Pulping and Bleaching PSE 476

Lecture #6 Kraft Pulping Chemicals

PSE 476: Lecture 6 1

Chemical Pulping 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

Kraft Pulping: Definition of Terms

• 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

Simplified Liquor Scheme

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

Typical Composition of Kraft Liquors

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

Typical White Liquor Composition

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

Definition of Terms (US)

• 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 S + Na 2 CO 3 1/2Na 2 SO 3 (as Na 2 O).

which contribute to active alkali.

+ » This is the sum of the sodium salts that contribute to or are converted during kraft cooking to chemicals • Active Alkali : Na 2 S + NaOH (as Na 2 O) 100g/L PSE 476: Lecture 6 7

Definition of Terms (US)

• 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

Definition of Terms (US)

• • • • Effective Alkali : NaOH + 1/2 Na 2 S (as Na 2 O) no more than 55 g/L 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

Why Na

2

O? (1)

• 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

Why Na

2

O? (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

Why Na

2

O? (3)

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

Kraft Pulping Liquor Sample Calculation

PSE 476: Lecture 6 13

Kraft Pulping Liquor

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

Kraft Pulping Liquor

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

Kraft Pulping Liquor

In-Class Example Calculations (3)

Sulfidity = Na 2 S 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

Kraft Pulping Liquor

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 NaOH = 7178/1200 ft 3 = 1.97 lbs./ft 3 = 5.98 lbs./ft 3 PSE 476: Lecture 6 17