(1.81?10?3)2dc?1.2243(2650?1.205)?1.205?5.73?10?5m?57.3?m

when Reynolds number ≥1000, the flow pattern follows Newton’s law and terminal velocity can be calculated by equation 3.2-19

ut?1.75gdp??p???? 1

critical Reynolds number is Ret??ut?dc??1000， 2

rearranging the equation 2 gives

ut?1000? 3 ??dccombination of equation 1 with equation 3

d?(???)g1000??1.74cS ??dc?(?S??)?solving for critical diameter

??32.33 dc

?2

(1.81?10?3)2??32.33dc(2650?1.205)?1.205?1.512?10?3m?1512um

3.3 It is desired to remove dust particles 50 microns in diameter from 226.5m3/min of air, using a settling chamber for the purpose. The temperature and pressure are 21oC and 1 atm. The particle density is 2403kg/m3. What minimum dimensions of the chamber are consistent with these conditions? (the maximum permissible velocity of the air is 3m/s) solution:

to calculate terminal velocity from the equation 3.2-16

ut?2??p???gdp18?

The density of 21℃ air ρ=1.205 kg/m3 and viscosity μ=1.81×10-5N·s/m2

ut?2??p???gdp18?(50?10?6)2(2403?1.205)?9.81＝?0.181m/s

18?1.81?10?5Q?BLut so BL?Q226.5??20.86m2 1 ut60?0.181from equation3.3-4

LH? uutthe maximum permissible velocity of the air is 3m/s

LH ?30.181L?16.58H 2

set B to be 3m, then from equation 1 L=7m And H=0.42m

3.4 A standard cyclone is to be used to separate the dust of density of 2300 kg/m3 from the gas. The flow rate of gas is 1000m3/h, the viscosity of the gas is 3.6?10-5N·s/ m2, and the density is 0.674 kg/ m3. If the diameter of cyclone is 400 mm, attempt to estimate the critical diameter. Solution: D=0.4m B=D/4=0.1m h=D/2=0.2m

ui?Q1000??13.9m/s hB3600?0.2?0.1

According to the equation3.3-12 for N=5：

9?Bdc???N(?p??)ui9(3.6?10?5)(0.1)?8?10?6m?8?m

5?(2300?0)?13.9

3.6 A filter press of 0.1m2 filtering area is used for filtering a sample of the slurry. The filtration is carried out at constant pressure with a vacuum 500mmHg.The volume of filtrate collected in the first 5min was one liter and, after a further 5min, an additional 0.6 liter was collected. How much filtrate will be obtained when the filtration has been carried out for 15min on assuming the cake to be incompressible?

Solution:

The equation for the constant-pressure filtration V?2VVm?KAt

5min 1l. 1?2Vm?K0.1?5 10min 1.6l. 1.6?2?1.6Vm?K?0.1?10 solving the equations above for Vm and K

222222Vm?0.7 and K=48

For t?15min V?2?0.7?V?48?0.1?15 Solving for V=2.073 l

3.7 The following data are obtained for a filter press of 0.0093 m2 filtering area in the test

Pressure difference/kgf/cm2 filtering time/s filtrate/ m3 1.05 50 2.27×10-3 660 9.10×10-3 3.50 17.1 2.27×10-3 233 9.10×10-3 Calculate:

(1) filtration constant K, Vm at the pressure difference of 1.05

(2) if the frame of the filter is filled with the cake at 660s, what is the final rate of filtration

22?dV??? ?dt?E(3) and what is the compressible constant of cake n? solution:

①from equation 3.4-19a

q2?2qqm?Kt For pressure differencep?1.05㎏/㎝2

?2.27?10?3?2.27?10?3??0.0093???2?0.0093qm?K?50 1 ???9.1?10?3?9.1?10?3??0.0093???2?0.0093qm?K?660 2 ??solving the equations 1 and 2 gives

22m3qm?0.03792mK?1.56?10?3m2/s

②

KA2?dV??63=7.14?10m/s ????dt?E2(V?Vm)

For pressure differencep?3.5㎏/㎝2

?2.27?10?3?2.27?10?3??K??1.71 3 ??0.0093???2?0.0093qm???9.1?10?3?9.1?10?3??K??233 4 ??0.0093???2?0.0093qm??solving the equations 3 and 4 gives

22m3??0.03092qmm????p?????????p??then

1?nK??4.37?10?3m2/s

4.37?10?3?3.5????1.56?10?3?1.05?1?n

1?n?ln2.8

ln3.33solving for n=0.142

3.8 A slurry if filtered by a filter press of 0.1m2 filtering area at constant pressure, the equation for a constant pressure filtration is as follows

(q?10)2?250(t?0.4)

where q=filtrate volume per unit filtering area,in l/m2, t= filtering time, in min calculate:

(1) how much filtrate will be gotten after249.6min?

(2) If the pressure difference is double and both the resistances of the filtration medium and cake are constant, how much filtrate will be obtained after249.6min? solution: (1)

(q?10)2?250(t?0.4)?250(249.6?0.4)?2502

q=240 l/m2 V=24 l