Answer:
Explanation:
From the given information:
weight of fiber [tex]w_f[/tex] = 3.0 g
weight of composite specimen [tex]w_c[/tex] = 4.0 g
specimen composite weight in water [tex]C_{wm}[/tex] = 2.0 g
specific gravity of fiber [tex]S_f[/tex] = 2.4
specific gravity of matrix [tex]S_m[/tex] = 1.3
The weight of the matrix = weight of the composite - the weight of fiber
⇒ (4.0 - 3.0) g
= 1.0 g
The theoretical density of the composite [tex]\rho_{ct}[/tex] can be determined by using the formula:
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{w_f}{w_cS_f}+ \dfrac{w_m}{w_cS_m}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{(4.0 \times 2.4)}+ \dfrac{1.0}{(4.0\times 1.3)}[/tex]
[tex]\dfrac{1}{\rho_{ct}} = \dfrac{3.0}{9.6}+ \dfrac{1.0}{5.2}[/tex]
[tex]\dfrac{1}{\rho_{ct}} =0.505\\[/tex]
[tex]\rho_{ct} =\dfrac{1}{0.505}[/tex]
[tex]\mathbf{\rho_{ct} = 1.980 \ g/cm^3}[/tex]
The experimental density [tex]\rho _{ce}[/tex] is determined by using the equation:
[tex]\rho _{ce} = \dfrac{w_f + w_c}{\dfrac{w_f }{S_f} + \dfrac{w_c }{S_m} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{\dfrac{3.0 }{2.4} + \dfrac{4.0 }{1.3} }[/tex]
[tex]\rho _{ce} = \dfrac{3.0 + 4.0}{1.250 +3.077 }[/tex]
[tex]\mathbf{\rho _{ce} = 1.620 \ g/cm^3}[/tex]
The void fraction is: [tex]= \dfrac{\rho_{ct}-\rho_{ce}}{\rho_{ct}}[/tex]
[tex]= \dfrac{1.980-1.620}{1.980}[/tex]
= 0.1818
