# Sulfuric Acid Plant Calculation

**Calculation:**

During the operation management of the sulfuric acid plant, or you need a huge amount of computational work at the time of new construction and renovation of facilities. Here I am not going to describe all of them. Please refer to a relatively opportunities are many, and so a brief description of some for a useful calculation.

- Dilution during the amount of water in the sulfuric acid
- Ditto acid temperature
- Heat loss from the equipment
- Emissions possible SOx amount (K value regulation)
- Etc.

**Of sulfuric acid dilution water :**

1) concentrated sulfuric acid (98.3% H2SO4) was diluted with water, Calculate the amount of water when making 62.5% H2SO4.

Traditionally, for each amount of sulfuric acid for obtaining a concentration of the target by mixing different concentrations of sulfuric acid, a convenient method of "square law" is famous. This is, each of sulfuric acid concentration A% before mixing and B%, and a sulfuric acid concentration C% of the target place as described below, and the subtraction in an oblique direction, respectively.

A C - B C B A - C

As a result, (CB) :( AC) = concentration A% of the amount of sulfuric acid: It becomes the concentration B% of the amount of sulfuric acid. For example, 98% H2SO4 (A) and 20% H2SO4 (B) are mixed with 94% H2SO4 when making the (C) is CB = 94-20 = 74 AC = 98-94 = 4 In other words, it is the result that should I you mixed the 98% sulfuric acid = 74 parts and 20% sulfuric acid = 4 parts. Therefore, the solution of the question

98.3 62.5 62.5 0 35.8

To 98.3% 62.5 parts of sulfuric acid, and it will may be added 35.8 parts of water (57.8% of concentrated sulfuric acid). * In the case of mixing of fuming sulfuric acid are each displayed in all SO3 amount, and then apply the rectangle method in the same manner. For example, "a mixture of fuming sulfuric acid and 94% sulfuric acid of 25% free-SO3, 98% get a sulfuric acid" case:

86.22 3.27 80.0 76.73 6.22

∴25% freeSO3 fuming sulfuric acid: 94% sulfuric acid = 3.27: 6.22 I will be mixed in a ratio of.

**At the time of diluted sulfuric acid temperature :**2) in the case of the above example, Calculate the temperature of the product 62.5% sulfuric acid. 98.3% sulfuric acid is 40 ℃, the water I and 25 ℃.

This is First, you must calculate the heat of dilution of sulfuric acid. There are several ways, engineering sulfuric acid (moat, Nakagawa AD, Inc., 1967) According to the, If you want to dilute the H2SO4 of 1 mol of water n mol, the dilution heat Q = 17860n / (n + 1.7983) ··· (1) Q: cal / mol-H2SO4 n: mol-H2O / mol-H2SO4 When generation to the sulfuric acid concentration = 100A% A = 98.08 / (18.02n + 98.08) Therefore, n = (98.08 / 18.02) (1 / A-1) When you assign this to the equation (1) Q = 97209 (1-A) / (5.4428 -3.6445A) Therefore, 100% H2SO4 → 98.3% H2SO4 dilution heat of: Q1 = 97209 * (1-0.983) / (5.4428 -3.6445 * 0.983) = 888cal / mol-H2SO4 100% H2SO4 → 62.5% H2SO4 dilution heat of: Q2 = 97209 * (1-0.625) / (5.4428 -3.6445 * 0.625) = 11,518cal / mol-H2SO4 ∴98.3% H2SO4 → 62.5% H2SO4 of dilution heat Î”Q = Q2 - Q1 = 10,630cal / mol-H2SO4 = 108 kcal / kg-H2SO4 98.3% When will the amount of the sulfuric acid and 1kg, Heat of dilution Hd = 108 * 0.983 = 106kcal 98.3% held by the sensible heat of sulfuric acid H1 = 0.344 * 1 * 40 = 14 kcal Water holding sensible heat H2 = 1 * 0.578 * 25 = 14 kcal ∴ held sensible heat of the generated 62.5% sulfuric acid H3 = 106 + 14 + 14 = 134kcal Assuming this temperature and 120 ℃, specific heat = 0.553kca / kg / ℃, generating 62.5% sulfuric acid = 1.578kg so Generated 62.5% sulfuric acid temperature = 134 / 1.578 / 0.553 = 153 ℃ Boiling point of 62.5% H2SO4 is so 145 ℃, it will be sulfuric acid boil. In such a case, and to cool the sulfuric acid immediately after mixing in a shell-and-tube cooler, it will be that the heat exchange is hindered because of the bubble of steam in the tube, and diluted and coolingis divided into two stagesHow it is taken. → "Various sulfuric acid production process", " dilute sulfuric acid See " For example, * first stage 98.3% → 80% * The second stage 80% → 62.5% As a method of calculating the heat of dilution, there is also a way to use a chart of Othmer and Fusullo et al. Numerical value of the data to be used is different in each subtle, practically no problem. (Of course, such as the heat transfer area of cooler you should look at a sufficient margin)