Chapter 8, Problem 8.101P

To determine

(a)

Terminal velocity.

Answer to Problem 8.101P

The required value of the terminal velocity is $1.95m/s$.

Explanation of Solution

Given Information:

$\begin{array}{l}SG=7.85\\ D=2cm\\ T={20}^{o}C\\ C_D=0.47\end{array}$

Formula used:

$\left(m+m_h\right)\frac{dV}{dt}=W_{net}-C_D\frac{\rho }{2}{V}^{2}A$

$A=\frac{\pi }{4}{D}^2$

$W_{net}=\left({\rho }_{steel}-{\rho }_{water}\right)g\frac{\pi }{6}{D}^3$

$V_f=\sqrt{\frac{2W_{net}}{C_D\rho A}}$

Calculation:

To solve this problem, for water we are taking the value of $\rho =998kg/m^3$.

We know the hydrodynamic mass to the differential equation, $\left(m+m_h\right)\frac{dV}{dt}=W_{net}-C_D\frac{\rho }{2}{V}^{2}A$ ;

And the area of the sphere is, $A=\frac{\pi }{4}{D}^2$ ;

And, $W_{net}=\left({\rho }_{steel}-{\rho }_{water}\right)g\frac{\pi }{6}{D}^3$ ;

Now separate the variables and integrate;

$\Rightarrow V=\sqrt{\frac{2W_{net}}{C_D\rho A}}\tanh \left(t\sqrt{\frac{W_{net}{C}_D\rho A}{2{\left(m+m_h\right)}^2}}\right)$

Since the terminal velocity is the coefficient of the tanh functions in the previous equation;

Hence, $V_f=\sqrt{\frac{2W_{net}}{C_D\rho A}}$ ;

Put the values in the above equation;

$V_f=\sqrt{\frac{2\left(7850-998\right)\left(9.81\right)\left(\pi /6\right){\left(0.02\right)}^3}{\left(0.47\right)\left(998\right)\left(\pi /4\right){\left(0.02\right)}^2}}$

$\Rightarrow 1.95m/s$.

To determine

(b)

Time to reach 99 percent of terminal velocity and examine the result.

Answer to Problem 8.101P

The required value of the time is $0.642s$

The value of the time that we have got is $6\%$ than when $m_h=0$.

Explanation of Solution

Given information:

$\begin{array}{l}\text{SG}\text{=}\text{7}\text{.85}\\ \text{D}\text{=}\text{2}\text{cm}\\ \text{T}\text{=}{\text{20}}^{\text{o}}\text{C}\\ {\text{C}}_{\text{D}}\text{=}\text{0}\text{.47}\\ {\text{V}}_{\text{terminal}}\text{=}\text{1}\text{.95}\text{m/s}\\ {\text{t}}_{\text{99\%}}\text{=}\text{0}\text{.605}\text{s}\end{array}$

Formula used:

$t\sqrt{\frac{W_{net}{C}_D\rho A}{2{\left(m+m_h\right)}^2}}=2.647$

Calculation:

We have note that, $\left(2.647\right)=0.99$, we find the time to approach $99\%$ of $V_f$ to be;

$t\sqrt{\frac{W_{net}{C}_D\rho A}{2{\left(m+m_h\right)}^2}}=2.647$

Put the values in the above equation;

$\Rightarrow t\sqrt{\frac{\left(7834-998\right)\left(9.81\right)\left(\pi /6\right){\left(0.02\right)}^3\left(0.47\right)\left(998\right)\left(\pi /4\right){\left(0.02\right)}^2}{2{\left[\left(7834+998/2\right)\left(\pi /6\right){\left(0.02\right)}^3\right]}^2}}=4.122t$

$\begin{array}{l}t=2.647/4.122\\ \Rightarrow 0.642s\end{array}$

Which is $6\%$ than when $m_h=0$.