Mass Balance Report(1)
Short Description
mass balance calculation...
Description
Mass Balance Hand calculation Mass Balance across Desulphurization Tank
S1 = 100 kmol/h
Desulphurization
S2 = 99.85 kmol/h
CH4,1 = 84.8725 kmol/h kmol/h
CH4,2 = 84.8725
CO2,1 = 14.9775 kmol/h kmol/h
CO2,2 = 14.9775
H2S,1 = 0.1500 kmol/h
Basis set, S1= 100 kmol/h CH4,1
(
)
CO2,1
(
)
H2S,1
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
Hot Water
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.
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 kmol/h
H2O,5 = 203.5261
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 MethaneSteam Reforming Reaction, At °C, we obtained Kp value for methanesteam reforming reaction is 1.578 * Using given equation, we get
.
( (
)( )
(
)( )
(
(
)(
(
)(
)
)(
)
(
) )(
(
)(
)(
(
( ) )
)( (
(
) )(
(
)
)
)( (
(Kp)(P2)
(
( ),
)( (
)(
)( )
)
(
(
)( )
)(
) )
) )
) (
)
(
) =0
Therefore,
(
)(
(
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) )
(
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.
)
Carbon 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 f1 f2 f3 f4 f5
are set. 1.350772222 1.5489 20.1500 658.59 179.505
2. An initial guess for the 5 solution were taken. (20,30,40,50,60 respectively) Component H20 CH4 CO2 CO H2
Reactor effluent 20 30 40 50 60
Total 3. An objective cell was set. ( objective cell
200 ) 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
nCO5
= 52.0762 kmol/h
nH2,5
= 244.1240 kmol/h
Mass Balance around Dryer
S6 Water
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 S12
S11
CO : 0.3a CO2 : 0.3b0.005S13 S10
CH4 : d0.005S13
REACTOR
S7
S8
H2 : c1.4a2.1b0.003S13
S9 SEPARATOR
zCO(7)=52.08kmol/h
CO
:a
CO : 0.3a
zO2(7)=36.95kmol/h
CO2
:b
CO2 : 0.3b
zH2(7)=244.12kmol/h H2
:c
H2 : c1.4a2.1b
zCH4(7)=10.82kmol/h CH4
:d
CH4 : d
CO2
: 0.005 x S13
CH3OH : 0.7a+0.7b
H2
: 0.003 x S13
H2O : 0.7b
S13 CH4
: 0.005 x S13
CH3OH : 0.7a+0.7b H2O
: 0.7b
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.
Unknowns set for solving are: S10, S13, a, b, c, d
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 CO balance:
0.3a S11 zCO( 7 ) a S10
mixing point,
0.3a 137 .588 52 .08 a S 10 41 .2764 a 52 .08 a S10
CO2 balance:
0.3b 0.005 S13 S11 zCO2 ( 7 ) b S10 0.3b 0.005 S13 137 .588 36 .95 b S10 41 .2764 b 0.68794 S13 36 .95 b S 10
CH4 balance:
d 0.005 S13 S11 zCH 4 ( 7 ) d S10 d 0.005 S13 137 .588 10 .82 d S10 137 .588 d 0.68794 S13 10 .82 d S 10
All the equations derived are rearranged such that they are equated to "0". c + d  0.4a 0.7a + 1.4b
0.4b  0.013S13  S10 = 0 41 .2764 a 52 .08 a 0 S10 + 0.013S13  S13 = 0
a + b + c + d  S8 = 0 41 .2764 b 0.68794 S13 36 .95 b 0 S10
137 .588 d 0.68794 S13 10 .82 d 0 S10
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
Another table for functions are set. The 6 equations derived are then typed into the cells from H1H7 by substituting the unknowns in the equation from the variable cells (B2B7). G
H
1
Function
Equations
2
F1
3
F2
4
F3
5
F4
6
F5
7
F6
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 B2B7 are taken for the column "By Changing Variable Cells" in the solver parameter. Cells from H2H7 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
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. Stream 8 Componen t
Flowrate (kmol/h)
Mole Fraction
CO
64.1696
CO2
45.1038
0.093662
H2
344.1224
0.714602
CH4
28.1622
0.058481
Total
481.558
1
0.133254
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
= 0.7a + 0.7b
76.4914
0.2328
= 0.7b
31.5727
0.09609
328.5752
1
CH3OH H2O Total
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
219.0879
1
Total
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
= 0.7a + 0.7b
76.4914
0.6896
= 0.7b
31.5727
0.2884
109.4874
1
CH3OH H2O Total
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
137.588
1
Total
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
81.49986
1
Total
Mass balance across Distillation Columns S17
S15
xCH3OH(17) : 0.9972
Light Gas
S13
S14
DC1
DC2
CH3OH : 76.4914kmol/h H2O : 31.5727kmol/h CO2 : 0.5474kmol/h H2 : 0.3285kmol/h
CH4 : 0.5474kmol/h
S16 xCH3OH(16) : 0.001
Distillation 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 Component mass balance
 (1)
: 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
180000 tonnes 1000 kg kmol year year tonne 32 .0007 kg 8000 hours kmol 703 .1089 hour
Scale up Factor, k (CH3OH)T
= k [zCH3OH(17)]
703.1089 = k (76.4603) k
= 9.1958
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 = 780.4682 kmol/h (scaled up)
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 1.87E08
8.03E12
Hydrogen
H2
2.016
Gas
0.02884
7.65E08
3.288E09
8.698E13
Methane
CH4
16.04
Gas
0.03431
0.00005469 3.661E09
1.1E11
Hydrogen Sulphate
H2S
34.08
Gas
0.03351
0.00001547 3.012E09
3.292E12
Carbon Monoxide
CO
28.01
Gas
0.02895
0.00000411 3.548E09
2.22E12
Carbon Dioxide
CO2
44.01
Gas
0.03611
0.00004233 2.887E08
7.464E12
Water
H2O
18.016
Gas
0.03346
0.00000688 7.604E09
3.593E12
Liquid
0.0754

c

E1 (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 (E2) To calculate the power requirement of heat exchanger, this formula is used ∫ While
By integrating the Cp, we can get
=(ToutTin) 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 E3 & E18 Few assumptions are made for E3 & E18:
Assume both are isothermal reactors
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 ( changing variable as Tout
)
and
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 E7, E8 & E9 Assumptions
No pressure drop across the unit, remain 23.5 bars Latent heat is taken into consideration to find the total heat
+
∫
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. Q=0
∫
Inter stage Cooling For Inter stage cooling, E12E14E13, 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 E14 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
h = 94 MJ/kmol CH3OH
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
h= 53 MJ/kmol CH3OH
CH3OH + H2O
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 𝑻𝒇
𝑯𝒘
𝒐 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏
∫ 𝒄𝒑 𝒅𝑻 𝑻𝒊
𝒂𝑻𝒇
𝒃(𝑻𝒇 )^𝟐 𝟐
+
𝒄(𝑻𝒇 )^𝟑 𝟑
+
𝒅(𝑻𝒇 )^𝟒 𝟒
𝐚𝑻𝒊
𝒃(𝑻𝒊 )^𝟐 𝒄(𝑻𝒊 )^𝟑 𝒅(𝑻𝒊 )^𝟒  𝟑  𝟒 𝟐
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.
(𝑎 (
(
(
)
)
𝑏(
)
+
)
𝑐(
+
)
𝑑(
(
)
(
)
(
)
(
)
(
)
(
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)
= 3463.20 (
(
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)
= 3505.521 (
(
)
(
)
(
)
(
)
(
)
= 4621.73
(
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(
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(
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)
27810.15 (
(
(
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)
(
)
(
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)
= 16369.33 (
(
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)
(
)
(
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)
= 17048.07 (
(
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(
)
(
)
(
)
)
)
)
= 25415.17 (
(
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)
(
)
(
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)
= 38465.40 (
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(
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)
= 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  (
Changing variable as T2, we get T2 = 578.53 °C. Hence, exit temperature for methanol reactor is 593.15 °C.
) ^2 to the value of 0,
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 Sg
Pressure 42.5 bars Hg Hf
3233 6.3885
2799.5 1105.25
Sg Sf
6.044 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 reaction is superheated.
, so this
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%, P out=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 2954
T2 350 3076.956 3088
T2 345.8791 And we assume that this turbine is isentropic, so s1 = s2' At 42.5 bars S2’ Hg Sg 3076.956 6.530523 6.530523
At 0.13 Sg Sf
0.7165 8.058
So, It is saturated reaction, interpolation is used to find out the value of h'
At 0.13 0.7165
6.530523
213.5 h' h'=
8.058 2593
2097.9198
These formulas are then used to find out the steam requirement of the turbine
(
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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.
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