Hot-Air Balloon — Temperature for Neutral Buoyancy: 04-BS-7, December 2019, Q2

Question 2: Hot-Air Balloon — Temperature for Neutral Buoyancy (15 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
Sphere diameter	18 m
Cone half-angle (to vertical axis)	45°
Total balloon volume, V (given)	3147 m³
Structure + payload mass (100+60+110+50+160)	480 kg
Ambient pressure / temperature	100 kPa / 15°C (288.15 K)
Gas constant for air, R (Constants, p.12)	287 J/kg·K

Check: the exam states the total volume (3147 m³) directly, so it is used as given rather than re-derived; a sphere-plus-cone decomposition from the stated 18 m diameter and 45° half-angle is a useful sanity sketch of the shape but is not required to solve the buoyancy balance below, which needs only the total volume and total mass.

Sphere (D=18 m) over an inverted cone (45° half-angle); total volume 3147 m³ given.

Find. The hot-air temperature inside the envelope for neutral buoyancy.

Approach. At neutral buoyancy the weight of ambient air displaced by the whole balloon exactly balances the total weight (structure/payload plus the hot air filling the envelope). Writing both densities from the ideal gas law at the SAME pressure (the envelope is open at the throat, so internal and ambient pressure are essentially equal) isolates the hot-air density, and hence its temperature.

Ambient air density. $$\rho_{amb}=\frac{p_0}{RT_{amb}}=\frac{100{,}000}{(287)(288.15)}=\boxed{1.2092\ \text{kg/m}^3}$$
Buoyancy balance. Neutral buoyancy requires the displaced-air weight to equal the total weight (structure/payload $m_s$ plus the hot air of density $\rho_{hot}$ filling the same volume $V$): $$\rho_{amb}Vg=m_sg+\rho_{hot}Vg\ \Rightarrow\ \rho_{hot}=\rho_{amb}-\frac{m_s}{V}$$ $$\rho_{hot}=1.2092-\frac{480}{3147}=1.2092-0.1525=\boxed{1.0567\ \text{kg/m}^3}$$
Hot-air temperature (same pressure as ambient). $$T_{hot}=\frac{p_0}{R\,\rho_{hot}}=\frac{100{,}000}{(287)(1.0567)}=329.7\ \text{K}$$ $$\boxed{T_{hot}=56.6^\circ\text{C}}$$

Quantity	Value
Ambient air density	1.2092 kg/m³
Required hot-air density	1.0567 kg/m³
Hot-air temperature for neutral buoyancy	329.7 K (56.6°C)

Concept & Theory

Topic. Buoyancy of a lighter-than-air balloon; density control via temperature (ideal gas law).

Key relations. Archimedes' principle $F_B=\rho_{amb}Vg$; ideal gas law $\rho=p/RT$.

Why this works. A hot air balloon carries no separate lifting gas; it works purely by heating the SAME air (open at the throat, so at ambient pressure) to a lower density than the surrounding atmosphere, so the buoyant force on the whole envelope volume exceeds the combined weight of the warm air inside plus the structure — only about a 13% density deficit (1.057 vs 1.209 kg/m³) is needed here because the volume is so large relative to the payload mass.

Common pitfall. Forgetting to subtract the structure/payload mass's density-equivalent ($m_s/V$) from the ambient density — treating $\rho_{hot}=\rho_{amb}$ alone would (incorrectly) require no heating at all.

Source. White, Fluid Mechanics, Ch. 2 (Buoyancy).

Question 2 of 13: Hot-Air Balloon — Temperature for Neutral Buoyancy

Question 2: Hot-Air Balloon — Temperature for Neutral Buoyancy (15 marks)