Question

A jet of water squirts out horizontally from a hole near the bottom of the tank shown in Figure P14.84. If the hole has a diameter of $\mathbf{3}.\mathbf{50}\mathbf{mm}$, what is the height$h$of the water level in the tank?

Short answer

The height $h$of the water level in the tank is $h=y_1-y_2=9.00cm$.

Step-by-step solution

Bernoulli's equation:

The sum of pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume have the same value for an ideal fluid at all points along the streamline. This result is summarized in Bernoulli's equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Here, $P$is the pressure, $\rho$is the density, $g$is the gravity, and $y$is the distance.

the height of the water level in the tank:

First, consider the path from the viewpoint of projectile motion to find the speed at which the water emerges from the tank.

From

$\Delta y=v_{yi}t+\frac{1}{2}{a}_y{t}^2$ ….. (1)

Here,

The velocity,$v_{yi}=0$,

Distance,$\Delta y=-1.00m$

Acceleration,$a_y=-g$

The distance,$\Delta x=0.600m$

Substitute these values into equation (1).

$\Delta y=0+\frac{1}{2}{a}_y{t}^2$

$$\begin{gathered} t=\sqrt{\frac{2\left(∆y\right)}{a_y}} \\ =\sqrt{\frac{2\times \left(-1.00m\right)}{\left(-g\right)}} \\ =\sqrt{\frac{\left(2.00m\right)}{9.8m/s^2}} \\ =0.452s \end{gathered}$$

From the horizontal motion, the speed of the water coming out of the hole is

$$\begin{gathered} v_2=v_{xi} \\ =\frac{∆x}{t} \\ =\frac{0.600m}{0.452s} \\ =1.33m/s \end{gathered}$$

Now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the hole. With $P_1=P_2=P_{atm}$and $v_1\approx 0$, this gives

$$\begin{gathered} P_1+\frac{1}{2}\rho v_1^2+\rho g{y}_1=P_2+\frac{1}{2}\rho v_2^2+\rho g{y}_2 \\ {P}_{atm}+0+\rho g{y}_1=P_{atm}+\frac{1}{2}\rho v_2^2+\rho g{y}_2 \\ \rho g{y}_1=\frac{1}{2}\rho v_2^2+\rho g{y}_2 \end{gathered}$$

$$\begin{gathered} h=y_1-y_2 \\ =\frac{v_2^2}{2g} \end{gathered}$$

$$\begin{gathered} h=\frac{{\left(1.33m/s\right)}^2}{2\times 9.8m/s^2} \\ =9.00\times {10}^{-2} \\ =9.00cm \end{gathered}$$