i
College of Mechanical Engineering Technology/Benghazi
Chemical Engineering Department
PROCESS DESIGN FOR A SYNTHESIS UNIT OF
METHANOL PRODUCTION
A PROJECT REPORT
Submitted by
SANAD JOUMAH HAMAD
132168
MOHANAD ABDULLAH ALSAITY
161036
Fall 2021
ii
College of Mechanical Engineering Technology / Benghazi
Department of Chemical Engineering Technology
PROCESS DESIGN FOR A SYNTHESIS UNIT OF
METHANOL PRODUCTION
A project submitted
in partial fulfillment of the requirements for the degree of Chemical
Engineering
SANAD JOUMAH HAMAD
MOHANAD ABDULLAH ALSAITY
Supervisor
Dr. Ali. H.ElHudayri
Fall 2021
132168
161036
iii
CERTIFICATE
The project work entitled:
PROCESS DESIGN FOR A SYNTHESIS UNIT OF METHANOL
PRODUCTION
Which is being submitted by students :

SANAD JOUMAH HAMAD
132168

MOHANAD ABDULLAH ALSAITY
161036
In partial fulfillment of the requirements for the award of BA. Tech degree in
Chemical Engineering has been carried out under my supervision and accepted for
Presentation / examination
Supervisor:
Date: ….../..…./………..
Signature: ……………..
Corrected approved copy for records
Examiner 1:
Examiner 2:
Date: ….../..…./………..
Date: ….../..…./………..
Signature: ………………
Signature: ………………
Supervisor:
head of chemical department
Date: ….../..…./………..
Date: ….../..…./………..
Signature: ………………
Signature: ………………
iv
ACKNOWLEDGEMENT
We dedicate our graduation to the father and mother who were the greatest
reasons for success and excellence in our academic and scientific career. We
also express our thanks and gratitude to our virtuous teacher who supervised
our project and contributed to its completion. We also thank all the teachers
and everyone who supported me in these years on the path to success. I
thank you all from the bottom of my heart, I ask May God Almighty reward
you on our behalf and bless you in your work.
‫‪v‬‬
‫شكر وتقدير‬
‫نهدد ت رجنا ددو لد ولدد دول ولدد ولدددعظم أونددو لنجدد للددوو ول سددون دولدردد ة وددع ميدد‪،‬نر و ول ولدد‪ ،‬دول د ‪،‬دد‬
‫أ دددو نجدددك ولالدددان دوعمد دددوف ل د دددو ولروىدددر ولدددعت ل دددن‬
‫ندددد مالدددندن و دلدددوف ودددع إر ومددد‬
‫دنالدددان‬
‫لظضدددو أووددد ول د ددد‪،‬م دأدددر مدددم لدددون نو ودددع فدددعي وليددد ‪،‬م ودددع ‪،‬نظدددر ول سدددون وع دددانأ ا ‪ ،‬دددو مدددم لن دددوة‬
‫قدوع للعل هللا ّ‬
‫نز دا ّر لف ظسزظا ن و خ‪،‬ن ولسزوء دلف ظوو ك لا وع ن دا ‪.‬‬
vi
ABSTRACT
Preliminary process design of synthesis unit for methanol , was established with a
production rate of 1000 ton/yr of crude methanol , in the first chapter a general
review of methods of methanol Production was presented , low pressure method was
adopted for this project , this methanol is used in sirte oil company , Brega.
A material and energy balance calculations for each unit was carried out , based on a
data from sirte oil company , Brega.
A process design for the cooler , was established based on the design results of the
cooler it is recommended to split the cooler into tow coolers at lest , either in series
or parallel.
‫‪vii‬‬
‫الخالصة‬
‫ظدض م ول الندع رص ‪ ،‬مو ئع ل ح رص ‪،‬ع ول ‪،‬ثون ل‬
‫ل إندوج ‪، 1000‬م‪/‬ل لد ‪،‬ثون ل ولجوم ص‬
‫نوم ر ردج‪،‬ك ولطنة لد ولدق ‪،‬وت ول يدج م وع فعو ول سول در حيو أر مم ول وزن ول ودظ دولحنو ظ‬
‫ل ح وت منحد ولدص ‪،‬ع در رص ‪ ،‬ول وودل ول وند دولعت ظق م دونظ غوزوت ول دج ولجو ج مم ول رونر‪.‬‬
‫ر ولدج وم ض ولو‪،‬ونوت مم مصود‬
‫وء ند ندوئج رص ‪ ،‬ول وند ظ ص‬
‫نأ لنت لد رط ول ص ع ولجوص إندوج ول ‪،‬ثون ل‬
‫دسزئ ول وند إل وث ‪،‬م لد لأثن ل وء ند ولد ولع لد ند ولد وزت‬
viii
TABLE OF CONTENTS
Title
i
CERTIFICATE
iii
ABSTRACT
vi
List of Figures
x
List of Tables
xi
List of Symbols
xii
Chapter 1 : Introduction
1.1 Introduction
1.1.1 History of Methanol
1
2
3
1.2 Methanol Uses
4
1.3 Production Processes
5
1.4 Methanol production technologies
6
1.5 Methanol synthesis loop
6
1.5.1 Process description
6
1.5.2 Process Flow Diagram
7
Chapter 2 : Material Balance
Chapter 3 : Energy Balance
3.1 Process design of cooler
Chapter 4 : Safety
4.1 Methanol Safety Data Sheet
2
11
40
27
45
4.1.1 Potential Health Effects
45
4.1.2 First Aid Measures
46
4.1.3 Accidental Release Measures
46
4.1.4 Storage And Handling
46
4.1.5 Exposure Controls/Personal Protection
47
ix
4.1.6 Stability And Reactivity
47
Reference
48
Appendices
49
x
List of Figures
Figure (1.1) : Global Methanol Fuel Demand ...............................................................4
Figure (1.2) : Basic steps of methanol production ........................................................5
Figure (1.3) : Process Flow Diagram ...........................................................................7
Figure (2.1) : Process Flow Diagram ............................................................................9
Figure (2.2) : mole balance for the reactor and separator ..........................................10
Figure (2.3) : Mole Balance for the 2nd separator SV2 .............................................14
Figure (2.4) : Balance at SV1 .....................................................................................14
Figure (2.5) : Balance at SV2 .....................................................................................14
Figure (3.1) : Energy Balance for the Mug compressor ..............................................30
Figure (3.2) : Energy Balance for the mixing point ....................................................31
Figure (3.3) : Energy Balance for the compressor K-02 ...........................................33
Figure (3.4) : Energy Balance for reactor ..................................................................34
Figure (3.5) : Energy Balance for Heat exchanger .....................................................36
Figure (3.6) : Energy Balance for the Cooler ..............................................................37
Figure (3.7) : Design of Cooler .................................................................................39
xi
List Of Tables
Table (1.1) history of methanol .....................................................................................3
Table (2.1) composition stream S1 ................................................................................9
Table (2.2) composition stream S9 ..............................................................................18
Table (2.3) Summary of all stream rate (Kmol/hr) .....................................................23
Table (2.4) Mole fraction of Components in each stream ..........................................24
Table (2.5) : actual stream flow rate each stream must be multiplied by the scale
factor : 124.67 ............................................................................................................25
xii
List of symbols
S : feed stream (mol/hr)
P : pressure (Pa)
V : volume (m3)
ρ : Density (Kg/m3)
μ : Viscosity (Pa.s)
K : thermal conductivity (W/m.K)
T : temperature (K°)
∆H : Enthalpy (KJ/hr)
Cp : Heat capacity (Kj/Kmol.K)
∆T : temperature difference (K°)
∆Tm : Log mean temperature difference (K)
A0 : Heat transfer area (m2)
a0 : Area of one tube (m2)
d0 : tube outside diameter (m)
di : inside diameter tube (m)
L : Length (m)
Nt : number of tube
Db : tube bundle diameter
Ds : shell-side diameter (m)
As : shell-side cross flow area (m2)
LB : baffle spacing (m)
de : triangle tube pitch, equivalent diameter (m)
Gs : shell-side mass velocity (Kg/s.m2)
xiii
Ws : mass flow rate of shell-side fluid (Kg/s)
us : shell-side velocity (m/s)
Re : Reynolds number
Pr : Prandtl number
At : flow area in tube side (m2)
Wt : mass flow rate of water in the tube (Kg/s)
ns : molar flow rate (mol/hr)
Z : mole fractions
Mwav : average molecular weight (Kg/Kmol)
Q cooler : Heat transferred in the cooler (KJ/hr)
∆H: Latent heat (KJ/hr)
W : watt (J/s)
Chapter 1
Introduction
2
1.1 Introduction
Methanol is an alcoholic chemical compound that is considered to be the simplest
alcohol. It is a light, colorless, flammable liquid at room temperature, and contains
less carbon and more hydrogen than any other liquid fuel. The chemical formula for
methanol is CH3OH (often abbreviated as MeOH). Methanol appears naturally in the
environment.
Methanol is often called wood alcohol because it was once produced chiefly as a
byproduct of the destructive distillation of wood. Today, the most common
feedstocks for its production are natural gas and coal. This simple alcohol can be
made from virtually anything that is, or ever was, a plant. This includes common
fossil fuels – like natural gas and coal – and renewable resources like biomass,
landfill gas.
To produce methanol, synthesis gas needs to be created from the feedstock first.
Through gasification, synthesis gas can be produced from anything that is or ever
was a plant, which includes biomass, agricultural and timber waste, solid municipal
waste, recycled carbon dioxide and a number of other feedstocks in addition to
natural gas and coal. In a typical facility, methanol production is carried out in two
steps. The first step is to convert the feedstock into a synthesis gas consisting of CO,
CO2, H2O and H2. This is usually accomplished by the catalytic reforming of feed
such as SMR (Steam methane reforming), POX (Partial oxidation), ATR (Auto
thermal reforming) and Combined Reforming. The second step is the catalytic
synthesis of methanol from the synthesis gas.
Since the 1800s, methanol has been widely used as an industrial chemical compound
to produce a variety of traditional chemical derivatives, including formaldehyde,
acetic acid, dimethyl terephthalate, methyl methacrylate, and methyl chloride to
manufacture a wide range of end products. With its diversity of feedstocks and array
of applications, methanol is one of the world’s most widely used industrial chemical.
Currently, approximately 60% of the global demand for methanol comes from the
industrial chemical market demand and 40% comes from fuels applications demand.
3
1.1.1 History Of Methanol
Methanol was first produced by destructive distillation of wood in 1830. The process
prevailed for about a century until the first synthetic methanol plant was introduced
by Badische Anilin und Soda Fabrik (BASF) in 1923. DuPont introduce the synthetic
methanol plan in the United States in 1927. In late 1940, natural gases replaced the
water gas as a source of syngas (i.e. CO and H2). Imperial Commercial Industry
(ICI) announced that a low pressure methanol process in 1966 using a copper based
catalyst. This operates at 5-10 Mpa (50-100 atm) compared with 35 Mpa (345.42314
atm) for the older high pressure process. The Arab oil embargo in 1973 first
generated much interest in methanol as an alternative automobile fuel.
In 1989, the us administration proposed a clean air regulations that would mandate
the use of cleaner alternative automobile fuels. The amended Clean Air Act, passed
in 1990, requires a reduction in ozone layer and carbon monoxide emissions,
although it does not mandate used of the alternative fuel. Currently, methyl-tert-butyl
ether derived from isobutene and methanol is the most widely used oxygenated in
reformulated gasoline, and automakers, and local government authorities have
announced plans to introduce methanol-fueled vehicles. The interest in methanol fuel
applications has shifted from the economic considerations in the 1970’sto
environmental considerations in the 1990s. this environmental impact will continue
into the next century and could have a strong effect on demand for methanol.
Furthermore, it was recently discovered that some crops treated with methanol or
nutrient supplement methanol showed significant increases in crop yields. This has
opened up another area of research and the development for the methanol and
provide another opportunity for future methanol growth.
Table (1.1) history of methanol
Year
1830
1905
1923
1927
Late 1940s
1966
1970
1973
1989
1990
Early 1990s
Events
First commercial methanol process by destructive distillation
Synthetic methanol route suggested by French chemist Paul
First synthetic methanol plant commercialized by BASF
Synthetic methanol process introduced in United States
Conversion from water gas to natural gas as source of synthetic methanol reactors
Low pressure methanol process announced by ICI
Acetic acids process by methanol carbonylation introduced
Arab oil-embargo reassessment for alternative fuels
Clean Air regulations proposed by Bush Administration
Passage of the amended Clean Air Act in the United States
Discovery of enhanced crop yields with methanol treatment
4
1.2 Methanol Uses
The major methanol end-uses in fuels applications include :

Gasoline Blend stock: In some countries, methanol is blended directly with
gasoline. Methanol has attractive properties as a fuel blending component; it
has a high octane number, combusts efficiently, emits fewer harmful byproducts than gasoline, and is relatively low-cost.

Dimethyl Ether (DME): DME can be used to blend into LPG4, as a
replacement for LPG fuel in heating and cooking applications, as a
transportation fuel in modified diesel engines, and for power generation.

Methyl Tert-Butyl Ether (MTBE): MTBE (an oxygenate) is added to

gasoline to increase its octane number. Use of MTBE in a gasoline blend also
allows for a cleaner combustion process, emitting less pollution.

Biodiesel: Biodiesel is the product obtained when vegetable oil or animal fat
is chemically reacted with an alcohol - methanol, for example - to produce a
compound known as a fatty acid alkyl ester. When methanol is used, the
product is fatty acid methyl ester (FAME).

Methanol-to-Gasoline (MTG): MTG refers to the process of making gasoline
via a methanol route. In some ways, MTG is a misnomer, because methanol
is often an intermediate in the process, and units are back-integrated to coal
or natural gas. Some MTG units in China may run on merchant methanol.

Others: Methanol can also be used as a fuel in power generation plants and in
methanol fuel cells.
Figure (1.1) : Global Methanol Fuel Demand
5
1.3 Production Processes
Methanol is manufactured using a proven technology based predominantly on natural
gas, with a small portion of the world capacity being based on petroleum fractions
and process off-gases. , when using natural gas as the feedstock, comprises the
following process steps :

desulphurization of natural gas

synthesis gas generation

synthesis gas compression

methanol synthesis

methanol distillation
Figure(1.2) : Basic steps of methanol production
(Figure 1.2) shows the various options available for the individual process steps. The
option selected as the most suitable and, consequently, the plant configuration
depends on the customer’s requirements and local conditions. In addition, high
pressure steam (HPS) and medium pressure steam (MPS) are usually produced by
cooling the high temperature synthesis gas and methanol synthesis reactor,
respectively.
The main reactions for methanol production are:
CO + 2 H2 → CH3OH
CO2 + 3 H2 → CH3OH + H2O
Therefore, the following ratio is ideal for methanol production :
(H2 - CO2) / (CO + CO2) = 2
6
The synthesis of carbon oxides to methanol is based on a CuO/ZnO based catalysts,
e.g. Synetix catalysts. This catalyst type has proved its performance in adiabatic as
well as in isothermal reactors. The conversion of carbon oxides to methanol is an
exothermic process which takes place at high pressures and low temperatures. The
synthesis unit operates at 40 - 110 bar and around 200 to 300 °C. The aim is to
achieve a relatively high carbon efficiency so as to minimize the amount of synthesis
gas to be processed.
Carbon eff. = [moles methanol produced / moles (CO + CO2) in syngas] × 100%
1.4 Methanol production technologies
there are several of Methanol production technologies in the world. All of the
processes have their own advantages and disadvantages. It is important to choose the
most efficient process in order to have a good and almost perfect production of
methanol. Here are the lists of the technologies in producing the methanol :

Methanol production based on one step reforming.

Methanol production based on two step reforming.

Methanol synthesis loop.

Methanol distillation
1.5 Methanol synthesis loop
1.5.1 Process description
Feed gas comprising of hydrogen & carbon monoxide is compressed to 3000-5000
psi, mixed with recycle gas, then fed to a high pressure converter. Internal preheat is
usually employed. The reactor is copper lined steel & contains a mixed catalyst of
zinc,chromium,manganese, or aluminium oxides.The temperature is maintained at
300-375 °C by proper space velocity & heat exchanger design. Exit gases from the
reactor are cooled by heat exchange with reactants, then with water. Methanol
condenses under full operating pressure to maximize yields-50%.The liquid
methanol is depressurized, then sent to the separator then to another tower to
minimize the water content.
7
1.5.2 Process Flow Diagram
Figure (1.3) : Process Flow Diagram
Chapter 2
Material Balance
8
Take a basic of 100 Kmol/hr in stream S1 , the composition of the make-up gas
stream S1 is given as following , based on sub oil
Table (2.1) composition stream S1
Composition
Mol %
CO
11.77
CO2
11.865
H2
69.885
H2O
0.408
CH4
5.804
N2
0.25
CH3OH
0.0216
Total
100
Figure (2.1) : Process Flow Diagram
The main reactions are taken place in the reactor :
CO + 2 H2 → CH3OH
CO2 + 3 H2 → CH3OH + H2O
9
The given date from sirte oil company is :
CO conversion is 54.77%
CO2conversion is 20.88%
The mole balance will be taken for the reactor and separator
Figure (2.2) : mole balance for the reactor and separator
According to the date from sirte oil company, the following information will be used:
1- The recycle stream SR is 76% of Stream S10
2- The purge stream is 24% of S10
3- The dissolved gases in stream S9 is 5 mol% of the exit stream from the reactor
except methanol and water and 95 mol% in S10
Balance of CO
Take Y as mole of CO in stream S5
Conversion of CO is 0.5477
CO in S6 = CO in S5 – CO converted
CO output from the reactor in S6 = Y – 0.5477Y
CO in S10 = 0.95 [Y – 0.5477Y]
CO in purge stream S11 = 0.24 S10
CO in S11 = 0.24 × 0.95[Y – 0.5477Y]
CO Recycled in SR = 0.76 × [Y – 0.5477Y]
CO in S9 = 0.05[Y – 0.5477Y]
Note the S9 is the bottom stream of the separator SV1
10
Balance at the mixing point B for CO will be :
Y = CO in S1 + CO SR
Y = 11.77 + 0.76 × [Y – 0.5477Y]
Solve for Y
Y = 11.77 + 0.722Y – 0.3954Y
Y = 36.037 Kmol/hr
CO in S10 = 0.95 [36.037 – 0.5477 × 36.037]
= 15.48 Kmol/hr
CO in purge stream S11 = 0.24 × 15.48
= 3.76 Kmol/hr
CO in SR = 0.76 × 15.48
= 11.768 Kmol/hr
CO in S9 = 0.05 (36.037 – 0.5477 × 36.037)
= 0.814 Kmol/hr
CO in S6 = 36.037 (1 – 0.547)
= 16.3 Kmol/hr
Balance of CO2
Make CO2 in S5 = X
CO2 (exit from the rector in S6 = X - CO2 converted
= X – 0.2088 × X
CO2in purge stream S11 = 0.24 × 0.95[X – 0.2088 × X]
CO2in SR = 0.76 × 0.95[X – 0.2088 × X]
CO2in S9 = 0.05 [X – 0.2088 × X]
Balance at the mixing point B for CO2 will be :
11
X = CO2in S1 + CO2 in SR
X = 11.865 + 0.76 × [X – 0.2088 × X]
X = 11.865 + 0.722 X – 0.15 X
Solve for X
X = 27.72 Kmol/hr
CO2 in S9 = 0.05 × 0.95[27.72 – 0.2088 × 27.72]
= 1.096 Kmol/hr
CO2 in S6 = 27.72 (1-0.2088)
= 21.9 Kmol/hr
CO2 in S11 = 0.24 × 0.95 (27.72 - 0.2088 × 27.72)
= 5 Kmol/hr
CO2 in S10 = 0.95 × [27.72 (1- 0.2088)]
= 20.83 Kmol/hr
CO2 in recycled SR = 15.8 Kmol/hr
Balance of H2
Make H2 in S5 = Z
Total hydrogen converted = H2 converted in rector 1 + H2 converted in rector 2
∴ Total hydrogen converted = 2 (0.5477Y) + 3(0.2088X)
Y = 36.037 , X = 27.72
= 2(0.5477 × 36.037) + 3(0.2088 × 27.72)
= 56.83 Kmol/hr
H2 in S6 (exit from the rector) = Z - H2 reacted = Z – 56.83
= 103.81 – 56.83 = 46.98 Kmol/hr
12
H2 in S10 = 0.95[Z – 56.83]
H2 in the purge stream S11 = 0.24 × 0.95[Z – 56.83]
H2 Recycled in SR = 0.76 × 0.95[Z – 56.83]
H2 in S9 = 0.05 [Z – 56.83]
Balance at the mixing point B for Hydrogen will be :
Z = H2 in S1 + H2 in SR
Z = 69.8855 + 0.76 × 0.95 [Z – 56.83]
Solve for Z
Z = 103.81 Kmol/hr
X = 27.72 Kmol/hr
H2 in S9 = 0.05 [103.81 – 56.83]
= 2.349 Kmol/hr
H2 in S10 = 0.95[103.81 – 56.83]
= 44.631 Kmol/hr
Balance for CH4
Make CH4 in S5 equal to M Kmol/hr
According to the given date , 95 mol% of gases leaving the reactor goes to S10 for
the 1st separator SV1 while 5mol% dissolve in the bottom stream S9
CH4 gas in considered as inert (it doesn't participate in the reaction )
CH4 in S10 = 0.95M
CH4 in S9 = 0.05M
CH4 in the purge stream S11 = 0.24 [ 0.95M]
CH4 Recycled in SR = 0.76 [ 0.95M]
CH4 in SR = 15.03 Kmol/hr
13
Balance at the mixing point B will be :
CH4 in S1 + CH4 in SR = S5
5.784 + 0.76 (0.95M) = M
Solve for M
M = 20.82 Kmol/hr
CH4 in S10 = 19.78 Kmol/hr
CH4 in S9 = 1.041 Kmol/hr
CH4 in S11 = 0.24 (0.95 × 20.82)
= 4.74 Kmol/hr
CH4 in S12 = 0.76 (0.95 × 20.82)
= 15.03 Kmol/hr
Balance for N2
Nitrogen is inert gas (it doesn't react)
Put N2 in S5 (Reactor feed = N)
N2 in S5 = N2 in S6
N2 in S10 = 0.95N
N2 in S9 = 0.05N
N2 in SR = 0.76 [0.95N]
N2 in S11 = 0.24 [0.95N]
Balance at the mixing point B will be :
N2 in S1 + N2 in SR = S5
0.25 + 0.76 (0.95N) = N
Solve for N
N = 0.899 ≈ 0.9 Kmol/hr
14
N2 in S10 = 0.855 Kmol/hr
N2 in S9 = 0.045 Kmol/hr
Calculation of the mole fraction of water and methanol in the gas stream of the
separator SV1 (S10) using the following relation :
PV = PT × y
PV : vapor pressure , PT : Total pressure
y : mole fraction of the component
temperate = 40 C°
using Antonio's equation , the vapor pressure of water and methanol
is obtained at 40 C°
ln 𝑃𝑉 = 𝐴 −
𝐵
𝐶+𝑇
A , B and C are constants
T : temperate in Kelvin
PV for water = 7375 Pa
PV for methanol = 39182 N/m2
𝑦𝑤𝑎𝑡𝑒𝑟 =
𝑃𝑉
7375
=
= 0.0013
𝑃𝑇
5591700
𝑦𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 =
𝑃𝑉
39182
=
= 0.007
𝑃𝑇
5591700
mole fractions of water + methanol will be 0.0013 + 0.007 = 0.0083
mole fractions of gases in S10 (except water and methanol) is :
1 – 0.0083 = 0.9917
Total molar flow rate in S10 will be :
𝑆10 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑔𝑎𝑠𝑒𝑠 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
𝑚𝑜𝑙𝑒𝑠 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑔𝑎𝑠𝑒𝑠 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑤𝑎𝑡𝑒𝑟 𝑎𝑛𝑑 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙
moles of gases without water and methanol
15
= CO moles + CO2 moles + H2 moles + CH4 moles + N2 moles
= 15.48 + 20.93 + 136.86 + 19.78 + 0.955
= 193.8 Kmol/hr
The mole fraction of the gases of CO + CO2 + H2 + CH4 + N2 is = 0.9917
Total mole in 𝑆10 =
193.8
0.9917
= 195.4 Kmol/hr
Balance for water for the reactor and separator units
H2O moles in S10 = 0.0013 * S10
= 0.0013 × 195.4 = 2.54 Kmol/hr
H2O moles in S11 (purge stream) = 0.24 (2.54)
= 0.609 Kmol/hr
H2O moles in SR (Recycle) = 0.79 (2.54)
= 1.93 Kmol/hr
From the reaction the produced water will be estimated as follows :
CO2 +3H2 → CH3OH +H2
H2O produced = CO2 reacted = (CO2 in S5) × conversion CO2
CO2 in S5 = X = 27.72
CO2 conversion = 0.2088
H2 produced = 27.72 × 0.2088
= 5.787 Kmol/hr
Balance at the mixing point B will be :
Water in S1 + Water in SR = Water in S5
0.408 + 1.93 = 2.338 Kmol/hr
Water in S6 (output from the reactor) = Water input to the reactor + Water produced
due to reaction
16
H2O in S6 = 2.338 + 5.787
= 8.125 Kmol/hr
Water in the bottom stream of the separator SV1 (S9) = H2O in S6 – H2 in S10
= 8.125 – 2.54
= 5.585 Kmol/hr
Balance for methanol (reactor and separator loop)
The main reactions for methanol Production is given :
CO + 2 H2 → CH3OH
CO2 + 3 H2 → CH3OH + H2O
CH3OH produced in 1st reaction :
= (CO in S5) * (CO conversion)
= 36.037 * 0.5477
= 19.737 Kmol/hr
CH3OH produced in 2nd reaction :
= (CO2 in S5) * (CO2 conversion)
= 27.72 × 0.2088
= 5.787 Kmol/hr
Total methanol produced = 19.737 + 5.787
= 25.5 Kmol/hr
17
Balance at the mixing point B :
Figure (2.4) : Balance at SV1
CH3OH in S5 = CH3OH in SR + CH3OH in S1
CH3OH in SR = S10 × 0.76 × 0.007
= 195.4 × 0.76 × 0.007
= 1.039 Kmol/hr
CH3OH in S5 = 1.039 + 0.0216
= 1.411 Kmol/hr
CH3OH in S6 = CH3OH in S5 + Total CH3OH produced in the reactions
CH3OH in S6 (output from the reactor) = 1.411 + 25.5
= 26.91 Kmol/hr
CH3OH Balance separator SV1 :
S6 = S9 + S10
CH3OH in S9 = S6 – S10 = 26.91 – (195.4 × 0.07)
= 13.23 Kmol/hr
CH3OH in S11 (purge) = CH3OH in S10 × 0.24
CH3OH in S11 (purge) = (195.4 × 0.0007) × 0.24
18
= 0.328 Kmol/hr
Mole Balance for the 2nd separator SV2 :
Figure (2.3) : Mole Balance for the 2nd separator SV2
The temperature and pressure data are obtained from sirte oil company , according to
the the previous calculation, component in S9 is given :
Table (2.2) : composition stream S9
composition
Kmol/hr
CO
0.814
CO2
1.069
H2
2.349
CH4
1.041
N2
0.045
H2O
5.585
CH3OH
13.23
Total (S9)
24.16
According to the date from sirte oil company , 99 mol% of gases in the feed S9 goes
into the vapor –phase S12 , and 1 mol% an dissolved in liquid-phase (S13)
The mole fractions of H2O and CH3OH in the vapor –phase will be obtained as
following :
𝒚 𝑯𝟐 𝑶 =
𝑷𝑽 (𝒇𝒐𝒓 𝑯𝟐𝑶)
𝑷𝑻𝒐𝒕𝒂𝒍
PV : for water vapor at the given Temperature and pressure (40 C° and 343350 Pa) =
7375 Pa
19
∴ 𝑦𝐻2 𝑂 (𝑚𝑜𝑙𝑒 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑖𝑛 𝑆12 ) =
7375
343350
𝒚𝑯𝟐 𝑶 = 0.02
mole fractions of methanol in the vapor –phase will be obtained :
𝑦𝐶𝐻3 𝑂𝐻 =
𝑃𝑉 (𝑓𝑜𝑟 𝐶𝐻3 𝑂𝐻)
𝑃𝑇𝑜𝑡𝑎𝑙
PV : for CH3OH at 40 C° and 39182.7 Pa
∴ 𝑦𝐶𝐻3 𝑂𝐻 =
39182.7
343350
∴ 𝑦𝐶𝐻3 𝑂𝐻 = 0.114
Mole fraction of the gases in S12 (except H2O and CH3OH)
= 1- (0.02 + 0.114) = 0.866
Total stream S12 =
S12 =
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 (CO + CO2 + H2 + CH4 + N2 ) ×0.99
0.866
(0.814+1.096+2.349+1.041+0.245) ×0.99
0.866
S12 = 6.11 Kmol/hr
Mole fraction of the composition in S12 will be :
𝑦𝐶𝑂 =
𝑦𝐶𝑂2 =
𝑦𝐻2 =
𝑦𝐶𝐻4 =
𝑦𝑁2 =
𝐶𝑂 𝑜𝑛 𝑆9 ×0.99
𝑆12
=
𝐶𝑂2 𝑜𝑛 𝑆9 ×0.99
𝑆12
𝐻2 𝑜𝑛 𝑆9 ×0.99
𝑆12
=
𝐶𝐻4 𝑜𝑛 𝑆9 ×0.99
𝑆12
𝑁2 𝑜𝑛 𝑆9 ×0.99
𝑆12
=
0.814 ×0.99
6.11
=
1.096 ×0.99
6.11
2.399 ×0.99
6.11
=
= 0.1318
= 0.3806
1.041 ×0.99
6.11
0.045 ×0.99
6.11
𝑦𝐻2 𝑂 = 0.02 , 𝒚𝑪𝑯𝟑 𝑶𝑯 = 0.114
= 0.1775
= 0.1686
= 0.0073
20
flow rate of the liquid-phase Stream S13 will obtained by the following balance :
Figure (2.5) : Balance at SV2
S9 = S12 + S13
S13 = S9 – S12 = 24.16 – 6.11
= 18.05 Kmol/hr
Composition flow in S13 will be :
H2O in S13 = H2O in S9 - H2O in S12
= 5.585 – (6.11 × 0.02)
= 5.46 Kmol/hr
CH3OH in S13 = CH3OH in S9 - CH3OH in S12
= 13.23 – (6.11 × 0.114)
= 12.533 Kmol/hr
Gases in S13 will be 1 mol% of the gases in S9 (except H2O and CH3OH) will be :
21
[CO + CO2 + H2 + CH4 + N2] × 0.01
= [0.814 + 1.196 + 2.349 + 1.041 + 0.045] × 0.01
= 0.05345 Kmol/hr
H2O in S13 = H2O in S9 - H2O in S12
= 5.585 – (6.11 × 0.02)
= 5.46 Kmol/hr
CH3OH in S13 = CH3OH in S9 - CH3OH in S12
= 13.23 – (6.11 × 0.119)
= 12.533 Kmol/hr
Total flow rate of S13 = 5.46 +12.533 + 0.05345
S13 = 18.04 Kmol/hr
Identical Value was obtained total balance
Component mole fractions in S13 will be :
𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑜
𝑦𝐶𝑂 =
𝑇𝑜𝑡𝑎𝑙
𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑂2
𝑦𝐶𝑂2 =
𝑦𝐻2 =
𝑇𝑜𝑡𝑎𝑙
𝑚𝑜𝑙𝑒 𝑜𝑓 𝐻2
𝑇𝑜𝑡𝑎𝑙
=
𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻4
𝑦𝐶𝐻4 =
𝑦𝑁2 =
=
𝑇𝑜𝑡𝑎𝑙
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑁2
𝑦𝐻2 𝑂 =
𝑇𝑜𝑡𝑎𝑙
𝑜𝑙𝑒 𝑜𝑓 𝐻2 𝑂
𝑇𝑜𝑡𝑎𝑙
𝑦𝐶𝐻3 𝑂𝐻 =
=
0.814 ×0.01
18.04
=
1.096 ×0.01
18.04
2.349 ×0.01
18.04
=
= 0.00045
= 0.0006
= 0.0013
1.041 ×0.01
18.04
0.045×0.01
18.04
= 0.00057
= 0.000024
5.46
= 18.04 = 0.302
𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝐻3 𝑂𝐻
𝑇𝑜𝑡𝑎𝑙
=
12.535
18.04
= 0.694
Calculation of the scale factor for methanol balance :
Required production rate for methanol is 4 × 108 Kg/year (based on 8000 working
hours per year)
22
The required rate =
4 ×108
8000
= 50000Kg/year
Molecular weight of CH3OH = 32 Kg/Kmol
production rate in Kmol/hr =
50000
32
= 1562.5 Kmol/hr
the production of methanol in S13 = 12.533 Kmol/hr
(based on 100 Kmol/hr of S1)
scale factor for material balance =
1562.5
12.533
= 124.67
This number must multiplied to all streams.
23
Table (2.3) : Summary of all stream rate (Kmol/hr)
Streams
S1
S2
S3
S4
S5
S6
S7
S8
S9
SR
S10
S11
S12
S13
CO
11.77
11.77
23.53
23.53
23.53
16.3
16.3
16.3
0.814
11.76
15.48
3.72
0.805
0.00814
CO2
11.865
11.865
27.695
27.695
27.695
21.9
21.9
21.9
1.096
15.83
20.83
4.99
1.085
0.01096
H2
69.885
69.885
103.785
103.785
103.785
46.98
46.98
46.98
2.349
33.9
44.631
10.71
2.325
0.02349
H2O
0.408
0.408
2.338
2.338
2.338
8.125
8.125
8.125
5.585
1.93
2.54
0.609
0.122
5.46
CH4
5.804
5.804
20.834
20.834
20.834
20.82
20.82
20.82
1.041
15.03
19.78
4.74
1.03
0.01041
N2
0.25
0.25
0.899
0.899
0.9
0.9
0.9
0.045
0.649
0.855
0.205
0.0445
0.00045
CH3OH
0.0216
0.0216
1.0606
0.9 ≈
0.9 ≈
26.91
26.91
26.91
13.23
1.039
1.37
0.328
0.696
12.533
Total
100
100
180.146
1.0606
1.0606
141.9
141.9
141.9
24.16
80.138
105.48
25.3
6.11
18.04
Component
0.899
≈ 0.9
24
Table (2.4) : Mole fraction of Components in each stream
Streams
S1
S2
S3
S4
S5
S6
S7
S8
S9
SR
S10
S11
S12
S13
CO
0.117
0.117
0.13
0.13
0.13
0.1148
0.1148
0.1148
0.0336
0.1467
0.146
0.1470
0.1317
0.0004
CO2
0.1186
0.1186
0.15
0.15
0.15
0.1543
0.1543
0.1543
0.045
0.1975
0.1974
0.1972
0.177
0.0006
H2
0.698
0.698
0.57
0.57
0.57
0.331
0.331
0.331
0.097
0.4231
0.4231
0.4233
0.3805
0.00130
H2O
0.004
0.004
0.0129
0.0129
0.0129
0.057
0.057
0.057
0.2311
0.0240
0.00240
0.0240
0.01996
0.3026
CH4
0.058
0.058
0.1156
0.1156
0.1156
0.146
0.146
0.146
0.0430
0.1875
0.1875
0.1873
0.1685
0.0005
N2
0.002
0.002
0.0049
0.0049
0.0049
0.006
0.006
0.006
0.00186
0.0081
0.0081
0.008
0.0072
0.00002
CH3OH
0.0002
0.0002
0.0058
0.0058
0.0058
0.1896
0.1896
0.1896
0.5475
0.0129
0.0129
0.0129
0.11391
0.6947
Component
25
Table (2.5) : actual stream flow rate each stream must be multiplied by the scale factor : 124.67
Streams
S1
S2
S3
S4
S5
S6
S7
S8
S9
SR
S10
S11
S12
S13
CO
1467.37
1467.37
2933.49
2933.49
2933.49
2032.1
2032.1
2032.1
101.48
1466.12
1929.89
463.77
100.36
1.0148
CO2
1479.2
1479.2
3452.7
3452.7
3452.7
2730.3
2730.3
2730.3
136.6
1973.5
2526.88
622.1
135.27
1.37
H2
8712.56
8712.56
12938.9
12938.9
12938.9
5856.99
5856.99
5856.99
292.8
4226.3
5564.15
1335.22
289.86
2.93
H2O
50.87
50.87
291.48
291.48
291.48
1012.94
1012.94
1012.94
696.28
240.6
316.66
75.92
15.21
680.7
CH4
723.58
723.58
2597.4
2597.4
2597.4
2595.6
2595.6
2595.6
129.78
1873.79
2465.97
590.9
128.4
1.298
N2
31.168
31.168
112.2
112.2
112.2
112.2
112.2
112.2
5.61
80.9
106.59
25.558
5.55
0.056
CH3OH
2.69
2.69
132.23
132.23
132.23
3354.87
3354.87
3354.87
1649.38
129.5
170.8
40.89
86.77
1562.49
Total
12467.44
12467.44
22458.395
22458.395
22458.395
17694.9997
17694.9997
17694.9997
3011.9
9990.71
13150.94
3154.358
761.42
2249.8
Component
Chapter 3
Energy Balance
27
Heat capacity data
Heat capacity for the mixture will be estimated an additional property.
𝑛
𝐶𝑝𝑚𝑖𝑥𝑡𝑢𝑟𝑒 = ∑ 𝑦𝑖 𝐶𝑝𝑖
𝑖=1
𝑦𝑖 : mole fraction for ith component
𝐶𝑝𝑖 : Heat capacity for ith component
An average value of Component heat capacity will be taken.
Enthalpy for each stream in Kj/hr will be :
𝐻𝑠𝑖 = 𝑆𝑖 . 𝐶𝑝𝑠𝑖 (𝑇𝑠𝑖 − 𝑇𝑅 )
𝑇𝑠𝑖 : temperature of ith stream.
𝐶𝑝𝑠𝑖 : Heat capacity for ith stream.
𝑇𝑅 : reference temperature = 298K
Composition
Cp (Kj/Kmol K) gas-phase
CO
29.5
CO2
38.07
H2
29
H2O
34.04
CH4
38.14
N2
29.5
CH3OH
44.76
Table (3.1) : average value of Component heat capacity
Heat capacity for liquid-phase for water and CH3OH will be :
𝐶𝑝𝐻2𝑂 (liquid) = 75 Kj/Kmol K
𝐶𝑝𝐶𝐻3𝑂𝐻 (liquid) = 90 Kj/Kmol K
𝐶𝑝 for gas-phase as a function of temperature
𝐶𝑝𝐶𝑂 = 30.8 − 0.013𝑇 + 2.79 × 10−5 𝑇 2
𝐶𝑝𝐶𝑂2 = 19.8 + 0.073𝑇 − 5.6 × 10−5 𝑇 2
28
𝐶𝑝𝐻2 = 14.5
𝐶𝑝𝐻2𝑂 = 32.2 + 0.013𝑇 + 1.92 × 10−3 𝑇 + 1.055 × 10−5 𝑇 2
𝐶𝑝𝐶𝐻4 = 19.2 + 0.052𝑇 + 1.2 × 10−5 𝑇 2
𝐶𝑝𝑁2 = 31.1 − 0.0135𝑇 + 2.68 × 10−5 𝑇 2
𝐶𝑝𝐶𝐻3𝑂𝐻 = 21.1 + 0.07𝑇 + 2.58 × 10−5 𝑇 2
Heat capacity for the mixture in each stream :
𝐶𝑝𝑠1 = 𝑦𝐶𝑂 . 𝐶𝑝𝐶𝑂 + 𝑦𝐶𝑂2 . 𝐶𝑝𝐶𝑂2 + 𝑦𝐻2 . 𝐶𝑝𝐻2 + 𝑦𝐻2𝑂 . 𝐶𝑝𝐻2𝑂 + 𝑦𝐶𝐻4 . 𝐶𝑝𝐶𝐻4
+ 𝑦𝑁2 . 𝐶𝑝𝑁2 + 𝑦𝐶𝐻3𝑂𝐻 . 𝐶𝑝𝐶𝐻3𝑂𝐻
Neglect the small value of mole fraction
𝐶𝑝𝑠1 = 0.1177 (29.5) + 0.1186 (38.07) + 0.696 (29)
𝐶𝑝𝑠1 = 26.07 Kj/Kmol K
𝐶𝑝𝑠3 = 0.13 (29.5) + 0.15 (38.07) + 0.57 (29)
𝐶𝑝𝑠3 = 26.17 Kj/Kmol K
𝐶𝑝𝑠6 = 0.1148 (29.5) + 0.1543 (38.07) + 0.331 (29) + 0.2 (44)
𝐶𝑝𝑠6 = 27 Kj/Kmol K
𝐶𝑝𝑠𝑅 = 0.1467 (29.5) + 0.1975 (38.07) + 0.4230 (29)
𝐶𝑝𝑠𝑅 = 24.11 Kj/Kmol K
𝐶𝑝𝑠10 = 0.146 (29.5) + 0.1975 (38.07) + 0.42310 (29)
𝐶𝑝𝑠10 = 14.24 Kj/Kmol K
𝐶𝑝𝑠11 = 0.1470 (29.5) + 0.1974 (38.07) + 0.4233(29)
𝐶𝑝𝑠11 = 24.13 Kj/Kmol K
𝐶𝑝𝑠12 = 0.1317 (29.5) + 0.1770 (38.07) + 0.38050 (29)
𝐶𝑝𝑠12 = 21.66 Kj/Kmol K
29
𝐶𝑝𝑠9 = 𝑦𝐻2𝑂 . 𝐶𝑝𝐻2𝑂 + 𝑦𝐶𝐻3𝑂𝐻 . 𝐶𝑝𝐶𝐻3𝑂𝐻
𝐶𝑝𝐻2𝑂 (liquid) = 75 Kj/Kmol K
𝐶𝑝𝐶𝐻3𝑂𝐻 (liquid) = 90 Kj/Kmol K
𝐶𝑝𝑠9 = 0.23 (75) + 0.547 (90)
𝐶𝑝𝑠9 9 = 66.48 Kj/Kmol K
𝐶𝑝𝑠13 = 0.3026 (75) + 0.6947 (90)
𝐶𝑝𝑠13 = 85.21 Kj/Kmol K
𝐶𝑝𝑠5 = 0.186 (29.5) + 0.143 (38) + 0.537 (29) + 0.012 (34) + 0.1 (38)
𝐶𝑝𝑠5 = 32 Kj/Kmol K
𝐶𝑝𝑠6 = 0.115 (29.5) + 0.15 (38) + 0.33 (29) + 0.057 (34.9) + 0.146 (38) + 0.2 (44)
𝐶𝑝𝑠6 = 35 Kj/Kmol K
30
Energy balance around the mug compressor :
Figure (3.1) : Energy balance around the mug compressor
∆ 𝐻𝑆1 = 𝑆1 𝐶𝑝𝑠1 (𝑇1 − 𝑇𝑅 ) , 𝑇𝑅 = 298 𝐾
∆ 𝐻𝑆1 = 1246.4 × 26.07 (313 − 298 )
∆ 𝐻𝑆1 = 4875376 Kj/hr
The out let stream temperature Tś2 will be calculated using the following relation for
a diabetic compression
𝐶𝑝
K = constant = 𝐶 = 1.4
𝑣
Data from sirte oil company :
Ps1 = 1716 KPa
Ps2 = 5791 KPa
For a diabetic compression , we have :
𝑇2
𝑇1
𝑃2
= (𝑃 )
1
𝛾−1
𝛾
,𝛾=
𝐶𝑝
𝐶𝑣
T1 = Ts1 = 313K
P1 = Ps1 = 1716 KPa
P1 = Ps1 = 1716 KPa
= 1.39
31
P2 = Pś2 = 5791 KPa
Tś2= T2 = unknown
𝑇2
5791
=(
)
313
1716
1.39 −1
1.39
Tś2= 440.3 K
Work done by compressor :
𝑊𝐾−01 = ∆ 𝐻ś2 − ∆ 𝐻𝑆1
∆ 𝐻ś2 = ś2 𝐶𝑝ś2 (𝑇ś 2 − 298)
∆ 𝐻ś2 = 12467.4 × 26.07 (440 − 298 )
∆ 𝐻ś2 = 46251074.29 𝐾𝑗/ℎ𝑟
∆ 𝐻𝑆1 = 𝑆1 𝐶𝑝𝑆1 (𝑇1 − 𝑇𝑅 ) = 4875376 𝐾𝑗/ℎ𝑟
𝑊𝐾−01 = 41375698 𝐾𝑗/ℎ𝑟
Energy balance around the mixing point as shown below :
TsR= 313 K
S3
Ts2= ? K
Tś2= 440.3 K
Figure (3.2) : Energy balance around the mixing point
Make energy balance will be :
∆ 𝐻𝑆𝑅 + ∆ 𝐻ś2 = ∆ 𝐻𝑆3
∆ 𝐻𝑆𝑅 = 𝑆𝑅 𝐶𝑝𝑅 (𝑇𝑆 𝑅 − 298)
32
∆ 𝐻ś2 = ś2 𝐶𝑝ś2 (𝑇ś 2 − 298)
∆ 𝐻𝑆3 = 𝑆3 𝐶𝑝3 (𝑇𝑆 3 − 298)
𝑆𝑅 𝐶𝑝𝑅 (𝑇𝑆 𝑅 − 298) + ś2 𝐶𝑝ś2 (𝑇ś 2 − 298) = 𝑆3 𝐶𝑝3 (𝑇𝑆 3 − 298)
𝑆𝑅 = 9990.7 𝐾𝑚𝑜𝑙/ℎ𝑟 , Ś2 = 𝑆1 = 12467 𝐾𝑚𝑜𝑙/ℎ𝑟 , 𝑆3 = 22458.4 𝐾𝑚𝑜𝑙/ℎ𝑟
𝐶𝑝𝑅 = 24.11 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾 , 𝐶𝑝ś2 = 𝐶𝑝𝑠1 = 26.07 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝐶𝑝𝑠3 = 26.17 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
Solve for 𝑇𝑆 3
3612883 + 46152085 = 58772690 (𝑇𝑆 3 − 298)
0.8467 = 𝑇𝑆 3 − 298
𝑇𝑆 3 = 382.9 𝐾
33
Energy balance around the compressor K-02 :
The mixed stream S3 is compressed from 5687.9 KPa to 6377.625 KPa.
Figure (3.3) : Energy balance around the compressor K-02
For a diabetic compression , we have :
𝑇𝑠 4
𝑇𝑠 3
𝑃𝑠 4
𝛾−1
𝛾
= (𝑃 )
𝑠3
, 𝛾 = 1.39
1.39 −1
1.39
𝑇𝑠 4
6377.625
=(
)
382.9
5687.8
𝑇𝑠 4 = 395.36 K
Work done by compressor K-02 will be :
𝑊𝐾−02 = ∆ 𝐻𝑠4 − ∆ 𝐻𝑠3
∆ 𝐻𝑠4 = 𝑆4 𝐶𝑝4 (𝑇𝑆 4 − 298)
∆ 𝐻𝑠4 = 22458.39 × 26.17 (395.36 − 298 )
∆ 𝐻𝑠4 = 57221983 𝐾𝑗/ℎ𝑟
∆ 𝐻𝑆3 = 𝑆3 𝐶𝑝3 (𝑇𝑆 3 − 298)
∆ 𝐻𝑆3 = 22458.39 × 26.17 (382.9 − 298 )
∆ 𝐻𝑆3 = 49898792 𝐾𝑗/ℎ𝑟
34
𝑊𝐾−02 = 57221983 − 49898792
𝑊𝐾−02 = 7323191 𝐾𝑗/ℎ𝑟
Energy balance around the reactor
Figure (3.4) : Energy balance around reactor
To calculate the temperature of the steam , we need to make energy balance
𝑄 = ∆ 𝐻𝑠6 − ∆ 𝐻𝑠5 + ∆ 𝐻𝑅1 (𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑂)
+ ∆ 𝐻𝑅2 (𝑐𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑚𝑜𝑙𝑒 𝑜𝑓 𝐶𝑂2 )
∆ 𝐻𝑅1 (𝑎𝑡 298𝐾) = −90700 𝐾𝑗/ℎ𝑟
∆ 𝐻𝑅2 (𝑎𝑡 298𝐾) = −49500 𝐾𝑗/ℎ𝑟
From material balance :
Consumed mole of CO = 4493 × 0.5477
= 2461 𝐾𝑚𝑜𝑙/ℎ𝑟
Consumed mole of CO2 = 3456 × 0.2088
= 721.6 𝐾𝑚𝑜𝑙/ℎ𝑟
35
∆ 𝐻𝑠6 = 𝑆6 𝐶𝑝6 (𝑇𝑆 6 − 298)
𝑆6 = 17695 𝐾𝑚𝑜𝑙/ℎ𝑟
𝐶𝑝6 = 35 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝑇𝑆 6 = 533𝐾
∆ 𝐻𝑠6 = 17695 × 35 (533 − 298 )
∆ 𝐻𝑠6 = 145541375 𝐾𝑗/ℎ𝑟
∆ 𝐻𝑠5 = 𝑆5 𝐶𝑝5 (𝑇𝑆 5 − 298)
𝑆5 = 24066 𝐾𝑚𝑜𝑙/ℎ𝑟
𝐶𝑝5 = 32 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝑇𝑆 5 = 473𝐾
∆ 𝐻𝑠5 = 24066 × 32 (473 − 298 )
∆ 𝐻𝑠6 = 134769600 𝐾𝑗/ℎ𝑟
Heat removed from the reactor :
𝑄 = 145541375 − 134769600 + (−90700)(2461) + (−49500)(721.6)
𝑄 = −248160125 𝐾𝑗/ℎ𝑟
36
Energy balance around the heat exchanger (E) :
Figure (3.5) : Energy balance around heat exchanger
𝐶𝑝𝑠4 − 𝐶𝑝𝑠5 = 35 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝐶𝑝𝑠6 − 𝐶𝑝𝑠7 = 35 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
To estimate 𝑇𝑠 7 , an energy balance will be taken :
𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = 𝑆4 𝐶𝑝4 (𝑇𝑆 5 − 𝑇𝑆 4 )
𝑄𝑔𝑎𝑖𝑛𝑒𝑑 = 24066 × 35 (473 − 395.3)
= 65447487 𝐾𝑗/ℎ𝑟
65447487 = 𝑆6 𝐶𝑝6 (533 − 𝑇𝑆 7 )
65447487 = 17695 × 35 (533 − 𝑇𝑆 7 )
105.67 = 533 − 𝑇𝑆 7
𝑇𝑆 7 = 427.3 𝐾
37
Energy balance around the cooler (C) :
Figure (3.6) : Energy balance around the cooler
Heat removed by Cooling water :
𝑄𝑐𝑜𝑜𝑙𝑒𝑟 = 𝑆7 𝐶𝑝7 (𝑇𝑆 7 − 𝑇𝑆 8 )
𝐶𝑝𝑠7 = 35 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝑄𝑐𝑜𝑜𝑙𝑒𝑟 = 17695 × 35 (427.3 − 313)
𝑄𝑐𝑜𝑜𝑙𝑒𝑟 = 70788847𝐾𝑗/ℎ𝑟
Cooling water flow rate 𝑚𝐻2 𝑂 = 𝐶𝑝
𝑄
𝐻2 𝑂 ∆𝑇
𝐶𝑝𝐻2 𝑂 = 75 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
∆𝑇 = 25 𝐶° = 25𝐾
𝑚𝐻2 𝑂 =
70788847
75 × 25
𝑚𝐻2 𝑂 = 37754 𝐾𝑚𝑜𝑙/ℎ𝑟 = 10.5 𝐾𝑚𝑜𝑙/𝑠
𝑚𝐻2 𝑂 = 188 𝐾𝑔/𝑠
Enthalpies of streams S10 , S9 , S12 and S13 will be calculated as following :
∆ 𝐻𝑠10 = 𝑆10 𝐶𝑝10 (𝑇𝑆10 − 298)
∆ 𝐻𝑠9 = 𝑆9 𝐶𝑝9 (𝑇𝑆 9 − 298)
∆ 𝐻𝑠12 = 𝑆12 𝐶𝑝12 (𝑇𝑆12 − 298)
38
∆ 𝐻𝑠13 = 𝑆13 𝐶𝑝13 (𝑇𝑆13 − 298)
𝑆10 = 13150 𝐾𝑚𝑜𝑙/ℎ𝑟
𝑆9 = 3012 𝐾𝑚𝑜𝑙/ℎ𝑟
𝑆12 = 760 𝐾𝑚𝑜𝑙/ℎ𝑟
𝑆13 = 2249.2 𝐾𝑚𝑜𝑙/ℎ𝑟
𝐶𝑝10 = 31.4 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝐶𝑝9 = 35.2 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝐶𝑝12 = 33 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
𝐶𝑝13 = 85.5 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾
Temperature of streams S10 , S9 , S12 and S13 at 313K
∆ 𝐻𝑠13 = 2249.2 × 85.9 (313 − 298)
∆ 𝐻𝑠10 = 13150 × 31.4 (313 − 298)
∆ 𝐻𝑠9 = 3012 × 35.2 (313 − 298)
∆ 𝐻𝑠12 = 760 × 33 (313 − 298)
Chapter 4
Process design of cooler
40
4.1 Process design of cooler
Figure (3.7) : Design of cooler
Counter – Current flow arrangement
Tube – Side fluid : cooling water.
Shell – Side fluid : Stream S7 and S8.
∆𝑇𝑙𝑚 =
(313 − 298) − (427.3 − 323) 15 − 104.3
=
313 − 298
15
ln (427.3 − 323)
ln (104.3)
∆𝑇𝑙𝑚 = 46 𝐾 = 46 𝐶°
Physical properties of tube – side fluid (water) :
Heat capacity 𝐶𝑝 = 75 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾 = 4.2 𝐾𝑗/𝐾𝑔 𝐾
Viscosity 𝜇 = 0.8 × 10−3 𝑃𝑎. 𝑠
Thermal conductivity 𝐾 = 0.6 𝑊/𝑚𝐶°
Density 𝜌 = 1000 𝐾𝑔/𝑚3
Mass flow rate 𝑚𝐻2𝑂 = 188 𝐾𝑔/𝑠𝑒𝑐
Physical properties of Stream S7 (shell – side fluid) :
𝐶𝑝 = 35 𝐾𝑗/𝐾𝑚𝑜𝑙 𝐾 = 1.85 𝐾𝑗/𝐾𝑔 𝐾
𝜇 = 1 × 10−5 𝑃𝑎. 𝑠
41
𝐾 = 0.11 𝑊/𝑚𝐶°
𝜌 = 34 𝐾𝑔/𝑚3
𝐴𝑣𝑒 𝑀𝑤𝑡 = 18.9 𝐾𝑔/𝐾𝑚𝑜𝑙
Physical properties are taken at an average temperature of shell – side fluid
=
427.3+313
2
= 370𝐾
Assume overall heat transfer coefficient
𝑢0 = 300 𝑊/𝑚2 𝐾
𝑄 = 𝑢0 𝐴0 ∆𝑇𝑙𝑚
𝑄 = 70788847 𝐾𝑗/ℎ𝑟 = 19663568 𝑗/𝑠
Heat transfer area 𝐴0 = 𝑢
𝑄
0 ∆𝑇𝑙𝑚
𝐴0 =
19663568
= 2137 𝑚2
300 × 46
Select tube dimensions :
Length 𝐿 = 6 m
𝑑𝑖 = 16 𝑚𝑚
𝑑0 = 20 𝑚𝑚
Area of one tube 𝑎0 = 𝜋𝑑0 𝐿
𝑎0 = 𝜋(0.02)(4.88) = 0.3066 𝑚2
𝑡𝑜𝑡𝑎𝑙 𝑎𝑟𝑒𝑎
Number of tubes 𝑁𝑡 = 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛 𝑡𝑢𝑏𝑒
2137
𝑁𝑡 = (3.14)(20×10−3 )(6) = 5671 𝑡𝑢𝑏𝑒
shell – side calculation :
𝑁
1
Bundle diameter 𝐷𝑏 = 𝑑0 (𝐾 𝑡 )𝑛1
1
For one tube pass and triangle pith
42
𝐾1 = 0.319 , 𝑛1 = 2.142 , Ref (4)
𝐷𝑏 = 20(
3782 1
)2.142
0.319
𝐷𝑏 = 1.6 𝑚
Cross –flow area 𝐴𝑠 =
(𝑃𝑡 −𝑑0 )𝐷𝑠 𝐿𝐵
𝑃𝑡
𝐿𝐵 : Baffle spacing
Choose baffle spacing 𝐿𝐵 =
𝐷𝑠
5
𝑃𝑡 = 12.5𝑑0 = 1.25 × 25 𝑚𝑚
𝐷𝑠 = 𝐷𝑏 + 20 𝑚𝑚
𝐷𝑠 = 1927 + 20 = 1947 𝑚𝑚 = 1.947 𝑚
𝐴𝑠 =
(𝑃𝑡 − 𝑑0 )
𝐷𝑠
. 𝐷𝑠 ×
𝑃𝑡
5
(25 − 20) (1.947)2
𝐴𝑠 =
.
25
5
𝐴𝑠 = 0.07788 𝑚2
Equivalent diameter
𝑑𝑒 =
1.1 2
(𝑃𝑡 − 0.917𝑑0 2 )
𝑑0
𝑑𝑒 =
1.1
(252 − 0.917(20)2 )
20
𝑑𝑒 = 14.201 𝑚𝑚 = 0.0142 𝑚
Mass velocity 𝐺 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑖𝑛 𝑠ℎ𝑒𝑙𝑙 𝑡𝑢𝑏𝑒 𝑆7
𝐴𝑠
𝑆7 = 17695 𝐾𝑚𝑜𝑙/ℎ𝑟 = 17695 × (𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑀𝑤𝑡)
= 17695 × 18.9
𝐴𝑣𝑒 𝑀𝑤𝑡 = 18.9 𝐾𝑔/𝐾𝑚𝑜𝑙
𝑆7 = 334435.5 𝐾𝑔/ℎ𝑟 = 92.9𝐾𝑔/𝑠𝑒𝑐
43
𝐺=
92.9
= 1192.8608 𝐾𝑔/𝑚2 . 𝑠
0.07788
Reynolds number for shell – side fluid
𝑅𝑒 =
𝐺𝑒 𝑑𝑒 × 0.0142
=
𝜇
1 × 10−5
𝑅𝑒 = 1694060
Prandtl number 𝑃𝑟 =
𝐶𝑝𝜇
𝐾
=
(1.85×10−5 )(1×10−5 )
0.11
𝑃𝑟 = 0.168
From fig (1) , shell – side heat transfer factor (for 25% baffle cut)
𝑅𝑒 = 1694060
𝑗ℎ = 7 × 10−4
Convection heat transfer coefficient
ℎ𝑠 𝑑𝑒
= 𝑗ℎ . (𝑅𝑒). (𝑃𝑟)0.33
𝐾
7 × 10−4 (1694060)(0.168)0.33 × 0.11
ℎ𝑠 =
0.0142
ℎ𝑠 = 5099 𝑊/𝑚2 𝐾
tube – side calculation :
𝜋
𝑎𝑡 total tube cross flow area = 4 𝑑𝑖 2 × 𝑁𝑡
=
𝜋
(0.016)2 × 5671
4
= 1.342 𝑚2
𝐺𝑡 =
𝑅𝑒 =
𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑡𝑎𝑡𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
𝑎𝑡
𝐺𝑡 𝑑𝑖 140.1 × 0.016
=
𝜇
0.8 × 10−3
𝑅𝑒 = 2802
188
= 1.342 = 140.1 𝐾𝑔/𝑚2 . 𝑠
44
Water linear velocity 𝑢𝑡 =
Prandtl number 𝑃𝑟 =
𝐶𝑝𝜇
𝐾
𝐺𝑡
𝜌
=
=
2808
1000
= 2.802 𝑚/𝑠
(4.2×103 )(0.8×10−3 )
0.6
𝑃𝑟 = 5.6
𝐿
6
=
= 375
𝑑𝑖
0.016
Convective heat transfer coefficient for tube - side :
ℎ𝑖 𝑑𝑖
= 𝑗ℎ . (𝑅𝑒). (𝑃𝑟)0.33
𝐾
𝐿
𝑗ℎ , from fig (2) , for (𝑑𝑖 ) = 375
𝑖
ℎ𝑖 =
𝑗ℎ . (𝑅𝑒). (𝑃𝑟)0.33 𝐾
𝑑𝑖
ℎ𝑖 =
3.9 × 10−3 (2802)(5.6)0.33 × 0.6
0.016
ℎ𝑖 = 482 𝑊/𝑚2 𝐶°
Overall heat transfer coefficient :
𝑑0
1
1
1 𝑑0 ln( 𝑑𝑖 ) 𝑑0 1
1 𝑑0
=
+ +
+ × + ×
𝑢0
ℎ𝑠 ℎ0
2𝐾𝑤
𝑑𝑖 ℎ𝑖 ℎ𝑖 𝑑𝑖
20
0.02 ln (16) 20
1
1
1
1
1
1.25
=
+
+
+
×
+
×
𝑢0
1.279 5000
2 × 50
16 5000 2129
20
1
= 340
𝑢0
Since the estimated u higher then the assumed value, then the heat transfer will be
taking place successfully.
Chapter 5
Safety
45
5.1 Methanol Safety Data Sheet
5.1.1 Potential Health Effects
 Eye : may cause painful sensitization to light. Methanol is a mild to moderate
eye irritant. Inhalation, ingestion or skin absorption of methanol can cause
significant disturbance in vision, including blindness.
 Skin : causes moderate skin irritation. May be absorbed through the skin in
harmful amounts. Prolonged and or repeated contact may cause defatting of
skin and dermatitis. Methanol can be absorbed through the skin, producing
systemic effects that include visual disturbances.
 Ingestion : may be fatal or cause blindness if swallowed. Aspiration hazard.
Cannot be made nonpoisonous. May cause gastrointestinal irritation with
nausea, vomiting and diarrhea. May cause systematic toxicity with acidosis.
May cause central nervous system depression, characterized by excitement,
followed by headache, dizziness, drowsiness, and nausea. Advanced stages
may cause collapse, unconsciousness, coma, and possible death due to failed
respiratory failure. May cause cardiopulmonary system effects.
 Inhalation : methanol is toxic and can very readily form extremely high
vapor concentrations at room temperature. Inhalation is the most common
route of occupational exposure. At first, methanol causes cns depression with
nausea, headache, vomiting, dizziness and incoordination. A time period with
no obvious symptoms follows (typically 8-24 hrs). This latent period is
followed by metabolic acidosis and severe visual effects which may include
reduced reactivity and/or increased sensitivity to light, blurred, doubt and/or
snowy vision, and blindness. Depending on the severity of exposure and the
promptness of treatment, survivors may recover completely or may have
permanent blindness, vision disturbances and/or nervous system effects.
 Chronic : prolonged or repeated skin contact may cause dermatitis. Chronic
exposure may cause effects similar to those of acute exposure. Methanol is
only very slowly eliminated from the body. Because of this slow elimination,
methanol should be regarded as a cumulative poison. Though a single
exposure may cause no effect, daily exposures may result in the accumulation
of a harmful amount.
46
5.1.2 First Aid Measures
 Eyes : in case of contact, immediately flush eyes with plenty of water for a t
least 15 minutes. Get medical aid.
 Skin : in case of contact, immediately flush skin with plenty of water for at
least 15 minutes while removing contaminated clothing and shoes. Get
medical aid immediately. Wash clothing before reuse.
 Ingestion : potential for aspiration if swallowed. Get medical aid
immediately. Do not induce vomiting unless directed to do so by medical
personnel. Never give anything by mouth to an unconscious person. If
vomiting occurs naturally, have victim lean forward.
 Inhalation : if inhaled, remove to fresh air. If not breathing, give artificial
respiration. If breathing is difficult, give oxygen. Get medical aid.
 Notes to physician : effects may be delayed.
 Antidote : ethanol may inhibit methanol metabolism.
5.1.3 Accidental Release Measures
 Spills/leaks : use water spray to disperse the gas/vapor. Remove all sources
of ignition. Absorb spill using an absorbent, non-combustible material such
as earth, sand, or vermiculite. Do not use combustible materials such as
sawdust. Use a spark-proof tool. Provide ventilation. A vapor suppressing
foam may be used to reduce vapors. Water spray may reduce vapor but may
not prevent ignition in closed spaces.
5.1.4 Storage And Handling
 Handling : wash thoroughly after handling. Remove contaminated clothing
and wash before reuse. Ground and bond containers when transferring
material. Use spark-proof tools and explosion proof equipment. Avoid
contact with eyes, skin, and clothing. Empty containers retain product
residue, (liquid and/or vapor), and can be dangerous. Keep container tightly
closed. Do not ingest or inhale. Do not pressurize, cut, weld, braze, solder,
drill, grind, or expose empty containers to heat, sparks or open flames. Use
only with.
47
 Storage : keep away from heat, sparks, and flame. Keep away from sources
of ignition. Store in a cool, dry, well-ventilated area away from incompatible
substances. Flammables-area. Keep containers tightly.
5.1.5 Exposure Controls/Personal Protection
 Engineering controls : use explosion-proof ventilation equipment. Facilities
storing or utilizing this material should be equipped with an eyewash facility
and a safety shower. Use adequate general or local exhaust ventilation to keep
airborne concentrations below the permissible exposure limits. Osha vacated
pels: methanol: 200 ppm twa; 260 mg/m3 twa
 Personal protective equipment :
Eyes : wear chemical splash goggles.
Skin : wear butyl rubber gloves, apron, and/or clothing
Clothing : wear appropriate protective clothing to prevent skin exposure.
Respirators : follow the osha respirator regulations found in 29 cfr 1910.134
or european standard en 149. Use a niosh/msha or european standard en 149
approved respirator if exposure limits are exceeded or if irritation or other
symptoms are experienced.
5.1.6 Stability And Reactivity
 Chemical stability : stable under normal temperatures and pressures
 Conditions to avoid : high temperatures, ignition sources, confined spaces.
 Incompability (materials to avoid) : oxidizing agents, reducing agents,
acids, alkali metals, potassium, sodium, metals as powders (e.g. Hafnium,
raney nickel), acid anhydrides, acid chlorides, powdered aluminum,
powdered magnesium.
 Hazardous decomposition products : carbon monoxide, irritating and toxic
fumes and gases, carbon dioxide, formaldehyde.
 Hazardous polymerization : will not occur.
Reference
1- Hansen, J.B. and Nielsen, P.E.H., ―Methanol Synthesis‖, in: Ertl, G., Knözinger, H.,
Schûth, F., Weitkamp, J., Handbook of Heterogeneous Catalysis, Second Edition
2008, Vol. 6, p. 2920, Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim.
2- Huang, P., Ju, H., Tan, S., Wang, H., and Zhao, T., The Future of Methanol Fuel,
Project Report, 2015
3- . Hansen, J. B., Methanol Production Technology, Methanol Workshop, Lund
University, March 2015.
4- Chemical Engineering , Volume 6 , Coulson and Richardson.
Figure (1). Shell-side heat-transfer factor, (for 25% baffle cut) , Ref (4)
Appendices
APPENDIX . A
Figure (2). Tube-side heat-transfer factor - Ref (4)
APPENDIX . B