Chapter 1, Problem 42P

Water at 20°C from a garden hose fills a 2.0 L container in 2.85 s. Using unity conversation ratios and showing all your work, calculate the volume flow rate in liters per minute (Lpm) and the mass flow rate in kg/s.

To determine

The terminal velocity of sphere in water.

Answer to Problem 42P

The terminal velocity of sphere in water is $0.153\text{m}/\text{sec}$.

Explanation of Solution

Given information:

The diameter of the plastic sphere is $6\text{mm}$ the density of sphere is $1150\text{kg}/{\text{m}}^{\text{2}}$, the acceleration due to gravity is $9.81\text{m}/{\text{sec}}^{\text{2}},$ the coefficient of drag is $0.5$.

Write the expression for the weight of the sphere.

$W={\rho }_{\text{sphere}}g\left(\frac{\pi D^3}{6}\right)$ …… (I)

Here, the density of the sphere is ${\rho }_{\text{sphere}}$, the diameter of the sphere is $D$, the acceleration due to gravity is $g$.

Write the expression for the frontal area of the sphere.

$A=\frac{\pi }{4}{d}^2$ …… (II)

Here, the total diameter of the sphere is $d$, the area of the sphere is $A$.

Write the expression for drag force.

$F_D=\frac{C_D\rho A{V}^2}{2}$ …… (III)

Here, the coefficient of drag is $C_D$, the density of fuel is $\rho$, the frontal area of sphere is $A$, the velocity of flow is $V$, the drag force is $F_D$.

Write the expression for the buoyant force.

$F_B={\rho }_{\text{water}}gv$ …… (IV)

Here, the density of water is ${\rho }_{\text{water}}$ the acceleration due to gravity is $g$, the volume of sphere submerged is $v,$ and the buoyant force is $F_B$.

Write the expression for the total weight.

$W=F_D+F_B$ …… (V)

Here, the drag force is $F_D,$ the buoyant force is $F_B,$ and the total weight is $W$.

Write the expression for the volume submerged.

$v=\frac{\pi }{6}\left(D^3\right)$ …… (VI)

Here, the diameter of sphere is $D$ and the volume submerged is $v$.

Calculation:

Substitute $1150\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{sphere}}$, $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$ and $6\text{mm}$ for $D$ in Equation (I).

$\begin{array}{c}W=1150\text{kg}/{\text{m}}^3\times 9.81\text{m}/{\text{sec}}^{\text{2}}\times \left(\frac{\pi }{6}\times {\left(6\text{mm}\right)}^3\right)\\ =1150\text{kg}/{\text{m}}^3\times 9.81\text{m}/{\text{sec}}^{\text{2}}\times \left(\frac{\pi }{6}\times {\left(6\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^3\right)\\ =0.001275\left(1\text{kg}\cdot \text{m}/{\text{sec}}^{\text{2}}\right)\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{sec}}^{\text{2}}}\right)\\ =0.001275\text{N}\end{array}$

Substitute $6\text{mm}$ for $d$ in Equation (II).

$\begin{array}{c}A=\frac{\pi }{4}\times {\left(6\text{mm}\right)}^2\\ =0.785\times 36{\text{mm}}^{\text{2}}\\ =28.27{\text{mm}}^{\text{2}}\end{array}$

Substitute $0.5$ for $C_D$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $28.27{\text{mm}}^{\text{2}}$ for $A$ in Equation(III).

$\begin{array}{c}{F}_D=\frac{1000\text{kg}/{\text{m}}^{\text{3}}\times 0.5\times 28.27{\text{mm}}^{\text{2}}\times {\left(V\right)}^2}{2}\\ =\frac{500\text{kg}/{\text{m}}^{\text{3}}\times 28.27{\text{mm}}^{\text{2}}\left(\frac{1{\text{m}}^{\text{2}}}{{\left(1000\right)}^2{\text{mm}}^{\text{2}}}\right)\times {\left(V\right)}^2}{2}\\ =\frac{0.014135}{2}\text{kg}/\text{m}{\left(V\right)}^2\\ =0.00706\text{kg}/\text{m}{\left(V\right)}^2\end{array}$

Substitute $6\text{mm}$ for $D$ in Equation (VI).

$\begin{array}{c}v=\frac{\pi }{6}\times {\left(6\text{mm}\right)}^3\\ =0.523\times 216{\text{mm}}^{\text{3}}\\ =113.09{\text{mm}}^{\text{3}}\end{array}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{sec}}^{\text{2}}$ for $g$, $113.09{\text{mm}}^{\text{3}}$ for $v$ in Equation (IV).

$\begin{array}{c}{F}_B=1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{sec}}^{\text{2}}\times 113.09{\text{mm}}^{\text{3}}\\ =1000\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{sec}}^{\text{2}}\times 113.09{\text{mm}}^3\left(\frac{\text{1}{\text{m}}^{\text{3}}}{{\left(1000\right)}^3{\text{mm}}^3}\right)\\ =0.001109\text{kg}\cdot \text{m}/{\sec }^2\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\sec }^2}\right)\\ =0.001109\text{N}\end{array}$

Substitute $0.001109\text{N}$ for $F_B$, $0.00706\text{kg}/\text{m}{\left(V\right)}^2$ for $F_D,$ and $0.001275\text{N}$ for $W$ in Equation (V).

$\begin{array}{l}0.001275\text{N}=0.00706\text{kg}/\text{m}{\left(V\right)}^2+0.001109\text{N}\\ 0.00706\text{kg}/\text{m}{\left(V\right)}^2=0.001275\text{N}-0.001109\text{N}\\ {\left(V\right)}^2=\frac{0.000166\text{N}}{0.00706\text{kg}/\text{m}}\left(\frac{\text{1}\text{kg}\cdot \text{m}/{\sec }^2}{1\text{N}}\right)\\ V=0.153\text{m}/\sec \end{array}$

Conclusion:

The terminal velocity of sphere in water is $0.153\text{m}/\text{sec}$.