DME production UITM

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 APPENDICES .......................................................................................................................................................... 63

1

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


NURUL SYAZMIN BT CHE JOHARI


SITI NOR ROUDAH BT HAIRUL ANUAR


2

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.

3

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.

4

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

Codes/Label Flammable

Class 2.1

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

5

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:

6

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

According to the information presently available to the New

hazard

Jersey Department of Health and Senior Services, dimethyl ether has not been tested for its ability to affect reproduction.

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

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.

7

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 8

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

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

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

COMPOSITION (kmol/h)

(Mole Fraction)

9

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

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

(Mole Fraction)

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

DME

0.0

71.757

3.672

Methanol

1.071

0.306

165.852

H2O

201.348

0.0

23.366

DME

0

0.9958

0.0212

Methanol

0.0053

0.0042

0.9593

H2O

0.9947

0

0.0195


(Mole Fraction)

APPENDIX A

10

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

11

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 6328630𝑔 ℎ𝑟

x

1 000 kg 1𝑡𝑜𝑛𝑛𝑒

x

1 𝑑𝑎𝑦 24 ℎ𝑟

1𝑚𝑜𝑙

x 46.07g = 137.37

1𝑦𝑒𝑎𝑟 𝑑𝑎𝑦

x

330 𝑑𝑎𝑦

=9099.88 kg/hr

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

= 5947.64 1𝑚𝑜𝑙

𝑘𝑔 hr

9099.88 kg/hr

Ratio:- 5947.64 kg/hr =1.53

Mixer F2=12806.1 kg/h F3=16049.7 x CH3OH=0.9904 x H2O=0.0075 xDme=0

F13=3258.9 kg/hr x CH3OH=0.9592 x H2O=0.01960 xDme=0.0212

kg/h Mixer

x CH3OH = 0.9838 x H2O = 0.0117 xDme = 0.0046 12

Overall Balanced F2+F13=F3 1280.61 kg/h +3258.9 kg/h=16049.7 kg/h Stream 2 13 3

Methanol (CH3OH) 397.3419𝑘𝑚𝑜𝑙 401.166 𝑘𝑚𝑜𝑙 1.989𝑘𝑚𝑜𝑙 101.439 𝑘𝑚𝑜𝑙 494.19𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙

=0.9904 =0.9592 =0.9838

Water (H2O) 3.825 𝑘𝑚𝑜𝑙 401.166 𝑘𝑚𝑜𝑙 97.308 𝑘𝑚𝑜𝑙 101.439 ℎ𝑟 5.814𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙

Dme 0

=0.0075 =0.01960 =0.0117

2.142𝑘𝑚𝑜𝑙 101.439 𝑘𝑚𝑜𝑙 2.295𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙

=0.0212 = 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 X CH3OH=0.1976 X CH3OH=0.9800 X H2O=0.01155

X H2O=0.4048 X (CH3)2O =0.3975

XCH3OCH3=0.0045

13

Overall Balanced F5 =F6 16049.7 kg/h =16049.7 kg/h

Stream 5

Methanol (CH3OH) =0.9800 502.299𝑘𝑚𝑜𝑙

Water (H2O) =0.01155 502.299 𝑘𝑚𝑜𝑙

502.299 𝑘𝑚𝑜𝑙

6

99.297𝑘𝑚𝑜𝑙

203.337 𝑚𝑜𝑙

199.665𝑘𝑚𝑜𝑙


502.299𝑘𝑚𝑜𝑙

=0.1976

5.814 𝑘𝑚𝑜𝑙

502.299𝑘𝑚𝑜𝑙

=0.4048


502.299𝑘𝑚𝑜𝑙

Dme =0.0045 =0.3975

First Separator in Distillation Column (T-201) 

The balance around the first separator 1:

F10 =9134.1 kgmol/h

F9=16049.7 kgmol/h

XCH3OH=0.0046 X H2O=0 XCH3OCH3=0.9954

XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976 T-100

F11=6915.6 kgmole/h X CH3OH=0. 3238 X H2O=0.6692 X CH3OCH3=0.00710

14

Overall Balanced F9=F11+F10 16049.7 kg/h =9134.1 kg/h+6915.6 kg/h Stream 9 10 11

Methanol (CH3OH) 99.297 𝑘𝑚𝑜𝑙

=0.1976

502.299𝑘𝑚𝑜𝑙 0.918𝑘𝑚𝑜𝑙

303.858𝑚𝑜𝑙

203.337 𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙

=0.4048 0

=0.0046

198.441𝑘𝑚𝑜𝑙 98.379𝑘𝑚𝑜𝑙

Water (H2O)

=0.3238

203.337𝑘𝑚𝑜𝑙 303.858𝑘𝑚𝑜𝑙

=0.6692

Dme 199.665 𝑘𝑚𝑜𝑙 502.299𝑘𝑚𝑜𝑙 2.142𝑘𝑚𝑜𝑙 198.441𝑘𝑚𝑜𝑙 2.142𝑘𝑚𝑜𝑙 303.858 𝑘𝑚𝑜𝑙

=0.3976 =0.9954 = 0.00710

Second Separator in Distillation Column (T-202) 

The balance around the second separator 2:

F13=3258.9 kgmole/h

F12=6915.6 kgmole/h

XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976

XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976 T-202

F14=3656.7 kgmole/h XCH3OH=0.1976 XH2O=0.4048 X(CH3)2O =0.3976

15

Overall Balanced F9=F11+F10 6915.6 kg/h =3258.9 kg/h+3656.7 kg/h

Stream 12 13 14

Methanol (CH3OH) 64.3𝑘𝑚𝑜𝑙

=0.3238

303.858 𝑚𝑜𝑙 97.308𝑘𝑚𝑜𝑙

=0.9593

101.439𝑘𝑚𝑜𝑙 1.071 𝑘𝑚𝑜𝑙 202.419𝑚𝑜𝑙

=0.005

Water (H2O) 203.𝑘𝑚𝑜𝑙 303.858𝑚𝑜𝑙

2.142 𝑘𝑚𝑜𝑙

=0.6693

1.989𝑘𝑚𝑜𝑙 101.439𝑘𝑚𝑜𝑙 201.348 𝑘𝑚𝑜𝑙 202.419𝑘𝑚𝑜𝑙

Dme 303.858𝑚𝑜𝑙

=0.0070

2.142𝑚𝑜𝑙

=0.0196

101.439𝑘𝑚𝑜𝑙

=0.0211 0

=0.995

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

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

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


(Mole Fraction)

16

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

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

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


(Mole Fraction)

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

DME

0.0

71.757

3.672

Methanol

1.071

0.306

165.852

H2O

201.348

0.0

23.366

DME

0

0.9958

0.0212

Methanol

0.0053

0.0042

0.9593

H2O

0.9947

0

0.0195


(Mole Fraction)

17

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

(Chemical Properties Handbook, Mc Graw Hills) 18

Compounds

Table : Heat capacities of compounds Cp = a + bT + cT2 + dT3 + eT4 (J/mol.K) 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3)

Methanol

110.100- 0.15747T+ 5.1853x10-4T2

Dme 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 (Chemical Properties Handbook, Mc Graw Hills)

19

Energy Balance For Reactors 

Energy Balance for Reactor 201:

Overall chemical equation 2CH3OH(CH3)2O+H2O Methanol Dimethyl ether+ Water ∆𝐻𝑟 ° = ∑𝑖 𝑣 ∆𝐻ofio = ∑𝑝𝑟𝑜𝑑𝑢𝑐𝑡|𝑣𝑖 | ∆Ȟ fio

∆𝐻𝑟 ° = (−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 T=250oC

F6=16049.7 kg/h Reactor 201

T=364oC

X CH3OH=15728.7 kg/h

X CH3OH=3171.42 kg/h

X H2O=185.37 kg/h

X H2O=6496.92 kg/h

XCH3OCH3=72.22 kg/h

X (CH3)2O =6379.76 kg/h

20

Reference H2O, DME,Methanol at 298oC and 1 atm Component Mole fraction Cpin (kJ/kg.K) Methanol Cp1 0.0046

Mole fraction 0.3975

DME

0.9838

Cp2

0.1977

Cp5

Water

0.0115

Cp3

0.4048

Cp6

Cpout(kJ/kg.K) Cp4

Calculation for inlet stream:523

Cp1= ∫298 C 𝑃 dt 523 = [21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3)] 298 = 73.63-47.7501 kJ/kg K = 25.8805 kJ/kg K 523

Cp2= ∫298 C 𝑃 dt 523 = [110.100- 0.15747T+ 5.1853x10-4T2] 298 = 169.5762—109.221 kJ/kg K =60.35kJ/kg K

523

Cp3= ∫298 C 𝑃 dt = [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:523

Cp4= ∫298 C 𝑃 dt 523 = [21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3)] 298 21

= 73.63 -47.7501 kJ/kg K = 25.8835 kJ/kg K

637

Cp5= ∫298 C 𝑃 dt 637 = [110.100- 0.15747T+ 5.1853x10-4T2] 298 = 220.195—109.221 kJ/kg K =110.974 kJ/kg K

637

Cp6= ∫298 C 𝑃 dt = [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

22

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.

T10=250 oC

Energy Balances for T-201

F10= 9134.1kg/h XDME=0.9954 Xmethanol=0.0046 Xwater=0 T-201

T9= 89 oC F9= 16049.7kg/h XDME=0.3975 Xmethanol=0.1977 Xwater=0.4048

T11=153 oC F11= 6915.6 kg/h XDME=0.0071 Xmethanol=0.3238 Xwater=0.6692

23

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


mwater=6496.91 kg/h

mwater=0 kg/h

Tref=89oC , Cp(DME)

523

:- ∫362

110.100- 0.15747T+ 5.1853x10-4T2

= 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 523

Cp Methanol :- ∫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

24

Inlet

Outlet (BOTTOM)

F9=16049.7 kg/h

F11= 6915.6 kg/h

mDme=6379.76 kg/h

mDme= 49.10 kg/h


mmethanol= 2239.27 kg/h

mwater=6496.91 kg/h

mwater=4627.92 kg/h

Tref=89oC , Cp(DME)

426

:- ∫362

110.100- 0.15747T+ 5.1853x10-4T2


426

:- ∫362

32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

= 1.653 kJ/kg.K 426


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 25

Heat Exchanger E201

T3= 45 oC

T4=154 oC

F3=16049.7 kg/h

F4=16049.7 kg/h

XDme=0.0046

XDme=0.0046

Xmethanol=0.9838

Xmethanol=0.9838

Xwater=0.0115

Xwater=0.0115

Tref=45oC , Cp(DME)

427

:- ∫318

110.100- 0.15747T+ 5.1853x10-4T2


427

:- ∫318

32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

=2.7895 kJ/kg.K 427


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

26

Heat Exchanger E202

T3= 154 oC

T4=250 oC

F3=16049.7 kg/h

F4=16049.7 kg/h

mDme=738.29 kg/h

mDme=738.29 kg/h



mwater=1845.72 kg/h

mwater=1845.72 kg/h

Tref=154oC ,

Cp(DME)

523

:- ∫427

110.100- 0.15747T+ 5.1853x10-4T2


523

:- ∫427

32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

=2.5709 kJ/kg.K 523


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

27

Cooler E-203

T7= 278 oC

T8=100 oC

F7=16049.7 kg/h

F8=16049.7 kg/h

mDme=6379.76 kg/h

mDme=6379.76 kg/h



mwater=3173.03 kg/h

mwater=3173.03 kg/h

Tref=100oC , Cp(DME)

551

:- ∫373

110.100- 0.15747T+ 5.1853x10-4T2


551

:- ∫337

32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

=4.745 kJ/kg.K 551


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

28

Cooler E-208

T14=167 oC

T15=50 oC

F14=3656.7 kg/h

F8=3656.7kg/h

mDme=0 kg/h

mDme=0 kg/h



mwater=3637.32kg/h

mwater=3637.32kg/h

Tref=323K , CP Water

440

:- ∫323

32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

=3.0 kJ/kg.K 440


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

29

Table of energy by equipment Equipment Reactor (R-101) Tower (T-201) Tower (T-202) Heat exchanger (E-202) Heat exchanger (E-203) Heat exchanger (E-208)

Energy ,Q (kJ/kg.K) 36.964 x106 -12.775 x106 10.800 x106 20.630 x106 -92.68 x106 -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 Methanol Dme

Cp = a + bT + cT2 + dT3 + eT4 (J/mol.K) 21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) 110.100- 0.15747T+ 5.1853x10-4T2

Water 32.24+0.01924T+(1.055x10-5)T2+(-3.596x10-9)T3

30

Calculation Cp for each stream :Stream 3:T3:-45oC Number Component 1 DME

Composition 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

31


Composition 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


Composition 0.3975

2

Methanol

0.1977

3

Water

0.4048 32

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


Composition 0.3975

2

Methanol

0.1977

3

Water

0.4048

Cp(DME)

:- ∫25

278

110.100- 0.15747T+ 5.1853x10-4T2

= -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 278


21.15+0.07902(T) + 2.587x10-5 (T2)+2.852x10-8(T3) 33

=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 1 DME

Composition 0

2

Methanol

0.0053

3

Water

0.9947

Cp(DME)

:- ∫25

167

110.100- 0.15747T+ 5.1853x10-4T2

= -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 167


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

34

stream

1

2

3

9.6152

2.1908

6.644

278

∑ 𝑚̇ 𝐶𝑝 ∆𝑇 kW/°C 268

A 167

1067.2872 157

B 164

276.3216 154

C 100

-4.446 90

D 55

-114.952 45

E 50

(kW)

40

-200.394

Step 2: The Temperature Interval Diagram

35

A 1067.2872 kW 1067.2872kW B 35.418 kW

COLD UTILITY

1102.7052 kW C 330.368 kW 1433.0732 kW D -200.394 kW 1232.6792 kW E 10.95 kW 1243.6292kW 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

36

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 ∆Tmin

= Tthreshold the hot utility demand is zero whereas the cold utility demand is 37

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 𝑄𝑐1 and 𝑄𝑐2 being equal to 𝑄𝑐 . 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

utility demand is constant and hot utility demand is zero which is shown in Figure (c)

38

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

Stream2 : 16.0 bar

T=25 oC

T=25 oC

39

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 Flow rate =

120806.1 kg/h 791.80 𝑘𝑔/𝑚3 𝑚3

efficiency of pump, ɛ= 0.60 %

= 152.57

Kw=(1.67)( 2.54 𝑚𝑖𝑛 ) x (

15 𝑏𝑎𝑟 0.60

𝑚3 ℎ

1ℎ

𝑚3

x 60 𝑚𝑖𝑛 = 2.54 𝑚𝑖𝑛

)

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:-

E-203

40

Rule 1 :- set F= 0.9 Fouling factor Rule 6: Δ T = 10 o.C minimum temperature approach is 10oC 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:-

E-201

41

Rule 1 :- set F= 0.9 Fouling factor Rule 6: Δ T = 10 o.C minimum temperature approach is 10oC 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) Rule 13 :- The value of a catalyst may improve the selectivity.

42

Volume of reactor. FA = 3171.28 mol/h FAo = 15728 mol/h Plug Flow Reactor CAo = 0.98 CAo = 0.0115 −𝐸𝑎

−𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 𝑘𝑜 exp ( 𝑅𝑇 ) 𝑃𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙

−𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 1.21𝑥106 exp (

−80.48 )1 8.314 𝑥 364

−𝑟 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 = 0.1936𝑥106

𝐹𝑎𝑜

0.8

𝑉=

∫ 1 𝑑𝑥 −𝑟𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 0

𝑉=

15728 𝑥 32.04 0.1936𝑥106

0.8

∫0 1 𝑑𝑥 𝑉 = 2122 𝐿

43

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.

44

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

Equipment

Type

Description

Reactor

Mixer /

K1

K2

K3

Min Size

Max Size

0.04

60

Units 4.7116

0.4479

0.0004

settler

log10 Cp0

Capacity

Volume, m

3

= 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

Reactor

Jacketed Agitated

4.0

45

Bare Module Cost, CBM CBM = C0p FBM = 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,CRM 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 ∗ 106 100 ∗ 106 ∗ 0.4

𝑘𝑔 𝑦𝑒𝑎𝑟

𝑈𝑆𝐷 𝑈𝑆𝐷 = 40 ∗ 106 𝐾𝐺 𝑦𝑒𝑎𝑟

Total raw material cost per year is 40 million USD.

46

Equipment Cost Calculation

Heat exchanger 1

𝐶𝑎 𝐴𝑎 𝑛 =( ) 𝐶𝑏 𝐴𝑏 Data available : A = 180m2 , USD = 15000usd, time = 2010 At 2010, A= 254.87

15000 180 0.59 =( ) 𝑥 254.87 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

47

Heat exchanger 2

A= 280.6

15000 180 0.59 =( ) 𝑥 280.6 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 180 0.59 =( ) 𝑥 300 x = USD 20275 in 2010 CEPCI 2010 = 551 2014 = 621 Therefore, cost for heat exchanger with A= 300 m2 C2 =C1

𝐼2

( ) 𝐼1

48

=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 (2)2 (15.8) 7 = 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

(10.6 + 1)2 + 0.00315 2(944)(0.9) − 0.6(10 + 1) 𝐹𝑝𝑡𝑜𝑤𝑒𝑟 = 0.0063 = 0.98 FBM = 2.22 + 1.82(0.98)(3.11) = 7.766 CBM = 37126 (7.766) = USD288357

49

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

Assume two identical operates, the total cost of tower = USD 835 883

50

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 =

22 7

(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 925.8

Cpᵒ at 2014 = 7405( 397 ) = 17268

𝐹𝑃(𝑣𝑒𝑠𝑠𝑒𝑙) =

14.7+1 +0.00315 2(944)(0.9)−(0.6)(14.7+1)

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 𝐶𝐵𝑀 = 3.367(17268) = 58141 USD

51

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,

𝒎𝑖𝑛 = 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

52

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: 𝑁𝑂𝐿 = (6.29 + 31.7𝑃 + 0.23𝑁𝑁𝑂 )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;

𝟑. 𝟗 ∗ 𝟑 ∗ 𝟏𝟐 ∗ 𝟗𝟎𝟎 = 𝟏𝟐𝟔 𝟑𝟔𝟎 𝑼𝑺𝑫.

53

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 Operating Labour

3 448. 00 126 360. 00

Raw material

40 000 000. 00

Miscellaneous

7 000 000. 00 Total : 60 829 808. 00 GRAND TOTAL: 69.14 million USD

54

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

55

Cumulative cash flow graph, without depreciation.

(50.9)

0

1

2

3

4

5

6

7

8

9

10

11

(-70.14)

56

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 investment Of Year 0 (1) 1 (5.3) 2 (60.83+4)=64.84 3 4 5 6 7 8 9 10 11 12 1+5.3 *(all number in millions)

dk

FCILƩ

R

COMd (R-C0Mdk)x(1-t) +dk

Cash flow*

Cumulative Cash flow*

4.3 2.15 2.43 1.075 -

9.3 9.3 9.3 5 2.85 0.42 0.00 0.00 0.00 0.00 0.00 0.00 0.00

20 20 20 20 20 20 20 20 20 30

5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3 5.3

(1) (5.3) (64.84) 15.25 17.7 10.405 9.93 14.7 14.7 14.7 14.7 14.7 21

(1) (6.3) (71.14) (55.89) (38.19) (27.85) (17.855) (3.155) 17.545 26.15 41.086 55.855 76.855

15.25 17.7 10.405 9.93 14.7 14.7 14.7 14.7 14.7 24.7

Profitability Criteria

(15.25 +17.7 + 10.405 + 9.93 + (6∗14.7))

ROROI = ( PVR =

10

15.25+17.7+10.405+9.93+(14.7∗6)+ 21 1+5.3+64.84

1

1

∗ 9.3) − 10 = 1.4213

= 2.284

57

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

(71.54 million USD)

58

Discounted Cash Flow Table

End of year

Nondiscounted cash flow

Discounted cash flow (million USD)

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

(1) (5.3) (64.84) 15.25 17.7 10.405 9.93 14.7 14.7 14.7 14.7 14.7 21

(1)= (5.3)/1.1=(4.8) (64.84)/ 1.12 =(53.58) 15.25/1.13=11.4 17.7/1.14 =12.09 10.405/1.15 =6.17 9.93/1.16 =8.72 14.7/1.17 =7.54 14.7/1.18 =6.86 14.7/1.19 =6.23 14.7/1.110 =5.667 14.7/1.111 =5.15 21/1.112 =6.69

Cumulative discounted cash flow (million USD) (1) (5.8) (59.38) (47.98) (35.89) (29.72) (21) (13.46) (6.6) (0.37) 5.3 10.447 17.137

59

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.

60

Proses Flow Manufacturing DME (Dimethy Ether)

61

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

62

APPENDICES

63

64

65

66

67

68

69

70

DME production UITM

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