The steam consumption per unit of soda production in our factory (please correct me if I’m wrong)
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The last edit to this post was made by sunjl1981 on 2013-1-7 at 00:01. Calculation of the steam consumption per unit volume in the evaporation section (Part 1): First evaporation stage: 1. Given: The concentration of the feed liquid in Stage I is approximately 10%, its specific gravity is 1.109, and the temperature ranges from 110 to 120°C; 115°C is used as the value； The effluent concentration is 15% ； The primary steam pressure is 5–6 kg/cm2; it is taken as 5.5 kg/cm2. The secondary steam pressure in the effector is 0.18 MPa ； The wall temperature of the effector is 150°C, while the temperature of the surrounding air is approximately 35°C℃ ； The volume of the liquid material output in one cycle is 35 m3, with a time duration of 20 minutes. The mass of the liquid output per batch is 38.8 tons, of which 3.88 tons is NaOH. Therefore, the time required to produce 1 ton of NaOH is 0.086 hours. The average specific heat of 10% NaOH solution is 0.906 kcal/kg ； At an evaporation pressure of 0.18 MPa, the latent heat of vaporization of water vapor is: 519.3 kcal/kg (Chemical Process Design Manual). The heat of solution for 10% NaOH is 298 kg, and that for 15% NaOH is 294 kg ； ΔHj: the crystallization heat of sodium chloride: approximately 19.9 kcal/kg (chlor-alkali production technology). The heat enthalpy of water at 154°C is approximately 155 kcal/kg, while that of water vapor at 154°C is approximately 657 kcal/kg (Chemical Process Design Manual). 2. Calculation: (1) Heat required to preheat the liquid to its boiling point: q1 = S0 × C × (t – t0) (Chlor-Alkali Production Technology, Volume 1). S0: Amount of feed liquid = 1000 ÷ 0.1 = 10,000 kg. C: Average specific heat of the liquid. t: Boiling point of the liquid in the evaporator. a: Boiling point of the electrolyte. Calculation formula: T = (1.0479 + 3.1633 × 10^-4 × Z) + 0.8207 × 10^-3 × Z^2 + 0.27444 × Z + 1.1744. Z = x1 × 100 / (1 – x1 – x2). x1: Concentration of NaOH ； x2: Salt concentration. (In chlor-alkali technology) At 100°C, a 10% electrolyte contains 19.96% NaCl, while at 15% it contains 16.2% (Volume 1 of Chlor-Alkali Production Technology). The calculations show that Z=14.2776 and T=128°C. The heat required is as follows: q1 = S0C(t-t0) = 10000 × 0.906 × 4.18 × (128 – 115) = 492320.4 kJ. (2) The heat required to evaporate the water: q2 = wγ, where γ is the latent heat of vaporization in Kcal/kg (Volume 1 of Chlor-Alkali Production Technology). The amount of water that evaporates in the evaporator: Initial water content X1: 10% = 1000 / (X1 + 1000 + 1996), so X1 = 8804 kg. Water content at the end of the process X2: 15% = 1000 / (X2 + 1000 + 1620), so X2 = 4047 kg. The amount of water that needs to be evaporated is 8804 – 4047 = 4757 kg. The heat required for evaporation is q2 = wγ = 519.3 × 4.18 × 4757 = 10325896.22 kJ. (3) Heat required for concentrating the solution: (for alkaline solutions) q3 = S0C2oΔHn (Volume 1 of Chlor-Alkali Production Technology). Here, S0C2o = 1000 kg, and ΔHn = 298 – 294 = 4 kcal/kg. Thus, q3 = S0C2oΔH = 4 × 4.18 × 1000 = 16720 kJ. (4) Heat released during crystal formation: q4 = WyΔHj (Volume 1 of Chlor-Alkali Production Technology). Wy represents the amount of salt that precipitates in the evaporator. Initial salt content in the solution: 1996 kg. Salt content at the end of the process: 1620 kg. Therefore, the amount of salt that precipitates is 1996 – 1620 = 376 kg. The heat of crystallization for sodium chloride: Since the precipitation of sodium chloride is an exothermic reaction, the value of ΔHj is negative. q4 = WyΔHj = -19.9 × 4.18 × 376 = -31276.4 kJ. (5) Heat loss from the equipment: q5 = αTFS(tw – tf). (From the first volume of Chlor-Alkali Production Technology) tw = Temperature of the equipment’s surface; tf = Temperature of the ambient air. α is the combined heat transfer coefficient that takes both convection and radiation into account. αT = 8.1 + 0.045(tw – tf) = 13.275. The heat dissipation area of the equipment: Based on the drawings of the evaporation unit, the heat dissipation area is calculated as FS = 106.8 m2. Therefore, q5 = αTFS(tw – tf) = 13.275 × 106.8 × (150 – 35) × 4.18 = 265711.94 kJ/h. The total heat loss is then 265711.94 × 0.086 = 22854.23 kJ. The total amount of heat required is: 492320.4 + 10325896.22 + 16720 – 31276.4 + 22854.23 = 10826514.45 kJ. (6) Steam consumption: q = D(I – θ). (From the first volume of Chlor-Alkali Production Technology) I is the enthalpy of the heating steam. At a pressure of 5.5 kg/cm2, the temperature of the steam is: T = 159.0504 + 27.3028ln(p/10) + 2.05224p + 1.25981/p = 154°C. The steam consumption is given by 10826514.45 = D × (657 – 155) × 4.18. From this, it can be calculated that D = 5159.5 kg/h &
It seems that our plant consumes a large amount of steam. I remember reading an article on the chlor-alkali industry; it said that their plant used only a little over 2 tons (I can’t remember exactly). Although we evaporate 45% alkali, that does seem like a rather high amount, doesn’t it? Please help and check where there are mistakes.
It’s the evaporation of diaphragm alkali; waste heat utilization isn’t taken into account, I guess. The fuel consumption won’t be this high
Yes, but that calculation only takes into account the heating in the first stage; the secondary steam generated in the first stage is used, and the condensate water from the steam produced in the first stage is used to heat the brine. In my calculations, I used the temperature of the brine after it has been heated, so I did not take into account the utilization of residual heat. However, I really haven’t considered whether the condensate water after heating the brine can still be utilized for its residual heat, nor the condensate water from the secondary steam. Thank you for pointing this out. I was wondering: apart from secondary steam, raw steam, and the condensate from secondary steam, do you know of any other areas where waste heat can be utilized?