Chapter 7, Problem 7.3CP

To determine

(a)

Estimate the force on sign board both orientations?

Answer to Problem 7.3CP

For position (A) $F_{drag}=36lbf$

For position (B) $F_{drag}=0.19lbf$

Explanation of Solution

Given information:

The sign board is $5ft$ and a height of $1.5ft$

Velocity is equal to $40mi/h$

The Reynolds’s number based on length $L$ is defined as,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }$

In above equation,

$\rho$ - Density of the fluid

$\mu$ - Dynamic viscosity

$V$ - Velocity

The drag co-efficient is defined as,

$C_D=\frac{0.031}{{\mathrm{Re}}_L{}^{1/7}}-\frac{1440}{{\mathrm{Re}}_L}$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

$A$ - Characteristic area

Assume, the air at $20°C$ will have,

$\rho =0.00233slug/f{t}^3$

$\mu =3.76\times {10}^{-7}slug/fts$

Calculation:

For position (A),

According to the definitions, the drag co-efficient is equal to,

$C_D=1.2$

Convert,

$40mi/h=58.67ft/s$

Calculate the drag force,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A=\left(1.2\right)\frac{\left(0.00233slug/f{t}^3\right)}{2}{\left(58.67ft/s\right)}^2\left(1.5ft\right)\left(5ft\right)=36lbf$

For position (B),

Calculate the Reynolds’s number,

${\mathrm{Re}}_L=\frac{\rho VL}{\mu }=\frac{\left(0.00233slug/f{t}^3\right)\left(58.67ft/s\right)\left(5ft\right)}{3.76\times {10}^{-7}slug/fts}=1817833.78$

Calculate the drag co-efficient,

$C_D=\frac{0.031}{{\mathrm{Re}}_L{}^{1/7}}-\frac{1440}{{\mathrm{Re}}_L}=\frac{0.031}{{\left(1817833.78\right)}^{1/7}}-\frac{1440}{1817833.78}=3.16\times {10}^{-3}$

Calculate the drag force,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A=\left(3.16\times {10}^{-3}\right)\frac{\left(0.00233slug/f{t}^3\right)}{2}{\left(58.67ft/s\right)}^2\left(1.5ft\right)\left(5ft\right)=0.095lbf$

Therefore, both sides

$\left(0.095lbf\right)\left(2\right)=0.19lbf$

Conclusion:

For position (A) $F_{drag}=36lbf$

For position (B) $F_{drag}=0.19lbf$.

To determine

(b)

Calculate the total drag for car-sign combination for both orientations?

Answer to Problem 7.3CP

For position (A) $F_{Totaldrag}=100.16lbf$

For position (B) $F_{Totaldrag}=64.352lbf$

Explanation of Solution

Given information:

The sign board is $5ft$ and a height of $1.5ft$

Velocity is equal to $40mi/h$

Drag co-efficient $0.4$

Frontal area is equal to $40f{t}^2$

The drag force is defined as,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A$

Where,

$C_D$ - Drag co-efficient

$\rho$ - Density of the fluid

$V$ - Velocity

$A$ - Characteristic area

Assume, the air at $20°C$ will have,

$\rho =0.00233slug/f{t}^3$

Calculation:

Calculate the drag force for car,

$F_{drag}=C_D\frac{\rho }{2}{V}^{2}A=\left(0.4\right)\frac{\left(0.00233slug/f{t}^3\right)}{2}{\left(58.67ft/s\right)}^2\left(40f{t}^2\right)=64.162lbf$

Therefore,

The total drag for position (A),

$F_{Totaldrag}=\left(64.162lbf+36lbf\right)=100.16lbf$

The total drag for position (B),

$F_{Totaldrag}=\left(64.162lbf+0.19lbf\right)=64.352lbf$

Conclusion:

For position (A) $F_{Totaldrag}=100.16lbf$

For position (B) $F_{Totaldrag}=64.352lbf$.

To determine

(c)

Calculate the horse power required for both orientations?

Answer to Problem 7.3CP

For position (A) $horsepower=14.95hp$

For position (B) $horsepower=11.13hp$

Explanation of Solution

Given information:

Velocity is equal to $40mi/h$

The rolling resistance is equal to $40lbf$

The power required $P$ is defined as,

$P=FV$

In above equation,

$F$ - Total force

$V$ - Velocity

Calculation:

Convert,

$40mi/h=58.67ft/s$

For position (A)

Calculate the power required,

$P=FV$

Substitute,

$P=\left(F_{Totaldrag}+F_{Rolling}\right)V=\left(100.16lbf+40lbf\right)\left(58.67ft/s\right)=8223.18ft.lbf/s$

Convert to horse power,

$P=\frac{8223.18ft.lbf/s}{550}=14.95hp$

For position (B)

Calculate the power required,

$P=FV$

Substitute,

$P=\left(F_{Totaldrag}+F_{Rolling}\right)V=\left(64.352lbf+40lbf\right)\left(58.67ft/s\right)=6122.33ft.lbf/s$

Convert to horse power,

$P=\frac{6122.33ft.lbf/s}{550}=11.13hp$

Conclusion:

For position (A) $horsepower=14.95hp$

For position (B) $horsepower=11.13hp$.

To determine

(d)

Calculate the fuel efficiency for both orientations in $mi/gal$ ?

Answer to Problem 7.3CP

For position (A) $mpg=26.75mi/gal$

For position (B) $mpg=35.94mi/gal$

Explanation of Solution

Given information:

Velocity is equal to $40mi/h$

Power delivered by engine per gallon of gasoline is equal to $10hp/h$

The fuel efficiency is defined as,

$mpg=\left(\frac{Speed\times Powerdeliveredpergallonofgasoline}{horsepower}\right)$

Calculation:

Calculate the fuel efficiency,

For position (A),

$mpg=\left(40mi/h\right)\left(10hp/h.gal\right)\left(\frac{1}{14.95hp}\right)=26.75mi/gal$

For position (B),

$mpg=\left(40mi/h\right)\left(10hp/h.gal\right)\left(\frac{1}{11.13hp}\right)=35.94mi/gal$

Conclusion:

For position (A) $mpg=26.75mi/gal$

For position (B) $mpg=35.94mi/gal$.