METHANOL SYNTHESIS LOOP

The total and component flowrates S7 are calculated. "z" represents the component flowrates.

As the recycle loop is operated by a recycle ratio of 0.4 (ratio of dry recycle gas to dry MUG) thus the total flowrates for S8 and S11 can be found by the equations below.

S11 = 0.4S7

S11 = 0.4 x 343.97S11 = 137.588 kmol/h

S8 = S7 + S11

S8 = 343.97 + 137.588S8 = 481.558 kmol/h

"S" represent total flowrate of each stream while "a", "b", "c" and "d" are terms used to represent the component flowrates of CO, CO2, H2 and CH4 respectively in Stream 8. The equations formed for component flowrates of other streams are termed with respect to "a", "b", "c" and "d" as shown in the diagram above.

The recycle loop is solved by using "Solver" from Microsoft Excel.6 equations are formed with 6 unknowns.

Unknowns set for solving are: S10, S13, a, b, c, d

CO2 : 0.005 x S13

H2 : 0.003 x S13

CH4 : 0.005 x S13

CH3OH : 0.7a+0.7b H2O : 0.7b

S10

S11 S12

CO : 0.3aCO2 : 0.3bH2 : c-1.4a-2.1bCH4 : dCH3OH: 0.7a+0.7bH2O : 0.7b

CO : aCO2 : bH2 : cCH4 : d

S13

CO : 0.3aCO2 : 0.3b-0.005S13

H2 : c-1.4a-2.1b-0.003S13

CH4 : d-0.005S13

S9S7 S8

zCO(7)=52.08kmol/hzO2(7)=36.95kmol/hzH2(7)=244.12kmol/hzCH4(7)=10.82kmol/h

REACTOR

SEPARATOR

Total flowrate S10: S10 = c + d - 0.4a - 0.4b - 0.013S13 --------------(1)

Total flowrate S13: S13= 0.7a + 1.4b + 0.013S13 -------------(2)

Total flowrate S8: S8 = a + b + c + d -------------(3)

Component balance at mixing point,

CO:

--------------- (4)

CO2:

------------(5)

CH4:

--------------(6)

aS

a

aSa

asCOSSa

08.522764.41

)08.52(588.1373.0

3.0

10

10

)7(1110

aS

a

aSa

azCOSSa

08.522764.41

08.52588.1373.0

3.0

10

10

)7(1110

bS

Sb

bS

Sb

bzCOSS

Sb

95.3668794.02764.41

95.36588.137005.03.0

005.03.0

10

13

10

13

)7(21110

13

dS

Sd

dS

Sd

dzCHSS

Sd

82.1068794.0588.137

82.10588.137005.0

005.0

10

13

10

13

)7(41110

13

All the equations derived are rearranged such that they are equated to "0".

c + d - 0.4a - 0.4b - 0.013S13 - S10 = 0 -------------(1)

0.7a + 1.4b + 0.013S13 - S13 = 0 -------------(2)

a + b + c + d - S8 = 0 -------------(3)

---------------(4)

---------------(5)

----------------(6)

All the unknowns to be calculated are put into a table with an initial guess of value for each term as shown below.

Another table for functions are set. The 6 equations derived are then typed into the cells from H1-H7 by substituting the unknowns in the equation from the variable cells (B2-B7).

A B1 Terms Values2 S10 2003 S13 1004 a 655 b 456 c 3407 d 30

G H1 Function Equations2 F13 F24 F35 F46 F57 F6

095.3668794.02764.41

10

13

bS

Sb

082.1068794.0588.137

10

13

dS

Sd

008.522764.41

10

a

Sa

Objective cells are then typed in with the formula=(H2^2)+(H3^2)+(H4^2)+(H5^2)+(H6^2)+(H7^2)

Open "Solver" and set all the parameters. In the set objective column the objective cell is inserted then setting the value of "0". The cells from B2-B7 are taken for the column "By Changing Variable Cells" in the solver parameter. Cells from H2-H7 are then taken for the column "Subject to the Constraints" where all the constraints are set to "= 0".

After it is solved, the values for all the unknown terms will be shown in the variable cells.

A B1 Terms Values2 S10 219.08793 S13 109.48744 a 64.16965 b 45.10386 c 344.12247 d 28.1622

The component flowrates for stream 8 are directly taken from the variable cells.

The component flowrates for S9, S10 and S13 are then calculates by substituting the known values of S10, S13, a, b, c, d into the equation in the diagram above.

Stream 9Component Equations Flowrate

(kmol/h)Mole Fraction

CO = 0.3a 19.2509 0.05859CO2 = 0.3b 13.5311 0.04118H2 = c - 1.4a - 2.1b 159.567 0.4856

CH4 = d 28.1622 0.08571CH3OH = 0.7a + 0.7b 76.4914 0.2328

H2O = 0.7b 31.5727 0.09609Total 328.5752 1

Stream 10Component Equations Flowrate

(kmol/h)Mole

FractionCO = 0.3a 19.2509 0.087868CO2 = 0.3b - (0.005S13) 12.9837 0.059263H2 = c - 1.4a - 2.1b - (0.003S13) 159.239 0.726825

CH4 = d - 0.005S13 27.6148 0.126044Total 219.0879 1

Stream 13Component Equations Flowrate

(kmol/h)Mole Fraction

CO2 = 0.005S13 0.5474 0.005H2 = 0.003S13 0.3285 0.003

Stream 8

Component Flowrate (kmol/h)

Mole Fraction

CO 64.1696 0.133254CO2 45.1038 0.093662H2 344.1224 0.714602

CH4 28.1622 0.058481Total 481.558 1

CH4 = 0.005S13 0.5474 0.005CH3OH = 0.7a + 0.7b 76.4914 0.6896

H2O = 0.7b 31.5727 0.2884Total 109.4874 1

The composition of S10, S11 and S12 are the same.Thus the component flowrates for Stream 11 can be calculated by the relationship,

[xY(10)]S11 = zY(11)

Where, "xY(10)" represents the mole fraction of component "Y" in Stream 10 while "zY(11)" represents the component flowrate "Y" in Stream 11

Stream 11Component Equations Flowrate

(kmol/h)Mole Fraction

CO =0.087868 x S11 19.2509 0.087868CO2 =0.059263 x S11 13.5311 0.059263H2 =0.726825 x S11 159.567 0.726825

CH4 =0.126044 x S11 28.1622 0.126044Total 137.588 1

At purge point, S10 = S11 + S12

S12 = S10 - S11

S12 = 219.0879 - 137.588S12 = 81.4999 kmol/h

Component flowrates for Stream 12 are also calculated the same way as Stream 11,

[xY(10)]S12 = zY(12)

Where "xY(10)" represents the mole fraction of component "Y" in Stream 10 while "zY(12)" represents the component flowrate "Y" in Stream 12.

Stream 12Component Equations Flowrate

(kmol/h)Mole Fraction

CO =0.087868 x S12 7.16128 0.087868

CO2 =0.059263 x S12 4.829903 0.059263H2 =0.726825 x S12 59.23612 0.726825

CH4 =0.126044 x S12 10.27256 0.126044Total 81.49986 1

Methanol Distillation

Distillation Column 1We assume all the gases (CO2, H2 and CH4) are distilled out from the crude methanol at the first distillation column. Thus, component flowrates for CO2, H2 and CH4 in Stream 13 and Stream 15 are the same while the component flowrates for CH3OH and H2O in Stream 13 and Stream 14 are the same.

Stream 15Component Flowrate

(kmol/h)Mole Fraction

CO2 0.5474 0.3846

xCH3OH(16) : 0.001

S13

S16

S17S15

S14

Light Gas

DC1 DC2CH3OH : 76.4914kmol/hH2O : 31.5727kmol/hCO2 : 0.5474kmol/hH2 : 0.3285kmol/hCH4 : 0.5474kmol/h

xCH3OH(17) : 0.9972

H2 0.3285 0.2308CH4 0.5474 0.3846Total 1.4233 1

Stream 14Component Flowrate

(kmol/h)Mole Fraction

CH3OH 76.4914 0.7078H2O 31.5727 0.2922Total 108.064 1

Distillation Column 2The design specification given is 0.28 mole % impurities thus 99.72 mole % methanol produced from S17. The flowrates for S16 and S17 are calculated using 2 equations simultaneously,

Overall mass balance: S14 = S16 + S17 -----------(7)Component mass balance : zCH3OH(14)=[xCH3OH(16)]S16 + [xCH3OH(17)]S17 ----------(8)

Where, "xCH3OH(16)" represents the mole fraction of CH3OH in Stream 16; "xCH3OH(17)" represents the mole fraction of CH3OH(in Stream 17;while "zCH3OH(14)" represents the CH3OH flowrate in Stream 14.

From equation (7),S16 = S14 - S17 ------------(9)

Substitute equation (9) into (8),zCH3OH(14) = [xCH3OH(16)]( S14 - S17 ) + [xCH3OH(17)]S17 -----------(10)76.4914 = 0.001 ( 108.064 - S17 ) + ( 0.9972 ) S17

76.4914 = 0.108064 - ( 0.001 ) S17 + ( 0.9972 ) S17

76.4914 - 0.108064 = ( 0.9962 ) S17

S17 = 76.675 kmol/h

Substitute S17 = 76.675 kmol/h into equation (7),S16 = S14 - S17

S16 = 108.064 - 76.675S16 = 31.389 kmol/h

Component flowrates for CH3OH in Stream 16 and Stream 17 are calculated by multiplying stream flowrate with mole fraction of CH3OH for the same stream.

zCH3OH(16) = [xCH3OH(16)]( S16 )= 0.001 ( 31.389 )= 0.3139 kmol/h

zCH3OH(17) = [xCH3OH(17)]( S17 )= 0.9972 ( 76.675 )

= 76.4603 kmol/h

Component flowrates for H2O in Stream 16 and Stream 17 are calculated by subtracting flowrate of CH3OH from stream flowrate.

zH2O(16) = S16 - zCH3OH(16)

= 31.389 - 0.3139= 31.358 kmol/h

zH2O(17) = S17 - zCH3OH(17)

= 76.6747 - 76.4603= 0.3137 kmol/h

Scale Up

Mr (CH3OH) = 32.04 kg/kmolMr (H2O) = 18.016 kg/kmol

Average molecular mass of crude methanol,= [ xCH3OH(17) x Mr(CH3OH)] + [xH2O(17) x Mr(H2O)]= (0.9972 x 32.04) + (18.016 x 0.0028)= 32.0007 kg/kmol

Methanol production,

(CH3OH)T

Scale up Factor, k (CH3OH)T = k [zCH3OH(17)] 703.1089 = k (76.4603) k = 9.1958

All the stream and componenet flowrates obtained from the basis are then multiplied by the scale up factor to get the scaled up figures.

For Stream flowrates, e .g: S1 = 100 kmol/h (basis)

= 100k= 100 x 9.1958= 919.58 kmol/h (scaled up)

For component flowrates, e.g: zCH4(1) = 84.8725 kmol/h (basis)

hourkmol

hoursyear

kgkmol

tonnekg

yeartonnes

1089.703

80000007.321000180000

= 84.8725k= 84.8725 x 9.1958= 780.4682 kmol/h (scaled up)