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04-BS-7 · December 2015

Question 8 of 13: Rate of Rise of a Helium Meteorological Balloon

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — December 2015 (National Examinations, three hours, closed book). Section A (Calculative) offers 9 questions and instructs "do seven"; Section B (Analytical) offers 4 questions and instructs "do three." Every question is answered below (13 of 13), so students can use the full paper as a study resource. Constants used throughout (from the paper's own Constants page): g = 9.81 m/s², ρwater = 1000 kg/m³, ρair = 1.19 kg/m³ (20°C) / 1.21 kg/m³ (15°C), μair = 1.8×10⁻⁵ N·s/m², Rair = 287 J/kg·K, Rhelium = 2077 J/kg·K, patm = 100 kPa.

Reference texts: F. M. White, Fluid Mechanics, 8th ed. (McGraw-Hill) — fluid statics and manometry (Ch. 2), control-volume momentum/energy and propulsion (Ch. 3), potential/inviscid flow around cylinders (Ch. 8), viscosity and Newtonian shear (Ch. 1), pipe friction and the Moody chart (Ch. 6), and drag on immersed bodies (Ch. 7).

Question 8: Rate of Rise of a Helium Meteorological 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.

QuantityValue
Balloon diameter600 mm
Balloon (envelope) mass80 g
Ambient temperature15°C
ρair (15°C)1.21 kg/m³
μair1.8×10⁻⁵ N·s/m²
Rhelium2077 J/kg·K

Find. Terminal rate of rise V.

Buoyancy 1.342 NW_He + W_material= 0.970 NRise V ≈ 2.2 m/sDrag balancesnet buoyancy at V_terminal
Fig. Q8 — free body of the rising balloon: buoyancy up, weight of helium + envelope material down; at terminal (constant) velocity the aerodynamic drag exactly balances the net upward force.

Approach. Compute the constant net upward force (buoyancy minus helium weight minus envelope weight); at terminal velocity this equals drag, CD·½ρV²A, but CD itself depends on V through the Reynolds number — so guess trial velocities, compute the required CD and the resulting Re for each, and find where those points meet the sphere drag curve on the attached chart.

  1. Net upward (excess buoyancy) force, independent of velocity. Helium density at ambient conditions: $$\rho_{He} = \frac{p_0}{R_{He}T} = \frac{100\,000}{2077(288.15)} = 0.167\ \text{kg/m}^3$$ Volume $V_{vol}=\frac{\pi}{6}D^3=\frac{\pi}{6}(0.6)^3=0.1131\ \text{m}^3$. Buoyancy, helium weight, and envelope weight: $$F_{buoy}=\rho_{air}gV_{vol}=1.342\ \text{N},\quad W_{He}=\rho_{He}gV_{vol}=0.185\ \text{N},\quad W_{mat}=mg=0.0800(9.81)=0.785\ \text{N}$$ $$F_{net} = 1.342 - 0.185 - 0.785 = \boxed{0.372\ \text{N}}$$
  2. Trial velocities: compute the required CD and the resulting Re for each (frontal area $A=\frac{\pi}{4}D^2=0.2827\ \text{m}^2$): $$C_{D,req}(V) = \frac{2F_{net}}{\rho_{air}AV^2}, \qquad Re(V) = \frac{\rho_{air}VD}{\mu_{air}}$$ For V = 2.0, 2.2, 2.4 m/s: CD,req = 0.55, 0.45, 0.38 and Re = 8.1×10⁴, 8.9×10⁴, 9.7×10⁴.
  3. Locate the intersection with the sphere drag curve. Plotting these (Re, CD,req) points against the attached sphere-drag curve (which sits at CD≈0.4–0.5 through this Reynolds range, well below the drag-crisis dip near Re≈3×10⁵), the curves cross close to V ≈ 2.2 m/s, CD≈0.45, Re≈8.9×10⁴.
  4. Report the terminal rate of rise. $$V_{terminal} = \sqrt{\frac{2F_{net}}{\rho_{air}AC_D}} = \sqrt{\frac{2(0.372)}{1.21(0.2827)(0.45)}} = \boxed{2.20\ \text{m/s}}$$
QuantityResult
Net upward force0.372 N
Reynolds number at terminal velocity≈ 8.9×10⁴
Drag coefficient (chart intersection)≈ 0.45
Rate of rise≈ 2.2 m/s