Octave code for calculating and verifying the given example. Take note of the step-by-step procedures and all variables required to arrive at an answer.
Octave code for calculating and verifying the given example. Take note of the step-by-step procedures and all variables required to arrive at an answer.
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Octave code for calculating and verifying the given example. Take note of the step-by-step procedures and all variables required to arrive at an answer.
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Fraction of Solutes Remaining in the Cake, R
0.8
0.6
0.4
0.2
1.5
2.0
Wash Ratio, W
Figure 19.30 Retention of solutes as a function of wash ratio.
0.0
0.5
1.0
2.5
amount of solutes in the effluent equals the solutes still
retained in the cake. Thus,
[² (²) aw - " ( ² ) aw =R * (3) aw
dWR
dW
R{W}=1
3.0
(19-35)
Rearranging (19-35), R as a function of W can be com-
puted by integrating, e.g., Curve 2 in Figure 19.29. The result
of that integration for Curves 1 and 2 is shown in Figure
19.30.
(3) aw .
TO)
dW
(19-36)
There are two approaches to designing wash cycles. The
first is to establish, by experiment, tables of data or plots like
Curve 2 in Figure 19.29, and then manipulate the data to find
the wash cycle that satisfies design objectives, which might
be reducing the concentration of dissolved solutes to a certain
level, displacing a certain amount of mother liquor with wash
liquid, or, if the wash takes place on a rotary vacuum filter,
determining the amount of wash liquid required to remove a
stipulated amount of material in a given amount of time.
Example 19.11 illustrates typical data manipulations. An
alternative is to devise general mathematical models for
washing in much the same way as leaching and drying opera-
tions are modeled [7].
EXAMPLE 19.11 Washing a Biomass Cake.
Following fermentation, biomass is separated from the broth, often
by centrifugation. Then the wet biomass is washed to recover the
occluded broth, if the desired product is in the broth, e.g., if the bio-
product is extracellular. A 1,000-kg biomass sample has been freed
of broth and found to weigh 450 kg. (a) What is the average porosity
of the biomass if the density of the broth is 1,050 kg/m³, the biomass
density is 1,150 kg/m³, and the wash liquid is water at 1,000 kg/m³?
(b) The following experimental wash data were obtained for S/S, as
a function of W:
W
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
S/So
1.000
1.000
1.000
1.000
1.000
1.000
0.998
0.979
0.897
0.725
0.500
0.295
0.151
0.068
0.028
0.010
0.004
0.001
0.000
0.000
0.000
$19.0
(3) aw
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.699
0.793
0.874
0.935
0.975
0.997
1.008
1.013
1.015
1.015
1.015
1.015
1.015
1.015
805
R
1.000
0.902
0.803
0.705
0.606
0.508
0.409
0.312
0.220
0.140
0.080
0.040
0.018
0.008
0.003
0.001
0.000
0.000
0.000
0.000
0.000
What fractions of broth are recovered for wash ratios of 0.5, 1.0, 1.5,
and 2.0? (c) What mass of wash water is required to recover 98% of
the broth?
Solution
(a) One thousand kg of wet cake consists of 550 kg broth and 450 kg
dry biomass. Broth volume = 550 kg/m³/1,050 kg/m³ = 0.524 m³
and biomass volume=450 kg/1,150 kg/m³ = 0.391 m². Thus, the
total volume is 0.915 m³ and volume fractions are biomass
0.427 and broth-0.573. The wash water replaces the broth, so
the void volume €₁vg=0.573. The volume of water for W=1
(0.573)(0.915 m³)=0.524 m², or a wash water mass of 0.524
(1,000)=524 kg.
(b) From (19-36), values of R can be computed by graphical inte-
gration for values of the wash ratio, W. Because the experimen-
tal data are closely spaced, reasonable accuracy is achieved
using the trapezoidal rule with a spreadsheet. For each interval
in W, the arithmetic-average value of S/S, is determined and
multiplied by AW. For example, in the interval of W from 0.80
to 0.90, the average value of S/S, is (0.897+0.725)/2=0.811,
which, multiplied by AW=0.1, is 0.0811. In the above table,
the change in the integral for this interval in W is (0.874 -
0.793)=0.081. Values of the integral and R for a sequence of
values of W are included in the above table. Equation (19-36)
requires the value of the integral for W=00. In the above table,
it is seen that by W=1.60, the integral is no longer changing; so
that value of 1.015 can be used.
(e) For recovery of 98% of the broth, the above table shows that a
wash ratio of 1.2 is sufficient, or a mass of wash water = 1.2
(524)=629 kg. In this example, washing is very efficient.
Often, it is not.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F492fa439-ea16-42d3-a6e8-df39b090d9be%2F508f7b12-01cf-4911-917d-ff3d9d11c869%2F5k7w8pc_processed.png&w=3840&q=75)
Transcribed Image Text:Է
Fraction of Solutes Remaining in the Cake, R
0.8
0.6
0.4
0.2
1.5
2.0
Wash Ratio, W
Figure 19.30 Retention of solutes as a function of wash ratio.
0.0
0.5
1.0
2.5
amount of solutes in the effluent equals the solutes still
retained in the cake. Thus,
[² (²) aw - " ( ² ) aw =R * (3) aw
dWR
dW
R{W}=1
3.0
(19-35)
Rearranging (19-35), R as a function of W can be com-
puted by integrating, e.g., Curve 2 in Figure 19.29. The result
of that integration for Curves 1 and 2 is shown in Figure
19.30.
(3) aw .
TO)
dW
(19-36)
There are two approaches to designing wash cycles. The
first is to establish, by experiment, tables of data or plots like
Curve 2 in Figure 19.29, and then manipulate the data to find
the wash cycle that satisfies design objectives, which might
be reducing the concentration of dissolved solutes to a certain
level, displacing a certain amount of mother liquor with wash
liquid, or, if the wash takes place on a rotary vacuum filter,
determining the amount of wash liquid required to remove a
stipulated amount of material in a given amount of time.
Example 19.11 illustrates typical data manipulations. An
alternative is to devise general mathematical models for
washing in much the same way as leaching and drying opera-
tions are modeled [7].
EXAMPLE 19.11 Washing a Biomass Cake.
Following fermentation, biomass is separated from the broth, often
by centrifugation. Then the wet biomass is washed to recover the
occluded broth, if the desired product is in the broth, e.g., if the bio-
product is extracellular. A 1,000-kg biomass sample has been freed
of broth and found to weigh 450 kg. (a) What is the average porosity
of the biomass if the density of the broth is 1,050 kg/m³, the biomass
density is 1,150 kg/m³, and the wash liquid is water at 1,000 kg/m³?
(b) The following experimental wash data were obtained for S/S, as
a function of W:
W
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
S/So
1.000
1.000
1.000
1.000
1.000
1.000
0.998
0.979
0.897
0.725
0.500
0.295
0.151
0.068
0.028
0.010
0.004
0.001
0.000
0.000
0.000
$19.0
(3) aw
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.699
0.793
0.874
0.935
0.975
0.997
1.008
1.013
1.015
1.015
1.015
1.015
1.015
1.015
805
R
1.000
0.902
0.803
0.705
0.606
0.508
0.409
0.312
0.220
0.140
0.080
0.040
0.018
0.008
0.003
0.001
0.000
0.000
0.000
0.000
0.000
What fractions of broth are recovered for wash ratios of 0.5, 1.0, 1.5,
and 2.0? (c) What mass of wash water is required to recover 98% of
the broth?
Solution
(a) One thousand kg of wet cake consists of 550 kg broth and 450 kg
dry biomass. Broth volume = 550 kg/m³/1,050 kg/m³ = 0.524 m³
and biomass volume=450 kg/1,150 kg/m³ = 0.391 m². Thus, the
total volume is 0.915 m³ and volume fractions are biomass
0.427 and broth-0.573. The wash water replaces the broth, so
the void volume €₁vg=0.573. The volume of water for W=1
(0.573)(0.915 m³)=0.524 m², or a wash water mass of 0.524
(1,000)=524 kg.
(b) From (19-36), values of R can be computed by graphical inte-
gration for values of the wash ratio, W. Because the experimen-
tal data are closely spaced, reasonable accuracy is achieved
using the trapezoidal rule with a spreadsheet. For each interval
in W, the arithmetic-average value of S/S, is determined and
multiplied by AW. For example, in the interval of W from 0.80
to 0.90, the average value of S/S, is (0.897+0.725)/2=0.811,
which, multiplied by AW=0.1, is 0.0811. In the above table,
the change in the integral for this interval in W is (0.874 -
0.793)=0.081. Values of the integral and R for a sequence of
values of W are included in the above table. Equation (19-36)
requires the value of the integral for W=00. In the above table,
it is seen that by W=1.60, the integral is no longer changing; so
that value of 1.015 can be used.
(e) For recovery of 98% of the broth, the above table shows that a
wash ratio of 1.2 is sufficient, or a mass of wash water = 1.2
(524)=629 kg. In this example, washing is very efficient.
Often, it is not.
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