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Jet plane lift and drag coefficients, 19920 N

Short answer

Lift equals weight. Drag is obtained from 80% of engine power: FD = P/V = 8474 N. Dynamic pressure q = ½ρSV² = 48,225 N. Then CL = W/q = 0.413 and CD = FD/q = 0.176.

A jet plane which weighs 19920 N has a wing area of 25 m². It is flying at a speed of 200 km/hr. When the engine develops 588.5 KW, 80% of this power is used to overcome the drag resistance of the wing. Calculate the coefficient of lift and coefficient of drag for the wing. Take density of air = 1.25 kg/m³.

In level flight lift equals weight. The power used against drag gives the drag force (P=FDVP = F_D V), and then both coefficients follow from F=C⋅12ρSV2F = C\cdot\tfrac12\rho S V^2.

Given

W=19 920W = 19\,920 N, S=25 m2S = 25\ \text{m}^2, V=200V = 200 km/h =55.56= 55.56 m/s, ρ=1.25 kg/m3\rho = 1.25\ \text{kg/m}^3, engine power =588.5= 588.5 kW, of which 80% overcomes drag.

Drag force

PD=0.8(588.5)=470.8 kW=470 800 WFD=PDV=470 80055.56=8474 N\begin{aligned} P_D &= 0.8(588.5) = 470.8\ \text{kW} = 470\,800\ \text{W} \\ F_D &= \frac{P_D}{V} = \frac{470\,800}{55.56} = 8474\ \text{N} \end{aligned}

Dynamic pressure term

12ρSV2=0.5(1.25)(25)(55.56)2=48 225 N\tfrac12\rho S V^2 = 0.5(1.25)(25)(55.56)^2 = 48\,225\ \text{N}

Coefficients

CL=W12ρSV2=19 92048 225=0.413CD=FD12ρSV2=847448 225=0.1757\begin{aligned} C_L &= \frac{W}{\tfrac12\rho S V^2} = \frac{19\,920}{48\,225} = 0.413 \\[4pt] C_D &= \frac{F_D}{\tfrac12\rho S V^2} = \frac{8474}{48\,225} = 0.1757 \end{aligned}

Answer: CL≈0.413C_L \approx 0.413 and CD≈0.176C_D \approx 0.176.

✓ Answer checked by Claude Opus 5.5 (high effort)

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