Chapter 9.1, Problem 1P

To determine

The coefficient of drag.

The coefficient of lift.

Answer to Problem 1P

The coefficient of drag is $1$.

The coefficient of lift is $0$.

Explanation of Solution

The below figure represent the equatorial triangular bar.

Figure-(1)

Write the expression for the drag force.

    $D={\displaystyle \underset{1}{\int }{P}_1\cos {\theta }_{1}dA}+{\displaystyle \underset{2}{\int }{P}_2\cos {\theta }_{2}dA}+{\displaystyle \underset{3}{\int }{P}_3\cos {\theta }_{3}dA}$        (I)

Here, the force at state 1 is $P_1$, the force at state 2 is $P_2$, the force at state 3 is $P_3$, the angle between force at state 1 applied with the horizontal is ${\theta }_1$, angle between force at state 2 applied with the horizontal is ${\theta }_2$, angle between force at state 3 applied with the horizontal is ${\theta }_3$ and the differential area is $dA$.

Write the expression for the coefficient of drag.

    $C_D=\frac{D}{\frac{1}{2}\rho V^{2}A}$        (II)

Here, the density is $\rho$, velocity of flow is $V$, the area is $A$ and the drag force is $D$.

Write the expression for the coefficient of lift.

    $C_L=\frac{L}{\frac{1}{2}\rho V^{2}A}$        (III)

Here, the lift force is $L$.

Conclusion:

Substitute $60°$ for ${\theta }_1$, $60°$ for ${\theta }_2$ and $0°$ for , $0.5\rho V^2$ for $P_1$, $0.5\rho V^2$ for $P_2$, $lb$ for $dA$ and $0.25\rho V^2$ for $P_3$ in Equation (I).

    $\begin{array}{c}D={\displaystyle \underset{1}{\int }0.5\rho V^2\cos 60°dA}+{\displaystyle \underset{2}{\int }0.5\rho V^2\cos 60°dA}+{\displaystyle \underset{3}{\int }-0.25\rho V^2\cos 0°dA}\\ =0.5\rho V^2\cos 60°lb+0.5\rho V^2\cos 60°lb-0.25\rho V^{2}lb\\ =0.25\rho V^{2}lb+0.25\rho V^{2}lb-0.25\rho V^{2}lb\\ =0.50\rho V^{2}lb\end{array}$

Refer to Table (B-1)- “Physical properties of water” to obtain the value of density at temperature of $50°$ as $1.94\text{slug}/{\text{ft}}^{\text{3}}$.

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $5\text{ft}/\text{sec}$ for $V$, $0.1\text{ft}$ for $l$ and $4\text{ft}$ for $b$ in Equation (I).

    $\begin{array}{c}D=0.50\times \left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right)\times {\left(5\text{ft}/\text{sec}\right)}^2\times 0.1\text{ft}\times 4\text{ft}\\ =\text{7}\text{.1295}\text{slug}\cdot \text{ft}/{\sec }^2\left(\frac{1\text{lb}}{\text{slug}\cdot \text{ft}/{\sec }^2}\right)\\ =\text{9}\text{.7}\text{lb}\end{array}$

Substitute $1.94\text{slug}/{\text{ft}}^{\text{3}}$ for $\rho$, $5\text{ft}/\text{sec}$ for $V$, $0.1\text{ft}$ for $l$, $4\text{ft}$ for $b$ and $\text{9}\text{.7}\text{lb}$ for $D$ in Equation (II).

    $\begin{array}{c}{C}_D=\frac{\text{9}\text{.7}\text{lb}}{\frac{1}{2}\left(1.94\text{slug}/{\text{ft}}^{\text{3}}\right){\left(5\text{ft}/\text{sec}\right)}^{2}0.1\text{ft}\times \text{4}\text{ft}}\\ =\frac{\text{9}\text{.7}\text{lb}}{0.2{\text{ft}}^2\times \left(1.94\text{slug}/{\text{ft}}^{\text{3}}\left(\frac{1\text{lb}}{\text{slug}\cdot \text{ft}/{\sec }^2}\right)\right)\times {\left(5\text{ft}/\text{sec}\right)}^2}\\ =1\end{array}$

Substitute $0$  for $L$ in Equation (III).

    $\begin{array}{c}{C}_L=\frac{0}{\frac{1}{2}\rho V^{2}A}\\ =0\end{array}$

The coefficient of drag is $1$.

The coefficient of lift is $0$.