Question 2 of 13: Hot Air Balloon — Neutral Buoyancy Temperature
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-May. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (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: Crowe, C.T., Elger, D.F. & Roberson, J.A., Engineering Fluid Mechanics (the exam's own Moody chart and drag-coefficient chart are reproduced from this text); Douglas, J.F., Gasiorek, J.M., Swaffield, J.A. & Jack, L.B., Fluid Mechanics; White, F.M., Fluid Mechanics.
Check — assumptions used across this paper:
Where a question does not restate an ambient temperature, air is taken at the ISA sea-level standard of 15°C, giving ρair = 1.21 kg/m³ (the Constants table's 15°C value) — used in Q5, Q6, Q7 and Q9.
Q8's Moody chart and Q9's sphere drag-coefficient chart are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 (the Moody chart is a graphical solution of Colebrook–White) and the Morrison (2013) curve-fit for sphere drag versus Reynolds number for Q9 (which reproduces the published "Sphere" curve to within a few percent over this Re range).
Q1, Q3 and Q11 depend on their figures; the relevant crop of the exam page is reproduced beside each, and every reading used is taken from it.
Question 2: Hot Air Balloon — Neutral Buoyancy Temperature (5 marks)
Given. Envelope volume $V=3147\ \text{m}^3$; ambient $P_0=100\ \text{kPa}$, $T_0=15^\circ\text{C}$ ($\rho_{amb}=1.21\ \text{kg/m}^3$ per the Constants table); hardware mass $=100+60+110+50+160=480\ \text{kg}$; gas constant for air $R=287\ \text{J/kg K}$.
Sphere-and-cone envelope; total volume and hardware mass are given directly.
Find. Temperature of the hot air inside the envelope for neutral buoyancy.
Approach. Neutral buoyancy requires the weight of ambient air displaced to equal the total weight (hardware + hot air inside); solve for the hot-air density, then back out its temperature from the ideal gas law at ambient pressure.
Buoyant force balance. $\rho_{amb} V g = (m_{hw} + \rho_{hot} V) g$, so $\rho_{hot} = \rho_{amb} - m_{hw}/V = 1.21 - 480/3147 = 1.21 - 0.1526 = 1.0575\ \text{kg/m}^3$.
Hot-air mass check. $m_{hot} = \rho_{hot}V = 1.0575\times3147 = 3327.9\ \text{kg}$, i.e. the envelope must hold 3327.9 kg of hot air to lift 480 kg of hardware plus itself.
Ideal gas law for the envelope temperature. The envelope is open at the base, so its internal pressure is essentially ambient, $P_0=100\ \text{kPa}$:
$$T_{hot} = \frac{P_0}{R\,\rho_{hot}} = \frac{100{,}000}{287\times1.0575} = 329.5\ \text{K}$$
$$\boxed{T_{hot} = 329.5\ \text{K} = 56.3^\circ\text{C}}$$
