Chapter 11, Problem 88P

A small aircraft has a wing area of 40 m², a lift coefficient of 0.4$ 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. (b) the wing loading, and (c) the required power to maintain a constant cruising speed of 360 km/h for a cruising drag coefficient of 0.035.

To determine

(a)

Take off speed at sea level at standard atmospheric conditions.

Answer to Problem 88P

  $V_{\text{takeoff}}=257.74\frac{\text{km}}{\text{hr}}$

Explanation of Solution

Given information:

  $\text{Total mass}=4,000\text{kg}$

  $\text{Wing area}=40{\text{m}}^{\text{2}}$

  $\text{Lift coefficient}=0.45$

Concept used:

  $F_L=C_{L}A\frac{\rho V^2}{2}$

  $\begin{array}{c}{F}_L\to \text{Lift force}\\ C_L\to \text{Maximum lift coefficient}\\ \rho \to \text{Density of air}\\ V\to \text{Velocity}\\ A\to \text{Wing area}\end{array}$

Calculation:

For standard air at sea level, the density is $\rho =1.225\frac{\text{kg}}{{\text{m}}^{\text{3}}}$.

  $\begin{array}{c}W=F_L\\ W=C_{L}A\frac{\rho V^2}{2}\\ V=\frac{2W}{C_{L}A\rho }\\ =\sqrt{\frac{2\times 4,000\times 9.81}{0.45\times 40\times 1.225}}\\ =59.66\times \frac{3600}{1000}\frac{\text{km}}{\text{hr}}\\ =214.78\frac{\text{km}}{\text{hr}}\end{array}$

  $\begin{array}{c}{V}_{\text{takeoff}}=1.2\times V\\ =1.2\times 214.78\\ =257.74\frac{\text{km}}{\text{hr}}\end{array}$

Conclusion:

The takeoff speed at sea level at standard atmospheric conditions is $257.74\frac{\text{km}}{\text{hr}}$.

To determine

(b)

Wing loading of the aircraft.

Answer to Problem 88P

  $F_{\text{Loading}}=981\frac{\text{N}}{{\text{m}}^{\text{2}}}$

Explanation of Solution

Given information:

  $\text{Total mass}=4,000\text{kg}$

  $\text{Wing area}=40{\text{m}}^{\text{2}}$

  $\text{Lift coefficient}=0.45$

Concept used:

  $F_{\text{Loading}}=\frac{W}{A}$

  $\begin{array}{c}{F}_{Loading}\to \text{Wing loading}\\ W\to \text{Weight of the aircraft}\\ A\to \text{Wing area}\end{array}$

Calculation:

  $\begin{array}{c}{F}_{\text{Loading}}=\frac{W}{A}\\ =\frac{mg}{A}\\ =\frac{4000\times 9.81}{40}\\ =981\frac{\text{N}}{{\text{m}}^{\text{2}}}\end{array}$

Conclusion:

The wing loading of the aircraft is $981\frac{\text{N}}{{\text{m}}^{\text{2}}}$.

To determine

(c)

Power required by the engine to maintain a given constant cruising speed $360\frac{\text{km}}{\text{hr}}$.

Answer to Problem 88P

  ${\dot{W}}_{\text{prop}}=857.5\text{kW}$

Explanation of Solution

Given information:

  $\text{Total mass}=4,000\text{kg}$

  $\text{Wing area}=40{\text{m}}^{\text{2}}$

  $\text{Drag coefficient}=0.035$

Concept used:

  $F_D=C_{D}A\frac{\rho V^2}{2}$

  ${\dot{W}}_{\text{prop}}=F_D\times V$

  $\begin{array}{c}{\dot{W}}_{\text{prop}}\to \text{Power}\\ F_D\to \text{Drag force}\\ C_D\to \text{Drag coefficient}\\ \rho \to \text{Density}\\ V\to \text{Velocity}\\ A\to \text{Wing area}\end{array}$

Calculation:

For standard air at sea level, the density is $\rho =1.225\frac{\text{kg}}{{\text{m}}^{\text{3}}}$.

  $\begin{array}{c}{F}_D=C_{D}A\frac{\rho V^2}{2}\\ =0.035\times 40\times \frac{1.225\times {\left(360\times \frac{1000}{3600}\frac{\text{m}}{\text{s}}\right)}^2}{2}\times \left(\frac{\text{1}\text{kN}}{\text{1000}\text{kg}\cdot {\text{m/s}}^{\text{2}}}\right)\\ =8.575\text{kN}\end{array}$

  $\begin{array}{c}{\dot{W}}_{\text{prop}}=F_D\times V\\ =8.575\times 360\times \frac{1000}{3600}\frac{\text{km}}{\text{hr}}\times \left(\frac{\text{1}\text{kW}}{\text{1}\text{kN}\cdot \text{m/s}}\right)\\ =857.5\text{kW}\end{array}$

Conclusion:

The power required by the engine to maintain a constant cruising speed of $360\frac{\text{km}}{\text{hr}}$ is $857.5\text{kW}$.