---
title: "Jet plane lift and drag coefficients, 19920 N"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:jet-plane-lift-drag-19920n"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/jet-plane-lift-drag-19920n
site: BCT Engineering
---

# Jet plane lift and drag coefficients, 19920 N

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2075 Asoj, 8 marks
- 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-19920n
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

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.

## 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.

## 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

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

#### Drag force

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

#### Dynamic pressure term

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

#### Coefficients

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

**Answer: $C_L \approx 0.413$ and $C_D \approx 0.176$.**

## 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 0.8 W
   C. Lift equals power/V
   D. Lift equals weight, 19,920 N
   Answer: D. Level flight means L = W. (Key point: Lift in level flight)

2. How is drag found?
   A. F_D = 588.5 kW/V
   B. F_D = 0.2 P/V
   C. F_D = 0.8 P / V = 470.8 kW/55.56 = 8474 N
   D. F_D = 0.8 P x V
   Answer: C. Only 80% of power goes to drag. (Key point: Drag force)

3. What is 1/2 rho S V^2 here?
   A. About 8,474 N
   B. About 48,225 N
   C. About 19,920 N
   D. About 482 N
   Answer: B. 0.5 x 1.25 x 25 x 55.56^2. (Key point: Reference force)

4. What are CL and CD?
   A. CL = 0.413 and CD = 0.176
   B. CL = 0.176 and CD = 0.413
   C. CL = 3.5 and CD = 0.15
   D. CL = 0.8 and CD = 0.2
   Answer: A. Each force divided by 48,225 N. (Key point: Coefficients)

## Close variants set in other exams

- [Jet plane lift and drag coefficients, 170 kN](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ajet-plane-lift-drag-170kn) — Fluid Mechanics, chapter 9, asked 2×