A $1.0-\mathrm{cm}-diameter$pipe widens to $2.0\mathrm{cm}$, then narrows to $5.0\mathrm{mm}$. Liquid flows through the first segment at a speed of$4.0m/s$.

a. What is the speed in the second and third segments?

b. What is the volume flow rate through the pipe?

According to continuity equation,

$A_1{v}_1=A_2{v}_2$

Here, $A_1$is the area of first segment, $A_2$is the area of second segment,$v_1$is the speed of first segment, and $v_2$is the speed of second segment.

(a)

The area of first segment is,

$A_1=\pi r_1^2$

Here, $r_1$is the radius of the first segment.

The area of second segment is,

$A_2=\pi r_2^2$

Here, $r$, is the radius of the first segment.

Substitute $\pi r_1^2$for $A_1$and $\pi r_2^2$for $A_2$in $A_1{v}_1=A_2{v}_2$.

$\left(\pi r_1^2\right)v_1=\left(\pi r_2^2\right)v_2$

$v_2=\frac{r_1^2{v}_1}{r_2^2}$

Substitute $0.5\times {10}^{-2}\mathrm{m}$for $r_{1_{}}$, $4.0\mathrm{m}/\mathrm{s}$for $v_1$, and $1.0\times {10}^{-2}\mathrm{m}$for $r_2$.

$v_2=\frac{{\left(0.5\times {10}^{-2}\mathrm{m}\right)}^2(4.0\mathrm{m}/\mathrm{s})}{{\left(1.0\times {10}^{-2}\mathrm{m}\right)}^2}$

$=1\mathrm{m}/\mathrm{s}$

The speed of the fluid along the third segment is,

$A_2{v}_2=A_3{v}_3$

$\pi r_2^2{v}_2=\pi r_3^2{v}_3$

$v_3=\frac{r_2^2{v}_2}{r_3^2}$

Substitute $1.0\times {10}^{-2}\mathrm{m}$for $r_2$,$1m/s$for $v_2$, and $0.25\times {10}^{-2}\mathrm{m}$for $r_3$,

$v_3=\frac{{\left(1.0\times {10}^{-2}\mathrm{m}\right)}^2(1\mathrm{m}/\mathrm{s})}{{\left(0.25\times {10}^{-2}\mathrm{m}\right)}^2}$

$v_3$$=16\mathrm{m}/\mathrm{s}$

(b)

The volume flow rate is common throughout the pipe. So.

$\frac{dv}{dt}=A_1{v}_1$

Substitute $\pi r_1^2$for $A_1$.

$\frac{dv}{dt}=\pi r_1^2{v}_1$

Substitute $0.5\times {10}^{-2}\mathrm{m}$for $r_1$and $4.0\mathrm{m}/\mathrm{s}$for $v_1$,

$\frac{dv}{dt}=(3.14){\left(0.5\times {10}^{-2}\mathrm{m}\right)}^2(4.0\mathrm{m}/\mathrm{s})$

$\frac{dv}{dt}$ $=3.14\times {10}^{-4}{\mathrm{m}}^3/\mathrm{s}$