---
title: "Aircraft allowable cargo and engine power"
subject: "Fluid Mechanics"
chapter: 9
key: "semester_3/fluid_mechanics:aircraft-cargo-weight-power"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/aircraft-cargo-weight-power
site: BCT Engineering
---

# Aircraft allowable cargo and engine power

Fluid Mechanics · Chapter 9: Flow past through submerged bodies

- Asked in: 2073 Shrawan, 3+2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/aircraft-cargo-weight-power
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-9

## Question

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.

## 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.

## Answer

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 $= 1000$ kN, $S = 220\ \text{m}^2$, $V = 300$ km/h $= 83.33$ m/s, $\rho = 1.2\ \text{kg/m}^3$, $C_L = 1.42$, $C_D = 0.17$ at $20^\circ$.

#### Dynamic pressure term

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

#### Lift and cargo

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

#### Drag and power

$$
\begin{aligned}
F_D &= C_D\cdot\tfrac12\rho S V^2 = 0.17(916\,667) = 155\,833\ \text{N} \\
P &= F_D V = 155\,833(83.33) = 1.2986\times10^{7}\ \text{W} \approx 12.99\ \text{MW}
\end{aligned}
$$

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

## Check yourself

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

1. At take-off, what is the lift force equal to?
   A. The empty weight only
   B. The drag force
   C. The total weight of the aircraft plus cargo
   D. The engine thrust
   Answer: C. For the aircraft to leave the ground, lift balances the total weight (empty plus cargo). Cargo is then lift minus empty weight. (Key point: Take-off condition)

2. With 0.5*rho*S*V^2 = 916,667 N and CL = 1.42, what is the lift and the allowable cargo weight?
   A. Lift 1301.7 kN, cargo 1301.7 kN
   B. Lift 1301.7 kN, cargo 301.7 kN
   C. Lift 155.8 kN, cargo 155.8 kN
   D. Lift 916.7 kN, cargo 0 kN (no cargo allowed)
   Answer: B. L = 1.42 x 916,667 = 1301.7 kN. Cargo = 1301.7 - 1000 = 301.7 kN. (Key point: Lift calculation)

3. How is the engine power found, and what is it approximately?
   A. P = drag x V, about 13.0 MW
   B. P = lift x V, about 108 MW
   C. P = drag x V, about 1.30 MW
   D. P = weight x V, about 13.0 MW
   Answer: A. Power overcomes drag at speed V. Drag = 0.17 x 916,667 = 155.8 kN, so P = 155,833 x 83.33 = about 13.0 MW. (Key point: Power required)

4. What speed is used for 300 km/h in the calculation?
   A. 300 m/s
   B. 108 m/s
   C. 83.33 m/s
   D. 30 m/s
   Answer: C. 300 km/h divided by 3.6 gives 83.33 m/s, which is needed because SI units are used. (Key point: Unit conversion)

## Related in Fluid Mechanics

- [Aeroplane angle of attack and take-off power](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Aaeroplane-angle-of-attack-power) — Fluid Mechanics, chapter 9

## Related in other subjects

- [Eccentric bolted bracket, ISMB 300, 220 kN](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abracket-bolted-ismb300-220kn) — Design of Steel and Timber Structure, chapter 4
- [Slab base for SC 220, 1000 kN, M30](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Aslab-base-sc220-m30) — Design of Steel and Timber Structure, chapter 6