Solution Exercise 4
Part 1:
[latex]pBulk\ = \ \left( p\min\ \ \lbrack 1 - n\rbrack \right) + \left( pH_{2}O\ \ n \right);[/latex]
[latex]p\min\ = \ pBulk - \frac{pH_{2}O\ n}{1 - n}[/latex]
[latex]\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( 2.61\frac{g}{{cm}^{3}} \right) - \ \frac{1\ (0.12)}{0.88}\ = \ 2.83\ \frac{g}{cm^{3}}[/latex]
Part 2:
a)
[latex]pBulk\ = \ \left( p\min\ \ \lbrack 1 - n\rbrack \right)[/latex]
[latex]n\ = \ 1 - \left( \frac{pBulk}{p\min} \right)[/latex]
[latex]1 - \frac{2,320\frac{kg}{m^{3}}}{2,650\frac{kg}{m^{3}}} = \ 0.125[/latex]
b) 25 percent:
[latex]\ 0.125\ \left( 1,000\ \frac{kg}{m^{3}}1\ m^{3}\ (0.25) \right)\ = \ 31.1\ kg\ + \ 2,320\ kg\ = \ 2,351\ kg[/latex]
75 percent: [latex]0.125\ \left( 1,000\ \frac{kg}{m^{3}}1\ m^{3}\ (0.75) \right)\ = \ 93.4\ kg\ + \ 2,320\ kg\ = \ 2,413\ kg[/latex]
100 percent: [latex]0.125\ \left( 1000\ \frac{kg}{m^{3}}1\ m^{3}\ (1.00) \right) = \ 124.5\ kg\ + \ 2,320\ kg\ = \ 2,445\ kg[/latex]
Part 3:
a) [latex]pBulk\ = \ \frac{mass}{volume} = \ \frac{753.7\ kg}{0.36\ m^{3}}\ = \ 2093.5\ \frac{kg}{m^{3}}[/latex]
[latex]\ \ pBulk\ = \ p\min\ \lbrack 1 - n\rbrack[/latex]
[latex]\ \ n\ = \ 1 - \frac{pBulk}{p\min}[/latex]
[latex]1 - \frac{2093.5\frac{kg}{m^{3}}}{2650\frac{kg}{m^{3}}}\ = \ 0.21[/latex]
b) [latex]\ \left( (0.21)\ 1000\ \frac{kg}{m^{3}}\ 0.36\ m^{3}\ (0.62) \right) + 753.7\ kg = \ 46.87\ kg\ 753.7\ kg\ = \ 800.6\ kg[/latex]