Chapter 5, Problem 103P

A pressurized 2-rn-diameter tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmosphere. The water level initially is 3 m above the outlet. The tank air pressure above the water level is maintained at 450 kPa absolute and the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine (a) how long it will take for half of the water in the tank to be discharged and (b) the water level in the tank after 10 s.

To determine

(a)

The time required for half discharge of water from tank.

Answer to Problem 103P

The time required for half discharge of water from tank is $21.8\text{s}$.

Explanation of Solution

Given information:

The pressurized tank diameter is $2\text{m}$, the diameter of orifice is $10\text{cm}$, the initial water level is $3\text{m}$, the absolute pressure of water level is $450\text{kPa,}$ and atmospheric pressure is $100\text{kPa}$.

Write the expression for the Bernoulli's Equation.

  $\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2$  .......(I)

Here, the density of water is $\rho$, the velocity at inlet is $V_1$, the initial water pressure is, the initial water pressure is $p_1$, the velocity at exit is $V_2$, the height of water level at inlet is $z_1$ the atmospheric pressure is $p_2,$ and the height of water level at exit is $z_2$.

Write the expression for the area of orifice.

  $A=\frac{\pi }{4}{D}^2$

Here, the diameter of orifice is $D$, the height of water in tank is h.

Write the expression for the volume flow rate through orifice.

  $V_2=\frac{\dot{v}}{A}$  .......(II)

Here, the time is $t$.

Substitute $\frac{\pi }{4}{D}^2$ for in Equation (II).

  $\begin{array}{l}{V}_2=\frac{\dot{v}}{\frac{\pi }{4}{D}^2}\\ \dot{v}=V_2\frac{\pi }{4}{D}^2\end{array}$  .......(III)

Write the expression for volume of water in tank.

  $v=\frac{\pi }{4}{D}_t^{2}h$  .......(IV)

Here, the diameter of tank is $D_t$ and the height of water in tank is $h$.

Write the expression for time required in half discharge.

  $t=\frac{\frac{v}{2}}{\dot{v}}$  .......(V)

Calculations:

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $450\text{kPa}$ for $p_1$, $100\text{kPa}$ for $p_2$, $3\text{m}$ for $z_1$, $0$ for $z_2$, $0$ for $V_1$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (I).

  $\left[\begin{array}{l}\frac{450\text{kPa}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{0^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+3\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{100\text{kPa}}{1000\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{V_2^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\end{array}\right]$

  $\left[\begin{array}{l}\frac{450\text{kPa}\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +3\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{100\text{kPa}\left(\frac{1000\text{N}/{\text{m}}^{\text{2}}}{1\text{kPa}}\right)\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{V_2^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right]$

  $45.87\text{m}+3\text{m}=10.19\text{m}+\frac{V_2^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}$

  $V_2^2=758.83{\text{m}}^2/{\text{s}}^{\text{2}}$

  $V_2=27.54\text{m}/\text{s}$

Substitute $10\text{cm}$ for $D$ and $27.54\text{m}/\text{s}$ for $V_2$ in Equation (III).

  $\begin{array}{c}\dot{v}=27.54\text{m}/\text{s}\frac{\pi }{4}{\left(10\text{cm}\right)}^2\\ =27.54\text{m}/\text{s}\frac{\pi }{4}{\left(10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =0.216{\text{m}}^3/\text{s}\end{array}$

Substitute $2\text{m}$ for $D_t$ and $3\text{m}$ for $h$ in Equation (IV).

  $\begin{array}{c}v=\frac{\pi }{4}{\left(2\text{m}\right)}^2\left(3\text{m}\right)\\ =\frac{\pi }{4}\left(4{\text{m}}^{\text{2}}\right)\left(3\text{m}\right)\\ =\text{9}\text{.42}{\text{m}}^3\end{array}$

Substitute $\text{9}\text{.42}{\text{m}}^3$ for $v$ and $0.216{\text{m}}^3/\text{s}$ for $\dot{v}$ in Equation (V).

  $\begin{array}{c}t=\frac{\frac{\text{9}\text{.42}{\text{m}}^3}{2}}{0.216{\text{m}}^3/\text{s}}\\ =21.8\text{s}\end{array}$

Conclusion:

The time required for half discharge of water from tank is $21.8\text{s}$.

To determine

(b)

The water level in tank after $10\text{s}$.

Answer to Problem 103P

The water level in tank after $10\text{s}$ is $2.13\text{m}$.

Explanation of Solution

Given information:

The pressurized tank diameter is $2\text{m}$, the diameter of orifice is $10\text{cm}$, the initial water level is $3\text{m}$, the absolute pressure of water level is $450\text{kPa,}$ and atmospheric pressure is $100\text{kPa}$.

Write the expression for volume discharge in $10\text{s}$.

  $v_d=\dot{v}t$  .......(VI)

Write the expression for the volume of water remaining after $10\text{s}$.

  $v_r=v-v_d$  .......(VII)

Write the expression for remaining volume in tank.

  $v_r=\frac{\pi }{4}{D}_t^2{h}_r$  .......(VIII)

Here, the height of remaining volume is $h_r$.

Calculations:

Substitute $10\text{s}$ for $t$ and $0.216{\text{m}}^3/\text{s}$ for $\dot{v}$ in Equation (VI).

  $\begin{array}{c}{v}_d=\left(0.216{\text{m}}^3/\text{s}\right)10\text{s}\\ =\text{2}\text{.16}{\text{m}}^3\end{array}$

Substitute $\text{9}\text{.42}{\text{m}}^3$ for $v$ and $\text{2}\text{.16}{\text{m}}^3$ for $v_d$ in Equation (VII).

  $\begin{array}{c}{v}_r=\text{9}\text{.42}{\text{m}}^3-\text{2}\text{.16}{\text{m}}^3\\ =7.26{\text{m}}^3\end{array}$

Substitute $2\text{m}$ for $D_t$ and $7.26{\text{m}}^3$

  $3\text{m}$ for $v_r$ in Equation (VIII).

  $\begin{array}{l}7.26{\text{m}}^3=\frac{\pi }{4}{\left(2\text{m}\right)}^2{h}_r\\ h_r=\frac{7.26{\text{m}}^3}{\frac{\pi }{4}{\left(2\text{m}\right)}^2}\\ h_r=2.13\text{m}\end{array}$

Conclusion:

The water level in tank after $10\text{s}$ is $2.13\text{m}$.