Chapter 11, Problem 68P

One of the popular demonstrations in science museums involves the suspension of a ping-pong ball by an upward air jet. Children are amused by the ball always coming back to the center when it is pushed by a finger to the side of the jet. Explain this phenomenon using the Bernoulli equation. Also determine the velocity of air if the ball has a mass of 3.1 g and a diameter of 4.2 cm. Assume the air is at 1 atm and 25°C.

FIGURE P11-68

To determine

The velocity of air and an explanation of the phenomenon using Bernoulli equation.

Answer to Problem 68P

The velocity of the air is $4.27370\text{m}/\text{s}$.

Explanation of Solution

Given information:

The mass of the ball is $3.1\text{g}$ and the diameter of the bass is $4.2\text{cm}$.

Write the expression for the drag force acting on the ball.

  $F_D=\frac{C_D{\rho }_a\pi D^2{V}^2}{8}$........ (I)

Here, the drag ball on the ball is $F_D$, the coefficient of the drag is $C_D$, the density of the air is ${\rho }_a$, the diameter of the ball is $D$ and the velocity of the air is $V$.

Write the expression for the buoyancy force acting on the ball.

  $F_B=\frac{{\rho }_{a}g\pi D^3}{6}$........ (II)

Here, the buoyancy force is $F_B$ and the acceleration due to gravity is $g$.

Write the expression for the weight of the body.

  $W=F_B+F_D$........ (III)

Here, the weight of the body is $W$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{VD}{v}$........ (IV)

Here, the Reynolds number is $\mathrm{Re}$ and the kinematics viscosity of the ball is $v$.

Write the expression for the weight of the ball.

  $W=mg$......... (V)

Here, the mass of the ball is $m$.

Write the expression for the Bernoulli's equation.

  $\frac{P_1}{\rho g}+\frac{{\left(V\right)}^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{{\left(V_2\right)}^2}{2g}+z_2$............ (VI)

Here, the pressure at the point 1 is $P_1$, the velocity at point 1 is $V$, the height of the datum at point 1 is $z_1$, the pressure at the point 2 is $P_2$, the velocity at point 2 is $V_2$ and the height of the datum at point 2 is $z_2$.

Calculation:

Substitute $4.2\text{cm}$ for $D$ and $1.184\text{kg}/{\text{m}}^3$ for ${\rho }_a$ and in Equation (I).

  $\begin{array}{c}{F}_D=\frac{C_D\left(1.184\text{kg}/{\text{m}}^3\right)\pi {\left(4.2\text{cm}\right)}^2{V}^2}{8}\\ =\frac{C_D\left(1.184\text{kg}/{\text{m}}^3\right)\pi {\left(4.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2{V}^2}{8}\\ =\left(C_D{V}^2\right)8.2018\times {10}^{-4}\text{kg}/\text{m}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$, $4.2\text{cm}$ for $D$ and $1.184\text{kg}/{\text{m}}^3$ for ${\rho }_a$ and in Equation (II).

  $\begin{array}{c}{F}_B=\frac{\left(1.184\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\pi {\left(4.2\text{cm}\right)}^3}{6}\\ =\frac{\left(1.184\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\pi {\left(4.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^3}{6}\\ =4.50575\times {10}^{-4}\text{kg}\cdot \text{m}/{\text{s}}^2\end{array}$

Substitute $3.1\text{g}$ for $m$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (V).

  $\begin{array}{c}W=\left(9.81\text{m}/{\text{s}}^2\right)\left(3.1\text{g}\right)\\ =\left(9.81\text{m}/{\text{s}}^2\right)\left(3.1\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\\ =0.030411\text{kg}\cdot \text{m}/{\text{s}}^2\end{array}$

Substitute $0.030411\text{kg}\cdot \text{m}/{\text{s}}^2$ for $W$, $4.50575\times {10}^{-4}\text{kg}\cdot \text{m}/{\text{s}}^2$ for $F_B$ and $\left(C_D{V}^2\right)8.2018\times {10}^{-4}\text{kg}/\text{m}$ for $F_D$ in Equation (III).

  $\begin{array}{l}0.030411\text{kg}\cdot \text{m}/{\text{s}}^2=4.50575\times {10}^{-4}\text{kg}\cdot \text{m}/{\text{s}}^2+\left(C_D{V}^2\right)8.2018\times {10}^{-4}\text{kg}/\text{m}\\ \left(C_D{V}^2\right)8.2018\times {10}^{-4}\text{kg}/\text{m}=0.02996\text{kg}\cdot \text{m}/{\text{s}}^2\\ C_D{V}^2=\frac{0.02996\text{kg}\cdot \text{m}/{\text{s}}^2}{8.2018\times {10}^{-4}\text{kg}/\text{m}}\\ C_D{V}^2=36.3529{\text{m}}^2/{\text{s}}^2\end{array}$

  $\begin{array}{l}{V}^2=\frac{36.3529{\text{m}}^2/{\text{s}}^2}{C_D}\\ V=\sqrt{\frac{36.3529{\text{m}}^2/{\text{s}}^2}{C_D}}\\ V=\frac{6.043937\text{m}/\text{s}}{\sqrt{C_D}}\end{array}$......... (VI)

Refer the Table-11.2, "Representative drag coefficients for various three dimensional bodies based on the frontal area for Reynolds number unless stated otherwise" to obtain the value of the coefficient of drag is $0.2$ corresponding to the sphere.

Substitute $0.2$ for $C_D$ in Equation (VI).

  $\begin{array}{c}V=\frac{6.043937\text{m}/\text{s}}{\sqrt{0.2}}\\ =\frac{6.043937\text{m}/\text{s}}{1.4142}\\ =4.27370\text{m}/\text{s}\end{array}$

Refer the Table-B-1, "Physical properties of air at standard atmospheric Pressure" to obtain the value of the kinematic viscosity is $1.56\times {10}^{-5}{\text{m}}^2/\text{s}$ corresponding to the temperature $25°\text{C}$.

Substitute $4.27370\text{m}/\text{s}$ for $V$, $1.56\times {10}^{-5}{\text{m}}^2/\text{s}$ for $v$ and $4.2\text{cm}$ for $D$ in Equation (IV).

  $\begin{array}{c}\mathrm{Re}=\frac{4.27370\text{m}/\text{s}\times \left(4.2\text{cm}\right)}{1.56\times {10}^{-5}{\text{m}}^2/\text{s}}\\ =\frac{4.27370\text{m}/\text{s}\times \left(4.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{1.56\times {10}^{-5}{\text{m}}^2/\text{s}}\\ =\frac{4.27370\text{m}/\text{s}\times \left(0.042\text{m}\right)}{1.56\times {10}^{-5}{\text{m}}^2/\text{s}}\\ =11506.13878\end{array}$

By using the Bernoulli's equation, the datum at the point 1 is lesser then the datum at the point 2. If the ball is pushed by the finger to the side of the jet, the ball will come back to the centre of the jet. This is so because, in the middle portion of the jet, the velocity is higher than the velocity at point 1, which is, $4.27370\text{m}/\text{s}$. As a result, the pressure will be lower compared to the region outside the jet. The ball tends to get attracted towards the region where the pressure is lower.

Conclusion:

The velocity of the air is $4.27370\text{m}/\text{s}$.