MATERIAL BALANCE AROUND REACTOR

Plant capacity =          600 Metric Tons/ Day.

                                    =          25253 Kg/hr.

Basis:                                      Unit hour operation

Further we will use the notations as under

Mono-ethylene Glycol                   MEG

Di-ethylene Glycol                         DEG

Tri-ethylene Glycol                         DEG

If we use 1:12 molar ratio of EO: H₂O and selectivity of different products is as under:

MEG = 85 %,              DEG = 14 %,               TEG = 1 %                  (Patent = US6605192)

Reactions:

C₂H₄O+H₂O                                    OHCH₂CH₂OH                                (MEG)

2C₂H₄O+H₂O                                 OHCH₂CH₂OCH₂CH₂OH                 (DEG)

3C₂H₄O+H₂O                                      OHCH₂CH₂OCH₂CH₂OCH₂CH₂OH  (TEG)

Suppose 1 % of MEG is lost during evaporation and distillation then

MEG                            =          1.01*25253     = 25508 Kg              

Total Mixture (MEG+DEG+TEG)                    =25508/0.85    =30009Kg

DEG                            =          30009*0.14     =4201Kg

TEG                             =          30009*0.01     =300Kg

  

Reaction 1: -

 

            EO                   +          H2O                                        MEG

            411.4 Kmole                411.4 Kmole                            411.4 Kmole

Reaction 2:-

           

2EO                 +          H2O                                        DEG

            80 Kmole                     40 Kmole                                 40 Kmole

Reaction 3:-

           

3EO                 +          H2O                                        TEG

            6Kmole                        2 Kmole                                   2 Kmole

EO required                 = 411.4 + 80 +6                      = 497.4 Kmole

                                                                                    =21885.6 Kg  

Total H2O Require       = 411.4+40+2                                     = 453.4 Kmole

                                                                                    = 8161.2 Kg

As         

                                                H2O/EO = 12/1

Total H2O Supplied     = 12 * EO                                = 12 * 497.4 = 5968.8 Kmole

                                                                                    = 107438.4 Kg

Excess H2O                 = 5968.8– 453.4                      = 5515.4 Kmole

                                                                                    = 99277.4 Kg

3.1.1:MATERIAL ENTERING THE REACTOR:

Component	Mass (kg)	Mass %	Mole	Mole %
Ethylene Oxide	21885.6	16.9	497.4	7.7
Water	107438.4	83.1	5968.8	92.3
Total	129324	100	6466.2	100

3.1.2 MATERIAL LEAVING THE REACTOR:

Component	Mass (kg)	Mass %	Mole	Mole %
Water	99277	77	5515.4	92.4
MEG	25508	19.7	411.4	6.9
DEG	4201	3.1	40	0.67
TEG	300	0.2	2	0.03
Total	129286	100	5968.8	100

MATERIAL BALANCE AROUND PRESSURE VESSEL

Component	Mass (kg) IN	Mass % IN	Mass (kg) OUT	Mass % OUT
Ethylene Oxide	21885.6	16.9	21885.6	16.9
Recycle Water	98910	76.5	-------	-------
Fresh Water	8528.4	6.6	107438.4	83.1
Total	129324	100	6466.2	100

MATERIAL BALANCE AROUND PREHEATER

Component	Mass (kg)	Mass %	Mole	Mole %
Ethylene Oxide	21885.6	16.9	497.4	7.7
Water	107438.4	83.1	5968.8	92.3
Total	129324	100	6466.2	100

MATERIAL BALANCE AROUND EVAPORATOR

Suppose MEG is concentrated from 19.73 % to 60 % in this long tube vertical falling film evaporator.

Feed:

H2O                            =99277KG

MEG                            =25508KG

DEG                            =4201KG

TEG                             =300KG

Product:

H2O                            =?

MEG                            =25508KG

DEG                            =4201KG

TEG                             =300KG

Total material out from evaporator= 25508 / 0.6

= 42513 kg

H2o at Outlet                           = 42513-25508-4201-300

= 12504 kg

H2O Evaporator                      = 99277-12504

= 86773 kg

D = D1+D2+D3                      =86773 Kg ________(1)

1st Effect:

ms λs                                       = FCP(T1 – T2) +  D1 λ2

Steam temp. Ts                        = 486 K

λs                                             =1886 kJ/kg

C_p                                            = 1.98 KJ/kg.k

m_s (1886)=129286(1.98)(455-468)+D1(202.2)

m_s = -1764.49 +D1(1.071)______(2)

2nd Effect:

D₁ λ₁    = (F-D1)Cp(T2-T1)+D2 λ2

λ₁                                 = 2020.2 kj/kg

λ₂                                 = 2179 kj/kg

C_p                                =4.23 kJ/kg

(2020.2)D₁ = (129286-D₁)*4.23*(405- 455)+D₂(2179)

D₁ = -15118.03+D₂*(1.205) ______(3)

3rd effect:

D₂ λ₂ = (F-D₁-D₂)C_p(T₃-T₂)+D₃ λ₃

λ₃                                            =2326 KJ/KG

λ₂                                            =2179 KJ/KG

C_p                                           =4.03 KJ/KG.K

(2179)D2 = (129286-D₁-D₂)*4.03*(350- 405)+D₃(2326)

From =1

D₃        = 86773-D₁-D₂

2179 *D₂ = (129386-D₁-D₂)(-221.65)+(86773-D₁-D₂)*2326

4283.35*D₂ = 173177756.1- 2104.35*D₁___________(4)

From = 3

4283.35*D₂ = 173177754.1 - 2104.35 *  (-15118.03+1.21*D2)

D₂        = 22194 kg/hr

From = 4

D₁           = 33927kg/hr

From = 1

D₃        = 33927 kg/hr

From = 2

ms        = 27931 kg/hr

Component	Mass (kg) IN	Mass % Inx	Mass(kg) OUT	Mass % OUT
Water	99277	76.8	12504	29.4
MEG	25508	19.73	25508	60.0
DEG	4201	3.25	4201	9.9
TEG	300	0.25	300	0.7
Total	129286	100.0	42513	29.4

MATERIAL BALANCE AROUND LIGHT END COLUMN

Component	Mass (kg) IN	Mass % IN	Mass(kg) OUT	Mass % OUT
Water	12504	29.4	10	0.033
MEG	25508	60.0	25408	84.923
DEG	4201	9.9	4201	14.041
TEG	300	0.7	300	1.002
Total	42513	100.0	29919	100.0

Distillate

            Component   Mass     Mass %

                   MEG       100      0.8 %

                   Water    12494    99.2 %

                    Total     12594    100 %

Material out                  =29919+12594            =42513(kg)

                    Material In =Material Out