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CLARKSON ME 437 - AEROSOL PARTICLE MOTION

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ME437/537 G. Ahmadi 1AEROSOL PARTICLE MOTION Equation of Motion Consider an aerosol particle in fluid flow as shown in Figure 1. The equation of motion of a spherical aerosol particle of mass m and diameter d is given as guuupfpm)(Cd3dtdmc+−πµ= (1) Here pu is the particle velocity, fu is the fluid velocity, g is the acceleration of gravity and the buoyancy effect in air is neglected. Here it is assume that the particle is away from walls and the Stokes drag is assumed. Figure 1. Schematics of an aerosol motion in a gas flow. Dividing Equation (1) by cCd3πµand rearranging, we find guuupfpτ+−=τ )(dtd (2) where the particle response (relaxation) time is defined as ν=µρ=πµ=τ18CSd18Cdd3mCc2cp2c, (3) where 6dmp3ρπ= , νis the kinematic viscosity of the fluid and fp/S ρρ= is the density ratio. In practice, for non-Brownian particles, 1≈cC and Drag GravityME437/537 G. Ahmadi 2 µρ≈τ18dp2 (4) Terminal Velocity For a particle starting from rest, the solution to (2) is given as )e1)((/tfp τ−−τ+= guu (5) where fu is assumed to be a constant vector. For 0=fu and large t, the terminal velocity of particle ut is given by µρ=τ=18gCdguc2pt (6) Table 7 – Relaxation time τ for a unit density particle in air (p = 1 atm, T = 20oC). Diameter, µm gutτ= τ sec Stop Distance uo= 1 m/s Stop Distance uo= 10 m/s 0.05 0.39 µm/s 4 × 108− 0.04 µm 4 × 104− mm 0.1 0.93 µm/s 9.15 × 108− 0.092 µm 9.15 × 104− mm 0.5 10.1 µm/s 1.03 × 106− 1.03 µm 0.0103 mm 1 35 µm/s 3.57 × 106− 3.6 µm 0.0357 mm 5 0.77 mm/s 7.86 × 105− 78.6 µm 0.786 mm 10 3.03 mm/s 3.09 × 104− 309 µm 3.09 mm 50 7.47 cm/s 7.62 × 103− 7.62 mm 76.2 mm Stopping Distance In the absence of gravity and fluid flow, for a particle with an initial velocity of upo, the solution to (2) is given by )e1(/t τ−−τ=popux (7) τ−=/teopuu (8) where px is the position of the particle. As t → ∞, up→ 0 and τ=popux (9)ME437/537 G. Ahmadi 3is known as the stopping distance of the particle. For an initial velocity of 1000 cm/s, the stop distance for various particles are listed in table 7. Particle Path For constant fluid velocity, integrating Equation (5), the position of the particle is given by )]e1(t)[()e1(/tf/t τ−τ−−τ−τ++−τ+= guuxxpopop (11) Here pox is the initial position of the particle. For a particle starting from rest, when the fluid velocity is in x-direction and gravity is in the negative y-direction, Equation (10) reduces to )]e1(/t[u/x/tfp τ−−−τ=τ (12) )]e1(/t[u/y/tfp τ−−−τα−=τ (13) where the ratio of the terminal velocity to the fluid velocity α is given by ττ=αfug (14) Figure 2 shows the variation of vertical position of the particle with time. -12 -10 -8 -6 -4 -2 0 y/utau0 1 2 3 4 5 6 t/tauα =0.1 α =1 α =2 Figure 2. Variations of the particle vertical position with time.ME437/537 G. Ahmadi 4 From Equations (12) and (13), it follows that ppxy α−= (15) That is the particle paths are straight lines. Figure 3 shows sample particle trajectories. Buoyancy Effects For small particles in liquids, the buoyancy effect must be included. Thus, Equation (1) is replaced by guuupfp)mm()(Cd3dtd)mm(fca−+−πµ=+ (16) where fmis the mass of the equivalent volume fluid given as 6dmf3fρπ= (17) and amis the apparent mass with ρf being the fluid density. For spherical particles, .m21mfa= -12 -10 -8 -6 -4 -2 0 y/utau0 1 2 3 4 5 6 x/utauα =0.1 α =1α =2Figure 3. Sample particle trajectories.ME437/537 G. Ahmadi 5 Keeping the same definition for particle relaxation time as given by (3), Equation (2) may be restated as )S11()(dtd)S211( −τ+−=τ+ guuupfp (18) The expression for the terminal velocity then becomes )1(18gCd)S11(gupfc2ptρρ−µρ=−τ= (18) Note that the Basset force and the memory effects are neglected in this

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