Role of Phosphoric Acid (H3 PO4 ) Dosing in Sugar Process with Online Calculator
Phosphoric acid occurs in the cane juice as both soluble phosphates and in combination as protein in the cell material. But both phosphates are not participate in reaction, only the soluble phosphates take part in the reaction of defecation.
If the phosphate content is less than 200 to 250 mg/P2O5 in juice, it is beneficial to make additions of inorganic phosphate to raise the juice phosphate level to above 250-300 mg/kg.
The phosphoric acid added to the juice precipitates part of the colloids and coloring matter that it contains. The precipitate formed with lime is mainly Tricalcium Phosphate.
a) Greater colloidal elimination
b) Less calcium in clear juice;
c) Faster mud floc formation, with more rapid settling;
d) Better clarification with lighter colored clarified juice.
Disadvantages of too much phosphate addition can be higher chemical costs, higher lime consumption and higher mud volume (and hence higher filter mud sugar loss).
Action Due to phosphoric Acid Addition:
The major chemical reaction is that of the Ca++ cation with the phosphate ion to form phosphate Intermediates and to precipitate tricalcium phosphate Ca 3(PO4)2.
We have to know the activity of calcium and phosphate ions and how the phosphate components behave in the adsorption and absorption of organic colloids and other non-sugars.
It is after reacting with the available Ca++ and then further with lime is one of the major reaction responsible for juice clarification. In the beginning the Ca++ ion reacts with HPO—4 to form CaHPO4. This is second order reaction & hence it is fast reaction,
Ca2++ + HPO– –4→ CaHPO4
Second stage phosphate ion reacts with free Ca ion to form the precipitate of tricalcium phosphate. This a higher order reaction, hence takes time to complete
3Ca++ + 2PO– —4 → Ca3 (PO4)2 TriCalcium Phosphate
3 Ca (OH) 2 + 2 H3 PO4 ———- Ca3 (PO4) 2 + 6 H2O
Steps in Calculation for required quantity of Phosphoric acid to add in Raw juice:
a) Crushing Rate ( M ) in T/hr
b) Percentage of mixed juice ( P ) = Take 100%
c) Density of juice ( D )(Generally take 1.06 gm/ml for raw juice)
d) Required phosphoric acid (in P2O5 Content) dosing in PPM ( R )to raw juice ( To know this analyze P2O5 content in mixed juice before addition than we can concluded to how much dosing required to raw juice).
e) Percentage of Ortho phosphoric acid chemical ( H ) ( i.e Percentage of H3PO4 purity in used chemical)
Calculate mixed juice quantity (Q ) in Ltrs/hr.
Q = ( M x 1000)/ D
P2O5 requirement in ( T ) Kg/hr (1ppm=1 mg/lit)
T = ( Q x R) / 1000000
Calculate P2 O5 % ( Y ) in given chemical H3PO4
2 H3PO4 → P2O5 + 3H2O
2(3+31+64) (2*31+5*16) + (3*2+3*16)
196 142 54
i.e 196 parts of H3PO4 gives 142 parts P205)
Y = ( H x 142)/196
H3PO4 required quantity in kg/Hr (T x 100)/Y
Phosphoric Acid (H3 PO4 ) Online Calculator :
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