Mass Balance Hand calculation
Mass Balance across Desulphurization Tank
S1 = 100 kmol/h S2 = 99.85 kmol/h CH4,1 = 84.8725 kmol/h CH4,2 = 84.8725 kmol/h CO2,1 = 14.9775 kmol/h CO2,2 = 14.9775 kmol/h H2S,1 = 0.1500 kmol/h
Basis set, S1= 100 kmol/h
CH4,1 ( ) CO2,1 ( ) H2S,1 Desulphurization
Since desulphurization is only meant to remove the H2S from the dry natural gas, therefore the flow rate of CH4 and CO2 remain unchanged from stream 1 to stream 2.
Mass Balance across Mixer
S2 = 99.85 kmol/h Mixer S3(a) = 235.2886 kmol/h CH4,2 = 84.8725 kmol/h CH4,3(a) = 84.8725 kmol/h CO2,2 = 14.9775 kmol/h CO2,3(a) = 14.9775 kmol/h H2O, 3(a) = 135.4386 kmol/h
Given relative humidity of S3(a) is 62%;
To find interpolation using values from the steam table has to be done, Taking 22 bar and 24 bar with their saturated temperatures as reference,
Pressure at S3(a) is 25 bar,
0.5758 Mole fraction of water in S3(a)= 0.5758 S3= 0.5758 * S3(a)
S3= 135.53 kmol/h
Therefore, the flow rate of hot water entering in is 135.53 kmol/h. Hot Water
Mass Balance around mixing point Steam
S4 = 164.1114 kmol/h
S3 (a) = 235.2886 kmol/h S4 (a) = 399.4 kmol/h H2O,3(a) = 135.4386 kmol/h H2O,4(a) = 299.55 kmol/h CH4,3(a) = 84.8725 kmol/h CH4,4(a) = 84.8725 kmol/h CO2,3(a) = 14.9775 kmol/h CO2,4(a) = 14.9775 kmol/h
After steam is added, the water to carbon molar ratio is 3.0.
n(H2O)= 3n(C)
CH4 and CO2 molecules contain one carbon atom each. n(H2O)= 3(n(CH4 + CO2)
= 3(84.873 + 14. 9775) = 299.55 kmol/h
Mass Balance around reformer
S4 (a) = 399.4 kmol/h Reformer S5 = 547.5001 kmol/h H2O,4(a) = 299.55 kmol/h H2O,5 = 203.5261 kmol/h
CH4,4(a) = 84.8725 kmol/h CH4,5 = 10.8224 kmol/h CO2,4(a) = 14.9775 kmol/h CO2,5 = 36.9513 kmol/h CO,5 = 52.0762 kmol/h H2,5 = 244.1240 kmol/h
Since there are 5 unknowns in stream 5, 5 equations are needed in order to solve the unknowns.
°C , °C ,
Pressure at equilibrium state= 23.5 bar Hence, °C.
Kp = K.10n
From Table 1: Equilibrium Constants for Methane-Steam Reforming Reaction,
At °C, we obtained Kp value for methane-steam reforming reaction is 1.578 * . Using given equation, we get
Therefore, ( )( )
( )( ) ( )
Using the same method, another equation can be formed using the Kp value of shift reaction.
Reaction temperature for shift reaction is °C ,
From Table 2: Equilibrium Constants for Carbon Monoxide Shift Reaction, We get Kp value for shift reaction is 0.8511,
For the remaining 3 equations, those are created using atomic balance of C, H and O respectively.
( )( ) ( )( ) ( )( ) ( )( )
( )( ) ( )( )
(
) ,
( )( ) ( )( )
(
)
(Kp)(P2)( )( ) ( )( )
(
)
( )( )
( )( ) ( )( )(
)
( )( ) ( )( ) ( )( ) ( ) = 0Carbon atom balance :
Hydrogen atom balance :
( ) ( ) ( )
Oxygen atom balance:
( )
Now with 5 unknowns and 5 equations obtained, the 5 unknowns can be determined by using a tool in Microsoft Excel called solver.
1. Cells for function and are set.
f1 -1.350772222 f2 -1.5489 f3 -20.1500 f4 658.59 f5 179.505
2. An initial guess for the 5 solution were taken. (20,30,40,50,60 respectively) Component Reactor effluent
H20 20
CH4 30
CO2 40
CO 50
Total 200 3. An objective cell was set. ( )
objective cell 466373.0793
4. The solver window was opened and all the information was set. ( constraints = 0 and objective function = 0)
5. Therefore using solver , we get nH2O,5 = 203.5261 kmol/h
nCH4,5 = 10.8224 kmol/h
nCO2,5 = 36.9513 kmol/h
nH2,5 = 244.1240 kmol/h
Mass Balance around Dryer S6
S5 = 547.5001 kmol/h
H2O,5 = 203.5261 kmol/h Dryer CH4,5 = 10.8224 kmol/h
CO2,5 = 36.9513 kmol/h S7 = 343.9740 kmol/h CO,5 = 52.0762 kmol/h CH4,7 = 10.8224 kmol/h H2,5 = 244.1240 kmol/h CO2,7 = 36.9513 kmol/h CO,5 = 52.0762 kmol/h
H2,5 = 244.1240 kmol/h
Since stream 5 passes through a dryer, it is assumed that all the water is removed from stream 5, meaning there is no water in stream 7. The flow rates of CH4, CO2, CO and H2 remains unchanged from stream 5 to stream 7.
S5= 203.526 kmol/h
Mass balance around methanol synthesis loop 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 : 0.3a CO2 : 0.3b H2 : c-1.4a-2.1b CH4 : d CH3OH : 0.7a+0.7b H2O : 0.7b CO : a CO2 : b H2 : c CH4 : d S13 CO : 0.3a CO2 : 0.3b-0.005S13 H2 : c-1.4a-2.1b-0.003S13 CH4 : d-0.005S13 S9 S7 S8 zCO(7)=52.08kmol/h zO2(7)=36.95kmol/h zH2(7)=244.12kmol/h zCH4(7)=10.82kmol/h REACTOR SEPARATOR
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 343.97 S11 = 137.588 kmol/h S8 = S7 + S11 S8 = 343.97 + 137.588 S8 = 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.
Total flowrate S10: S10 = c + d - 0.4a - 0.4b - 0.013S13
Total flowrate S13: S13= 0.7a + 1.4b + 0.013S13 Total flowrate S8: S8 = a + b + c + d
Component balance at mixing point,
CO balance: CO2 balance: CH4 balance: a S a a S a a sCO S S a 08 . 52 2764 . 41 ) 08 . 52 ( 588 . 137 3 . 0 3 . 0 10 10 ) 7 ( 11 10
b S S b b S S b b zCO S S S b 95 . 36 68794 . 0 2764 . 41 95 . 36 588 . 137 005 . 0 3 . 0 005 . 0 3 . 0 10 13 10 13 ) 7 ( 2 11 10 13
d S S d d S S d d zCH S S S d 82 . 10 68794 . 0 588 . 137 82 . 10 588 . 137 005 . 0 005 . 0 10 13 10 13 ) 7 ( 4 11 10 13
a S a a S a a zCO S S a 08 . 52 2764 . 41 08 . 52 588 . 137 3 . 0 3 . 0 10 10 ) 7 ( 11 10All the equations derived are rearranged such that they are equated to "0". c + d - 0.4a - 0.4b - 0.013S13 - S10 = 0 0.7a + 1.4b + 0.013S13 - S13 = 0 a + b + c + d - S8 = 0
All the unknowns to be calculated are put into a table with an initial guess of value for each term as shown below.
A B 1 Terms Values 2 S10 200 3 S13 100 4 a 65 5 b 45 6 c 340 7 d 30 0 95 . 36 68794 . 0 2764 . 41 10 13 b S S b 0 82 . 10 68794 . 0 588 . 137 10 13 d S S d 0 08 . 52 2764 . 41 10 a S a
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).
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".
G H 1 Function Equations 2 F1 3 F2 4 F3 5 F4 6 F5 7 F6
After it is solved, the values for all the unknown terms will be shown in the variable cells. A B 1 Terms Values 2 S10 219.0879 3 S13 109.4874 4 a 64.1696 5 b 45.1038 6 c 344.1224 7 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 9 Componen t Equations Flowrate (kmol/h) Mole Fraction CO = 0.3a 19.2509 0.05859 CO2 = 0.3b 13.5311 0.04118 H2 = c - 1.4a - 2.1b 159.567 0.4856 CH4 = d 28.1622 0.08571 CH3OH = 0.7a + 0.7b 76.4914 0.2328 H2O = 0.7b 31.5727 0.09609 Total 328.5752 1 Stream 8 Componen t Flowrate (kmol/h) Mole Fraction CO 64.1696 0.133254 CO2 45.1038 0.093662 H2 344.1224 0.714602 CH4 28.1622 0.058481 Total 481.558 1
Stream 10 Componen t Equations Flowrate (kmol/h) Mole Fraction CO = 0.3a 19.2509 0.087868 CO2 = 0.3b - (0.005S13) 12.9837 0.059263 H2 = c - 1.4a - 2.1b - (0.003S13) 159.239 0.726825 CH4 = d - 0.005S13 27.6148 0.126044 Total 219.0879 1 Stream 13 Componen t Equations Flowrate (kmol/h) Mole Fraction CO2 = 0.005S13 0.5474 0.005 H2 = 0.003S13 0.3285 0.003 CH4 = 0.005S13 0.5474 0.005 CH3OH = 0.7a + 0.7b 76.4914 0.6896 H2O = 0.7b 31.5727 0.2884 Total 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 11
Component Equations Flowrate
(kmol/h) Mole Fraction CO =0.087868 x S11 19.2509 0.087868 CO2 =0.059263 x S11 13.5311 0.059263 H2 =0.726825 x S11 159.567 0.726825 CH4 =0.126044 x S11 28.1622 0.126044 Total 137.588 1 At purge point, S10 = S11 + S12 S12 = S10 - S11 S12 = 219.0879 - 137.588 S12 = 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 12
Component Equations Flowrate
(kmol/h) Mole Fraction CO =0.087868 x S12 7.16128 0.087868 CO2 =0.059263 x S12 4.829903 0.059263 H2 =0.726825 x S12 59.23612 0.726825 CH4 =0.126044 x S12 10.27256 0.126044 Total 81.49986 1
Mass balance across Distillation Columns xCH3OH(16) : 0.001 S13 S16 S17 S15 S14 Light Gas
DC1
DC2
CH3OH : 76.4914kmol/h H2O : 31.5727kmol/h CO2 : 0.5474kmol/h H2 : 0.3285kmol/h CH4 : 0.5474kmol/h xCH3OH(17) : 0.9972Distillation Column 1
We 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 15 Component Flowrate (kmol/h) Mole Fraction CO2 0.5474 0.3846 H2 0.3285 0.2308 CH4 0.5474 0.3846 Total 1.4233 1 Stream 14 Component Flowrate (kmol/h) Mole Fraction CH3OH 76.4914 0.7078 H2O 31.5727 0.2922 Total 108.064 1
Distillation Column 2
The 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 --- (1)
Component mass balance : zCH3OH(14)=[xCH3OH(16)]S16 +[xCH3OH(17)]S17 --- (2)
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 (1),
S16 = S14 - S17 --- (3)
Substitute equation (3) into (2),
zCH3OH(14) = [xCH3OH(16)]( S14 - S17 ) +[xCH3OH(17)]S17 ---(4) 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 (1), S16 = S14 - S17
S16 = 108.064 - 76.675 S16 = 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
Scaling up
Mr (CH3OH) = 32.04 kg/kmol Mr (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 hour kmol hours year kg kmol tonne kg year tonnes 1089 . 703 8000 0007 . 32 1000 180000
All the stream and component 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) = 84.8725k
= 84.8725 x 9.1958
Appendix on Energy Balance
Assumptions made on calculating Energy Balance
1. Ideal gas is assumed in our calculation. Hence, volume fraction equals to molar fraction can be used in our calculation. Ideal gas is assumed due to the high reaction temperature in our plant.
2. For turbines and compressors, we take
which means the required work of compression or expansion is equal to the enthalpy rise or drop of the fluid. This can be proven using the First Law of Thermodynamic for the steady state open system.
Where and is negligible as the initial fluid velocity is low and the equipment are at the same datum level as the fluid. Q is negligible because the flow and power terms are large in comparison to the surface area of the machine, so heat transfer is negligible.
3. For compressors, as we assume ideal condition,
-W isothermal < -W polytropic < -W isentropic relationship is taken into account. 4. The steady flow performance of turbines and compressors are idealized by
assuming that in an ideal case, the working fluid does work reversibly by expanding or compressing at a constant entropy. This is to provide a basis for analyzing the performance of equipment.
5. Plant is well insulated from disturbances is assumed. Hence, for an example from this assumption, outlet temperature of E1 equals to inlet temperature of E2. 6. No boundary work done is assumed.
7. Mixer, sulphur guard, knock out pot, separator have relatively low difference in heat. Hence Q required or released is negligible by assuming they are isothermal devices.
8. For the reformer part, the operating temperature for the reforming and shift reactions to be at 860 °C. Hence, superheated liquid to be assumed as the outlet temperature is 860 °C.
9. When there’s mixture of steam with other gases, we assume ideal mixing taking place. Hence, condensation point of the steam to be the saturated temperature at that particular temperature.
10. At mixing points, assumption of no heat loss to the surrounding is made. Hence, relationship of can be used in our calculation.
11. Methanol synthesis catalytic reactor is assumed to be adiabatic.
12. The composition of the reformer gas is calculated by assuming chemical equilibrium at the exit where the temperature is 858 °C and pressure at 23.5 bar.
13. It is assumed that only water and methanol is in the feed for the distillation column (II) as all light gas has been removed in distillation column (I).
Table 3. Cp values for components involved
Components Mr State a b c d
Methanol CH3OH 32.04 Liquid 0.07586 0.0001683 - -
Gas 0.04293 0.00008301 1.87E-08 8.03E-12
Hydrogen H2 2.016 Gas 0.02884 7.65E-08 3.288E-09 -8.698E-13 Methane CH4 16.04 Gas 0.03431 0.00005469 3.661E-09 -1.1E-11 Hydrogen
Sulphate
H2S 34.08 Gas 0.03351 0.00001547 3.012E-09 -3.292E-12 Carbon
Monoxide
CO 28.01 Gas 0.02895 0.00000411 3.548E-09 -2.22E-12 Carbon
Dioxide
CO2 44.01 Gas 0.03611 0.00004233 -2.887E-08 7.464E-12 Water H2O 18.016 Gas 0.03346 0.00000688 7.604E-09
-3.593E-12
E-1 (Compressor)
To solve the compressors, solver method is used.
The two equations below are the constraints for the solver )
( ) ) ( )
From substituting the both equations, we can know that
By setting the 1st and 2nd equations as constraints, the objective function of the solver will be
( ) (
)
And the changing variable will be our unknown variable, which is outlet temperature, T2 Below picture is the example of solver
By getting the outlet temperature, T2 the formula below is used to find out the power requirement for the compressor.
( ) ( * While = 3031.32 kW
Heat exchanger (E-2)
To calculate the power requirement of heat exchanger, this formula is used ∫
While
By integrating the Cp, we can get
=(Tout-Tin)
Since we have the values of outlet temperature and inlet temperature, by substituting the both values inside the equations, we can get the value of q.
By getting the value q, the formula below is used to get the power requirement.
= 1426.25 kW
m=molar flowrate
E-3 & E-18
Few assumptions are made for E-3 & E-18:
At mixing point for reformer feed Few assumptions are made
All the heat released by stream 4 is absorbed by stream 3
The pressure of stream is assumed to be 42.5 bars
Steam in stream 4 assumed to be saturated steam rather than superheated steam because by mixing superheated steam with another fluid, heat transfer is bad as the heat transfer coefficient is very low. Hence, relationship of ‘All the heat released by stream 4 is absorbed by stream 3’ cannot be applied.
The temperature of stream 4 will be taken as 253.0˚C which the value is taken from steam table
Solver method is used to find the outlet temperature of mixing point
Below are the constraints equations for solver
∫ ∫ ∫ ∫ After integration, the formulas become,
( )
( * ( *
( )
Having the two constraints formula, the objective formula is set to ( ) and changing variable as Tout
Using the solver, we can find out that Tout is 232.95˚C
Then these two formulas are used again by substituting the Tout to find out the power of both streams.
( * ( * ( )
Reformer
At this part the formulas below are used
∫
Since we have the inlet temperature and outlet temperature which is 232.95˚C and 860˚C respectively, Q for outlet of mixing point and Q for stream 5 are calculated. After that, the total Q is calculated by finding the difference of Q of the both streams.
Then, formula below is used because the heat of reactions must be considered in order to get the total heat of the reformer reactions.
∫ ( )
The Δh for the reforming and shift reactions are given which are and .
Hence Q= (moles of CH4 consumed)* +(moles of CO consumed )* + ∫ = 65307.66 kJ/s
Heat Exchanger E-7, E-8 & E-9 Assumptions
No pressure drop across the unit, remain 23.5 bars
Same formulas as shown in below to use to find out the heat for each component except water
∫
It is because from temperature change from 860 ˚C to 35 ˚C, phase change has occurred for water. So, latent heat of water must be considered in order to get the total heat of water.
∫ ∫ Hence, Q total= -61343.30 kJ/s
Knock out pot Assumptions:
It is assumed to be isothermal.
Inter stage Cooling
For Inter stage cooling, E12-E14-E13, the way of finding the work done for E12 is similar to the previous compressor.
Inlet temperature, T1= 35 °C Inlet pressure, P1
Outlet pressure, P2
Using formula ( ) k= 1.60
Using solver method, We get T2= 110.57 °C Average Cp =0.030318 kJ/mol.K ( ) ( ( ) ( ) ( ) ( ) ) = 80.01 kJ/s ( ) ( ( ) ( ) ( ) ( ) ) = 1360.225 kJ/s ( ) ( ( ) ( ) ( ) ( ) ) = 294.25 ( ) ( ( ) ( ) ( ) ( ) ) = 278.38 kW
Total Heat for E12= Total Work (proven in assumption part) = 2012.87 kJ/s
Since it involves inter cooling, hence, both compressors E12 and E13 should share the similar work done.
Using solver, we get T2’ = 126.6 °C Using formula ( ) k’= 1.73
Both k values of the compressors have almost the same value. Hence, it complies with the requirement of inter stage cooling technique.
( ) ( ( ) ( ) ( ) ( ) ) = 91.72 kJ/s ( ) ( ( ) ( ) ( ) ( ) ) = 1647.80 kJ/s ( ) ( ( ) ( ) ( ) ( ) ) = 35460 ( ) ( ( ) ( ) ( ) ( ) ) = 324.72 kJ/s
Total Heat= Total Work (proven in assumption part) = 2418.80 kJ/s
Heat Exchanger E-14
Having the both value of inlet and outlet temperatures, we can use this formula to get total heat ∫ ∫ ∫ ∫ ( * ( * ( * ( * = -2012.87 kJ/s
Hence, heat loads for inter stage compressor cooling load =( 2012.87 -2012.87 +2418.80) = 2418.80 kJ/s
Methanol synthesis reactor
Assumption: The catalytic reactor is assumed to be adiabatic Inlet temperature given: 145 °C
Outlet temperature: unknown T2
For adiabatic reaction:
Reactions that are involved in this reactor:
CO + 2H2 CH3OH h = -94 MJ/kmol CH3OH Methanol produced in this reaction= moles of CO used
= 590.09 kmol/h- 177.03 kmol/h = 413.06 kmol/h
for this reaction = = 10785.46 kW
CO2 + 3H2 CH3OH + H2O h= -53 MJ/kmol CH3OH Methanol produced in this reaction= moles of CO2 used
= 414.6 kmol/h- 124.43 kmol/h = 290.17 kmol/h
for this reaction = 4271.95 kJ/s
for both reactions= 10785.46 kW + 4271.95 kW = 15059.9 kJ/s 𝒂𝑻𝒇 𝒃(𝑻𝒇)^𝟐 𝟐 + 𝒄(𝑻𝒇)^𝟑 𝟑 + 𝒅(𝑻𝒇)^𝟒 𝟒 𝐚𝑻𝒊 𝒃(𝑻𝒊)^𝟐 𝟐 - 𝒄(𝑻𝒊)^𝟑 𝟑 - 𝒅(𝑻𝒊)^𝟒 𝟒 𝑯𝒘 𝒐 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 ∫ 𝒄𝒑𝒅𝑻 𝑻𝒇 𝑻𝒊
We take reference temperature at 25 °C in our calculation.
Values of a, b, c and d for each components can be obtained from Table 1 above.
( ( ) ( ) ( ) ( ) ) = 3463.20 ( ( ) ( ) ( ) ( ) ) = 3505.521 ( ( ) ( ) ( ) ( ) ) = 4621.73 ( ( ) ( ) ( ) ( ) ) 27810.15 ( ( ) ( ) ( ) ( ) ) = 16369.33 ( ( ) ( ) ( ) ( ) ) = 17048.07 ( ( ) ( ) ( ) ( ) ) (𝑎 ( ) 𝑏( ) +𝑐( ) +𝑑( ) )
= 25415.17 ( ( ) ( ) ( ) ( ) ) = 38465.40 ( ( ) ( ) ( ) ( ) ) = 20220.23 = (258.97*28055.03 + 1467.34* 16164.43 +177.03 *16878.02 + 124.43*25404.40 + 703.40* 39060.46 + 290.33 *20060.45 – 3164.47 *3464.80 – 258.97* - 590.09 * 3519.26 – 414.76 *4736.60 ) / 3600 (kW) = 15059.89 kJ/s
Applying solver method to get T2,
We set objective function as ( )^2 - ( ) ^2 to the value of 0, Changing variable as T2, we get T2 = 578.53 °C.
Hence, exit temperature for methanol reactor is 593.15 °C.
Turbine
1st Turbine
We know that the efficiency of the turbine is 65% and the formula below is use to find out the h2 of the turbine
By given inlet and outlet pressure and temperature, we can use interpolation method to find out the h' value. Below are the ways of doing it
Inlet pressure=105 bars Outlet pressure= 42.5 bars Temperature= 450˚C Pressure 105 bars Hg 3233 Sg 6.3885 Pressure 42.5 bars Hg 2799.5 Sg 6.044 Hf 1105.25 Sf 2.8295
We assume that the turbine is isentropic, so s1 = s2'
But from the table above, we can know that the value of entropy is , so this reaction is superheated.
Interpolation to find out h' will be using the enthalpy value of superheated steam from steam table.
At 42.5
6.326 6.3885 6.55075
2954 h' 3094
2992.9321
After getting the h' value, h2 can be found using formula above and below formula will be used to find out the steam requirement of turbine, m
( ) ( ) 2nd Turbine
From information given, we know that, this turbine efficiency is 75%, Pout=0.13bars and Pin=42.5bars.
Since the stream from 1st turbine is connected to this turbine we can find the temperature of the stream using interpolation
300 T2 350
2954 3076.956 3088
T2 345.8791
And we assume that this turbine is isentropic, so s1 = s2'
At 42.5 bars Hg Sg S2’ 3076.956 6.530523 6.530523 At 0.13 Sg 0.7165 Sf 8.058 So,
At 0.13
0.7165 6.530523 8.058
213.5 h' 2593
h'= 2097.9198
These formulas are then used to find out the steam requirement of the turbine ( ) ( ) ( )
Reference list
1. Richard M Felder and Ronald W Rousseau (1999), Elementary Principles of Chemical Processes, Third Edition, published by John Wiley and Sons INC. (1999).
2. Rao, YVC Rao, Chemical Engineering Thermodynamic (1997), published by University Press (India) Private Limited (1997).
3. RK Sinnott (1999), Coulson and Richardson’s Chemical Engineering, Third Edition, published by Buterworth Heinemann (1999).
4. Robert H.Perry and Don W. Green (1997), Perry’s Chemical Engineers’ Handbook, Seventh Edition, published by RR Donnelley and Sons’ Company.
