Fundarmentalof two-phase flow 1. 1 Free sedimentation terminal velocit When spherical particle sediments in resting liquid, the forces act to It are gravity: w y 6 buoyancy: Wh=5r resistance: R=ypd=u nen the kinetic equation of grain w-w-p du substituting into above three force formula then 2(P,-0)-Wmdn2=a dt du p-p yup S (10-7) d or =8 (10-7a)21
21 1.1 Free sedimentation terminal velocity When spherical particle sediments in resting liquid, the forces act to it are: 3 3 2 2 3 3 2 2 gravity : 6 buoyancy : 6 resistance: then the kinetic equation of grain , substituting into above three force formula then ( ) 6 6 s s s b s b s s s s s d W d W R d u du W W R m dt d d d u g g r r r r r 2 0 , 6 (10-7) r (10-7a) s s s s R du dt du u g dt d du o g a dt r r r r r
祖流动 1、自由沉降末速 球形颗粒在静止流体中沉降时,所受到的作用力有: 重力W y 浮力Wb=6 6 y 阻力R=vod2ln2 则颗粒的运动方程为W-W-R=mm 代入三个力的表达式 得x(p-p)-m2=m du 6°dlt 则如=2.-g-5mp (10-7) 或简写为dt 80-ak (10-7a)22 t
22 1、自由沉降末速 球形颗粒在静止流体中沉降时，所受到的作用力有： (10 - 7a) (10 - 7) 6 , 6 ( ) 6 , 6 6 0 2 3 2 2 3 2 2 3 3 R s s s s s s s s s b s s b s s g a dt du d u g dt du dt d du d u d dt du W W R m R d u d W d W 或简写为 则 得 则颗粒的运动方程为 代入三个力的表达式 阻力 浮力 重力 r r r r r r r r r r g g
Fundaientalof two-phase low where: gravity acceleration only connect with density ar-resistance acceleration, connect with density and dimension, direct ratio with rate square When particle starts to sediment, velocity and resistance are zero, acceleration Is max as sedimentation velocity augmenting the resistance acceleration increasing after some time. the outside forces acted to particle attain balance, particle sediment velocity invariable this velocity is called sediment terminal velocity express with i Using Eq. (10-7) we have u, (u=o) Td(p-p 1 6yp g (10-8) For spherical particle, resistance coefficient changing along with Re, due to ( 10-3),(10-5),(10-6),obtain y substituting into (10-8),then
23 w 0 here gravity acceleration , only connect with density; resistance acceleration ,connect with density and dimension, direct ratio with rate square R g a ： — — When particle starts to sediment, velocity and resistance are zero, acceleration is max; as sedimentation velocity augmenting, the resistance acceleration increasing,after some time,the outside forces acted to particle attain balance, particle sediment velocity invariable, this velocity is called sediment terminal velocity, express with ut Using Eq.（10－7）we have ut ( 0) dt du (10 - 8) 6 ( ) g d u s s t r r r For spherical particle , resistance coefficient changing along with Re ,due to（10－3）,（10－5）,（10－6）,obtain , substituting into （10－8）, then
祖流塑基 式中:80颗粒的重力加速度,只与密度有关; aa一阻力加速度,与密度和尺寸均有关系, 且与速度平方成正比 颗粒开始下降时,速度为零,阻力亦为零,加速度为最 大值,随着下降速度增大,阻力加速度增大,经一定时间, 作用在颗粒上的外力达到平衡,颗粒等速下降,这个速度称 之为沉降末速,以4表示。 由式(10-7)可解得a1(a=0) Id(e-p u 68 (10-8) 对于球形颗粒,阻力系数随Re变化,由(10-3),(10-5) (10-6)三式中,解出v值,代入(10-8)式中,则分别得 24
24 且与速度平方成正比。 —阻力加速度，与密度和尺寸均有关系， 式中： 0 — 颗粒的重力加速度，只与密度有关； aR g 颗粒开始下降时，速度为零，阻力亦为零，加速度为最 大值，随着下降速度增大，阻力加速度增大，经一定时间， 作用在颗粒上的外力达到平衡，颗粒等速下降，这个速度称 之为沉降末速，以 ut 表示。 由式（10－7）可解得 ut ( 0) dt du (10 - 8) 6 ( ) g d u s s t r r r 对于球形颗粒，阻力系数随 Re 变化，由（10－3），（10－5）， （10－6）三式中，解出 值，代入（10－8）式中，则分别得：
Fundarmentalof two-phase flow For re <1, stokes sediment terminal velocity formula u g(p、-P)d 1811 (10-9) For re=25 500, A-Lian sediment terminal velocity ormula n2=d,(2g2 (10-10) For re=500-104, Newton- Reutlingen sediment terminal velocity formula 8 (10-11) 5
25 For Re <1 , Stokes sediment terminal velocity formula (10 -9) 18 ( ) 2 r s r s t g d u For Re=25~500 , A-Lian sediment terminal velocity formula g ) (10 -10) 15 2 3 ( 3 s 2 r r r r t s u d For Re=500~10 4 ,Newton- Reutlingen sediment terminal velocity formula (10 -11) ( ) 3 8 u d g s t s r r r