1. PSE 476: Lecture 61 Pulping and Bleaching PSE 476 Lecture #6 Kraft Pulping Chemicals Lecture #6 Kraft Pulping Chemicals

2. PSE 476: Lecture 62 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? 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?

3. PSE 476: Lecture 63 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). 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).

4. PSE 476: Lecture 64 Simplified Liquor Scheme 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. Digester White Liquor Black Liquor Recovery Furnace Lime Kilm Green Liquor

5. PSE 476: Lecture 65 Typical Composition of Kraft Liquors * Median concentrations as g/l as Na 2 O

6. PSE 476: Lecture 66 Typical White Liquor Composition Notes Page

7. PSE 476: Lecture 67 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). »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 S + NaOH (as Na 2 O) 100g/L 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). »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 S + NaOH (as Na 2 O) 100g/L

8. PSE 476: Lecture 68 Definition of Terms (US) Sulfidity: 24-28% Causticity: Sulfidity: 24-28% Causticity: Na 2 S Sulfidity = NaOH + Na 2 S * 100% NaOH Causticity = NaOH + Na 2 S * 100%

9. PSE 476: Lecture 69 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% Reduction Efficiency: 95% 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% Reduction Efficiency: 95% NaOH Causticizing eff. = NaOH + Na 2 CO 3 * 100% Na 2 S Reduction eff. = Na 2 S + Na 2 SO 4 + Na 2 SO 3 + Na 2 S 2 O 3 * 100%

10. PSE 476: Lecture 610 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 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

11. PSE 476: Lecture 611 Why Na 2 O? (2)

12. PSE 476: Lecture 612 Why Na 2 O? (3)

13. PSE 476: Lecture 613 Kraft Pulping Liquor Sample Calculation Kraft Pulping Liquor Sample Calculation

14. PSE 476: Lecture 614 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) 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)

15. PSE 476: Lecture 615 Kraft Pulping Liquor In-Class Example Calculations (2) 50 tons chips 2000 lbs./ton 0.5 (m.c.) = 50,000 lbs. o.d. wood EA = NaOH + 1/2 Na 2 S = 13% on od wood. NaOH = 6500 lbs. - 1/2 Na 2 S Step 1: Calculate the amount of oven dry 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 NaOH + 1/2 Na 2 S = 0.13 50,000 = 6500 lbs.

16. PSE 476: Lecture 616 Kraft Pulping Liquor In-Class Example Calculations (3) Na 2 S Na 2 S + (6500 - 1/2Na 2 S) = 0.252 = Sulfidity = Na 2 S Na 2 S + NaOH 100 = 25.2% 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)

17. PSE 476: Lecture 617 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 = 1.97 lbs./ft 3 NaOH = 7178/1200 ft 3 = 5.98 lbs./ft 3