---
title: "Jet plane lift and drag coefficients, 170 kN"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:jet-plane-lift-drag-170kn"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-plane-lift-drag-170kn
site: BCT Engineering
---

# Jet plane lift and drag coefficients, 170 kN

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2076 Asoj, 8 marks
- Video: [Drag, Coefficient of Drag, L/D ratio](https://www.youtube.com/watch?v=dXOm_Bt1ujw) — Planes Over Head (POH)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-plane-lift-drag-170kn
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

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.

## 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.

## Answer

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

#### Given

$W = 170$ kN, $S = 25\ \text{m}^2$, $V = 200$ km/h $= 55.56$ m/s, $\rho = 1.25\ \text{kg/m}^3$, engine power $= 580$ kW, of which 70% overcomes drag.

#### Drag force

$$
\begin{aligned}
P_D &= 0.7(580) = 406\ \text{kW} = 406\,000\ \text{W} \\
F_D &= \frac{P_D}{V} = \frac{406\,000}{55.56} = 7308\ \text{N}
\end{aligned}
$$

#### Dynamic pressure term

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

#### Coefficients

$$
\begin{aligned}
C_L &= \frac{W}{\tfrac12\rho S V^2} = \frac{170\,000}{48\,225} = 3.525 \\[4pt]
C_D &= \frac{F_D}{\tfrac12\rho S V^2} = \frac{7308}{48\,225} = 0.1515
\end{aligned}
$$

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What gives lift?
   A. Lift equals drag
   B. Lift equals engine power/V
   C. Lift equals 0.7 W
   D. Lift equals weight, 170 kN
   Answer: D. Level flight means L = W. (Key point: Lift in level flight)

2. How is the drag found?
   A. F_D = 580 kW/V
   B. F_D = 0.7 P x V
   C. F_D = 0.7 P / V = 406 kW/55.56 m/s = 7308 N
   D. F_D = 0.3 P/V
   Answer: C. Only 70% of power overcomes drag. (Key point: Drag force)

3. What speed should be used?
   A. 200 m/s
   B. 55.56 m/s (200 km/h converted)
   C. 20 m/s
   D. 720 m/s
   Answer: B. 200/3.6 = 55.56. (Key point: Speed unit)

4. What are CL and CD?
   A. CL = 3.53 and CD = 0.152
   B. CL = 0.152 and CD = 3.53
   C. CL = 0.35 and CD = 0.015
   D. CL = 1.0 and CD = 0.3
   Answer: A. Both divide by 1/2 rho S V^2 = 48,225 N. (Key point: Coefficients)

## Close variants set in other exams

- [Jet plane lift and drag coefficients, 19920 N](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ajet-plane-lift-drag-19920n) — Fluid Mechanics, chapter 9, asked 2×

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