Chapter 15, Problem 93RQ

(a)

To determine

The minimum safe speed for takeoff and landing with and without extending the flaps.

Explanation of Solution

Given:

Total mass is 150000 lbm.

Wing area is $1800{\text{ft}}^2$.

Cruising speed is $550\text{mi/h}$.

Altitude is 38000 ft.

Density of air is $0.0208\text{lbm}/{\text{ft}}^3$.

Density of air on the ground is $0.075\text{lbm}/{\text{ft}}^3$.

Calculation:

Refer figure 15-45 and obtain the maximum lift coefficient with and without flaps as follows:

  $\begin{array}{l}{C}_{L\max ,1}=3.48\\ C_{L\max ,2}=1.52\end{array}$

During takeoff of the aircraft,

  $\begin{array}{l}W=F_L\\ W=\frac{1}{2}{C}_L\rho A{V}^2\\ V=\sqrt{\frac{2W}{C_L\rho A}}\end{array}$

Calculate the stall speed with flaps.

  $\begin{array}{c}{V}_{\min ,1}=\sqrt{\frac{2W}{C_{L\max ,1}\rho A}}\\ =\sqrt{\frac{2\left(150000\text{ lbf}\right)}{3.48\left(0.075\text{lbm}/{\text{ft}}^3\right)\left({\text{1800 ft}}^2\right)}}\\ =143\text{ft}/\text{s}\end{array}$

Calculate the stall speed without flaps.

  $\begin{array}{c}{V}_{\min ,2}=\sqrt{\frac{2W}{C_{L\max ,2}\rho A}}\\ =\sqrt{\frac{2\left(150000\text{ lbf}\right)}{1.52\left(0.075\text{lbm}/{\text{ft}}^3\right)\left({\text{1800 ft}}^2\right)}}\\ =217\text{ft}/\text{s}\end{array}$

Calculate the safe speed with flaps.

  $\begin{array}{c}{V}_1=1.2V_{\min ,1}\\ =1.2\left(143\text{ft}/\text{s}\right)\left(\frac{1\text{ mi/h}}{1.4667\text{ft}/\text{s}}\right)\\ =117\text{mph}\end{array}$

Thus, the safe speed with flaps is $117\text{mph}$.

Calculate the safe speed without flaps.

  $\begin{array}{c}{V}_2=1.2V_{\min ,2}\\ =1.2\left(217\text{ft}/\text{s}\right)\left(\frac{1\text{ mi/h}}{1.4667\text{ft}/\text{s}}\right)\\ =178\text{mph}\end{array}$

Thus, the safe speed without flaps is $178\text{mph}$.

(b)

To determine

The angle of attack.

Explanation of Solution

Calculation:

The lift coefficient is,

  $\begin{array}{c}{C}_L=\frac{F_L}{\frac{1}{2}\rho A{V}^2}\\ =\frac{150000\text{ lbf}}{\frac{1}{2}\left(0.0208\text{lbm}/{\text{ft}}^3\right)\left({\text{1800 ft}}^2\right){\left(550\times 1.4667\text{ ft/s}\right)}^2}\left(\frac{32.2\text{lbm}\cdot {\text{ft/s}}^2}{1\text{lbf}}\right)\\ =0.40\end{array}$

From Figure 15-45, the angle of attack corresponding to the lift coefficient is $3.5°$.

Thus, the angle of attack is $3.5°$.

(c)

To determine

The power required to overcome drag.

Explanation of Solution

Calculation:

From Figure 15-45, the drag coefficient corresponding to the lift coefficient is $0.015$.

The drag force is,

  $\begin{array}{c}{F}_D=C_D\rho A\frac{V^2}{2}\\ =0.015\left(0.0208\text{lbm}/{\text{ft}}^3\right)\left({\text{1800 ft}}^2\right)\frac{{\left(550\times 1.4667\text{ ft/s}\right)}^2}{2}\left(\frac{1\text{ lbf}}{32.2\text{lbm}\cdot {\text{ft/s}}^2}\right)\\ =5675\text{ lbf}\end{array}$

The power required to overcome the drag is,

  $\begin{array}{c}{\dot{W}}_{drag}=F_{D}V\\ =\left(5675\text{lbf}\right)\left(550\text{mi/h}\times \frac{5280\text{ft}}{1\text{mi}}\times \frac{1\text{h}}{3600\text{s}}\right)\left(\frac{1\text{ kW}}{737.56\text{ lbf}\cdot \text{ft/s}}\right)\\ =6200\text{kW}\end{array}$

Thus, the power required to overcome the drag is $6200\text{ kW}$.