PSE 476
Lecture #6
Kraft Pulping Chemicals
PSE 476: Lecture 6 1

Agenda
• Basic Description of Liquors & Process
» White, Black & Green Liquors
• Definition of Terms
» Total alkali, Effective Alkali, Sulfidity, etc.
• Why is everything on a Na
2
O basis?
PSE 476: Lecture 6 2

• White liquor.
» Fresh pulping liquor for the kraft process containing
NaOH, Na
2
S, and a variety of impurities.
• Black liquor.
» The waste liquor from the kraft pulping process.
Contains most of the original inorganic components
(most in different forms) and a high concentration of dissolved organics.
• Green liquor.
» Partially recovered kraft liquor (intermediate liquor in recovery sequence).
PSE 476: Lecture 6 3

White Liquor
Digester
Black Liquor
Lime
Kilm
Green Liquor
Recovery
Furnace
This is a very simplified diagram. There are several steps between each box. We will discuss this whole sequence in depth in a later lecture.
PSE 476: Lecture 6 4

Chemical White*
NaOH
Black*
95 (53%) 1.4 (7%)
Green*
15 (8%)
Na
2
S
Na
2
CO
3
38 (21%)
26 (15%)
4.2 (19%)
7.8 (36%)
37 (20%)
107 (60%)
Na
2
SO
3
Na
2
SO
4
4.8 (3%)
9.1 (5%)
2 (9%)
2.8 (13%)
6.1 (3%)
11 (6%)
Na
2
S
2
O
3
6 (3%)
Organics None
* Median concentrations as g/l as Na
2
O
3.4 (16%)
Lots
PSE 476: Lecture 6
5.5 (3%)
None
5

Chemical Amount
(as Na
2
O)
Class* Source of unwanteds:
(Recovery System)
NaOH 81 - 120 g/l Active NA
Na
2
S 30 - 40 g/l Active NA
Na
2
CO
3
11 - 44 g/l Inactive Incomplete Caustizing
Na
2
SO
3
2 - 6.9 Inactive Incomplete Reduction
Na
2
SO
4
4.4 - 18 g/l Inactive Incomplete Reduction
Na
2
S
2
O
3
4 - 8.9 g/l Inactive Oxidation of Sulfide
Notes Page
PSE 476: Lecture 6 6

• All chemicals are reported as concentrations in liquor (g/l) or as charge (%) on dry wood.
• Total Chemical : All sodium salts (as Na
2
O).
• Total Alkali : NaOH + Na
2
1/2Na
2
SO
3
(as Na
2
O).
S + Na
2
CO
3
+
» This is the sum of the sodium salts that contribute to or are converted during kraft cooking to chemicals which contribute to active alkali.
• Active Alkali : Na
2
100g/L
S + NaOH (as Na
2
O)
PSE 476: Lecture 6 7

• Sulfidity: 24-28%
Sulfidity =
Na
2
S
NaOH + Na
2
S
• Causticity:
Causticity =
NaOH
NaOH + Na
2
S
* 100%
* 100%
PSE 476: Lecture 6 8

• Effective Alkali : NaOH + 1/2 Na
2 than 55 g/L
S (as Na
2
O) no more
• Activity : % ratio of Active to Total Alkali
• Causticizing Efficiency: 78-80%
NaOH
Causticizing eff. =
NaOH + Na
2
CO
3
• Reduction Efficiency: 95%
* 100%
Na
2
S
Reduction eff. =
Na
2
S + Na
2
SO
4
+ Na
2
SO
3
+ Na
2
S
2
O
3
* 100%
PSE 476: Lecture 6 9

2
• Expressions such as sulfidity, causticity, effective alkali, etc “best” describe the conditions in a kraft cook.
» These expressions contain information on the amounts (g/liter or %) of different chemicals such as NaOH, Na
2
S, etc which have different degrees of effectiveness
» Reporting on a Na
2
O basis indicates the actual chemical relationship between these chemicals
PSE 476: Lecture 6 10

2
Na
2
O + H
2
O = 2NaOH
1 mole Na
2
O creates 2 moles of NaOH
62 grams Na
2
O is equivalent to 80 grams of NaOH
Therefore, 1 mole of NaOH is equal to 1/2 mole of Na
2
O
40 g/l NaOH = 0.5 * 62 g/l Na
2
O = 31 g/l on Na
2
O basis
PSE 476: Lecture 6 11

2
Na
2
O + H
2
S Na
2
S + H
2
O (hypothetical equation)
1 mole Na
2
O creates 1 moles of Na
2
S
Therefore, 1 mole of Na
2
S is equal to 1 mole of Na
2
O
62 grams Na
2
O is equivalent to 78.1 grams of Na
2
S
78.1 g/l Na
2
S = 62 g/l Na
2
O
Na
Na
Na
2
2
2
O + H
O + H
O + H
2
2
2
CO
SO
SO
3
4
3
Na
2
CO
3
+ H
2
O
Na
2
SO
4
+ H
2
O
Na
2
SO
3
+ H
2
O
PSE 476: Lecture 6 12

PSE 476: Lecture 6 13

In-Class Example Calculations (1)
• 50 Tons Chips
• 50% Moisture Content
• Liquor Charge to Digester:
» 1200 ft 3 white liquor
- EA = 13% (alkali charge on OD wood as Na
2
O)
- Sulfidity = 25.2%
» 1300 ft 3 black liquor
• Question: How many lbs./ft 3 of NaOH and
Na
2
S were charged to the digester in the white liquor?
(assume no chemical contribution from black liquor)
PSE 476: Lecture 6 14

In-Class Example Calculations (2)
Step 1: Calculate the amount of oven dry wood
50 tons chips • 2000 lbs./ton • 0.5 (m.c.) = 50,000 lbs. o.d. wood
Step 2: Calculate the amount of NaOH and Na
2
S as Na
2
O in the white liquor using the EA and Sulfidity numbers
EA = NaOH + 1/2 Na
2
S = 13% on od wood.
NaOH + 1/2 Na
2
S = 0.13 • 50,000 = 6500 lbs.
NaOH = 6500 lbs. - 1/2 Na
2
S
PSE 476: Lecture 6 15

In-Class Example Calculations (3)
Na
2
S
Sulfidity =
Na
2
S + NaOH
• 100 = 25.2%
Na
2
S
Na
2
S + (6500 - 1/2Na
2
S)
= 0.252 =
Na
2
S
0.5 Na
2
S + 6500 lbs.
Na
2
S = 0.126 Na
2
S + 1638 lbs.
0.874 Na
2
S = 1638 lbs.
Na
2
S = 1874 lbs. (Na
2
O)
NaOH = 6500 lbs. - (0.5)(1874 lbs.) = 5563 lbs. (Na
2
O)
PSE 476: Lecture 6 16

In-Class Example Calculations (4)
Step 3: Convert NaOH and Na
2
S values from Na
2
O
Na
2
O = 62 g/mole or lbs./mole for this exercise
NaOH = 40 g/mole
Na
2
S = 78.1 g/mole
As we discussed in class, these calculations are based on an equivalence in sodium (Na). This means that Na
2
S and NaOH are equivalents but that NaOH is equal to 1/2 Na
2
O.
Na
2
S = 1874 lbs.
(Na
2
O)
• 1mole/62 lbs. • 78.1 lbs./mole = 2360.6 lbs.
NaOH = 5563 lbs. • 1 mole/62 lbs. • 2.0 • 40 lbs./mole = 7178 lbs.
So: Na
2
S = 2360.6/1200 ft 3 = 1.97 lbs./ft 3
NaOH = 7178/1200 ft 3 = 5.98 lbs./ft 3
PSE 476: Lecture 6 17