5.3 Application of Bernoulli's theorem

It is a device which is used for measuring the speed and rate of flow of incompressible & non-viscous liquid through pipes.

The schematic diagram of venturimeter is as shown below,

From the equation of continuity we can write, $${A_1}{v_1} = {A_2}{v_2}\quad ...(i)$$

We know, \[\left. \begin{gathered} {P_1} = {P_O} + \rho g{h_1} \hspace{1em} \\ {P_2} = {P_O} + \rho g{h_2} \hspace{1em} \\ \end{gathered} \right\}\quad ...(ii)\]

Applying Bernoulli's equation at point $A$ and $B$ we get, $$\begin{equation} \begin{aligned} {P_1} + 0 + \frac{1}{2}\rho v_1^2 = {P_2} + 0 + \frac{1}{2}\rho v_2^2 \\ {P_1} - {P_2} = \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right)\quad ...(iii) \\\end{aligned} \end{equation} $$

From equation $(i)$, $(ii)$ and $(iii)$ we get, $$\begin{equation} \begin{aligned} \left( {{P_O} + \rho g{h_1}} \right) - \left( {{P_O} + \rho g{h_2}} \right) = \frac{1}{2}\rho \left[ {v_2^2 - v_2^2{{\left( {\frac{{{A_2}}}{{{A_1}}}} \right)}^2}} \right] \\ \rho g\left( {{h_1} - {h_2}} \right) = \frac{1}{2}\rho v_2^2\left[ {1 - {{\left( {\frac{{{A_2}}}{{{A_1}}}} \right)}^2}} \right] \\ \rho gh = \frac{1}{2}\rho v_2^2\left[ {1 - {{\left( {\frac{{{A_2}}}{{{A_1}}}} \right)}^2}} \right]\quad \quad (As,\;h = {h_1} - {h_2}) \\ {v_2} = \sqrt {\frac{{2gh}}{{1 - {{\left( {\frac{{{A_2}}}{{{A_1}}}} \right)}^2}}}} \\\end{aligned} \end{equation} $$

If ${A_1} > > {A_2}$ then, $${v_2} = \sqrt {2gh} $$Rate of flow of liquid is given by, $$\begin{equation} \begin{aligned} \frac{{dV}}{{dt}} = {A_2}{v_2} \\ \frac{{dV}}{{dt}} = {A_2}\sqrt {\frac{{2gh}}{{1 - {{\left( {\frac{{{A_2}}}{{{A_1}}}} \right)}^2}}}} \\\end{aligned} \end{equation} $$

Aircraft wings are designed in such a way, so that the speed of air at the top of the wing is greater than the speed of the air beneath the wing.

This difference in speed creates a pressure difference which provides an upward thrust force and allows the aircraft to fly.

According to Bernoulli's equation we get,

$$\begin{equation} \begin{aligned} {P_1} + 0 + \frac{1}{2}\rho v_1^2 = {P_2} + 0 + \frac{1}{2}\rho v_2^2 \\ {P_1} = {P_2} + \frac{1}{2}\rho \left( {v_2^2 - v_1^2} \right) \\\end{aligned} \end{equation} $$

As, ${v_1} > {v_2}$ So, ${P_2} > {P_1}$

So, higher the value of $P_2$

Torricelli's law states that velocity of efflux i.e. the velocity with which the liquid flows out of an orifice is equal to that which a freely falling body would acquire in falling through a vertical distance equal to the depth of the orifice below the free surface of the liquid.

$v$ is the velocity of an efflux.

From the equation of continuity, $${A_1}{v_1} = {A_2}{v_2}$$ So, $${v_1} = {v_2}\left( {\frac{{{A_2}}}{{{A_1}}}} \right)$$

As, ${A_1} > > {A_2}$. So, ${v_1} \approx 0$

Applying Bernoulli's equation at point $A$ and $B$ we get, $${P_O} + \rho gh + \frac{1}{2}\rho v_1^2 = {P_O} + 0 + \frac{1}{2}\rho v_2^2$$

As, ${v_1} = 0$ So, $$\begin{equation} \begin{aligned} \rho gh = \frac{1}{2}\rho v_2^2 \\ {v_2} = \sqrt {2gh} \\\end{aligned} \end{equation} $$

Time taken by liquid to reach the ground, $$\begin{equation} \begin{aligned} {s_y} = {u_y}t + \frac{1}{2}{a_y}{t^2} \\ - (H - h) = (0 \times t) + \frac{1}{2}( - g){t^2} \\ t = \sqrt {\frac{{2(H - h)}}{g}} \\\end{aligned} \end{equation} $$

Range of a liquid flowing out of an orifice is, $$\begin{equation} \begin{aligned} R = {v_2}t \\ R = \sqrt {2gh} \times \sqrt {\frac{{2(H - h)}}{g}} \\ R = 2\sqrt {h(H - h)} \\\end{aligned} \end{equation} $$

with liquid of density $\rho $ as shown in the figure. A hole of cross sectional area $A_2$ is made at the bottom the tank.

From the equation of continuity, $${A_1}{v_1} = {A_2}{v_2}$$

Let an infinitesimally small section of liquid of thickness $dx$ which flows out of an orifice in time $dt$ as shown in the figure.

Volume of the infinitesimally small section, $$dV = {A_1}dx$$

Volume of the liquid flows out of an orifice, $${A_2}{v_2}dt$$

So, $$ - {A_1}dx = {A_2}{v_2}dt$$ -ve sign is because the volume in the tank is decreasing

$$dt = - \frac{{{A_1}}}{{{A_2}}}\frac{{dx}}{{\sqrt {2gx} }}$$

Integrating the above equation with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_0^t {dt} = - \frac{{{A_1}}}{{{A_2}}}\int\limits_H^0 {\frac{{dx}}{{\sqrt {2gx} }}} \\ t = \frac{{{A_1}}}{{{A_2}}}\sqrt {\frac{{2H}}{g}} \\\end{aligned} \end{equation} $$

A siphon is an inverted U shaped pipe which draws liquid from a tank and delivers it to the another tank due to the pressure difference.

Area of cross-section of the tank and pipe is $A_1$ and $A_2$ respectively.