The flowsheet shown in Figure represents the process for the production of titan
ID: 514553 • Letter: T
Question
The flowsheet shown in Figure represents the process for the production of titanium dioxide (TIO2) used by Canadian Titanium Pigments at Varennes. Quebec. Sorel stag of the following analysis: is fed to a digester and reacted with H2S04, which enters as 67% by weight H2S04 in a water solution. The reactions in the digester are as follows: NO_2 + H_2SO_3 rightarrow nOSO_4 + H_2O Fe + 1/2 O_2 + H_2SO_2 rightarrow FeSO_4 + H_2O Both reactions are complete. The theoretically required amount of H2SO4 for the Sorel slag is fed. Pure oxygen is fed in the theoretical amount for all of the Fe in the Sorel slag. Scrap iron (pure Fe) is added to the digester to reduce the formation of ferric sulfate to negligible amounts. Thirty-six pounds of scrap iron are added per pound of Sorel slag. The products of the digester are sent to the clarifier, where all of the inert silicates and unreacted Fe are removed. The solution of TiOSO4 and FeSO4 from the clarifier is cooled, crystallizing the FeSO4, which is completely removed by a filter. The product TiOSO4 solution from the filter is evaporated down to a slurry that is 82% by weight TiOSO4. The slurry is sent to a dryer from which a product of pure hydrate, TiOSO4 + H2O, is obtained, he hydrate crystals are sent to a direct-fired rotary kiln, where the pure TiO is produced according to the following reaction: TiOSO4 + H2O rightarrow TiO2 + H2SO4 Reaction (3) is complete. On the basis of 100 lib of Sorel slag lead, calculate (a) the pounds of water removed by the evaporator; (b) the exit pounds of H2O per pound of dry air from the dryer if the air enters having 0.036 moles of H2O per mole of dry air and the air rate is 18 lib mol of dry air per 100 lb of Sorel slag: (c) the pounds of product TiO2 produced.Explanation / Answer
Sorel slag feed reactions in digester:
TiO2 + H2SO4 ------------------> TiOSO4 + H2O..............eq.1
1 eq 1 eq 1 eq 1 eq
Mol wts 79.87 98 160 18
Mol wts(lb) = 0.176 0.216 0.353 0.04
Feed amount = 100 lbs
Fe + 1/2 O2 + H2SO4 -------------> FeSO4 + H2O .........eq.2
1 eq 0.5 eq 1 eq 1 eq 1 eq
Mol wts(g) 55.84g 16g 98g 152g 18 g
Mol wts(lb) = 0.123 0.035 0.216 0.335 0.04
Feed amounts = Fe 36 lb/lb of slag
TiOSO4 + H2O ---------------------> TiO2 + H2SO4..........eq.3
1 eq 1 eq 1 eq 1 eq
Mol wts 160 18 79.87 98
Mol wts(lb) = 0.353 0.04 0.176 0.216
(a) Exit pounds of water removed by the evaporator
(b) Exit pounds of water per pound of dry air from the dryer if the air enters having 0.036 moles of H2O per mole of dry air and the air rate is 18 lb mol of dry air per 100 lb of Sorel slag
(c) the pounds of product TiO2 produced
Answer for question (1)
Water produced for 100 lb of Slag
According to eq.1 total amount of water produced
For 0.176 lb TiO2 slag 0.04 lb water produced
For 100 lb TiO2 slag = 100 x 0.04 / 0.176 = 4/0.176 = 22.72 lb water produced
Since water is also produced according to eq.2 this also actually needs to be considered for overall water
production during the reaction. However at this stage it's not calculated as it's not known how much scrap iron has reacted with H2SO4.
Total amount of TiOSO4 produced as a slurry
From 0.176 lb TiO2 slag 0.353 lb of TiOSO4 is produced according to eq.1
So from 100 lb of TiO2 slag = 100 x 0.353 / 0.176 = 35.3 / 0.176 = 200.56 lb TiOSO4 is produced.
It's assumed that all of the H2SO4 is consumed and water is released during the reaction. 200.56 lb TiOSO4 is suspended in 22.72 lb water which is produced during the reaction apart from the water produced in eq.2.
Total slurry (TiOSO4 + water) produced = 200.56 lb + 22.72 lb = 223.28 lb
223.28 lb slurry is reduced to its 82% weight slurry which = 223.28 x 82 / 100 = 183.089 lb slurry by wt
Which is equal to 223.28 - 183.089 = 40.191 lb water removed by evaporation.
Answer for question (2)
Dry air rate 18 lb / 100 lb slag
Total slurry of TiOSO4 = 200.56 lb TiOSO4 + 22.72 lb water = 223.28 lb slurry
Total dry air for 223.28 lb slurry = 223.38 x 18 / 100 = 40.20 lb dry air
Total water in dry air entering into evaporator = 40.20 x 0.036 moles water = 1.45 mole (26.1 g, 0.057 lb) water
Total water removed = 40.191 lb water + 0.057 lb water = 40.25 lb
Exit pounds of water/ lb of dry air
Total dry air = 40.20 lb
Total water removed = 40.25 lb
Exit pounds of water/ lb of dry air = 40.25/40.20 = 1 lb of water/1 lb of dry air
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