Chapter 5, Problem 55P

A pressurized tank of water has a 10-cm-diameter orifice at the bottom, where water discharges to the atmosphere. The water level is 2.5 m above the outlet. The tank air pressure above the water level is 250 kPa (absolute) while the atmospheric pressure is 100 kPa. Neglecting frictional effects, determine the initial discharge rate of water from the tank. Answer: 0.147 m³/s

To determine

The initial discharge rate of water from the tank.

Answer to Problem 55P

The initial discharge rate of water from the tank is $0.1467{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The diameter of the orifice is $10\text{cm}$, the water level above from the outlet is $2.5\text{m}$, the pressure at the top of the water surface is $250\text{KPa}$, the atmospheric pressure is $100\text{KPa,}$ and the height of the water column is $2.5\text{m}$.

The figure below shows the different sections of the tank.

  

  Figure-(1)

Write the expression for Bernoulli's equation between the point $1$ and 2.

  $\frac{P_1}{\rho g}+\frac{V_1{}^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+Z_2$   .......(I)

Here, the pressure at section $1$ is $P_1$, the pressure at section $2$ is $P_2$, the density of flowing fluid is $\rho$, acceleration due to gravity is $g$, the velocity at section $1$ is $V_1$, the velocity at section $2$ is $V_2$, the datum head at section $1$ is $Z_1,$ and the datum head at section $2$ is $Z_2$.

Write the expression discharge of liquid through orifice.

  $\dot{Q}=A_{orifice}{V}_2$   .......(II)

Here, the area of the orifice is $A_{orifice}$.

Write the expression for area of the orifice.

  $A_{orifice}=\frac{\pi }{4}{d}_0{}^2$

Here, the diameter of the orifice is $d_0$.

Substitute $\frac{\pi }{4}{d}_0{}^2$ for $A_{orifice}$ in Equation (II).

  $\dot{V}=\left(\frac{\pi }{4}{d}_0{}^2\right)V_2$   .......(III)

Calculation:

The section $2$ is taken as datum line.

Substitute $250\text{KPa}$ for $P_1$, $100\text{KPa}$ for $P_2$, $0\text{m}/\text{s}$ for $V_1$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.5\text{m}$ for $Z_1,$ and $0\text{m}$ for $Z_2$ in Equation (I).

  $\left[\begin{array}{l}\frac{250\text{KPa}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ +\frac{0\text{m}/\text{s}}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+2.5\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{100\text{KPa}}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ +\frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\text{m}\end{array}\right]$

  $\left[\begin{array}{l}\frac{250\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+\\ \frac{0\text{m}/\text{s}}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+2.5\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{100\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}}+\\ \frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\text{m}\end{array}\right]$

  $\frac{25.484\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\left(\frac{1\text{N}/{\text{m}}^{\text{3}}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)}+2.5\text{m}=\frac{10.1936\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{1\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}\left(\frac{1\text{N}/{\text{m}}^{\text{3}}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)}+\frac{V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}$

  $\begin{array}{l}17.790\text{m}=\frac{V_2{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ V_2=\sqrt{17.790\text{m}\times 19.62\text{m}/{\text{s}}^{\text{2}}}\\ V_2=18.682\text{m}/\text{s}\end{array}$

Substitute $18.682\text{m}/\text{s}$ for $V_2$ and $10\text{cm}$ for $d_0$ in Equation (III).

  $\begin{array}{c}\dot{Q}=\left(\frac{\pi }{4}{\left(10\text{cm}\right)}^2\right)18.682\text{m}/\text{s}\\ =\frac{\pi }{4}{\left(10\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^{2}18.682\text{m}/\text{s}\\ =\frac{\pi }{4}{\left(0.1\text{m}\right)}^{2}18.682\text{m}/\text{s}\\ =0.1467{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The initial discharge rate of water from the tank is $0.1467{\text{m}}^{\text{3}}/\text{s}$.