Chapter 11, Problem 104EP

A commercial airplane has a total mass of 150.000 Ibm and a wing planform area of 1700 ft2. The plane has a cruising speed of 625 mi1 and a cruising altitude of 38.000 ft where the air density is 0.0208 Ibm/fl³. The plane has double-slotted flaps for use during takeoff and landing. but it cruises with all flaps retracted. Assuming the lift and drag characteristics of the wings can be approximated by NACA 23012, detennine (a) the minimum safe speed for takeoff and landing with and without extending the flaps. (b) the anzle of attack to cruise steadily at the cruising altitude, and (c) the power that needs to be supplied to provide enough thrust to overcome drag. Take the air density on the ground to be 0.075 Ibm/ft³.

To determine

(a)

The minimum safe speed for takeoff and landing with the flaps.

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

Answer to Problem 104EP

The minimum safe speed for takeoff and landing with the flaps is $267.91\text{ft}/\text{s}$.

The minimum safe speed for takeoff and landing without the flaps is $177.06\text{ft}/\text{s}$.

Explanation of Solution

Given information:

The total mass of the airplane is $150000\text{lbm}$, the wing platform area is $1700{\text{ft}}^{\text{2}}$, the cursing speed of the plane is $625\text{mi}/\text{h}$, the cursing altitude is $38000\text{ft}$, the density of the air at the altitude of $38000\text{ft}$ is $0.0208\text{lbm}/{\text{ft}}^{\text{3}}$ and the density of air at ground is $0.075\text{lbm}/{\text{ft}}^{\text{3}}$.

Write the expression for the initial velocity of the aircraft.

  $V_1=\sqrt{\left(\frac{2W}{{\rho }_a{C}_{L1}A}\right)}$........ (I)

Here, the weight of the aircraft is $W$, the density of the air is ${\rho }_a$, the frontal area is $A,$ and the lift coefficient at initial state is $C_{L1}$.

Write the expression for the final velocity of the aircraft.

  $V_2=\sqrt{\left(\frac{2W}{{\rho }_a{C}_{L2}A}\right)}$........ (II)

Here, the lift coefficient at final state is $C_{L2}$.

Write the expression for the weight of the aircraft.

  $W=mg$........ (III)

Here, the mass of the airplane is $m$ and the acceleration due to gravity is $g$.

Write the expression for the safe velocity with flaps.

  $V_{s1}=1.2V_1$........ (IV)

Write the expression for the safe velocity without flaps.

  $V_{s2}=1.2V_2$........ (V)

Calculation:

Substitute $150000\text{lbm}$ for $m$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (III).

  $\begin{array}{c}W=150000\text{lbm}\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\\ =4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$

Refer to Figure 15-45 titled "Effect of flaps on the lift and drag coefficient of an airfoil" to obtain maximum lift coefficient with flaps as $1.52$ and without flaps as $3.48$.

Substitute $1.52$ for $C_{L1}$, $4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for $W$, $0.075\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ and $1700{\text{ft}}^2$ for $A$ in Equation (I).

  $\begin{array}{c}{V}_1=\sqrt{\left(\frac{2\left(4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\right)}{1.52\left(0.075\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(1700{\text{ft}}^2\right)}\right)}\\ =\sqrt{\left(\frac{\left(9660000{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}{\left(193.8\right)}\right)}\\ =\sqrt{\left(49845.20{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}\\ =223.26\text{ft}/\text{s}\end{array}$

Substitute $3.48$ for $C_{L2}$, $4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for $W$, $0.075\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ and $1700{\text{ft}}^2$ for $A$ in Equation (II).

  $\begin{array}{c}{V}_2=\sqrt{\left(\frac{2\left(4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\right)}{3.48\left(0.075\text{lbm}/{\text{ft}}^{\text{3}}\right)\left(1700{\text{ft}}^2\right)}\right)}\\ =\sqrt{\left(\frac{\left(9660000{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}{\left(443.7\right)}\right)}\\ =\sqrt{\left(21771.46{\text{ft}}^2/{\text{s}}^{\text{2}}\right)}\\ =147.55\text{ft}/\text{s}\end{array}$

Substitute $223.26\text{ft}/\text{s}$ for $V_1$ in Equation (IV).

  $\begin{array}{c}{V}_{s1}=1.2\left(223.26\text{ft}/\text{s}\right)\\ =267.91\text{ft}/\text{s}\end{array}$

Substitute $147.55\text{ft}/\text{s}$ for $V_2$ in Equation (V).

  $\begin{array}{c}{V}_{s2}=1.2\left(147.55\text{ft}/\text{s}\right)\\ =177.06\text{ft}/\text{s}\end{array}$

Conclusion:

The minimum safe speed for takeoff and landing with the flaps is $267.91\text{ft}/\text{s}$.

The minimum safe speed for takeoff and landing without the flaps is $177.06\text{ft}/\text{s}$.

To determine

(b)

The angle of attack to cruise steadily at the cruising altitude.

Answer to Problem 104EP

The angle of attack to cruise steadily at the cruising altitude is $\alpha =3.3°$.

Explanation of Solution

Write the expression for the coefficient of the lift force.

  $C_L=\frac{F_L}{\frac{1}{2}{\rho }_a{V}^{2}A}$........ (VI)

Here, the lift force is $F_L$, the velocity of the airplane is $V,$ and the density of air at the given altitude is $\rho$

Calculation:

The lift force is equals to weight of the airplane that is $F_L=4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$.

Substitute $4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for $F_L$, $0.0208\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $1700{\text{ft}}^2$ for $A$ and $625\text{mi}/\text{h}$ for $V$ in Equation (VI).

  $\begin{array}{c}{C}_L=\frac{4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}{\frac{1}{2}\left(0.0208\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(625\text{mi}/\text{h}\right)}^2\left(1700{\text{ft}}^2\right)}\\ =\frac{4830000\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}{0.5\left(0.0208\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(625\text{mi}/\text{h}\left(\frac{1.466\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)}^2\left(1700{\text{ft}}^2\right)}\\ =\frac{4830000}{14842608.63}\\ =0.325\end{array}$

Refer to Figure 15-45 "Effect of flaps on the lift and drag coefficients of an airfoil" to obtain the angle of attack of a clean flap as $\alpha =3.3°.$

Conclusion:

The angle of attack to cruise steadily at the cruising altitude is $\alpha =3.3°$.

To determine

(c)

The power to be supplied to present enough thrust to overcome the drag.

Answer to Problem 104EP

The power to be supplied the enough thrust to overcome the drag is $8589.38\text{kW}$.

Explanation of Solution

Write the expression for the drag force.

  $F_D=C_D\frac{1}{2}{\rho }_a{V}^{2}A$........ (VII)

Here, the drag force coefficient is $F_D$.

Write the expression for the drag force and the velocity of the airplane.

  $P=F_D\times V$........ (VIII)

Calculation:

Refer to Figure 15-45 "Effect of flaps on the lift and drag coefficients of an airfoil" to obtain the drag coefficient as $0.015$ at the lift coefficient $C_D=0.325$.

Substitute $0.015$ for $C_D$, $0.0208\text{lbm}/{\text{ft}}^{\text{3}}$ for $\rho$, $1700{\text{ft}}^2$ for $A$ and $625\text{mi}/\text{h}$ for $V$ in Equation (VII).

  $\begin{array}{c}{F}_D=\left(0.015\right)\frac{1}{2}\left(0.0208\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(625\text{mi}/\text{h}\right)}^2\left(1700{\text{ft}}^2\right)\\ =\left(0.0075\right)\left(0.0208\text{lbm}/{\text{ft}}^{\text{3}}\right){\left(625\text{mi}/\text{h}\left(\frac{1.466\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)}^2\left(1700{\text{ft}}^2\right)\\ =222639.12\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}\left(\frac{1\text{lbf}}{32.2\text{lbm}\cdot \text{ft}/{\text{s}}^{\text{2}}}\right)\\ =6914.25\text{lbf}\end{array}$

Substitute $6914.25\text{lbf}$ for $F_D$ and $625\text{mi}/\text{h}$ for $V$ in Equation (VIII).

  $\begin{array}{c}p=\left(6914.25\text{lbf}\right)\left(625\text{mi}/\text{h}\right)\\ =\left(6914.25\text{lbf}\right)\left(625\text{mi}/\text{h}\left(\frac{1.466\text{ft}/\text{s}}{1\text{mi}/\text{h}}\right)\right)\\ =6335189.51\text{lbf}\cdot \text{ft}/\text{s}\left(\frac{1\text{kW}}{737.56\text{lbf}\cdot \text{ft}/\text{s}}\right)\\ =8589.38\text{kW}\end{array}$

Conclusion:

The power to be supplied the enough thrust to overcome the drag is $8589.38\text{kW}$.