Chapter 4 (dp1)- Energy Balance.pdf
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Chapter 4 (dp1)- Energy Balance.pdf as PDF for free.
More details
- Words: 10,489
- Pages: 58
EH2207YY
DESIGN PROJECT 1 (SEPTEMBER 2017) PRODUCTION OF 212,499 METRIC TONNES OF METHANOL PER YEAR
CHAPTER 4: ENERGY BALANCE GROUP MEMBERS MUHAMMAD AMIR ASYRAF BIN ABD MANAP MIRDZA FAROUK BIN MURHAN MUKHOYIDDIN NUR IZYAN IZZATI BINTI AMIRUDDIN NABIHAH BINTI AZRON NURAIN SYAFIQAH BINTI MAHAZIR NURUL NAZAHA BINTI SHAMSURI
2015655852 2015636698 2014235224 2014479476 2014422778 2015635412
SUPERVISOR MADAM NURUL ASYIKIN BINTI MD ZAKI FACULTY OF CHEMICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA UITM SHAH ALAM
TABLE OF CONTENTS CHAPTER 4: ENERGY BALANCE 4.0
Introduction
4-1
Process Flow Diagram
4-4
4.1
Energy Equation
4-5
4.1.1
First Law of Thermodynamics
4-5
4.1.2
Enthalpy
4-6
4.2
Equation Relate to Energy Balance
4-7
i. Equation for Reactive Process
4-7
ii. Equation for Non-Reactive Process
4-7
iii. Equation for Heat Capacity
4-7
Thermodynamics Properties
4-8
i. Physical Properties
4-8
ii. Specific Heat Capacity Constants for Liquid
4-8
iii. Specific Heat Capacity Constant for Gas
4-8
iv. Heat of Vaporisation
4-9
v. Heat of Formation
4-9
Energy Balance Calculation
4-10
i. Procedure of Energy Balance
4-10
ii. General Assumptions for Manual Calculation
4-10
Compressor (C-101& C-102)
4-11
Cooler/Heater (E-101, E-102, E-103, E-104 & E-105)
4-20
Heat Exchanger (E-106)
4-38
Reactor (R-101)
4-39
Separator (S-101)
4-44
Flash Drum (S-102)
4-47
Distillation Column (T-101)
4-51
Distillation Column (T-102)
4-53
Table of Summary of Energy Balance and Error Percentage
4-55
4.5
Conclusion
4-56
4.6
References
4-56
4.3
4.4
CHAPTER 4
ENERGY BALANCE
4.0
Introduction This energy balance calculation is for the production of 125,000 metric tonnes of
Methanol per year. The methanol production is from produced from the syngas that supplied by the supplier. The syngas can be catalytically converted into methanol via overall exothermic reaction at medium temperatures of 150-270˚C. The formation of methanol from synthesis gas is taking place according to the following equation. There are two equation for the methanol synthesis, equation (1) the methanol is produced from carbon monoxide and hydrogen while equation (2) the methanol is produces from carbon dioxide and hydrogen. CO + 2H2
CH3OH
∆H˚298= -90.8 kJ/mol
(1)
CO2 + 3H2
CH3OH + H2O
∆H˚298= -49.6 kJ/mol
(2)
The process flow diagram shown below is the process flow for the methanol production from syngas. The temperatures and pressure conditions for every equipment are shown at every stream in the process flow diagram. The temperature and pressure operating condition for the methanol synthesis reactor is 150˚C and 50 atm. The highest temperature in this process is 180 ˚C, the outlet stream of the reactor. There are many equipment involve in this methanol production plant such as compressor, cooler, heater, mixer, heat exchanger, reactor, separator, flash drum and distillation tower The syngas enters the process in the stream 1 at 50˚C and 25 atm. The syngas is mostly contain hydrogen, carbon monoxide and carbon dioxide, but it also contains small amount of methane, nitrogen and water. The inert gas components that present in the syngas must be purged out from the system. The syngas is then compressed in two-stage compressor system in order to achieved 50 atm, the optimum condition to produced methanol are found to be reactor pressure at 50 atm. The production methanol from syngas may occur at low or high pressure. The high pressure process operates normally at 200 atm while the low pressure process operates
1
at 50-100 atm (Amin, Hassan, Das, Yeasmin, Rahman & Hossain, n.d). The low pressure process of producing methanol has such economical and operational advantages (Klier, 1982). After the syngas is compressed in the first compressor (C-101), the syngas will flow in to the cooler (E-101) to be cooled from 71 ˚C to 68 ˚C before entering the second compressor (C-102). The syngas need to be cooled down first before entering the second compressor to make sure that the temperature of the syngas do not increase drastically after being compressed in the compressor. The function of the compressor is to compress the gas and to increase the pressure of gas. When the pressure increase, the temperature also increase. The compressed gas at stream 4 will be mixed with the recycle stream (stream 12) from the separator (S-101) in the mixer (M-101). The gas will be cooled in E-102 from 78 ˚C to 31 ˚C before it entering the heat exchanger (E-106). The gas will be heated to 150 ˚C in E-106 before entering the reactor. The gas leaving stream the E-106 at 150 ˚C and 50 atm and will enter the reactor (R-101). The operating condition for the reactor are set to be 150 ˚C and 50 atm. The stream leaving from the reactor (stream 8) are at 180 ˚C and 48 atm and will be flow into the heat exchanger to cooled down the temperature from 180 ˚C to 140 ˚C. Then, it is cooled until the temperature is 41 ˚C and partially condensed in a water-cooler heat exchanger (E-103). Stream 10 enter the separator (S-101) at 41 ˚C and 48 atm. The gases component will be separated from the methanol and water mixture. The mixture of methanol and water will flow at the bottom stream (stream 13) while the gas will flow at the above stream and recycled back into the process. The liquid from the S-101 (stream 13) contains significant amount of light components and inert components. If this stream is fed directly to the distillation column, these lights components would build up in the condenser. Either a low temperature or high pressure needed in the condenser, it may require the use of expensive refrigeration. Therefore, a flash drum (S102) is used to vent the light components and the inert from the liquid before feeding into the distillation column. The inert are allowed to build up so that the losses of the reactants (hydrogen, carbon monoxide and carbon dioxide) are kept small. The flash drum operated at 0.2 atm.
2
The liquid from the flash drum (stream 16) is flow into the distillation column (T-101).. The column will separate methanol and water. The top product (stream 17) of the distillation column is methanol with purity about 98.9% while the bottom product (stream 18) is water. The temperature of the top stream is 52 ˚C while the temperature of the bottom stream is 71 ˚C. The bottom product from the distillation column will be flow into the second distillation column for further purification. .
3
50 25
Syngas
R-101
1
68 29.7
C-102
31 50
C-101
150
180
50
48.5
7 3
71 30
2
77 32.5
E-101
77 32.5
4
6
8
E-106
5 50 47.8
M-101
9
E-102
140
48.5
12 40 48.8
E-103 11
40 48.8
10
E-104 30 0.24
S-101
Vent gas
15
38 47
40 48.8
13
KEY
14
S-102
51 0.64
1 50667.48607 4993 50 25 11 20187.95341 2219.000 40 48.8
60 0.95
30 0.24
16
PFD and Stream Table Stream Mass flowrate (kg/h) Mole flowrate (Kmol/h)) Temperature (°C) Pressure (atm) Stream Mass flowrate (kg/h) Mole flowrate (Kmol/h)) Temperature (°C) Pressure (atm)
MeOH
17
E-105
Stream number Temperature (° C) Pressure (atm)
2 50667.48607 4993 71 30 12 20187.95341 2219.000 50 47.8
3 50667.48607 4993 68 29.7 13 30486.10459 1030.000 40 48.8
4 50667.48607 4993 77 32.5 14 30486.10459 1030.000 38 47
5 50667.48607 4993 77 32.5 15 1077.81980 41.160 30 0.24
MeOH
19
T-101 71 0.9
6 50667.48607 4993 31 50 16 29395.53457 988.401 30 0.24
7 50667.48607 4993 150 50 17 14581.37770 456.300 51 0.64
8 50486.63402 3970.034 180 48.5 18 14816.23938 532.200 71 0.9
9 50486.63402 3970.034 140 48.5 19 (MeOH) 11848.83718 370.3 60 0.95
10 50486.63402 3970.034 40 48.8 20 (Water) 2967.14745 161.900 99 1
T-102 18
99 1
20
Water
4
4.1
Energy Equation
4.1.1 First Law of Thermodynamics The First Law of Thermodynamic stated that the energy cannot be created or destroyed, but it can be converted from one form to another form with the interaction of heat, work and internal energy. The general equation for the First Law of Thermodynamics for closed system can be express as in Equation 4.1 ∆𝐸𝑇 = 𝑄 − 𝑊 (Equation 4.1) Where: ∆ET = the total energy change transfer to the system Q = heat transfer to system from its surrounding W= work done by the system on its surrounding The total energy changes (∆ET) that are being transferred to the system consist of internal energy (∆U), kinetic energy (∆Ek), and potential energy (∆EP). For the whole calculation, the kinetic and potential energy can be neglected because the system is not accelerating and not rising or falling, thus the Equation 4.2 turn to Equation 4.3 ∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊 (Equation 4.2) ∆𝑈 = 𝑄 − 𝑊 (Equation 4.3) Work done on or by a system is accomplished by the movement of the system boundary against a resisting force or the passage of an electrical current or radiation across the system. If there are no moving parts or electrical currents or radiation at the system, then work done is equal to zero, thus the Equation 4.3 can be simplified to Equation 4.4. ∆𝑈 = 𝑄(Equation 4.4)
5
The internal energy of a system is depends on the chemical composition, state of aggregation (solid, liquid or gas) and the temperature of the system materials. It is independent of pressure for ideal gases system and nearly independent of pressure for solids and liquid. If there is no temperature changes, phase changes or no chemical reaction occur in a system and if the change in pressure is less than a few atmosphere, then the internal energy is zero, thus the Equation 4.3 become Equation 4.5. 0 = 𝑄 − 𝑊 (Equation 4.5) The First Law of Thermodynamic equation for open system resembles the equation for closed system (Equation 4.2): ∆U̇ + ∆Ėk + ∆Ėp = Q̇ - Ẇ (Equation 4.6) Where: ∆U̇ = the rate transfer of internal energy (kJ/s) ∆Ėk = the rate transfer of kinetic energy (kJ/s) ∆Ėp = the rate transfer of potential energy (kJ/s) Q̇ = the rate heat transfer to system from its surrounding Ẇ= the rate work done by the system on its surrounding 4.1.2 Enthalpy Enthalpy is a measurement of energy in a thermodynamic system. When a process occurs at constant pressure, the heat evolved (either being released or absorbed) is equal to the change in the enthalpy. Enthalpy is state of function which depends on the state of function of temperature (T), pressure (P) and internal energy (U). Enthalpy is usually expressed as the change in the enthalpy (∆H) for a process between initial and final state ∆𝐻 = ∆𝑈 + ∆𝑃𝑉 (Equation 4.7) Where H is the enthalpy, U is the internal energy, P is the pressure and V is volume. For the liquid system, PV is neglect. Substitute Equation 4.7 into Equation 4.4 𝑄 = ∆𝐻 (Equation 4.8) Equation 4.8 also can be written as: For closed system, 𝑄 = ∑𝑛𝑖 ,𝑜𝑢𝑡 Ĥ𝑖 ,𝑜𝑢𝑡 − ∑𝑛𝑖 ,𝑖𝑛 Ĥ𝑖 ,𝑖𝑛 6
For open system, Q̇ = ∑ṅI ,out Ĥi,out − ∑ṅi ,in Ĥi ,in
Where: ni = molar of the i component (mol) ṅi = molar flowrate of the i component (mol/s) Ĥi = Specific enthalpy of i component (kJ/mol) 4.2
Equation Relate to Energy Balance i. Equation for Reactive Process: a) Heat of reaction method (preferable when there is a single reaction for which ∆Ĥ˚𝑟 is known). 𝑄 = ∆𝐻 = 𝜉∆Ĥ˚𝑟 + ∑𝑛𝑖 ,𝑜𝑢𝑡 Ĥ𝑖 ,𝑜𝑢𝑡 − ∑𝑛𝑖 ,𝑖𝑛 Ĥ𝑖 ,𝑖𝑛 ∆Ĥ˚𝑟 =
∑ |𝑣𝑖 |∆Ĥ˚𝑓𝑖 − 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
∑
|𝑣𝑖 |∆Ĥ˚𝑓𝑖
𝑟𝑒𝑎𝑎𝑐𝑡𝑎𝑛𝑡𝑠
b) Heat of formation method (preferable for multiple reactions and single reaction for which ∆Ĥ𝑟 is not readily available). Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in 𝑇2
Ĥi = ∆ Ĥ˚𝑓𝑖 + ∫ 𝐶𝑃𝑖 𝑑𝑇 𝑇1
Where: ∆ Ĥ˚𝑓𝑖 = standard heat of formation of i specie ∆Ĥ˚𝑟 = standard heat of reaction ii. Equation for Non-Reactive Process Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in iii. Equation for Heat Capacity 𝑇2
Ĥ = ∫ (𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + 𝐷𝑇 3 ) 𝑑𝑇 𝑇1
7
4.3
Thermodynamics Properties All the thermodynamics properties were taken from Elementary Principles of
Chemical Processes, 2005. i. Physical Properties Compound
Mol. Wt.
SG
Melting Point
Boiling Point
(g/mol)
(20⁰/4⁰)
(Tm)
(Tb)
Carbon monoxide
28.01
-
-205.1
-191.5
Carbon Dioxide
44.01
-
-56.6
-78
Hydrogen
2.016
-
-259.19
-252.76
Methane
16.04
-
-182.5
-161.5
Nitrogen
28.02
-
-210.00
-195.8
Water
18.016
1.004⁰
0.00
100.0
Methanol
32.04
0.792
-97.9
64.7
Table 4.3.1: Compounds Physical Properties
ii. Specific Heat Capacity Constants for Liquid Compound
A
B
C
D
Water
75.4x10-3
-
-
-
Methanol
75.86x10-3
16.83x10-5
-
-
Table 4.3.2: Specific Heat Capacity Constants for Liquid
iii. Specific Heat Capacity Constants for Gas Compound
A
B
C
D
Water
33.46x10-3
0.6880x10-5
0.760x10-8
-3.593x10-12
Methanol
42.03x10-3
8.301x10-5
-1.87x10-8
-8.03x10-12
Carbon monoxide 28.95x10-3
0.4110x10-5
0.3548x10-8
-2.220x10-12
Carbon Dioxide
36.11x10-3
4.233 x10-5
-2.887x10-8
7.464x10-12
Hydrogen
28.84x10-3
0.00765x10-5 0.3288x10-8
-0.8698x10-12
Methane
34.31x10-3
5.469x10-5
0.3661x10-8
-11.00x10-12
Nitrogen
29.00x10-3
0.2199x10-5
0.5723x10-8
-2.871x10-12
Table 4.3.3: Specific Heat Capacity Constants for Gas
8
iv. Heat of vaporisation Compound
Heat of vaporisation Hv (kJ/mol)
Carbon Dioxide
304.2
Carbon Monoxide
6.042
Hydrogen
0.904
Nitrogen
5.577
Methane
8.179
Water
40.656
Methanol
35.27
Note: The enthalpy of formation is referred at 25 ˚C and 1 atm Table 4.3.4: Compounds Heat of Vaporization
v. Heat of formation Compound
Heat of formation Hf (kJ/mol)
Water (l)
-285.84
Water (g)
-241.83
Methanol (l)
-238.61
Methanol (g)
-201.2
Carbon monoxide
-110.52
Carbon Dioxide
-393.5
Hydrogen
0
Methane
-74.85
Nitrogen
0
Note: The enthalpy of formation is referred at 25 ˚C and 1 atm Table 4.3.5: Compounds Heat of Formation
9
4.4
Energy Balance Calculation Before starting on the energy balance calculation, the method of the calculation should be known and some assumptions should be made for ease of calculation. i. Procedure of Energy Balance: 1. Perform all the required material calculations. 2. Write the appropriate form of the energy balance and delete any of the terms that are either zero or negligible for the given process system. 3. Choose the reference state-phase, temperature, and pressure for each species involved in the process. 4. For an open system, construct a table with columns for inlet and outlet steam component flow rates and specific enthalpies relative to the chosen reference states. 5. Calculate all required values and insert the values in the appropriate places in the table. 6. Solve the energy balance for whichever variable is unknown. ii. General Assumptions for Manual Calculations: 1. The system is assumed to be an open system. 2. The system operated at steady state. 3. There are no moving parts in the system (no shaft work, Ws = 0) 4. The linear velocities of all stream are identical (no kinetic energy, ΔEk = 0) 5. The elevation are at constant height (no potential energy, Δep = 0) 6. Effect of pressure on total enthalpy is negligible if the change of pressure, ΔP is lower than 50 atm. 7. The system is ideal solution system. 8. All the gas in the system is assumed to be ideal gas. 9. Heat of mixing is negligible (Hmix= 0)
10
COMPRESSOR (C-101 & C-102) Stream 1 Tin = 50˚ C Pin = 25 atm
C-101 Stream 2 Tout = 71˚ C Pout = 30 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 1 Stream 2 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 159.776 159.776 Ĥ1 = 0.941 Ĥ7 = 1.751 748.95 748.95 Ĥ2 = 0.728 Ĥ8 = 1.341 9.986 9.986 Ĥ3 = 1.885 Ĥ9 = 3.468 3065.702 3065.702 Ĥ4 = 0.721 Ĥ10 = 1.327 9.986 9.986 Ĥ5 = 0.727 Ĥ11 = 1.339 998.6 998.6 Ĥ6 = 0.909 Ĥ12 = 1.699 Table 4.4.1: Molar Flowrate and Specific Enthalpy for C-101
Path
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 50˚ C, 25 atm)
CO2 (g, 50˚ C, 1 atm)
Ĥ1𝑏
Calculation: 50
Ĥ1 = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 0.941 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 50˚ C, 25 atm)
CO (g, 50˚ C, 1 atm)
Ĥ2𝑏
11
Calculation: 50
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.728 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 50˚ C, 25 atm)
H2O (l, 50˚ C, 1 atm)
Ĥ3𝑏
Calculation: 50
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 1.885 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 50˚ C, 25 atm)
H2 (g, 50˚ C, 1 atm)
Ĥ4𝑏
Calculation: 50
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.721 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 50˚ C, 25 atm)
N2 (g, 50˚ C, 1 atm)
Ĥ5𝑏
Calculation: 50
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.727 kJ/mol
12
Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 50˚ C, 25 atm)
CH4 (g, 50˚ C, 1 atm)
Ĥ6𝑏
Calculation: 50
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.909 kJ/mol
Path: CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO2 (g, 71˚ C, 30 atm)
CO2 (g, 71˚ C, 1 atm)
Ĥ7𝑏
Calculation: 71
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.751 kJ/mol
Path:
CO (g, 25˚ C, 1 atm)
Ĥ8𝑎
Ĥ8
CO (g, 71˚ C, 30 atm)
CO (g, 71˚ C, 1 atm)
Ĥ8𝑏
13
Calculation: 71
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.341 kJ/mol
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 71˚ C, 30 atm)
H2O (l, 71˚ C, 1 atm)
Ĥ9𝑏
Calculation: 71
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.468 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
H2 (g, 71˚ C, 30 atm)
H2 (g, 71˚ C, 1 atm)
Ĥ10𝑏
Calculation: 71
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.327 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 71˚ C, 30 atm)
N2 (g, 71˚ C, 1 atm)
Ĥ11𝑏
Calculation: 71
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.339 kJ/mol
14
Path: CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 71˚ C, 30 atm)
CH4 (g, 71˚ C, 1 atm)
Ĥ12𝑏
Calculation: 71
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.699 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(159976)(1.751)+(748950)(1.341)+(9986)(3.468)+(3065702)(1.327)+(9986)(1.339)+(99860 0)(1.699)] [(159976)(0.941)+(748950)(0.728)+(9986)(1.885)+(3065702)(0.721)+(9986)(0.727)+(99860 0)(0.909)] = 3257153.592 kJ/h x 1h/3600s = 904.76488 kW
Stream 3 Tin = 68˚ C Pin = 29.7 atm
C-102
Stream 4 Tin = 77˚ C Pin = 32.5 atm
15
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 3 Stream 4 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 159.776 Ĥ1 = 1.635 Ĥ7 =1.986 748.95 748.95 Ĥ2 = 1.253 Ĥ8 =1.517 9.986 9.986 Ĥ3 = 3.242 Ĥ9 =3.921 3065.702 3065.702 Ĥ4 = 1.241 Ĥ10 =1.509 9.986 9.986 Ĥ5 = 1.252 Ĥ11 =1.515 998.6 998.6 Ĥ6 = 1.586 Ĥ12 =1.930 Table 4.4.2: Molar Flowrate and Specific Enthalpy for C-102
Path CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 68˚ C, 29.7 atm) Ĥ1𝑏
CO2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.635 kJ/mol
Path:
Ĥ2
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
CO (g, 68˚ C, 29.7 atm)
CO (g, 68˚ C, 1 atm)
Ĥ2𝑏
Calculation: 68
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.253 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 68˚ C, 29.7 atm)
H2O (l, 68˚ C, 1 atm)
Ĥ3𝑏
16
Calculation: 68
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.242 kJ/mol
Path:
Ĥ4
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
H2 (g, 68˚ C, 29.7 atm) Ĥ4𝑏
H2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.241 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 68˚ C, 29.7 atm)
N2 (g, 68˚ C, 1 atm)
Ĥ5𝑏
Calculation: 68
Ĥ5 = ∫ 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.252 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 68˚ C, 29.7 atm)
CH4 (g, 668˚ C, 1 atm)
Ĥ6𝑏
Calculation: 68
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.586 kJ/mol
17
Path: Ĥ7
CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
CO2 (g, 77˚ C, 32.5 atm)
CO2 (g, 77˚ C, 1 atm)
Ĥ7𝑏
Calculation: 77
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.986 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO (g, 77˚ C, 32.5 atm)
CO (g, 77˚ C, 1 atm)
Ĥ8𝑏
Calculation: 77
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.517 kJ/mol
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 77˚ C, 32.5 atm)
H2O (l, 77˚ C, 1 atm)
Ĥ9𝑏
Calculation: 77
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.921 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
H2 (g, 77˚ C, 32.5 atm)
H2 (g, 77˚ C, 1 atm)
Ĥ10𝑏
18
Calculation: 77
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.509 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 77˚ C, 32.5 atm) Ĥ11𝑏
N2 (g, 77˚ C, 1 atm)
Calculation: 77
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.515 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 77˚ C, 32.5 atm)
CH4 (g, 77˚ C, 1 atm)
Ĥ12𝑏
Calculation: 77
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.930 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (8061198.5 – 6632860.976) kJ/mol = 396.76042 kW
19
COOLER/HEATER (E-101, E-102, E-103, E-104, E-105) Stream 2 Tin = 71˚ C Pin = 30 atm
Stream 3 Tout = 68˚ C Pout = 29.7 atm
E-101 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Path
Stream 2 Stream 3 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 159.776 159.776 Ĥ1 = 1.751 Ĥ7 = 1.635 748.95 748.95 Ĥ2 = 1.341 Ĥ8 = 1.253 9.986 9.986 Ĥ3 = 3.468 Ĥ9 = 3.242 3065.702 3065.702 Ĥ4 = 1.327 Ĥ10 = 1.241 9.986 9.986 Ĥ5 = 1.339 Ĥ11 = 1.252 998.6 998.6 Ĥ6 = 1.699 Ĥ12 = 1.586 Table 4.4.3: Molar Flowrate and Specific Enthalpy for E-101
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 71˚ C, 30 atm) Ĥ1𝑏
CO2 (g, 71˚ C, 1 atm)
Calculation: 71
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.751 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 71˚ C, 30 atm)
CO (g, 71˚ C, 1 atm)
Ĥ2𝑏
Calculation: 71
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.341 kJ/mol 20
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 71˚ C, 30 atm) Ĥ3𝑏
H2O (l, 71˚ C, 1 atm)
Calculation: 71
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.468 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 71˚ C, 30 atm)
H2 (g, 71˚ C, 1 atm)
Ĥ4𝑏
Calculation: 71
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.327 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 71˚ C, 30 atm)
N2 (g, 71˚ C, 1 atm)
Ĥ5𝑏
Calculation: 71
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.339 kJ/mol
21
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 71˚ C, 30 atm)
CH4 (g, 71˚ C, 1 atm)
Ĥ6𝑏
Calculation: 71
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.699 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO2 (g, 68˚ C, 29.7 atm)
CO2 (g, 68˚ C, 1 atm)
Ĥ7𝑏
Calculation: 68
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.635 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO (g, 68˚ C, 29.7 atm)
CO (g, 68˚ C, 1 atm)
Ĥ8𝑏
Calculation: 68
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.253 kJ/mol
22
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 68˚ C, 29.7 atm)
H2O (l, 68˚ C, 1 atm)
Ĥ9𝑏
Calculation: 68
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.242 kJ/mol
Path:
Ĥ10
H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
H2 (g, 68˚ C, 29.7 atm) Ĥ10𝑏
H2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.241 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 68˚ C, 29.7 atm)
N2 (g, 68˚ C, 1 atm)
Ĥ11𝑏
Calculation: 68
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.252 kJ/mol
23
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 68˚ C, 29.7 atm)
CH4 (g, 668˚ C, 1 atm)
Ĥ12𝑏
Calculation: 68
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.586 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (6632860.976 – 7096920.382) kJ/mol = - 128.90540 kW
Stream 5 Tin = 77˚ C Pin = 32.5 atm
Stream 6 Tout = 31˚ C Pout = 50 atm
E-102
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 5 Stream 6 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 159.776 Ĥ1 = 1.986 Ĥ8 =0.224 748.95 748.95 Ĥ2 = 1.516 Ĥ9 =0.174 9.986 9.986 Ĥ3 = 3.921 Ĥ10 =0.452 3065.702 3065.702 Ĥ4 = 1.500 Ĥ11 =0.173 9.986 9.986 Ĥ5 = 1.515 Ĥ12 =0.174 998.6 998.6 Ĥ6 = 1.930 Ĥ13 =0.215 Ĥ7 = 39.18 Ĥ14 =0.483 Table 4.4.4: Molar Flowrate and Specific Enthalpy for E-102
24
Path: Ĥ1
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
CO2 (g, 77˚ C, 32.5 atm) Ĥ1𝑏
CO2 (g, 77˚ C, 1 atm)
Calculation: 77
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.986 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 77˚ C, 32.5 atm)
CO (g, 77˚ C, 1 atm)
Ĥ2𝑏
Calculation: 77
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.516 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 77˚ C, 32.5 atm) Ĥ3𝑏
H2O (l, 77˚ C, 1 atm)
Calculation: 77
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.921 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 77˚ C, 32.5 atm)
H2 (g, 77˚ C, 1 atm)
Ĥ4𝑏
25
Calculation: 77
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.500 kJ/mol
Ĥ5
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
N2 (g, 77˚ C, 32.5 atm)
N2 (g, 77˚ C, 1 atm)
Ĥ5𝑏
Calculation: 77
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.515 kJ/mol
Path:
Ĥ6
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
CH4 (g, 77 C, 32.5 atm)
CH4 (g, 77˚ C, 1 atm)
Ĥ6𝑏
Calculation: 77
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.930 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ7
Ĥ7𝑎
CH3OH (g, 77˚ C, 32.5 atm) Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
26
Calculation: 64.7
77
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 0.599 kJ/mol = 39.18 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO2 (g, 31˚ C, 50 atm)
CO2 (g, 31 ˚C, 1 atm)
Ĥ8𝑏
Calculation: 30
Ĥ₈ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 0.224 kJ/mol
Path: CO (g, 25˚C, 1 atm) Ĥ9𝑎
Ĥ9
CO (g, 31˚C, 50 atm) Ĥ9𝑏
CO (g, 31˚C, 1 atm)
Calculation: 31
Ĥ9 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.174 kJ/mol
Path: H2O (l, 25˚ C, 1 atm)
Ĥ10𝑎
Ĥ10 H2O (l, 31˚C, 50 atm)
H2O (l, 31˚C, 1 atm)
Ĥ10𝑏
Calculation: 31
Ĥ10 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 0.452 kJ/mol 27
Path:
Ĥ11
H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
H2 (g, 31˚ C, 50 atm) Ĥ11𝑏
H2 (g, 31˚ C, 1 atm)
Calculation: 31
Ĥ11 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.173 kJ/mol
Path: N2 (g, 25˚C, 1 atm) Ĥ12𝑎
Ĥ12 N2 (g, 31˚C, 50 atm)
N2 (g, 31⁰C, 1 atm)
Ĥ12𝑏
Calculation: 31
Ĥ12 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.174 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm)
Ĥ13𝑎
Ĥ13
CH4 (g, 31˚ C, 50 atm)
CH4 (g, 31˚ C, 1 atm)
Ĥ13𝑏
Calculation: 31
Ĥ13 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.215 kJ/mol
Path:
CH3OH (l, 25 ˚C, 1 atm) Ĥ14𝑎
Calculation:
Ĥ14
CH3OH (l, 31 ˚C, 50 atm)
CH3OH (l, 31 ˚C, 1 atm)
Ĥ14𝑏
31
Ĥ14 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.483 kJ/mol 28
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (917423.806 – 8032858.232) kJ/mol = -1976.50956 kW
E-103
Stream 9 Tin = 140˚ C Pin = 48.5 atm
Stream 10 Tout = 40˚ C Pout = 48.8 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 9 Stream 10 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 131.011 131.011 Ĥ1 = 4.529 Ĥ8 = 0.562 269.962 269.962 Ĥ2 = 3.371 Ĥ9 = 0.436 27.790 27.790 Ĥ3 = 47.686 Ĥ10 = 1.131 2024.717 2024.717 Ĥ4 = 3.320 Ĥ11 = 0.433 11.910 11.910 Ĥ5 = 3.361 Ĥ12 = 0.436 1000.449 1000.449 Ĥ6 = 4.467 Ĥ13 = 0.541 508.164 508.164 Ĥ7 = 42.437 Ĥ14 = 1.220 Table 4.4.5: Molar Flowrate and Specific Enthalpy for E-103
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 140˚ C, 48.5 atm)
CO2 (g, 140˚ C, 1 atm)
Ĥ1𝑏
Calculation: 140
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 4.529 kJ/mol Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 140˚ C, 48.5 atm)
CO2 (g, 140˚ C, 1 atm)
Ĥ2𝑏
29
Calculation: 140
Ĥ2 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.371 kJ/mol Ĥ3
H2O (l, 25˚ C, 1 atm)
Path:
H2O (g, 140˚ C, 48.5 atm)
Ĥ3𝑎
Ĥ3𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
140
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 5.655 kJ/mol + 40.656 kJ/mol + 1.375 kJ/mol = 47.686 kJ/mol Ĥ4
Path: H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
H2 (g, 140˚ C, 48.5 atm) Ĥ4𝑏
H2 (g,140 C, 1 atm)
Calculation: 140
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.320 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5 N2 (g, 140˚C, 48.5 atm)
N2 (g, 140˚ C, 1 atm)
Ĥ5𝑏
Calculation: 140
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.361 kJ/mol Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6 CH4 (l, 140˚C, 48.5 atm)
CH4 (g, 140˚ C, 1 atm)
Ĥ6𝑏 30
Calculation: 140
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 4.467 kJ/mol
Ĥ7
Path: CH3OH (l, 25˚ C, 1 atm)
CH3OH (g, 140 ˚C, 48.5 atm)
Ĥ7𝑎
Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
140
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 3.856 kJ/mol = 42.437 kJ/mol
Path: Ĥ8
CO2 (g, 25˚ C, 1 atm) Ĥ8𝑎
CO2 (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ8𝑏
Calculation: 40
Ĥ8 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
CO (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ9𝑏
Calculation: 40
Ĥ9 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
31
Path:
Ĥ10
H2O (l, 25˚ C, 1 atm) Ĥ10𝑎
H2O (l, 40˚ C, 48.8 atm) Ĥ10𝑏
H2O (l, 40˚ C, 1 atm)
Calculation: 40
Ĥ10 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ11𝑏
Calculation: 40
Ĥ11 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
N2 (g, 40˚ C, 48.8 atm)
N2 (g, 40˚ C, 1 atm)
Ĥ12𝑏
Calculation: 40
Ĥ12 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol Path: CH4 (g, 25˚ C, 1 atm) Ĥ13𝑎
Ĥ13
CH4 (l, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ13𝑏
32
Calculation: 40
Ĥ13 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
Path: CH3OH (l, 25˚ C, 1 atm)
Ĥ14𝑎
Ĥ14
CH3OH (l, 38˚ C, 48.8 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ14𝑏
Calculation: 40
Ĥ14 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (2265860.314 – 35624635.96) kJ/h = -9266.32657 kW
Stream 11 Tin = 40˚ C Pin = 48.8 atm
Stream 12 Tout = 50˚ C Pout = 47.8 atm
E-104 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 11 Stream 12 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 6.657 6.657 Ĥ1 =0.562 Ĥ7 =0.9414 31.066 31.066 Ĥ2 =0.436 Ĥ8 =0.7277 1164.975 1164.975 Ĥ3 =0.433 Ĥ9 =0.7212 11.095 11.095 Ĥ4 =0.436 Ĥ10 =0.7273 987.455 987.455 Ĥ5 =0.541 Ĥ11 =2.8516 19.971 19.971 Ĥ6 =1.22 Ĥ12 =2.0543 Table 4.4.6: Molar Flowrate and Specific Enthalpy for E-104
33
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1815510.354– 1085135.38) kJ/h = 202.88193 kW
Stream 13 Tin = 40˚ C Pin = 48.8 atm
Stream 14 Tout = 38˚ C Pout = 47 atm
E-105 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 13 Stream 14 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 1.03 1.03 Ĥ1 = 0.562 Ĥ6 = 0.486 162.74 162.74 Ĥ2 = 1.131 Ĥ7 = 0.980 2.06 2.06 Ĥ3 = 0.433 Ĥ8 = 0.375 10.3 10.3 Ĥ4 = 0.541 Ĥ9 = 0.468 853.87 853.87 Ĥ5 = 1.220 Ĥ10 = 1.055 Table 4.4.7: Molar Flowrate and Specific Enthalpy for E-105
34
Path: CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 40˚ C, 48.8 atm) Ĥ1𝑏
CO2 (g, 40˚ C, 1 atm)
Calculation: 40
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol
Path: Ĥ2
H2O (l, 25˚ C, 1 atm) Ĥ2𝑎 Calculation:
H2O (l, 40˚ C, 48.8 atm)
H2O (l, 40˚ C, 1 atm)
Ĥ2𝑏
40
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ3𝑏
Calculation: 40
Ĥ3 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
CH4 (g, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ4𝑏
Calculation: 40
Ĥ4 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
35
Path:
Ĥ5
CH3OH (l, 25˚ C, 1 atm) Ĥ5𝑎
CH3OH (l, 40˚ C, 48.8 atm) Ĥ5𝑏 CH3OH (l, 40˚ C, 1 atm)
Calculation: 40
Ĥ5 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Path: CO2 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CO2 (g, 38˚ C, 47 atm)
CO2 (g, 38˚ C, 1 atm)
Ĥ6𝑏
Calculation: 38
Ĥ6 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.486 kJ/mol
Path: H2O (g, 25˚ C, 1 atm) Ĥ7𝑎
Calculation:
Ĥ7
H2O (g, 38˚ C, 47 atm) Ĥ7𝑏
H2O (g, 38˚ C, 1 atm)
38
Ĥ8 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.980 kJ/mol
Path: H2 (l, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
H2 (l, 38˚ C, 47 atm)
H2 (l, 38˚ C, 1 atm)
Ĥ8𝑏
Calculation: 38
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.375 kJ/mol
36
Path: CH4 (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
CH4 (g, 38˚ C, 47 atm)
CH4 (g, 38˚ C, 1 atm)
Ĥ9𝑏
Calculation: 38
Ĥ9 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.468 kJ/mol Path: CH3OH (l, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
CH3OH (l, 38˚ C, 47 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ10𝑏
Calculation: 38
Ĥ10 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.055 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1065639.03 – 1231931.5) kJ/h = -46.19235 kW
37
Heat Exchanger (E-106) Stream 8 Tin = 180˚ C Pin = 48.5 atm
Stream 6 Tin = 31˚ C Pin = 50 atm
Stream 7 Tout = 150˚ C Pout = 50 atm E-106
Stream 9 Tout = 140˚ C Pout = 48.5 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 6 Stream 8 Stream 7 ṅin ṅin ṅ𝑜𝑢𝑡 Ĥ𝑖𝑛 Ĥ𝑖𝑛 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) (kmol/hr) (kJ/mol) (kmol/hr) (kJ/mol) 159.776 0.224 131.011 6.2156 159.776 4.9453 748.950 0.174 269.962 4.5588 748.950 3.6674 9.986 0.452 27.790 49.076 9.986 48.033 3065.702 0.173 2024.717 4.4776 3065.702 3.6094 9.986 0.174 11.910 4.5403 9.986 3.6551 998.6 0.215 1000.449 6.1911 998.6 4.8896 0 0.483 508.164 44.665 0 42.983 Table 4.4.8: Molar Flowrate and Specific Enthalpy for FEHE
Stream 9 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 131.011 4.529 269.962 3.371 27.790 47.686 2024.717 3.32 11.910 3.361 1000.449 4.467 508.164 42.437
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(19999920.9+35624635.96) – (917423.806 + 41420622.12) kJ/h = 3690.69747 kW
38
REACTOR (R-101)
R-101 Stream 8 Tout = 180˚ C Pout = 48.5 atm
Stream 7 Tin = 150˚ C Pin = 50 atm
Reference state: C(s), O2 (g), H2 (g), N2 (g) at 25˚C and 1 atm Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 7 Stream 8 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 131.011 Ĥ1 = -388.555 Ĥ8 = -387.284 748.950 269.962 Ĥ2 = -106.853 Ĥ9 = -105.961 9.986 27.790 Ĥ3 = -237.807 Ĥ10 = -236.764 3065.702 2024.717 Ĥ4 = 3.609 Ĥ11 = 4.478 9.986 11.910 Ĥ5 = 3.655 Ĥ12 = 4.450 998.6 1000.449 Ĥ6 = -69.96 Ĥ13 = -68.659 0 508.164 Ĥ7 = -195.617 Ĥ14 = -193.935 Table 4.4.9: Molar Flowrate and Specific Enthalpy for R-101
Path: CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 150˚ C, 50 atm)
CO2 (g, 150˚ C, 1 atm)
Ĥ1𝑏
Calculation: 150
Ĥ1 = ∆ Ĥ˚𝑓 + ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = -393.5 kJ/mol + 4.945 kJ/mol = -388.555 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 150˚ C, 50 atm)
CO (g, 150˚ C, 1 atm)
Ĥ2𝑏
39
Calculation: 150
Ĥ2 = ∆ Ĥ˚𝑓 + ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -110.52 kJ/mol +3.667 kJ/mol = -106.853 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm)
Ĥ3
H2O (g, 150˚ C, 50 atm)
Ĥ3𝑎
Ĥ3𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
150
Ĥ3 = ∆ Ĥ˚𝑓 + ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -285.84 kJ/mol + 5.655 kJ/mol + 40.656 kJ/mol + 1.722 kJ/mol = -237.807 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 150˚ C, 50 atm)
H2 (g, 150˚ C, 1 atm)
Ĥ4𝑏
Calculation: 150
Ĥ4 = ∆ Ĥ˚𝑓 + ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +3.609 kJ/mol = 3.609 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 150˚ C, 50 atm)
N2 (g, 150˚ C, 1 atm)
Ĥ5𝑏
40
Calculation: 150
Ĥ5 = ∆ Ĥ˚𝑓 + ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +3.655kJ/mol = 3.655 kJ/mol
Path:
Ĥ6
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
CH4 (g, 150˚ C, 50 atm)
CH4 (g, 150˚ C, 1 atm)
Ĥ6𝑏
Calculation: 150
Ĥ6 = ∆ Ĥ˚𝑓 + ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -74.85 kJ/mol +4.890 kJ/mol = -69.96 kJ/mol
Path:
Ĥ7
CH3OH (l, 25˚ C, 1 atm)
CH3OH (g, 150˚ C, 50 atm)
Ĥ7𝑎
Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
150
Ĥ7 = ∆ Ĥ˚𝑓 + ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -238.6 kJ/mol + 3.311 kJ/mol + 35.27 kJ/mol + 4.402 kJ/mol = -195.617 kJ/mol Path: CO2 (g, 25˚ C, 1 atm)
Ĥ8𝑎
Ĥ8
CO2 (g, 180˚ C, 48.5 atm)
CO2 (g, 180˚ C, 1 atm)
Ĥ8𝑏
Calculation: 180
Ĥ8 = ∆ Ĥ˚𝑓 + ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = -393.5 kJ/mol + 6.216 kJ/mol 41
= -387.284 kJ/mol
Path:
Ĥ9
CO (g, 25˚ C, 1 atm) Ĥ9𝑎
CO (g, 180˚ C, 48.5 atm)
CO (g, 180˚ C, 1 atm)
Ĥ9𝑏
Calculation: 180
Ĥ9 = ∆ Ĥ˚𝑓 + ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -110.52 kJ/mol +4.559 kJ/mol = -105.961 kJ/mol Path:
H2O (l, 25˚ C, 1 atm)
Ĥ10
H2O (g, 180˚ C, 48.5 atm)
Ĥ10𝑎
Ĥ10𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
180
Ĥ10 = ∆ Ĥ˚𝑓 + ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -285.84 kJ/mol + 5.655 kJ/mol + 40.656 kJ/mol + 2.765 kJ/mol = -236.764 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
H2 (g, 180˚ C, 48.5 atm)
H2 (g, 180˚ C, 1 atm)
Ĥ11𝑏
Calculation: 180
Ĥ11 = ∆ Ĥ˚𝑓 + ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +4.478 kJ/mol = 4.478 kJ/mol
42
Path: N2 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
N2 (g, 180˚ C, 48.5 atm)
N2 (g, 180˚ C, 1 atm)
Ĥ12𝑏
Calculation: 180
Ĥ12 = ∆ Ĥ˚𝑓 + ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +4.540 kJ/mol = 4.540 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ13𝑎
Ĥ13
CH4 (g, 150˚ C, 48,5 atm)
CH4 (g, 180˚ C, 1 atm)
Ĥ13𝑏
Calculation: 180
Ĥ13 = ∆ Ĥ˚𝑓 + ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -74.85 kJ/mol + 6.191 kJ/mol = -68.659 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ14
Ĥ14𝑎
CH3OH (g, 180˚ C, 48.5 atm) Ĥ14𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
180
Ĥ14 = ∆ Ĥ˚𝑓 + ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -238.6 kJ/mol + 3.311 kJ/mol + 35.27 kJ/mol + 6.084 kJ/mol = -193.935 kJ/mol
43
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = -2440434438.3- (-203245497.4) = -11332.7614 kW
SEPARATOR (S-101) Stream 11 Tout = 40˚ C Pout = 48.8 atm Stream 10 Tin = 40˚ C Pin = 48.8 atm
S-101
Stream 13 Tout = 40˚ C Pout = 48.8 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
Stream 10 ṅin (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) 131.011 Ĥ1 = 0.562 269.962 Ĥ2 = 0.436 27.790 Ĥ3 = 1.131 2024.717 Ĥ4 = 0.433
Stream 13 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 1.030 Ĥ14 = 0.562 162.740 Ĥ15 = 1.131 2.06 Ĥ16 = 0.433 11.910 11.095 Ĥ5 = 0.436 Ĥ11 = 0.436 1000.449 987.455 Ĥ6 = 0.541 Ĥ12 = 0.541 10.3 Ĥ17 = 0.541 508.164 19.971 Ĥ7 = 1.220 Ĥ13 = 1.220 853.870 Ĥ18 = 1.220 Table 4.4.11: Molar Flowrate and Specific Enthalpy for S-101
CO2 CO H2O H2 N2 CH4 CH3OH
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
Stream 11 ṅout Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 6.657 Ĥ8 = 0.562 31.066 Ĥ9 = 0.436 1164.975 Ĥ10 = 0.433
CO2 (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ1𝑏
Calculation: 40
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol 44
Path:
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ2𝑏
Calculation: 40
Ĥ2 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 40˚ C, 48.8 atm)
H2O (l, 40˚ C, 1 atm)
Ĥ3𝑏
Calculation: 40
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ4𝑏
Calculation: 40
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm)
Ĥ5𝑎
Ĥ5
N2 (g, 40˚ C, 48.8 atm)
N2 (g, 40˚ C, 1 atm)
Ĥ5𝑏
Calculation: 40
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
45
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (l, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ6𝑏
Calculation: 40
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CH3OH (l, 38˚ C, 48.8 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ7𝑏
Calculation: 40
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(1085135.38)+ 1231931.5)] – (2265860.314) = 14.22405 kW
46
Flash Drum (S-102)
Stream 15 Tout = 30˚ C Pout = 0.24 atm
Stream 14 Tin = 38˚ C Pin = 47 atm
S-102
Stream 16 Tout = 30˚ C Pout = 0.24 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
Stream 14 Stream 15 Stream 16 ṅin (kmol/hr) Ĥ𝑖𝑛 ṅout ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kmol/hr) (kJ/mol) (kJ/mol) (kJ/mol) 1.030 Ĥ1 = 0.486 0.576 Ĥ6 = 0.186 0.0823 Ĥ7 = 0.145 162.740 Ĥ2 = 0.980 0.741 Ĥ8 = 0.377 162.098 Ĥ13 = 0.377 2.060 Ĥ3 = 0.375 2.428 Ĥ9 = 0.144 0.0823 Ĥ10 = 0.145 10.30 Ĥ4 = 0.468 10.249 Ĥ11 = 0.179 853.870 Ĥ5 = 1.055 27.001 Ĥ12 = 0.402 826.303 Ĥ14 = 0.402 Table 4.4.12: Molar Flowrate and Specific Enthalpy for S-102
CO2 CO H2O H2 N2 CH4 CH3OH
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 38˚ C, 47 atm)
CO2 (g, 38˚ C, 1 atm)
Ĥ1𝑏
Calculation: 38
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.486 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
H2O (l, 38˚ C, 47 atm)
H2O (l, 38˚ C, 1 atm)
Ĥ2𝑏
47
Calculation: 38
Ĥ2 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.980 kJ/mol
Path:
Ĥ3
H2 (g, 25˚ C, 1 atm) Ĥ3𝑎
H2 (g, 38˚ C, 47 atm) Ĥ3𝑏
H2 (g, 38˚ C, 1 atm)
Calculation: 38
Ĥ3 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.375 kJ/mol
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
CH4 (g, 38˚ C, 47 atm)
CH4 (g, 38˚ C, 1 atm)
Ĥ4𝑏
Calculation: 38
Ĥ4 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.468 kJ/mol
Path:
Ĥ5
CH3OH (l, 25˚ C, 1 atm) Ĥ5𝑎
CH3OH (l, 38˚ C, 47 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ5𝑏
Calculation: 38
Ĥ5 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.055 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CO2 (g, 30˚ C, 0.24 atm)
CO2 (g, 30˚ C, 1 atm)
Ĥ6𝑏
Calculation: 30
Ĥ6 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.186 kJ/mol
48
Path:
CO (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO (g, 30˚ C, 0.24 atm)
CO2 (g, 30˚ C, 1 atm)
Ĥ7𝑏
Calculation: 30
Ĥ7 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.145 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
H2O (l, 30˚ C, 0.24 atm)
H2O (l, 30˚ C, 1 atm)
Ĥ8𝑏
Calculation: 30
Ĥ8 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 30˚ C, 1 atm)
Ĥ9𝑏
Calculation: 30
Ĥ9 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.144 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
N2 (g, 30˚ C, 0.24 atm)
N2 (g, 30˚ C, 1 atm)
Ĥ10𝑏
Calculation: 30
Ĥ10 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.145 kJ/mol
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
CH4 (l, 30˚ C, 0.24 atm)
CH4 (g, 30˚ C, 1 atm)
Ĥ11𝑏
49
Calculation: 30
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.179 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH3OH (l, 30˚ C, 0.24 atm)
CH3OH (l, 30˚ C, 1 atm)
Ĥ12𝑏
Calculation: 30
Ĥ12 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.402 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (13099.33+393284.752) – 1065639.03) kJ/h = -183.12637 kW
50
Distillation Column (T-101)
Stream 17 Tout = 51˚ C Pout = 0.64 atm
Stream 16 Tin = 30˚ C Pin = 0.24 atm
T-101
Stream 18 Tout = 71˚ C Pout = 0.9 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 16 Stream 17 Stream 18 ṅin ṅ ṅ Ĥ𝑖𝑛 Ĥ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 out 𝑜𝑢𝑡 (kmol/hr) (kmol/hr) (kmol/hr) (kJ/mol) (kJ/mol) (kJ/mol) 162.098 Ĥ1 = 0.377 2.7378 Ĥ3 = 1.960 159.66 Ĥ5 = 3.468 826.303 453.5622 372.54 Ĥ2 = 0.402 Ĥ4 = 2.139 Ĥ6 = 3.861 Table 4.4.13: Molar Flowrate and Specific Enthalpy for T-101
H2O (l, 25˚ C, 1 atm)
Path:
Ĥ1𝑎
Ĥ1
H2O (l, 30˚ C, 0.24 atm)
H2O (l, 30˚ C, 1 atm)
Ĥ1𝑏
Calculation: 30
Ĥ1 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ2𝑎
Calculation:
Ĥ2
CH3OH (l, 30˚ C, 0.24 atm)
CH3OH (l, 30˚ C, 1 atm)
Ĥ3𝑏
30
Ĥ2 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.402 kJ/mol 51
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 51˚ C, 0.6 atm) Ĥ3𝑏
H2O (l, 51˚ C, 1 atm)
Calculation: 51
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.960 kJ/mol
Path:
Ĥ4
CH3OH (l, 25˚ C, 1 atm) Ĥ4𝑎
CH3OH (l, 51˚ C, 0.6 atm) Ĥ4𝑏
CH3OH (l, 51˚ C, 1 atm)
Calculation: 51
Ĥ4 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 2.139 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
H2O (l, 71˚ C, 0.9 atm) Ĥ5𝑏
H2O (l, 71˚ C, 1 atm)
Calculation: 71
Ĥ5 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 3.468 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH3OH (l, 71˚ C, 0.9 atm)
CH3OH (l, 71˚ C, 1 atm)
Ĥ6𝑏
Calculation: 71
Ĥ6 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 3.861 kJ/mol Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (975535.6338 + 1992077.82) – 393284.752 = 715.09131 kW
52
Distillation Column (T-102)
Stream 19 Tout = 60˚ C Pout = 0.95 atm
Stream 18 Tin = 71˚ C Pin = 0.9 atm
T-102
Stream 20 Tout = 99˚ C Pout = 1 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 18 ṅin Ĥ𝑖𝑛 (kmol/hr) (kJ/mol) 159.66 Ĥ1 = 3.468
Stream 20 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 158.3544 Ĥ5 = 5.5796 372.540 Ĥ2 = 3.861 368.1891 Ĥ4 = 2.905 3.54561 Ĥ6 = 3.861 Table 4.4.13: Molar Flowrate and Specific Enthalpy for T-101
H2O (l, 25˚ C, 1 atm)
Path:
Ĥ1𝑎
Ĥ1
Stream 19 ṅout Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 1.1109 Ĥ3 = 2.639
H2O (l, 71˚ C, 0.9 atm)
H2O (l, 71˚ C, 1 atm)
Ĥ1𝑏
Calculation: 71
Ĥ1 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol Path: CH3OH (l, 25˚ C, 1 atm) Ĥ2𝑎 Calculation:
Ĥ2
CH3OH (l, 71˚ C, 0.9 atm)
CH3OH (l, 71˚ C, 1 atm)
Ĥ3𝑏
71
Ĥ2 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 53
= 3.861 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 60˚ C, 0.95 atm) Ĥ3𝑏
H2O (l, 60˚ C, 1 atm)
Calculation: 60
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 2.639 kJ/mol
Path:
Ĥ4
CH3OH (l, 25˚ C, 1 atm) Ĥ4𝑎
CH3OH (l, 60˚ C, 0.95 atm)
CH3OH (l, 60˚ C, 1 atm)
Ĥ4𝑏
Calculation: 60
Ĥ4 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 2.905 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm)
Ĥ5
H2O (l, 99˚ C, 1 atm)
Calculation: 99
Ĥ5 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 5.5796 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ6
CH3OH (g, 99˚ C, 1 atm)
Ĥ6𝑎
Ĥ4𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
99
Ĥ6 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 1.701kJ/mol = 40.282 kJ/mol
54
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1072521.001 + 1026378.42) – 1992077.82 = 29.67266 kW
Table 4.4 14: Summary of Energy Balance. Equipment
Q calc (kW)
C-101 C-102 E-101 E-102 E-103 E-104 E-105 FEHE R-101 S-101 S-102 T-101 T-102
904.76489 396.76042 -128.90539 -1976.50956 -9266.32657 202.88194 -46.19235 3690.69747 -11332.7614 14.22405 -183.12637 715.09131 29.67266
55
4.5
Conclusion From the calculation the value of heat transfer, Q for equipment was not same. This
happens due to the different inlet and outlet components and the needs of energy used. The heat transfer with positive value represents the heat absorb which is as endothermic reaction. Then negative value represent heat release and as exothermic reaction.
REFERENCES
Amin, D. R., Hassan, I., Das, A., Yeasmin, R., Rahman, T., & Hossain, T. (n.d.). Simulation of Methanol Production from. 1-14. Felder, R. M. (2005). Elementary Principles of Chemical Processes. United States of America: John Wiley & Sons, Inc. Klier, K. (1982). Methanol Synthesis. Advances in Catalyst, 243-310. Luyben, W. L. (2010). Design and Control of a Methanol Reactor/Column Process. Industrial Engineering Chemical, 6150-6163.
56
DESIGN PROJECT 1 (SEPTEMBER 2017) PRODUCTION OF 212,499 METRIC TONNES OF METHANOL PER YEAR
CHAPTER 4: ENERGY BALANCE GROUP MEMBERS MUHAMMAD AMIR ASYRAF BIN ABD MANAP MIRDZA FAROUK BIN MURHAN MUKHOYIDDIN NUR IZYAN IZZATI BINTI AMIRUDDIN NABIHAH BINTI AZRON NURAIN SYAFIQAH BINTI MAHAZIR NURUL NAZAHA BINTI SHAMSURI
2015655852 2015636698 2014235224 2014479476 2014422778 2015635412
SUPERVISOR MADAM NURUL ASYIKIN BINTI MD ZAKI FACULTY OF CHEMICAL ENGINEERING UNIVERSITI TEKNOLOGI MARA UITM SHAH ALAM
TABLE OF CONTENTS CHAPTER 4: ENERGY BALANCE 4.0
Introduction
4-1
Process Flow Diagram
4-4
4.1
Energy Equation
4-5
4.1.1
First Law of Thermodynamics
4-5
4.1.2
Enthalpy
4-6
4.2
Equation Relate to Energy Balance
4-7
i. Equation for Reactive Process
4-7
ii. Equation for Non-Reactive Process
4-7
iii. Equation for Heat Capacity
4-7
Thermodynamics Properties
4-8
i. Physical Properties
4-8
ii. Specific Heat Capacity Constants for Liquid
4-8
iii. Specific Heat Capacity Constant for Gas
4-8
iv. Heat of Vaporisation
4-9
v. Heat of Formation
4-9
Energy Balance Calculation
4-10
i. Procedure of Energy Balance
4-10
ii. General Assumptions for Manual Calculation
4-10
Compressor (C-101& C-102)
4-11
Cooler/Heater (E-101, E-102, E-103, E-104 & E-105)
4-20
Heat Exchanger (E-106)
4-38
Reactor (R-101)
4-39
Separator (S-101)
4-44
Flash Drum (S-102)
4-47
Distillation Column (T-101)
4-51
Distillation Column (T-102)
4-53
Table of Summary of Energy Balance and Error Percentage
4-55
4.5
Conclusion
4-56
4.6
References
4-56
4.3
4.4
CHAPTER 4
ENERGY BALANCE
4.0
Introduction This energy balance calculation is for the production of 125,000 metric tonnes of
Methanol per year. The methanol production is from produced from the syngas that supplied by the supplier. The syngas can be catalytically converted into methanol via overall exothermic reaction at medium temperatures of 150-270˚C. The formation of methanol from synthesis gas is taking place according to the following equation. There are two equation for the methanol synthesis, equation (1) the methanol is produced from carbon monoxide and hydrogen while equation (2) the methanol is produces from carbon dioxide and hydrogen. CO + 2H2
CH3OH
∆H˚298= -90.8 kJ/mol
(1)
CO2 + 3H2
CH3OH + H2O
∆H˚298= -49.6 kJ/mol
(2)
The process flow diagram shown below is the process flow for the methanol production from syngas. The temperatures and pressure conditions for every equipment are shown at every stream in the process flow diagram. The temperature and pressure operating condition for the methanol synthesis reactor is 150˚C and 50 atm. The highest temperature in this process is 180 ˚C, the outlet stream of the reactor. There are many equipment involve in this methanol production plant such as compressor, cooler, heater, mixer, heat exchanger, reactor, separator, flash drum and distillation tower The syngas enters the process in the stream 1 at 50˚C and 25 atm. The syngas is mostly contain hydrogen, carbon monoxide and carbon dioxide, but it also contains small amount of methane, nitrogen and water. The inert gas components that present in the syngas must be purged out from the system. The syngas is then compressed in two-stage compressor system in order to achieved 50 atm, the optimum condition to produced methanol are found to be reactor pressure at 50 atm. The production methanol from syngas may occur at low or high pressure. The high pressure process operates normally at 200 atm while the low pressure process operates
1
at 50-100 atm (Amin, Hassan, Das, Yeasmin, Rahman & Hossain, n.d). The low pressure process of producing methanol has such economical and operational advantages (Klier, 1982). After the syngas is compressed in the first compressor (C-101), the syngas will flow in to the cooler (E-101) to be cooled from 71 ˚C to 68 ˚C before entering the second compressor (C-102). The syngas need to be cooled down first before entering the second compressor to make sure that the temperature of the syngas do not increase drastically after being compressed in the compressor. The function of the compressor is to compress the gas and to increase the pressure of gas. When the pressure increase, the temperature also increase. The compressed gas at stream 4 will be mixed with the recycle stream (stream 12) from the separator (S-101) in the mixer (M-101). The gas will be cooled in E-102 from 78 ˚C to 31 ˚C before it entering the heat exchanger (E-106). The gas will be heated to 150 ˚C in E-106 before entering the reactor. The gas leaving stream the E-106 at 150 ˚C and 50 atm and will enter the reactor (R-101). The operating condition for the reactor are set to be 150 ˚C and 50 atm. The stream leaving from the reactor (stream 8) are at 180 ˚C and 48 atm and will be flow into the heat exchanger to cooled down the temperature from 180 ˚C to 140 ˚C. Then, it is cooled until the temperature is 41 ˚C and partially condensed in a water-cooler heat exchanger (E-103). Stream 10 enter the separator (S-101) at 41 ˚C and 48 atm. The gases component will be separated from the methanol and water mixture. The mixture of methanol and water will flow at the bottom stream (stream 13) while the gas will flow at the above stream and recycled back into the process. The liquid from the S-101 (stream 13) contains significant amount of light components and inert components. If this stream is fed directly to the distillation column, these lights components would build up in the condenser. Either a low temperature or high pressure needed in the condenser, it may require the use of expensive refrigeration. Therefore, a flash drum (S102) is used to vent the light components and the inert from the liquid before feeding into the distillation column. The inert are allowed to build up so that the losses of the reactants (hydrogen, carbon monoxide and carbon dioxide) are kept small. The flash drum operated at 0.2 atm.
2
The liquid from the flash drum (stream 16) is flow into the distillation column (T-101).. The column will separate methanol and water. The top product (stream 17) of the distillation column is methanol with purity about 98.9% while the bottom product (stream 18) is water. The temperature of the top stream is 52 ˚C while the temperature of the bottom stream is 71 ˚C. The bottom product from the distillation column will be flow into the second distillation column for further purification. .
3
50 25
Syngas
R-101
1
68 29.7
C-102
31 50
C-101
150
180
50
48.5
7 3
71 30
2
77 32.5
E-101
77 32.5
4
6
8
E-106
5 50 47.8
M-101
9
E-102
140
48.5
12 40 48.8
E-103 11
40 48.8
10
E-104 30 0.24
S-101
Vent gas
15
38 47
40 48.8
13
KEY
14
S-102
51 0.64
1 50667.48607 4993 50 25 11 20187.95341 2219.000 40 48.8
60 0.95
30 0.24
16
PFD and Stream Table Stream Mass flowrate (kg/h) Mole flowrate (Kmol/h)) Temperature (°C) Pressure (atm) Stream Mass flowrate (kg/h) Mole flowrate (Kmol/h)) Temperature (°C) Pressure (atm)
MeOH
17
E-105
Stream number Temperature (° C) Pressure (atm)
2 50667.48607 4993 71 30 12 20187.95341 2219.000 50 47.8
3 50667.48607 4993 68 29.7 13 30486.10459 1030.000 40 48.8
4 50667.48607 4993 77 32.5 14 30486.10459 1030.000 38 47
5 50667.48607 4993 77 32.5 15 1077.81980 41.160 30 0.24
MeOH
19
T-101 71 0.9
6 50667.48607 4993 31 50 16 29395.53457 988.401 30 0.24
7 50667.48607 4993 150 50 17 14581.37770 456.300 51 0.64
8 50486.63402 3970.034 180 48.5 18 14816.23938 532.200 71 0.9
9 50486.63402 3970.034 140 48.5 19 (MeOH) 11848.83718 370.3 60 0.95
10 50486.63402 3970.034 40 48.8 20 (Water) 2967.14745 161.900 99 1
T-102 18
99 1
20
Water
4
4.1
Energy Equation
4.1.1 First Law of Thermodynamics The First Law of Thermodynamic stated that the energy cannot be created or destroyed, but it can be converted from one form to another form with the interaction of heat, work and internal energy. The general equation for the First Law of Thermodynamics for closed system can be express as in Equation 4.1 ∆𝐸𝑇 = 𝑄 − 𝑊 (Equation 4.1) Where: ∆ET = the total energy change transfer to the system Q = heat transfer to system from its surrounding W= work done by the system on its surrounding The total energy changes (∆ET) that are being transferred to the system consist of internal energy (∆U), kinetic energy (∆Ek), and potential energy (∆EP). For the whole calculation, the kinetic and potential energy can be neglected because the system is not accelerating and not rising or falling, thus the Equation 4.2 turn to Equation 4.3 ∆𝑈 + ∆𝐸𝑘 + ∆𝐸𝑝 = 𝑄 − 𝑊 (Equation 4.2) ∆𝑈 = 𝑄 − 𝑊 (Equation 4.3) Work done on or by a system is accomplished by the movement of the system boundary against a resisting force or the passage of an electrical current or radiation across the system. If there are no moving parts or electrical currents or radiation at the system, then work done is equal to zero, thus the Equation 4.3 can be simplified to Equation 4.4. ∆𝑈 = 𝑄(Equation 4.4)
5
The internal energy of a system is depends on the chemical composition, state of aggregation (solid, liquid or gas) and the temperature of the system materials. It is independent of pressure for ideal gases system and nearly independent of pressure for solids and liquid. If there is no temperature changes, phase changes or no chemical reaction occur in a system and if the change in pressure is less than a few atmosphere, then the internal energy is zero, thus the Equation 4.3 become Equation 4.5. 0 = 𝑄 − 𝑊 (Equation 4.5) The First Law of Thermodynamic equation for open system resembles the equation for closed system (Equation 4.2): ∆U̇ + ∆Ėk + ∆Ėp = Q̇ - Ẇ (Equation 4.6) Where: ∆U̇ = the rate transfer of internal energy (kJ/s) ∆Ėk = the rate transfer of kinetic energy (kJ/s) ∆Ėp = the rate transfer of potential energy (kJ/s) Q̇ = the rate heat transfer to system from its surrounding Ẇ= the rate work done by the system on its surrounding 4.1.2 Enthalpy Enthalpy is a measurement of energy in a thermodynamic system. When a process occurs at constant pressure, the heat evolved (either being released or absorbed) is equal to the change in the enthalpy. Enthalpy is state of function which depends on the state of function of temperature (T), pressure (P) and internal energy (U). Enthalpy is usually expressed as the change in the enthalpy (∆H) for a process between initial and final state ∆𝐻 = ∆𝑈 + ∆𝑃𝑉 (Equation 4.7) Where H is the enthalpy, U is the internal energy, P is the pressure and V is volume. For the liquid system, PV is neglect. Substitute Equation 4.7 into Equation 4.4 𝑄 = ∆𝐻 (Equation 4.8) Equation 4.8 also can be written as: For closed system, 𝑄 = ∑𝑛𝑖 ,𝑜𝑢𝑡 Ĥ𝑖 ,𝑜𝑢𝑡 − ∑𝑛𝑖 ,𝑖𝑛 Ĥ𝑖 ,𝑖𝑛 6
For open system, Q̇ = ∑ṅI ,out Ĥi,out − ∑ṅi ,in Ĥi ,in
Where: ni = molar of the i component (mol) ṅi = molar flowrate of the i component (mol/s) Ĥi = Specific enthalpy of i component (kJ/mol) 4.2
Equation Relate to Energy Balance i. Equation for Reactive Process: a) Heat of reaction method (preferable when there is a single reaction for which ∆Ĥ˚𝑟 is known). 𝑄 = ∆𝐻 = 𝜉∆Ĥ˚𝑟 + ∑𝑛𝑖 ,𝑜𝑢𝑡 Ĥ𝑖 ,𝑜𝑢𝑡 − ∑𝑛𝑖 ,𝑖𝑛 Ĥ𝑖 ,𝑖𝑛 ∆Ĥ˚𝑟 =
∑ |𝑣𝑖 |∆Ĥ˚𝑓𝑖 − 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠
∑
|𝑣𝑖 |∆Ĥ˚𝑓𝑖
𝑟𝑒𝑎𝑎𝑐𝑡𝑎𝑛𝑡𝑠
b) Heat of formation method (preferable for multiple reactions and single reaction for which ∆Ĥ𝑟 is not readily available). Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in 𝑇2
Ĥi = ∆ Ĥ˚𝑓𝑖 + ∫ 𝐶𝑃𝑖 𝑑𝑇 𝑇1
Where: ∆ Ĥ˚𝑓𝑖 = standard heat of formation of i specie ∆Ĥ˚𝑟 = standard heat of reaction ii. Equation for Non-Reactive Process Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in iii. Equation for Heat Capacity 𝑇2
Ĥ = ∫ (𝐴 + 𝐵𝑇 + 𝐶𝑇 2 + 𝐷𝑇 3 ) 𝑑𝑇 𝑇1
7
4.3
Thermodynamics Properties All the thermodynamics properties were taken from Elementary Principles of
Chemical Processes, 2005. i. Physical Properties Compound
Mol. Wt.
SG
Melting Point
Boiling Point
(g/mol)
(20⁰/4⁰)
(Tm)
(Tb)
Carbon monoxide
28.01
-
-205.1
-191.5
Carbon Dioxide
44.01
-
-56.6
-78
Hydrogen
2.016
-
-259.19
-252.76
Methane
16.04
-
-182.5
-161.5
Nitrogen
28.02
-
-210.00
-195.8
Water
18.016
1.004⁰
0.00
100.0
Methanol
32.04
0.792
-97.9
64.7
Table 4.3.1: Compounds Physical Properties
ii. Specific Heat Capacity Constants for Liquid Compound
A
B
C
D
Water
75.4x10-3
-
-
-
Methanol
75.86x10-3
16.83x10-5
-
-
Table 4.3.2: Specific Heat Capacity Constants for Liquid
iii. Specific Heat Capacity Constants for Gas Compound
A
B
C
D
Water
33.46x10-3
0.6880x10-5
0.760x10-8
-3.593x10-12
Methanol
42.03x10-3
8.301x10-5
-1.87x10-8
-8.03x10-12
Carbon monoxide 28.95x10-3
0.4110x10-5
0.3548x10-8
-2.220x10-12
Carbon Dioxide
36.11x10-3
4.233 x10-5
-2.887x10-8
7.464x10-12
Hydrogen
28.84x10-3
0.00765x10-5 0.3288x10-8
-0.8698x10-12
Methane
34.31x10-3
5.469x10-5
0.3661x10-8
-11.00x10-12
Nitrogen
29.00x10-3
0.2199x10-5
0.5723x10-8
-2.871x10-12
Table 4.3.3: Specific Heat Capacity Constants for Gas
8
iv. Heat of vaporisation Compound
Heat of vaporisation Hv (kJ/mol)
Carbon Dioxide
304.2
Carbon Monoxide
6.042
Hydrogen
0.904
Nitrogen
5.577
Methane
8.179
Water
40.656
Methanol
35.27
Note: The enthalpy of formation is referred at 25 ˚C and 1 atm Table 4.3.4: Compounds Heat of Vaporization
v. Heat of formation Compound
Heat of formation Hf (kJ/mol)
Water (l)
-285.84
Water (g)
-241.83
Methanol (l)
-238.61
Methanol (g)
-201.2
Carbon monoxide
-110.52
Carbon Dioxide
-393.5
Hydrogen
0
Methane
-74.85
Nitrogen
0
Note: The enthalpy of formation is referred at 25 ˚C and 1 atm Table 4.3.5: Compounds Heat of Formation
9
4.4
Energy Balance Calculation Before starting on the energy balance calculation, the method of the calculation should be known and some assumptions should be made for ease of calculation. i. Procedure of Energy Balance: 1. Perform all the required material calculations. 2. Write the appropriate form of the energy balance and delete any of the terms that are either zero or negligible for the given process system. 3. Choose the reference state-phase, temperature, and pressure for each species involved in the process. 4. For an open system, construct a table with columns for inlet and outlet steam component flow rates and specific enthalpies relative to the chosen reference states. 5. Calculate all required values and insert the values in the appropriate places in the table. 6. Solve the energy balance for whichever variable is unknown. ii. General Assumptions for Manual Calculations: 1. The system is assumed to be an open system. 2. The system operated at steady state. 3. There are no moving parts in the system (no shaft work, Ws = 0) 4. The linear velocities of all stream are identical (no kinetic energy, ΔEk = 0) 5. The elevation are at constant height (no potential energy, Δep = 0) 6. Effect of pressure on total enthalpy is negligible if the change of pressure, ΔP is lower than 50 atm. 7. The system is ideal solution system. 8. All the gas in the system is assumed to be ideal gas. 9. Heat of mixing is negligible (Hmix= 0)
10
COMPRESSOR (C-101 & C-102) Stream 1 Tin = 50˚ C Pin = 25 atm
C-101 Stream 2 Tout = 71˚ C Pout = 30 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 1 Stream 2 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 159.776 159.776 Ĥ1 = 0.941 Ĥ7 = 1.751 748.95 748.95 Ĥ2 = 0.728 Ĥ8 = 1.341 9.986 9.986 Ĥ3 = 1.885 Ĥ9 = 3.468 3065.702 3065.702 Ĥ4 = 0.721 Ĥ10 = 1.327 9.986 9.986 Ĥ5 = 0.727 Ĥ11 = 1.339 998.6 998.6 Ĥ6 = 0.909 Ĥ12 = 1.699 Table 4.4.1: Molar Flowrate and Specific Enthalpy for C-101
Path
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 50˚ C, 25 atm)
CO2 (g, 50˚ C, 1 atm)
Ĥ1𝑏
Calculation: 50
Ĥ1 = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 0.941 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 50˚ C, 25 atm)
CO (g, 50˚ C, 1 atm)
Ĥ2𝑏
11
Calculation: 50
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.728 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 50˚ C, 25 atm)
H2O (l, 50˚ C, 1 atm)
Ĥ3𝑏
Calculation: 50
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 1.885 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 50˚ C, 25 atm)
H2 (g, 50˚ C, 1 atm)
Ĥ4𝑏
Calculation: 50
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.721 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 50˚ C, 25 atm)
N2 (g, 50˚ C, 1 atm)
Ĥ5𝑏
Calculation: 50
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.727 kJ/mol
12
Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 50˚ C, 25 atm)
CH4 (g, 50˚ C, 1 atm)
Ĥ6𝑏
Calculation: 50
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.909 kJ/mol
Path: CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO2 (g, 71˚ C, 30 atm)
CO2 (g, 71˚ C, 1 atm)
Ĥ7𝑏
Calculation: 71
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.751 kJ/mol
Path:
CO (g, 25˚ C, 1 atm)
Ĥ8𝑎
Ĥ8
CO (g, 71˚ C, 30 atm)
CO (g, 71˚ C, 1 atm)
Ĥ8𝑏
13
Calculation: 71
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.341 kJ/mol
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 71˚ C, 30 atm)
H2O (l, 71˚ C, 1 atm)
Ĥ9𝑏
Calculation: 71
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.468 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
H2 (g, 71˚ C, 30 atm)
H2 (g, 71˚ C, 1 atm)
Ĥ10𝑏
Calculation: 71
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.327 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 71˚ C, 30 atm)
N2 (g, 71˚ C, 1 atm)
Ĥ11𝑏
Calculation: 71
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.339 kJ/mol
14
Path: CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 71˚ C, 30 atm)
CH4 (g, 71˚ C, 1 atm)
Ĥ12𝑏
Calculation: 71
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.699 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(159976)(1.751)+(748950)(1.341)+(9986)(3.468)+(3065702)(1.327)+(9986)(1.339)+(99860 0)(1.699)] [(159976)(0.941)+(748950)(0.728)+(9986)(1.885)+(3065702)(0.721)+(9986)(0.727)+(99860 0)(0.909)] = 3257153.592 kJ/h x 1h/3600s = 904.76488 kW
Stream 3 Tin = 68˚ C Pin = 29.7 atm
C-102
Stream 4 Tin = 77˚ C Pin = 32.5 atm
15
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 3 Stream 4 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 159.776 Ĥ1 = 1.635 Ĥ7 =1.986 748.95 748.95 Ĥ2 = 1.253 Ĥ8 =1.517 9.986 9.986 Ĥ3 = 3.242 Ĥ9 =3.921 3065.702 3065.702 Ĥ4 = 1.241 Ĥ10 =1.509 9.986 9.986 Ĥ5 = 1.252 Ĥ11 =1.515 998.6 998.6 Ĥ6 = 1.586 Ĥ12 =1.930 Table 4.4.2: Molar Flowrate and Specific Enthalpy for C-102
Path CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 68˚ C, 29.7 atm) Ĥ1𝑏
CO2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.635 kJ/mol
Path:
Ĥ2
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
CO (g, 68˚ C, 29.7 atm)
CO (g, 68˚ C, 1 atm)
Ĥ2𝑏
Calculation: 68
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.253 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 68˚ C, 29.7 atm)
H2O (l, 68˚ C, 1 atm)
Ĥ3𝑏
16
Calculation: 68
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.242 kJ/mol
Path:
Ĥ4
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
H2 (g, 68˚ C, 29.7 atm) Ĥ4𝑏
H2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.241 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 68˚ C, 29.7 atm)
N2 (g, 68˚ C, 1 atm)
Ĥ5𝑏
Calculation: 68
Ĥ5 = ∫ 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.252 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 68˚ C, 29.7 atm)
CH4 (g, 668˚ C, 1 atm)
Ĥ6𝑏
Calculation: 68
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.586 kJ/mol
17
Path: Ĥ7
CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
CO2 (g, 77˚ C, 32.5 atm)
CO2 (g, 77˚ C, 1 atm)
Ĥ7𝑏
Calculation: 77
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.986 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO (g, 77˚ C, 32.5 atm)
CO (g, 77˚ C, 1 atm)
Ĥ8𝑏
Calculation: 77
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.517 kJ/mol
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 77˚ C, 32.5 atm)
H2O (l, 77˚ C, 1 atm)
Ĥ9𝑏
Calculation: 77
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.921 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
H2 (g, 77˚ C, 32.5 atm)
H2 (g, 77˚ C, 1 atm)
Ĥ10𝑏
18
Calculation: 77
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.509 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 77˚ C, 32.5 atm) Ĥ11𝑏
N2 (g, 77˚ C, 1 atm)
Calculation: 77
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.515 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 77˚ C, 32.5 atm)
CH4 (g, 77˚ C, 1 atm)
Ĥ12𝑏
Calculation: 77
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.930 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (8061198.5 – 6632860.976) kJ/mol = 396.76042 kW
19
COOLER/HEATER (E-101, E-102, E-103, E-104, E-105) Stream 2 Tin = 71˚ C Pin = 30 atm
Stream 3 Tout = 68˚ C Pout = 29.7 atm
E-101 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Path
Stream 2 Stream 3 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 159.776 159.776 Ĥ1 = 1.751 Ĥ7 = 1.635 748.95 748.95 Ĥ2 = 1.341 Ĥ8 = 1.253 9.986 9.986 Ĥ3 = 3.468 Ĥ9 = 3.242 3065.702 3065.702 Ĥ4 = 1.327 Ĥ10 = 1.241 9.986 9.986 Ĥ5 = 1.339 Ĥ11 = 1.252 998.6 998.6 Ĥ6 = 1.699 Ĥ12 = 1.586 Table 4.4.3: Molar Flowrate and Specific Enthalpy for E-101
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 71˚ C, 30 atm) Ĥ1𝑏
CO2 (g, 71˚ C, 1 atm)
Calculation: 71
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.751 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 71˚ C, 30 atm)
CO (g, 71˚ C, 1 atm)
Ĥ2𝑏
Calculation: 71
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.341 kJ/mol 20
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 71˚ C, 30 atm) Ĥ3𝑏
H2O (l, 71˚ C, 1 atm)
Calculation: 71
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.468 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 71˚ C, 30 atm)
H2 (g, 71˚ C, 1 atm)
Ĥ4𝑏
Calculation: 71
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.327 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 71˚ C, 30 atm)
N2 (g, 71˚ C, 1 atm)
Ĥ5𝑏
Calculation: 71
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.339 kJ/mol
21
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (g, 71˚ C, 30 atm)
CH4 (g, 71˚ C, 1 atm)
Ĥ6𝑏
Calculation: 71
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.699 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO2 (g, 68˚ C, 29.7 atm)
CO2 (g, 68˚ C, 1 atm)
Ĥ7𝑏
Calculation: 68
Ĥ₇ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.635 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO (g, 68˚ C, 29.7 atm)
CO (g, 68˚ C, 1 atm)
Ĥ8𝑏
Calculation: 68
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.253 kJ/mol
22
Path: H2O (l, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2O (l, 68˚ C, 29.7 atm)
H2O (l, 68˚ C, 1 atm)
Ĥ9𝑏
Calculation: 68
Ĥ9 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.242 kJ/mol
Path:
Ĥ10
H2 (g, 25˚ C, 1 atm) Ĥ10𝑎
H2 (g, 68˚ C, 29.7 atm) Ĥ10𝑏
H2 (g, 68˚ C, 1 atm)
Calculation: 68
Ĥ10 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.241 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
N2 (g, 68˚ C, 29.7 atm)
N2 (g, 68˚ C, 1 atm)
Ĥ11𝑏
Calculation: 68
Ĥ11 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.252 kJ/mol
23
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH4 (g, 68˚ C, 29.7 atm)
CH4 (g, 668˚ C, 1 atm)
Ĥ12𝑏
Calculation: 68
Ĥ12 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.586 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (6632860.976 – 7096920.382) kJ/mol = - 128.90540 kW
Stream 5 Tin = 77˚ C Pin = 32.5 atm
Stream 6 Tout = 31˚ C Pout = 50 atm
E-102
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 5 Stream 6 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 159.776 Ĥ1 = 1.986 Ĥ8 =0.224 748.95 748.95 Ĥ2 = 1.516 Ĥ9 =0.174 9.986 9.986 Ĥ3 = 3.921 Ĥ10 =0.452 3065.702 3065.702 Ĥ4 = 1.500 Ĥ11 =0.173 9.986 9.986 Ĥ5 = 1.515 Ĥ12 =0.174 998.6 998.6 Ĥ6 = 1.930 Ĥ13 =0.215 Ĥ7 = 39.18 Ĥ14 =0.483 Table 4.4.4: Molar Flowrate and Specific Enthalpy for E-102
24
Path: Ĥ1
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
CO2 (g, 77˚ C, 32.5 atm) Ĥ1𝑏
CO2 (g, 77˚ C, 1 atm)
Calculation: 77
Ĥ₁ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 1.986 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 77˚ C, 32.5 atm)
CO (g, 77˚ C, 1 atm)
Ĥ2𝑏
Calculation: 77
Ĥ2 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.516 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 77˚ C, 32.5 atm) Ĥ3𝑏
H2O (l, 77˚ C, 1 atm)
Calculation: 77
Ĥ3 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 3.921 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 77˚ C, 32.5 atm)
H2 (g, 77˚ C, 1 atm)
Ĥ4𝑏
25
Calculation: 77
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.500 kJ/mol
Ĥ5
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
N2 (g, 77˚ C, 32.5 atm)
N2 (g, 77˚ C, 1 atm)
Ĥ5𝑏
Calculation: 77
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 1.515 kJ/mol
Path:
Ĥ6
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
CH4 (g, 77 C, 32.5 atm)
CH4 (g, 77˚ C, 1 atm)
Ĥ6𝑏
Calculation: 77
Ĥ6 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 1.930 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ7
Ĥ7𝑎
CH3OH (g, 77˚ C, 32.5 atm) Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
26
Calculation: 64.7
77
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 0.599 kJ/mol = 39.18 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
CO2 (g, 31˚ C, 50 atm)
CO2 (g, 31 ˚C, 1 atm)
Ĥ8𝑏
Calculation: 30
Ĥ₈ = ∫ 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 25
= 0.224 kJ/mol
Path: CO (g, 25˚C, 1 atm) Ĥ9𝑎
Ĥ9
CO (g, 31˚C, 50 atm) Ĥ9𝑏
CO (g, 31˚C, 1 atm)
Calculation: 31
Ĥ9 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.174 kJ/mol
Path: H2O (l, 25˚ C, 1 atm)
Ĥ10𝑎
Ĥ10 H2O (l, 31˚C, 50 atm)
H2O (l, 31˚C, 1 atm)
Ĥ10𝑏
Calculation: 31
Ĥ10 = ∫ 75.4 𝑥 10−3 𝑑𝑡 25
= 0.452 kJ/mol 27
Path:
Ĥ11
H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
H2 (g, 31˚ C, 50 atm) Ĥ11𝑏
H2 (g, 31˚ C, 1 atm)
Calculation: 31
Ĥ11 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.173 kJ/mol
Path: N2 (g, 25˚C, 1 atm) Ĥ12𝑎
Ĥ12 N2 (g, 31˚C, 50 atm)
N2 (g, 31⁰C, 1 atm)
Ĥ12𝑏
Calculation: 31
Ĥ12 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.174 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm)
Ĥ13𝑎
Ĥ13
CH4 (g, 31˚ C, 50 atm)
CH4 (g, 31˚ C, 1 atm)
Ĥ13𝑏
Calculation: 31
Ĥ13 = ∫ 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.215 kJ/mol
Path:
CH3OH (l, 25 ˚C, 1 atm) Ĥ14𝑎
Calculation:
Ĥ14
CH3OH (l, 31 ˚C, 50 atm)
CH3OH (l, 31 ˚C, 1 atm)
Ĥ14𝑏
31
Ĥ14 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.483 kJ/mol 28
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (917423.806 – 8032858.232) kJ/mol = -1976.50956 kW
E-103
Stream 9 Tin = 140˚ C Pin = 48.5 atm
Stream 10 Tout = 40˚ C Pout = 48.8 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 9 Stream 10 ṅin (kmol/hr) ṅ (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) Ĥ𝑜𝑢𝑡 (kJ/mol) 𝑜𝑢𝑡 131.011 131.011 Ĥ1 = 4.529 Ĥ8 = 0.562 269.962 269.962 Ĥ2 = 3.371 Ĥ9 = 0.436 27.790 27.790 Ĥ3 = 47.686 Ĥ10 = 1.131 2024.717 2024.717 Ĥ4 = 3.320 Ĥ11 = 0.433 11.910 11.910 Ĥ5 = 3.361 Ĥ12 = 0.436 1000.449 1000.449 Ĥ6 = 4.467 Ĥ13 = 0.541 508.164 508.164 Ĥ7 = 42.437 Ĥ14 = 1.220 Table 4.4.5: Molar Flowrate and Specific Enthalpy for E-103
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 140˚ C, 48.5 atm)
CO2 (g, 140˚ C, 1 atm)
Ĥ1𝑏
Calculation: 140
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 4.529 kJ/mol Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 140˚ C, 48.5 atm)
CO2 (g, 140˚ C, 1 atm)
Ĥ2𝑏
29
Calculation: 140
Ĥ2 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.371 kJ/mol Ĥ3
H2O (l, 25˚ C, 1 atm)
Path:
H2O (g, 140˚ C, 48.5 atm)
Ĥ3𝑎
Ĥ3𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
140
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 5.655 kJ/mol + 40.656 kJ/mol + 1.375 kJ/mol = 47.686 kJ/mol Ĥ4
Path: H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
H2 (g, 140˚ C, 48.5 atm) Ĥ4𝑏
H2 (g,140 C, 1 atm)
Calculation: 140
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.320 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5 N2 (g, 140˚C, 48.5 atm)
N2 (g, 140˚ C, 1 atm)
Ĥ5𝑏
Calculation: 140
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.361 kJ/mol Path: CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6 CH4 (l, 140˚C, 48.5 atm)
CH4 (g, 140˚ C, 1 atm)
Ĥ6𝑏 30
Calculation: 140
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 4.467 kJ/mol
Ĥ7
Path: CH3OH (l, 25˚ C, 1 atm)
CH3OH (g, 140 ˚C, 48.5 atm)
Ĥ7𝑎
Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
140
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 3.856 kJ/mol = 42.437 kJ/mol
Path: Ĥ8
CO2 (g, 25˚ C, 1 atm) Ĥ8𝑎
CO2 (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ8𝑏
Calculation: 40
Ĥ8 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol
Path:
CO (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
CO (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ9𝑏
Calculation: 40
Ĥ9 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
31
Path:
Ĥ10
H2O (l, 25˚ C, 1 atm) Ĥ10𝑎
H2O (l, 40˚ C, 48.8 atm) Ĥ10𝑏
H2O (l, 40˚ C, 1 atm)
Calculation: 40
Ĥ10 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ11𝑏
Calculation: 40
Ĥ11 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
N2 (g, 40˚ C, 48.8 atm)
N2 (g, 40˚ C, 1 atm)
Ĥ12𝑏
Calculation: 40
Ĥ12 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol Path: CH4 (g, 25˚ C, 1 atm) Ĥ13𝑎
Ĥ13
CH4 (l, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ13𝑏
32
Calculation: 40
Ĥ13 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
Path: CH3OH (l, 25˚ C, 1 atm)
Ĥ14𝑎
Ĥ14
CH3OH (l, 38˚ C, 48.8 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ14𝑏
Calculation: 40
Ĥ14 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (2265860.314 – 35624635.96) kJ/h = -9266.32657 kW
Stream 11 Tin = 40˚ C Pin = 48.8 atm
Stream 12 Tout = 50˚ C Pout = 47.8 atm
E-104 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 11 Stream 12 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 6.657 6.657 Ĥ1 =0.562 Ĥ7 =0.9414 31.066 31.066 Ĥ2 =0.436 Ĥ8 =0.7277 1164.975 1164.975 Ĥ3 =0.433 Ĥ9 =0.7212 11.095 11.095 Ĥ4 =0.436 Ĥ10 =0.7273 987.455 987.455 Ĥ5 =0.541 Ĥ11 =2.8516 19.971 19.971 Ĥ6 =1.22 Ĥ12 =2.0543 Table 4.4.6: Molar Flowrate and Specific Enthalpy for E-104
33
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1815510.354– 1085135.38) kJ/h = 202.88193 kW
Stream 13 Tin = 40˚ C Pin = 48.8 atm
Stream 14 Tout = 38˚ C Pout = 47 atm
E-105 Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 13 Stream 14 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 1.03 1.03 Ĥ1 = 0.562 Ĥ6 = 0.486 162.74 162.74 Ĥ2 = 1.131 Ĥ7 = 0.980 2.06 2.06 Ĥ3 = 0.433 Ĥ8 = 0.375 10.3 10.3 Ĥ4 = 0.541 Ĥ9 = 0.468 853.87 853.87 Ĥ5 = 1.220 Ĥ10 = 1.055 Table 4.4.7: Molar Flowrate and Specific Enthalpy for E-105
34
Path: CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 40˚ C, 48.8 atm) Ĥ1𝑏
CO2 (g, 40˚ C, 1 atm)
Calculation: 40
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol
Path: Ĥ2
H2O (l, 25˚ C, 1 atm) Ĥ2𝑎 Calculation:
H2O (l, 40˚ C, 48.8 atm)
H2O (l, 40˚ C, 1 atm)
Ĥ2𝑏
40
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ3𝑏
Calculation: 40
Ĥ3 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
CH4 (g, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ4𝑏
Calculation: 40
Ĥ4 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
35
Path:
Ĥ5
CH3OH (l, 25˚ C, 1 atm) Ĥ5𝑎
CH3OH (l, 40˚ C, 48.8 atm) Ĥ5𝑏 CH3OH (l, 40˚ C, 1 atm)
Calculation: 40
Ĥ5 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Path: CO2 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CO2 (g, 38˚ C, 47 atm)
CO2 (g, 38˚ C, 1 atm)
Ĥ6𝑏
Calculation: 38
Ĥ6 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.486 kJ/mol
Path: H2O (g, 25˚ C, 1 atm) Ĥ7𝑎
Calculation:
Ĥ7
H2O (g, 38˚ C, 47 atm) Ĥ7𝑏
H2O (g, 38˚ C, 1 atm)
38
Ĥ8 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.980 kJ/mol
Path: H2 (l, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
H2 (l, 38˚ C, 47 atm)
H2 (l, 38˚ C, 1 atm)
Ĥ8𝑏
Calculation: 38
Ĥ8 = ∫ 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 25
= 0.375 kJ/mol
36
Path: CH4 (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
CH4 (g, 38˚ C, 47 atm)
CH4 (g, 38˚ C, 1 atm)
Ĥ9𝑏
Calculation: 38
Ĥ9 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.468 kJ/mol Path: CH3OH (l, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
CH3OH (l, 38˚ C, 47 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ10𝑏
Calculation: 38
Ĥ10 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.055 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1065639.03 – 1231931.5) kJ/h = -46.19235 kW
37
Heat Exchanger (E-106) Stream 8 Tin = 180˚ C Pin = 48.5 atm
Stream 6 Tin = 31˚ C Pin = 50 atm
Stream 7 Tout = 150˚ C Pout = 50 atm E-106
Stream 9 Tout = 140˚ C Pout = 48.5 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 6 Stream 8 Stream 7 ṅin ṅin ṅ𝑜𝑢𝑡 Ĥ𝑖𝑛 Ĥ𝑖𝑛 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) (kmol/hr) (kJ/mol) (kmol/hr) (kJ/mol) 159.776 0.224 131.011 6.2156 159.776 4.9453 748.950 0.174 269.962 4.5588 748.950 3.6674 9.986 0.452 27.790 49.076 9.986 48.033 3065.702 0.173 2024.717 4.4776 3065.702 3.6094 9.986 0.174 11.910 4.5403 9.986 3.6551 998.6 0.215 1000.449 6.1911 998.6 4.8896 0 0.483 508.164 44.665 0 42.983 Table 4.4.8: Molar Flowrate and Specific Enthalpy for FEHE
Stream 9 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 131.011 4.529 269.962 3.371 27.790 47.686 2024.717 3.32 11.910 3.361 1000.449 4.467 508.164 42.437
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(19999920.9+35624635.96) – (917423.806 + 41420622.12) kJ/h = 3690.69747 kW
38
REACTOR (R-101)
R-101 Stream 8 Tout = 180˚ C Pout = 48.5 atm
Stream 7 Tin = 150˚ C Pin = 50 atm
Reference state: C(s), O2 (g), H2 (g), N2 (g) at 25˚C and 1 atm Compound CO2 CO H2O H2 N2 CH4 CH3OH
Stream 7 Stream 8 ṅin (kmol/hr) ṅ𝑜𝑢𝑡 (kmol/hr) Ĥ𝑜𝑢𝑡 (kJ/mol) Ĥ𝑖𝑛 (kJ/mol) 159.776 131.011 Ĥ1 = -388.555 Ĥ8 = -387.284 748.950 269.962 Ĥ2 = -106.853 Ĥ9 = -105.961 9.986 27.790 Ĥ3 = -237.807 Ĥ10 = -236.764 3065.702 2024.717 Ĥ4 = 3.609 Ĥ11 = 4.478 9.986 11.910 Ĥ5 = 3.655 Ĥ12 = 4.450 998.6 1000.449 Ĥ6 = -69.96 Ĥ13 = -68.659 0 508.164 Ĥ7 = -195.617 Ĥ14 = -193.935 Table 4.4.9: Molar Flowrate and Specific Enthalpy for R-101
Path: CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 150˚ C, 50 atm)
CO2 (g, 150˚ C, 1 atm)
Ĥ1𝑏
Calculation: 150
Ĥ1 = ∆ Ĥ˚𝑓 + ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = -393.5 kJ/mol + 4.945 kJ/mol = -388.555 kJ/mol
Path: CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 150˚ C, 50 atm)
CO (g, 150˚ C, 1 atm)
Ĥ2𝑏
39
Calculation: 150
Ĥ2 = ∆ Ĥ˚𝑓 + ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -110.52 kJ/mol +3.667 kJ/mol = -106.853 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm)
Ĥ3
H2O (g, 150˚ C, 50 atm)
Ĥ3𝑎
Ĥ3𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
150
Ĥ3 = ∆ Ĥ˚𝑓 + ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -285.84 kJ/mol + 5.655 kJ/mol + 40.656 kJ/mol + 1.722 kJ/mol = -237.807 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 150˚ C, 50 atm)
H2 (g, 150˚ C, 1 atm)
Ĥ4𝑏
Calculation: 150
Ĥ4 = ∆ Ĥ˚𝑓 + ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +3.609 kJ/mol = 3.609 kJ/mol
Path: N2 (g, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
N2 (g, 150˚ C, 50 atm)
N2 (g, 150˚ C, 1 atm)
Ĥ5𝑏
40
Calculation: 150
Ĥ5 = ∆ Ĥ˚𝑓 + ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +3.655kJ/mol = 3.655 kJ/mol
Path:
Ĥ6
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
CH4 (g, 150˚ C, 50 atm)
CH4 (g, 150˚ C, 1 atm)
Ĥ6𝑏
Calculation: 150
Ĥ6 = ∆ Ĥ˚𝑓 + ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -74.85 kJ/mol +4.890 kJ/mol = -69.96 kJ/mol
Path:
Ĥ7
CH3OH (l, 25˚ C, 1 atm)
CH3OH (g, 150˚ C, 50 atm)
Ĥ7𝑎
Ĥ7𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
150
Ĥ7 = ∆ Ĥ˚𝑓 + ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -238.6 kJ/mol + 3.311 kJ/mol + 35.27 kJ/mol + 4.402 kJ/mol = -195.617 kJ/mol Path: CO2 (g, 25˚ C, 1 atm)
Ĥ8𝑎
Ĥ8
CO2 (g, 180˚ C, 48.5 atm)
CO2 (g, 180˚ C, 1 atm)
Ĥ8𝑏
Calculation: 180
Ĥ8 = ∆ Ĥ˚𝑓 + ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = -393.5 kJ/mol + 6.216 kJ/mol 41
= -387.284 kJ/mol
Path:
Ĥ9
CO (g, 25˚ C, 1 atm) Ĥ9𝑎
CO (g, 180˚ C, 48.5 atm)
CO (g, 180˚ C, 1 atm)
Ĥ9𝑏
Calculation: 180
Ĥ9 = ∆ Ĥ˚𝑓 + ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -110.52 kJ/mol +4.559 kJ/mol = -105.961 kJ/mol Path:
H2O (l, 25˚ C, 1 atm)
Ĥ10
H2O (g, 180˚ C, 48.5 atm)
Ĥ10𝑎
Ĥ10𝑏
H2O (l, 100˚ C, 1 atm)
H2O (g, 100˚ C, 1 atm) Ĥ𝑣
Calculation: 100
180
Ĥ10 = ∆ Ĥ˚𝑓 + ∫25 75.4 𝑥 10−3 𝑑𝑡 + Ĥ𝑣 + ∫100 33.46 𝑥 10−3 + 0.6880 𝑥 10−5 𝑇 + 0.7604 𝑥 10−8 𝑇 2 + (−3.593 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -285.84 kJ/mol + 5.655 kJ/mol + 40.656 kJ/mol + 2.765 kJ/mol = -236.764 kJ/mol
Path: H2 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
H2 (g, 180˚ C, 48.5 atm)
H2 (g, 180˚ C, 1 atm)
Ĥ11𝑏
Calculation: 180
Ĥ11 = ∆ Ĥ˚𝑓 + ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +4.478 kJ/mol = 4.478 kJ/mol
42
Path: N2 (g, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
N2 (g, 180˚ C, 48.5 atm)
N2 (g, 180˚ C, 1 atm)
Ĥ12𝑏
Calculation: 180
Ĥ12 = ∆ Ĥ˚𝑓 + ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0 kJ/mol +4.540 kJ/mol = 4.540 kJ/mol
Path: CH4 (g, 25˚ C, 1 atm) Ĥ13𝑎
Ĥ13
CH4 (g, 150˚ C, 48,5 atm)
CH4 (g, 180˚ C, 1 atm)
Ĥ13𝑏
Calculation: 180
Ĥ13 = ∆ Ĥ˚𝑓 + ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -74.85 kJ/mol + 6.191 kJ/mol = -68.659 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ14
Ĥ14𝑎
CH3OH (g, 180˚ C, 48.5 atm) Ĥ14𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
180
Ĥ14 = ∆ Ĥ˚𝑓 + ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = -238.6 kJ/mol + 3.311 kJ/mol + 35.27 kJ/mol + 6.084 kJ/mol = -193.935 kJ/mol
43
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = -2440434438.3- (-203245497.4) = -11332.7614 kW
SEPARATOR (S-101) Stream 11 Tout = 40˚ C Pout = 48.8 atm Stream 10 Tin = 40˚ C Pin = 48.8 atm
S-101
Stream 13 Tout = 40˚ C Pout = 48.8 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
Stream 10 ṅin (kmol/hr) Ĥ𝑖𝑛 (kJ/mol) 131.011 Ĥ1 = 0.562 269.962 Ĥ2 = 0.436 27.790 Ĥ3 = 1.131 2024.717 Ĥ4 = 0.433
Stream 13 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 1.030 Ĥ14 = 0.562 162.740 Ĥ15 = 1.131 2.06 Ĥ16 = 0.433 11.910 11.095 Ĥ5 = 0.436 Ĥ11 = 0.436 1000.449 987.455 Ĥ6 = 0.541 Ĥ12 = 0.541 10.3 Ĥ17 = 0.541 508.164 19.971 Ĥ7 = 1.220 Ĥ13 = 1.220 853.870 Ĥ18 = 1.220 Table 4.4.11: Molar Flowrate and Specific Enthalpy for S-101
CO2 CO H2O H2 N2 CH4 CH3OH
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
Stream 11 ṅout Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 6.657 Ĥ8 = 0.562 31.066 Ĥ9 = 0.436 1164.975 Ĥ10 = 0.433
CO2 (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ1𝑏
Calculation: 40
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.562 kJ/mol 44
Path:
CO (g, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
CO (g, 40˚ C, 48.8 atm)
CO2 (g, 40˚ C, 1 atm)
Ĥ2𝑏
Calculation: 40
Ĥ2 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 40˚ C, 48.8 atm)
H2O (l, 40˚ C, 1 atm)
Ĥ3𝑏
Calculation: 40
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.131 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 40˚ C, 1 atm)
Ĥ4𝑏
Calculation: 40
Ĥ4 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.433 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm)
Ĥ5𝑎
Ĥ5
N2 (g, 40˚ C, 48.8 atm)
N2 (g, 40˚ C, 1 atm)
Ĥ5𝑏
Calculation: 40
Ĥ5 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.436 kJ/mol
45
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH4 (l, 40˚ C, 48.8 atm)
CH4 (g, 40˚ C, 1 atm)
Ĥ6𝑏
Calculation: 40
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.541 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CH3OH (l, 38˚ C, 48.8 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ7𝑏
Calculation: 40
Ĥ7 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.220 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = [(1085135.38)+ 1231931.5)] – (2265860.314) = 14.22405 kW
46
Flash Drum (S-102)
Stream 15 Tout = 30˚ C Pout = 0.24 atm
Stream 14 Tin = 38˚ C Pin = 47 atm
S-102
Stream 16 Tout = 30˚ C Pout = 0.24 atm Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
Stream 14 Stream 15 Stream 16 ṅin (kmol/hr) Ĥ𝑖𝑛 ṅout ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kmol/hr) (kJ/mol) (kJ/mol) (kJ/mol) 1.030 Ĥ1 = 0.486 0.576 Ĥ6 = 0.186 0.0823 Ĥ7 = 0.145 162.740 Ĥ2 = 0.980 0.741 Ĥ8 = 0.377 162.098 Ĥ13 = 0.377 2.060 Ĥ3 = 0.375 2.428 Ĥ9 = 0.144 0.0823 Ĥ10 = 0.145 10.30 Ĥ4 = 0.468 10.249 Ĥ11 = 0.179 853.870 Ĥ5 = 1.055 27.001 Ĥ12 = 0.402 826.303 Ĥ14 = 0.402 Table 4.4.12: Molar Flowrate and Specific Enthalpy for S-102
CO2 CO H2O H2 N2 CH4 CH3OH
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ1𝑎
Ĥ1
CO2 (g, 38˚ C, 47 atm)
CO2 (g, 38˚ C, 1 atm)
Ĥ1𝑏
Calculation: 38
Ĥ1 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.486 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ2𝑎
Ĥ2
H2O (l, 38˚ C, 47 atm)
H2O (l, 38˚ C, 1 atm)
Ĥ2𝑏
47
Calculation: 38
Ĥ2 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.980 kJ/mol
Path:
Ĥ3
H2 (g, 25˚ C, 1 atm) Ĥ3𝑎
H2 (g, 38˚ C, 47 atm) Ĥ3𝑏
H2 (g, 38˚ C, 1 atm)
Calculation: 38
Ĥ3 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.375 kJ/mol
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ4𝑎
Ĥ4
CH4 (g, 38˚ C, 47 atm)
CH4 (g, 38˚ C, 1 atm)
Ĥ4𝑏
Calculation: 38
Ĥ4 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.468 kJ/mol
Path:
Ĥ5
CH3OH (l, 25˚ C, 1 atm) Ĥ5𝑎
CH3OH (l, 38˚ C, 47 atm)
CH3OH (l, 38˚ C, 1 atm)
Ĥ5𝑏
Calculation: 38
Ĥ5 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 1.055 kJ/mol
Path:
CO2 (g, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CO2 (g, 30˚ C, 0.24 atm)
CO2 (g, 30˚ C, 1 atm)
Ĥ6𝑏
Calculation: 30
Ĥ6 = ∫25 36.11 𝑥 10−3 + 4.233 𝑥 10−5 𝑇 + (−2.887 𝑥 10−8 )𝑇 2 + 7.464 𝑥10−12 𝑇 3 𝑑𝑡 = 0.186 kJ/mol
48
Path:
CO (g, 25˚ C, 1 atm) Ĥ7𝑎
Ĥ7
CO (g, 30˚ C, 0.24 atm)
CO2 (g, 30˚ C, 1 atm)
Ĥ7𝑏
Calculation: 30
Ĥ7 = ∫25 28.95 𝑥 10−3 + 0.4110 𝑥 10−5 𝑇 + 0.3548 𝑥10−8 𝑇 2 + (−2.220 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.145 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ8𝑎
Ĥ8
H2O (l, 30˚ C, 0.24 atm)
H2O (l, 30˚ C, 1 atm)
Ĥ8𝑏
Calculation: 30
Ĥ8 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol
Path:
H2 (g, 25˚ C, 1 atm) Ĥ9𝑎
Ĥ9
H2 (g, 40˚ C, 48.8 atm)
H2 (g, 30˚ C, 1 atm)
Ĥ9𝑏
Calculation: 30
Ĥ9 = ∫25 28.84 𝑥 10−3 + 0.00765 𝑥 10−5 𝑇 + 0.3288 𝑥 10−8 𝑇 2 + (−0.8698 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.144 kJ/mol
Path:
N2 (g, 25˚ C, 1 atm) Ĥ10𝑎
Ĥ10
N2 (g, 30˚ C, 0.24 atm)
N2 (g, 30˚ C, 1 atm)
Ĥ10𝑏
Calculation: 30
Ĥ10 = ∫25 29.00 𝑥 10−3 + 0.2199 𝑥 10−5 𝑇 + 0.5723 𝑥 10−8 𝑇 2 + (−2.871 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.145 kJ/mol
Path:
CH4 (g, 25˚ C, 1 atm) Ĥ11𝑎
Ĥ11
CH4 (l, 30˚ C, 0.24 atm)
CH4 (g, 30˚ C, 1 atm)
Ĥ11𝑏
49
Calculation: 30
Ĥ6 = ∫25 34.31 𝑥 10−3 + 5.469 𝑥 10−5 𝑇 + 0.3661 𝑥 10−8 𝑇 2 + (−11.00 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 0.179 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ12𝑎
Ĥ12
CH3OH (l, 30˚ C, 0.24 atm)
CH3OH (l, 30˚ C, 1 atm)
Ĥ12𝑏
Calculation: 30
Ĥ12 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.402 kJ/mol
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (13099.33+393284.752) – 1065639.03) kJ/h = -183.12637 kW
50
Distillation Column (T-101)
Stream 17 Tout = 51˚ C Pout = 0.64 atm
Stream 16 Tin = 30˚ C Pin = 0.24 atm
T-101
Stream 18 Tout = 71˚ C Pout = 0.9 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 16 Stream 17 Stream 18 ṅin ṅ ṅ Ĥ𝑖𝑛 Ĥ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 out 𝑜𝑢𝑡 (kmol/hr) (kmol/hr) (kmol/hr) (kJ/mol) (kJ/mol) (kJ/mol) 162.098 Ĥ1 = 0.377 2.7378 Ĥ3 = 1.960 159.66 Ĥ5 = 3.468 826.303 453.5622 372.54 Ĥ2 = 0.402 Ĥ4 = 2.139 Ĥ6 = 3.861 Table 4.4.13: Molar Flowrate and Specific Enthalpy for T-101
H2O (l, 25˚ C, 1 atm)
Path:
Ĥ1𝑎
Ĥ1
H2O (l, 30˚ C, 0.24 atm)
H2O (l, 30˚ C, 1 atm)
Ĥ1𝑏
Calculation: 30
Ĥ1 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ2𝑎
Calculation:
Ĥ2
CH3OH (l, 30˚ C, 0.24 atm)
CH3OH (l, 30˚ C, 1 atm)
Ĥ3𝑏
30
Ĥ2 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 0.402 kJ/mol 51
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 51˚ C, 0.6 atm) Ĥ3𝑏
H2O (l, 51˚ C, 1 atm)
Calculation: 51
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 1.960 kJ/mol
Path:
Ĥ4
CH3OH (l, 25˚ C, 1 atm) Ĥ4𝑎
CH3OH (l, 51˚ C, 0.6 atm) Ĥ4𝑏
CH3OH (l, 51˚ C, 1 atm)
Calculation: 51
Ĥ4 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 2.139 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ5𝑎
Ĥ5
H2O (l, 71˚ C, 0.9 atm) Ĥ5𝑏
H2O (l, 71˚ C, 1 atm)
Calculation: 71
Ĥ5 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 3.468 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm) Ĥ6𝑎
Ĥ6
CH3OH (l, 71˚ C, 0.9 atm)
CH3OH (l, 71˚ C, 1 atm)
Ĥ6𝑏
Calculation: 71
Ĥ6 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 3.861 kJ/mol Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (975535.6338 + 1992077.82) – 393284.752 = 715.09131 kW
52
Distillation Column (T-102)
Stream 19 Tout = 60˚ C Pout = 0.95 atm
Stream 18 Tin = 71˚ C Pin = 0.9 atm
T-102
Stream 20 Tout = 99˚ C Pout = 1 atm
Reference state: CO2, CO, H2, N2, CH4 (gas, 25˚ C, 1 atm), H2O (liquid, 25˚ C, 1atm) Compound
CO2 CO H2O H2 N2 CH4 CH3OH
Stream 18 ṅin Ĥ𝑖𝑛 (kmol/hr) (kJ/mol) 159.66 Ĥ1 = 3.468
Stream 20 ṅ𝑜𝑢𝑡 Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 158.3544 Ĥ5 = 5.5796 372.540 Ĥ2 = 3.861 368.1891 Ĥ4 = 2.905 3.54561 Ĥ6 = 3.861 Table 4.4.13: Molar Flowrate and Specific Enthalpy for T-101
H2O (l, 25˚ C, 1 atm)
Path:
Ĥ1𝑎
Ĥ1
Stream 19 ṅout Ĥ𝑜𝑢𝑡 (kmol/hr) (kJ/mol) 1.1109 Ĥ3 = 2.639
H2O (l, 71˚ C, 0.9 atm)
H2O (l, 71˚ C, 1 atm)
Ĥ1𝑏
Calculation: 71
Ĥ1 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 0.377 kJ/mol Path: CH3OH (l, 25˚ C, 1 atm) Ĥ2𝑎 Calculation:
Ĥ2
CH3OH (l, 71˚ C, 0.9 atm)
CH3OH (l, 71˚ C, 1 atm)
Ĥ3𝑏
71
Ĥ2 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 53
= 3.861 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm) Ĥ3𝑎
Ĥ3
H2O (l, 60˚ C, 0.95 atm) Ĥ3𝑏
H2O (l, 60˚ C, 1 atm)
Calculation: 60
Ĥ3 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 2.639 kJ/mol
Path:
Ĥ4
CH3OH (l, 25˚ C, 1 atm) Ĥ4𝑎
CH3OH (l, 60˚ C, 0.95 atm)
CH3OH (l, 60˚ C, 1 atm)
Ĥ4𝑏
Calculation: 60
Ĥ4 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 = 2.905 kJ/mol
Path:
H2O (l, 25˚ C, 1 atm)
Ĥ5
H2O (l, 99˚ C, 1 atm)
Calculation: 99
Ĥ5 = ∫25 75.4 𝑥 10−3 𝑑𝑡 = 5.5796 kJ/mol
Path:
CH3OH (l, 25˚ C, 1 atm)
Ĥ6
CH3OH (g, 99˚ C, 1 atm)
Ĥ6𝑎
Ĥ4𝑏
CH3OH (l, 64.7˚ C, 1 atm)
CH3OH (g, 64.7˚ C, 1 atm) Ĥ𝑣
Calculation: 64.7
99
Ĥ6 = ∫25 75.86 𝑥 10−3 + 16.83 𝑥 10−5 𝑇 𝑑𝑡 + Ĥ𝑣 + ∫64.7 42.93 𝑥 10−3 + 8.301 𝑥 10−5 𝑇 + (−1.87 𝑥 10−8 ) 𝑇 2 + (−8.03 𝑥10−12 ) 𝑇 3 𝑑𝑡 = 3.311 kJ/mol + 35.27 kJ/mol + 1.701kJ/mol = 40.282 kJ/mol
54
Q̇ = ∆Ḣ = ∑ṅi ,out Ĥi,out − ∑ṅi ,in Ĥi ,in = (1072521.001 + 1026378.42) – 1992077.82 = 29.67266 kW
Table 4.4 14: Summary of Energy Balance. Equipment
Q calc (kW)
C-101 C-102 E-101 E-102 E-103 E-104 E-105 FEHE R-101 S-101 S-102 T-101 T-102
904.76489 396.76042 -128.90539 -1976.50956 -9266.32657 202.88194 -46.19235 3690.69747 -11332.7614 14.22405 -183.12637 715.09131 29.67266
55
4.5
Conclusion From the calculation the value of heat transfer, Q for equipment was not same. This
happens due to the different inlet and outlet components and the needs of energy used. The heat transfer with positive value represents the heat absorb which is as endothermic reaction. Then negative value represent heat release and as exothermic reaction.
REFERENCES
Amin, D. R., Hassan, I., Das, A., Yeasmin, R., Rahman, T., & Hossain, T. (n.d.). Simulation of Methanol Production from. 1-14. Felder, R. M. (2005). Elementary Principles of Chemical Processes. United States of America: John Wiley & Sons, Inc. Klier, K. (1982). Methanol Synthesis. Advances in Catalyst, 243-310. Luyben, W. L. (2010). Design and Control of a Methanol Reactor/Column Process. Industrial Engineering Chemical, 6150-6163.
56
Related Documents
Chapter 4 (dp1)- Energy Balance.pdf
November 2019 0
Dp1
May 2020 0
Sugar Cane's Energy - Chapter 4
April 2020 3
Dp1.docx
June 2020 5
L17-dp1.pdf
June 2020 0
Matriks Dp1.docx
October 2019 32More Documents from "irma"
Chapter 4 (dp1)- Energy Balance.pdf
November 2019 0
1 Step: The Spread Baking Soda
November 2019 0
Kemahiran 1 Konsep Nombor.docx
December 2019 12
Tutorial 2 Ppi.docx
December 2019 8
Keselamatan Mesin Automatik.docx
November 2019 18