Chapter 10 Unit Operation Problems
1. Filtering of wine
Pressure
= 350kPa gauge
Rate of flow = 450lh1
Filter area
= 0.82m2
Assume 
= 1000kgm-3
dV/dt =
0.125 =
R
=
=
In larger plant:
Area
=
Rate of flow =
=
=
dV/dt =
1.736 =
P
=
=
=
= 350 x 103 Pa
= 0.125ls-1
(A P)/ R
(0.82 x 350x103)/R
2296 x 103
2.296 x 106 kPasm-1
6.5 + 0.82
= 7.32 m2
500 hectolitres per 8 hours (1 hectolitre = 100 litres)
500 x 100 litres/8 hours
1.736 litres s-1
(A P)/ R
(7.32 P)/(2.296 x 10 6)
1.736/7.32 (2.296 x 10 6)
0.5445 x 10 6Pa
544 kPa
2. Cake resistance in wine filtering

= 1.8x 10-3 Nsm-2
= 1.8 x 10 kg m-1s-1
Resistance
R
=
r(Lc + L)
Lc
=
wV/A
=
(0.04 x 1.736)/(7.32)
=
0.00949
=
9.49 x 10-3m
=
9.5mm
Ignoring L, which is very small.
r
=
(2.296 x 106)/ (1.8 x 10-3 x 9.49 x 10-3)
=
0.1343 x 1012
3. Compressibility of cake in wine making
Flow rate
=
2800lh-1
=
0.78 ls-1
P
=
544 kPa
=
544 x 103 Pa
dV/dt =
(A P)/ R2
R2
=
(7.32 x 544 x 10 3)/ 0.78
=
5105 x 103
=
5.105 x 106
Lc
=
(0.04 x 0.78)/7.32
=
0.00426
=
4.26 x 10-3
r2
=
(5.105 x 106)/ (1.8 x 10-3 x 4.25 x 10-3 )
=
0.667 x 10 12
r
=
r’Ps
r/ r2
=
Ps /P2s
0.1343 x 1012 /0.667 x 1012
0.201
s
=
=
=
(544 kPa/544 kPa)s
1S
-1.7?? something wrong
or
=
=
=
=
1.013 x 105 Pa
0.000250 x 1014
2.5 x 1010
r’Ps
1atms
r’
r
=
0.667 x 10 12 =
(544 x 103)s =
=
3
s ln(544 x 10 )
=
s
=
=
(2.5 x 1010 ) (544 x 103)s
(0.667 x 10 12 )/(2.5 x 1010 )
26.68
ln 26.68
3.28/13.21
0.25
4. Settling of sand particles in salt solution
Sand
Diameter of sand particles = 0.2mm = 2 x 10-4 m
Density
= 2010 kgm 3
Salt solution
22%
Density
= 1240kg m 3
Viscosity
= 2.7 x 10-3 Nsm-2
=
D2g(p - f) /18
=
[(2 x 10-4 )2 x 9.81 x (2010 –1240)] / (18 x 2.7 x 10 -3)
=
4 x 10-8 x 9.81 x 770 / (18 x 2.7 x 10 -3)
=
621.7 x 10-5
=
6.217 x 10-3 ms-1
Checking the Reynolds number
(Re)
=
(Dv/)
=
(2 x 10-4 x 6.217 x 10-3 x 1240)/ 2.7 x 10-3
=
5710 x 10-4
=
0.571
vm
5. Continuous sedimentation of sand in salt solution
Sand
Diameter of sand particles
= ?
Density
= 2010 kgm 3
Salt solution
22%
Density
= 1240kg m 3
Viscosity
= 2.7 x 10-3 Nsm-2
Flow rate through trough
= 0.01ms -1
Trough length
=
0.8m
Time for brine to move along trough =
80s
=
maximum time for sand to settle
Assuming that the depth of the outlet below the surface is 0.20m
Rate of sedimentation
vm
vm
2.5 x 10-3
D2
D
=
=
=
=
=
=
=
=
=
=
=
0.20/80
0.0025ms-1
2.5 x 10-3 ms-1
D2g(p - f) /18
D2 9.81(2010 - 1240) /18 x 2.7 x 10-3
[2.5 x 10-3 x 18 x 2.7 x 10-3]/9.81 x 770
121.5x10-6 /7.55x103
16.09 x10-9
0.127 x 10-3m
0.127mm
1.27microns
6. Centrifugal force in small centrifuge
Fc
=
6000g
r
=
9cm =
0.09m
2
Fc/ Fg =
0.011 rN /9.81
=
0.011 x 0.09 N2 /9.81
=
6000
N2
=
[600 x 9.81]/[0.011 x 0.09]
=
5.945 x 106
N
=
2438 revs per minute
7. Centrifuging of oil from brine
Feed In
Brine
Oil
r0
rb = radius brine
Mean radius for brine
Mean radius for oil
rn
rb
ro
=
=
= radius oil
( rn + rb)/2
( rn + r0)/2
Volume of brine
=
( ro2 h -  rn2 h )
Volume of oil
=
( rn2 h -  rb2 h )
And these are equal, so
( ro2 h -  rn2 h )/ ( rn2 h -  rb2 h )
=
1
(r02 - rn2 )/ ( rn2 - rb2 )
=
1
2
2
(ro - rn ) =
( rn2 - rb2 )
2
2
(rb + ro )
=
2 rn 2
Discharge radius for oil = 0.5cm = 0.005 m =
ro
(rb2 + 0.0052) =
2 rn 2
2
So
rb
=
2 rn2 -0.0052
Density of oil o = 1070 kg m3 b
=
900 kg m3
rn 2
=
(oro2 - b rb2)/ (o- b)
=
[1070(0.005)2 – 900(2 rn2 -0.0052)]/ [1070 – 900]
=
[0.02675 - 1800 rn2 + 0.000025]/170
=
[0.0268 -1800 rn2 ]/ 170
=
1.58x10-4- 10.59 rn2
2
11.59 rn
=
1.58 x 10-4
2
rn
=
0.1363 x10-4
rn
=
0.37 x 10-2
=
0.37cm
rb 2
=
2 rn2 -0.0052
=
2 x 0.1363 x10-4 - 0.0052
=
0.2726 x10-4 - 0.25 x10-4
=
0.0226 x 10-4
rb
=
0.15 x 10-2
=
0.15cm
Brine = 0.15cm, neutral(feed) = 0.37cm, oil = 0.5cm
8. Gravity settling compared with centrifugal separating
For a gravity settler: vm  g
A  1/v
So if we replace g by mv/r = mr 2 = (2N/60) mr
= 0.011 mrN 2
So we are comparing mg with 0.011 mrN 2
Therefore area 100m2 will be reduced to 100 x (g/0.011 rN 2 )
If N = 2000rpm and r = 0.5m
Then we have 9.81/(0.011 x 0.5 x (2000)2 ) = 4.46 x 10-4
i.e to 0.045m2 = 1/20 m2
9. Olive oil emulsion separating
D = 50 m
= 50 x 10-6 m
Water:  = 1,000kgm-3  = 1 x 10-3 Nsm-2
Oil:  = 910 kgm-3
r centrifuge= 5cm = 0.05 m
Radial travel of droplets is 0.03 m in 1 sec so velocity must be 0.03 m/s
For 5m droplets, Stokes Law and using eqn. 10.8
0.03
= vm = D2 N2 r (p - f)/1640
= [(5 x 10-5)2 x N2]/(1640 x 1.8 x 10-3)
2
and so N
= 4.27 x 106
and N
= 2065 rpm
10. Sieve analysis
Sieve size
Wt. retained
mm
m
1.00
1,000
0
0.50
500
64
0.250
250
324
0.125
125
240
0.063
63
48
Through 0.063
24
Total
700
% retained
% less than size
100.0
90.0
44.6
10.3
3.4
9.1
46.3
34.3
6.9
3.4
100.0
If these are graphed, size m against % less than size, the graph for the percentage of
given size can be determined.
11. Dust in cyclone collector
Size
Collection efficiency
m
%
%
< 0.5
20
(80)
3
60
(40)
6
75
(25)
10
90
(10)
15
95
(5)
25
100
In exit < 0.5m
3m
6m
Wt of particle
g
0.2
0.7
0.4
0.2
0.1
= 0.16kg
= 0.28kg
= 0.10kg
Wt. leaving
g
0.16
0.28
0.10
0.002
0.015