Chapter 3, Problem 26P

Water from a reservoir is raised in a vertical tube of internal diameter D =30 cm under the influence of the pulling force F of a piston. Determine the force needed to raise the water to a height of h = 1.5 m above the free surface. What would your response be for h = 3 m? Also, taking the atmospheric pressure to be 96 kPa, plot the absolute water pressure at the piston face as h varies from 0 to 3 m.

FIGURE P3-26

To determine

The force needed to raise the water to a height of $h=1.5\text{m}$ above the free surface.

The force needed to raise the water to a height of $h=3\text{m}$ above the free surface.

The force needed to raise the water to a height of $h=0\text{m}$ above the free surface.

Answer to Problem 26P

The force needed to raise the water to a height of $h=1.5\text{m}$ above the free surface is $1.04\text{kN}$.

The force needed to raise the water to a height of $h=3\text{m}$ above the free surface is $2.08\text{kN}$.

The force needed to raise the water to a height of $h=0\text{m}$ above the free surface is $0\text{kN}$.

Explanation of Solution

Given information:

The internal diameter of the piston is $30\text{cm}$, and atmospheric pressure is $96\text{kPa}$.

Write the expression for the pressure exerted by a fluid

  $P=\rho gh$   ...... (I)

Here, the pressure exerted by a fluid is $P$, and acceleration due to gravity is $g$, and density of the fluid is $\rho$, height of the column is $h$.

Write the expression for area.

  $A=\frac{\pi }{4}{D}^2$

Here, cross-section area of the piston is $A$, internal diameter of the piston is $D$.

  $W=AP$   ...... (II)

Here, weight exerted by the fluid column is $W$.

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

  $W=\left(\frac{\pi }{4}{D}^2\right)P$   ...... (III)

Write the expression for net vertical force.

  $\begin{array}{l}F-W=0\\ F=W\end{array}$   ...... (IV)

Here, force required to raise the water is $F$

Write the expression for absolute pressure acting on the piston face.

  $P_{abs}=P_{atm}-P$   ...... (V)

Here, absolute pressure is $P_{abs}$, atmospheric pressure is $P_{atm}$.

Calculation:

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1.5\text{m}$ for $h$ in Equation (I).

  $\begin{array}{c}P=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(1.5\text{m}\right)\\ \text{=14}\text{.72}\times \text{1000}\left(\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(1/{\text{m}}^2\right)\\ =14.72\times \text{1000}\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\end{array}$

Substitute $30\text{cm}$ for $D$, $14.72\text{kPa}$ for $P$ in Equation (III).

  $\begin{array}{c}W=\frac{\pi }{4}{\left(30\text{cm}\right)}^2\left(14.72\text{kPa}\right)\\ =\frac{\pi }{4}{\left(\left(30\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right)}^2\left(14.72\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^2}{1\text{kPa}}\right)\\ =1.04\text{kN}\end{array}$

Substitute $1.04\text{N}$ for $W$ in Equation (IV).

  ${\left(F\right)}_{h=1.5}=1.04\text{kN}$

Here, the force required to raise the water is at height of $1.5\text{m}$ is ${\left(F\right)}_{h=1.5}$.

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for $g$, and $3\text{m}$ for $h$ in Equation (I).

  $\begin{array}{c}P=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(3\text{m}\right)\\ =29.43\times \text{1000}\left(\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(1/{\text{m}}^2\right)\\ =29.43\times \text{1000}\text{Pa}\left(\frac{1\text{kPa}}{1000\text{Pa}}\right)\\ =29.43\text{kPa}\end{array}$

Substitute $30\text{cm}$ for $D$, $29.43\text{kPa}$ for $P$ in Equation (III).

  $\begin{array}{c}W=\frac{\pi }{4}{\left(30\text{cm}\right)}^2\left(29.43\text{kPa}\right)\\ =\frac{\pi }{4}{\left(\left(30\text{cm}\right)\left(\frac{0.01\text{m}}{1\text{cm}}\right)\right)}^2\left(29.43\text{kPa}\right)\left(\frac{1\text{kN}/{\text{m}}^2}{1\text{kPa}}\right)\\ =2.08\text{kN}\end{array}$

Substitute $2.08\text{kN}$ for $W$ in Equation (IV).

  $F=2.08\text{kN}$

Here, the force required to raise the water is at height of $3\text{m}$ is ${\left(F\right)}_{h=3}$.

Substitute $1000\text{kg}/{\text{m}}^3$ for $\rho$, $\text{9}\text{.81}\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0\text{m}$ for $h$ in Equation (I).

  $\begin{array}{c}P=\left(1000\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left(0\text{m}\right)\\ =0\text{kPa}\end{array}$

Substitute $30\text{cm}$ for $D$, $0\text{kPa}$ for $P$ in Equation (III).

  $\begin{array}{c}W=\frac{\pi }{4}{\left(30\text{cm}\right)}^2\left(0\text{kPa}\right)\\ =0\text{kN}\end{array}$

Substitute $0\text{kN}$ for $W$ in Equation (IV).

  $F=0\text{kN}$

Here, the force required to raise the water is at height of $0\text{m}$ is ${\left(F\right)}_{h=0}$.

Substitute $96\text{kPa}$ for $P_{atm}$ and $14.72\text{kPa}$ for $P$ in Equation (V).

  $\begin{array}{c}{\left(P_{abs}\right)}_{h=1.5}=96\text{kPa}-14.72\text{kPa}\\ \text{=81}\text{.28}\text{kPa}\end{array}$

Here, absolute pressure at height $1.5\text{m}$ is ${\left(P_{abs}\right)}_{h=1.5}$.

Substitute $96\text{kPa}$ for $P_{atm}$ and $29.43\text{kPa}$ for $P$. in Equation (V).

  $\begin{array}{c}{\left(P_{abs}\right)}_{h=3}=96\text{kPa}-29.43\text{kPa}\\ =\text{66}\text{.57}\text{kPa}\end{array}$

Here, absolute pressure at height $3\text{m}$ is ${\left(P_{abs}\right)}_{h=3}$.

Substitute $96\text{kPa}$ for $P_{atm}$ and $0\text{kPa}$ for $P$. in Equation (V).

  $\begin{array}{c}{\left(P_{abs}\right)}_{h=0}=96\text{kPa}-0\text{kPa}\\ =96\text{kPa}\end{array}$

Here, absolute pressure at height $0\text{m}$ is ${\left(P_{abs}\right)}_{h=0}$.

The following Table-(1) shows the water pressure at different height of water level in the tube.

Water level $h\left(\text{m}\right)$	Absolute Pressure $P\left(\text{Pa}\right)$
$1.0$	$96000$
$0.5$	$91095$
$1.0$	$86190$
$1.5$	$81285$
$2.0$	$76380$
$2.5$	$71475$
$3.0$	$66570$

The following figure variation of absolute water verse height of water in the tube.

  

Figure-(1)

Conclusion:

The force needed to raise the water to a height of $h=1.5\text{m}$ above the free surface is $1.04\text{kN}$.

The force needed to raise the water to a height of $h=3\text{m}$ above the free surface is $2.08\text{kN}$.

The force needed to raise the water to a height of $h=0\text{m}$ above the free surface is $0\text{kN}$.