Chapter 10, Problem 48P

To determine

The speed of fluid leaving the opening at the bottom is $v_1=\sqrt{\frac{2gh}{1-{\left(\frac{A_1}{A_2}\right)}^2}}$ , that should be proved and also, under the given condition the flow remains steady and laminar is to be proved also.

Explanation of Solution

Introduction:

According to the Bernoulli’s principle, the pressure inside the fluid system inversely deals with the volume of the system. The Pressure is decreased with the increase in the volume of the fluid system and vice-versa. When there is a pressure difference across the body, it causes a push on the body from the region of high pressure to that of low pressure.

The Bernoulli’s equation is written as,

  $P_1+\rho g{h}_1+\frac{1}{2}\rho v_1{}^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2{}^2$

To prove:

The situation which sown in the figure, $v_{1}and{v}_2$ are the speed of water coming out from the opening and top surfaces respectively. The pressure above the top surfaces is $P_0$ atmospheric.

Now according to the Bernoulli’s equation,

  $\begin{array}{l}{P}_0+\rho g{h}_1+\frac{1}{2}\rho v_1{}^2=P_0+\rho g{h}_2+\frac{1}{2}\rho v_2{}^2\\ P_0+\frac{1}{2}\rho v_2{}^2+\rho g\left(h_2-h_1\right)=P_0+\frac{1}{2}\rho v_1{}^2\mathrm{.....}\left(1\right)\end{array}$

From the equation of continuity,

  $\begin{array}{l}\\ A_1{v}_1=A_2{v}_2\\ v_2=\frac{A_1}{A_2}{v}_1\mathrm{....}\left(2\right)\end{array}$

From the equations (1) and (2), the resulted expression is

  $\begin{array}{l}{P}_0+\rho g{h}_1+\frac{1}{2}\rho v_1{}^2=P_0+\rho g{h}_2+\frac{1}{2}\rho v_2{}^2\\ P_0+\frac{1}{2}\rho v_2{}^2+\rho g\left(h_2-h_1\right)=P_0+\frac{1}{2}\rho v_1{}^2\\ \frac{1}{2}\rho v_1{}^2-\frac{1}{2}\rho v_2{}^2=\rho gh\\ \frac{1}{2}{v}_1{}^2-\frac{1}{2}{v}_2{}^2=gh\\ \frac{1}{2}{v}_1{}^2-\frac{1}{2}{\left(\frac{A_1}{A_2}{v}_1\right)}^2=gh\\ v_1{}^2\left(1-{\left(\frac{A_1}{A_2}\right)}^2\right)=2gh\end{array}$

  $\begin{array}{l}{v}_1{}^2=\frac{2gh}{1-{\left(\frac{A_1}{A_2}\right)}^2}\\ v_1=\sqrt{\frac{2gh}{1-{\left(\frac{A_1}{A_2}\right)}^2}}\end{array}$

Where, $h=y_2-y_1$

By assuming the given condition,

  $\begin{array}{l}{A}_1<<,A_2\\ \frac{A_1}{A_2}<<<1\end{array}$

So, the ratio of area is neglected as compared to the one. So, the seed of fluid becomes,

  $v_1=\sqrt{2gh}$

Hence, the flow remains nearly laminar and steady.

Conclusion:

Hence, the speed of fluid leaving the opening at the bottom $v_1=\sqrt{\frac{2gh}{1-{\left(\frac{A_1}{A_2}\right)}^2}}$ is proved and hence the flow remains nearly laminar and steady.