Introduction
The industrial process to manufacture sodium carbonate is also known as ammonium soda process. Sodium carbonate (Na2CO3) common name is soda ash, and it has a wide economic importance. According to Garrett (1992), this product is commonly used in the manufacture of chemicals, glass, paper, detergents and other products. Brine, a natural sodium chloride solution contains impurities in form of calcium and magnesium ions. Hence, brine is extracted and purified through filtration to remove those impurities. Ammoniacal brine is carbonated to obtain a suspension of sodium bicarbonate. After the filtration of Sodium carbonate the residual is treated with slaked slime (hydroxide solution), this assist in recovering of ammonia from ammonium salts which includes carbonate, ammonium chloride, bicarbonate and other salts. At the absorption stage, the recovered ammonia is recycled into the brine. After filtration, the sodium bicarbonate yielded is washed, dried, and through calcinations sodium carbonate also known as soda ash is obtained (Garrett 1992).
Sodium Carbonate
Sodium carbonate is mostly prepared by a process called ammonia soda. The process starts when carbon dioxide is bubbled through a brine solution with high concentration of ammonia. The result is the formation of sodium hydrogen carbonate. The formed sodium hydrogen carbonates precipitates and filtered to get sodium carbonate through ignition. This also yield carbon dioxide and water. In the first reaction it is done in carbonators in presences of CO2. Sodium bicarbonate precipitates out of the solution as crystal. In the second reaction, sodium carbonate is recovered from the sodium bicarbonate by heating and driving off both carbon dioxide and water. Some assumptions have been considered to facilitate the solving of some problems. The first assumption is that there is no material lost during the production process. Actually, some material are lost in the process through units, but in this project this fact was neglected. In addition, it is assumed there is no energy lost, through heat exchangers and other units. Carbonation system is adiabatic system. Nevertheless, we still find that heat is added or removed from the steam leaving the precarbonator before being fed to the first carbonator. Furthermore, the temperature of the brine, which leaves carbonators, needs to be cooled before being passed to filter unit, thus it is cooled on three stages and enter to filtrate unit. Hence, some assumptions in the calculations to solve problems in the project have been adopted.
In conclusion, there exist many ways by which the sodium carbonate can be manufactured. In this project, soda ash is produced from the brine. This process is considered to be economical and hence very common in manufacturing sodium carbonate. This process also produces other by products, and carbon dioxide produced is recycled, and this reduces cost. The process involved in this process undergoes complex energy balances, which show how much material, and flow rate is needed to enter and exit from each unit. The running cost of the plant is dictated by the value of energy required for heat exchanger and other units. There is energy required for pumping the brine to the storage. In addition, the flow rate of sodium carbonate with water (Na2CO3.H2O) in the last step of production soda ash is180563.8 Kg/hr. Due to high late of cooling, the plant is set near a river where large supply of water for cooling units is needed. The steam produced is also used to generate power.
Recommendations
From this process, we have found that there is no difference between temperatures from the outlet precarbonator and carbonators. Hence, when considering cooling precautions, it is not necessary to in the carbonator.
Questions
Q1. Construct a detailed flowchart of the process, showing known compositions, flow rates and a final flowchart to be prepared when you have finished the case study ?
Q2. Why is sodium carbonate converted to sodium bicarbonate in the process, only to be reconverted later back to sodium carbonate ?
Q3. Obtain a recent value of the selling price for soda ash and estimate total sales in the United States ?
Q4. Estimate the rate (kg/h) at which sodium bicarbonate exits the second carbonation reactor in both crystalline and solution form. Assume the filtrate leaving the first filtration stage is saturated with sodium bicarbonate and estimate the crystal production rate in kg/h ?
Q5. What are the production rates (kg/h) of light ash and dense ash that can be expected from this process? Convert your answers to metric tons per year. What is the yield of sodium carbonate as a percentage of that fed to the process?
Q6. At what rate (kmol/hr) is CO2 generated in the reaction of sodium bicarbonate to soda ash? At what rate (kmol/h) must fresh CO2 be supplied to the process.
Q7. Determine the rate (kmol/h) at which carbon dioxide is absorbed in each of the carbonators and in the precarbonator. You may assume that all of the CO2 absorbed reacts with sodium carbonate. What is feed rate (kmol/h) of carbon dioxide to the carbonators?
Q8. Suppose that the fresh CO2 is produced by the combustion of coal, whose ultimate composition is given in Table 4. Estimate the rate at which the coal must be burned to completion to generate the required amount of carbon dioxide. Determine the rate at Which SO2 is emitted from the combustion process in kilograms per metric ton of coal and in metric tons per year.
Table 4
Ultimate analysis of coal used in steam
Generation
Element Wt%
Carbon 52.5
Hydrogen 4.8
Oxygen 18.3
Nitrogen 1.2
Sulfur 0.6
Ash 22.6
Q9. Determine the volume of the carbonation reactors. Note that entertainment of liquid in the gas exiting the reactor is directly related to the velocity of that gas. Calculate the velocity (m/s) of the gas flowing through the free space above the threephase mixture in the reactor as a function of reactor diameter.
Q10. Estimate the composition of the liquor or slurry streams leaving the precarbonator and each carbonator. Using the solubility data for sodium bicarbonate, indicate whether each of these streams is a liquid or a slurry. If your calculations of the show that a stream is a slurry, determine the solids fraction and the composition of the liquid portion of the stream.
Q11. Prepare plots of weight fractions of NaCl and Na2SO4 in liquor adhering to the sodium bicarbonate crystals as a function of the number of wash stages. How many stages are required to reduce the sodium chloride content of the filter cake to less than 0.12 kg NaCl per 100 kg NaHCO3? Determine the amount of wash liquor required to wash the sodium bicarbonate crystals, expressing your answer as kg solution /kg NaHCO3 crystals. Suggest possible sources of the wash liquor. How is the production rate of soda ash changed if some of the sodium bicarbonate is used to produce the 85% – saturated wash liquor? Obtain a revised estimate of the required feed of fresh CO2.
Q12. Determine the feed rate (kg/h) of fresh water to the sodium carbonate monohydrate crystallizer, assuming no losses by evaporation.
Q13. Estimate the requirements (kW) for pumping the brine from its subterranean source to storage tanks whose height is 10 m. Assume the pumps have an efficiency of 55% that the friction loss is 10% of the potential energy change, and that kinetic energy changes can be neglected. (Hint: Perform a mechanical energy balance, taking the first point to be at the upper surface of the underground lake at 1 atm and the second point at the upper surface of the storage tank) .
Q14. Determine the temperature of the brine as it leaves each of the exchangers in the threestage crystallization operation.
Q15. At what rate (kg/h) does water leave in the gas discharge from the precarbonator if the gas is saturated at the temperature of the liquid entering this vessel?
Q16. As indicated, the carbonation reactors are operated so that the slurry leaving the second unit is at 60^{o}C. Recalling that the precarbonator and each carbonation reactor function adiabatically, to what temperature must the feed to the first carbonation reactor be adjusted? At what rate must heat be added to or removed from the stream leaving the precarbonator before it is fed to the first carbonation reactor? What is the temperature of the product from the first carbonation reactor? Determine the amount of steam or cooling water required using the following description of the utilities: If heat addition is required, assume that the heat is supplied by saturated steam at 3 bars absolute; if cooling is required, use cooling water at 30^{o}C and allow a temperature increase in the water of 15^{o}C.
Q17. Suppose both carbonation reactors were to operate at 60^{o}C. To what temperature would the liquid feed to each have to be adjusted? At what rate would heat have to be removed from or added to these feed streams to obtain the required temperature?.
The following Assumptions has been applied in solving the questions
 We assume that the conversion of Na_{2}CO_{3 }to NaHCO_{3} _{ }is 100%.
 The exit of any moisture in the precarbonator is negligible.
 In the first filtration stage all the filtrate is saturated with sodium bicarbonate .
 In the carbonator the amount of NaHCO_{3} solids dissolved is unknown
 All impurities are removed
 In the filter cake is the washing process is constant at1.97% wt.
 All of the CO2 absorbed reacts with sodium carbonate.
 The system is assumed to be closed
 Kinetic energy changes are neglected in the equation given below
 Friction loss is 10% of the potential energy change.
 The pumps have an efficiency of 55%
 There is presence of Steady state in the system
 The exit of any moisture in the precarbonator is negligible.
 The percentage of dissolved NaHCO_{3 }is negligible.
 There is presence of steady state.
 Enthalpy changes due to crystallization is withdrawn or added to the solution depending on the amount of crystal formed.
 Heat flow rate of mother liquor is equal to the heat flow rate of the brine
 Sodium chloride is totally dissolved in the water.
Answers to Question 2
The reason why Sodium carbonate is converted to sodium bicarbonate in the presence of CO_{2} and later be converted back to sodium carbonate is to precipitate sodium bicarbonate out of the solution as crystals and produce other byproducts, which are recycled in the process and hence reduce cost. The process is reversed by calcination, which drives off carbon dioxide and water.
Answer to Question 3
Soda ash prices in U.S.A are unstable and vary throughout the year. This is because of law of demand and supply in the USA and global markets. In 2011 March, the price of soda ash in America varied from $260 to $265 per ton, courtesy of Australia chemical consultancy company. The average total price of American soda ash if USA produces eight million tons would be;
8,000,000 * ($265+$260)/2= 8,000,000*262.5
=2,100,000,000 $/ tons
Answer to question 4
The constants at are given as;
Let Volume be v, Specific gravity be gr, Mass flow rate be m
Hint; Mass flow rate of brine = density x volume
(v) = 38 m^{3}/min
Specific gravity of the brine = 1.25
The density of brine = 1.25 x 1000 kg/m^{3 }=1250 kg/m^{3}
Mass flow rate of brine (m^{.}) = (38 m^{3}/min) x (1250 kg/m^{3} )
= 47500 kg/min
Weight of compound (Wt) = Mass flow rate x percentage weight
= 47500 x 6.2 =294,500.
The constants are given below for moles calculation
component  Wt%  Wt(kg/min)  MW(kg/kgmol)  Kgmol/min 
Na_{2}CO_{3}  6.2  2945  106  27.78 
NaHCO_{3}  0.6  285  84  3.39 
NaCl  19  9025  58. 5  154.3 
Na_{2}SO_{4}  4  1900  142  13.38 
H_{2}O  70.2  33345  18  1852.5 
From reaction 1.
 Given that 1 kgmole of Na_{2}CO_{3 }is 27.78, then in 2 kgmole of NaHCO3 assuming that the conversion is 100% then
27.78 x 2 = 55.56 kgmole/min of sodium carbonate produced.
Total Sodium bicarbonate out = 55.56+3.39 = 58.95 kg mole/min
 Total water consumed 1 kg mole of H_{2}O is 2 kg mole of NaHCO_{3}
= 55.56/2
= 27.78 kg mole/min
Moles of H_{2}O in brine (1852.5) (from the above diagram) – moles of H_{2}O consumed (27.78)
Therefore, 1852.527.78 = 1824.72 kg mole/ min
 The number of moles of Carbon dioxide Absorbed
1 kg mole of CO_{2} = 2 kg mole of NaHCO_{3}
= 55.56/2 = 27.78 kg mole/ min
Mass flow rate of NaHCO_{3} = 58.95 x 84 x 60
Mass flow rate of NaHCO_{3} = 297108 kg/hr
 The solubility of sodium bicarbonate in brine at 60°C in the second carbonator is 2.3 Wt%, then
Mass flow rate of water/hr = 1970697.6 kg/hr
Mass flow rate of sodium bicarbonate = 297108 kg/hr
Soluble sodium bicarbonate = y /(y + 1970697.6) = 0.023
Making y the subject of the above equation, then we get;
y = 46393.085 kg/hr
y is the soluble sodium bicarbonate.
 Crystalline of sodium bicarbonate = 297,108 46,393.08
= 250,714.92 kg/hr.
 At 27 °C (third crystallizer) = 1.97
Soluble sodium bicarbonate flow mass rate is
= y / (y+1,970,697.6) = 0.0197
Making y the subject of the formula we get y= 39,602.92 kg/hr
 Crystals of sodium bicarbonate
= 29,7108 – 39,602.92 = 257505.08 kg/hr
total solution of NaHCO_{3} , NaCl , Na_{2}SO_{4} , H_{2}O
Total solution flow rate = 39602.92 + 1970697.6 + (114,000) + (541500)
= 2665800.5 kg/hr
X 100 % soln
Z = 2,021,225.52 kg/hr Filter Y
Crystal = 257,505.08 kg/hr 91% solid Crystal.
9% Solution
 Value of Y is given as; 257505.08 = 0.91Y
Y = 257505.08/ 0.91 = 282972.6 kg/hr
Material balance for solution will be given by the calculation,
2,021,225.52 = X + 0.09Y
2,021,225.52 = X + (0.09 x 282972.6 )
X = 1995757.99 kg/hr
Flow rate of crystals
= 282972.6 x 0.91 =257505.07 kg/hr
Flow rate of solution
= 282972.6 x 0.09= 25467.53 kg/hr
Answers to question 5
Liquor 85% Saturated Steam
NaHCO_{3 }
Dryer

Sol.= 25467.53 kg/hr Washing 27°C Bleaching Na_{2}CO_{3 } Na_{2}CO_{3}.H_{2}O Facility Facility
Cry. = 257505.07 kg/hr
Steam
Washed H_{2}O Purge & 55% 90% CO2 off gas
Solution of NaHCO_{3 }
2 NaHCO_{3} (s) _{ } Na_{2}CO_{3 }(s) + H_{2}O (g) + CO_{2 }(g)
Moles of NaHCO_{3} = kg/hr we divided by MW
= 257505.07/ 84 (kg/kgmol)
= 3065.5 kgmole/hr
Mass flow rate = Moles x MW
=3065.5 x 84
= 257502 kg/hr
2 kgmole of NaHCO_{3 }(s) = 1 kgmole of Na_{2}CO_{3}
Moles of Na_{2}CO_{3}
= 3065.5/ 2 = 1532.8 kgmole/ hr
Mass flow rate of Na_{2}CO_{3} left= Moles of Na_{2}CO_{3 }x MW (Na_{2}CO_{3 })
= 1532.8 x 106
= 162476.8 kg/hr
1 kgmole of Na_{2}CO_{3 }=1 kgmole of Na_{2}CO_{3}.H_{2}O
The ratio is 1:1 hence, 1 kgmole of Na_{2}CO_{3}.H_{2}O is also1532.8 kgmole/ hr
Flow rate of Na_{2}CO_{3}.H_{2}O =Moles x MW
= 1532.8 x 124 = 190067.2 kg/hr
95% of Na_{2}CO_{3}.H_{2}O will be sent to the dryer
= (95/ 100) x 190067.2 = 180563.8 kg/hr
The number of moles of Na_{2}CO_{3}.H_{2}O will be given by 180563.84/124 = 1456.16 kgmole/hr
Na_{2}CO_{3}.H_{2}O _{ } Na_{2}CO_{3 }+ H_{2}O
moles ratio of Na_{2}CO_{3}.H_{2}O Na_{2}CO_{3 }is 1:1 we get
Mass flow rate of anhydrous Na_{2}CO_{3} left = Moles x MW
=1456.16 x106 = 154352.96 kg/hr
Hence, the mass flow rate of Na_{2}CO_{3}.H_{2}O left in tons = moles x duration
=180563.8 kg/hr x 24hr x 365day
= 1581738.9 ton/ year
Mass of anhydrous Na_{2}CO_{3} left = (154352.96 kg/hr x 24hr x 365
= 1352131.9 ton/ year
Yield = Mass flow rate of anhydrous Na_{2}CO_{3} left divided by Wt in percentage
=154352.96 / 2945 x 60) % = 87.4%
Answers to Question 6
NB
Inlet 100% CO_{2 }
Out let 10%
NaHCO3 at 27.78 kgmol/min_{ }
Hint: There is 91 wt% solid crystals and 9 wt% solution in the filter cake.
Mass flow rate of crystals from filter cake
= 282972.6 x 91 % = 257505 kg/hr
Mass flow rate of dissolved crystals in the solution
= 282972.6 x 9 % = 25467 kg/hr
The Solubility of sodium bicarbonate is 1.97% at 27 degrees.
Mass flow rate of NaHCO_{3}
= 25467 x 1.97 % = 501.7 kg/hr
The total mass flow rate of NaHCO_{3} in the dryer
Mass of crystal + flow rate of NaHCO_{3}
_{ }= 501.7+257505=258006.7 kg/hr
Moles of NaHCO_{3 }is flow rate divided by MW
_{ } = 258006.7/84 = 3071.5 kgmole/hr
From the equation we find that
2 kgmole of NaHCO_{3 }= 1 kgmole of CO_{2}
1 kgmole NaHCO_{3} is 3071.5 kgmole/hr
1 kgmole of CO_{2} is 3071.5/ 2 = 1536 kgmole/hr
There is 10% of CO_{2 }is lost, so 1536 x 10% = 153.6
Remaining is moles
= 1536 153.6 = 1382.4
Then, Total of CO_{2 } fed =
=1852.2– 1382 = 469.8 kgmole/hr
Answers to question 7
In pre carbonator
 Intake co2 is 10%
 Outlet CO2 is 15 %
Carbonators
 CO2 is 85 %
 From reaction 1.
The moles of Na_{2}CO_{3} to CO_{2 }is 1:2
Hence, the number of CO_{2} is 27.78 kgmole/min
The number of moles of CO_{2 }in the feed
= 27.78 is 90%
% = 30.87 kgmole/min
10% of CO_{2} is lost to the surroundings and the remaining 90% from the given data, then
Number of CO_{2} moles lost is 30.87 x 10 % = 3.09 kgmole/min.
At 85% wash liquor is saturated with sodium bicarbonate at 27°C and refiltered.
CO_{2 }absorbed in both carbonator = 30.87 x 85 % = 26.24kgmole/min
For each carburetor it is 26.24/2 = 13.12 kgmole/min
= 13.12 x 60 = 787.2 kgmol/hr
Since 5% is moles of CO_{2 }absorbed in the precarbonator
= 30.87x 5% =1.54kgmol/hr
Per hr = 1.54 kgmole/min x 60min/1 hr
= 92.4 kgmol/hr
Answers to Question 8
Table 4
Ultimate analysis of coal used in steam
Generation
Element Wt%
Carbon 52.5
Hydrogen 4.8
Oxygen 18.3
Nitrogen 1.2
Sulfur 0.6
Ash 22.6
From the data given above the two reactions takes place
S+ O_{2} SO_{2 } (1)
C+ O_{2} CO_{2 } (2)
The ratio of S to SO_{2 }is 1:1 hence,
Number of moles SO_{2} =0.001875 kg mole/hr
Mass flow rate of SO_{2}
=0.0001875 x 64 = 0.012 kg/hr
In question 6 above the number of fresh CO_{2 }required is 469.8 kg moles/hr
Hence, 1 kg of coal is 0.043 kg of CO_{2}
Therefore, 496.8 kg/hr of coal is yielded
Mass flow rate of coal required is 10925.6 kg/hr
Coal required in metric ton/ year
= 15891.8 kg/hr x 24hr x 365
= 139212.168 metric ton/ year
The number of moles SO_{2} emitted. The ratio of coal to SO2 is 1: 0.012
10925.6 kg mole/hr of coal = 0.012 x 10925.6 of SO_{2}
= 131.1 kg mole/hr
Mass flow rate SO_{2} emitted
= 131.1 x 64 = 8390.4kg /hr
Metric ton of coal = (10925.6 kg/hr x 24 hr x 365
= 95708.3 metric ton/year
Rate of SO2 emitted from the combustion process in (kg/ metric ton) of coal= (1149312kg/year of SO_{2})/ (95708.3 metric ton/year coal) = 12 kg SO2/ metric ton coal
Mass flow rate of SO_{2} emitted (kg/year)
= 131.1 kg/hr x 24 x 365
= 1148436 kg/yr
Mass of carbon (C)
= 52.5/ 100 = 0.52 kg/hr
The number of moles (C)
= 0.52/ 12 = 0.043 kg mole/hr
The ratio of C to CO_{2 }is 1:1 hence, no of moles of the number of moles CO_{2} = 0.043 kg mole/hr
Mass flow rate of CO_{2 }
_{ }= 0.043 x 44=1.892 kg/hr
Mass flow rate of Sulfur (S)
= 0.6/ 100 = 0.006 kg/hr
The number of moles (S)
= 0.006/32= 0.000188 kg mole/hr
Answers to question 9
Hint: resident time for (slurry volume / slurry volumetric flow rate) = 45 min
: Volumetric flow rate = 38 m^{3} /min
: The volume of three phases occupies 1.25 times the two phases
: 20% of the volume is free
Total volume in the reactor is
= 38 m^{3} x 45 min x 1.2 x 1.25 = 2565 m^{3}
Flow rate of unabsorbed CO_{2} in each carbonator
= 30.87/2 – 13.12 kg mole/min x 1min/60sec
= 0.0386 kg mole/sec
From our assumption we have the ideal gas i.e. PV=Zn RT , and Z = 1 (ideal gas)
PV= n RT
Universal gas constant (R) = 8.31 kP m^{3}/kg mole.k
Velosity = volumetric flow rate divided by the area
Pressure (p) = 2 x 100 kpa
=200 kpa
Moles = 0.0386 kgmole/sec
Universal gas constant (R) = 8.31 kP m^{3}/kg mole.k
The temperature of the second carbonator = 60°C +273=333K
volume of the gas
(V) = nRT /P
By replacing the constants
Volumetric flow rate of gas (V)
= (0.0386 x 8.31 x333 K)/ 200
= 0.534 m^{3}/ sec
We know that the Velocity of gas = volumetric flow rate/Area
By calculation, we get Velocity of gas
= (0.534 x 4)/ π r^{2}
= 0.6713 r^{2} / ms
Answers to Question 10
From reaction 1.
The following reaction takes place.
Na_{2}CO_{3} (aq) _{ }+ CO_{2} (g) +H_{2}O (L) 2 NaHCO_{3 }(aq) _{ }
_{ }
15% of CO_{2}
Precarbonator
5% absorbed

Na_{2}CO_{3}, NaHCO_{3 }NaHCO_{3} , H_{2}O
H_{2}O, Nacl, Na_{2}SO_{4} Nacl, Na_{2}SO_{4}
10% of CO_{2}
Given that, the number of moles CO_{2} is 30.87 kgmole/ min
15 % of CO2 =30.57 x 0.15
= 4.63 kgmole/min
10 % CO_{2} out = 30.57 x 0.1
= 3.06 kgmole/ min
The Stoichiometric coefficient of CO_{2} = 1
= (3.06 4.63)/1 = 1.54
Material balance of water (H_{2}O) will be given as
=1852.5 1.54
= 1850.96 kgmole/min
Material balance of Na_{2}CO_{3 }
= 27.781.54
= 26.24 kgmole/min
Material balance of NaHCO_{3} in solid state.
The number of moles Na_{2}CO_{3} out (n _{Na2CO3})
= 2 x 1.54
= 3.08 kg mole/ min
Mass flow rate of water
=1850.96 x 18.016
= 33346.896 kg/ min
Dissolved sodium bicarbonate
= x /(x +33346.896)] = 0.023
= 785.034 kg/ min
By calculation the dissolved sodium bicarbonate
= 785.034/ 84
= 9.3456 kg mole/ min
The Mass flow rate of Na_{2}CO_{3 }solids
= 26.24 x 106
= 2781 kg/ min
Mass flow rate of NaHCO_{3} solids
= 3.08 x 84
= 258.72 kg/ min
Weight percent
The % of Na_{2}CO_{3 } = 2781/(2781+258.72)]
= 0.9148 x 100
= 91.48%
The % NaHCO_{3 }= 258.72/(258.72+2781)
= 0.08511x 100
= 8.51%
The % of Impurities (Nacl, Na_{2}SO_{4})
= (1900+9024)/(1900+9024+33346.896)
= 0.24675 x100
=24.675%
% H_{2}O = (33346.896/44270) x 100
=75.326 %
Mass flow rate of water
= mass of water Mass flow rate of (NaCl)
= 33346.896 9.35
= 33337.546 kg/min
This illustration below indicates how the above processes take place.
 CO_{2}5% at inlet
 Na_{2}CO_{3} = 26.24 kgmole
 H_{2}O, Nacl, Na_{2}SO_{4} at 96 kgmol
 CO_{2} at 7.5% at outlet
Explanation
Total CO_{2} entering to each carbonator is given by 42.5% , then
The number of moles of CO_{2} = 0.425 x 30.87
= 13.12 kgmole/min
The ratio of CO2 to H2O is 1:1
moles of water reacted =13.12 kgmole/ min
The number of moles of water entering
=1850.96 kgmole/ min
The number of moles of water getting out.
=1850.96 – 13.12 = 1837.86 kgmole/ min
For Na_{2}CO_{3}:Ratio of CO2 to NaCO3 is 1:2
No. of moles of Na_{2}CO_{3} reacted
= 2 x 13.12
= 26.24 kgmole/ min
Moles of Na_{2}CO_{3} out
=26.2413.12 = 13.12 kgmol/min
For NaHCO_{3}: the ratio is CO2 to NaHCO_{3 }is 1:2
No. of moles of NaHCO_{3} reacted
= 2 x 13.12
= 26.24 kgmol/ min
No. of moles of NaHCO_{3} in
= 3.08 kgmol/ min
No. of moles of NaHCO_{3 }out
= 3.08 +26.24 = 29.32kgmol/ min
Mass flow rate of water H_{2}O
= 1837.84 x 18 = 33095.4 kg/ min
Moles of X in dissolved in NaHCO_{3 }
= x/(x + 33099.5) = 0.023 , then
The number of moles of NaHCO_{3 }dissolved
= 779.21 kgmole/ min
Mass flow rate of NaHCO_{3 }dissolved
=779.21/ 84 = 9.276 kgmol/ min
Solids composition of the slurry out of carbonator 1
Mass flow rate of NaHCO_{3 } out
_{ }= 13.12 x 106 = 1390.72 kg/ min
% wt of Na_{2}CO_{3 } = (1390.72/(1390.72 +2462.8)) x 100
= 36.082 %
Mass flow rate of NaHCO_{3 }
_{ }= 29.32 x 84 = 2462.88 kg/ min
% NaHCO_{3 }= 100% – 36.1 % = 63.9 % wt_{ }
The weight percent of water = Mass flow rate of water/ total mass flow rate of liquid
Mass flow rate of water =33095 kg/min
% Water = [33095/ (33095 +9024+1900)]
= 0.752 x 100 = 75.2 %
% Impurities = 100% – 75.2% = 24.8%
No. of moles of Slurry out =13.12 –13.12=0 hence no slurry out
Moles of NaHCO_{3} entering = 29.32 kg mole/min
Moles of NaHCO_{3} reactant = 26.24 kgmol/min
Mole out = 29.32 +26.24
= 55.56 kg mole/min
Weight percent of water = 75.2 %
Impurities = 100% – 75.2% = 24.8%
No. of moles of water out in the liquid (mixture) = Moles (in) – Moles (reactant)
No. of moles of water out
=183713.12=1823.88 kgmole/min
Mass flow rate of water
=1823.88 x18 =32829.84 kg/ min
Mass flow rate of water
= 32829.84 x 60 = 1969790.48 kg/ hr
Answer to Question 11
0 
0.1 
0.2 
0 
1 
2 
3 
4 
5 
6 
Number of
Wash stages 
NaCl & Na_{2}SO_{4} in solution 
Figure 2 Number of wash stages required to reduce the sodium chloride.
Figure 2. Shows the number of stages required to reduce the sodium chloride content of the filter cake to less than 0.12 kg NaCl per 100kg NaHCO_{3}.
In the figure, it is evident that the fraction of NaCl and Na_{2}SO_{4} decrease gradually with increasing the number of wash stages until becomes zero at stage five.
Hint: Specific graphite of sodium bicarbonate filtrate is given1.21 kg/m^{3}
Part of dissolved NaHCO_{3 }in 9% solution given the Specific graphite = 1.21 kg/m^{3}, is negligible given that solubility is 1.97% wt at 27°C
Number of wash stages required
Data provided indicates that, wt% of water and the impurities = 70.2 +19+4 = 93.2%
weight fraction of sodium chloride
= 9025/(33345+9025+1900) = 0.204
Mass of sodium chloride = 0.204 x 25467
= 5095.268 kg per 257505 kg of baking soda
5093.4 kg of NaCl / Since 55% remaining in the filter cake (data provide), then
Each wash is reduced to 45%
Calculating the number of wash
Wash one = 45% x 5191
=2335.95 kg left per 257,505kg of baking soda
Wash two = 45% x 2335 = 1050.75 kg left
Wash three = 45% x 1050.75 = 472.837 kg left
Wash four = 45% x 472.837
= 212.78 kg
The last wash is four (4) because the left is 212.78 kg and it is less than 309 kg
From the data provided
The volume of wash liquor = 2 volume of solution
The density of solution (ρ) = 1000 kg/m^{3}
Calculating the volume of the solution we get
= 25467/1000 = 25.467 m^{3}
The volume of wash liquor = 25.467 x 2
= 50.93 m^{3}
The mass of the liquor
Mass of the liquor = 50.93 X 1000 = 50930 kg
Hence, 50930 kg liquor/ 257505 kg of baking soda
Therefore, the production of baking soda will be decreased if some of baking soda is used as wash liquor since the wash liquor is cleared out.
Answer to Question 12
Flow rate of fresh Water is given by (F) = P – F
_{ }=180563.8162476.8 = 18087.8 kg/hr
Answers to Question 13
Steady state energy balance equation is
∆H + KE + PE = W + Q
Energy required to pump the brine (E)
PE = mg∆h
= 791.67 x 9.81 x 10
= 77662.827 W 10% of the energy is lost because of friction, then
Brine = E 10% E = 77662.827
Hence E is = 86292.03 W
The pump efficiency is 55%
Energy required (E) = 86292.03 x 55% = 47460.6 W
Answer to question 14
Given that
∆H = CP∆T, CP of brine = 0.77 cal/g°C, H _{brine}= H _{NaHCON3 }filtrate CP_{ NaHCON3 filtrate} = 0.81 cal/g°C
Outlet brine temperature
For Crystallizer 3
= 0.77( T_{2} – 25) = 0.81(3827)
= 36.5714°C
For crystallizer 2
0.77(T_{2}– 36.5714 ) =0.81 (4938)
= 48.1428C° which is the outlet brine temperature
For crystallized 1
0.77(T_{2}– 48.1428) = 0.81 (6049.14)
= 59.5669°C
Answers to Question 15
From reaction 2.
2 NaHCO_{3} (s) _{ } Na_{2}CO_{3 }(s) + H_{2}O (g) + CO_{2 }(g)
The ratio of moles is 2:1
Moles H_{2}O (X) = 3065.5/ 2
= 1532.75 kgmole/ hr
Mass flow rate of H_{2}O
= 1532.75 x 18
= 27589.5 kg/hr
There is 10% lost, then
Mass flow rate of H_{2}O = 27589.5 – (27589.5 x 10%)
= 27589.5 – 2758.95
= 24830.55 kg /hr
Overall material balance
=2000700 +24830.55
= 1970697.6 + water waste
Mass flow rate of water waste = 2000700 +24830.55 1970697.6
= 54832.95kg/hr
Answers to Question 16
Heat capacity of brine = 0.77 cal/g°C
Heat capacity of NaHCO_{3} = 0.25 cal/g°C
Heat of reaction 1 at 60°C = 6.61kcal/mole
For carbonator 1.
∆H = 26.24 x 6.61= 173.446 kcal/min
For carbonator1 outlet
∆H = mCP∆H _{solids} + mCP∆H _{solution}
= (2462 x 0.25 x35.8) + (44798.52 x 0.77 x35.8)
=1256950.902 kcal/min
For carbonator 1 inlet
∆H = mCP∆H _{solids} + mCP∆H _{solution}
= [258 x 0.25 x (T25)] + (44775 x 0.77 x(T25))
∆H _{in} = 173.4 + 1256950.9 = 1257124.3 kcal/min
=1257124.3 = [258 x 0.25 x(T25)] +[44775 x 0.77 x (T25)]
T= 61.386 C°
For pre carbonator.
∆H = 3.08 x 6.61 = 20.359 kcal/min
For precarbonator inlet
∆H = mCP∆H_{solids} + mCP∆H_{solution },
mCP∆H_{solids} = 0
= 47500 x 0.77 x(59.5625)]
=1264032 kcal/min
For pre carbonator Outlet
∆H = mCP∆H _{solids} + mCP∆H _{solution}
= [258 x 0.25 x (T25)] +[45039 x 0.77(T25)]
∆H _{in} = ∆H _{carbonator1} +∆H _{outlet}
= 1264032 + 20.405 = 1264052.4 kcal/min
1264052.4 = [258 * 0.25 * (T25)] + [45039 * 0.77(T25)]
= 61.3764C°
For Carbonator 2
∆H = kgmoles of NaHCO_{3} reacted is
= 26.24 x 6.61
=173.4464 kcal/min
For carbonator 2 outlet
∆H = mCP∆T _{solids} + mCP∆T _{solution}
= (4178 x 0.25 x 35) + (44543 x 0.77 x 35)
= 123699.35 kcal/min
For carbonator 2 inlet
∆H = mCP∆H solids + mCP∆H solution
= [ 0.25 x 2462 x (T25)] + [44798.52 x 0.77 x(T25)]
∆H _{in} = ∆H _{carbonator2} + ∆H_{out}
∆H _{in }= 173.4 +1236991
= 1237164.4 kcal/min
1237164.4 = [2462 x 0.25 x (T25)] +[44798.57 x 0.77 x (T25)]
= 60.774°C
Answers to Question 17
For carbonator 1
∆H = 26.24 x 6.61= 173.4464 kcal/kgmol
For carbonator1 outlet
∆H = mCP∆T _{solids} + mCP∆T _{solution}
=2462 x 0.25 x 35.8 + (44798.52 x 0.77 x 35.8)
=1256950.902 kcal/min
For carbonator 1 inlet
∆H = mCP∆T _{solids} + mCP∆T _{solution}
= 258 x 0.25 x (T60)] +[44775 x0.77 x (T60)]
But ∆H _{in} = ∆H _{carbonator1} +∆H _{outlet}
∆H _{in} = 173.4 + 1256950.9 kcal/min
1257124.3 = 258 x 0.25 x (T60)] +[44775 x 0.77 x (T60)]
T= 94.763C°
For Carbonator 2
∆H = kgmoles of NaHCO_{3} reacted x 6.61
= 26.24 x 6.61=173.446 kcal/min
For carbonator 2 outlet
∆H = mCP∆T _{solids} + mCP∆T _{solution}
= 4178 x 0.25 x35) + (44543 x 0.77 x 35)
=123699.35 kcal/min
For carbonator 2 inlet
∆H = mCP∆T _{solids} + mCP∆T_{ solution}
= (0.25 x 44798.52 x 0.77 x (T60)) + [0.25 x 2462 x (T60)]
To get T we consider that ∆H _{in} = ∆H _{carbonator2} + ∆H_{out}
∆H _{in }= 173.4 +1236991= 1237164.4 kcal/min
Hence 1237164.4 = (0.25 x 44798.52 x 0.77 x (T60)) + [0.25 x 2462 x (T60)]
1237164.4 = 8623.7151 T 517422.906 + 615.5T 36930
1791517.306 = 9239.2151T
T=193.9°C
End of Calculations
Reference List
Chemical Australia Consultants, 2007. http://www.chemlink.com.au/soda.htm
(Accessed may 1, 2011)
Garrett, D., 1992. Natural soda ash: Occurrence, processing, and use. New York: Van No strand Reinhold.
Himmelblau, D. 1996. Basics principles and calculations in chemical engineering. New Jersey: Prentice Hall PTR
Kostick, D. n d.U.S Geological Survey, Mineral Commodity Summaries. 2000
http://minerals.er.usgs.gov/minerals/pubs/commodity/soda_ash/610300.pdf
(Accessed may 1, 2011)