Neutral-Buoyancy Temperature of a Hot-Air Balloon: 04-BS-7, December 2016, Q1

Question 1: Neutral-Buoyancy Temperature of a Hot-Air Balloon (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Total envelope volume V	3147 m³
Structure + payload mass (100+60+110+50+160)	480 kg
Ambient pressure p₀	100 kPa
Ambient temperature T₀	15°C = 288.15 K
Gas constant for air R	287 J/kg·K

Find. The temperature of the hot air inside the envelope, T_hot, for neutral buoyancy.

Approach. Neutral buoyancy requires the weight of ambient air displaced by the whole balloon to equal the total weight (structure + payload + the hot air filling the envelope); with the ideal-gas law this fixes the hot-air density, and hence its temperature at the same (atmospheric) pressure.

Ambient air density from the ideal gas law. $$\rho_{amb} = \frac{p_0}{R\,T_0} = \frac{100{,}000}{287(288.15)} = 1.209\ \text{kg/m}^3$$
Force balance for neutral buoyancy. Buoyant force equals total weight: $$\rho_{amb} g V = (m_{struct} + \rho_{hot} V)\,g \quad\Rightarrow\quad \rho_{hot} = \rho_{amb} - \frac{m_{struct}}{V}$$ $$\rho_{hot} = 1.209 - \frac{480}{3147} = 1.209 - 0.1525 = \boxed{1.057\ \text{kg/m}^3}$$
Hot-air temperature at the same (atmospheric) pressure. $$T_{hot} = \frac{p_0}{R\,\rho_{hot}} = \frac{100{,}000}{287(1.057)} = 329.8\ \text{K} = \boxed{56.6\,{}^{\circ}\text{C}}$$

Quantity	Result
Ambient air density	1.209 kg/m³
Required hot-air density	1.057 kg/m³
Hot-air temperature for neutral buoyancy	56.6°C (329.8 K)

Topic: Buoyancy of a lighter-than-ambient gas envelope (Archimedes' principle + ideal gas law).

Key relations: Buoyant force F_B = ρ_amb g V; ideal gas law p = ρRT.

Why this works. A hot air balloon carries no separate lifting gas — it relies entirely on heating the enclosed air to reduce its density below the surrounding air, so the displaced-air weight exceeds the balloon's own weight. Because the hot air and the ambient air share the same pressure (the envelope is open at the bottom), the density deficit converts directly to a temperature difference via the ideal gas law.

Common pitfall. Forgetting that the structure/payload mass must be divided by the FULL envelope volume (not just the "envelope" component's own 100 kg) to get the equivalent density deficit — the whole airborne mass is supported by the whole displaced volume.

Source: White, Fluid Mechanics, Ch. 2 (buoyancy).

Question 1 of 13: Neutral-Buoyancy Temperature of a Hot-Air Balloon

Question 1: Neutral-Buoyancy Temperature of a Hot-Air Balloon (5 marks)