Suppose the radius at A is R and it decreases uniformaly to...

Suppose the radius at $A$ is $R$ and it decreases uniformaly to $r$ at $B$ where $S=\pi R^{2}$ and $s=\pi r^{2} .$ Assume also that the semi vectical angle at 0 is $\alpha$. Then $$ \frac{R}{L_{2}}=\frac{r}{L_{1}}=\frac{y}{x} $$ So $$ y=r+\frac{R-r}{L_{2}-L_{1}}\left(x-L_{1)}\right. $$ where $y$ is the radius at the point $P$ distant $x$ from the vertex $O$. Suppose the velocity with which the liquid flows out is $V$ at $A, v$ 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 $$ v=\sqrt{2 g h} $$ and Bernoulli's theorem gives $$ p_{p}+\frac{1}{2} \rho u^{2}=p_{0}+\frac{1}{2} \rho v^{2} $$ 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 is then $F=\int\left(p_{\rho}-p_{0}\right) \sin \theta 2 \pi y d s$ We have subtracted $p_{0}$ which is the force due to atmosphenic pressure the factor $\sin \theta$ gives horizontal component of the force and $d s$ is the length of the element of nozzle surface, $d s=d x \sec \theta$ and $$ \tan \theta=\frac{R-r}{L_{2}-L_{1}} $$ Thus $$ F=\int_{L}^{L_{2}} \frac{1}{2}\left(v^{2}-u^{2}\right) \rho 2 \pi y \frac{R-r}{L_{2}-L_{1}} d x $$$$ \begin{aligned} &=\pi \rho \int_{r}^{R} v^{2}\left(1-\frac{r^{4}}{y^{4}}\right) y d y \\ &=\pi \rho v^{2} \frac{1}{2}\left(R^{2}-r^{2}+\frac{r^{4}}{R^{2}}-r^{2}\right)=\rho g h\left(\frac{\pi\left(R^{2}-r^{2}\right)^{2}}{R^{2}}\right) \end{aligned} $$ $=\rho g h(S-s)^{2} / S=6.02 \mathrm{~N}$ on putting the values. Note : If we try to calculate $F$ from the momentum change of the liquid flowing out wi 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 $F$ fron Newton's law is difficult.

Suppose the radius at $A$ is $R$ and it decreases uniformaly to $r$ at $B$ where $S=\pi R^{2}$ and $s=\pi r^{2} .$ Assume also that the semi vectical angle at 0 is $\alpha$. Then
$$
\frac{R}{L_{2}}=\frac{r}{L_{1}}=\frac{y}{x}
$$
So
$$
y=r+\frac{R-r}{L_{2}-L_{1}}\left(x-L_{1)}\right.
$$
where $y$ is the radius at the point $P$ distant $x$ from the vertex $O$. Suppose the velocity with which the liquid flows out is $V$ at $A, v$ 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
$$
v=\sqrt{2 g h}
$$
and Bernoulli's theorem gives
$$
p_{p}+\frac{1}{2} \rho u^{2}=p_{0}+\frac{1}{2} \rho v^{2}
$$
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 is then $F=\int\left(p_{\rho}-p_{0}\right) \sin \theta 2 \pi y d s$
We have subtracted $p_{0}$ which is the force due to atmosphenic pressure the factor $\sin \theta$ gives horizontal component of the force and $d s$ is the length of the element of nozzle surface, $d s=d x \sec \theta$ and
$$
\tan \theta=\frac{R-r}{L_{2}-L_{1}}
$$
Thus
$$
F=\int_{L}^{L_{2}} \frac{1}{2}\left(v^{2}-u^{2}\right) \rho 2 \pi y \frac{R-r}{L_{2}-L_{1}} d x
$$$$
\begin{aligned}
&=\pi \rho \int_{r}^{R} v^{2}\left(1-\frac{r^{4}}{y^{4}}\right) y d y \\
&=\pi \rho v^{2} \frac{1}{2}\left(R^{2}-r^{2}+\frac{r^{4}}{R^{2}}-r^{2}\right)=\rho g h\left(\frac{\pi\left(R^{2}-r^{2}\right)^{2}}{R^{2}}\right)
\end{aligned}
$$
$=\rho g h(S-s)^{2} / S=6.02 \mathrm{~N}$ on putting the values. Note : If we try to calculate $F$ from the momentum change of the liquid flowing out wi 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 $F$ fron Newton's law is difficult.

Key Concepts

Continuity Equation

The continuity equation in fluid mechanics expresses the conservation of mass for an incompressible fluid. It states that the product of the cross-sectional area and the velocity of the fluid remains constant along a streamline, ensuring that the mass flow rate is conserved as the fluid moves through regions of varying cross-sectional dimensions.

Bernoulli's Principle

Bernoulli's principle is based on the conservation of energy in fluid flow. It relates the pressure, kinetic energy, and potential energy along a streamline, indicating that an increase in a fluid's velocity leads to a decrease in its static pressure, and vice versa. This principle is key in analyzing how pressure differences arise due to velocity variations.

Pressure Distribution in Fluids

Understanding the pressure distribution in a fluid is crucial for analyzing the forces it exerts on surfaces. As fluid flows through varying cross-sections or around objects, the pressure changes according to alterations in velocity and elevation. This distribution is fundamental when calculating net forces exerted by the fluid on structures or components within the system.

Integration of Pressure Forces Over a Surface

Calculating the net force on a surface immersed in or in contact with a fluid requires integrating the pressure forces acting over that surface. This process takes into account the varying pressure magnitudes and directions at different points, allowing for the determination of the resultant force that contributes to phenomena such as thrust or drag.

Momentum Analysis in Fluid Flow

Momentum analysis involves applying Newton's laws to a fluid system to relate the change in momentum to the forces acting on the fluid. While this approach is powerful, it can become complex in scenarios with significant pressure gradients and velocity variations. Care must be taken, as directly applying momentum conservation without considering pressure contributions may lead to incorrect or incomplete results.

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