Question 4 of 13: Head Loss in a Long Water Supply Pipeline
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 4: Head Loss in a Long Water Supply Pipeline (5 marks)
Find. Head loss hL between the two measurement points.
Approach. Apply the general energy equation between the two points; since the pipe diameter (hence velocity) is the same at both ends, the velocity-head terms cancel and hL follows directly from the pressure and elevation data (the stated flow rate confirms the same-diameter assumption but otherwise drops out).
Energy equation, same-diameter pipe. $\dfrac{p_1}{\rho g}+z_1+\dfrac{V_1^2}{2g} = \dfrac{p_2}{\rho g}+z_2+\dfrac{V_2^2}{2g}+h_L$, and $V_1=V_2$ (constant D, same Q), so the velocity heads cancel:
$$h_L = \frac{p_1-p_2}{\rho g} + (z_1-z_2)$$
