Question

An open large tank with a nozzle attached contains three immiscible, inviscid fluids as shown. Assuming that the changes in $h_1,h_2$ and $h_3$ are negligible, the instantaneous discharge velocity is

• A. $\sqrt{2g{h}_3(1+\frac{{\rho }_1}{{\rho }_3}\frac{h_1}{h_3}+\frac{{\rho }_2}{{\rho }_3}\frac{h_2}{h_3})}$
• B. $\sqrt{2g(h_1+h_2+h_3)}$
• C. $\sqrt{2g\left(\frac{{\rho }_1{h}_1+{\rho }_2{h}_2+{\rho }_3{h}_3}{{\rho }_1+{\rho }_2+{\rho }_3}\right)}$
• D. $\sqrt{2g\left(\frac{{\rho }_3{h}_2{h}_3+{\rho }_2{h}_3{h}_1+{\rho }_3{h}_1{h}_2}{{\rho }_1{h}_1+{\rho }_2{h}_2+{\rho }_3{h}_3}\right)}$

Answer 1

The correct option is A $\sqrt{2g{h}_3(1+\frac{{\rho }_1}{{\rho }_3}\frac{h_1}{h_3}+\frac{{\rho }_2}{{\rho }_3}\frac{h_2}{h_3})}$

Let the instantaneous discharge velocity is $V$.
Applying the Bernoulli's equation between states $1$ and $2$, we get

$\frac{P_1}{{\rho }_{3}g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{{\rho }_{3}g}+\frac{V_2^2}{2g}+z_2$

Considering the atmospheric pressure as reference pressure and referring figure we have,
$z_1=z_2$
$V_1=0$
$V_2=V$
$P_2=P_0$
$P_1=P_0+{\rho }_{3}g{h}_3+{\rho }_{2}g{h}_2+{\rho }_{1}g{h}_1$

Substituting the values in the Bernoulli's equation, we get

$\frac{{\rho }_{3}g{h}_3+{\rho }_{2}g{h}_2+{\rho }_{1}g{h}_1}{{\rho }_{3}g}+0+0=0+\frac{V^2}{2g}+0$

$\Rightarrow h_3+\frac{{\rho }_2}{{\rho }_3}{h}_2+\frac{{\rho }_1}{{\rho }_3}{h}_1=\frac{V^2}{2g}$

$\Rightarrow V^2=2g(h_3+\frac{{\rho }_2}{{\rho }_3}{h}_2+\frac{{\rho }_1}{{\rho }_3}{h}_1)$

$\Rightarrow V=\sqrt{2g(h_3+\frac{{\rho }_2}{{\rho }_3}{h}_2+\frac{{\rho }_1}{{\rho }_3}{h}_1)}$

$\therefore V=\sqrt{2g{h}_3(1+\frac{{\rho }_2}{{\rho }_3}\frac{h_2}{h_3}+\frac{{\rho }_1}{{\rho }_3}\frac{h_1}{h_3})}$

$\begin{array}{c}\text{Why This Question?}\\ \text{To understand the application of Bernoulli's equation which can be applied between two points under same liquid.}\end{array}$