91Ó°ÊÓ

The given data can be listed below as,

• The liquid speed is,$v_0$.
• The radius of the stream of liquid is, $r_0$.
• The speed in vertical pipe is, 1.20 m/s .
• The radius one half of original radius of the stream.

In any steady-state process, the product of the velocity of the fluid and the cross-sectional area of the pipe at any point of the fluid flow is constant.

Part (a)

In the given below figure assume point 1 is the end of the pipe and point 2 is in the stream of liquid at a distance $y_2$beneath the end of pipe.

Consider the free fall of a liquid Take positive to be downward.

The third equation of motion is expressed as,

$\begin{array}{l}{v}_2^2=v_1^2+2gh\\ v_2^2=v_1^2+2g{y}_2\\ v_2=\sqrt{v_1^2+2g{y}_2}\end{array}$ ...(i)

Here $v_1$ and $v_2$are the speed at point 1 and point 2, g is the gravitational acceleration.

Substitute the initial speed $v_1=v_0$ and a distance $y_2=y$.in the equation (ii), so it is expressed as,

$v_2=\sqrt{v_0^2+2gy}$

Hence theequation for the speed of the liquid as a function of the distance yis.

$\sqrt{v_0^2+2gy}$

The continuity equation at points 1 and 2 is expressed as,

$\begin{array}{l}{A}_1{v}_1=A_2{v}_2\\ \mathrm{Ï€}{r}_1^2{\mathrm{v}}_1=\mathrm{Ï€}{r}_2^{2}v\end{array}$

$v_2=\frac{r_1^2}{r_2^2}{v}_1$ ...(ii)

Here $r_1$and $r_2$ are the radius at point 1 and 2.

Substitute the value of initial speed $v_1=v_0$, the radius at point 1 $r_1=r_0$ and the radius at point 2 $r_2=r$ in the equation (ii). So it is expressed as,

$\begin{array}{r}\frac{r_0^2}{r^2}{v}_0=\sqrt{v_0^2+2gy}\\ r=\frac{r_0\sqrt{v_0}}{{\left(v_0^2+2gy\right)}^{\frac{1}{4}}}\end{array}$

Hence, the radius of the stream as a function of y is, $\frac{r_0\sqrt{v_0}}{{\left(v_0^2+2gy\right)}^1}$.

Part (b)

Theequation for theradius of the streamas a function of the distance y is, expressed as,

$\begin{array}{r}r=\frac{r_0\sqrt{v_0}}{{\left(v_0^2+2gy\right)}^{\frac{1}{4}}}\\ r^4\left(v_0^2+2gy\right)=r_0^4{v}_0^2\\ y=\frac{\left({\left(\frac{r_0}{r}\right)}^4−1\right)}{2g}{v}_0^2\end{array}$

The value of r that gives, $r=\frac{1}{2}{r}_0$, then $r_0=2r$.

Substitute 2r for $r_0$, 1.20 m/s for $v_0$ , and $9.81m/s^2$ for g in the above equation.

$\begin{array}{r}y=\frac{\left({\left(\frac{2r}{r}\right)}^4−1\right)}{2×9.81\mathrm{m}/{\mathrm{s}}^2}×(1.20\mathrm{m}/\mathrm{s}{)}^2\\ =1.1\mathrm{m}\end{array}$

Hence the distance below the outlet is, .