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Pulping and Bleaching PSE 476. Lecture #6 Kraft Pulping Chemicals. 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?.

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## Pulping and Bleaching PSE 476

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**Pulping and BleachingPSE 476**Lecture #6 Kraft Pulping Chemicals PSE 476: Lecture 6**Chemical PulpingAgenda**• Basic Description of Liquors & Process • White, Black & Green Liquors • Definition of Terms • Total alkali, Effective Alkali, Sulfidity, etc. • Why is everything on a Na2O basis? PSE 476: Lecture 6**Kraft Pulping:Definition of Terms**• White liquor. • Fresh pulping liquor for the kraft process containing NaOH, Na2S, 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**Lime**Kilm White Liquor Green Liquor Digester Recovery Furnace Black Liquor 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. PSE 476: Lecture 6**Typical Composition of Kraft Liquors*** Median concentrations as g/l as Na2O PSE 476: Lecture 6**Typical White Liquor Composition**Notes Page PSE 476: Lecture 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 Na2O). • Total Alkali: NaOH + Na2S + Na2CO3 + 1/2Na2SO3 (as Na2O). • 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: Na2S + NaOH (as Na2O) 100g/L PSE 476: Lecture 6**Na2S**Sulfidity = * 100% NaOH + Na2S NaOH Causticity = * 100% NaOH + Na2S Definition of Terms (US) • Sulfidity: 24-28% • Causticity: PSE 476: Lecture 6**NaOH**Causticizing eff. = * 100% NaOH + Na2CO3 Na2S Reduction eff. = * 100% Na2S + Na2SO4 + Na2SO3 + Na2S2O3 Definition of Terms (US) • Effective Alkali: NaOH + 1/2 Na2S (as Na2O) no more than 55 g/L • Activity: % ratio of Active to Total Alkali • Causticizing Efficiency: 78-80% • Reduction Efficiency: 95% PSE 476: Lecture 6**Why Na2O? (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, Na2S, etc which have different degrees of effectiveness • Reporting on a Na2O basis indicates the actual chemical relationship between these chemicals PSE 476: Lecture 6**Why Na2O? (2)**PSE 476: Lecture 6**Why Na2O? (3)**PSE 476: Lecture 6**Kraft Pulping Liquor**Sample Calculation PSE 476: Lecture 6**Kraft Pulping LiquorIn-Class Example Calculations (1)**• 50 Tons Chips • 50% Moisture Content • Liquor Charge to Digester: • 1200 ft3 white liquor • EA = 13% (alkali charge on OD wood as Na2O) • Sulfidity = 25.2% • 1300 ft3 black liquor • Question: How many lbs./ft3 of NaOH and Na2S were charged to the digester in the white liquor? (assume no chemical contribution from black liquor) PSE 476: Lecture 6**Kraft Pulping LiquorIn-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 Na2S as Na2O in the white liquor using the EA and Sulfidity numbers EA = NaOH + 1/2 Na2S = 13% on od wood. NaOH + 1/2 Na2S = 0.13 • 50,000 = 6500 lbs. NaOH = 6500 lbs. - 1/2 Na2S PSE 476: Lecture 6**Na2S**0.5 Na2S + 6500 lbs. Na2S • 100 = 25.2% Sulfidity = Na2S + NaOH Na2S = 0.252 = Na2S + (6500 - 1/2Na2S) Kraft Pulping LiquorIn-Class Example Calculations (3) Na2S = 0.126 Na2S + 1638 lbs. 0.874 Na2S = 1638 lbs. Na2S = 1874 lbs. (Na2O) NaOH = 6500 lbs. - (0.5)(1874 lbs.) = 5563 lbs. (Na2O) PSE 476: Lecture 6**Kraft Pulping LiquorIn-Class Example Calculations (4)**Step 3: Convert NaOH and Na2S values from Na2O Na2O = 62 g/mole or lbs./mole for this exercise NaOH = 40 g/mole Na2S = 78.1 g/mole As we discussed in class, these calculations are based on an equivalence in sodium (Na). This means that Na2S and NaOH are equivalents but that NaOH is equal to 1/2 Na2O. Na2S = 1874 lbs. (Na2O) • 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: Na2S = 2360.6/1200 ft3 = 1.97 lbs./ft3 NaOH = 7178/1200 ft3 = 5.98 lbs./ft3 PSE 476: Lecture 6

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