Question 10.23: Figure 10.30a illustrates the effects of flaps (no flap, sin...

(a)–(b) The minimum takeoff and landing speed is determined by applying Equation 10.25 F_{L,mim} = W = C_{ L,max} \frac{1}{2} \rho.V^{2}_{min} . A. The lift force must equal the weight of the aircraft at takeoff and landing. Furthermore, the maximum lift coefficient at takeoff and landing is read from Figure 10.30a for the given angle of attack of 10^{\circ } .
(a) The minimum takeoff and landing speed for wings with no flaps is determined as follows:

slug: = 1 lb \frac{sec^{2}}{ft}                           b: = 25 ft                           c: = 20 ft                           A_{plan}: = 2.b.c = 1 \times 10^{3} ft^{2}

\rho : = 0.0023768 \frac{slug}{ft^{3}}                           \alpha := 10 deg                           W: = 30000 lb                           F_{Lmin}: = W = 3 \times 10^{4} lb

Guess value:                           V_{min}: = 100 \frac{ft}{sec}

Given

F_{Lmin} = C_{Lmax} \frac{1}{2} \rho .V_{min}^{2} . A_{plan}
V_{min} : =Find (V_{min}) = 142 .11 \frac{ft}{s}                           V_{min} = 96.893 mph

Furthermore, since this is not the stall speed, it does not have to be adjusted.
(b) The minimum takeoff and landing speed for wings with slotted flaps is determined as follows:

Guess value:                           V_{min}: = 100 \frac{ft}{sec}

Given

F_{Lmin} = C_{Lmax} \frac{1}{2} \rho .V_{min}^{2} . A_{plan}
V_{min} : =Find (V_{min}) = 100.487 \frac{ft}{s}                           V_{min} = 68.514 mph

Furthermore, since this is not the stall speed, it does not have to be adjusted (increased). And, finally, for the given angle of attack, α of 10^{\circ } , the use of slotted flaps increases the lift coefficient, C_{L} from 1.25 to 2.5 and thus allows a decrease in the corresponding minimum takeoff and landing speed from 96.893 mph to 68.514mph.