Hw06A

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Homework 6
6.1.- Assume generation rate of 4 lb/p d  Total generation = 400,000 lb/d
Assuming sorting rate of 1,000 lb/p hr  Total daily man-hrs required for sorting = 400 hr/p
(i.e. 50 people on 8 hrs shift). At \$10/hr wage  \$4,000 /d = \$1.46 M/yr
6.3.- No effect on n_c but reduced horsepower.
6.5- From Appendix B
rho_Al –chips = 17.5 lb/ft^3 = 280.3 kg/m^3 ;
rho_Glass-crushed =45 lb/ft^3 = 720.8 kg/m^3
rho_Air = 0.0749 lb/ft^3 = 1.2 kg/m^3
d_Al = 1 in = 0.0254 m; d_Glass1 = 0.5 in = 0.0125 m ; d_Glass2 = 0.005 in = 0.00125 m
v = {(4 (rho – rho_Air) g d)/(3 C_D rho_Air)]^1/2
v_Al = 5.55 m/s ; v_Glass1 = 6.31 m/s ; v_Glass2 = 1.99 m/s
So 5.5 5&lt; v &lt; 6.3 will get the Al chips in the extract, also some of the fine glass but not the
largest glass particles.
6.7.- In this case v ~ 6.3 m/s = V_u
Also A_available = 0.6 A_unclogged and
Q _unclogged = A_u V_u = Q_free = A_a V_a  V_a = 10.5 m/s
6.9.- Assume Feed ~ 100 lb (80 lb organics + 20 lb inorganics)
Manufacturer A : X1 = 0.8 (80) = 64 lb ; Y1=4 lb ; Y2 = 0.8 (20) = 16 ; X2=16 lb
 Purity = X1/(X1+Y1) = (64/68) = 0.94 ; Overall Recovery = (80/100) = 0.8
E_A = [(X1 Y2)/(X0 Y0)]^1/2 = [64*16/80*20]^1/2 = 0.8
OR_A = (X1+Y2)/(X0+Y0) = (64+16)/(100) = 0.8
6.11.- n_c = [9.81/(4 (3.1416)^2 (1.2)]^1/2 = 0.45 rev/s = 27.3 rpm  n/n_c ~ 0.15 and
Fig. 6-8 (p.219) F ~ 0.05  particles probably cataracting
6.13.- For plastic B
Recovery R = 35/38 = 0.92 ; Purity P = 35/(35+5+18) = 0.6
v_Stokes = (0.00025)^2 (9.81)*(900-1200)/(18*0.0015) = 0.006 m/s
and time to reach the top = 2/0.006 = 293 s ~ 5 min
6.14.a) Sequence 1  2
Feed: A0 = 10 , B0 = 4
Recovery after Operation 1: A1=0.8*A0 = 8 ; B1=0.6*B0 = 2.4
Rejection after Operation1 : A01=2 ; B01=1.6
Recovery after Operation 2: A2=0.5*A1 = 4; B2=0.2*B1 = 0.48
Rejection after Operation2 : A02=4 ; B02=1.12
P_A = 4/4.48 = 0.83
R_A = 4/10 = 0.4
Sequence 2  1
Feed: A0 = 10 , B0 = 4
Recovery after Operation 2: A1=0.5*A0 = 5 ; B1=0.2*B0 = 0.8
Rejection after Operation1 : A01=5 ; B01=3.2
Recovery after Operation 2: A2=0.8*A1 = 4 ; B2=0.6*B1 = 0.32
Rejection after Operation2 : A02=1 ; B02=0.48
P_A = 4/4.32 = 0.92
R_A = 4/10 = 0.4
c)
E_x,y(1-2) = [(4*1.92)/(10*4)]^1/2 = 0.43
E_x,y(2-1) = [(4*0.48)/(10*4)]^1/2 = 0.21
6.16.Since v = d^2 g (rho_s – rho_Air).(18*mu)  d = [18*mu*v/(g*(rho_s-rho_Air))]^1/2
With mu = 5e-5 lb s/ft^2 = 2.4e-5 kg/m s and v = 1500 ft/s = 7.62 m/s  d = 4.6e-4 m
6.18.a) Complete the data by adding columns for x2, y2 then plot to see that the desired purity of 85%
is obtained when v is somewhere between 50 and 60. Linear interpolation yields v = 53.3.
Another linear interpolation yields Rx1 = 0.76  x1 = 13.67 at that velocity.
b) Linear interpolation yields y2 = 17.5  E_x,y = [(13.67*17.5)/(18*20)]^1/2 = 0.815
Another linear interpolation yields x2 = 4.33  Py2 = 17.5/(4.33+17.5) = 0.8
6.19.- See Excel worksheet
6.23.-
6.25.-
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