Chapter 15, Problem 97RQ

To determine

The maximum speed of the rig.

Explanation of Solution

Given:

Total mass of the rig is $17000\text{ kg}$.

Frontal area of the rig is $9.2{\text{ m}}^2$.

The coefficient of drag is $0.96$.

The coefficient of rolling resistance is $0.05$.

The bearing friction resistance is $350\text{N}$.

Maximum speed is $110\text{km}/\text{h}$.

Density of air is $1.25\text{kg}/{\text{m}}^3$.

The new coefficient of drag is $0.76$.

Calculation:

The rolling resistance is,

  $\begin{array}{c}{F}_{RR}=C_{RR}W\\ =\left(0.05\right)\left(17000\text{ kg}\times \text{9}\text{.81}\text{m}/{\text{s}}^2\right)\\ =8339\text{ N}\end{array}$

The drag force is,

  $\begin{array}{c}{F}_D=C_{D}A\frac{\rho V^2}{2}\\ =\left(0.96\right)\left(9.2{\text{ m}}^2\right)\frac{\left(1.25\text{kg}/{\text{m}}^3\right){\left(110\text{km}/\text{h}\times \frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)}^2}{2}\\ =5154\text{}\text{N}\end{array}$

The power required to overcome the resistances is,

  $\begin{array}{c}{\dot{W}}_{total}={\dot{W}}_{drag}+{\dot{W}}_{RR}+{\dot{W}}_{friction}\\ =\left(F_D+F_{RR}+F_{friction}\right)V\\ =\left(5154\text{ N}+8339\text{}\text{N}+350\text{ N}\right)\left(110\text{km}/\text{h}\times \frac{5\text{m}/\text{s}}{18\text{km}/\text{h}}\right)\\ =423\text{ kW}\end{array}$

Calculate the new maximum speed when the drag coefficient is changed.

  $\begin{array}{l}{\dot{W}}_{total}={\dot{W}}_{drag}+{\dot{W}}_{RR}+{\dot{W}}_{friction}\\ 423\text{ kW}=\left(F_D+F_{RR}+F_{friction}\right)V_{new}\\ 423\text{ kW}=\left(C_{D,new}A\frac{\rho V_{new}{}^2}{2}+8339\text{}\text{N}+350\text{ N}\right)V_{new}\\ 423000\text{ W}=\left[\left(0.76\right)\left(9.2{\text{ m}}^2\right)\frac{\left(1.25\text{kg}/{\text{m}}^3\right)V_{new}{}^2}{2}+8339\text{}\text{N}+350\text{ N}\right]V_{new}\\ 423000=8689V_{new}+4.37V_{new}{}^3\\ V_{new}=36.9\text{m}/\text{s}\times \frac{18\text{km}/\text{h}}{5\text{m}/\text{s}}\\ V_{new}=133\text{km}/\text{h}\end{array}$

Thus, the maximum speed of the rig is $133\text{km}/\text{h}$.