Chapter 2, Problem 1P

To determine

The power required to overcome the aerodynamics drag.

Answer to Problem 1P

The power required to overcome the aerodynamics drag is $\text{57}\text{.19hp}$

Explanation of Solution

Given:

Drag Coefficient of sports car, ${\text{C}}_{\text{D}}$ = $0.30$

Front area of the car, ${\text{A=21ft}}^2$

Speed of the car, $\text{V}$ = $\text{110mi/h}$

Density, $\text{ρ}$ = $\text{0}{\text{.002378slugs/ft}}^{\text{2}}$

Formula Used:

The power required to overcome the aerodynamic drag is

  $\text{P = F × V}$

Where,

Force = $\text{F }$

Velocity = $\text{V}$

Calculation:

The required power is calculated by power equation

  $\text{P = F × V}$ ..................... $\left(\text{1}\right)$

We know that

Force applied on caris given by

  $\text{F = }\frac{\text{1}}{\text{2}}{\text{ρAC}}_{\text{D}}{\text{V}}^{\text{2}}$ ...................... $\left(2\right)$

Where,

Density = $\text{ρ}$

Front Area of the Car = $\text{A}$

Drag coefficient = ${\text{C}}_{\text{D}}$

Rearranging the equation $\left(\text{1}\right)$ as

  $\text{P = F × V}$

  $\text{P = }\left(\frac{\text{1}}{\text{2}}{\text{ ρAC}}_{\text{D}}{\text{×V}}^{\text{2}}\right)\text{ ×}\left(\text{V}\right)$

  $\text{P = }\left(\frac{\text{1}}{\text{2}}{\text{ ρAC}}_{\text{D}}\right){\text{×V}}^3$ .................. $\left(3\right)$

To calculate power in $\left(\text{lb-ft}/\text{r}\right)$, we have to convert the velocity from $\text{mi/h}$ to $\text{ft / sec}$

  $\text{V = 110 mi/h = }\frac{\text{110×5280}}{\text{3600}}\text{ft/s}$

As -

  $\text{1mi/h = }\frac{\text{5280 ft}}{\text{3600 sec}}$

Put the given values in equation $\left(3\right)$

  $\begin{array}{l}\text{P = }\left(\frac{\text{1}}{\text{2}}{\text{ ρAC}}_{\text{D}}\right){\text{×V}}^{\text{3}}\\ \text{P = }\left(\frac{\text{1}}{\text{2}}\text{×0}\text{.002378×21×0}\text{.3}\right)\text{×}{\left(\frac{\text{110×5280}}{\text{3600}}\right)}^{\text{3}}\end{array}$

  $\text{P = 31455}\text{.364 lb-ft/s}$ .......................... $\left(4\right)$

But we want the power in $\text{hp}$ -

We know that - $\text{1 lb-ft/s = }\frac{1}{550}\text{hp }$

So, the calculated power of equation $\left(4\right)$ is converted into $\text{hp}$ as

  $\begin{array}{l}\text{P = }\frac{\text{31455}\text{.364}}{\text{550}}\text{hp}\\ \text{P = 57}\text{.19 hp}\end{array}$

  $$

Conclusion:

Thus, the power required to overcome the aerodynamics drag is $\text{57}\text{.19hp}$