Chapter 11, Problem 104P

To determine

(a)

The additional power required by the Recreational vehicle at given condition.

Answer to Problem 104P

The additional power required by the recreational vehicle, ${\dot{W}}_{\text{drag}}=1.05\text{kW}$

Explanation of Solution

Given information:

  $\text{Length}=3\text{m}$

  $\text{Diameter}=0.5\text{m}$

  $\text{Atmospheric pressure}=87\text{kPa}$

  $\text{Temperature}={20}^{\circ }\text{C}=295\text{K}$

  $\text{Velocity}=80\frac{\text{km}}{\text{hr}}$

Concept used:

  $\rho =\frac{P}{RT}$

  $F_D=C_{D}A\frac{\rho V^2}{2}$

  ${\dot{W}}_{\text{drag}}=F_D\times V$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density of air}\\ V\to \text{Velocity}\\ A\to \text{Area}\\ {\dot{W}}_{\text{drag}}\to \text{Power}\end{array}$

Calculation:

The gas constant is, $R=0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

  $\begin{array}{c}\rho =\frac{P}{RT}\\ =\frac{87}{0.287\times 295}\\ =1.028\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

Refer the table 11.2 from the textbook, for horizontal cylinder of $\frac{L}{D}=\frac{3}{0.5}=6$, we get $C_D=0.95$.

  $\begin{array}{c}{F}_D=C_{D}A\frac{\rho V^2}{2}\\ =0.95\times \left[\frac{\pi }{4}\times {\left(0.5\right)}^2\right]\times \frac{1.028\times {\left(80\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =47.35\text{N}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{drag}}=F_D\times V\\ =47.35\times \left(80\times \frac{1000}{3600}\right)\times \left(\frac{\text{1}\text{kW}}{\text{1000}\text{N}\cdot \text{m/s}}\right)\\ =1.05\text{kW}\end{array}$

Conclusion:

The additional power required by the recreational vehicle is $1.05\text{kW}$.

To determine

(b)

The additional power required by the recreational vehicle at given conditions.

Answer to Problem 104P

The additional power required by the recreational vehicle, ${\dot{W}}_{\text{drag}}=6.77\text{kW}$

Explanation of Solution

Given information:

  $\text{Length}=3\text{m}$

  $\text{Diameter}=0.5\text{m}$

  $\text{Atmospheric pressure}=87\text{kPa}$

  $\text{Temperature}={20}^{\circ }\text{C}=295\text{K}$

  $\text{Velocity}=80\frac{\text{km}}{\text{hr}}$

Concept used:

  $\rho =\frac{P}{RT}$

  $F_D=C_{D}A\frac{\rho V^2}{2}$

  ${\dot{W}}_{\text{drag}}=F_D\times V$

  $\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density of air}\\ V\to \text{Velocity}\\ A\to \text{Area}\\ {\dot{W}}_{\text{drag}}\to \text{Power}\end{array}$

Calculation:

The gas constant is, $R=0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$.

  $\begin{array}{c}\rho =\frac{P}{RT}\\ =\frac{87}{0.287\times 295}\\ =1.028\frac{\text{kg}}{{\text{m}}^{\text{3}}}\end{array}$

Refer the table 11.2 from the textbook, for vertical cylinder of $\frac{L}{D}=\frac{3}{0.5}=6$, we get $C_D=0.95$.

  $\begin{array}{c}{F}_D=C_{D}A\frac{\rho V^2}{2}\\ =0.8\times \left(3\times 0.5\right)\times \frac{1.028\times {\left(80\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{N}}{\text{1}\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =304.6\text{N}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{drag}}=F_D\times V\\ =304.6\times \left(80\times \frac{1000}{3600}\right)\times \left(\frac{\text{1}\text{kW}}{\text{1000}\text{N}\cdot \text{m/s}}\right)\\ =6.77\text{kW}\end{array}$

Conclusion:

The additional power required by the recreational vehicle, ${\dot{W}}_{\text{drag}}=6.77\text{kW}$