First, from the formula given we solve the equation for D:
[tex]\begin{gathered} 4PD=D^2LN\pi, \\ \frac{4PD}{LN\pi}=D^2, \\ \sqrt{\frac{4PD}{LN\pi}}=D^{}. \end{gathered}[/tex]
Now, substituting the given data:
(a) PD=405 cu in, L=4.7 in, and N=7,
in the above equation we get:
[tex]D=\sqrt[]{\frac{4\times405}{4.7\times7\times\pi}}in=\sqrt[]{15.6736175}in\approx3.96\text{ in}[/tex]
(b) PD=399.4 cu in, L=2 in, and N=6,
in the above equation we get:
[tex]D=\sqrt[]{\frac{4\times399.4}{2\times6\times\pi}}in=\sqrt[]{42.37765618}in\approx6.51\text{ in}[/tex]
Answer:
(a) 3.96 in.
(b) 6.51 in.
