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Aircraft allowable cargo and engine power

Short answer

Lift equals total weight: compute q = ½ρSV² = 916,667 N. Lift = CL·q = 1.42×916,667 = 1301.7 kN, so cargo = 1301.7 − 1000 = 301.7 kN. Drag = CD·q = 0.17×916,667 = 155,833 N; power = drag × velocity = 155,833×83.33 ≈ 13.0 MW.

An aircraft weighing 1000 KN when empty has a wing area of 220 m². It is to take off at a velocity of 300 Km/hr and a 20° angle of attack. Determine the allowable weight of cargo and power required for the engine. Take density of air as 1.2 kg/m³. Assume coefficient of lift for the wing at 20°, angle of attack as 1.42 and coefficient of drag as 0.17.

At take-off, lift equals the total weight (aircraft plus cargo). The allowable cargo is the lift minus the empty weight. The power equals drag times speed.

Given

Empty weight =1000= 1000 kN, S=220 m2S = 220\ \text{m}^2, V=300V = 300 km/h =83.33= 83.33 m/s, ρ=1.2 kg/m3\rho = 1.2\ \text{kg/m}^3, CL=1.42C_L = 1.42, CD=0.17C_D = 0.17 at 20∘20^\circ.

Dynamic pressure term

12ρSV2=0.5(1.2)(220)(83.33)2=916 667 N\tfrac12\rho S V^2 = 0.5(1.2)(220)(83.33)^2 = 916\,667\ \text{N}

Lift and cargo

L=CL⋅12ρSV2=1.42(916 667)=1 301 667 N=1301.7 kNWtotal=L=1301.7 kNWcargo=1301.7−1000=301.7 kN\begin{aligned} L &= C_L\cdot\tfrac12\rho S V^2 = 1.42(916\,667) = 1\,301\,667\ \text{N} = 1301.7\ \text{kN} \\ W_{total} &= L = 1301.7\ \text{kN} \\ W_{cargo} &= 1301.7 - 1000 = 301.7\ \text{kN} \end{aligned}

Drag and power

FD=CD⋅12ρSV2=0.17(916 667)=155 833 NP=FDV=155 833(83.33)=1.2986×107 W≈12.99 MW\begin{aligned} F_D &= C_D\cdot\tfrac12\rho S V^2 = 0.17(916\,667) = 155\,833\ \text{N} \\ P &= F_D V = 155\,833(83.33) = 1.2986\times10^{7}\ \text{W} \approx 12.99\ \text{MW} \end{aligned}

Answer: allowable cargo weight ≈301.7\approx 301.7 kN (about 302 kN); engine power required ≈13.0\approx 13.0 MW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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