Examination for Process
9
BUKU PINTAR PROCESS GROUP 1. Structure of Sucrose Sucrose ( Saccharose, Cane sugar, Beet sugar ) is a carbohydrate of the brutto formula C 12 H 22 O 11 . It is a disaccharide, consisting of two mono – saccharidic components : Exact chemical name of sucrose is α– D – glucopyranosyl –β– D – fructofuranoside. ( Glucose Component ) ( Fructose Component ). Sucrose In plants the carbohydrates ( sugars, starch, cellulose ) are formed by the photosynthetic assimilation process. 6CO 2 + 6H 2 O + 675cal = C 6 H 12 O 6 + 6O 2 . The process is catalyzed by the chlorophyll. The carbon dioxide taken from the air is equivalent to the oxygen given to the air. The necessary energy per molecule of formed oxygen corresponds with air least three quanta of the orange light absorbed by the chlorophyll. It seems that first phosphoglyceric acid, these phosphates, and hexose phosphates are formed, and that the sucrose is UNIT PHYSICAL OPERATION
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Transcript of Examination for Process
BUKU PINTAR
PROCESS GROUP
1. Structure of Sucrose
Sucrose ( Saccharose, Cane sugar, Beet sugar ) is a carbohydrate of the brutto formula C 12
H22 O11. It is a disaccharide, consisting of two mono – saccharidic components :
Exact chemical name of sucrose is α– D – glucopyranosyl –β– D – fructofuranoside.
( Glucose Component ) ( Fructose Component ).
Sucrose
In plants the carbohydrates ( sugars, starch, cellulose ) are formed by the photosynthetic
assimilation process.
6CO2 + 6H2O + 675cal = C6H12O6 + 6O2.
The process is catalyzed by the chlorophyll. The carbon dioxide taken from the air is
equivalent to the oxygen given to the air. The necessary energy per molecule of formed
oxygen corresponds with air least three quanta of the orange light absorbed by the
chlorophyll.
It seems that first phosphoglyceric acid, these phosphates, and hexose phosphates are
formed, and that the sucrose is the first free carbohydrate in the plant, formed prior to the
free fucose and fructose.
The formation takes plants in the green parts of the plants, but sucrose is found also in
stall, roots, fruits.
Sucrose is by for the most widespread, sugar present in the sap of plant.
More than 17% of sucrose are contained in the sap of sugar cane.
UNIT PHYSICAL OPERATION
Carbohydrate
The simplest carbohydrates are carbon chains, containing alcohol group
( H– C– OH ) and one aldehyde ( H - C = 0 )
2. Physical Properties of sucrose1) Molecule
The molecule weight of sucrose is 342.296.The calculation from grams of sucrose to moles of sucrose.
Grams of sucrose Moles of sucrose Moles of sucrose Grams of sucrose 100 0.292145 1 342.296 200 0.584290 2 684.592 500 1.46072 5 1701.48 The normal entropy (at 25°C and 760 mm pressure) of sucrose is 86.1 cal/° absolute per mole.
Content of heat (at 25°C and 760 mm pressure) is 530.8 Kcal per mole.
The enthalpy of combustion is 1351.3 Kcal per mole, accordingly the heat of combustion is 1351.3 Kcal per mole or 3.949 Kcal per gram of sucrose.
2) Followings formula are recommended to use computer calculation.(Baikow second revision Appendix III).
Specific gravity 20/20 = Refer to the tableDensity = Refer to the table
True weight (in vacuo) = apparent weight
Temperature correction factor = Refer to the table
3. General Formula for chemical calculation
a. Specific heat =
Approx = 1 – 0.6 BxHeat of crystallization (BTU lb-1) =
Total Heat (BTU lb-1) =
Solubility (Bx)
=
Boiling point rise (°C) =
b. Self Evaporation (lb per 1000 lb of solution)
=
Specific Gravity
=
4. Relation between brix, pol & purity
water
Non sugar Total weight
Sucrose Solid(pol)
Pol : polarization = sucroseSolid : non – sugar (+) sucrose (pol)
5. How to Calculate Chemical Balance in Boiling (To prepare for PTPN 13 in June 2008, referring is as same as same step)1) Sugar Recovery by SJM Formula (Deerr)
γ = Massecuite purityM = Molasses purity after centrifugal separationS = Sugar purity after centrifugal separation
Solid % in sugar and molasses in massecuite
Solid % in molasses per solid in massecuite = 100 – x % -------- (3)
2) Cobenze diagramRequired massecuite purity, Pme = 65 boiled with syrup (purity, Ps = 85) and molasses (purity, Pm = 55), what is solid ratio in syrup and molasses.
Ps (Pm –Pme) 85 10
Pme 65
Pm (Ps – Pme) 55 20
Syrup and molasses mixing ratio is 10:20Or syrup solid 10 and molasses solid 20.
…….
/////////////////////////////////
SYRUP85
A. MASS.85
A. SUGAR98
A. MOL
70
B. MASS.75
B. SUGAR98
B. MOL
55
C. MASS.60
C. SUGAR87
C. MOL
30
Figure are all of purity,A sugar, B Sugar & C Sugar are as product.
3) Sample Chemical Balance in Boiling House
A-Massecuite
B-Massecuite
Syrup solid % in B-Massecuite = 100 – 20 = 80 %
Syrup solid % in B-massecuite is 11.7 % in A-massecuite solid
Total solid of B-Massecuite is 58.2 % in A-massecuite
11.7 % + 46.5 = 58.2 %
C – Massecuite
B-Mol % of C-Massecuite = 100 – 16.7 = 83.3 %
Solid % of B-molasses per A-massecuite solid (100)
= 58.2 x 0.58 = 33.8 %
Syrup solid % in B massecuite solid =
Syrup solid % of B massecuite per A-massecuite solid (100)
= 33.8 x 0.2 = 6.7 %
Total solid of C-massecuite per A-massecuite solid (100)
= 33.8 + 6.7 = 40.5 %
Total solid in total massecuite per A-massecuite solid (100)
From syrup, 100 (A-Mass) + 11.7 (B-Mass) + 6.7 (C-Mass) = 118.4
From A-Mol = 46.5
From B-Mol = 33.8
Total = 198.7
Total solid in total massecuite per syrup solid (100)
A-Massecuite solid per syrup solid (100)
From syrup = 100
From A-Mol
From B-Mol
Total = 166.9
B-Massecuite solid per syrup solid (100)
84.5 x 0.582 = 49 %
C-Massecuite solid = 84.5 x 0.405 = 34 %
C-Massecuite solid per syrup solid (100)
84.5 x 0.405 x 0.473 = 16.2 %
In case C-Sugar remelt or use as seed
So total solid in total massecuite per total solid in syrup (100)
166.9 + 18 = 185
4) Pol Recovery (Chemical Balance)Pol recovery will be 96.8 % in good practice factory.Losses are 3.2 % and details are as follows :-
In Filter Cake ------ 0.1 %In Final Molasses ------ 1.6 %In Unknown Losses (others) ------ 1.5 %
Total 3.2 %5) Weight Yield
Based on above it should be approx 95%.
6. Mixture of Liquid / Solid
1) To mix molasses 100 kg having brix 60 and 100 kg of sweet water having brix 20.(Answer)
Brix Solid Water TotalMolasses 60 60 40 100Sweet Water 20 20 80 100
80 120 200
= 40
2) To operate fine liquor evaporator, flow rate is 100 ton/hr and brix 60. Now require to increase brix to 72. How much water to be removed ?(Answer)
= 16.67 ton/hr3) Following Labo data received, how much sugar content in R1 massecuite. To
calculate by cobenzs diagram.
Bx = 90 Pty = 98.5
R1 MOL R1 Sugar
R1 Massecuite
Centrifugal
Solid = 73 Solid = 99.9Pty = 97.4 pty = 99.5
(Answer)R1 sugar 99.5 1.1 52.4%
98.5R1 MOL 97.4 1.0 47.6 %
2.1 100 %
4) Continue centrifugal machines are operating for E-massecuite. Following data received from labo.
Final Molasses purity = 32E – Sugar Purity = 84E – Massecuite = 62
What is solid balance in final molasses solid of 20.85 tonsTo calculate by coenz’s diagram.(Answer)
30 (+) 22 52
62
84 32 E-sugar Final mol
Final molasses solid = 20.85 tons
E-massecuite solid = 20.85 x = 49.28 ton
Ton pol E-massecuite = 49.28 x 0.62 = 30.55 tonTon solid E-sugar = 49.28 – 20.85 = 28.43 tonTon pol E-sugar = 28.43 x 0.84 = 23.88 ton
Brix = 49.28 Pol = 30.55
(57.7%) (42.3%)
E-sugar Final MolassesBrix : 28.43 Brix : 20.85Pol : 23.88 pol : 6.67
E - Massecuite
Centrifugal
