Chapter 3, Problem 67P

A cylindrical tank is folly filled with water (Fig. P3-67). In order to increase the flow from the tank, an additional pressure is applied to the water surface by a compressor. For $P_0=0,P_0=5$ bar, and $P_0=10$ bar, calculate the hydrostatic force on the surface A exerted by water.

To determine

The hydrostatic force on the surface (A) exerted by water.

Answer to Problem 67P

The hydrostatic force on the surface (A) exerted by water for air pressure $0\text{bar}$ is $0.832\text{kN}$

The hydrostatic force on the surface (A) exerted by water for air pressure $5\text{bar}$ is $142.18\text{kN}$

The hydrostatic force on the surface (A) exerted by water for air pressure $10\text{bar}$ is $283.53\text{kN}$

Explanation of Solution

Given information:

The pressure applied to the water by compressor are $0\text{bar}$, $5\text{bar,}$ and $10\text{bar}$.

Write the expression for the pressure exerted by the fluid.

  $P_w=\rho gh$...... (I)

Density of the fluid is $\rho$, the acceleration due to gravity is $g,$ and the height of the fluid is $h$.

Substitute $\frac{D}{2}$ for $h$ in equation (I).

  $P_w=\rho g\left(\frac{D}{2}\right)$..... (II)

Here, diameter of the pipe is $D$

Write the expression for total pressure exerted on the surface.

  $P_T=P_o+P_w$...... (III)

Here, air pressure is $P_o$, pressure exerted by the water is $P_w$.

Substitute $\rho g\left(\frac{D}{2}\right)$ for $P_w$ in equation (III).

  $P_T=P_o+\rho g\left(\frac{D}{2}\right)$...... (IV)

Write the expression for the area of the surface $A$.

  $A=\frac{\pi }{4}{D}^2$

Here the diameter of the pipe is $D$.

Write the expression for the force acting on the surface

  $F=P_{T}A$..... (V)

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

  $F=P_T\left(\frac{\pi }{4}{D}^2\right)$...... (VI)

Calculation:

For air pressure $0\text{bar}$.

Substitute $0$ for $P_o$

  $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $60\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}{P}_T=\left(0\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}}{2}\right)\\ =\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\right)\\ =2943\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =2.943\text{kPa}\end{array}$

Substitute $2.943\text{kPa}$ for $P_T$ and and $60\text{cm}$ for $D$ in Equation (VI)

  $\begin{array}{c}F=\left(2.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\right)}^2\right)\\ =\left(2.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\\ =\left(2.943\text{kPa}\right)\left(0.2827{\text{m}}^{\text{2}}\right)\\ =0.832\text{kN}\end{array}$

For air pressure $5\text{bar}$.

Substitute $5\text{bar}$ for $P_o$

  $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $60\text{cm}$ for $D$ in Equation (VI).

  $\begin{array}{c}{P}_T=\left(5\text{bar}\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}}{2}\right)\\ =\left(5\text{bar}\left(\frac{100\text{kPa}}{1\text{bar}}\right)\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\right)\\ =502.943\text{kPa}\end{array}$

Substitute $502.943\text{kPa}$ for $P_T$ and $60\text{cm}$ for $D$ in Equation (VI).

  $\begin{array}{c}F=\left(502.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\right)}^2\right)\\ =\left(502.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\\ =\left(502.943\text{kPa}\right)\left(0.2827{\text{m}}^{\text{2}}\right)\\ =142.18\text{kN}\end{array}$

For air pressure $10\text{bar}$.

Substitute $10\text{bar}$ for $P_o$

  $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$

  $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $60\text{cm}$ for $D$ in Equation (VI).

  $\begin{array}{c}{P}_T=\left(10\text{bar}\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}}{2}\right)\\ =\left(10\text{bar}\left(\frac{100\text{kPa}}{1\text{bar}}\right)\right)+\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(\frac{60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}{2}\right)\\ =1002.943\text{kPa}\end{array}$

Substitute $1002.943\text{kPa}$ for $P_T$ and and $60\text{cm}$ for $D$ in Equation (VI)

  $\begin{array}{c}F=\left(1002.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\right)}^2\right)\\ =\left(1002.943\text{kPa}\right)\left(\frac{\pi }{4}{\left(60\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\right)\\ =\left(1002.943\text{kPa}\right)\left(0.2827{\text{m}}^{\text{2}}\right)\\ =283.53\text{kN}\end{array}$

Conclusion:

The hydrostatic force on the surface (A) exerted by water for air pressure $0\text{bar}$ is $0.832\text{kN}$

The hydrostatic force on the surface (A) exerted by water for air pressure $5\text{bar}$ is $142.18\text{kN}$

The hydrostatic force on the surface (A) exerted by water for air pressure $10\text{bar}$ is $283.53\text{kN}$