Chapter 11, Problem 85P

A small aircraft has a wing area of 35 $m^2$ a lift coefficient of 0.45 at takeoff settings, and a total mass of 4000 kg. Determine (a) the takeoff speed of this aircraft at sea level at standard atmospheric conditions, (h) the wing loading, and (c) the required power to maintain a constant cruising speed of 300 km/h for a cruising drag coefficient of 0.035.

To determine

The takeoff speed.

Answer to Problem 85P

The takeoff speed is $63.77\text{m}/\text{s}$.

Explanation of Solution

Given information:

The area of aircraft is $35{\text{m}}^{\text{2}}$, the lift coefficient is $0.45$, mass of aircraft is $4000\text{kg}$, the velocity of aircraft is $300\text{km}/\text{h},$ and the cursing drag coefficient is $0.035$.

Concept used:

Write the expression for the lift force.

$F_L=C_{L}A\left(\frac{\rho V^2}{2}\right)$ ……. (I)

Here, the density of water is $\rho$, viscosity of water is $\mu$, the diameter of pipe is $D$, the lift coefficient is $C_L,$ and average velocity is $V.$

Since the weight of airplane should be equal to the lift force for takeoff of airplane.

$F_L=W$ ……. (II)

Write the expression for the weight of airplane.

$W=mg$

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

Substitute $C_{L}A\left(\frac{\rho V^2}{2}\right)$ for $F_L$ and $mg$ for $W$ in (II).

$\begin{array}{l}{C}_{L}A\left(\frac{\rho V^2}{2}\right)=mg\\ V^2=\frac{2mg}{\rho C_{L}A}\\ V=\sqrt{\frac{2mg}{\rho C_{L}A}}\end{array}$ ……. (III)

Write the expression for the drag force.

$F_D=C_{d}A\left(\frac{\rho V^2}{2}\right)$ ……. (II)

Here, the discharge coefficient is $C_d$

Write the expression for frontal area of flow.

$A=LD$

Here, the length of pipe is $L$.

Substitute $LD$ for $A$ in Equation (II).

$F_D=C_{d}LD\left(\frac{\rho V^2}{2}\right)$ ……. (III)

Calculations:

Substitute $1.225\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $4000\text{kg}$ for $m$ and $35{\text{m}}^{\text{2}}$ for $A$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g,$ and $0.45$ for $C_L$ in Equation (III).

$\begin{array}{c}V=\sqrt{\frac{2\left(4000\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{1.225\text{kg}/{\text{m}}^{\text{3}}\left(0.45\right)\left(35{\text{m}}^{\text{2}}\right)}}\\ =\sqrt{4067.63{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ =63.77\text{m}/\text{s}\end{array}$

Conclusion:

The takeoff speed is $63.77\text{m}/\text{s}$.

To determine

The wing loading.

Answer to Problem 85P

The wing loading is $1121.14\text{Pa}$.

Explanation of Solution

Given information:

The area of aircraft is $35{\text{m}}^{\text{2}}$, the lift coefficient is $0.45$, mass of aircraft is $4000\text{kg}$, the velocity of aircraft is $300\text{km}/\text{h},$ and the cursing drag coefficient is $0.035$.

Concept used:

Since the weight of airplane should be equal to the lift force for takeoff of airplane.

$F_L=W$ ……. (IV)

Write the expression for the weight of airplane.

$W=mg$

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

Substitute $mg$ for $W$ in (IV).

$F_L=mg$

Write the expression for the load on wings.

$F=\frac{F_L}{A}$ ……. (V).

Substitute $mg$ for $F_L$ in Equation (V).

$F=\frac{mg}{A}$ …….. (VI)

Calculations:

Substitute $4000\text{kg}$ for $m$ and $35{\text{m}}^{\text{2}}$ for $A$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VI).

$\begin{array}{c}F=\frac{\left(4000\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}{35{\text{m}}^{\text{2}}}\\ =1121.14\text{kg}\cdot \text{m}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\right)\\ =1121.14\text{N}/{\text{m}}^{\text{2}}\left(\frac{1\text{Pa}}{1\text{N}/{\text{m}}^{\text{2}}}\right)\\ =1121.14\text{Pa}\end{array}$

Conclusion:

The wing loading is $1121.14\text{Pa}$.

To determine

The power required.

Answer to Problem 85P

The power required is $3295.89\text{kW}$.

Explanation of Solution

Given information:

The area of aircraft is $35{\text{m}}^{\text{2}}$, the lift coefficient is $0.45$, mass of aircraft is $4000\text{kg}$, the velocity of aircraft is $300\text{km}/\text{h},$ and the cursing drag coefficient is $0.035$.

Concept used:

Write the expression for power required.

$P=F_D\text{×}V$ …….. (VII)

Write the expression for the drag force.

$F_D=C_{d}A\left(\frac{\rho V^2}{2}\right)$ ……. (VIII)

Here, the discharge coefficient is $C_d$

Calculations:

Substitute $300\text{km}/\text{h}$ for $V$, $1.225\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $35{\text{m}}^{\text{2}}$ for $A$, and $0.035$ for $C_d$ in Equation (VIII).

$\begin{array}{c}{F}_D=\left(0.035\right)\left(35{\text{m}}^{\text{2}}\right)\left(\frac{1.225\text{kg}/{\text{m}}^{\text{3}}{\left(300\text{km}/\text{h}\right)}^2}{2}\right)\\ =\left(0.035\right)\left(35{\text{m}}^{\text{2}}\right)\left(\frac{1.225\text{kg}/{\text{m}}^{\text{3}}{\left(300\text{km}/\text{h}\left(\frac{\frac{\text{5}}{\text{8}}\text{m}/\text{s}}{\text{1}\text{km}/\text{h}}\right)\right)}^2}{2}\right)\\ =17578.125\text{N}\end{array}$

Substitute $300\text{km}/\text{h}$ for $V$ and $17578.125\text{N}$ for $F_D$ in Equation (VII).

$\begin{array}{c}P=17578.125\text{N×}\left(300\text{km}/\text{h}\right)\\ =17578.125\text{N×}\left(300\text{km}/\text{h}\right)\left(\frac{\frac{\text{5}}{\text{8}}\text{m}/\text{s}}{\text{1}\text{km}/\text{h}}\right)\\ =3295898.43\text{W}\left(\frac{\text{1}\text{kW}}{\text{1000}\text{W}}\right)\\ =3295.89\text{kW}\end{array}$

Conclusion:

The power required is $3295.89\text{kW}$.