In Fig. 10-54, take into account the speed of the top surface of the tank and show that the speed of fluid leaving an opening near the bottom is \({{\bf{v}}_{\bf{1}}}{\bf{ = }}\sqrt {\frac{{{\bf{2gh}}}}{{\left( {{\bf{1 - A}}_{\bf{1}}^{\bf{2}}{\bf{/A}}_{\bf{2}}^{\bf{2}}} \right)}}} \),

where \({\bf{h = }}{{\bf{y}}_{\bf{2}}} - {{\bf{y}}_{\bf{1}}}\), and \({{\bf{A}}_{\bf{1}}}\) and \({{\bf{A}}_{\bf{2}}}\) are the areas of the opening and of the top surface, respectively. Assume \({{\bf{A}}_{\bf{1}}}{\bf{ < < }}{{\bf{A}}_{\bf{2}}}\) so that the flow remains nearly steady and laminar.

Figure 10-54

The multiplication of area of the pipe and the speed of flow at any point of the pipe is always constant, stated by the equation of continuity.

The formula for equation of continuity,

\(\begin{array}{c}{A_1}{v_1} = {A_2}{v_2}\\{v_2} = \frac{{{A_1}{v_1}}}{{{A_2}}}\end{array}\)

The expression fo the Bernoulli’s equation is given as,

\({P_1} + \frac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_2} + \frac{1}{2}\rho v_2^2 + \rho g{h_2}\)

As the flow of fluid remains steady. Therefore,

\({P_1} = {P_2}\).

Substitute the values in the above equation,

\(\begin{array}{c}{P_1} + \frac{1}{2}\rho v_1^2 + \rho g{h_1} = {P_1} + \frac{1}{2}\rho v_2^2 + \rho g{h_2}\\\frac{1}{2}\rho \left( {v_1^2 - v_2^2} \right) = \rho g\left( {{h_2} - {h_1}} \right)\\v_1^2 - v_2^2 = 2g\left( {{h_2} - {h_1}} \right)\end{array}\)

Above equation can be further solved as,

\(\begin{array}{c}v_1^2 - {\left( {\frac{{{A_1}{v_1}}}{{{A_2}}}} \right)^2} = 2gy\\v_1^2\left( {1 - \frac{{A_1^2}}{{A_2^2}}} \right) = 2gy\\v_1^2 = \frac{{2gy}}{{\left( {1 - \frac{{A_1^2}}{{A_2^2}}} \right)}}\\{v_1} = \sqrt {\frac{{2gy}}{{1 - \frac{{A_1^2}}{{A_2^2}}}}} \end{array}\)

Therefore, the velocity of fluid at the bottom is \({v_1} = \sqrt {\frac{{2gh}}{{\left( {1 - A_1^2/A_2^2} \right)}}} \).