chapter 4 -material balance
TRANSCRIPT
CHAPTER IV
MATERIAL BALANCE
The raw materials to be utilized in this study are the fiber found in the rice straw. The
assumptions of the amounts of the substances entering in some of the equipment are based on
data gathered by the researchers wherein actual experiments were conducted.
Table 4.1 Chemical Composition of Rice Straw
Cellulose
(w%)
Hemicellulose
(w%)
Lignin
(w%)
Ash
(w%)
Rice Straw 41% 33% 12% 14%
Source: Bledzki and Gassan 1999; Bledzki et al. 2002; Lu et al. 2006; Marti-Ferrer et al. 2006;
I. Material Balance around the Shredder
Forty tons of rice straws are feed in the shredder for size reduction to turn the raw material
into segments about 1-2 inches.
Mass Balance
40 tons Raw Material 40 tons Raw Material
Input = Output
40 tons Raw Material = 40 tons Raw Material
II. Mass Balance around the Cooker
The fibers are cooked at 80oC for 120 minutes where lignocellulosic materials are
treated with an 80% aqueous solution of acetic acid and 0.6% HCl. (Jahan, M., Lee, Z., Jin, Y.,
2005).
SHREDDER
Basis: 40 tons dry fiber
Mass Balance
AcOH HCl
40 tons Fiber 35.0128 tons cooked fiber
+ 204.9872 tons Black Liquor
Assumptions:
i.
mass liquormass fiber = 5 (Haas and Lang, 1971)
ii. Kappa # = 45.8 = % lignin left after 120 minutes cooking
(Jahan, M., Lee, Z., Jin, Y., 2005)
iii. 57.4% of the ash is retained in the cooking (source)
iv. 80% of the total liquor is CH3COOH (Jahan, M., Lee, Z., Jin, Y., 2005)
v. 0.6% of the total Liquor is HCl (Jahan, M., Lee, Z., Jin, Y., 2005)
COOKER
Amount of lignin in the fiber = 40 tons fibers X 0.12 lignin
= 4.8 tons lignin
Amount of Lignin left in the cooked fiber = amount of lignin in the fiber X kappa #
= 4.8 tons lignin X 0.458
= 2.1984 tons lignin
Amount of Lignin removed = 4.8 tons – 2.1984 tons
= 2.6016 tons lignin
Amount of ash in the fiber = 40 tons fibers X 0.14 ash
= 5.6 tons ash
57.4% of ash is retained in the cooking
5.6 tons Ash X 0.574 = 3.2144 tons Ash
Amount of Ash removed = amount of ash in the fiber – amount of ash retained
= 5.6 tons – 3.2144 tons
= 2.3856 tons Ash
Input:
Tons of fiber = 40 tons
Mass of Liquor = 5 (40 tons) = 200 tons
Tons of CH3COOH = 200 tons X 0.80 = 160 tons CH3COOH
Tons of HCl = 200 tons X 0.006 = 1.2 tons HCl
Tons of H2O = mass liquor – tons CH3COOH – tons HCl
Tons of H2O = 200 tons - 160 tons – 1.2 tons = 38.8 tons H2O
Output:
Fibers = feed Fiber – extracted Lignin – extracted Ash
Fibers = 40 tons – 2.6016 tons – 2.3856 tons = 35.0128 tons of cooked fiber
Black Liquor = 200 tons + 2.6016 tons + 2.3856 tons = 204.9872 tons black liquor
Overall Material Balance:
Fiber + Liquor = cooked Fiber + Black Liquor
40 tons + 200 tons = 35.0128 tons + 204.9872 tons
III. Mass Balance around Filter
The pulp slurry from the cooking process will pass through a filter wherein the liquor is
separated from the pulp.
240 tons slurry 35.0128 tons cooked fiber
204.9872 tons black liquor
Mass Balance
Input = Output
240 tons pulp slurry = 35.0128 tons cooked fiber + 204.9872 tons black liquor
IV. Mass Balance around the Washer
The fibers are washed with 80% acetic acid and with large quantities of water.
Mass Balance
Acetic Acid and Water
35.0128 tons cooked fiber 35.0128 tons cooked fiber
Washing
Input = Output
35.0128 tons impregnated fiber = 35.0128 tons impregnated fiber
V. Mass Balance around the Extractor
It is known that organic acid pulping gives rigid and brittle fibers and this pulp has lower
strength properties. In order to improve the strength properties of organic acid treated pulp by
alkali. It is known that the following 3 reactions take place with alkali treatment: 1)
solubilization of hemicellulose and residual lignin in pulp, 2) deacetylation of pulp, and 3)
solubilization of residual silica (ash) in pulp.
70.0256 tons Liquor
35.0128 tons 27.5449904 tons pulp
Cooked fiber
77.4934096 tons extracted Solution
Mass Balance:
Input = Output
Assumption:
i.
mass liquormass pulp = 2 (Jahan, M., Lee, Z., Jin, Y., 2005)
ii. Kappa # = 32.7 = % Lignin left (Jahan, M., Lee, Z., Jin, Y., 2005)
iii. 30% of Hemicellulose is desired to be dissolved.
iv. 36.9% of the original Ash content (5.6 tons) is retained in the
pulp (Pan, Xue-Jun, et. al., 1999)
v. 10 % NaOH (Jahan, M., Lee, Z., Jin, Y., 2005)
Input:
Amount of liquor = 2 x 35.0128 tons pulp = 70.0256 tons Liquor
Amount of NaOH = 70.0256 tons pulp x 0.10 = 7.00256 tons NaOH
Amount of H2O = 70.0256 – 7.00256 = 63.02304 tons H2O
Amount of Lignin left in the pulp = amount of lignin(left after cooking) X kappa #
= 2.1984 tons lignin X 0.327
= 0.7188768 tons lignin
Amount Lignin dissolved = amount of lignin(left after cooking) - amount of Lignin left in the pulp
= 2.1984 tons lignin - 0.7188768 tons lignin
= 1.4795232 tons lignin
Amount of Hemicellulose (from Raw material) = Mass of Raw material
X % hemicellulose of the Raw material
Amount of Hemicellulose(from Raw material) = 40 tons x 0.33
= 13.2 tons Hemicellulose
Desired Hemicellulose to Dissolved = Amount of Hemicellulose(from Raw Materials)
X % desired to be dissolved.
Desired Hemicellulose to Dissolved = 13.2 tons x 0.3 = 3.96 tons Hemicellulose
Ash Retained = 3.2144 tons ash x 0.369 ash Retained = 1.1861136 tons Ash
Ash Removed = 3.2144 tons ash(left after cooking) – 1.1861136 tons ash
= 2.0282864 tons Ash
Output:
Tons of pulp = Feed – (amount lignin dissolved + desired hemicellulose to dissolved + Ash removed)
= 35.0128 tons – (1.4795232 tons + 3.96 tons + 2.0282864 tons)
= 27.5449904 tons pulp
Overall Material Balance:
35.0128tons pulp + 70.0256 tons NaOH sol’n = extracted mat’ls + 27.5449904 tons pulp
Extracted material = 77.4934096 tons
VI. Mass Balance around Filter
The pulp slurry coming from the extractor will enter the filter to collect the solid materials
and separate the solution containing the dissolved lignin, hemicellulose and ash content.
77.4934096 tons extracted mat’l
+ 27.5449904 tons pulp 27.5449904 tons pulp
77.4934096 tons Extracted mat’ls
Mass Balance
Input = Output
Extracted mat’ls + 27.5449904 tons pulp = Extracted mat’ls + 27.5449904 tons pulp
VII. Mass Balance around Screener
The reject fiber and dirt from the pulp are removed. 8% of reject materials will be separated
from the pulp. The separation mechanism of the equipment is through the vibrating action of
the screen.
Basis: 27.5449904 tons pulp
27.5449904 tons pulp
2.203599232 tons reject fiber
and dirt 25.34139117 tons pulp
Mass Balance
Input = output
Input = 27.5449904 tons pulp
Reject: 27.5449904 tons (0.08) = 2.203599232 tons reject fibers and dirt
Output = 27.5449904 tons – 2.203599232 tons = 25.34139117 tons pulp
VIII. Mass balance around Refiner
This is where the cellulose fibers pass through a refining process which is vital in the art
of papermaking. Before refining, the fibers are stiff, inflexible and form few bonds. The stock is
pumped through a conical machine which consists of a series of revolving discs. The violent
abrasive and bruising action has the effect of cutting, opening up and declustering the fibers
and making the ends divide.
25.34139117 tons pulp
25.34139117 tons pulp
Mass Balance
Input = Output
25.34139117 tons pulp = 25.34139117 tons pulp
IX. Material Balance around Paper Machine
The bleached pulp is prepared for paper making with a standard consistency of 0.5 % pulp for
thin sheets paper. The sheet of paper will be formed on a travelling wire or cylinder dewatered under
rollers, dried by heated rolls, and finished by calendar rolls. The machine is expected to produce
paperboards with a GMS of 200 g/m2 which is based on the specification of the equipment.
25.34139117 tons pulp(0.5% pulp)
+ 26.95892678 tons
5042.936843 tons H2O per roll of paper
5041.41636 tons H2O
Mass Balance
Input = Output
Assumption:
i. 0.5% bone dry pulp (http://www.Paper_machine.htm)
ii. 6% moisture in the produced paper
(http://www.Paper_machine.htm)
Input:
Feed = 25.34139117 / 0.005 = 5068.278234 tons pulp slurry
Mass of added water =5068.278234 – 25.34139117 = 5042.936843 tons water
Output:
25.34139117 tons pulp / 0.94 = 26.95892678 tons per roll of paper
Rejected H2O = 5042.936843 – (25.34139117 x 0.06) = 5041.41636 tons H2O