Chapter 8, Problem 8.45P

To determine

(a)

The value of K for which the cavitation will begin at surface.

Answer to Problem 8.45P

The value of K for which the cavitation will begin at surface is $K_{cavitation}=4.4{\text{m}}^{\text{2}}\text{/s}$.

Explanation of Solution

Given information:

Free stream velocity $U_{\infty }=6\text{m/s}$

The value of pressure outside the boundary layer $p_{\infty }=200\text{kpa}$

The density of water at 20° C = $\rho =998\frac{kg}{m^3}$

The vapor pressure of water at 20° C = $p_{vap}=2337pa$

Calculation:

The expression of velocity at the surface is given by

$V_{surf}=-2U_{\infty }\sin \theta +\frac{K}{0.5}$

Putting the values,

$V_{surf}=-2\times 6\sin \theta +\frac{K}{0.5}$

At the stagnation conditions the velocity becomes zero, at the stagnation point the pressure is $p_o=218000pa$ . At $\theta ={270}^o$, applying Bernoulli’s equation.

$p_0=p_{vap}+\frac{\rho }{2}{(V_{surf})}^2$

Putting the values in the equation:

$218000=2337+\frac{998}{2}\left(-2\times 6\times \sin \left(270\right)+\frac{K}{0.5}\right)$

From the above equation, the value of K is $K_{cavitation}=4.4{\text{m}}^{\text{2}}\text{/s}\text{.}$

Conclusion:

Thus, the value of K for which the cavitation will begin at surface is $K_{cavitation}=4.4{\text{m}}^{\text{2}}\text{/s}$.

To determine

(b)

The point where the cavitation will begin.

Answer to Problem 8.45P

The cavitation will begin at the point where the pressure is lowest.

Explanation of Solution

A point in the fluid flow where pressure drops to such a value that it becomes equal to the atmospheric pressure is called the point of cavitation. The pressure is lowest at this point in the fluid flow.

In present fluid flow the pressure drops to the lowest value at the bottom most shoulder. The value of angle at this point is $\theta ={270}^o$.

To determine

(c)

The place of stagnation points.

Answer to Problem 8.45P

The stagnation points lie at two locations ${\theta }_{stag}={47}^o\text{and}{133}^o$

Explanation of Solution

Given information:

Free stream velocity $U_{\infty }=6\text{m/s}$

The value of pressure outside the boundary layer $p_{\infty }=200\text{kpa}$

The density of water at 20° C = $\rho =998\frac{kg}{m^3}$

The vapor pressure of water at 20° C = $p_{vap}=2337pa$

Calculation:

The expression of velocity at the surface is given by:

$V_{surf}=-2U_{\infty }\sin \theta +\frac{K}{0.5}$

Putting the values,

$V_{surf}=-2\times 6\sin \theta +\frac{K}{0.5}$

At the stagnation conditions the velocity becomes zero, at the stagnation point the pressure is $p_o=218000pa$ . At $\theta ={270}^o$, applying Bernoulli’s equation:

$p_0=p_{vap}+\frac{\rho }{2}{(V_{surf})}^2$

Putting the values in the equation:

$218000=2337+\frac{998}{2}\left(-2\times 6\times \sin \left(270\right)+\frac{K}{0.5}\right)$

From the above equation, the value of K is $K_{cavitation}=4.4{\text{m}}^{\text{2}}\text{/s}$

Now, from the value of K, the location of stagnation points can be found from the following equation:

$\sin {\theta }_{stag}=\frac{K}{2U_{\infty }a}$

Now, putting values in the above equation gives:

$\sin {\theta }_{stag}=\frac{4.4}{2\times 6\times 0.5}=0.732$

This gives two values of angle ${\theta }_{stag}={47}^o\text{and}{133}^o$.

Conclusion:

Thus, stagnation points lie at two locations ${\theta }_{stag}={47}^o\text{and}{133}^o$.