A gas stream is passed through a cyclone separator which has 35 weight percent p

Question

A gas stream is passed through a cyclone separator which has 35 weight percent particles with diameter 1 m, 30 weight percent particles with diameter 5 m and 35 weight percent particles with diameter 10 um. The cyclone has an inlet width, Wi -0.175 m with = 18 m/s, N = 5, Air- 1.8×10-5 kg/(m-s) and particle -2000 kg/m. Determine the cut diameter and the overall collection efficiency of this cyclone. QS (a) |-ou,part,part 9W, Hair Given: =1-exp (8 marks) The sulphur dioxide concentrations at the downwind distance of 1.2 km, and at a point 100 m to the side of and 25 m below the centerline of the stack plume is 600 g/m3 The stack has a diameter of 0.75 m, an exit velocity of 15 m/s and an exit temperature of 450 K. The top is 3 m/s. The ambient air temperature class is assumed to be B. Determine: (b) stack height is 50 m. The wind speed at the stack is 300 K. The Pasquill-Gifford stability (4 marks) (i) the effective stack height; the concentration rate of sulphur dioxide (e/s) emitted at the stack exit. (8 marks) (i) Given: For unstable-neutral condition: F= (gvd7)(47,), m'/s' H-21 .425F3/4/Mh, tn, if F is less than 55 mts H-38.71 F3/5/m, m, if F is equal to or greater than 55 m%" AH-3dvhh, m Buoyant flux, Buoyant rise, Momentum rise, Stability category for "T” T = 24.167-2.5334 ln (x) T 18.333-1.8096 In (x) T 12.5-1.0857 In (x) T = 8.3333-0.72382 ln (x) Table Q5(b).1 Pasquill-Gifford's horizontal parameter,o (1000x tan Ty2.15, m.

Explanation / Answer

a) Cut diameter, Dpart = 35/100*1+30/100*5+35/100*10=5.35 um

efficiency = 1-exp(-3.14*5*18*(5.35*10^-6)^2*2000/(9*.175*1.8*10^-5))=43%

b) f=10*15*.75^2*(450-300)/4/300=10.54

Buoyant rise = 21.42*10.54^.75/3=41.76 m

momentum rise = 3*.75*15/3=11.25 m

=> Effective stack height = 50+41.76+11.25=103 m

b) x=1.2 km

for b class, T=18.3-1.8*log(1.2)=18.15

y=(1000*1.2*tan(18.15))/2.15=300 m

z=109*1.2^1.09=132 m

z=H-25 = 103-25 = 78 m

H = 103 m

C(x,y,z,H) = C(1.2,100,78,103)=600=E/2/3.14/15/300/132*exp(-0.5*(100/300)^2)*(exp(-.5*((78-103)/132)^2+exp(-.5*((78+103)/132)^2)

=>E=1629 g/s

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