Question

A liquid flowing from a vertical pipe has a very 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 v0 and the radius of the stream of liquid is r0.
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.

Answer 1

(a). The equation for the speed of the liquid after it leaves the pipe in terms o the distance covered is $\boxed{v=\sqrt\left(v_o^2+2gy\right)}$.  

(b). The expression for the radius of the stream in terms of the distance covered by liquid is $\boxed{r=\dfrac{r_o\sqrt{v_o}}{(v_o^2+2gy)^{1/4}}}$.

Explanation:  

Part (a):  

As the water falls from the pipe, it simply falls freely under the acceleration due to gravity of the Earth.

Consider the final speed of the water after covering a distance $y$ is $v$.

Write the expression for the final velocity of a body under acceleration:  

$v_f^2=v_i^2+2gS$                                     ...... (1)

Here, $v_f$ is the final velocity of water at the end, $S$ is the distance through which water falls and $v_i$ is the initial velocity of the water.

Substitute the value of the initial velocity and the distance covered by water in above expression.

$\begin{aligned}v^2&=v_o^2+2gy\\v&=\sqrt{v_o^2+2gy}\end{aligned}$

Thus, the equation for the speed of the liquid after it leaves the pipe in terms o the distance covered is $\boxed{v=\sqrt\left(v_o^2+2gy\right)}$.

Part (b):

The equation of continuity states that the rate of mass entering the cross-section area will be same as the rate of mass coming out of the area.

Write the expression for the equation of continuity:

$A_1v_{1}=A_2v_{2}$

Here, $A_1\&v_1$ are the initial area and velocity and $A_2\&v_2$ are the final area and velocity.

Substitute the values of area and velocity in above expression.

$\begin{aligned}(\pi{r_o}^2)v_o&=(\pi{r}^2)v\\r&=\sqrt{\dfrac{r_o^2v_o}{v}}\end{aligned}$

Substitute $\sqrt\left(v_o^2+2gy\right)$ for $v$ in above expression.

$r=\dfrac{r_o\sqrt{v_o}}{(v_o^2+2gy)^{1/4}}$.

Thus, the expression for the radius of the stream in terms of the distance covered by liquid is $\boxed{r=\dfrac{r_o\sqrt{v_o}}{(v_o^2+2gy)^{1/4}}}$.

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Fluid Mechanics

Keywords:

liquid flowing, vertical pipe, equation of continuity, radius of stream, under gravity, speed of the water, leaves the pipe, function of y.

Answer 2

Calculating speed as function of distance y is fairly easy. Once water leaves pipe it is under free fall which means that it is accelerating with gravitational acceleration "a".
V=Vo + Vff   where Vff is speed gained due to free fall.
$Vff = t*a = \sqrt{2ay}$
$V = Vo + \sqrt{2ay}$

Calculating radius of stream is a little bit more complicated. Because water is accelerating as it falls it has to lower its radius. The reason for this is that water flow must be the same as one at the exit of pipe. Water flow is expressed in  liters/meter^3
Because of this we write condition:
V1/V2 = A2/A1 where  A2 and A1 are sections of water stream.
let V1 and A1 be the speed and section of water stream at the end of pipe.
A2 = V1*A1/V2

$r^{2} = \frac{Vo* r_{0} ^{2} }{Vo+ \sqrt{2ay} }$