Question

A sealed tank containing seawater to a height of 11.0 m also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

Short answer

The velocity of the moving water is $28.3\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Given Data

The height is${\mathrm{y}}_1=11\mathrm{m}$ .

The gauge pressure is $\mathrm{P}=3\mathrm{atm}$.

Understanding the Bernoulli’s equation

In this problem, the velocity of the water leaving from the small hole of the tank is estimated by using the relation of velocity obtained from Bernoulli’s equation.

Determining the velocity of the moving water

The relation obtained from Bernoulli’s equation can be written as:

$\mathrm{v}=\sqrt{2{\mathrm{gy}}_1+\frac{2\mathrm{P}}{\mathrm{p}}}$

Here,$p$ is the density of seawater and g is the gravitational acceleration.

Substitute 3 atm for P, $1030\mathrm{kg}/{\mathrm{m}}^3$for $p$, 11 m for ${\mathrm{y}}_1$and $9.80\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$for g in the above relation.

$$\begin{gathered} \mathrm{v}=\sqrt{2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)\left(11\mathrm{m}\right)+\frac{2\left(3\mathrm{atm}\times {\displaystyle \frac{1.013\times {10}^5\mathrm{N}/{\mathrm{m}}^2}{1\mathrm{atm}}}\right)}{\left(1030\mathrm{kg}/{\mathrm{m}}^3\right)}} \\ \mathrm{v}=28.3\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Thus, the velocity of the water is $28.3\mathrm{m}/{\mathrm{s}}^2$.