SIM, Tristan James G.

2013-01032

CHE 311 (ChE Calculations I)

Engr. Clydelle Rondaris

Culminating Activity

Chloride process for the manufacture of the titanium dioxide (TiO2) pigment

7550 kg of rutile ore is fed into the fluidized-bed reactor with 6420 kg of feed 1 (F1). Stream

1 is then brought to a cyclone for dust separation of metal impurities (generally metal chlorides in

the industry). Stream 2, consisting of TiCl4, oxides of carbon, and inerts, proceeds to the

condenser and is then vaporized prior to rectification (Stream 4). Rectification purifies the TiCl4

(with inerts present), and as a result, the bottom products consist of carbon oxides. Stream 5 goes

into an oxidation reactor to produce the pigment, Cl2, and some inerts after cooling.

Assumption:

During phase changes (condensation, vaporization), the mass of each of the

components is conserved.

If the yield of TiO2 is 0.838, calculate the following:

(a) amount of product, P

(b) amount of carbon oxides and

metal impurities discharged

(c) amount of the recycle

stream, R

(d) amount of the distillate, D

(e) amount of inerts in stream 5

(f) fraction of the recycle stream

to feed 1 (F1)

(g) amount of stream 6

Solution:

(a)

P

0.8489F2

= P

0.8489(7550) = 0.838

P = 5370.91 kg

(b)

OMB: F1 + F2 = metal imp. +B (carbon oxides) + P

metal imp. bal at cyclone :0.1511F2 = 0.1511(7550) = metal imp.

metal imp. = 1140.81 kg

B (carbon oxides) = 7458.28 kg

(c)

OMB at fluidized: F1 + F2 + R = S1

13970 + R = S1

Cl2 in TiCl4 = MWCl2

MWTiCl4

= 70.9

189.67

Cl2 bal at fluidized: 0.50F1 +0.9733R = (70.9

189.67) 0.8512S1

3210 + 0.9733R = (70.9

189.67) 0.8512S1

Solving both simultaneous equations, we obtain

S1 = 15855.22 kg

R = 1885.22 kg

(d)

D + B = S4 = S3 = S2

S2 = S1 + metal imp. = 15855.22 + 1140.81

S2 = 16696.03 kg

D + 7458.28 = 16696.03

D = 9537.75 kg

(e)

D = S5

inerts in S5 = 0.0157(9537.75)

inerts in S5 = 149.74 kg

(f)

R

F1

= 1885.22

6420 = 0.29

(g)

S6 = P + R

S6 = 5370.91 + 1885.22

S6 = 7256.13 kg