Chapter 2.8, Problem 124RP

To determine

The power required to move car through the air.

The area of the effective flow channel behind the car.

Answer to Problem 124RP

The power required to move car through the air is $\underset{\_}{\text{4}\text{.41}\text{kW}}$.

The area of the effective flow channel behind the car is $\underset{\_}{3.29{\text{m}}^{\text{2}}}$.

Explanation of Solution

Convert the absolute pressure of the air from mm Hg to kPa.

$\begin{array}{c}P=\left(\text{700}\text{mm}\text{Hg}\right)\left(\frac{0.1333\text{kPa}}{1\text{mm}\text{Hg}}\right)\\ =93.31\text{kPa}\end{array}$

Write the specific volume of the air.

$\nu =\frac{RT}{P}$ (I)

Here, gas constant is R and temperature while traveling a car at 90 km/h is T.

Write the mass flow ate through the control volume.

$\dot{m}=\frac{A_1{V}_1}{\nu }$ (II)

Here, the area and velocity at section 1 is $A_1$ and $V_1$ respectively.

Write the power required to move car through the air.

$\dot{W}=\dot{m}\frac{\left(V_1^2-V_2^2\right)}{2}$ (III)

Calculate the outlet area of the effective flow channel behind the car.

$A_2=\frac{\dot{m}\nu }{V_2}$ (IV)

Conclusion:

Substitute $0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}$ for R, $20°\text{C}$ for T, and 93.31 kPa for P in Equation (I).

$\begin{array}{c}\nu =\frac{\left(0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)20°\text{C}}{93.31\text{kPa}}\\ =\frac{\left(0.287\frac{\text{kPa}\cdot {\text{m}}^{\text{3}}}{\text{kg}\cdot \text{K}}\right)\left(20+273\right)\text{K}}{93.31\text{kPa}}\\ =0.9012{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitute $3{\text{m}}^{\text{2}}$ for $A_1$, 90 km/hr for $V_1$, and $0.9012{\text{m}}^{\text{3}}/\text{kg}$ for $\nu$ in Equation (II).

$\begin{array}{c}\dot{m}=\frac{\left(3{\text{m}}^{\text{2}}\right)90\text{km/hr}}{0.9012{\text{m}}^{\text{3}}/\text{kg}}\\ =\frac{\left(3{\text{m}}^{\text{2}}\right)\left(90\frac{\text{km}\times \text{1000}\frac{\text{m}}{1\text{km}}}{\text{hr}\times 3600\frac{\text{s}}{\text{hr}}}\right)}{0.9012{\text{m}}^{\text{3}}/\text{kg}}\\ =83.22\text{kg/s}\end{array}$

Substitute 83.22 kg/s for $\dot{m}$, 90 km/hr for $V_1$, and 82 km/hr for $V_2$ in Equation (III).

$\begin{array}{c}\dot{W}=\left(83.22\text{kg/s}\right)\frac{\left({\left(90\frac{\text{km}}{\text{hr}}\right)}^2-{\left(82\frac{\text{km}}{\text{hr}}\right)}^2\right)}{2}\\ =83.22\text{kg/s}\frac{\left[{\left(90\frac{\text{km}\times \text{1000}\frac{\text{m}}{\text{km}}}{\text{hr}\times 3600\frac{\text{s}}{1\text{hr}}}\right)}^2-{\left(82\frac{\text{km}\times \text{1000}\frac{\text{m}}{\text{km}}}{\text{hr}\times 3600\frac{\text{s}}{1\text{hr}}}\right)}^2\right]}{2}\\ =83.22\text{kg/s}\frac{\left(625-518.8\right){\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{2}\\ =4417.852\frac{\text{kg}\cdot {\text{m}}^{\text{2}}}{{\text{s}}^{\text{3}}}\times 1\frac{\text{W}\times \frac{1\text{kW}}{1000\text{W}}}{\frac{\text{kg}\cdot {\text{m}}^{\text{2}}}{{\text{s}}^{\text{3}}}}\end{array}$

$=\text{4}\text{.41}\text{kW}$

Thus, the power required to move car through the air is $\underset{\_}{\text{4}\text{.41}\text{kW}}$.

Substitute $83.22\text{kg/s}$ for $\dot{m}$, $0.9012{\text{m}}^{\text{3}}/\text{kg}$ for $\nu$, and 82 km/hr for $V_2$ in Equation (IV).

$\begin{array}{c}{A}_2=\frac{83.22\text{kg/s}\left(0.9012{\text{m}}^{\text{3}}/\text{kg}\right)}{82\text{km/hr}}\\ =\frac{83.22\text{kg/s}\left(0.9012{\text{m}}^{\text{3}}/\text{kg}\right)}{82\frac{\text{km}\times \text{1000}\frac{\text{m}}{1\text{km}}}{\text{hr}\times \text{3600}\frac{\text{s}}{1\text{hr}}}}\\ =3.29{\text{m}}^2\end{array}$

Thus, the area of the effective flow channel behind the car is $\underset{\_}{3.29{\text{m}}^{\text{2}}}$.