Chapter 11, Problem 108P

To determine

The maximum speed of the rig with the fairing installed to the tractor.

Answer to Problem 108P

  $V_{2,\max }=133\frac{\text{km}}{\text{hr}}$

Explanation of Solution

Given information:

  $\text{Mass of tractor}=17,000\text{kg}$

  $\text{Frontal area}=9.2{\text{m}}^{\text{2}}$

  $\text{Rolling resistance coefficient}=0.05$

  $\text{Bearing friction resistance}=350\text{N}$

  $\text{Density of air}=1.25\frac{\text{kg}}{{\text{m}}^{\text{3}}}$

  $V_{1,\max }=110\frac{\text{km}}{\text{hr}}$

  $\begin{array}{c}{C}_{D1}=0.96\\ C_{D2}=0.76\end{array}$

Concept used:

  $F_{RR}=C_{RR}\times W$

  $F_D=C_{D}A\frac{\rho V_{1,\max }^2}{2}$

  ${\dot{W}}_{\text{total}}={\dot{W}}_{\text{bearing}}+{\dot{W}}_{\text{drag}}+{\dot{W}}_{\text{RR}}$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density}\\ A\to \text{Area}\\ V\to \text{Velocity}\\ {\dot{W}}_{\text{bearing}}\to \text{Power needed to overcome bearing friction}\\ {\dot{W}}_{\text{drag}}\to \text{Power needed to overcome drag force}\\ {\dot{W}}_{\text{RR}}\to \text{Power needed to overcome rolling resistance}\end{array}$

Calculation:

  $\begin{array}{c}{F}_{RR}=C_{RR}\times W\\ =C_{RR}\times \left(mg\right)\times \left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =0.05\times 17,000\times 9.81\\ =8338.5\text{N}\end{array}$

  $\begin{array}{c}{F}_{D1}=C_{D1}A\frac{\rho V_{1,\max }^2}{2}\\ =0.96\times 9.2\times \frac{1.25\times {\left(110\times \frac{1000}{3600}\right)}^2}{2}\times \left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =5153.7\text{N}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_{\text{bearing}}+{\dot{W}}_{\text{drag}}+{\dot{W}}_{\text{RR}}\\ =\left(F_{\text{bearing}}+F_{D1}+F_{\text{RR}}\right)\times V\\ =\left(350+8338.5+5153.7\right)\times \left(110\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\right)\times \left(\frac{\text{1}\text{kW}}{\text{1000}\text{N}\cdot \text{m/s}}\right)\\ =422.96\text{kW}\end{array}$

After installing the fairing,

  $\begin{array}{c}{F}_{D2}=C_{D2}A\frac{\rho V_{2,\max }^2}{2}\\ =0.76\times 9.2\times \frac{1.25\times V_{2,\max }^2}{2}\\ =4.37V_{2,\max }^2\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{total}}={\dot{W}}_{\text{bearing}}+{\dot{W}}_{\text{drag}}+{\dot{W}}_{\text{RR}}\\ 422.96\text{kW}=\left(F_{\text{bearing}}+F_{D2}+F_{\text{RR}}\right)\times V_2\\ 422.96\text{kW}=\left(350+4.37V_{2,\max }^2+5153.7\right)\times V_{2,\max }\\ 422.96\text{kW}=350V_{2,\max }+4.37V_{2,\max }^3+5153.7V_{2,\max }\\ 422960=4.37V_{2,\max }^3+5503.7V_{2,\max }\end{array}$

By solving, we get

  $V_{2,\max }=36.9\times \frac{3600}{1000}=133\frac{\text{km}}{\text{hr}}$

Conclusion:

The maximum speed of the rig with the fairing is $133\frac{\text{km}}{\text{hr}}$.