Question

A side wall of a wide open tank is provided with a narrowing tube (figure shown above) through which water flows out. The cross-sectional area of the tube decreases from $S=3.0c{m}^2$ to $s=1.0c{m}^2$. The water level in the tank is $h=4.6m$ higher than that in the tube. Neglecting the viscosity of the water, find the horizontal component of the force tending to pull the tube out of the tank (in $N$). Round off to the closest integer.

Answer 1

Consider narrowing tube made off rings arranged across it's length.

where $p_p$ is the pressure at $P$ and $p_0$ is the atmospheric pressure which is the pressure just outside of $B$. The force on the nozzle tending to pull it out from diagram is

$dF=P\times dAsin\theta ;P=p_p-p_O;dA=2\pi yds$

$F=\int (p_p-p_O)\sin \theta 2\pi yds$

where $y$ is the radius at the point $P$ distant $x$ from the vertex $O$.We have subtracted $p_0$ which is the force due to atmospheric pressure the factor $\sin \theta$ gives horizontal component of the force and $ds$ is the length of the element of nozzle surface, $ds=dx\sec \theta$

Suppose the radius at $A$ is $R$ and it decreases uniformly to $r$ at $B$ where $S=\pi R^2$ and $s=\pi r^2$. Assume also that the semi vertical angle at 0 is $\theta$. Then

$\frac{R}{L_2}=\frac{r}{L_1}=\frac{y}{x}$

So $y=r+\frac{R-r}{L_2-L_1}(x-L_1)$

Suppose the velocity with which the liquid flows out is $V$ at $A$, at $B$ and $u$ at $P$. Then by the equation of continuity

$\pi R^{2}V=\pi r^{2}v=\pi y^{2}u$

The velocity $v$ of efflux is given by

$v=\sqrt{2gh}$

and Bernoulli's theorem gives

$p_p+\frac{1}{2}\rho u^2=p_0+\frac{1}{2}\rho v^2$

$\tan \theta =\frac{R-r}{L_2-L_1}$

Thus

$F={\int }_{L_1}^{L_2}\frac{1}{2}(v^2-u^2)\rho 2\pi y\frac{R-r}{L_2-L_1}dx$

$=\pi \rho {\int }_r^R{v}^2(1-\frac{r^4}{y^4})ydy$

$=\pi \rho v^2\frac{1}{2}(R^2-r^2+\frac{r^4}{R^2}-r^2)=\rho gh\left(\frac{\pi {(R^2-r^2)}^2}{R^2}\right)$

$=\rho gh\frac{{(S-s)}^2}{S}=6.02N$ on putting the values.

Note: If we try to calculate $F$ from the momentum change of the liquid flowing out will be wrong even as regards the sign of the force.

There is of course the effect of pressure at $S$ and $s$ but quantitative derivation of $F$ from Newton's law is difficult.