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Jet plane lift and drag coefficients, 170 kN

Short answer

Equating lift to weight gives C_L = 2W/(ρSV²) = 3.53. Drag power is 0.7×580 = 406 kW, drag force = 406000/55.56 = 7308 N, so C_D = 2F_D/(ρSV²) = 0.152.

A jet plane which weighs 170 KN has a wing area of 25 m². It is flying at a speed of 200 km/hr. When the engine develops 580 KW, 70% 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=170W = 170 kN, 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 =580= 580 kW, of which 70% overcomes drag.

Drag force

PD=0.7(580)=406 kW=406 000 WFD=PDV=406 00055.56=7308 N\begin{aligned} P_D &= 0.7(580) = 406\ \text{kW} = 406\,000\ \text{W} \\ F_D &= \frac{P_D}{V} = \frac{406\,000}{55.56} = 7308\ \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=170 00048 225=3.525CD=FD12ρSV2=730848 225=0.1515\begin{aligned} C_L &= \frac{W}{\tfrac12\rho S V^2} = \frac{170\,000}{48\,225} = 3.525 \\[4pt] C_D &= \frac{F_D}{\tfrac12\rho S V^2} = \frac{7308}{48\,225} = 0.1515 \end{aligned}

Answer: CL≈3.53C_L \approx 3.53 and CD≈0.152C_D \approx 0.152. (The high CLC_L is due to the low speed, so it implies flaps or high-lift devices.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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