Chapter 11, Problem 41EP

The drag coefficient of a vehicle increases when its windows are rolled down or its sunroof is opened. A sports car has a frontal area of 18 ft2 and a drag coefficient of 0.32 when the windows and sunroof are closed. The drag coefficient increases to 0.41 when the sunroof is open. Determine the additional power consumption of the car when the sunroof is opened at (a) 35 mi/h and (b) 70 mi/h. Take the density of be to be 0.075 Ibm/ft³.

To determine

(a)

The additional power consumption of the car when the roof opened at $35\text{mi/hr}$.

Answer to Problem 41EP

$0.35\text{kW}$

Explanation of Solution

Given information:

$\text{Frontal area}=18{\text{ft}}^{\text{2}}$

$\text{When sunroof is closed drag coefficient}=0.32$

$\text{When sunroof is opened drag coefficient}=0.41$

$\text{Density of air}=0.075\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$

$\text{Velocity}=35\frac{\text{mi}}{\text{hr}}$

Concept used:

$F_D=C_{D}A\frac{\rho V^2}{2}$

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

${\dot{W}}_{\text{extra}}={\left({\dot{W}}_{\text{extra}}\right)}_{\text{closed}}-{\left({\dot{W}}_{\text{extra}}\right)}_{\text{opened}}$

$\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

$\begin{array}{c}{\left(F_D\right)}_{\text{opened}}=C_{D}A\frac{\rho V^2}{2}\\ =0.32\times 18\times \frac{0.075\times {\left(35\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{lbf}}{\text{32}\text{.2}\text{lbm}\cdot {\text{ft/s}}^{\text{2}}}\right)\\ =17.68\text{lbf}\end{array}$

$\begin{array}{c}{\left({\dot{W}}_{\text{drag}}\right)}_{\text{opened}}=F_D\times V\\ =17.68\times \left(35\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)\times \left(\frac{\text{1}\text{kW}}{\text{737}\text{.56}\text{lbf}\cdot \text{ft/s}}\right)\\ =1.23\text{kW}\end{array}$

$\begin{array}{c}{\left(F_D\right)}_{\text{closed}}=C_{D}A\frac{\rho V^2}{2}\\ =0.41\times 18\times \frac{0.075\times {\left(35\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{lbf}}{\text{32}\text{.2}\text{lbm}\cdot {\text{ft/s}}^{\text{2}}}\right)\\ =22.65\text{lbf}\end{array}$

$\begin{array}{c}{\left({\dot{W}}_{\text{drag}}\right)}_{\text{closed}}=F_D\times V\\ =22.65\times \left(35\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)\times \left(\frac{\text{1}\text{kW}}{\text{737}\text{.56}\text{lbf}\cdot \text{ft/s}}\right)\\ =1.58\text{kW}\end{array}$

$\begin{array}{c}{\dot{W}}_{\text{extra}}={\left({\dot{W}}_{\text{extra}}\right)}_{\text{closed}}-{\left({\dot{W}}_{\text{extra}}\right)}_{\text{opened}}\\ =1.58-1.23\\ =0.35\text{kW}\end{array}$

Conclusion:

The additional power consumption of the car when the roof opened at $35\text{mi/hr}$ is $0.35\text{kW}$.

To determine

(b)

The additional power consumption of the car when the roof opened at $70\text{mi/hr}$.

Answer to Problem 41EP

$2.77\text{kW}$

Explanation of Solution

Given information:

$\text{Frontal area}=18{\text{ft}}^{\text{2}}$

$\text{When sunroof is closed drag coefficient}=0.32$

$\text{When sunroof is opened drag coefficient}=0.41$

$\text{Density of air}=0.075\frac{\text{lbm}}{{\text{ft}}^{\text{3}}}$

$\text{Velocity}=70\frac{\text{mi}}{\text{hr}}$

Concept used:

$F_D=C_{D}A\frac{\rho V^2}{2}$

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

${\dot{W}}_{\text{extra}}={\left({\dot{W}}_{\text{extra}}\right)}_{\text{closed}}-{\left({\dot{W}}_{\text{extra}}\right)}_{\text{opened}}$

$\begin{array}{c}{F}_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density}\\ A\to \text{Area}\\ V\to \text{Velocity}\end{array}$

Calculation:

$\begin{array}{c}{\left(F_D\right)}_{\text{opened}}=C_{D}A\frac{\rho V^2}{2}\\ =0.32\times 18\times \frac{0.075\times {\left(70\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{lbf}}{\text{32}\text{.2}\text{lbm}\cdot {\text{ft/s}}^{\text{2}}}\right)\\ =70.71\text{lbf}\end{array}$

$\begin{array}{c}{\left({\dot{W}}_{\text{drag}}\right)}_{\text{opened}}=F_D\times V\\ =70.71\times \left(70\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)\times \left(\frac{\text{1}\text{kW}}{\text{737}\text{.56}\text{lbf}\cdot \text{ft/s}}\right)\\ =9.84\text{kW}\end{array}$

$\begin{array}{c}{\left(F_D\right)}_{\text{closed}}=C_{D}A\frac{\rho V^2}{2}\\ =0.41\times 18\times \frac{0.075\times {\left(70\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{lbf}}{\text{32}\text{.2}\text{lbm}\cdot {\text{ft/s}}^{\text{2}}}\right)\\ =90.59\text{lbf}\end{array}$

$\begin{array}{c}{\left({\dot{W}}_{\text{drag}}\right)}_{\text{closed}}=F_D\times V\\ =90.59\times \left(70\times \frac{5280}{3600}\frac{\text{ft}}{\text{s}}\right)\times \left(\frac{\text{1}\text{kW}}{\text{737}\text{.56}\text{lbf}\cdot \text{ft/s}}\right)\\ =12.61\text{kW}\end{array}$

$\begin{array}{c}{\dot{W}}_{\text{extra}}={\left({\dot{W}}_{\text{extra}}\right)}_{\text{closed}}-{\left({\dot{W}}_{\text{extra}}\right)}_{\text{opened}}\\ =12.61-9.84\\ =2.77\text{kW}\end{array}$

Conclusion:

The additional power consumption of the car when the roof opened at $70\text{mi/hr}$ is $2.77\text{kW}$.