Chapter 3, Problem 15P

To determine

(a)

The additional horse power required by a passenger vehicle on the curve.

Answer to Problem 15P

  $P=43.66\text{hp}$

Explanation of Solution

Given information:

Speed $=55\text{mi/hr}$

Weight of car $=2500\text{lb}$

Cross sectional area of car $=30{\text{ft}}^{\text{2}}$

Radius of the curve $=850\text{ft}$

Concept used:

The curve resistance is calculated when the road surface has a horizontal curve. Then the power is calculated using the curve resistance force.

  $R_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)$

  $P=\frac{1.47R_{c}u}{550}$

  $\begin{array}{c}{R}_c\to \text{Curve resistance}\\ W\to \text{Weight of the vehicle}\\ R\to \text{Radius of the curve}\\ P\to \text{Power of vehicle}\end{array}$

Calculation:

  $\begin{array}{c}{R}_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)\\ =0.5\left(\frac{2.15\times {55}^2\times 2500}{32.2\times 850}\right)\\ =297.03\text{lb}\end{array}$

The power of the passenger vehicle is determined by

  $\begin{array}{c}P=\frac{1.47R_{c}u}{550}\\ =\frac{1.47\times 297.03\times 55}{550}\\ =43.66\text{hp}\end{array}$

Conclusion:

Therefore, the additional horsepower developed by a passenger vehicle on the curve is $43.66\text{hp}$.

To determine

(b)

The total resistance force developed on the vehicle.

Answer to Problem 15P

  $R=430.43\text{lb}$

Explanation of Solution

Given information:

Speed $=55\text{mi/hr}$

Weight of car $=2500\text{lb}$

Cross sectional area of car $=30{\text{ft}}^{\text{2}}$

Radius of the curve $=850\text{ft}$

Concept used:

The total resistance is the sum of air resistance, rolling resistance and curve resistance of the passenger vehicle.

  $R_a=0.5\left(\frac{2.15p{C}_{D}A{u}^2}{g}\right)$

  $R_r=\left(C_{rs}+2.15C_{rv}{u}^2\right)W$

  $R_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)$

  $\begin{array}{c}{R}_a\to \text{Air resistance}\\ C_D\to \text{Aerodynamic drag coefficient}\\ A\to \text{Cross sectional area}\\ u\to \text{Speed}\\ p\to \text{Density of air}=0.0766{\text{lb/ft}}^{\text{3}}\end{array}$

  $\begin{array}{c}{R}_r\to \text{Rolling resistance}\\ C_{rs}\to \text{Constant for cars}=0.012\\ C_{rv}\to \text{Constant for cars}=0.65\times {10}^{-6}{\text{sec}}^{\text{2}}{\text{/ft}}^{\text{2}}\\ W\to \text{Weight of vehicle}\end{array}$

Calculation:

Take aerodynamic drag coefficient for passenger car as $0.4$

  $\begin{array}{c}{R}_a=0.5\left(\frac{2.15p{C}_{D}A{u}^2}{g}\right)\\ =0.5\left(\frac{2.15\times 0.0766\times 0.4\times 30\times {55}^2}{32.2}\right)\\ =92.83\text{lb}\end{array}$

Rolling resistance force is calculated as

  $\begin{array}{c}{R}_r=\left(C_{rs}+2.15C_{rv}{u}^2\right)W\\ =\left(0.012+2.15\times 0.65\times {10}^{-6}\times {55}^2\right)\left(2500\right)\\ =40.57\text{lb}\end{array}$

Curve resistance force is calculated as

  $\begin{array}{c}{R}_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)\\ =0.5\left(\frac{2.15\times {55}^2\times 2500}{32.2\times 850}\right)\\ =297.03\text{lb}\end{array}$

The total resistance force is calculated as

  $\begin{array}{c}R=R_a+R_r+R_c\\ =92.83+40.57+297.03\\ =430.43\text{lb}\end{array}$

Conclusion:

The total resistance force on the vehicleis $430.43\text{lb}$.

To determine

(c)

The total horse power developed by the vehicle.

Answer to Problem 15P

  $P_{\text{total}}=63.27\text{hp}$

Explanation of Solution

Given information:

Speed $=55\text{mi/hr}$

Weight of car $=2500\text{lb}$

Cross sectional area of car $=30{\text{ft}}^{\text{2}}$

Radius of the curve $=850\text{ft}$

Concept used:

The total resistance is the sum of air resistance, rolling resistance and curve resistance of the passenger car. Using the total resistance the power of the passenger vehicle is calculated. The power is calculated individually for the straight road surface and for the curve surface. Finally the total power is determined by adding both the powers.

  $R_a=0.5\left(\frac{2.15p{C}_{D}A{u}^2}{g}\right)$

  $R_r=\left(C_{rs}+2.15C_{rv}{u}^2\right)W$

  $R_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)$

  $P=\frac{1.47R_{c}u}{550}$

  $\begin{array}{c}{R}_a\to \text{Air resistance}\\ C_D\to \text{Aerodynamic drag coefficient}\\ A\to \text{Cross sectional area}\\ u\to \text{Speed}\\ p\to \text{Density of air}=0.0766{\text{lb/ft}}^{\text{3}}\end{array}$

  $\begin{array}{c}{R}_r\to \text{Rolling resistance}\\ C_{rs}\to \text{Constant for cars}=0.012\\ C_{rv}\to \text{Constant for cars}=0.65\times {10}^{-6}{\text{sec}}^{\text{2}}{\text{/ft}}^{\text{2}}\\ W\to \text{Weight of vehicle}\\ P\to \text{Power of vehicle}\end{array}$

Calculation:

Take aerodynamic drag coefficient for passenger car as $0.4$.

  $\begin{array}{c}{R}_a=0.5\left(\frac{2.15p{C}_{D}A{u}^2}{g}\right)\\ =0.5\left(\frac{2.15\times 0.0766\times 0.4\times 30\times {55}^2}{32.2}\right)\\ =92.83\text{lb}\end{array}$

Rolling resistance force is calculated as

  $\begin{array}{c}{R}_r=\left(C_{rs}+2.15C_{rv}{u}^2\right)W\\ =\left(0.012+2.15\times 0.65\times {10}^{-6}\times {55}^2\right)\left(2500\right)\\ =40.57\text{lb}\end{array}$

Curve resistance force is calculated as

  $\begin{array}{c}{R}_c=0.5\left(\frac{2.15u^{2}W}{gR}\right)\\ =0.5\left(\frac{2.15\times {55}^2\times 2500}{32.2\times 850}\right)\\ =297.03\text{lb}\end{array}$

The total horse power is calculated as

  $\begin{array}{c}{P}_{\text{straight}}=\frac{1.47\left(R_a+R_r\right)u}{550}\\ =\frac{1.47\times \left(92.83+40.57\right)\times 55}{550}\\ =19.61\text{hp}\end{array}$

  $\begin{array}{c}{P}_{\text{curve}}=\frac{1.47R_{c}u}{550}\\ =\frac{1.47\times 297.03\times 55}{550}\\ =43.66\text{hp}\end{array}$

  $\begin{array}{c}{P}_{\text{total}}=P_{\text{straight}}+P_{\text{curve}}\\ =19.61+43.66\\ =63.27\text{hp}\end{array}$

Conclusion:

The total horsepower is $63.27\text{hp}$.