$\beta ={55.6}^{\circ }$ at which streamlines cross the circle at $r=40m$

Explanation of Solution

Given:

$v_r=49.5\frac{m}{s}$

$v_{\theta }=33.8\frac{m}{s}$

With circumferential and radial velocity known, the streamline angle ß is:

$\beta ={\tan }^{-1}\left(\frac{v_r}{v_{\theta }}\right)={\tan }^{-1}\left(\frac{49.5}{33.8}\right)$

$\beta \approx {55.6}^{\circ }$.