DME production UITM

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Contents

ABSTRACT ... 2 INTRODUCTION ... 3 PROCESS DESCRIPTION ... 8 EQUIPMENT DESCRIPTION ... 11 MATERIAL BALANCE ... 12 ENERGY BALANCES... 18 PINCH TECHNOLOGY ... 30 HEURISTICS ... 39 ECONOMIC ANALYSIS ... 45

CONCLUSIONS AND RECOMMENDATIONS ... 60

REFERENCES ... 62

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ABSTRACT

The purpose of this project is to design the process for production of Dimethyl Ether from methanol. This is down by using approaches direct method. The requires capacity is

72071.05 metric tonne/year and the desired purity of the product was 99 wt%.Throughout this project we as a student are be able to learn and apply knowledge how to design, how to structure and determine economic analysis, cost ,profit for the plant. The other

considerations in this project are maximizing production and minimizing the raw material consumption.

DME is used primarily as a propellant. DME is miscible with most organic solvents and it has a high solubility in water [1]. Recently, the use of DME as a fuel additive for diesel engines has been investigated due to its high volatility (desired for cold starting) and high cetane number.

AIZAQ SYAZWAN B ABDULLAH ZAWAWI Cost, Material and energy balances, Heuristics, Pinch FAIZ SHAFIQ B ZAILI Cost, Material and energy balances, Heuristics, Pinch NURUL SYAZMIN BT CHE JOHARI Cost, Material and energy balances, Heuristics, Pinch SITI NOR ROUDAH BT HAIRUL ANUAR Cost, Material and energy balances, Heuristics, Pinch

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INTRODUCTION

DME (dimethyl ether) is a clean, colorless gas that is easy to liquefy and transport. It has remarkable potential for increased use as an automotive fuel, for electric power generation, and in domestic applications such as heating and cooking.

DME can be derived from many sources, including renewable materials (biomass, waste and agricultural products) and fossil fuels (natural gas and coal).

DME has been used for decades in the personal care industry (as a benign aerosol

propellant), and is now increasingly being exploited for use as a clean burning alternative to LPG (liquefied petroleum gas), diesel and gasoline.

Like LPG, DME is gaseous at normal temperature and pressure, but changes to a liquid when subjected to modest pressure or cooling. This easy liquefaction makes DME easy to transport and store. This and other properties, including a high oxygen content, lack of sulfur or other noxious compounds, and ultra clean combustion make DME a versatile and

promising solution in the mixture of clean renewable and low-carbon fuels under consideration worldwide.

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How is DME Produced?

DME can be produced from a variety of abundant sources, including natural gas, coal, waste from pulp and paper mills, forest products, agricultural by-products, municipal waste and dedicated fuel crops such as switch grass.

World production today is primarily by means of methanol dehydration, but DME can also be manufactured directly from synthesis gas produced by the gasification of coal or biomass, or through natural gas reforming. Among the various processes for chemical conversion of natural gas, direct synthesis of DME is the most efficient.

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Identification

Dimethyl ether identification in the commercial industry is listed as below in Table 1

The physical and chemical properties of chlorobenzene can be concluded in the Table 1.1.

Properties Value

Molecular Weight 46.07 g

Solubility in water 20oC (moderate)

Vapor Pressure At 20oC, 520 kPa abs

Normal Boiling Point, At 1 atm, -24.84oC

Normal melting point At 1 atm, -141.49 oC

Liquid Density At 25oC and 1 atm, 655 kg/m3

Vapor Density At 21.1oC and 1 atm, 1.908 kg/m3

Chemical Name Dimethyl ether

Molecular Structure

Synonyms Methyl ether, methyl oxide, wood ether,

oxybismethane

IUPAC Name Dimethyl ether

Classification Ether

UN Identification Number UN1033

Hazardous Waste ID No. D001

Formula C2H6O

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USES

Due to its good ignition quality, with a high cetane number, DME can be used in diesel engines as a substitute for conventional diesel fuel. However, compared to diesel fuel DME has a lower viscosity (insufficient), and poor lubricity. Like LPG for gasoline engines, DME is stored in the liquid state under relatively low pressure of 0.5 MPa. This helps to limit the number of modifications required to the engine. Still, some slight engine modifications are necessary, primarily relating to the injection pump and the installation of a pressure tank, similar to that for LPG. The fuel line must also be adapted with specific elastomers. DME in diesel engine burns very cleanly with no soot.The infrastructure of LPG can be used for DME. As part of the FP7 project BioDME, under the leadership of the Volvo Group, DME production is being optimized, especially for use as a transport fuel.

HEALTH

Acute health effects

The following acute (short- term) health effects may occur immediately or shortly after exposure to dimethyl ether:

 Vapor can cause eye, nose and throat irritation.

 High exposure can cause headache, dizziness, lightheadedness, and even loss of consciousness.

 Skin contact with liquid dimethyl ether can cause severe frostbite Chronic health effects

The following chronic (long-term) health effects can occur at some time after exposure to dimethyl ether and can last for months or years:

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Cancer hazard According to the information presently available to the New Jersey Department of Health and Senior Services, dimethyl ether has not been tested for its ability to cause cancer in animals.

Reproductive hazard

According to the information presently available to the New Jersey Department of Health and Senior Services, dimethyl ether has not been tested for its ability to affect reproduction. Other long- term

effects

Dimethyl ether has not been tested for other chronic (long-term) helath effects.

HANDLING AND STORAGE

 DME is not compatible with ozone, oxidizing agents (such as perchlorates, peroxides, permanganate, chlorates, nitrates, chlorine, bromine and fluorine), strong acids (such as hydrochloric, sulfuric and nitric) and halogens.

 Store in tightly closed containers in a cool, well- ventilated area, and prevent air from entering container.

 Sources of ignition, such as smoking and open flames, are prohibited where DME is used, handled or stored.

 Metal containers involving the transfer of DME should be grounded and bonded.

 Use only non-sparking tools and equipment, especially when opening and closing containers of DME.

 Peroxide formation may occur in containers that have been opened and remain in storage for more than six months. Peroxide can be denoted by friction, impact or heating.

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PROCESS DESCRIPTION

A PFD of the process shown in the Figure B.1.1 and the belongings stream conditions are given in Appendix A. The essentials operations in the process are the preheating of the raw material (nearly pure Methanol), dehydration of Methanol from DME, product separation and Methanol separation and recycle.

The liquid Methanol pumped up from 1 bar to 15.5 bar. The stream 3 preheated with stream 13. The stream 4 flow through reactor cooler,E-202 before entering reactor, R-201. There is slightly difference in the pressure after leaving the reactor, which is the pressure at stream 5 is 14.7 bar meanwhile pressure at stream 6 which is 13.9 bar. Stream 6 with the temperature of 364K enters the DME Cooler and at the stream 7, the temperature becomes 278K. Stream 9 flows entering DME Tower, T-201 which separating Methanol and other components. DME

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Tower splits into two streams which are stream 10 and stream 11. Stream 10 rich with DME which is the composition of DME is 2.142 kmol/h meanwhile stream 11 rich with water, which the composition of water in that stream is 203.337 kmol/h. The separation does not stop there, stream 12 enters the second tower which is Methanol Tower, T-202 and splits into two streams which is stream 13 and stream 14. Stream 13 rich in Methanol which containing 97.308 kmol/h of Methanol and stream 13 recycle back to stream 2. Stream 14 rich in water which is the composition of the water in that stream is 201.348 kmol/h. Stream 14 enters wastewater cooler, E-208 which makes the temperature is 167K TO 50K at stream 15.

Stream Number 1 2 3 4 5 6 7

Temperature °C 25 25 45 154 250 364 278

Pressure [bar] 1.0 15.5 15.2 15.1 14.7 13.9 13.8

Vapor fraction 0.0 0.0 0.0 1.0 1.0 1.0 1.0

Mass flow [kg/h] 12806.1 12806.1 16049.7 16049.7 16049.7 16049.7 16049.7 Mole Flow [kmole/h] 401.166 401.166 502.299 502.299 502.299 502.299 502.299 COMPOSITION (kmol/h) DME 0.0 0.0 2.295 2.295 2.295 199.665 199.665 Methanol 397.341 397.341 494.19 494.19 494.19 99.297 99.297 H2O 3.825 3.825 5.814 5.814 5.814 203.337 203.337 (Mole Fraction) DME 0 0 0.0046 0.0046 0.0046 0.3975 0.3975 Methanol 0.9904 0.9904 0.9838 0.9838 0.9838 0.1977 0.1977 H2O 0.0096 0.0096 0.0115 0.0115 0.0115 0.4048 0.4048 Stream Number 8 9 10 11 12 13 14 Temperature °C 100 89 46 153 139 121 167 Pressure [bar] 13.4 10.4 11.4 10.5 7.4 15.5 7.6 Vapor 0.0798 0.148 0.0 0.0 0.04 0.0 0.0 Mass flow [kg/h] 16049.7 16049.7 9134.1 6915.6 6915.6 3258.9 3656.7 Mole Flow [kmole/h] 502.299 502.299 198.441 303.858 303.858 101.439 202.419

10 COMPOSITION (kmol/h) DME 199.665 199.665 2.142 2.142 2.142 2.142 0.0 Methanol 99.297 99.297 0.918 98.379 98.379 97.308 1.071 H2O 203.337 203.337 0.0 203.337 203.337 1.989 201.348 (Mole Fraction) DME 0.3975 0.3975 0.9954 0.0071 0.0070 0.0212 0 Methanol 0.1977 0.1977 0.0046 0.3238 0.3238 0.9592 0.0053 H2O 0.4048 0.4048 0 0.6692 0.6692 0.0196 0.9947 Stream Number 15 16 17 Temperature °C 50 46 121 Pressure [bar] 1.2 11.4 7.3 Vapor fraction 0 0 0 Mass flow [kg/h] 3656.7 3320.1 5538.6 Mole Flow [kmole/h] 202.419 72.063 172.89 COMPOSITION (kmol/h) DME 0.0 71.757 3.672 Methanol 1.071 0.306 165.852 H2O 201.348 0.0 23.366 (Mole Fraction) DME 0 0.9958 0.0212 Methanol 0.0053 0.0042 0.9593 H2O 0.9947 0 0.0195 APPENDIX A

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EQUIPMENT DESCRIPTION

All of the equipments used in the plant are chosen to be constructed of stainless steel. This is due to corrosive water in the streams and high pressure.

REACTOR

For cost estimations, the Packed Bed Reactor (PBR) is assumed to be a process vessel. The capacity needed for the cost estimations is the volume of the reactor. This was found the same way as for the pre-reformer and ATR in the direct method. The same cost estimation method as in the direct method is used here.

STORAGE TANKS

For the cost estimations all of the storage tanks are assumed to be API, fixed roof tanks made of stainless steel. The cost calculations for the tanks needed to store Methanol and Water are done in the same way as for the storage tanks in the direct method.

COMPRESSOR

The compressor used in the plant is assumed to be a centrifugal compressor made of stainless steel. It is assumed that the energy required to compress a gas is the energy found in the workbook. In reality, there is an efficiency factor involved. This factor has not been counted in when estimating the electric power needed for the plant.

HEAT EXCHANGERS

All of the heat exchangers in the plant are assumed to be shell and tube, floating head and stainless steel. The reason for this choice is the same as for the direct plant. The estimation of the cost for the heat exchanger is done the same way as for the exchangers in the direct plant.

PUMP

The pump used in the plant is assumed to be centrifugal, electric drive and made of stainless steel. Pump selection is based on flow rate and head required. In addition special care

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should be made when considering corrosion (in this process some water is present, and the pressure is high).

TOWER

There are two towers in this plant. The towers in the plant are distillation columns with sieve trays. The ideal numbers of trays were found by scaling up the number of trays. The first distillation column known as DME Tower meanwhile the second distillation column known as Methanol Tower.

MATERIAL BALANCE

Basis being used: 330 days/year of operation.

In which, it is required to produce 72071.05 metric tonne/year .

72071.05 𝑡𝑜𝑛𝑛𝑒 𝑦𝑒𝑎𝑟 x 1 000 kg 1𝑡𝑜𝑛𝑛𝑒 x 1 𝑑𝑎𝑦 24 ℎ𝑟 x 1𝑦𝑒𝑎𝑟 𝑑𝑎𝑦 330 𝑑𝑎𝑦 =9099.88 kg/hr 6328630𝑔 ℎ𝑟 x 1𝑚𝑜𝑙 46.07g = 137.37 𝑘𝑚𝑜𝑙 hr

From Table B.1.1 Stream table for unit 200, the total production of DME is 129.1kmol/h.

129.1𝑘𝑚𝑜𝑙 ℎ𝑟 x 46.07g 1𝑚𝑜𝑙= 5947.64 𝑘𝑔 hr Ratio:- 9099.88 kg/hr_{5947.64 kg/hr} =1.53 Mixer F2=12806.1 kg/h F3=16049.7 kg/h Mixer xCH3OH = 0.9838 x H2O = 0.0117 xDme = 0.0046 xCH3OH=0.9904 x H2O=0.0075 xDme=0 F13=3258.9 kg/hr xCH3OH=0.9592 x H2O=0.01960 xDme=0.0212

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Overall Balanced F2+F13=F3

1280.61 kg/h +3258.9 kg/h=16049.7 kg/h

Stream Methanol (CH3OH) Water (H2O) Dme

2 397.3419𝑘𝑚𝑜𝑙 401.166 𝑘𝑚𝑜𝑙 =0.9904 3.825 𝑘𝑚𝑜𝑙 401.166 𝑘𝑚𝑜𝑙 =0.0075 0 13 1.989𝑘𝑚𝑜𝑙 101.439 𝑘𝑚𝑜𝑙 =0.9592 97.308 𝑘𝑚𝑜𝑙 101.439 ℎ𝑟 =0.01960 2.142𝑘𝑚𝑜𝑙 101.439 𝑘𝑚𝑜𝑙 =0.0212 3 494.19𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.9838 5.814𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.0117 2.295𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 = 0.0046 Reactor R-201

The reaction that occurred around the reactor is as follow:

2CH3OH(CH3)2O+H2O conversion 80% percent of reactant(methanol)

F5=16049.7 kg/h F6=16049.7 kg/h R-101 XCH3OH=0.1976 X H2O=0.4048 X (CH3)2O =0.3975 XCH3OH=0.9800 X H2O=0.01155 XCH3OCH3=0.0045

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Overall Balanced F5 =F6

16049.7 kg/h =16049.7 kg/h

Stream Methanol (CH3OH) Water (H2O) Dme

5 494.19𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.9800 5.814 𝑘𝑚𝑜𝑙 502.299 𝑘𝑚𝑜𝑙 =0.01155 2.295𝑘𝑚𝑜𝑙 502.299 𝑘𝑚𝑜𝑙 =0.0045 6 99.297𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.1976 203.337 𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.4048 199.665𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.3975

First Separator in Distillation Column (T-201)

 The balance around the first separator 1: T-100 F9=16049.7 kgmol/h XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976 F10 =9134.1 kgmol/h XCH3OH=0.0046 X H2O=0 XCH3OCH3=0.9954 F11=6915.6 kgmole/h XCH3OH=0. 3238 X H2O=0.6692 X CH3OCH3=0.00710

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Overall Balanced F9=F11+F10

16049.7 kg/h =9134.1 kg/h+6915.6 kg/h

Stream Methanol (CH3OH) Water (H2O) Dme

9 99.297 𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.1976 203.337 𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.4048 199.665 𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 =0.3976 10 0.918𝑘𝑚𝑜𝑙 198.441𝑘𝑚𝑜𝑙 =0.0046 0 2.142𝑘𝑚𝑜𝑙 198.441𝑘𝑚𝑜𝑙 =0.9954 11 98.379𝑘𝑚𝑜𝑙 303.858𝑚𝑜𝑙 =0.3238 203.337𝑘𝑚𝑜𝑙 303.858𝑘𝑚𝑜𝑙 =0.6692 2.142𝑘𝑚𝑜𝑙 303.858 𝑘𝑚𝑜𝑙 = 0.00710

Second Separator in Distillation Column (T-202)

 The balance around the second separator 2: T-202 F12=6915.6 kgmole/h XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976 F13=3258.9 kgmole/h XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976 F14=3656.7 kgmole/h XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976

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Overall Balanced F9=F11+F10

6915.6 kg/h =3258.9 kg/h+3656.7 kg/h

Stream Methanol (CH3OH) Water (H2O) Dme

12 64.3𝑘𝑚𝑜𝑙 303.858 𝑚𝑜𝑙 =0.3238 203.𝑘𝑚𝑜𝑙 303.858𝑚𝑜𝑙 =0.6693 2.142 𝑘𝑚𝑜𝑙 303.858𝑚𝑜𝑙 =0.0070 13 97.308𝑘𝑚𝑜𝑙 101.439𝑘𝑚𝑜𝑙 =0.9593 1.989𝑘𝑚𝑜𝑙 101.439𝑘𝑚𝑜𝑙 =0.0196 2.142𝑚𝑜𝑙 101.439𝑘𝑚𝑜𝑙 =0.0211 14 1.071 𝑘𝑚𝑜𝑙 202.419𝑚𝑜𝑙 =0.005 201.348 𝑘𝑚𝑜𝑙 202.419𝑘𝑚𝑜𝑙 =0.995 0 Stream Number 1 2 3 4 5 6 7 Temperature °C 25 25 45 154 250 364 278 Pressure [bar] 1.0 15.5 15.2 15.1 14.7 13.9 13.8 Vapor fraction 0.0 0.0 0.0 1.0 1.0 1.0 1.0 Mass flow [kg/h] 12806.1 12806.1 16049.7 16049.7 16049.7 16049.7 16049.7 Mole Flow [kmole/h] 401.166 401.166 502.299 502.299 502.299 502.299 502.299 COMPOSITION (kmol/h) DME 0.0 0.0 2.295 2.295 2.295 199.665 199.665 Methanol 397.341 397.341 494.19 494.19 494.19 99.297 99.297 H2O 3.825 3.825 5.814 5.814 5.814 203.337 203.337 (Mole Fraction) DME 0 0 0.0046 0.0046 0.0046 0.3975 0.3975 Methanol 0.9904 0.9904 0.9838 0.9838 0.9838 0.1977 0.1977 H2O 0.0096 0.0096 0.0115 0.0115 0.0115 0.4048 0.4048 Stream Number 8 9 10 11 12 13 14 Temperature °C 100 89 46 153 139 121 167

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Pressure [bar] 13.4 10.4 11.4 10.5 7.4 15.5 7.6

Vapor 0.0798 0.148 0.0 0.0 0.04 0.0 0.0

Mass flow [kg/h] 16049.7 16049.7 9134.1 6915.6 6915.6 3258.9 3656.7 Mole Flow [kmole/h] 502.299 502.299 198.441 303.858 303.858 101.439 202.419 COMPOSITION (kmol/h) DME 199.665 199.665 2.142 2.142 2.142 2.142 0.0 Methanol 99.297 99.297 0.918 98.379 98.379 97.308 1.071 H2O 203.337 203.337 0.0 203.337 203.337 1.989 201.348 (Mole Fraction) DME 0.3975 0.3975 0.9954 0.0071 0.0070 0.0212 0 Methanol 0.1977 0.1977 0.0046 0.3238 0.3238 0.9592 0.0053 H2O 0.4048 0.4048 0 0.6692 0.6692 0.0196 0.9947 Stream Number 15 16 17 Temperature °C 50 46 121 Pressure [bar] 1.2 11.4 7.3 Vapor fraction 0 0 0 Mass flow [kg/h] 3656.7 3320.1 5538.6

Mole Flow [kmole/h] 202.419 72.063 172.89

COMPOSITION (kmol/h) DME 0.0 71.757 3.672 Methanol 1.071 0.306 165.852 H2O 201.348 0.0 23.366 (Mole Fraction) DME 0 0.9958 0.0212 Methanol 0.0053 0.0042 0.9593 H2O 0.9947 0 0.0195

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ENERGY BALANCES

Method of calculation

Energy balance are necessary in order to determine energy that needed in a process such as for heating and cooling, as well as power that needed in process design. In manual calculation done in design project, calculation was done by using equation from previous lesson such as in Elementary Principle of Chemical Processes. First law of thermodynamics also applied which states that energy cannot be created or destroyed. In forming manual energy balance calculations, some assumptions are made as below:

i. Pure reactant are used

ii. Values calculated up to 3 decimal place

iii. Energy out = Energy in + Generation – Consumption – Accumulation There are some other assumptions regarding to the equipment itself which are:

i. The potential and kinetic energy of streams are neglected, there are only enthalpy changes are considered

ii. For standard enthalpy, the standard reference used are; ΔH = 0, P0 = 1atm, T0 = 298

K

iii. Equipment is assumed working in ideal condition iv. Equipment is assumed perfectly insulated

Equations Used in Energy Balance

Equation used in calculation shown in table below: Table : List of formula used

Denotation Formula

Heat Capacity, Cp (kJ/mol.˚C) a + bT + cT2 + dT3

Heat Load (kJ/hr) Q = m.λ

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Table : Heat capacities of compounds

Compounds Cp = a + bT + cT2 + dT3 + eT4 (J/mol.K) Methanol 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) Dme 110.100- 0.15747T+ 5.1853x10-4T2 Water 32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3 (KNOVEL Database)

Table : Heat of Formation of Compounds

Compund Heat of Formation, ∆Hf at Tref = 298.15 K

Dimethyl Ether (Gas) -184.1 kJ/kmol

Water (Liquid) -285.84 kJ/kmol

(Gas) -241.83 kJ/kmol

Methanol (Liquid) - 238.655kJ/kmol

(Gas) -200.94 kJ/kmol

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Energy Balance For Reactors

 Energy Balance for Reactor 201: Overall chemical equation

2CH3OH(CH3)2O+H2O

Methanol Dimethyl ether+ Water ∆𝐻_𝑟° = ∑ 𝑣 _𝑖 ∆𝐻o

fio = ∑𝑝𝑟𝑜𝑑𝑢𝑐𝑡|𝑣𝑖| ∆Ȟ fio

∆𝐻_𝑟° = (−18.41) + (−241.83) − (−200.94) 𝑘𝐽/𝑘𝑚𝑜𝑙 = -59.3 kJ/kmol (exothermic)

Amount of DME formed during reaction = 130.5 kmol/hr Thus, heat of reaction = 130.5 kmol/hr × -59.3 kJ/kmol

= -7738.65 kJ/hr F5=16049.7 kg/h F6=16049.7 kg/h T=250oC T=364oC XCH3OH=3171.42 kg/h X H2O=6496.92 kg/h X (CH3)2O =6379.76 kg/h XCH3OH=15728.7 kg/h X H2O=185.37 kg/h XCH3OCH3=72.22 kg/h Reactor 201

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Reference H2O,DME,Methanol at 298oC and 1 atm

Component Mole fraction Cpin (kJ/kg.K) Mole fraction Cpout(kJ/kg.K)

Methanol 0.0046 Cp1 0.3975 Cp4

DME 0.9838 Cp2 0.1977 Cp5

Water 0.0115 Cp3 0.4048 Cp6

Calculation for inlet stream:- Cp1= ∫ C 𝑃 523 298 dt = [21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3)]523₂₉₈ = 73.63-47.7501 kJ/kg K = 25.8805 kJ/kg K Cp2= ∫ C 𝑃 523 298 dt = [110.100- 0.15747T+ 5.1853x10-4T2]523₂₉₈ = 169.5762—109.221 kJ/kg K =60.35kJ/kg K Cp3= ∫₂₉₈523C 𝑃dt = [32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3]523 298 = 44.6738—38.8152 kJ/kg K =5.8586 kJ/kg K

Calculation for Outlet stream:-

Cp4= ∫₂₉₈523C 𝑃dt

22 = 73.63 -47.7501 kJ/kg K = 25.8835 kJ/kg K Cp5= ∫₂₉₈637C 𝑃dt = [110.100- 0.15747T+ 5.1853x10-4T2]637₂₉₈ = 220.195—109.221 kJ/kg K =110.974 kJ/kg K Cp6= ∫₂₉₈637C 𝑃dt = [32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3]637 298 = 47.847—38.8152 kJ/kg K = 9.0318 kJ/kg K Cp inlet Cp of mixture = ∑ xiCpi =(25.8805 × 0.0046 ) + (0.9838 × 60.35) + (0.0115 × 5.8586 ) = 59.56 kJ/kg.˚K

Heat from inlet stream = mCp∆T = 16049.7 x 59.56 x 523 = 499935771.8 kJ/hr. Cp outlet Cp of mixture = ∑ xiCpi =(0.3975 × 25.8835 ) + (0.4048 × 9.0318) + (0.1977 × 110.974 ) = 35.88 kJ/kg.˚K

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Heat from OUTLET stream = mCp∆T = 16049.7 x 35.88 x 637

=366869085.3 kJ/hr.

Heat input + Heat of reaction – Heat Output = Rate of Accumulation 499935771.8 + (-7738.65 ) – 366869085.3= Rate of Accumulation

Rate of Accumulation = 133058947.9kJ/hr = 36960.82 kW = 36.96 MW

Therefore, heat supply needed to sustain the reactor temperature at 364 oC is approximately 36.964 MW.

Energy Balances for T-201

T-201 T10=250 oC F10= 9134.1kg/h XDME=0.9954 Xmethanol=0.0046 Xwater=0 T11=153 o C F11= 6915.6 kg/h XDME=0.0071 Xmethanol=0.3238 Xwater=0.6692 T9= 89 oC F9= 16049.7kg/h XDME=0.3975 Xmethanol=0.1977 Xwater=0.4048

24 Tref=89oC , Cp(DME) :- ∫ 523 362 110.100- 0.15747T+ 5.1853x10 -4 T2 = 48.53 kJ/kg.K CP Water :- ∫ 523 362 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 = 4.25 kJ/kg.K Cp Methanol :- ∫ 523 362 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =19.13 kJ/kg.K

Qin = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (6496.91 x 161 x 4.25 ) + (48.53x 161 x 6379.76) + (19.13 x 3173.03 x161 ) = 64.065 x 10 6kJ/kg.K

Q out = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (0 x 161 x 4.25 ) + (48.53x 161 x 9092.08 ) + (19.13 x 42.02 x161 ) = 83.98 x 10 6kJ/kg.K

Inlet Outlet (TOP)

F9=16049.7 kg/h F10=9134.1 kg/h

mDme=6379.76 kg/h mDme=9092.08 kg/h

mmethanol=3173.03 kg/h mmethanol=42.02 kg/h

25 Tref=89oC , Cp(DME) :- ∫ 426 362 110.100- 0.15747T+ 5.1853x10 -4 T2 = 16.07 kJ/kg.K CP Water :- ∫ 426 362 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 = 1.653 kJ/kg.K Cp Methanol :- ∫ 426 362 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =7.212 kJ/kg.K

Q in = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (6496.91 x 64 x 1.653 ) + (16.07 x 64 x 6379.76) + (7.212 x 3173.03 x 64 ) = 8.71 x 10 6kJ/kg.K

Qout = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (4627.92 x 64 x 1.653 ) + (16.07 x 64 x 49.10) + (7.212 x 2239.27 x 64 ) = 1.57 x 10 6kJ/kg.K Q in = Q out 64.065 x 10 6kJ/kg.K + 8.71 x 10 6kJ/kg.K = 1.57 x 10 6kJ/kg.K + 83.98 x 10 6kJ/kg.K 72.775 x 10 6 Kj/kg.K - 85.55 x 10 6 kJ/kg.K=0 ∆Q= -12.775 x 10

Inlet Outlet (BOTTOM)

F9=16049.7 kg/h F11= 6915.6 kg/h

mDme=6379.76 kg/h mDme= 49.10 kg/h

mmethanol=3173.03 kg/h mmethanol= 2239.27 kg/h

26

Heat Exchanger E201

Tref=45oC , Cp(DME) :- ∫ 427 318 110.100- 0.15747T+ 5.1853x10 -4 T2 = 24.943 kJ/kg.K CP Water :- ∫ 427 318 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =2.7895 kJ/kg.K Cp Methanol :- ∫ 427 318 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =6.126 kJ/kg.K

Q = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (2.7885 x 109 x 184.572 ) + (24.943 x 109 x 73.829 ) + (15789.69 x 6.126 x 109 ) = 10.80 x 106 kJ/kg.K T3= 45 oC F3=16049.7 kg/h XDme=0.0046 Xmethanol=0.9838 Xwater=0.0115 T4=154 oC F4=16049.7 kg/h XDme=0.0046 Xmethanol=0.9838 Xwater=0.0115

27 Heat Exchanger E202

Tref=154oC , Cp(DME) :- ∫ 523 427 110.100- 0.15747T+ 5.1853x10 -4 T2 = 32.177 kJ/kg.K CP Water :- ∫ 523 427 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =2.5709 kJ/kg.K Cp Methanol :- ∫ 523 427 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =11.8012 kJ/kg.K

Q = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (1845.72 x 96 x 2.5709 ) + (32.177 x 96 x 738.29) + (15791.30 x 11.8012 x 96 ) = 20.63 x 10 6kJ/kg.K T3= 154 oC F3=16049.7 kg/h mDme=738.29 kg/h mmethanol=15791.30 kg/h mwater=1845.72 kg/h T4=250 oC F4=16049.7 kg/h mDme=738.29 kg/h mmethanol=15791.30 kg/h mwater=1845.72 kg/h

28 Cooler E-203 Tref=100oC , Cp(DME) :- ∫ 551 373 110.100- 0.15747T+ 5.1853x10 -4 T2 = 57.25 kJ/kg.K CP Water :- ∫ 551 337 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =4.745 kJ/kg.K Cp Methanol :- ∫ 551 373 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =21.61 kJ/kg.K

Q = [m water x Cpwater x ΔT] + [mdme ×CpDME × ΔT]+ [mmethanol ×Cpmethanol × ΔT]

= (3173.03 x 178 x 4.745 ) + (57.25 x178 x 6379.76) + (6496.92 x 21.61 x 178 ) = -92.68 x 10 6kJ/kg.K T7= 278 oC F7=16049.7 kg/h mDme=6379.76 kg/h mmethanol=6496.92 kg/h mwater=3173.03 kg/h T8=100 oC F8=16049.7 kg/h mDme=6379.76 kg/h mmethanol=6496.92 kg/h mwater=3173.03 kg/h

29 Cooler E-208 Tref=323K , CP Water :- ∫ 440 323 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =3.0 kJ/kg.K Cp Methanol :- ∫ 440 323 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =13.0 kJ/kg.K

Q = [m water x Cpwater x ΔT] + [mmethanol ×Cpmethanol × ΔT]

= (3637.32kg x 117 x 3.0 ) + (19.38 x 117 x 13.0 ) = -1.31 x 10 6kJ/kg.K T14=167 oC F14=3656.7 kg/h mDme=0 kg/h mmethanol=19.38 kg/h mwater=3637.32kg/h T15=50 oC F8=3656.7kg/h mDme=0 kg/h mmethanol=19.38 kg/h mwater=3637.32kg/h

30

Table of energy by equipment

Equipment Energy ,Q (kJ/kg.K)

Reactor (R-101) 36.964 x106

Tower (T-201) -12.775 x106

Tower (T-202) 10.800 x106

Heat exchanger (E-202) 20.630 x106

Heat exchanger (E-203) -92.68 x106

Heat exchanger (E-208) -1.310 x106

PINCH TECHNOLOGY

CALCULATE CP FOR PINCH CALCULATION.

Stream Condition 𝒎̇Cp (Kw/ oC) Tin (oC) Tout (oC)

1 Hot 9.6152 278 100

2 Hot 2.1908 167 50

3 Cold 6.644 45 154

Step 1: The minimum approach temperature is chosen to be 10℃

Compounds Cp = a + bT + cT2 + dT3 + eT4 (J/mol.K)

Methanol 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3)

Dme 110.100- 0.15747T+ 5.1853x10-4T2

Water

31

Calculation Cp for each stream :-

Stream 3:- T3:-45oC

Number Component Composition

1 DME 0.0046 2 Methanol 0.9838 3 Water 0.0115 Cp(DME) :- 110.100- 0.15747T+ 5.1853x10-4T2 = -2.43 kJ/kg.oC CP Water :- 32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3 =0.402 kJ/kg.oC Cp Methanol :- 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) =1.62 kJ/kg.oC Cp of mixture = ∑ xiCpi = ( 0.0046 x -2.43) + (0.402 x 0.0115 ) + (1.62 x 0.9838 ) = 1.59 kJ/kg.oC

32

Calculation Cp for each stream :-

Stream 4:- T4:-154oC

Number Component Composition

1 DME 0.0046 2 Methanol 0.9838 3 Water 0.0115 Cp(DME) :- 110.100- 0.15747T+ 5.1853x10-4T2 = 98.15- 106.49 = -8.34 kJ/kg.oC CP Water :- 32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3 =35.44- 32.73=2.71 kJ/kg.oC Cp Methanol :- 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) =34.04-23.14= 10.90 kJ/kg.oC Cp of mixture = ∑ xiCpi = ( 0.0046 x -8.34) + (2.71 x 0.0115 ) + (10.90 x 0.9838 ) = 10.72 kJ/kg

Calculation Cp for each stream :-

Stream 6:- T6:-364oC

Number Component Composition

1 DME 0.3975

2 Methanol 0.1977

33 Cp(DME) :- 110.100- 0.15747T+ 5.1853x10-4T2 = 15.0 kJ/kg.oC CP Water :- 32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3 =7.74 kJ/kg.oC Cp Methanol :- 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) =9 kJ/kg.oC Cp of mixture = ∑ xiCpi = (0.3975x 15.0 ) + (7.74 x 0.4048) + (0.1977x 9) = 10.87 kJ/kg

Calculation Cp for each stream :-

Stream 7:- T7:-278oC

Number Component Composition

1 DME 0.3975 2 Methanol 0.1977 3 Water 0.4048 Cp(DME) :- ∫ 278 25 110.100- 0.15747T+ 5.1853x10 -4 T2 = -0.053 kJ/kg.oC CP Water :- ∫ 278 25 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =5.60 kJ/kg.oC Cp Methanol :- ∫ 278 25 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3)

34

=21.05 kJ/kg.oC Cp of mixture = ∑ xiCpi

= (0.3975x -0.053) + (0.1977x 21.05) + (0.4048x 5.60 ) = 6.41 kJ/kg.oC

Calculation Cp for each stream :-

Stream 14:- T12:-167 oC

Number Component Composition

1 DME 0 2 Methanol 0.0053 3 Water 0.9947 Cp(DME) :- ∫ 167 25 110.100- 0.15747T+ 5.1853x10 -4 T2 = -8.23 kJ/kg.oC CP Water :- ∫ 167 25 32.24+0.01924T+(1.055x10 -5 )T2+(-3.596x10-9)T3 =3 kJ/kg.oC Cp Methanol :- ∫ 167 25 21.15+0.07902(T) + 2.587x10 -5 (T2)+2.852x10-8(T3) =12.06 kJ/kg.oC Cp of mixture = ∑ xiCpi = (0.0 x -8.23) + (0.9947 x 3) + (0.0053 x 12.06 ) = 3.05 kJ/kg.oC

35 stream 1 2 3 ∑ 𝑚̇ 𝐶_𝑝∆𝑇 9.6152 2.1908 6.644 kW/°C (kW) 278 268 A 1067.2872 167 157 B 276.3216 164 154 C -4.446 100 90 D -114.952 55 45 E -200.394 50 40

36 A 1067.2872 kW 1067.2872kW B 35.418 kW 1102.7052 kW C 330.368 kW 1433.0732 kW D -200.394 kW 1232.6792 kW E 10.95 kW 1243.6292 kW

Step 3: The Cascade Diagram

There is no pinch happened in this system since this system have only one heater and two coolers. Pinch do not happened because of the mixture which Methanol which had the highest number of mole in the stream entered the heater after that the Methanol entered the first cooler. After came out from the first cooler, the reactant which is Water that contain the highest mole number of component in the mixture entered the second cooler. Because of this system, the temperature of the mixture became lower and since the mixture did not enter any

COL D U T IL IT Y 1243.6292kW

37

other heater than can increase the heat of the mixture, thus it considered does not has a pinch.

This type of problem that happened known as special category and this category known as threshold problems which do not have a pinch to divide the problem into two parts. Threshold problems only need a single thermal utility (either hot or cold but not both) over a range of minimum temperature difference ranging from zero to threshold temperature.

Figure (a) shows a threshold problem for which hot utility is zero. It only demands cold utility up to Tthreshold.

Figure (b) shows the effect of energy demand in terms of cold and hot utilities if the cold composite curve is shifted horizontally to positions “A” and “C”. At position “B” which is at

38

𝑄𝑐. When the cold composite is shifted to position “A” where ∆Tmin < Tthreshold it

demands 𝑄𝑐1 cold utility at a higher level and 𝑄𝑐2 cold utility at a lower level. Where, the sum

of 𝑄_𝑐1 and 𝑄_𝑐2 being equal to 𝑄_𝑐. For the position “C” where ∆Tmin > Tthreshold the process demands both cold and hot utilities. Thus in this case also for ∆Tmin ≤ Tthreshold the cold

39

HEURISTICS

1. Vessel (V-201) Based on table 11.7

Rule 1:- Acceptable, temperature design above maximum operating temperature T =25 oC in between 25 -30 oC to 345o C .

Rule 2:- Acceptable, where design operating pressure is 10 % or 0.69-1.7 bar (10-25psi)over the maximum operating pressure whichever is greater. The max. operating in turn is taken as 1.0 bar in range between 0.69 – 1.7 bar.

Rule 3:- Acceptable , Design pressures of vessels operating at 0-0.69 bar ( 0 -10 psig) where operating 1.1 bar in range of design operating pressure.

Rule 5:- Minimum wall thickness for rigidity use in this manufacturing Dimethyl Ether in range of less than 8.1mm for 1.07-1.52 m.

Rule6:-Design of vessel must have 3.8mm (0.15 in) for noncorrosive streams in order to prevent occurring corrosion inside vessel .

Rule 7 :-Use Carbon steel in order to allowable working stresses.(Low cost ,readily available, resist abrasion, standard fabrication, resists alkali.)

2. Pump (P-201 A/B)

Stream 1 : 1.0 bar T=25 oC

Stream2 : 16.0 bar T=25 oC

40

From Table 11.9 :- use the following heuristics: 1.Rule 1:-Acceptable:- Power for pumping liquid :Kw ΔP = 15.5 bar – 1.0 bar = 15 bar

Total p=m1p1+ m2p2 ρmethanol= 791.80 kg/m3 efficiency of pump, ɛ= 0.60 % Flow rate = 120806.1 kg/h_{791.80 𝑘𝑔/𝑚3} = 152.57 𝑚 3 ℎ x 1 ℎ 60 𝑚𝑖𝑛 = 2.54 𝑚3 𝑚𝑖𝑛 Kw=(1.67)( 2.54 _𝑚𝑖𝑛𝑚3 ) x ( 15 𝑏𝑎𝑟_0.60 )

Fluid Pumping Power = 106.045 kW x0.60

= 63.627 kW ( The total pump delivery required at P is more than 64 kW) Pump actual =106.045  107 kW if need to buy for to pumping the liquid flow.

Rule 4-7 :- pump choices are carbon steel ,electric drive, centrifugal. Choose reciprocating to be consistent with data given. Typical, ɛ= 0.60

Power shaft = 63.627 /0.60= 106.045 Kw required. From table 11.3 page 325-P 107 < 150 kW thus we use type of pump which is .Rotary and positive Displacement Centrifugal . ( compared with table b.13 major equipment summary for unit 200) is not suitable because the requirement is more than Power= 5.2 kw had given in table because after scale up).

3. Heat Exhanger ( E-203) Cooler

From Table 11.11 ,use the following heuristic:-

41

Rule 1 :- set F= 0.9 Fouling factor

Rule 6: Δ T = 10 o.C minimum temperature approach is 10o

C for fluids Rule 7:- by ΔT LMTD Calculation :-

Cold side Temperature in :- 30.0 oC Cold side Temperature out:- 55.0 oC

Thus based on followed the rule accepted where temperature inlet cooling water is 30.0 oC , temperature outlet is 55 oC which is above 45 oC(maximum).

Rule 8:- U=850 W/m2 oC For the moment,we find A by

Δ Tlm= [(278-55)-(100-30)]/[ln(278-55)/(100-30))]=132.04 oC

Q= -92.68 x 10 6 Kj/kg.K from calculation Q Heat Exchanger E-203 above = 92,680MJ/h=25.74x106 W

A= Q/U Δ Tlm F= 25.74 k /(850)(0.9)(132.04)]=254.87 m2

Area cooler need for E-203 is 254.87 m2

4. Heat Exhanger ( E-201) Heater

From Table 11.11 ,use the following heuristic:-

42

Rule 1 :- set F= 0.9 Fouling factor

Rule 6: Δ T = 10 o.C minimum temperature approach is 10o

C for fluids Rule 7:- by ΔT LMTD Calculation :-

Cold side Temperature in :- 30.0 oC Cold side Temperature out:- 45.0 oC

Thus based on followed the rule accepted where temperature inlet cooling water is 30.0 oC , temperature outlet is 45 oC

Rule 8:- U=850 W/m2 oC For the moment,we find A by

Δ Tlm= [(154-45)-(250-30)]/[ln(154-45)/(250-30))]=158.06 oC

Q= 13.11 x 10 6 Kj/kg.K from calculation Q Heat Exchanger E-201 above = 10800 MJ/h= 3.0 x106 W

A= Q/U Δ Tlm F= 3.0 k /(850)(0.9)(158.06)]= 280.6m2

Area heater need for E-201 is 280.6 m2

5. Reactor (R-201) Table 11.17

Rule 1 :- The rate of reaction in very instance is established on the laboratory. Rule 2 :- Dimensions of catalyst in packed bed is 2-5 mm (powder)

43 Volume of reactor.

−𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 𝑘𝑜 exp (−𝐸𝑎_𝑅𝑇 ) 𝑃𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 −𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 1.21𝑥106_{exp (} −80.48 8.314 𝑥 364) 1 −𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 0.1936𝑥106

𝑉 =

𝐹𝑎𝑜 −𝑟𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙

∫ 1 𝑑𝑥

0.8 0

𝑉 =

15728 𝑥 32.04_{0.1936𝑥106}

∫ 1 𝑑𝑥

₀0.8

𝑉 = 2122 𝐿

Plug Flow Reactor FAo = 15728 mol/h

FA = 3171.28 mol/h

CAo = 0.98 CAo = 0.0115

44

6. Tower (T-201)

From table 11.13 heuristics for tower.

Rule 9 :- A safety factor of 10% of the number of trays calculated by the best means is advisable.

Rule 14:- Limit the tower height to about 53m (175ft) max. because of wind load and foundation consideration. An additional criterion is that L/D be less than 30 (20< L/D< 30 often will require special design.)

L= 15.8m D= 2m L/D= 7.9

From table 11.14 heuristics for trays

Rule 1:- For reasons of accessibility, tray spacing are made 0.5-0.6m(20-24 in). Spacing of trays in T-201 is 24 in.

45

ECONOMIC ANALYSIS

CALCULATION 4.1.1 Capital Cost Reactor 1 Diameter = 0.72 m Height = 10 m

Maximum pressure rating of 14.7 bar Volume = 2122 m3

Purchased equipment cost, From table A.1,

Equipment Type Equipment Description K1 K2 K3 Capacity Units

Min Size Max Size

Reactor Mixer / settler 4.7116 0.4479 0.0004 Volume, m3 0.04 60

log10 Cp0 = K1 + K2 log10 A + K3 (log10 A)2

= 4.7116 +0.4479 (log10 2122) + 0.0004 (log10 2122)2

= 6.20

Cp0 = USD 1.607x 106

From table A.7,

Equipment Type Equipment Description Bare Module Factor, FBM

46 Bare Module Cost, CBM

CBM = C0pFBM = USD 1.607x 106 (4.0) = USD 251134.47 C1 = CBM = USD 6.4289x 10 6

Refer to CEPCI to find cost for 2013: Year 2011, CEPCI = 582

Year 2014, CEPCI= 656

C2 = C1 (I2/I1)

= USD 6.4289x 10 6 (656/582) = USD 7.2462 x 10 6

Cost of raw materials,C

RM

The only one reactant used in this reaction which is methanol. The price for methanol is 0.4USD/kg. Retrieved from: http://www.alibaba.com/product-detail/LGB-good-quality-dimethyl-ether-dme-_2013674779.html

From material balance, 12683kg/h of methanol is fed to the system. 12683 ∗ 24 ∗ 330 = 100 ∗ 106 𝑘𝑔 𝑦𝑒𝑎𝑟 100 ∗ 106_{∗ 0.4}𝑈𝑆𝐷 𝐾𝐺 = 40 ∗ 106 𝑈𝑆𝐷 𝑦𝑒𝑎𝑟 Total raw material cost per year is 40 million USD.

47

Equipment Cost Calculation

Heat exchanger 1

𝐶𝑎

𝐶𝑏

= (

𝐴𝑎

𝐴𝑏

)

𝑛 Data available :

A = 180m2 , USD = 15000usd, time = 2010

At 2010, A= 254.87

15000

𝑥

= (

180

254.87

)

0.59 x = USD 17847 in 2010 CEPCI 2010 = 551 2014 = 621

Therefore, cost for heat exchanger with A= 254.87m2

C2 =C1

(

𝐼2 𝐼1

)

=17847

(

621 551

)

=

USD 37956

48 Heat exchanger 2 A= 280.6

15000

𝑥

= (

180

280.6

)

0.59 x = USD 19467 in 2010 CEPCI 2010 = 551 2014 = 621

Therefore, cost for heat exchanger with A= 280.6 m2

C2 =C1

(

𝐼2 𝐼1

)

=19467

(

621 551

)

=

USD 21940 Heat exchanger 3 A= 300

15000

𝑥

= (

180

300

)

0.59 x = USD 20275 in 2010 CEPCI 2010 = 551 2014 = 621

Therefore, cost for heat exchanger with A= 300 m2

C2 =C1

(

𝐼2 𝐼1

)

49

=20275

(

621 551

)

=

USD 22850.77 Total cost for heat exchanger:

= USD 37956 + USD 21940 + USD 22850.77 = USD 82746.77

Equipment cost for tower. Material, carbon steel.

h = 15.8m d = 2m tray = 22 SS P = 10.6 bar Vessel calculation: V= πD2 L/4 = 22 7 (2)2 (15.8) 4 = 49.6 m2 appx. 50m2

log 𝐶𝑝0_{(2010) = 3.4974 + 0.4485 log(50) + 0.1074 (log(50)}2 log 𝐶𝑝0(2010) = 4.57 𝐶𝑝0_{(2010) = USD 37126.7} 𝐹𝑝𝑡𝑜𝑤𝑒𝑟 = (10.6 + 1)2 2(944)(0.9) − 0.6(10 + 1) + 0.00315 0.0063 = 0.98 FBM = 2.22 + 1.82(0.98)(3.11) = 7.766 CBM = 37126 (7.766) = USD288357

50 Tray calculation A = πD2 /4 = 22 7 (2)2 4 = 3.14m2

log 𝐶𝑝0_{(2011) = 2.9949 + 0.4465 log(3.14) + 0.3961 (log(3.14)}2 = 3.313

𝐶𝑝0_{(2011) = 2059} CBM = Cp N FBM fq

N= 22 , fq = 1.0 since tray > 20 For ss sieve tray, id num = 61 From figure A.9, FBM = 1.8 CBM = USD(2059)(22)(1.8)(1.0)

= USD 81536.4

Total cost at 2011= 81536 + 288357 = USD 369893

From appendix B, CEPCI at 2014=657.6 2011 = 582

= 369893

(

657 582

)

=

USD 417941

51

Equipment cost for pump P = 7.2Kw

From http://www.matche.com/equipcost/PumpCentr.html, The cost of electric centrifugal pump with power rating 7.2kW 9.2kW is USD 4300 USD

There are 2 pump installed. One as backup pump ∴ total cost of pump = USD 8600 USD

Equipment cost for Vessel P = 14.7 bar L = 5m D = 1.14m Time = 2001 Volume = 22₇ (5)(1.14)2/4 = 5.105 ≃ 5.12

log Cpᵒ = 3.4974 + 0.4485 log (5.12) + 0.1074(log 5.12)2 = 3.8695 Cpᵒ = 7405 USD Cpᵒ at 2014 = 7405(925.8₃₉₇) = 17268 𝐹𝑃(𝑣𝑒𝑠𝑠𝑒𝑙) = 14.7+1 2(944)(0.9)−(0.6)(14.7+1)

+0.00315

0.063

= 0.1975

ID Number for SS vertical vessel = 20, Fm = 3.11 𝐹𝐵𝑀 = 2.25 + 1.82 (0.197)(3.11)

= 3.367

𝐶_𝐵𝑀= 3.367(17268) = 58141 USD

52 UTILITY COST:

Electric :

Electric : USD = 0.06 kwh, 16.8 USD/GJ There only a pump used with P = 7.2kW 1 x 7.2KJ/s x 3600s/h x 24h/d x 330 d/y =2.052864 x 109_{J/year x 16.8 USD/GJ}

=3448.8 USD/year

Steam :

For Lower Pressure stream, (LPS) From energy and mass balance,

𝒎̇ for LPS = 40,600 KJ/h 40600 x 24 x 330 = 321552000 kg/year

Price per year = 321 553 000 kg/year x USD 27.70/1000kg = 8.9 x 106 _USD/year. For MPS ; 10 barg From MEB, 𝒎_𝑖𝑛= 21446 𝑘𝑔/ℎ 21446 x 330 x 24 = 1,698,522,320 kg/year 1,698,522,320 kg/year x 28.31 USD/1000 kg = 4.8 x 106 _USD/year

53 Operating labor cost, COL

The technique used for estimating labour cost is based on date obtained from five chemical companies and correlated by Alkhayat and Gerrard. According to this method, the operating labour requirement for chemical processing plant is given by:

𝑁_𝑂𝐿 = (6.29 + 31.7𝑃 + 0.23𝑁_𝑁𝑂 )0.5

Where Nnp , is summation of total equipments. The total number of equipment in this plant is

9.

Number of operator per shift; NOL= (6.29 + 0.23 ∗ 9)0.5 = 3.6

The plant operates 24hours per day and 330 day per year. The operator is paid at rate of 900USD/month. Thus, the total operating labour per year is;

54

Total Capital Cost

Equipment cost (USD)

Heat Exchangers 161 284. 00 Reactor 7 246 000. 00 Distillation Towers 835 883. 00 Pump 4 300. 00 Vessel 58 153. 00 Total: 8 305 620. 00 Working Capital for 1st year of operation

Utility : Steam LPS 8 900 000. 00 Steam MPH 4 800 000. 00 Electricity 3 448. 00 Operating Labour 126 360. 00 Raw material 40 000 000. 00 Miscellaneous 7 000 000. 00 Total : 60 829 808. 00 GRAND TOTAL: 69.14 million USD

55

Cumulative cash flow table, without depreciation.

Year Cash Flow (USD) (from Discrete CFD)

(USD)

Cumulative Cash Flow (USD) 2014 -20 000 000. 00 - 20 000 000. 00 2015 -30 000 000. 00 -50 000 000. 00 2016 -20 140 000. 00 - 70 140 000. 00 2017 13 457 000. 00 -56 683 000. 00 2018 13 457 000. 00 -43 226 000. 00 2019 13 457 000. 00 -29 769 000. 00 2020 13 457 000. 00 -16 312 000. 00 2021 13 457 000. 00 -2 855 000. 00 2022 13 457 000. 00 10 602 000. 00 2023 13 457 000. 00 24 059 000. 00 2024 13 457 000. 00 37 516 000. 00 2025 13 457 000. 00 50 972 000. 00

56

Cumulative cash flow graph, without depreciation.

(50.9)

0 1 2 3 4 5 6 7 8 9 10 11

57

Non-discounted After Tax Flow Cost of land, L = 1 million USD

Total fix capital investment, FCI = 9.3 million USD

Fixed capital investment during year 1 = 5.3 million USD Fixed capital investment during year 2 = 4.0 million USD Plant start-up at year 2

Working capital = 60.83 million USD

End Of Year investment dk FCIL -Ʃ R COMd (R-C0M-dk)x(1-t) +dk Cash flow* Cumulative Cash flow* 0 (1) - 9.3 - - - (1) (1) 1 (5.3) - 9.3 - - - (5.3) (6.3) 2 (60.83+4)=64.84 - 9.3 - - - (64.84) (71.14) 3 - 4.3 5 20 5.3 15.25 15.25 (55.89) 4 - 2.15 2.85 20 5.3 17.7 17.7 (38.19) 5 - 2.43 0.42 20 5.3 10.405 10.405 (27.85) 6 - 1.075 0.00 20 5.3 9.93 9.93 (17.855) 7 - - 0.00 20 5.3 14.7 14.7 (3.155) 8 - - 0.00 20 5.3 14.7 14.7 17.545 9 - - 0.00 20 5.3 14.7 14.7 26.15 10 - - 0.00 20 5.3 14.7 14.7 41.086 11 - - 0.00 20 5.3 14.7 14.7 55.855 12 1+5.3 - 0.00 30 5.3 24.7 21 76.855

*(all number in millions)

Profitability Criteria

ROROI = ((15.25 +17.7 + 10.405 + 9.93 + (6∗14.7))₁₀ ∗_9.31) −₁₀1 = 1.4213 PVR = 15.25+17.7+10.405+9.93+(14.7∗6)+ 21_1+5.3+64.84 = 2.284

58

The cash flow diagram for non-discounted after tax is shown below,

(76.85 million USD)

0 1 2 3 4 5 6 7 8 9 10 11 12

59

Discounted Cash Flow Table

End of year Nondiscounted

cash flow

Discounted cash flow (million USD) Cumulative discounted cash flow (million USD) 0 (1) (1)= (1) 1 (5.3) (5.3)/1.1=(4.8) (5.8) 2 (64.84) (64.84)/ 1.12=(53.58) (59.38) 3 15.25 15.25/1.13=11.4 (47.98) 4 17.7 17.7/1.14=12.09 (35.89) 5 10.405 10.405/1.15=6.17 (29.72) 6 9.93 9.93/1.16=8.72 (21) 7 14.7 14.7/1.17=7.54 (13.46) 8 14.7 14.7/1.18=6.86 (6.6) 9 14.7 14.7/1.19=6.23 (0.37) 10 14.7 14.7/1.110=5.667 5.3 11 14.7 14.7/1.111=5.15 10.447 12 21 21/1.112=6.69 17.137

60

CONCLUSIONS AND RECOMMENDATIONS

The total investment needed for putting up the direct plant is 69.16 million USD and the profit made at the end of the plant life will be 76.855 million USD . The payback time for this plant is 5.7 years and the return on investment is 1.42%. It is clear that there are huge risks related to the direct method plant investment. The sensitivity analysis results shows that the product price for DME is the most sensitive parameter in this project (DME price = 1020 USD/ton). The project depends on a good and stable price for DME. The variation in utility price had minimum effect on payback time.

The heat integration part was done first after we calculated the plant cost. If we had done the cost calculations based on a better heat integration in the operating course would decrease.

61

62

REFERENCES

1. H. Scott Fogler, Elements of Chemical Reaction Engineering, Fourth Edition.

2. Elementary Principle Of Chemical Process, Richard M. Felder, Wiley Publisher, Third Edition, 2005

3. Perrys-Chemical-Engineers-handbook-1999.pdf

4. Retrieved2014, http://files.rushim.ru/books/spravochniki/Perrys-Chemical-Engineers-handbook-1999.pdf

5. R. Turton, Analysis Synthesis and Design of Chemical Process

6. Chemical Evaluation and Research Institute of Japan. May 2007. “Hazard Assesment Report – Chlorobenzene.”

7. Retrieved and adapted from

http://www.cerij.or.jp/ceri_en/hazard_assessment_report/pdf/en_108_90_7.pdf. 8. http://www.alibaba.com/product-detail/shell-tube-heat-exchanger-price_1961236719.html 9. http://www.alibaba.com/product-detail/shell-tube-heat-exchanger-price_1961236719.html 10. http://www.matche.com/equipcost/PumpCentr.html 11. http://www.matche.com/equipcost/Reactor.html 12. http://aabi.uitm.edu.my/dp/ 13. https://www.scribd.com/doc/916251/Process-Design-Course-DME-Autumn-2009

63

APPENDICES