Question 3 of 13: Helium Balloon — Mass and Altitude Volume Change
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2017-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.
Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and capillarity (Ch. 2), integral analysis and buoyancy (Ch. 3), viscous flow in ducts, pipe friction and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), open-channel flow and the hydraulic jump (Ch. 10).
Check — assumptions used across this paper:
Q2's figure shows the vertical leaf of the gate rising above the free surface, so water acts over the full depth x = 2.0 m; both "0.3 m" dimensions locate the centre of gravity (0.3 m right of the vertical leaf, 0.3 m above the arm). Reservoir water also fills the space beneath the 1.2 m arm (the vertical lines below O are dimension extension lines, not a wall), so the arm carries uplift at pressure ρgx, and the arm tip bears up against a lip on the spillway crest — the gate can only open by rotating clockwise, vertical leaf toward the spillway.
Q4's flow coefficient K ≈ 0.69 is read from the attached VDI chart on the Do/D1 = 0.50 curve (third from the top, the one that levels off at 0.62) at the approach Reynolds number R ≈ 2550. The printed curve gives 0.695 at R = 2000 and 0.684 at R = 3000; a reading of 0.68–0.70 moves the answer by only ±0.3 kPa.
Q7 and Q8 friction factors are obtained from the Colebrook–White/Haaland equation the Moody chart itself plots (smooth-wall case, as both problems specify or imply smooth surfaces).
Q9's sphere drag coefficient vs. Reynolds number is obtained from the Schiller–Naumann correlation $C_D=\tfrac{24}{Re}\left(1+0.15\,Re^{0.687}\right)$, which reproduces the published sphere-drag curve (the same curve reprinted in the attachment) to within a few percent for $Re<1000$ — used here because the resulting Reynolds number (≈19–20) is well above the range where the simple Stokes'-law formula ($C_D=24/Re$) alone is valid.
Question 3: Helium Balloon — Mass and Altitude Volume Change (5 marks)
Given. Ground level: $V_1=9350\ \text{m}^3$, $p_1=100\ \text{kPa}$, $T_1=15^\circ\text{C}=288.15\ \text{K}$. At 10,000 m: $p_2=24\ \text{kPa}$, $T_2=-54^\circ\text{C}=219.15\ \text{K}$. Gas constant for helium $R_{He}=2077\ \text{J/kg K}$ (Constants table). Sun-heated case: $T_3=T_2+15=234.15\ \text{K}$, at the same ambient $p_2$.
Find. (a) Mass of helium; volume $V_2$ at 10,000 m. (b) Percentage increase in volume when heated to $T_3$.
Approach. Use the ideal gas law to find the (constant) helium mass at ground conditions, then the combined gas law ($pV/T=\text{const}$ for that fixed mass) to track volume with altitude and, separately, with solar heating at constant altitude pressure.
Mass of helium at ground level.
$$m=\frac{p_1V_1}{R_{He}T_1}=\frac{100{,}000\times9350}{2077\times288.15}=\boxed{1562.3\ \text{kg}}$$
Volume at 10,000 m (mass fixed). Combined gas law $p_1V_1/T_1=p_2V_2/T_2$:
$$V_2=V_1\times\frac{p_1}{p_2}\times\frac{T_2}{T_1}=9350\times\frac{100}{24}\times\frac{219.15}{288.15}=\boxed{29{,}629\ \text{m}^3}$$
Volume after solar heating (constant $p_2$, mass fixed). $V_3=V_2\,(T_3/T_2)$:
$$V_3=29{,}629\times\frac{234.15}{219.15}=31{,}657\ \text{m}^3$$
Percentage increase. At constant pressure, $V_3/V_2=T_3/T_2$, so the percentage increase is exactly $(T_3-T_2)/T_2$:
$$\frac{V_3-V_2}{V_2}\times100=\frac{15}{219.15}\times100=\boxed{6.84\%}$$
