Question 2 of 13: Viscous Heating in the Crankshaft Main Bearings
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 2: Viscous Heating in the Crankshaft Main Bearings (5 marks)
Find. Rate of viscous heat generation (power dissipated), in J/s, per bearing and for all 5 main bearings.
Approach. Treat the thin annular oil film as a Couette (linear) shear flow; Newton's law of viscosity gives the shear stress on the shaft surface, and shear force × surface speed gives the viscous power, scaled to all five identical bearings.
Radial clearance and surface speed. $c=(60.100-60.000)/2=0.050\ \text{mm}=5.0\times10^{-5}\ \text{m}$; shaft surface speed $V=\pi D N = \pi(0.060)(2800/60)=\boxed{8.796\ \text{m/s}}$.
Shear stress. Assuming a linear velocity profile across the thin film, $\tau=\mu V/c = 0.1\times8.796/(5.0\times10^{-5})=17{,}593\ \text{Pa}$.
Shear force and power, one bearing. Wetted area $A=\pi D L=\pi(0.060)(0.0225)=4.241\times10^{-3}\ \text{m}^2$, so $F=\tau A=17{,}593\times4.241\times10^{-3}=74.6\ \text{N}$, and the viscous power dissipated is
$$\dot{P}_{\text{one}} = F V = 74.6\times8.796 = \boxed{656.3\ \text{J/s}}$$
All five main bearings. With identical clearance and speed at each of the five bearings,
$$\dot{P}_{\text{total}} = 5\times656.3 = \boxed{3282\ \text{J/s} \approx 3.28\ \text{kW}}$$
