Question 5 of 13: Wind-Induced Ventilation of an Outhouse Pit
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-7 Mechanics of Fluids — National Examination, 2013-Dec. 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.
Air density is taken from the paper's own Constants table at the temperature each question states: 1.19 kg/m³ at 20°C (Q5's wind, Q7's inlet air).
Q1's touching-rod array is modelled as a repeating square unit cell of four mutually tangent rods (pitch = rod diameter, per the question's own "closely packed" wording), giving a curvilinear-square pore whose perimeter/area ratio drives the capillary rise.
Q8's Moody diagram and Q9's drag-coefficient diagram are supplied as attachments. Both are solved via the equations the charts themselves plot: the Colebrook–White equation for Q8 and the Morrison (2013) curve-fit for sphere drag vs. Reynolds number for Q9.
Question 5: Wind-Induced Ventilation of an Outhouse Pit (5 marks)
Given. Wind speed V = 25 km/hr blowing across the vent-pipe opening; air temperature 20°C, so ρair = 1.19 kg/m³ (Constants table); pit air essentially stagnant (V = 0).
Find. Differential pressure between the still pit air and the wind-swept vent-top opening that continuously draws air up and out.
Approach. Apply Bernoulli's equation along a streamline from the still, far-field atmosphere (pressure = local atmospheric, V=0) to the vent opening, where the wind blows across at speed V; the resulting local pressure drop at the vent top is the suction that keeps air moving up from the pit.
Bernoulli, same elevation, still air upstream vs. wind-swept vent top. $p_{atm}+0 = p_{top}+\tfrac12\rho_{air}V^2$, so the pressure reduction at the vent mouth is
$$\Delta p = p_{atm}-p_{top} = \tfrac12\rho_{air}V^2 = \tfrac12(1.19)(6.944)^2 = \boxed{28.7\ \text{Pa}}$$
Interpretation. Since pit air is otherwise close to atmospheric (connected to the outside air via slow percolation through the ground), this 28.7 Pa drop at the vent top is exactly the differential pressure available to draw pit air continuously up the vent pipe and out into the moving wind stream.
