An ideal fluid flows through a pipe made of two sections with diameters of 1.0 and 3.0 inches, respectively. The speed of the fluid flow through the 3.0-inch section will be what factor times that through the 1.0-inch section

Respuesta :

Answer:

[tex](\frac{r_1}{r_2})^2=\frac{1}{9}[/tex]

Explanation:

From the question we are told that:

Diameter 1 [tex]d_1=1.0[/tex]

Diameter 2 [tex]d_2=3.0[/tex]

Generally the equation for Radius is mathematically given by

At Diameter 1

[tex]r_{1}=\frac{1}{2} inch[/tex]

At Diameter 2

[tex]r_{2}=\frac{3}{2} inch[/tex]

Generally the equation for continuity is mathematically given by

 [tex]A_1V_1=A_2V_2[/tex]

Therefore

[tex](\frac{r_1}{r_2})^2=(\frac{1/2}{3/2})^2[/tex]

[tex](\frac{r_1}{r_2})^2=\frac{1}{9}[/tex]