Rate of Rise of a Hydrogen Balloon (Drag Chart): 04-BS-7, May 2013, Q9

Question 9: Rate of Rise of a Hydrogen Balloon (Drag Chart) (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. Sphere $D=600\ \text{mm}=0.6\ \text{m}$; negligible structural mass; internal pressure $=$ ambient $=100\ \text{kPa}$ at $15^\circ\text{C}$ ($T=288.15\ \text{K}$); $R_{H_2}=4120\ \text{J/kg K}$ (Constants table); $\rho_{air}=1.21\ \text{kg/m}^3$; air viscosity $\mu_{air}=1.8\times10^{-5}\ \text{Ns/m}^2$.

At terminal (constant) rise velocity, net buoyant force equals drag.

Find. Terminal (constant) rate of rise through the atmosphere.

Approach. Compute the net buoyant force (air displaced minus hydrogen contained, structure mass negligible), then solve iteratively for the velocity at which sphere drag (from the standard sphere drag-coefficient curve versus Reynolds number) balances that force.

Check: Cd(Re) is obtained from the Morrison (2013) curve-fit, which reproduces the published sphere-drag curve (the same one reprinted in the attachment) to within a few percent over the relevant Reynolds-number range.

Balloon volume and area. $V=\pi D^3/6=\pi(0.6)^3/6=0.1131\ \text{m}^3$; $A=\pi D^2/4=\pi(0.6)^2/4=0.2827\ \text{m}^2$.
Hydrogen density (ideal gas). $\rho_{H_2}=\dfrac{P}{R_{H_2}T}=\dfrac{100{,}000}{4120\times288.15}=0.0842\ \text{kg/m}^3$.
Net buoyant force. Structure mass negligible, so $$F_{net} = (\rho_{air}-\rho_{H_2})Vg = (1.21-0.0842)\times0.1131\times9.81 = 1.249\ \text{N}$$
Terminal velocity from drag balance. Solving $F_{net}=C_D(Re)\,\tfrac12\rho_{air}V_t^2A$ iteratively (Cd from the sphere drag curve) converges at $$Re = \frac{V_tD}{\nu_{air}} \approx 1.65\times10^5,\qquad C_D\approx0.434$$ $$\boxed{V_t \approx 4.10\ \text{m/s}}$$

Quantity	Value
Hydrogen density	0.0842 kg/m³
Net buoyant force	1.249 N
Terminal Reynolds number	1.65×10⁶
Drag coefficient	0.434
Rate of rise	4.10 m/s

Topic: Terminal velocity of a buoyant sphere (drag/buoyancy balance).

Key relations: buoyancy $F_B=(\rho_{air}-\rho_{gas})Vg$; drag $F_D=C_D\tfrac12\rho V^2A$; sphere drag curve $C_D(Re)$.

Why this works: a rising balloon accelerates only until drag grows to match the constant buoyant excess force, after which it rises at constant (terminal) velocity; because $C_D$ itself depends on the unknown velocity through Re, the balance must be solved iteratively (or read off the chart at successive trial velocities) rather than algebraically in one step.

Common pitfall: forgetting that the balloon's own gas mass (however light) still displaces buoyancy-reducing weight — only the balloon's structure is stated as negligible, not the hydrogen inside it.

Source: Crowe, Elger & Roberson, Engineering Fluid Mechanics, Ch. 10 (forces on immersed bodies, drag).

Question 9 of 13: Rate of Rise of a Hydrogen Balloon (Drag Chart)

Question 9: Rate of Rise of a Hydrogen Balloon (Drag Chart) (5 marks)