Lecture 7

Question 1

What is the physical setup of the journal bearing problem in lubrication theory?

Answer 1

A cylinder of radius 𝑎 rotates inside a housing of radius 𝑎(1+𝜀). The eccentricity 𝜆𝑎 measures how off-center the cylinder is, where 𝜆=0 is concentric and 𝜆=1 is contact.

Question 2

What assumption makes this a lubrication problem?

Answer 2

ε≪1, so there’s a clear separation between the radial gap and axial length scale. This lets us treat the flow as nearly unidirectional with dominant viscous forces.

Question 3

What does the gap height ℎ(𝜃) look like for a nearly concentric journal bearing?

Answer 3

h(θ)=aεH(θ)=aε(1−λcosθ+O(ε)). It varies with angular position 𝜃θ.

Question 4

Why is the velocity profile 𝑈 a combination of two terms in journal bearing flow?

Answer 4

One term comes from pressure-driven (Poiseuille-like) flow and the other from shear-driven (Couette-like) motion. Both effects coexist in narrow gap geometries.

Question 5

How is the load supported by the bearing computed?

Answer 5

Using lubrication pressure 𝑝(𝜃), integrate the vertical component of force around the bearing:
𝐹=12𝜋𝜇Ω𝑎^2/𝜀^2 ⋅ 𝜆/((2+𝜆^2)sqrt(1−𝜆^2))

Question 6

What is the physical situation of squeeze flow?

Answer 6

A sphere of radius 𝑎 moves normally toward a wall, squeezing fluid out of a narrowing gap. It is axisymmetric in cylindrical coordinates (𝑟,𝑧)

Question 7

What determines the characteristic length scale in squeeze flow?

Answer 7

The radial extent where the gap changes significantly is ℓ∼sqrt(𝑎ℎ_0(𝑡), where ℎ0(𝑡) is the instantaneous minimum gap.

Question 8

What is the leading-order expression for gap height ℎ(𝑟,𝑡)?

Answer 8

h(r,t)≈h_0(t)(1+(r^2/2ah_0(t))
This comes from geometry: the shape of the sphere near the wall.

Question 9

How does pressure behave in squeeze flow?

Answer 9

p(r,t)=const.− 3μVa/h_0^3 (1+r^2/2ah_0)^−2 It peaks at 𝑟=0 and decays with 𝑟, showing flow confinement.

Question 10

What is the expression for the force on the sphere due to lubrication pressure?

Answer 10

F=−6πμV a^2/h_0(t)
This shows the force grows rapidly as the gap ℎ0h0 becomes small.

Question 11

How does the minimum gap ℎ0(𝑡) evolve for a settling sphere with negligible inertia?

Answer 11

h_0(t)=h_0(0)exp(−2gaΔρt/9μ)
The gap decreases exponentially—settling is very slow due to lubrication resistance.