A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed ${\mathit{v}}_{\mathbf{0}}$ and the radius of the stream of liquid is ${\mathit{r}}_{\mathbf{0}}$ . (a) Find an equation for the speed of the liquid as a function of the distance y it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of y. (b) If water flows out of a vertical pipe at a speed of 1.2 km/s , how far below the outlet will the radius be one-half the original radius of the stream?

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,

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

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

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

Hence the distance below the outlet is, .