GAS TO LIQUIDS PLANT DESIGN
Prepared For
Dr. J.D. PALMER
Professor
Chemical Engineering Department
Louisiana Tech University
By
Bajracharya Prabuddha
Enaohwo Enakeme
Obe Akeem
Cmen 432 Students
May 5nd, 2011
1
EXECUTIVE SUMMARYThis report discusses the details of how we designed a Fischer-Tropsch Reaction unit (FTR) to convert syngas
(CO and H2) to hydrocarbons. The purpose of the design is to produce liquid fuels and reduce our over-
dependence on importation of oil from foreign countries.
In achieving our goal of effectively designing a Fischer-Tropsch Reaction unit to produce liquid fuel, design
objectives were adhered to strictly. The design was made to ensure that it does not pose any environmental,
health or safety hazards by ensuring that appropriate control valves were placed in needed areas. Rigorous
research was also done on the material of construction of the equipment to ensure that no reaction between the
components and the material of construction occur. This may lead to an unwanted product that could greatly
affect the purity and specification of our final product.
Also, rigorous economic analysis was done to ensure that the design is feasible. In the economic analysis, the
designed capital investment, expense costs of the air separation unit, syngas unit, FTR unit, and hydro-
isomerization unit were reflected. After an economic analysis on the design process, a net present value of 4.96
billion dollars was achieved under the process conditions defined in the report.
2
INTRODUCTIONSynthetic fuels production is one of the promising alternatives to petroleum fuels. Synthetic fuels can be
produced by converting natural gas to liquid petroleum products. One of these gases to liquid processes is
Fischer-Tropsch synthesis. Fischer-Tropsch is one of the oldest and mature processes to produce alternative fuel
source. This process converts carbon monoxide and hydrogen gas in combination called synthetic gas into
liquid hydrocarbon fuels like diesel and jet fuel.
The objective of this project is to provide preliminary design package for a grass roots Fischer-Tropsch reactor
unit with reactor effluent separation facilities. We are to design a safe, environmentally clean, thermally
integrated FTR unit operating cost utilization. The project should also reflect that the optimal cost/benefit
balance is achieved.
Some of the design specifications are:
Effective heat integration
Environmental, health or safety hazards properly mitigated
No continuous flaring/venting of hydrocarbons
The economic constraints for the plant are:
15 year project life, depreciation is SL over 15 years.
Tax is 33%
3% projected yearly inflation
Total capital investment estimated by multiplying equipment cost by 4.8
Operating expenses above and beyond utilities is estimated using 3% of the total capital investment
FTR unit will have 1 month turnaround to coincide with catalyst replacement
3
SUMMARYThe effective and efficient production of liquid fuel from syngas involves three processes. These processes are
conducted in the syngas unit (SU), Fischer-Tropsch Reaction (FTR) unit, and the hydro-isomerization (HI) unit.
In the syngas unit (SU), methane gas, steam, carbon dioxide, and oxygen are reacted in the syngas unit to
produce carbon monoxide and hydrogen gas. The syngas unit was designed to convert 500 MSCF/D (Actual
conditions: 500 PSIG, 100F) of clean methane to syngas. In the syngas unit, three reactions occur. The first
reaction is a steam reforming reaction between methane gas and water to produce syngas; the second reaction is
a partial oxidation reaction between methane gas and oxygen to produce carbon monoxide, and water. It should
be noted that the heat required to drive the endothermic auto thermal reforming reaction is supplied in-situ by
partial oxidation of methane.
In the Fischer-Tropsch reaction unit, the carbon monoxide and hydrogen gas from the syngas unit are fed into
two tubular fixed reactors arranged in series. In these reactors, different hydrocarbons were formed at specific
temperatures and pressures. The products from the FTR unit are sent to distillation columns to separate products
based on differences on their boiling points.
In the hydro-isomerization unit, the distillate and heavier fractions (material with greater than 350 F boiling
point) are fed to a catalytic hydro-isomerization reactor, where paraffin’s are isomerized and wax is converted
to lighter products. A catalyst was used to reduce the activation energy of the reaction. It should be noted that
this catalyst is sensitive to water and carbon monoxide; therefore the liquid feed to the hydro-isomerization
reactor was free from water above the solubility limit and the makeup gas has a carbon monoxide content no
greater than 0.1 mol%.
In order to check the viability of the design, an economic analysis is done to obtain a net present value that
would justify the investment to this process.
4
DISCUSSIONSELECTION OF THERMODYNAMIC MODEL
To ensure a good process simulation, an accurate thermodynamic model must be chosen. Choosing an accurate
thermodynamic model will depend on the components of the incoming feed stream and the products.
In order to simulate portions of the GTRL plant, we have to have an appropriate thermodynamic model. Choosing a
thermodynamic model will depend on the data available and the most cumbersome compounds. We consulted a decision
tree to narrow down the choice of thermodynamic models and we came up with SRK, and PR or their variants. We were
able to find data for different binary system and we used that data to compare with the data given by chemcad model. To
identify the best fit model that simulated the data we used sum of least square method and found out that MSRK produced
the best fit. The following diagrams show the graphical fitting for one of our component system
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 160
80
100
120
140
160
180
200
n-Butane/n-Heptane TP-xy Experimental
Exp LiquidExp vapor
n-Butane Mole Fraction
Temp
eratu
re C
In order to find the best thermodynamic model, the TP-XY for n-butane and n-heptane were plotted for SRK, PR, PSRK,
and MSRK. They were plotted alongside the experimental data and a regression analysis was done on the data to
determine what model was the best fit. The plots are shown below:
5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 160
80
100
120
140
160
180
200
SRK/TP-xy Experimental
Exp LiquidExp vaporSRK liquidSRK vapor
n-Butane Mole Fraction
Tem
pera
ture
C
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 160
80
100
120
140
160
180
200
Peng Robinson/TP-xy Experimental
Exp LiquidExp vaporPR LiquidPRvapor
n-Butane Mole Fraction
Temp
eratu
re C
6
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 160
80
100
120
140
160
180
200
PSRK/TP-xy Experimental
Exp LiquidExp vaporPSRK liquidPSRK vapor
n-Butane Mole Fraction
Tem
pera
ture
C
The graphs all seem similar and we can decide to just go with one of them, but to achieve even more accuracy in our
simulation, a regression analysis is done on the data for the vapor phase:
SRK PR PSRK MSRK0.700000000000001
0.750000000000001
0.800000000000001
0.850000000000001
0.900000000000001
0.950000000000001
1
R2 value vs. Thermodynamic Models
Hexane Vs HeptaneMethane vs. DecaneMethane vs. ButaneButane vs Heptane
Thermodynamic Models
R2 V
alue
Based on the analysis, we found out that some were better than others for different binary systems but for a global
thermodynamic model we found out that MSRK was the right choice. So MSRK was chosen as our thermodynamic
model.
7
MATERIAL OF CONSTRUCTIONWhen choosing the material of construction for any process, the corrosive effects of the components in the process must
be factored in, because it make the difference between having to replace equipment in the future and not having to do that
at all. The material of construction for the different equipment’s will depend on the contents of the feed and exit streams.
In order to choose the material of construction for our process, the NACE corrosion database was consulted. We were able
to find corrosion data for each major component:
Column1 CH4 Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-225 <22 SS 12Cr 100 25-475 <23 Al alloy 100 25-225 <24 SS 304. 100 25-475 <25 SS 316. 100 25-475 <26 (Ni-30Mo) alloy 100 25-225 <2
Column1 Steam Column2 Column3 Column4# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 212-475 <22 SS 12Cr 100 212-475 <23 Al alloy 100 212-475 <204 SS 304. 100 212-475 <25 SS 316. 100 212-475 <26 (Ni-30Mo) alloy 100 212-475 <2
Column1 H2 Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-475 <22 SS 12Cr 100 25-500 <23 Al alloy 100 25-500 <24 SS 304. 100 25-500 <25 SS 316. 100 25-500 <26 (Ni-30Mo) alloy 100 275-500 <2
Column1 CO Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-500 <22 SS 12Cr 100 25-500 <23 Al alloy 100 25-500 <24 SS 304. 100 25-500 <25 SS 316. 100 25-500 <26 (Ni-30Mo) alloy 100 275-500 <2
8
Column1 C2H6 Column2 Column3 Column4# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-500 <22 SS 12Cr 100 25-500 <23 Al alloy 100 25-255 <24 SS 304. 100 25-500 <25 SS 316. 100 25-500 <26 (Ni-30Mo) alloy 100 25-500 <2
The corrosion data for C3 is similar to that of C4, so it is not shown.
Column1 C4H10 Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-225 <22 SS 12Cr 100 25-225 <23 Al alloy 100 25-255 <24 SS 304. 100 25-255 <25 SS 316. 100 25-255 <26 (Ni-30Mo) alloy 100 25-225 <2
Column1 Naphtha (C5-C10) Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-225 <22 SS 12Cr 100 25-225 <23 Al alloy 100 175-225 <24 SS 304. 100 25-225 <25 SS 316. 100 25-225 <26 (Ni-30Mo) alloy 100 175-225 <2
Column1 C12H26 Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)1 CS 100 25-225 <202 SS 12Cr 100 25-225 <203 Al alloy 100 25-225 <204 SS 304. 100 25-225 <205 SS 316. 100 25-225 <206 (Ni-30Mo) alloy 100 25-225 <27 tantalum 100 25-225 <2
9
The table below shows a suitable material for the syngas unit reactor, but not the corrosion rate.
Column1 CO2/N2/O2/H20 Column2 Column3 Column4
# exposure medium Conc % temp (F) Corrosion rate(mpy)
1 (chlorendic unsaturated polyester, glass fiber reinforced) 100 120
Based on the data collected in the tables above, a combination of materials has to be used for the process. For example, we
can use carbon steel for the naphtha reboiler, we also have to use a Nickel alloy for the equipment that will be separating
the diesel from naphtha. These are just examples of how the equipment will be chosen; safety handbooks will also be
consulted to see if the contents of the entry stream have any adverse reactions with our chosen material.
EQUIPMENT TYPE AND SIZING
10
In this FT synthesis processes utilized different equipment like heat exchangers, distillation columns. The brief
description of general equipment sizing and detailed description of specialized equipment sizing is shown
below.
1. Heat Exchangers : There are multiple heat exchangers integrated in our design .All of the heat
exchangers were sized using the following equation
Q=UA ∆T lm
Where “Q” is the duty, “U” is the overall heat transfer coefficient, “A” is the heat transfer area and ∆Tlm
is the log mean temperature difference. The overall heat transfer coefficient for the condenser is 850
W/m2°C, and for the reboiler is 1140 W/m2°C.
2. Distillation columns: Three distillation columns are used to meet the purity specifications of the
products. The tower height can be calculated with the help of number of trays and tray spacing .The
number of trays can be changed as the tower is optimized.
Therefore, the tower height (H) is given by,
H=¿of trays∗tray spacing
Volume of the tower is given by
V=π r2h
3. Reflux tank: The reflux tank was sized according to hold uptime and volumetric flow rate. The holdup
time for the reflux tanks was 10 min.
Reflux Tanks can be sized using the following relation:
V=2∗M∗tρ
Where V is the volume of the drum, M is the mass flow rate; t is the holdup time and ρis the liquid
density.
Diameter of the drum for totally condensed liquid is given by4 the following relation which considers
diameter to be 4 times the height.
D=[Vπ ]1 /3
4. Pumps: Centrifugal pumps were used because they are cost efficient and can be easily controlled using
controlled valves. Each of the pump was sized according the formula below;11
Power (Hp )=(Flow(gpm)∗∆ P( psi))/1714 /efficiency
5. Tray efficiency
Based on 41 sets of performance data for different sets of trays Drickamer and Bradford correlated the
overall stage efficiency. The data was collected for mainly hydrocarbons mixtures and few water and
miscible organics4.
The correlation produces the following empherical equation
Eo=13.3−66.8 log(μ)
Where μ is the viscosity of liquid in centipoise, Eo is the overall efficiency in percent.
SPECIALIZED EQUIPMENT DESIGN
Syngas Reactor12
Syngas reactor was one of the major equipment for our design .It is the reactor where methane feed with
addition of carbon dioxide, oxygen and steam is converted into carbon monoxide and hydrogen. For the syngas
reactor different reactions were given in the problem statement and are as follows:
CH 4+H 2O↔CO+3 H2
CO+H 2O↔CO2+H2
CH 4+32O
2→CO+2H 2
Here the first reaction is the partial oxidation reaction. This was modeled offline and the two reactions that
follow are equilibrium reaction these reactions were modeled in chemcad in an equilibrium reactor. The partial
oxidation reaction is exothermic and two equilibrium reactions are exothermic. The reactors were aligned in
series in such a way that the exothermic partial oxidation would provide the heat needed for the endothermic
equilibrium reaction. The reactor had to be optimized in such a way that the steam to methane ratio would be
0.5 mol/mol so as to prevent coking and hydrogen to carbon monoxide molar ratio out of the syngas reactor was
to be optimal at 2:1 molar ratio. The following optimization was done to achieve this.
0 500 1000 1500 2000 2500 3000 3500 4000 45001.95
22.052.1
2.152.2
2.25H2/CO ratio 1000ºF
preheat500ºF
Molar Flow rate CO2 (kmol/h)
H2/C
O m
olar
ratio
13
From the optimization we found out that the preheat temperature did not make any difference. The best ratio
was achieved at 2.01:1 molar ratio of hydrogen to carbon monoxide. The molar flow rates of components are
given in the table.
Best ratio at 3300 and 3400
T. (ºF) CO (kmol/hr) H2(kmol/hr)
500 8664 17410
750 8664 17410
1000 8664 17410
The above diagram shows the PFD of the syngas section. Methane, steam and carbon dioxide are sent in as
feed. These components are then sent to a preheater which heats the feed to 600 °F .The preheated feed is then
mixed with oxygen, compressed and sent to the syngas reactor. The product that comes out of the syngas reactor
is mostly carbon monoxide and hydrogen. The stream also contains some water and unreacted carbon dioxide
and oxygen. The stream coming out of the syngas reactor is then cooled by a train of heat exchangers and sent
to a phase separator. In the phase separator water is separated out as liquid and sent to waste treatment whereas
the gas is sent to FTR section of the plant.
14
FISHER TROPSCH REACTOR DESIGN
The FTR reactor was the most important piece of equipment in the project. Therefore a proper design of this
equipment was necessary. It was also one of the most difficult parts of the design. FTR reactor is the reactor
where conversion of syngas to different hydrocarbon chains takes place. The product selectivity of different
hydrocarbon chains was based on different equation as given below:
Conversion of syngas
xH 2+CO→H 2O+(CH2)nH 2
Product Selectivity
SC H4=r CH 4
−rCO=0.03∗T 3
SCn=0.04∗SCH 4n=2,3,4
The selectivity of C5+ was based on Anderson Shulz Flory probability distribution
W n
n=[ (1−α )2
α ]∗α n
The calculations for the FTR reactor had to be done by simultaneous solving of material, hydraulic and heat
balance around the reactor. The following equations were used in solving the reactor.
The material balance was given by the general mole balance around a packed bed reactor.
dXdW
=−r A
FAO∗ρb
Where the rate equation was given by Langmuir Hinshelwood form
−r A=k∗T 1∗ph2∗pco
(1+k2∗T 2∗pco )2
The rate equation here was heavily based on the partial pressures of the components and the temperature.
The hydraulic balance around the reactor was achieved using the Ergun equation relating catalyst weight to the
pressure drop.
dPdW
=−α∗T∗Po∗FT
2∗T o∗PPo
∗F¿
The heat balance around the reactor was achieved using:
dTdW
=
U∗aρ b
∗(T a−T )+r A∗∆ H rxn
FAO(Σθi∗C pi+∆C pX )
The overall heat transfer coefficient was given in the problem statement as:
15
U=0.385∗G0.8
D0.2
Where G is the inlet mass velocity and D is the tube diameter.
The equations above were solved simultaneously using math lab and we found out that the optimum conversion
was 30% and the pressure drop was 38.0 psi. The temperature profile down the reactor seemed to increase at the
end of the reactor and was calculated to be 650 K.
As seen in the figure above the conversion profile down the reactor increased until it reached a plateau.
Different boiler feed water temperatures were used to achieve optimum conversion. The optimum conversion
achieved was 30% at boiler feed water temperature of 450 K and catalyst weight of 45000 lbm. Above BFW
temperature of 450 K the conversion did increase but was unstable and led to a runaway reaction.
The figure above shows the temperature profile down the length of the reactor related by catalyst weight. The
conversion profile showed us that the temperature increased as the catalyst weight/length down the reactor
16
increased and around 50000 lbm of catalyst there was a runaway reaction. Therefore use of 45000 lbm of
catalyst which was below the runaway conditions was used and optimized.
The following summarizes the different values that we got for the FTR reactor.
Conversion (X) = 30%
Pressure Drop (ΔP) = 38.9 psi
GHSV = 100/hr
11 reactors in parallel in 1st train
8 reactors in parallel in 2nd train
1.4 in diameter tubes
1 in tube spacing
3857 tubes per reactor.
The figure above is the PFD of the FTR section. Syngas containing carbon monoxide, hydrogen and carbon
dioxide enters the FTR section from Syngas section. The stream is first sent to an absorber where carbon
dioxide is absorbed off using MEA. From the absorber we have almost pure carbon monoxide and hydrogen
coming out. This pure syngas is first sent to a heat exchanger where it is heated and then to a compressor. From
the compressor the syngas is then sent to the FTR reactor where conversions into different hydrocarbons take
place. The hydrocarbons out of the reactor are then sent to a heat exchanger and separation section for further
processing.
17
PRODUCT SELECTIVITY
After calculating the reactor conversion, we were able to calculate the product selectivity for the different
alkanes. Methane is dependent on temperature over the range of 390-450oF, the equation is below:
SC H4=rC H 4
−rCO=0.03T 3(
molsmol
)
T 3=exp [−10000( 1T
− 1473 )]
T is the reactor operating temperature in degree K.
After calculating the selectivity of methane, we could calculate the selectivity of the C2-C4 alkanes. They are
linked to that of methane by the following equations:
SCn=0.04 SCH 4(molmol )n=2,3,4.
The selectivity for C5+ can be calculated using the Anderson –Shultz-Flory equation:
M n=(1−α )α n−1
Mn is the relative mol fraction of carbon n.
α is the chain growth parameter, which is slightly temperature dependent and can be calculated using the
following formula:
α=0.93T 4
T 4=exp [250( 1T
− 1473 )]
Using these equations, the relative mole fraction of the hydrocarbons in the feed was calculated, and the actual
molar flow rates are gotten by multiplying the relative mole fractions by the converted molar flow rate of CO
into the system.
18
ABSORBER SIZING
Number of Equilibrium Stages
Absorption is a process in which a gas mixture is contacted with a liquid (the absorbent or solvent) to
selectively dissolves one or more components by mass transfer from the liquid. In our project two absorption
columns were used in different stages of our design. In the first stage of our design, an absorption column was
used to remove CH4 from the incoming gas stream. In the second stage of our design, an absorption column was
used to remove CO, H2, and CH4.
1 st Absorption column
From Seader’s Separation process principles,
fractionof solute absorbed= AN+1−AAN+1−1
Where,
A is the absorption factor; N is the number of equilibrium trays
The absorption factor is given as,
A= LKV
Where,
L is the volumetric flow rate of solute-free absorbent; V is volumetric flow rate of solute-free gas; K is the
vapor-liquid equilibrium ratio.
The incoming gas stream has the following composition.
Components Molar flow rate (kmol/h)
CO2 5.98
CH4 0.26
H2O 0.20
H2 53.71
O2 5.27
N2 0.18
CO 26.07
19
Total(V) 91.7
L=137.51kmolh
The K values of the different components were obtained at a pressure of 1.88 atm and a temperature of 25°C.
Components K-values
CO2 0.83
CO 0.023
H2 0.019
N2 0.015
O2 0.032
CH4 4.45
Components A-values
CO2 1.80
CO 64.54
H2 78.66
N2 100.54
O2 47.17
CH4 0.34
The fraction of each component that is absorbed at different number of stages is presented in a table below.
Components 1 2 3 4
CO2 0.64 0.83 0.92 0.96
CO 0.98 1.00 1.00 1.00
H2 0.99 1.00 1.00 1.00
N2 0.99 1.00 1.00 1.00
O2 0.98 1.00 1.00 1.00
CH4 0.25 0.31 0.33 0.33
20
The number of equilibrium stages needed in the first absorption column is 4
2 nd Absorption Column
T = 95 °F
P = 342.43 psia
Components kmol/h K A
H2 6.124 27.77 0.05402
CO 11.5 15.38 0.09753
H2O 0.1011
CH4 0.4399 6.65 0.22556
C2H5 0.0176 1.64 0.91463
C3H7 0.0176 0.584 2.56849
N-Butane 0.0176 0.195 7.69231
N-Pentane 0.7639 0.0713 21.0379
N-Hexane 0.71032 0.02 75
N-Heptane 0.66049 0.008 187.5
N-Octane 0.61416 0.0001 15000
N-Nonane 0.57107 0.00001 150000
N-Decane 0.531 0.000001 1500000
N-Undecane 0.4938 0.0000001 1.5E+07
N-Dodecane 0.45911 0.0000001 1.5E+07
N-Tridecane 0.4269 0.0001 15000
N-Tetradecane 0.397 1E-10 1.5E+10
N-Pentadecane 0.3691 1E-11 1.5E+11
N-Hexadecane 0.3432 1E-12 1.5E+12
N-Heptadecane 0.31913 1E-13 1.5E+13
N-Octadene 0.2967 1E-14 1.5E+14
N-Nonadecane 0.2759 1E-15 1.5E+15
N-Eicosane 0.2566 1E-16 1.5E+16
Total 25.7062
21
The fraction of each component that is absorbed at different number of stages is presented in a table below
Components 1 2 3 4 5 6 7 8
H2 0.05 0.05 0.05 0.05 0.05 0.05 0.05 0.05
CO 0.09 0.10 0.10 0.10 0.10 0.10 0.10 0.10
CH4 0.18 0.22 0.22 0.23 0.23 0.23 0.23 0.23
C2H5 0.48 0.64 0.72 0.76 0.79 0.82 0.83 0.85
C3H7 0.72 0.90 0.96 0.99 0.99 1.00 1.00 1.00
N-Butane 0.88 0.99 1.00 1.00 1.00 1.00 1.00 1.00
N-Pentane 0.95 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Hexane 0.99 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Heptane 0.99 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Octane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Nonane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Decane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Undecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Dodecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Tridecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Tetradecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Pentadecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Hexadecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Heptadecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Octadene 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Nonadecane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
N-Eicosane 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
The number of equilibrium stages needed in the first absorption column is 8.
Overall Efficiency
Using the correlation by O’Connell (Seader), the overall efficiency is given as;
log Eo=1.597−0.199 log( KM LµL
ρ )−0.0896( log( K M LµL
ρ ))2
22
Where,
Ki = K-value of species being absorbed
ML =molecular weight of the liquid, lb./lbmol
µL = viscosity of the liquid, cP
ρL = density of the liquid, lb/ft3
1st Absorption column
Component Overall Stage Efficiency, %
CO 50.06
H2 50.85
2nd Absorption column
Component Overall Stage Efficiency, %
CO 6.83
H2 7.49
Tower Diameter
1 st Absorption Column
The flood velocity, Uf is given as;
U f=C ( ρL−ρVρV
)0.5
Where,
C = capacity parameter of Souders and Brown.
ρL = density of the absorbing liquid = 1000 kg/m3
ρV = density of the gas stream = 10.5 kg/m3
C=FST FF FHACF
Where,
FST = surface tension factor = (σ/20)0.5 =1.87
FF = foaming factor
FHA = 1.0 for Ah/Aa >/0.10 and 5(Ah/Aa) + 0.5 for 0.06 < Ah/Aa <0.1
σ = liquid surface tension = 70 dyne/cm
For no foaming systems, FF = 1.0; for many absorbers, FF may be 0.75 or even less (Seader).
23
FLV=( LM L
V MV)( ρV
ρL )0.5
=( 137.51∗1891.67∗13.97 )∗( 10.5
1000 )0.5
=0.1981
From Fig 6.24(Seader), with a tray spacing of 24inches, CF = 0.33ft/s
Because FLV < 0.1, Ad/A = 0.1 and FHA = 1.0. Thus,
C=(1.87 ) (0.75 ) (1 ) (0.33 )=0.46 fts
U f=0.46 (1000−10.510.5 )
0.5
=4.47 fts
DT=( 4V MV
f U f π (1− Ad
A ) ρV )0.5
=(4∗91.67
3600 ∗13.97
0.80∗( 4.473.28 )∗3.14∗0.9∗10.5 )
0.5
=0.66m
2 nd Absorption Column
FLV=( LM L
V MV)( ρV
ρL)
0.5
=( 38.56∗1825.71∗15 )∗( 71
1000 )0.5
=0.4796
From Fig 6.24(Seader), with a tray spacing of 24inches, CF = 0.23ft/s
Because FLV < 0.1, Ad/A = 0.1 and FHA = 1.0. Thus
C=(1.87 ) (0.75 ) (1 ) (0.23 )=0.32 fts
U f=0.32( 1000−7171 )
0.5
=1.16 fts
DT=( 4V MV
f U f π (1− Ad
A ) ρV )0.5
=(4∗25.72
3600 ∗15
0.80∗( 1.163.28 )∗3.14∗0.9∗71 )
0.5
=0.087m
24
25
PROCESS FLOW DIAGRAM
26
The figure above is the overall process flow diagram of the whole process. The FTR products from the reactor
are first sent to a stripping section where leftover methane, hydrogen and oxygen are stripped of the
hydrocarbon stream. The hydrocarbons are then sent to a settler where due to their density differences the water
can be separated off of the hydrocarbons. The hydrocarbons are then sent to a flash drum where tail gas is
separated and sent to the preheater as fuel. The bottom stream from the flash is then sent to another flash drum
where 50% LPG is recovered as gas and sold. The bottom stream of the 2nd flash is sent to a series of distillation
column where products are separated as ethane, LPG, and Naphtha. The bottom stream of the last distillation
column is then sent to hydro- isomerization unit where hydrogen is introduced and the heavier hydrocarbons are
broken into LPG, naphtha and diesel.
27
UNIT CONTROL AND INSTRUMENTATION
For Unit control and instrumentation a relief valve was sized for the Fischer Tropsch reactor for vapor service. Shown Below is the calculations and equations used for relief valve sizing.
In sizing reliefs in vapor service, the downstream pressure must be checked to ensure that it is less than the
choked pressure (Daniel, 2008).
Qm=Co∗A∗P √ ¿
Where
Qm is the discharge mass flow
Co is the discharge coefficient
A is the area of the discharge
P is the absolute upstream pressure
ɣ is the heat capacity ratio for the gas
gc is the gravitational constant
M is the molecular weight of the gas
Rg is the ideal gas constant
T is the absolute temperature of the discharge
The area of the relief vent given a specified mass flow rate Qm is;
A=Qm
C o∗P∗√ T
M/(( ɣ∗gc
Rg)∗( 2
ɣ+1 )ɣ+1ɣ−1)
Χ=√ ¿
The required relief vent area for an ideal gas is;
A=Qm
Co∗Χ∗P∗√ T
M
Χ=519.5∗√ ¿
The pressure, P is the maximum absolute relieving pressure. It is given for the
fixed-unit case by,
P=Pmax+14.7
From ASME Boiler and Pressure Vessel Code, the following guidelines are recommended for vapor reliefs.
28Pmax = 1.1Ps for unfired pressure vessels,Pmax = 1.2 Ps for vessels exposed to fire,Pmax = 1.33 Ps for piping
Using the input streams to the FT R, the area of the relief valve can be calculated.
From Chemcad simulation
Qm = 1169.75 lbm/h
T = 392 °F =
P = 362.59 psia
Cp/Cv =ɣ = 1.3683
M = 27.54 lbm/lb-mol
Χ=519.5∗√ ¿
Χ=519.5∗(1.3683∗( 21.3683+1 )
6.43)12 =352.92
Using the fire scenario for calculation
Pmax=1.2∗P s=1.2∗362.59=435.1 psia=P
Co = 0.975
A=Qm
C o∗Χ∗P∗√ T
M= 1169.75
0.975∗352.92∗435.1∗√ 851.67
27.54=0.0434 ¿2
D=( 4∗Aπ )
12=( 4∗0.0434 ¿2
π )12=0.2351∈¿
For our unit control we focused on the control of the FTR reactor as it is the most dangerous piece of equipment
in all of the plant. We came up with the following process control.
29
In the figure above there are many ways safety controls can be triggered. First of all the flow of boiler feed water is
controlled by temperatures from different parts of the reactor specifically the hot spots. From the calculations we found
out that the hotspots existed around the middle and at the end of the reactor. Next control is the liquid level of the utility
fluid. This is to ensure that there is no decrease in the liquid level and hotspot formation. The differential pressure sensors
also work in conjunction with the temperature sensor at the exit stream of the boiler feed water. The information from
these sensors are sent to an indicator alarm/DIC which then decides to close or open the incoming feed valve if there is
any spike in temperature or pressure difference. The closing of feed valve triggers the opening of a PSV in the feed stream
so as to direct the feed to an emergency flare. The flow of gas in the emergency flare then triggers the nitrogen purge into
the system. This flow of nitrogen then triggers the valve in the exit process stream to clos and PSV activates to direct the
contaminated process stream to a bypass vessel. This is done so as to avoid any contamination downstream.
SAFETY AND ENVIRONMENTAL SUMMARY
30
HAZARD ANALYSIS
The following table shows the potential hazards related to some chemicals in the process. The table is based on SAX’s dangerous properties of industrial materials.Materia
l
Health Hazards Fire/Explosion Hazard Corrosion/Reactivity Hazard
Methane
A simple asphyxiant High concentrations can cause nausea
and unconsciousness due to lack of oxygen
Very dangerous fire and explosion hazard when exposed to heat or flame.
Explosive in form of vapor when exposed to heat or flame.
Reacts violently with powerful oxidizers (e.g.: chlorine trifluoride, chlorine, fluorine)
Incompatible with halogens and interhalogens.
Carbon
monoxid
e
Toxic by inhalation can cause systemic changes like nausea, changes in blood pressure etc.
Prolonged exposure can cause heart disorders, nerve damage, reproductive effects, brain damage etc.
Blurred vision reported when in contact with eye
Very dangerous fire and explosion hazard when exposed to heat or flame.
Violent or explosive on contact with certain materials like iron oxide, chlorine dioxide, bromine trifluoride and liquid oxygen.
Can react with oxidizing materials, halogens, metal oxides, metals and lithium
Carbon
dioxide
Inhalation in large concentrations can cause rapid circulatory insufficiency and death.
An asphyxiant Contact with solid carbon dioxide
snow can cause burns
Minimal fire hazard (only reacts with certain materials like magnesium, zirconium, titanium etc and explodes).
Incompatible with acryladehyde, metal acetylides, sodium peroxide.
Hydroge
n
A simple asphyxiant Exposure to higher concentration may
lead to oxygen deficiency and may cause dizziness, drowsiness, nausea, loss of consciousness and death.
Highly flammable and explosive when exposed to heat, flame or oxidizers.
Flammable or explosive when mixed with air, oxygen and chlorine
Can react vigorously with oxidizing metals.
Forms explosive mixtures with bromine, chlorine, chlorine dioxide, liquid nitrogen etc.
Oxygen
Breathing high concentrations (< 75) can cause hyperoxia.
Liquid form can severely burn tissue due to extreme cold.
Noncombustible itself but essential to combustion
Can react with all flammable materials.
If proper conditions met can react with secondary alcohols, aluminum alloys, alkali metals.
C2-C4
Liquid form may cause frostbite An asphyxiant. Moderate
concentrations may cause headache, drowsiness and dizziness.
Lack of oxygen due to high concentrations of fuel gas may cause death.
Inhalation of butane may produce anesthetic effects and feeling of euphoria.
Highly flammable gas.(C2-C3) Forms explosive mixtures
with air and oxidizing agents. Gas may leak, spread and
create and explosive region. Flammable when combine
with air (C4).
The C2-C3 is normally stable. Avoid oxidizing agents,
chlorine dioxide, and chlorine.
31
Materia
l
Health Hazards Fire/Explosion Hazard Corrosion/Reactivity Hazard
Naphtha
Products
May cause lung damage if swallowed. Minor skin irritant can cause redness
or inflammation Inhalation may cause irritation to
respiratory system Mildly irritating to eyes
Highly Flammable Vapors can ignite rapidly
when exposed to ignition source.
Stable under normal conditions of use.
Avoid heat, flames and sparks and strong oxidizing agents
Diesel
Products
Contact with liquid or vapor may cause mild irritation
Causes skin irritation with prolonged contact
Aspiration may result in chemical pneumonia.
Can cause gastrointestinal disturbances, nausea and vomiting.
Can also effect the brain and severe cases can cause death.
Moderate fire hazard. Vapors can ignite rapidly
when exposed to ignition source.
Runoff to sewers may cause fire or explosion hazard
Mostly stable Incompatible with strong
oxidizers Can form hazardous
decomposition products.
32
The following table shows the safety and first aid measures related to the above mentioned hazards.
Material First Aid Measures Fire/Explosion Safety Environmental Safety
Methane Remove to fresh air if inhaled. If not breathing administer
artificial respiration and oxygen.
Wear a full face positive pressure self-contained breathing apparatus.
Use water spray to stop escaping gas and stop the flow of gas.
Follow proper procedures for waste disposal following federal, state and local environmental control.
Return in shipping container properly labeled to your supplier.
Carbon
monoxide
Remove to fresh air if inhaled If not breathing administer
artificial respiration and oxygen. If eye in contact with liquid
immediately flush eyes with plenty of water.
To fight fire use foam. Dry chemical and water to blanket fire.
Control containing vessels with water jets to reduce pressure and explosion
Do not dispose to sewers or water sources Follow proper procedures for disposal following
federal, state and local environmental control.
Carbon
dioxide
Remove to fresh air if inhaled If not breathing administer
artificial respiration and oxygen Flush with plenty of water if
exposed to skin or eye
N/A (noncombustible) Follow proper procedures for waste disposal following federal, state and local environmental control.
Return in shipping container properly labeled to your supplier.
Do not dispose to sewers or water sources
Hydrogen
Remove to fresh air if inhaled Provide adequate oxygen if
exposed to high concentrations
Use dry chemical powders to control small fires.
Use water spray, fog or foam to control large fire.
Follow proper procedures for disposal following federal, state and local environmental control.
Oxygen Remove to fresh air if inhaled in high concentrations
Provide assisted respiration if unconscious.
Exclude itself from fire area of fire
Return in shipping container properly labeled to your supplier.
Follow proper procedures for disposal following federal, state and local environmental control.
C2-C4
If skin or eye exposed to liquid immediately warm the frostbite area with warm water.
In inhaled immediately remove to fresh air.
Provide adequate oxygen if exposed to high concentrations
If digested do not induce vomiting and contact a physician
Immediately spray the fire with water from maximum distance.
Remove ignition source if without risk.
Evacuate all personnel from danger area.
May use Dry chemical extinguisher or water for butane fires
Follow proper procedures for waste disposal following federal, state and local environmental control.
Return in shipping container properly labeled to your supplier.
Do not dispose to sewers or water sources
33
Material First Aid Measures Fire/Explosion Safety Environmental Safety
Naphtha
Products
Remove the affected person to fresh air.
If not breathing apply artificial respiration.
Flush the contaminated skin with large amounts of water and use soap.
Flush eye with plenty of water if contaminated.
Isolate from heat source and naked flames.
Use foam, water spray or CO2 dry chemical powder for controlling fires
Recover and recycle if possible.
Contain spill with sand, earth or absorbent material.
Follow proper procedures for waste disposal following federal, state and local environmental control.
Diesel
products
Immediately flush with clean water if exposed to eyes.
In inhaled immediately remove to fresh air.
Provide adequate oxygen if exposed to high concentrations
If digested do not induce vomiting and contact a physician immediately.
Remove contaminated clothing and wash the contaminated area with soap and water
Use any extinguisher suitable for small class B fires, dry chemical, water spray, firefighting foam etc.
Use water spray, fog or firefighting foam to fight large fires
Follow proper procedures for waste disposal following federal, state and local environmental control.
Return in shipping container properly labeled to your
supplier. Do not dispose to sewers or water sources
Hazard Analysis for Unsafe Operation (HAZOP)
34
Equipment ProcessParameter
Guide Words
Deviation Causes Consequences Actions
Reactor
Level
More Less
Control valve for one of the incoming streams malfunctions
Clogged relief valveExplosions due to over pressurization, Process downtime
A pressure gauge should be installed between the spring relief valve and rupture disk
MoreLess
Pump FailurePump malfunctionNo Power
No control over feed flow to the reactor.Possible runaway reaction
Install flow indicator in the pipes leaving and entering the pump
Temperature MoreLess
Temperature of the reactor fluctuates
Too much or too less flow in utility stream.
Explosion of reactorRunaway reactionDisintegration of products
Install cascade of temperature and flow controllers in the reactor
Pressure MoreLess
Pressure inside the reactor deviates
Downstream compressor malfunctions, side reactions occurringTubes in reactor corrodes
Reduced conversion,Explosion hazardContamination of product
Install back up compressor, Install flow control valves and pressure controllers.
Distillation Column
Level More Less
Level indicator malfunctions or insufficient logging of feed data
Wear and tear in control valve.Lack of communication between operators
Adding too much feed to the process and causing over pressurization in the tower or adding too little and damaging the reboiler.
All indicators should be maintained and updated regularly and proper protocol should be followed when logging feed data
Temperature MoreLess
Temperature of the tower deviates from the set point
Failure of reboiler utility flow or failure of the condenser
Temperature in column reaches unsafe levels.Separation not achieved
Install control valves on both utility streams of condenser and reboiler
Pressure MoreLess
Overpressure in distillation column
Too much flow rate into the tower
Flooding the tower,Separation not achieved
Install control valves for the flow in and out of the unit op.
Heat Exchangers Flow
More Less
Supply too little or too much cooling/heating fluid
Wear and tear of control valves
Excess water is sent to the S/L separator
Install backup controllers
Equipment ProcessParameter
Guide Word
Deviation Causes Consequences Actions
35
s
Pumps Level MoreLess
Liquid level rises rapidly or decreases rapidly
Interlocks malfunctioning
Reduces the amount of feed produced.Too much flow or too less flow to unit op.
Making sure all interlocks are appropriately timed
Reflux Drums Level More
Less Tank LevelToo high or too flow rate to or from the reflux drums
Overflowing of tank and or over pressure.Cooling not achieved.
Relief valve should be installed, and an appropriate means of diverting excess fluid should be implemented
Absorbers
Level More Less Output flow rate
Wear and tear of control valve, or inappropriate dilution of corrosive compounds
Rapid corrosion of MOC, loss in HCl revenue
Care should be taken in calculating the required amount of water for dilution, and the appropriate MOC should be used
Temperature MoreLess
High liquid/gas flow temperatureLow liquid/ gas flow temperature
Temperature of the tower too high or too low
Low absorption.More/less utility flow rate.
Install temperature control indicator, set low and high alarms
Settler
Flow MoreLess
High or low flow rate into the system
Flow rate not controlledFlow control valves malfunction
Settling velocity is not met.Phase separation not achieved and product is contaminated
Install appropriate flow control valves upstream and downstream
Level MoreLess
High or low flow rate into the system
Flow rate not controlledFlow control valves malfunctionLevel controller malfunctions
Separation not achieved.Product may be contaminated by heavies or lights
Install appropriate level controllers in the settler.
36
ENVIRONMENTAL IMPACT
An environmental impact analysis was done on the waste water streams and the compounds in our process. By
utilizing the environmental impact tool on Chemcad we were able to evaluate each compound and its impact on
the environment. The tables showing the impacts are below:
Raw Component FactorsComponent Name Ozone Depletion Global
WarmingSmog Formation Acid Rain
Carbon Dioxide 0 1 0 0Methane 0 11 0.007 0
Water 0 0 0 0Hydrogen 0 0 0 0Oxygen 0 0 0 0Nitrogen 0 0 0 0
Carbon Monoxide 0 0 0 0Ethane 0 0 0.082 0
Propane 0 0 0.42 0n-Butane 0 0 0.41 0n-Pentane 0 0 0.408 0n-Hexane 0 0 0.421 0n-Heptane 0 0 0.529 0n-Octane 0 0 0.493 0n-Nonane 0 0 0.469 0n-Decane 0 0 0.464 0
n-Undecane 0 0 0.436 0n-Dodecane 0 0 0.412 0n-Tridecane 0 0 0 0
n-Tetradecane 0 0 0 0n-Pentadecane 0 0 0 0n-Hexadecane 0 0 0 0n-Heptadecane 0 0 0 0n-Octadecane 0 0 0 0n-Nonadecane 0 0 0 0
n-Eicosane 0 0 0 0
Raw Component FactorsComponent Name Human Toxicity Eco Toxicity
OSHA PEL LD 50 LC 50 LD 50
37
Carbon Dioxide 9000 0 24 0Methane 0 0 38.7995 0Water 0 0 0 0Hydrogen 0 0 0 0Oxygen 0 0 0 0Nitrogen 0 0 0 0Carbon Monoxide 55 0 15 0Ethane 0 0 95.7331 0Propane 1800 0 74.2984 0n-Butane 1900 1226.95 11.6925 1226.95n-Pentane 2950 1523.13 6.45919 1523.13n-Hexane 1800 28710 2.5 28710n-Heptane 2000 2115.28 1.96554 2115.28n-Octane 2350 2411.46 1.96735 2411.46n-Nonane 1050 2707.64 0.477075 2707.64n-Decane 0 3003.61 0.22227 3003.61n-Undecane 0 3300.01 0.167683 3300.01n-Dodecane 0 3595.76 0.330536 3595.76n-Tridecane 0 3892.16 0.593462 3892.16n-Tetradecane 0 4145.69 0.334588 4145.69n-Pentadecane 0 4484.31 0.194356 4484.31n-Hexadecane 0 4780.49 0 4780.49n-Heptadecane 0 5076.67 0 5076.67n-Octadecane 0 5372.64 0 5372.64n-Nonadecane 0 5668.82 0 5668.82n-Eicosane 0 5965 0 5965
Below, the environmental impacts of one of our waste water streams are shown:
38
Water Stream AnalysisComponent Name Flow
rate (kg/hr)
Ozone Depletion (impact/hr)
Global Warming (impact/hr)
Smog Formation (impact/hr)
Acid Rain (impact/hr)
Carbon Dioxide 0.01 0 4.76E-06 0 0Methane 0 0 5.41E-07 2.23E-06 0Water 187.54 0 0 0 0Carbon Monoxide 0.02 0 0 0 0
Water Stream AnalysisComponent Name Flow rate
(kg/hr)OSHA PEL (impact/hr)
LD50 (impact/hr)
LC50 (impact/hr)
LD50 (impact/hr)
Carbon Dioxide 0.01 4.85E-07 0 0.00135 0Methane 0 0 0.00 8.66E-06 0Water 187.54 0 0 0 0Carbon Monoxide 0.02 8.66E-05 0 2.36E-03 0
SAFETY CONSIDERATIONSIt is not enough to be able to design a plant that would be able to produce the desired products correctly; the
safety of the processes should also be put into consideration. In ensuring a safe process conditions, control
valves would be placed on all equipment to control pressure, temperature, and flow rate when needed.
Additional process modifications will also be carried out on important equipment. These modifications are
shown below;
Reactor
• Install a high-temperature alarm to alert the operator in the event of cooling function loss
• Install a high-temperature shut-down system .This system would automatically shut down the process in the event of a high reactor temperature
• Install a check valve in the cooling line to prevent reverse flow
• Study the cooling water source to consider possible contamination and interruption of supply
• Install a cooling water flow meter and low-flow alarm ( which will provide an immediate indication of cooling loss)
• Isolate Reactor with explosion walls
Drums
• Provide dip legs in all drums to prevent the free fall of solvent resulting in the generation and accumulation of static charge
All Equipment 39
• Ground all equipment
• Check the room periodically with colorimetric tubes to determine if any leaked vapors are present
• Increasing the ventilation with blowers
• Isolating process lines that contain hazardous streams
• Regular leak detection would be carried out
It is important to quantitatively carry out a risk analysis. The Chemical Process Quantitative Risk Analysis
(CQPRA) is a methodology designed to provide management with a tool to help evaluate process safety in the
chemical process industry. The steps1 involved are shown below;
In the event of leak from a vessel, the discharge rate can be quantified for liquid and gaseous flow.
From a general discharge rate model
Where
P= pressure (kPa)
ρ = density ( kg/m3)
g = acceleration due to gravity (m/s2)
gc = gravitational constant (N/kg-m/s2)
z = vertical height (m)40
v = fluid velocity (m/s)
f = frictional loss term
Ws = shaft work (J/s)
m = mass flow rate (kg/s)
ef = frictional loss term
Kf = excess head loss due to the pipe (dimensionless)
f = fanning friction factor
L = flow path length
D = flow path diameter
Using the 2k-Method to Estimate Kf
K f=k1
Nℜ+k∞∗(1+ 25.4
IDmin )Where, IDmin = internal diameter in mm
For pipe entrances and exits
K f=k1
Nℜ+k∞
For a normal entrance
k1 = 160
k∞ = 0.50
Border type entrance
k1 = 0
k∞ = 1.0
For high Reynolds number
NRe > 10000
Kf = k∞
For low Reynolds number
41
NRe < 50 ; Kf = k1/NRe
The fanning, f factor is calculated from the equation below;
1√ f
=−4 log( εD
3.7065−5.0452logA
N ℜ)
A={( εD )1.1098
2.8257+( 7.149
N ℜ)
0.8981}Where, ε = pipe roughness
For large Reynolds number
1√ f
=−4 log ( εD
3.7065 )For liquid discharge from a pipe assuming the flow is frictionless and that there is no shaft work would result in
a Bernoulli equation.
(P¿¿2−P1)1ρ+ ggc
∗( z2−z1 )+ 12∗gc
∗(v22−v1
2 )=0¿
The discharge of pure liquids through a sharp edged orifice
m=A∗CD∗√2 ρ∗gc∗(P1−P2 )Where,
m˙ = liquid discharge rate (kg/s)
A = area of hole (m2)
CD = discharge coefficient
ρ = density of fluid (kg/m3)
gc = gravitational constant (N/kg-m/s2)
P1 = pressure upstream of the hole (N/m2)
P2 = pressure of downstream of the hole (N/m2)
Tank Drainage Time
t= 1A∗C D∗√2g
∗∫V 2
V 1 dV (h )√h
Assumptions used in the equation above are;
Constant area
Constant discharge coefficient
Where,
42
t = time to drain tank from volume V2 to Volume V1 (s)
V = liquid volume in the tank above the tank (m3)
h = height of the liquid above the leakage (m)
Mass discharge rate of liquid from a hole in a tank
m=ρ∗v∗A=ρ∗A∗CD∗√2∗( gc∗Pg
ρ+g¿hL)
Where,
Pg = gauge pressure at the top of the tank (N/m2)
hL = height of liquid above the hole (m)
It should be noted that the mass discharge rate decreases with time as the liquid level drops. Therefore, the
maximum discharge rate happens when the leak first occurs.
Mass discharge rate of gas from a hole in a tank
Assumptions
Ideal gas
No heat transfer
No external shaft work
m=CD∗A∗P1∗√ 2∗gc
Rg∗M
T 1∗k
k−1∗{(P2
P1)
2k−( P2
P1)k−1k }
Where,
k = heat capacity ratio
Rg = ideal gas constant
T1 = initial upstream temperature of the gas
The mass discharge rate at choked flow is given as;
m=CD∗A∗P1∗√ k∗gc∗MRg∗T 1
∗( 2k+1 )(
k+1k−1 )
P choked
P1=( 2
k+1 )k
k +1
For ideal gas flow for both sonic and non-sonic conditions is represented by the Darcy formula and is given as;
m=Y∗A∗√ 2gc∗ρ1∗(P1−P2 )∑K f
Y = gas expansion factor (unitless)
43
Tempered Reaction
Q=m∗C v∗dT
dt+
(m∗V∗hfg )m∗v fg
Where,
Q = heat generation by reaction
m = mass within reactor vessel (kg)
Cv = heat capacity at constant volume (J/kg-°C)
T= absolute temperature of the reacting material (°C)
t = time (s)
hfg = enthalpy difference below liquid and vapor (J/kg-°C)
V = reactor vessel (m3)
vfg = specific volume difference between liquid and vapor
Heat release rate during overpressure
q=Cp
2∗{( dTdt ) , s+ dT
dt,m}
Where,
s = set conditions
m = turnaround conditions
For the quantitative analysis of the safety we considered a situation where there is a leak in oxygen power plant
and calculated the effects of the gas leak
Safety Calculation
Assuming that there is a leakage of gas in the Air separation unit, the discharge rate of the can be calculated to
determine if the leakage can be temporarily fixed or a shutdown of the system is needed.
The oxygen flow conditions are provided below;
P = 500 psig =34.77 bar
T = 75°F
Heat capacity = 1.37
P choked
P1=( 2
k+1 )k
k +1
P choked
P1=( 2
1.37+1 )1.37
1.37+ 1=0.9065
Pchoked=0.9065∗34.77=31.52 ¿44
Assume the diameter of the hole that is causing the leakage is 10mm
A=π∗D2
4=π∗(10mm )2
4=7.85∗10−5m2
( 2k+1 )
k +1k−1 =( 2
1.37+1 )1.37+11.37−1=0.3371
Assume CD = 0.85
mchoked=CD∗A∗P1∗√ k∗gc∗MRg∗T1
∗( 2k+1 )
k +1k−1
mchoked=.85∗7.85∗10−5m2∗34.77∗105 Pa∗√ 1.37∗ 31.96 kgkg−mole
8314 Pa∗m3
kg∗mole∗K∗296.89 K
∗0.3371
mchoked=0.5673kg
s
Initial flow rate of gas = 651040.8 kg/h
% lost=
1.5758∗10−4 kgh
651040.8 kgh
∗100 %=2.42∗10−8%
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10
10
20
30
40
50
60
Mass flowrate Vs Diameter
Diameter(m)
disc
harg
e m
ass fl
owra
te(k
g/s)
As we can see from the figure the quantity of mass discharge can be quantifies. According to our calculations
we found out that at around 0.018 m diameter 1% of the mass is lost .This mass lost can be significantly high
taking in account the total amount of mass flow rate through the system.
45
HEAT AND MATERIAL BALANCE Mass balance was done according to kg/hr basis taking in consideration the molecular weight of each
component in the entire system. The following table summarizes the overall mass balance of the system.
46
Feed Kg/hr Syngas RKg/hr HE
Kg/hr Absorber 1
Kg/hr CO2 out Kg/hr
CO2 44.01 CO244.0
1 CO244.0
1 CO2 0.02 CO243.99
5
Water237.1
7 Water208.
7 Water21.1
6 Water 10.6 Water10.58
2
Oxygen393.5
9 Oxygen1.81
4 Oxygen1.81
4 Oxygen 0 Oxygen1.813
6
Nitrogen 3.446 Nitrogen3.44
6 Nitrogen3.44
6 Nitrogen 0 Nitrogen3.445
7
47
CO 0 CO730.
1 CO 730 CO 730 sum59.83
6Hydrogen
0 Hydrogen108.
3 Hydrogen108.
3 Hydrogen 108
Methane422.4
1 Methane4.22
4 Methane4.22
4 Methane 4.22
sum1100.
6 sum 1101 sum 913 sum 853
Out HEKg/hr
Water187.
5
FTR Kg/hr Absorber 2Kg/hr Top Out
kg/hr Settler
kg/hr Sett Out kg/hr
CO2 0 CO2 0 CO2 0 CO2 0 CO2 0
Water10.58
2 Water10.5
8 Water 0 Water 10.6 Water 10.58Oxygen 0 Oxygen 0 Oxygen 0 Oxygen 0 Oxygen 0
Nitrogen 0 Nitrogen 0 Nitrogen430.
8 Nitrogen 0 Nitrogen 0
CO430.7
6 CO430.
8 CO63.6
3 CO 0 CO 0
Hydrogen 63.63 Hydrogen63.6
3 Hydrogen37.4
6 Hydrogen 0 Hydrogen 0
C1 37.46 C137.4
6 C1 0 C1 0 C1 0
C2 0.388 C20.38
8 C20.01
6 C2 0.37 C2 0
C3 0.569 C30.56
9 C30.02
3 C3 0.55 C3 0
C4 0.75 C4 0.75 C40.03
1 C4 0.72 C4 0
C5 40.44 C540.4
4 C5 1.65 C5 38.8 C5 0
C6 44.91 C644.9
1 C6 1.84 C6 43.1 C6 0
C7 48.56 C748.5
6 C7 1.99 C7 46.6 C7 0
C8 51.47 C851.4
7 C8 2.11 C8 49.4 C8 0
C9 53.74 C953.7
4 C9 2.2 C9 51.5 C9 0
C10 55.43 C1055.4
3 C10 2.27 C10 53.2 C10 0
C11 56.63 C1156.6
3 C11 2.32 C11 54.3 C11 0
C12 57.38 C1257.3
8 C12 2.35 C12 55 C12 0
C13 57.75 C1357.7
5 C13 2.36 C13 55.4 C13 0
C14 57.78 C1457.7
8 C14 2.36 C14 55.4 C14 0
C15 57.53 C1557.5
3 C15 2.35 C15 55.2 C15 0
C16 57.02 C1657.0
2 C16 2.33 C16 54.7 C16 0
C17 56.31 C1756.3
1 C17 2.3 C17 54 C17 0
C18 55.41 C1855.4
1 C18 2.27 C18 53.1 C18 0
C19 54.36 C1954.3
6 C19 2.22 C19 52.1 C19 0
C20 53.19 C2053.1
9 C20 2.22 C20 51 C20 0
sum1402.
1sum
1402 sum567.
1 sum 835 sum 10.58
48
AsTo
49
s
Colmn 1 kg/hr Dist 1 kg/hr Dist 2 kg/hr Dist 3kg/hr
CO2 0 C20.35
6 CO2 0 CO2 0
Water 0 C3 0 Water 0 Water 0
Oxygen 0 C4 0 Oxygen 0 Oxygen 0
Nitrogen 0 C5 0 Nitrogen 0 Nitrogen 0
CO 0 C6 0 CO 0 CO 0Hydrogen 0 sum
0.356
Hydrogen 0 Hydrogen 0
C1 0 C1 0 C1 0
C2 0.37 Colm 2 kg/hr C20.01
7 C2 0
C3 0.55 CO2 0 C30.54
6 C3 0
C4 0.72 Water 0 C40.41
3 C4 0.31
C5 38.78 Oxygen 0 C50.00
1 C5 38.8
C6 43.08 Nitrogen 0 sum0.97
7 C6 43.1
C7 46.57 CO 0 C7 46.6
C8 49.37Hydrogen 0 Colm 3
kg/hrC8 49.4
C9 51.54 C1 0 CO2 0 C9 51.5
C10 53.17 C20.01
7 Water 0 C10 52.8
C11 54.31 C30.54
6 Oxygen 0 C11 2.33
C12 55.03 C40.71
9 Nitrogen 0 C12 0.01
C13 55.39 C538.7
8 CO 0 sum 285
C14 55.42 C643.0
7Hydrogen 0
C15 55.17 C746.5
7 C1 0 Botms 3kg/hr
C16 54.69 C849.3
7 C2 0 C10 0.36
C17 54 C951.5
4 C3 0 C11 52
C18 53.14 C1053.1
7 C40.30
7 C12 55
C19 52.14 C1154.3
1 C538.7
8 C13 55.4
C20 51.01 C1255.0
3 C643.0
7 C14 55.4
sum824.4
5 C1355.3
9 C746.5
7 C15 55.2
C1455.4
2 C849.3
6 C16 54.7
C1555.1
7 C951.5
4 C17 54
C1654.6
9 C1053.1
7 C18 53.1
C17 54 C1154.3
1 C19 52.1
C1853.1
4 C1255.0
3 C20 51
C1952.1
4 C1355.3
8 sum 538
C2051.0
1 C1455.4
1
sum824.
1 C1555.1
7
C1654.6
9C17 54
C1853.1
4 Kg/hr
C1952.1
3 Tot out 1203.9
C2051.0
2 Tot in1100.6
2sum 625
To HI
ENERGY BALANCE
The energy balance around the reactor was achieved by considering the enthalpy of all the incoming streams
and outgoing streams. The following table summarized the energy balance.
Overall Energy Balance MJ/h (Input) MJ/h (Output)
Feed streams -6.12E+06
Product -1.38E+07
Total Heating 2.05E+06
Total Cooling -1.00E+07
Power Added 5.54E+05
Power Generated -4.87E+04
Total -1.43E+07 -1.38E+07
From the table we calculated the % closure.
% closure=−1.38−1.43
∗100 %=96.51 %
CONCLUSION50
OPTIMIZATIONS AND RESULTS
The optimizations were done using the parametric optimization process. In the parametric optimization process
important decision variables were varied in order to obtain the most positive net present value in dollars. The
following variables were varied in order to achieve optimum conditions.
Feed preheat optimization
The feed preheat temperature entering the syngas reactor was optimized in this optimization. This was done
until an optimum value was achieved.
350 450 550 650 750 850 950 1050
-$3.85
-$3.80
-$3.75
-$3.70
-$3.65
-$3.60
-$3.55
NPV vs. Feed Preheat Temp
Preheat Temp. (ºF)
NPV
(Billi
ons o
f Doll
ars)
600 °F (-3.65 Billion)
From the optimization we found out that the optimum feeds preheat temperature was 600ºF at net present value
of -3.65 billion dollars.
Initial breakdown of NPV
51
The following pie chart shows the initial breakdown of COMd.
Cwt0%
Cut21%
Crm68%
Col0%FCI11%
Initial Breakdown of COMd
COM 4.29E+09
CWT 4.18E+06
CUT 9.10E+08
CRM 2.90E+09
COL 2.80E+06
FCI 4.70E+08
Revenue 4.72E+09
NPV -4.77E+09
From the initial breakdown of COMd we found out that the most expensive contributing factor was cost of raw
materials. As we had no control over the price of the raw material, we decided to optimize the consumption of
it. We also decide to optimize the FTR unit.
Syngas Unit Optimization
To ensure the optimum usage of raw material the syngas unit was optimized using different pressure
temperature and oxygen rate. This was done until optimum conditions were achieved.
52
1500 1550 1600 1650 1700 1750 1800 1850 1900 1950 2000
-5.00-4.00-3.00-2.00-1.000.001.002.003.004.00
NPV vs. SU (T/P/O2 rate)
Pressure 300 psig 2:1 molar ratio400 psig500 psig
Syngas Temp (ºF)
NP
V (
Bill
ions
of
$)
P (psig) T (°F) CO (kmol/h) H2(kmol/h) O2( kmol/h)
300 1600 15314 31806 27000
500 1950 20058 41076 20000
From the optimization we found out that the optimum condition was achieved at the pressure of 500 psig,
optimum temperature of 1950 F and oxygen flow rate of 20000 kmol/hr. The net present value obtained was
optimum at 2.63 billion dollars.
FTR unit optimization
The FTR unit reactor entering the FTR reactor was optimized in this optimization. This was done until optimum
conditions were achieved.
200 220 240 260 280 300 320 340 360
-3-2-1012345
FT Reactor Temperature
25 atm30 atm20 atm
Reactor Temp (⁰C)
NP
V (
$ B
illio
ns)
53
From the FTR unit optimization we found out that the optimum entering temperature into the FTR reactor was
251.7 ⁰C. This bumped out net present value to 4.2 billion dollars.
Final breakdown of NPV
The following pie chart shows the final breakdown of COMd.
Cwt0%
Cut35%
Crm49%
Col0%
FCI16%
Break down of COMd after SU/FT unit op-timization
COM 5.95E+09
CWT 7.64E+06
CUT 2.06E+09
CRM 2.36E+09
COL 2.80E+06
FCI 9.79E+08
Revenue 7.52E+09
NPV 4.95E+09
From syngas and FTR unit optimization as we can see the cost of raw material decreased but cost of utilities
increase. Nevertheless the NPV value also went up.
54
Effect if reactor trains in series
12
0
1
2
3
4
Effect of reactor trains in series
# of Reactors
NPV (
Billio
ns of
$) X= 30 %
X= 60 %
The above graph shows the effect of reactor trains. We increased the reactor train from just one to two in series
this made out conversion increase up to 60% and the optimum NPV obtained was 3.9 Billion dollars
Taking credit for utilities
Credit for utilitiesNo credit forutilities
3.84
4.24.44.64.8
5
Taking credit for utilities
NPV(
Billio
ns of
$)
We also added the credit for different utility streams and tail gas streams according to their heating values. This
drastically increased our NPV from 4.2 billion dollars to 4.96 billion dollars
55
Number of Trays/Feed Tray location optimization
The numbers of trays/feed tray locations were varied in the three distillation columns in order to achieve the
most positive net present value. This was done until an optimum was achieved.
4 5 6 7 8 9 10 11 12 13$3,863.60
$3,864.00
$3,864.40
$3,864.80
$3,865.20
$3,865.60
$3,866.00
$3,866.40
T-101 Optimization
151617182014
Feed Tray Location
NPV
($X1
06)
For T-101 a net present value of $3866.33X106 was obtained at 15 trays and feed tray location of 7.
4 5 6 7 8 9 10 11 12 13 14$3,891.00
$3,891.10
$3,891.20
$3,891.30
$3,891.40
$3,891.50
$3,891.60
T-102 Optimization
161718
Feed Tray Location
NPV
($X1
06)
For T-102 a net present value of $3891.566X106 was obtained at 16 trays and feed tray location of 10.
56
4 6 8 10 12 14 16$3,891.56
$3,891.57
$3,891.58
T-103 Optimization
1920212223
Feed Tray Location
NPV
($X1
06)
For T-101 a net present value of $3891.573X106 was obtained at 22 trays and feed tray location of 10.
Tie Spec optimization
After concluding on the working conditions for the separation process, we changed the recovery rate of the
various organics from each column. To achieve this tie spec for the reboiler of the distillation columns were
optimized by also varying the tie spec for the condenser.
0.88 0.9 0.92 0.94 0.96 0.98 1 1.023889.6
3890
3890.4
3890.8
3891.2
3891.6
3892
T-101:Tie Spec Optimization
Tie spec Reboiler 0.90.990.9990.9999
Tie spec for reboiler 1( MCB-TCB)
NPV
($X
106)
A maximum NPV value of $3891.554X106 was obtained at reboiler tie spec 0.99 mole fraction of and
condenser tie spec of 0.9999 mole fraction.
57
0.88 0.9 0.92 0.94 0.96 0.98 1 1.023890
3890.4
3890.8
3891.2
3891.6
3892
T-102:Tie Spec Optimization
Tie spec Reboiler 0.999
Tie spec for reboiler 1( MCB-TCB)
NPV
($)
A maximum NPV value of $3891.554X106 was obtained at reboiler tie spec 0.999 mole fraction of and
condenser tie spec of 0.99 mole fraction.
0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 23889.4
3889.6
3889.8
3890
3890.2
3890.4
3890.6
3890.8
T-103:Tie Spec Optimization
Tie spec Reboiler 0.9
0.99
0.999
Tie spec for reboiler 1( MCB-TCB)
NPV
($)
A maximum NPV value of $3890.722X106 was obtained at reboiler tie spec 0.9 mole fraction of and condenser
tie spec of 0.3 reflux ratio.
58
ENERGY EFFICIENCY
We calculated the mass of carbon in the feed stream and exiting LPG, Naphtha, and Diesel streams by doing a
species balance on the carbon and estimating the closure with the following formula:
xkmol alkanehr
∗y kmolC
1kmolalkane∗12kgC
1kmolC
Hydrocarbon Carbon MassFeed CH4 2.99E+05LPGC3 6.90E+03C4 9.17E+03NaphthaC5 6.05E+03C6 6.97E+03C7 7.28E+03C8 2.43E+03C9 1.27E+03C10 6.85E+02DieselC11 2.06E+04C12 2.20E+04C13 2.29E+04C14 2.38E+04C15 2.45E+04C16 2.51E+04C17 2.56E+04C18 2.61E+04C19 2.65E+04C20 2.68E+04
Mass of carbon kgin 2.99E+05out 2.85E+05energy efficiency 95.23
The energy efficiency of our system is approximately 95%; we have assumed that the rest of the carbon is used up as fuel in the power generation unit or the unaccounted recycle stream..
59
RESULTS
We calculated the capital and operating cost of a GTL plant that utilizes a low temperature FT reactor. After performing all the optimizations we found out that these are the best conditions for our GTL plant. The following table summarizes the best case conditions:
Economics Summary
Best case $ Cost of Raw materials ($/yr.) 2,360,000,000
NPV 4,960,000,000 Methane 339,000,000
FCI 5,200,000,000 Oxygen 543,000,000
Top 5 equipment (name/cost) Carbon Dioxide 9,300,000
1 FTR 33,207,000 Steam 7,900,000
2 E-109 2,800,000 Catalyst 40,260,000
3 E-105 1,560,000 MEA 879,000,000
4 T-102 900,000 Waste treatment ($/yr.) 6,220,000
5 T-103 832,000 Revenue Total ($/yr.) 7,520,000,000
COM ($/yr.) 5,903,400,000 LPG 526,000,000
Top 5 utilities (name/dollars) Naphtha 1,410,000,000
1 FTR 290,000,000 Diesel 5,589,000,000
2 Oxygen
Plant
217,000,000 Credit For Utilities ($/yr) 18,287,000
3 E-101 183,000,000 1 Fuel gas 1,370,000
4 E-109 61,000,000 2 lps 4,800,000
5 E-105 60,300,000 3 mps 11,800,000
4 Steam Condensate 407,000
Operating Labor ($/yr) 1,240,000 Thermodynamics MSRK
60
After performing all the optimizations and calculation our final NPV value obtained was $ 4.96 Billion. The pie chart below shows different utility breakdown; which is followed by equipment specifics of final design.
E-10116% E-102
3%
E-1037%
E-1040%
E-1055% E-106
1%E-107
2%E-108
0%E-109
5%
R-10126%
R-10215%
H-1011%
C-1010%
P-101 A/B0%
P-102 A/B0%
P-103 A/B0%
C-1020%
C-SU10%
O2 plant19%
Utility Breakdown
The most contributing utility in our final design were the two reactors followed by oxygen plant and syngas cooling heat exchanger train.
NPV $4.96E+9
Feed Ratio (Hydrogen/Carbon
monoxide) 2:1
Reactor 1
Volume 533 m3
Pressure 25 atm
Temperature 500 of
Reactor 2
Volume 533 m3
Tower 101
Number of Trays 15
Feed tray location 7
Tower 102
Number of Trays 16
Feed tray location 10
Tower 103
Number of Trays 22
Feed tray location 10
Tie spec for T-101
Distillate Comp mole fraction 0.9999
Bottoms Comp mole fraction 0.99
61
Tie spec for T-102
Distillate Comp mole fraction 0.99
Bottoms Comp mole fraction 0.999
Tie spec for T-103
Distillate Reflux Ratio 0.3
Bottoms Comp mole fraction 0.9
RECOMMENDATIONS
We used a distillation column to separate the tail gas (ethane) from the rest of the FT products. This method
proved cumbersome, because we either had to operate the column at pressures above 300 psia, or use a
refrigerant at -50ºC. Both alternatives were expensive; we would recommend that a different means of
removing the ethane be researched. Another alternative to our process will be to operate the FT reactor at
temperatures that favor the product that is in demand. For instance, the FT reactor produces more diesel at
temperatures between 200 and 250ºC and more of the lighter alkanes at higher temperatures. Also we have
evidence that suggests the separation of LPG from Naphtha, and Diesel can be done in a fractional distillation
column. The product selling price in the problem statement is not current. The increase of price of the product
to current values will positively impact the NPV. However as the fuel prices increase there will inevitably
increase in utility and raw materials but from the following graph we see that 25% increase in selling price will
increase the NPV by 79%.This will definitely impact NPV positively even with the increase in utility and raw
materials.
0.30.4
0.00 5.00
10.00 15.00 20.00 25.00
25% increase in Product Selling Price
Selling Price ($/lb)
NP
V (
Bill
ions
of
$)
Naphtha $100/bblDiesel $118/bbl 79% Increase in NPV (18.82 Billion)
62
Also investigating production of oxygen with the plant is another option to consider. As oxygen was the most
expensive utility production of oxygen will drastically reduce raw materials cost. Also there are different
methods of syngas production which are more efficient. From our research we found out one of them is auto
thermal reforming .This can also impact NPV. Another option is investigating catalyst life. From a quantitative
analysis we found out that doubling the catalyst life increase the NPV by 6%.The graph below shows the
relation.
4 8
4.50
5.00
5.50 Effect of Doubling Catalyst Life
Catalyst Life
NP
V(B
illio
ns o
f $)
6% increase in NPV (317 million)
63
64
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