Question 4 of 6: Step (Rayleigh) Bearing — Lubrication Theory
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams – May 2017, 16-Mec-A6 Advanced Fluid Mechanics. Open book, 3 hours. Six questions of equal value (20 marks); any five constitute a complete paper. All six are solved here as a study resource.
Reference texts: Fox, Pritchard & McDonald, Introduction to Fluid Mechanics, 10th ed.; F. M. White, Fluid Mechanics, 8th ed.; J. D. Anderson, Modern Compressible Flow, 3rd ed. (Q1); I. G. Currie, Fundamental Mechanics of Fluids (Q2); Munson, Young & Okiishi, Fundamentals of Fluid Mechanics (Q5).
Note on the subject. The 16-Mec-A6 sittings (2017–2019) are the Advanced Fluid Mechanics syllabus — compressible flow, potential flow, integral momentum, lubrication theory, dimensional analysis and boundary layers — a different exam from the 07-Mec-A6 pump/turbine papers.
Question 4: Step (Rayleigh) Bearing — Lubrication Theory (20 marks)
Given. Steady, incompressible, isothermal creeping flow of a Newtonian oil in a thin two-step gap; inertia negligible; pressure rises linearly from $0$ to $P_s$ over $L_1$ then falls linearly to $0$ over $L_2$.
Given data (part d)
Quantity
Value
Slider speed $U$
0.5 m/s
Oil viscosity $\mu$
3.85 N·s/m²
Section 1: $h_1,\ L_1$
1 mm, 30 mm
Section 2: $h_2,\ L_2$
2 mm, 100 mm
Find. (a) the reduced momentum equations; (b) $q$ and $\Delta p$ in a section; (c) $p(x)$; (d) the load per unit width $W$.
Figure 4: Step bearing (slider over stepped bearing) and the resulting triangular pressure distribution, peaking at $P_s$ over the step.
Approach. Apply the lubrication (thin-film) scaling to Navier–Stokes, integrate the velocity profile with the stated no-slip conditions, impose continuity of the flow rate across the step to fix $P_s$, then integrate the pressure to obtain the load.
(a) Reduced equations. With $h\ll L$ and $Re\to0$, inertia and the streamwise diffusion term drop out. The $x$- and $y$-momentum equations reduce to
$$\frac{dp}{dx}=\mu\frac{\partial^{2}u}{\partial y^{2}},\qquad \frac{\partial p}{\partial y}=0,$$
so pressure is a function of $x$ only and the velocity profile is locally parabolic.
(b) Velocity profile and flow rate. Integrating twice with $u(0)=0$, $u(h)=-U$,
$$u(y)=\frac{1}{2\mu}\frac{dp}{dx}\big(y^{2}-hy\big)-\frac{U y}{h}.$$
The flow rate per unit width is
$$q=\int_0^{h}u\,dy=-\frac{h^{3}}{12\mu}\frac{dp}{dx}-\frac{U h}{2}.$$
Since $q$ and $h$ are constant within a section, $dp/dx$ is constant, so the pressure change over a length $\ell$ is $\Delta p=\dfrac{dp}{dx}\,\ell=-\left(\dfrac{12\mu q}{h^{3}}+\dfrac{6\mu U}{h^{2}}\right)\ell$ — linear in $x$.
(c) Match the flow rate across the step. With $dp/dx=+P_s/L_1$ in section 1 and $-P_s/L_2$ in section 2, equal flow rate $q_1=q_2$ gives
$$-\frac{h_1^{3}}{12\mu}\frac{P_s}{L_1}-\frac{U h_1}{2}=+\frac{h_2^{3}}{12\mu}\frac{P_s}{L_2}-\frac{U h_2}{2},$$
$$\boxed{P_s=\frac{6\mu U\,(h_2-h_1)}{\dfrac{h_1^{3}}{L_1}+\dfrac{h_2^{3}}{L_2}}.}$$
The pressure field is then $p=P_s\,x_1/L_1$ in section 1 and $p=P_s(1-x_2/L_2)$ in section 2 — the triangular profile of Figure 4.
(d) Peak pressure and load. Substituting the data ($h_1^3/L_1=3.33\times10^{-8}$, $h_2^3/L_2=8.0\times10^{-8}$ m²),
$$P_s=\frac{6(3.85)(0.5)(0.001)}{1.133\times10^{-7}}=1.019\times10^{5}\ \text{Pa}=102\ \text{kPa}.$$
The load per unit width is the area under the triangular pressure profile,
$$W=\tfrac12 P_s\,(L_1+L_2)=\tfrac12(1.019\times10^{5})(0.130),$$
$$\boxed{W=6.62\times10^{3}\ \text{N/m}\ (\approx6.62\ \text{kN per metre of width}).}$$
Check: the oil density is not needed — creeping flow is inertia-free, so $\rho$ only enters through the (negligible) Reynolds number. The sign of $P_s$ is positive because, with the stated $u(h)=-U$ convention, the entraining surface drags oil from the thick film ($h_2$) toward the thin film ($h_1$), a converging wedge that generates positive load.
