ME 433 Professor John M. Cimbala Lecture 27 Today, we will: • Continue to discuss inertial separation (particles in curved flows) • Discuss the analogy between gravimetric settling and inertial separation • Compare laminar vs. well-mixed inertial separation in curved ducts

Consider a particle in a round duct with air moving in circular motion in the duct:

Assumptions, Approximations, and Analysis: 1. We ignore gravity (assume that the inertial separation forces are much greater than

gravitational forces). 2. The air pollution particle moves tangentially at same tangential speed as the air,

Uθ = U, i.e., vθ = Uθ = U, as sketched. 3. The air pollution particle veers towards the outer wall at radial speed vr due to inertial

separation, as sketched. 4. The centripetal acceleration is aair = Uθ 2/rair on an imaginary air particle at the same

location as the actual particle, where rair is the radius of curvature of the air streamline, as sketched in blue.

5. The centripetal force on this imaginary air particle is thus Fair = mair Uθ 2/rair.

6. Now replace the imaginary air particle with the actual solid (or liquid) air pollution particle. The actual particle experiences the same centripetal force, Fparticle = mparticle Uθ 2/rparticle = Fair = mair Uθ 2/rair, where rparticle is the radius of curvature of the particle path, as sketched in red.

7. Solving this equation for the ratio of the two radii of curvature, particle particle particle

air air air

pr mr m

ρ ρρ ρ

= = = . Since the actual particle is heavier than the air particle,

the radius of curvature of the actual particle is greater than that of the air particle, as sketched (the particle veers outward).

8. After some further analysis (not shown), the radially outward “centrifugal force” felt by the actual air pollution particle is

( ) ( )32 2

centrifugal air 6p

p p

DU UF m mr rθ θπ ρ ρ= − = −

.

9. Meanwhile, as it moves outward, the aerodynamic drag force on the actual particle is 2 2 2

drag1 18 8

D Dp r r p r

C CF D v v D vC C

ρ π ρ π= − = −

. This force is radially inward since the

particle is moving outward relative to the air. (This is the same aerodynamic drag force that we used previously.)

10. Finally, when these two forces are equal (and opposite), the air pollution particle moves radially outward at some constant speed vr.

Analogy between inertial separation and gravimetric settling: Consider the simplest case in which vr is constant, and the two above forces must balance: The radial “settling” speed in a curved duct due to inertial separation is thus

243

pr p

D

U Cv Dr Cθ

ρ ρρ−

=

Compare this to the gravimetric settling speed in quiescent air (from a previous lecture):

43

pt p

D

CV gDC

ρ ρρ−

=

Analogy between inertial separation and gravimetric settling:

Example: Comparison of Centrifugal and Gravitational Settling Given: Dusty air enters a curved duct at

average speed U. Aerosol particles of a certain diameter Dp have a terminal settling speed of Vt = 0.00025 m/s in quiescent air. At the instant of time shown, a particle of diameter Dp is at radius r = 0.32 m.

To do: Assume laminar inertial settling (no turbulent mixing). Calculate the air speed U such that the radial velocity vr of the particle is the same as its terminal settling speed. Give your answer in m/s to three significant digits.

Solution:

Laminar vs. well-mixed settling: We discussed this twice previously:

1. Gravimetric settling in a room or container 2. Gravimetric settling in a horizontal duct

Now we apply the same principles to 3. Inertial separation in a curved duct.

1. Gravimetric settling in a room or container:

Initial state (t = 0)

H y(Dp)

cj(Dp) = cj(Dp)0

cj(Dp) = 0

cj(Dp) < cj(Dp)0 cj(Dp) = cj(Dp)0

Laminar settling (t = t1) Well-mixed settling (t = t1)

2. Gravimetric settling in a horizontal duct:

H

x

z

U

cj(Dp)in

cj(Dp) = cj(Dp)in

Laminar settling

cj(Dp) = 0

H

x

z

U

cj(Dp)in cj(Dp) < cj(Dp)in Well-mixed settling

3. Inertial separation in a curved duct:

Example: Removal efficiency of a curved duct as a function of particle diameter Given: Monodisperse aerosol particles of unit density (ρp = 1000 kg/m3) and Dp = 10.0 microns enter a 180o curved duct at average speed U = 10.3 m/s. The air is at SATP (ρ = 1.184 kg/m3, µ = 1.849 × 10-5 kg/(m s)). The middle duct radius is rm = 0.500 m and the duct width is W = 22.5 mm (0.0225 m). We assume laminar settling, and we assume that the particles stick to the outer wall when they hit, and are therefore removed from the flow.

To do: Calculate your best estimate of the predicted removal efficiency η of these particles (% to three significant digits).

Solution: The Cunningham correction factor is C = 1.0168, and Stokes flow is appropriate (CD = 24/Re).

Equations: 24

3p

r pD

U Cv Dr Cθ

ρ ρρ−

= , Re r pv Dρµ

= , 24ReDC = ,

Laminar settling model: 1(in)

j

j c

c xc L

η = − = , where cr

UL Wvθ= , mx r θ= .

U

cj (in)

U

cj (out)

rm

W