University of Basrah

Engineering College
Chemical Engineering Department
2021-2022

Production of diethyl ether

By

Ahmed Raad Abdul AL-jabbar (1)

Hussein Abdul Alsalam Kadpan (6)
Fatima Shaker salem (35)
Fatima Mohammed fadhel (40)

Project supervision of Dr. Ghadeer Jassim

TABLE OF CONTENTS

CHAPTER

1 INTRODUCTION 5

1.1 Introduction: ............................................................................................. 6

1.2 Physical properties of diethyl ether: ........................................................ 7
1.3 Chemical properties of diethyl ether: ...................................................... 8
1.4 Uses of dietyl ether ..................................................................................10
1.5 production ................................................................................................11
1.6 Chemical reaction of diethyl ether: ........................................................13
1.1.6 Halogenation .......................................................................................................... 13
2.1.6 Formation of peroxide: ............................................................................................ 14
3.1.6 Combustion .............................................................................................................. 14
4.1.6 Dehydration by (Al2O3) catalytic............................................................................ 14
5.1.6 Oxidation by (K2Cr2O7 + H2SO4) catalytic ......................................................... 14
1.7 Effects of diethyl ether ............................................................................15

|Page2
 2 MATERIAL BALANCE 16

2.1 Mass balance around reactor(R-1201): ...................................................17

2.2 Mass balance around flash vessel(v-1202) ...............................................21
2.3 Mass balance around distillation(T-1201) ...............................................27
2.4 Mass balance around distillation(T-1202) ...............................................34

3 ENERGY BALANCE 41

3.1 Energy balance around of feed drum(V-1201): .......................................44

3.2 Energy balance around of feed heater(E-1201) .......................................46
3.3 Energy balance around of reactor(R-1201) .............................................47
3.4 Energy balance around of Effluent cooler (E-1202) ...............................50
3.5 Energy balance around of Flash sepreator (V-1202) ..............................51
3.6 Energy balance around of DEE Column (T-1201) .................................53
3.7 Energy balance around of DEE Condenser (E-1203) .............................56
3.8 Energy balance around of DEE Reboiler (E-1204) ................................59
3.9 Energy balance around of Ethanol Column (T-1202) ............................62
3.10 Energy balance around of Ethanol Condenser (E-1206) ......................65
3.11 Energy balance around of Ethanol Reboiler (E-1207) .........................68

|Page3
 4 CHEMICAL DESIGN 71

4.1 Chemical design of Heat Exchanger I (E-1201): .....................................72

4.2 Chemical design of Reactor(R-1201)........................................................83
4.3 Chemical design of Heat Exchanger II (E-1202).....................................89
4.4 Chemical design of Distillation column(T-1202) ...................................103

5 MECHANICAL DESIGN 118

5.1 Mechanical design of Heat Exchanger I (E-1201): ...............................119

5.2 Mechanical design of Reactor(R-1201) ..................................................128
5.3 Mechanical design of Heat Exchanger II (E-1202) ...............................140
5.4 Mechanical design of Distillation column(T-1202)................................148

REFRENCES 163

|Page4
 CHAPTER I
INTRODUCTION

|Page5
1.1 INTRODUCTION
Diethyl ether is a colorless, highly volatile flammable liquid and organic
compound. Diethyl ether is also called ethoxyethane, ethyl ether, sulfuric ether,
or ether. It is part of the ether class, meaning it contains an ether group or an
oxygen atom connected to two alkyl or aryl groups. The general formula for the
ether group is R-O-R. Ethers are typical in organic and biochemistry, because
they are common links in carbohydrates and lignin. Technical-grade ether and
anesthetic ether are the two grades available for commercial use. Technical-grade
ether standardly does not contain peroxides, though no worldwide parameters
exist. Anesthetic diethyl ether must pass the tests for aldehydes and peroxides,
have a natural reaction, contain no foreign odors, and have a water content of
0.2% or below.

|Page6
1.2 PHYSICAL PROPERTIES OF DIETHYL ETHER –
(C2H5)2O:
TABLE (1.1) : physical properties of D.E.E

Property Value

Density at 20°C 0.708 g/mL

Boiling point 34.6 °C

Freezing point -116 °C

Flash point -45 °C (closed cup)

Critical temperature 193.55 °C

Critical pressure 36.4 bar

Critical volume 281 cm3 /mol

Heat of formation -65.2 kcal/mol

Viscosity at 20°C, 0.23 mPas

Color Colorless

Solubility in water 6.9ml/100ml

Auto ignition temperature 180 °C

Surface tension at 20°C, dyne/cm 17.3 dyne/cm

Heat of vaporization at 25°C 27140 j /mol

Vapor pressure at -79°C 0.57959 bar

|Page7
 TABLE(1.2): physical properties

Odour Sweet, Pungent odour

Appearance Colourless liquid

Vapour Pressure 439.8 mm Hg at 20 deg C

Solubility Ethers are polar in nature, so they are more

soluble in water than alkanes.

1.3 CHEMICAL PROPERTIES OF DIETHYL ETHER –

(C2H5)2O:

TABLE (1.3) : chemical properties of D.E.E

Melting point -116 °C

Boiling point 34.6 °C(lit.)
density 0.714
vapor density 2.6 (vs air)
vapor pressure 28.69 psi ( 55 °C)
refractive index n20/D 1.3530(lit.)
Fp -40 °F
storage temp. Store at RT.
solubility Soluble in water, miscible with ethanol (96 per cent),
with methylene chloride and with fatty oils. It is highly
flammable.

|Page8
 form Liquid
color max. 10
Specific Gravity 0.714 (20/4℃) ; 0.712 (25℃)
Relative polarity 2.9
Odor Pungent odor detectable at 0.33 ppm
explosive limit 1.7-36%(V)
Water Solubility 69 g/L (20 ºC)
FreezingPoint -116.3℃
Merck 14,3806
-4 3
Henry's Law Constant 12.50(x 10 atm?m /mol at 25 °C) (Signer et al., 1969)
Exposure limits TLV-TWA 1200 mg/m3 (400 ppm) (ACGIH and
OSHA); STEL 1500 mg/m3 (500 ppm) (ACGIH).
Stability: Stable, but light-sensitive, sensitive to air. May contain
BHT (2,6-di-tert-butyl-4-methylphenol) as a stabilizer.
Substances to be avoided include zinc, halogens,
halogen-halogen compounds, nonmetals, nonmetallic
oxyhalides, strong oxidizing agents, chromyl chloride,
turpentine oils, turps substitutes, nitrates, metallic
chlorides. Extre

|Page9
1.4 USES OF DIETHYL ETHER
1.Diethyl ether is a common laboratory solvent. It has limited solubility in water
(6.05 g/100 mL at 25°C) and dissolves in 1.5 g/100 mL of water at 25°C.
Therefore, it is commonly used for liquid liquid extraction. When used with an
aqueous solution, the organic layer is on top because diethyl ether has a density
lower than that of water. . It is also a common solvent for the Grignard reaction
as well as other reactions involving organometallic reagents.

2.ethyle ether has a wide range of uses in chemical industry

3.It is a good solvent or extractant for fats, waxes, oils, perfumes, resins, dyes, .
gums and alkaloids.

4.When mixed with ethanol, ethyl ether becomes an excellent solvent for
cellulose nitrate in the manufacture of guncotton (see Explosives and propellants)
collodion solutions (see Membrane technology), and pyroxylin plastics (see
Cellulose derivatives, esters).

5.Another important use is as an extractant for acetic acid as well as other organic
acids, eg, in the cellulose acetate (qv) and plastic industries to recover acetic acid
from dilute aqueous systems.

6.Ethyl ether is also used as a denaturant in several denaturant alcohol formulas.

7.It has been used as a starting fuel for diesel engines and as an entrainer for
dehydration of ethanol and isopropyl alcohol.

8.It may be used as an anhydrous, inert reaction medium for the Grignard and
Wurtz-Fittig reactions.

9.Ethyl ether is used as a general anesthetic in surgery.

| P a g e 01
1.5 PRODUCTION
Most diethyl ether is produced as a byproduct of the vapor-phase hydration of
ethylene to make ethanol. This process uses solid-supported phosphoric acid
catalysts and can be adjusted to make more ether if the need arises. Vapor-phase
dehydration of ethanol over some alumina catalysts can give diethyl ether yields
of up to 95%.

Diethyl ether can be prepared both in laboratories and on an industrial scale by

the acid ether synthesis. Ethanol is mixed with a strong acid, typically sulfuric
acid, H2SO4. The acid dissociates in the aqueous environment producing
hydronium ions, H3O+. A hydrogen ion protonates the electronegative oxygen
atom of the ethanol, giving the ethanol molecule a positive charge:

CH3CH2OH + H3O+ → CH3CH2OH2+ + H2O

A nucleophilic oxygen atom of unprotonated ethanol displaces a water molecule

from the protonated (electrophilic) ethanol molecule, producing water, a
hydrogen ion and diethyl ether.

CH3CH2OH2+ + CH3CH2OH → H2O + H + CH3CH2OCH2CH3

This reaction must be carried out at temperatures lower than 150 °C in order to
ensure that an elimination product (ethylene) is not a product of the reaction. At
higher temperatures, ethanol will dehydrate to form ethylene. The reaction to
make diethyl ether is reversible, so eventually an equilibrium between reactants
and products is achieved. Getting a good yield of ether requires that ether be
distilled out of the reaction mixture before it reverts to ethanol, taking advantage
of Le Chatelier's principle.

Another reaction that can be used for the preparation of ethers is the Williamson
ether synthesis, in which an alkoxide (produced by dissolving an alkali metal in
the alcohol to be used) performs a nucleophilic substitution upon an alkyl halide.

| P a g e 00
Figure 1.1 unit 1200 diethyl ether process

| P a g e 02
TABLE(1.4) : united states production and price of ETHYL ETHER, 1940_1977

Year Production ,thousands Unit price, $/kg

of metric tons
1940 7.8 0.24
1950 17.2 0.18
1960 42.5 0.24
1970 47.6 0.26
1971 37.2 0.18
1972 34.0 0.18
1973 32.8 0.18
1974 28.4 0.18
1975 12.5 0.44

1.6 CHEMICAL REACTIONS OF DIETHYL ETHER

1. Halogenation:

When diethyl ether treated with Cl2 or Br2 in the dark, substitution product are
obtained. The extent of substitution depends upon the reaction condition.
Halogenations preferentially take place at ά- carbon atoms.

CH3CH2-O-CH2CH3 + Cl2 → CH3CH(Cl)-O-CH2CH3 + HCl

Diethyl ether 1- chloro diethyl ether

CH3CH(Cl)-O-CH2CH3+ Cl2→ CH2(Cl)CH(Cl) -O-CH2CH3 + HCl

1- chloro diethyl ether 1,1- dichloro diethyl ether

| P a g e 03
2. Formation of peroxide:

Ether combine with atmospheric oxygen to form peroxide.

CH3CH2-O-CH2CH3 + O2→ CH3C(OOH)H-O-CH2CH3

Diethyl ether peroxide of diethyl ether

CH3CH2-O-CH2CH3 + PCl5 → 2 CH3CH2Cl + POCl3

Diethyl ether ethyl chloride

3. Combustion:

CH3CH2-O-CH2CH3 + 6O2 → 4CO2 + 5H2O

Diethyl ether

4. Dehydration by (Al2O3) catalytic

CH3CH2-O-CH2CH3 → 2CH2=2CH2 + H2O

Diethyl ether 360ºC ethylene

5. Oxidation by (K2Cr2O7 + H2SO4) catalytic

CH3CH2-O-CH2CH3 + O2→ 2CH3CH2O → CH3COOH

Diethyl ether acetaldehyde acetic acid

| P a g e 04
1.7 EFFECTS OF DIETHYL ETHER
Health Effects

diethyl ether is a severe eye and moderate skin irritant. Irregular breathing,
hypothermia, and bradycardia have been noted as early effects in industrial
exposures. Excessive salivation has been reported. Prolonged exposure to high
vapor concentrations may cause eye injury. Exhaustion, dizziness, excitation,
headache, CNS depression, and seizures have occurred.

Ethyl ether is moderately toxic to humans by ingestion. It is poisonous

experimentally by the subcutaneous route. It is moderately toxic by
intraperitoneal and intravenous routes and mildly toxic by inhalation. Olfactory
changes resulted from human inhalation.

High exposure can cause unconsciousness and even death. High exposure may
affect the kidneys. Repeated or prolonged skin contact can cause drying, scaling
and cracking of the skin. Diethyl Ether is a highly flammable liquid and a serious
fire hazard.

Effects on the environment

Diethyl ether is highly flammable (NFPA rating = 4) and is one of the most
dangerous common fire hazards in the laboratory, due to its volatility and
extremely low ignition temperature. Ether vapor can be ignited by hot surfaces
such as hot plates and static electricity discharge, and since vapor is heavier than
air, it may travel a great distance to the source of ignition and flash back. Ether
vapor forms an explosive mixture with air in concentrations from 1.9 to 36% (by
volume). Carbon dioxide or dry chemical extinguishers should be used for ether
fires. Diethyl ether forms unstable peroxides upon exposure to air in a reaction
enhanced by light; The presence of these peroxides may cause an explosive
residue upon distillation.

Behavior in Fire: Steam is heavier than air and may travel a great distance to the
source of ignition and flash back. It decomposes violently when heated.

| P a g e 05
 Chapter II
MATERIAL BALANCE

Figure 2.1 Unit 1200 Diethyl Ether Process

| P a g e 06
2.1 MASS BALANCE AROUND REACTOR (R-1201)

Selectivity = 80

X = 0.8

Input Output

2 3

FA°? FA ?

FB°? FB+FD1+FD2 ?

FC1-FC2 ?

FF ?

( )

ethanol diethyl ether

| P a g e 07
 Abbreviations used in the reactor:

FC: molar flow rate of Diethyl Ether

FA: molar flow rate of Ethanol output

FD: molar flow rate of Water output

FF: molar flow rate of Ethylene

FA°: molar flow rate of Ethanol input

FB°: molar flow rate of Water input

Basis:

Feed:70% ethanol

30% water

The productivity of the project was determined diethyl ether is 25000 ton/year

FC = (25000 ton)/years*(10^6 g)/(1 ton)*(1 years)/(330 days)*(1 day)/(24

hour)*(1 hour)/(60 min)*(1 min)/(60 sec)

FC = ( 876.82 ( g)⁄sec)/(74.12 g⁄mol)= 11.829 mole/sec

FC=FA°(𝜃c + )

FA°= = 29.5725 mole/sec

FB°= mole/sec

FA= FA° (1-X) = 29.5725 (1-0.8) = 5.9145 mole/sec

FD= FA° (𝜃d+ X) =29.5725( +( ))

| P a g e 08
 ( )

Diethyl ether ethylene

C 2F + D

( )
Selectivity = ( )

C D + 2F

1 1 2

? ? 0.1478

2X = 0.1478

X(FD) = = 0.0739mole/sec

X(FC) = = 0.0739mole/sec

| P a g e 09
INPUT :

FA°*Mw = 29.5725 * 46.07 =1362.405 g/sec

FB°*Mw= 12.673 * 18.01 =228.24 g/sec

Sum=1590.5

OUTPUT :

FA*Mw= 5.9145 * 46.07 =272.48g/sec

FB= (FD1*Mw) +( FD2*Mw) = (24.502*18 )+ (0.0739*18)

=442.611g/sec

FC1-FC2 = (11.829*74.12) – (0.0739*74.12) = 871.288 g/sec

FF*Mw = 0.1478 * 28.05 = 4.145 g/sec

Sum=1590.5

TABLE (2.1) : Summary of material balance of reactor

Component Input Output

Steam 2 Steam 3
Molar flow rate Mass flow Molar flow rate Mass flow
mol/sec rate g/sec mol/sec rate g/sec
Diethyl ether 0 0 11.7551 871.288

Ethanol 29.5725 1362.405 5.9145 272.48

Water 12.673 228.24 24.5759 442.611

Ethylene 0 0 0.1478 4.145

Sum 42.2455 1590.5 42.2933 1590.5

| P a g e 21
2.2 MASS BALANCE AROUND FLASH VESSEL ( V-1202 )

Yi
4
Ethylene

Water

D.E.E

Ethanol
3

Zi

5
Xi

Water

Ethanol

Ethylene

D.E.E

T = 40℃ P=1120 Kpa

| P a g e 20
Assume ℃

TABLE (2.2) : HEAT CAPACITY OF COMPONENTS

Component A B C
Ethanol 16.8958 3795.17 230.918
Ethylene 6.74756 585 255
D.E.E 14.0735 2511.29 231.200
Water 16.3872 3885.70 230.170

( )
℃

| P a g e 22
10.489*0.171+109.39*2.60* +86.406*0.547 + 4.271*0.27 =

= 50.495 Kpa

Assume ℃

TABLE (2.3) : HEAT CAPACITY OF COMPONENTS

Component A B C
Ethanol 16.8958 3795.17 230.918
Ethylene 6.74756 585 255
D.E.E 14.0735 2511.29 231.200
Water 16.3872 3885.70 230.170

| P a g e 23
 ℃

29.517*0.171+125.085*2.6* +171.057*0.547+12.403*0.27=

| P a g e 24
- Assume ( 0.002 ) from D.E.E , ethanol and water out with ethylene from the top

- Assume ( 0.001 ) from ethylene out with D.E.E and water from the bottom

Input

= 272.48

= 871.288

= 442.611

= 4.145

SUM.= 1590.5

Top product

( )

SUM.=7.3134

| P a g e 25
Bottom product

SUM.=1583.2099

TABLE (2.4) : SUMMARY OF MATERIAL BALANCE OF FLASH VESSEL

Output
input
Output
Steam
Steam Steam
3
4 5
Mass flow Molar flow Mass flow Molar flow Mass flow Molar
Component rate g/sec rate rate g/sec rate rate g/sec flow rate
mol/sec mole/sec mol/sec
D.E.E 871.288 11.7550 1.7425 0.0235 869.545 11.7315

Ethanol 272.48 5.9144 0.5449 0.0118 271.935 5.9026

Ethylene 4.14 0.1475 4.1408 0.1476 4.145* 0.0001

Water 443.611 24.6313 0.8852 0.0491 441.7258 24.5266

Sum 1590.5 42.4482 7.3134 0.232 1583.2099 42.1608

| P a g e 26
 2.3 MASS BALANCE AROUND DISTILATION ( T_1201)

Ethylene
6
D.E.E

Water

Ethanol

Ethylene

Ethanol

D.E.E

Water
5

Ethylene

Water

D.E.E

Ethanol

| P a g e 27
(The feed)

FDEE =

= 11.7316

Fwater =

= 24.5266 mole/sec

Fethanol =

= 5.9026

Fethylene =

= 1.4777 mole/sec

∑= 42.1610 mole/sec

Composition:-

XDEE = 0.2782

Xwater = 0.5817

Xethanol = 0.1400

Xethylene = 3.5048

∑= 0.9999

| P a g e 28
TBP = ∑ Xi * TBPi

TBP = (0.2782 * 34) + (0.5817 * 100) + (0.14000 * 78) + (1.4777* * (-

103.7))

TBP = 78.5334 C

TBP = 351.5334 K

ᶯDEE = 0.99 =
DDEE = 0.99 * 11.7316 = 11.6142 mole/sec

ᶯWater = 0.01 =
Dwater = 0.01 * 24.5267 = 0.2452 mole/sec

( )
℃

| P a g e 29
 ( )

( )

( )

( )
( )

Nm = 4.2

( )
ᶯethanol =
( ( ) )

( )
ᶯethanol = (( ) )
= 0.2464

( )
ᶯethylene =
( ( ) )

( )
ᶯethylene = (( ) )
= 0.6109

| P a g e 31
(THE TOP PRODUCT)

Dethanol = ᶯethanol * Fethanol

Dethanol = 0.2464 * 5.9026 = 1.4544 mole/sec

Dethylene = ᶯethylene * Fethylene

Dethylene = 0.6109 * 1.4777 *

DDEE = 11.6142 mole/sec

Dwater = 0.2452 mole/sec

∑= 13.3138 mole/sec

Composition:-

XDEE = 0.8723

Xwater = 0.0184

Xethanol = 0.1092

Xethylene = 6.7803 *

∑= 0.9999

| P a g e 30
)THE BOTTOM PRODUCT)
W=F–D

WDEE = 11.7316 – 11.6142 = 0.1174 mole/sec

Wwater = 24.5267 – 0.2452 = 24.2815 mole/sec

Wethanol = 5.9026 – 1.4544 = 4.4482 mole/sec

Wethylene =(1.4777 * ) ( )= 5.7498 *

∑= 28.8471 mole/sec

Composition:-

XDEE = 4.0697 *

Xwater = 0.8417

Xethanol = 0.1541

Xethylene = 3.03*

∑= 0.9999

| P a g e 32
 TABLE (2.5) : SUMMARY OF MATERIAL BALANCE OF DISTILATION
COLUMN

Component Steam Steam 6 Steam

5 7
Molar flow Mass flow Molar flow Mass Molar Mass
rate rate rate flow rate flow rate flow rate
D.E.E 11.7316 869.5461 11.6142 860.844 0.1174 8.70168

Ethanol 5.9026 271.9327 1.4544 67.0042 4.4482 204.9285

Ethylene 1.4777 4.144* 0.00253 5.7498 0.00161

*
Water 24.5267 441.7258 0.2452 4.4160 24.2815 437.3098

Sum
1583.2087 1583.2234

| P a g e 33
2.4 MASS BALANCE AROUND DISTILATION( T-1202 )

D.E.E

Ethanol

Ethylene

Water

D.E.E

Ethanol 7

Ethylene

Water

D.E.E

Ethanol

Ethylene

Water

| P a g e 34
 ‫)‪(The feed‬‬
‫‪FDEE = 0.1174‬‬

‫‪Fwater = 24.2815‬‬

‫‪Fethanol = 4.4482‬‬

‫‪Fethylene = 5.7498‬‬
‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 28.8471‬‬

‫‪Composition:-‬‬

‫‪XDEE = 4.0697‬‬

‫‪Xwater = 0.8417‬‬

‫‪Xethanol = 0.1541‬‬

‫‪Xethylene = 1.9931‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 0.9999‬‬

‫‪TBP = ∑ Xi * TBPi‬‬

‫‪TBP = (4.0697‬‬ ‫‪) + (0.8417 * 100) + (0.1541 * 78) + (1.9931‬‬

‫(‬ ‫))‬

‫‪TBP = 96.3279 C‬‬

‫‪TBP = 369.4749 K‬‬

‫‪| P a g e 35‬‬
ᶯethanol = 0.99 =
Dethanol = 0.99 * 4.4482 = 4.4037 mole/sec

ᶯWater = 0.01 =
Dwater = 0.01 * 24.2815 = 0.2428 mole/sec

( )

-The volatility of ethylene is calculated from the scheme used for flash process
thermodynamic:

| P a g e 36
 ( )

( )

( )

( )
( )

Nm = 11.389

ᶯDEE =
( )

( ( ) )

ᶯDEE = (
(
)
)
= 0.9999

ᶯEthylene =
( )

( ( ) )

ᶯDEE = (
(
)
)
=1

| P a g e 37
‫)‪)THE TOP PRODUCT‬‬

‫‪DDEE = ᶯDEE * FDEE‬‬

‫‪Dethanol = 0.9999 * 0.1174 = 0.1173 mole/sec‬‬

‫= ‪Dethylene‬‬ ‫‪ᶯethelenel * Fethylene‬‬

‫* ‪Dethylene = 1 * 5.7498‬‬

‫‪Dwaterl = 0.2428 mole/sec‬‬

‫‪Dethanol = 4.4037 mole/sec‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 4.76385 mole/sec‬‬

‫‪Composition:-‬‬

‫‪XDEE = 0.0246‬‬

‫‪Xwater = 0.0509‬‬

‫‪Xethanol = 0.9243‬‬

‫‪Xethylene = 1.2069‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 0.9998‬‬

‫‪| P a g e 38‬‬
‫)‪) THE BOTTOM PRODUCT‬‬
‫‪W=F–D‬‬

‫‪WDEE = 0.1174 – 0.1173 = 0.0001 mole/sec‬‬

‫‪Wwater = 24.2815 – 0.2428 = 24.0387 mole/sec‬‬

‫‪Wethanol = 4.4482 – 4.4037 = 0.0445 mole/sec‬‬

‫* ‪Wethylene =(5.7498‬‬ ‫)‬ ‫(‬ ‫‪)= 0‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 24.0833 mole/sec‬‬

‫‪Composition:-‬‬

‫‪XDEE = 4.1522‬‬

‫‪Xwater = 0.9981‬‬

‫‪Xethanol = 1.8477‬‬

‫‪Xethylene = 0‬‬

‫ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ‬

‫‪∑= 0.9999‬‬

‫‪| P a g e 39‬‬
 TABLE (2.6) : SUMMARY OF MATERIAL BALANCE OF DISTILATION
COLUMN

Stream Stream 8 Stream

component 7 9

Molar Mass flow Molar flow Mass flow Molar Mass

flow rate rate rate rate flow rate flow rate

D.E.E 0.1174 8.7016 0.1173 8.6942 0.0001 0.007412

Ethylene 5.7498 * 1.6128* 1.6128* 0 0

Ethanol 4.4482 204.9285 4.4037 202.8784 0.0445 2.05011

Water 24.2815 437.3098 0.2428 4.3728 24.0387 432.9369

Sum 650.9415 650.9414

| P a g e 41
 CHAPTER III
ENERGY BALANCE

Figure 1: Unit 1200 Diethyl Ether Process

| P a g e 40
GENERAL EQUATION FOR ENERGY
CONSERVATION

Ĥ=∫

H=n∫

CP = A + BT + C D +E (for gas)

H= ( − 𝑓) + /2 ( − 𝑓 ) + /3 ( − 𝑓 ) + /4 ( − 𝑓
)+ /5 ( − 𝑓 ) (for gas)

TABLE (3.1) :Heat capacity constant for gas components (J/mol.k)

Component A B C D E
DEE 35.979 2.8444 * -1.2673 * -1.0128 * 3.4529 *

Ethanol 27.091 1.1055 * 1.0957 -1.5046 * 4.6601 *

*
Water 33.933 -8.4186 * 2.9906 * -1.7825 * 3.6934 *

Ethylene 32.083 -1.4831 * 2.4774 * -2.3766 * 6.8274 *

| P a g e 42
CP = A + BT + C +D (for liquid)

H= ( − 𝑓) + /2 ( − 𝑓 ) + /3 ( − 𝑓 )

+ /4 ( − 𝑓 ) (for liquid)

TABLE (3.2) : Heat capacity constant for liquid components (J/mol.k)

Component A B C D
DEE 75.939 7.7335 * -2.7936 4.4383 *
*
Ethanol 59.342 3.6358 * -1.2164 * 1.8030 *

Water
Ethylene 25.597 5.7078 * -3.3620 * 8.4120 *

ΔH = ∑Hout -

Q = ΔH = (∑n * cp) *ΔT

Ĥ : specific enthalpy in joule per mole (J/mol).

Cp : heat capacities of components (J/mol.K).

A,B,C,D,E : Heat capacity constants.

T : temperature, k

n : mole flowrate (mol/sec ).

| P a g e 43
3.1 ENERGY BALANCE AROUND OF FEED DRUM ( V-1201 )

=298 k
1

v-1201

29
99
2 =?
2
=116.1499 k

∑H1 +∑H29 =∑H2

Temperature of stream (29) = 116.1499 °C = 389.1499 k

Temperature of stream (1) = 25 °C = 298 k

| P a g e 44
 TABLE (3.3) : Energy Flow on Streams (1) and (29)

Stream1 Stream29
Stream n H N H
component (mol/sec) (J/mol) (J/sec) (mol/sec) (J/mol) (J/sec)

DEE 0 0 0 0.0001 17602.17 1.760217

Ethanol
29.528 0 0 0.0445 10412.7883 463.3690

Water -11.3657 6866.8587

0 0 24.0387 165070.3562
Ethylene
0 0 0 0 15465.8207 0
Total
18.1623 0 0 24.0833 50347.6374 165535.4855

∑H2 = 0 + 165535.4855 = 165535.4855 J/sec

By trial and error

T3 = 336.5 K = 63.5 °C

| P a g e 45
3.2 ENERGY BALANCE AROUND OF FEED HEATER
( E-1201)

2 3

Q = ΔH = ∑H3 - ∑H2

Temperature of stream (2) = 63.5 °C = 336.5 k

Temperature of stream (3) = 177 °C = 450 K

TABLE (3.4) : Energy Flow on Streams (2) and (3)

Stream2 Stream3
N H N H
(mol/sec) (J/mol) (J/sec) (mol/sec) (J/mol) (J/sec)

0 .......... 0 0 ........... 0
29.5725 4230.9331 125119.2691 29.5725 11659.74486 344807.805
12.673 2897.0454 36714.25635 12.673 5188.1268 65749.13094
0 ............ 0 0 ........... 0
42.2455 7127.9785 161833.5255 42.2455 16847.87166 410556.9359

Q = ΔH = ∑H3 - ∑H2

Q = 410556.9359 – 161833.5255

Q = 2487723.4104 J/sec

| P a g e 46
 3.3 ENERGY BALANCE AROUND OF REACTER( R-1201 )

ΔHout - ΔHin + ΔH298 = 0

ΔHout = ΔHin - ΔH298

𝛥 298 = ∑ⱱ𝛥

ΔH298 = Standard heat of reaction.

ΔHF = Standard heat of formation.

ⱱ = Stoichiometric Coefficient.

| P a g e 47
 TABLE (3.5) : heat of formation of components

Components DEE Ethanol Water Ethylene

Heat of
formation -252.7 -234.81 -241.8 52.3
(KJ/mole)

2C2H5OH → (C2H5)2O + H2O

(Ethanol) (DEE)

ΔH(298)1 = (1*(-241.8)) + (1*(-252.7))⦌ - (2*(-234.81))⦌

ΔH(298)1 = -24.88 KJ/sec

(C2H5)2O → 2C2H4 + H2O

(DEE) )ethylene)

ΔH(298)2 = (52.3*2) + (1*(-241.8))⦌ - (1*(-252.7))⦌

ΔH(298)2 = 115.5 kJ/sec

ΔH(298) = ΔH(298)1 + ΔH(298)2

ΔH(298) = -24.88 + 115.5 = 90.62 kJ/sec = 90620 J/sec

Temperature of stream (3) = 177 °C = 450 K

| P a g e 48
 TABLE (3.6) : Energy Flow on Streams (3)

Stream N H
Component (mol/sec) (J/mol) (J/sec)

DEE 0 ............ 0
Ethanol 29.5725 11659.74486 344807.805
Water 12.673 5188.1268 65749.13094
Ethylene 0 ............ 0
Total 42.2455 410556.9359

ΔHout = ΔHin - ΔH298

ΔHout = 410556.9359 – 90620

ΔHout = 319936.9359 J/sec

By trial and error

T4 = 410.6 K = 137.6 °C

| P a g e 49
 3.4 ENERGY BALANCE AROUND OF EFFLUENT COOLER
( E-1202 )

4 5

Temperature of steam (4) = 137.6 ℃ = 410.6 k

Temperature of steam (5) = 40℃ = 313 k

TABLE (3.7) : Energy flow on stream (4) and (5)

Stream4 Stream5
n H n H
Component (mol/sec) (J/mol) (J/sec) (mol/sec) )J/mol( )J/sec(

D.E.E 11.7551 14925.03 175445.22 11.7551 2670.6071 31393.2535

Ethanol 5.9145 8336.8612 49308.365 5.9145 1624.5718 9608.5299
Water 24.5759 3826.6312 94042.9057 24.5759 1131.0359 27796.225
Ethylene 0.1478 5459.0823 806.8523 0.1478 1891.8261 279.6118
Sum 42.2455 32547.6047 319603.343 42.2455 7318.0409 69077.6205

Q = 69077.6205 – 319603.4108

Q = - 250525.7903 ⁄

| P a g e 51
3.5 ENERGY BALANCE AROUND OF FLASH SEPRTER
( V-1202 )

5 V-12023
5

( )

( )

| P a g e 50
 TABLE (3.8) : Energy Flow on Streams (6) and (7)

Stream 6 Stream 7

N H N H
Component
(mol/sec) (J/sec) (J/mol) (mol/sec) (J/sec) (J/mol)

D.E.E 11.7315 31330.2272 2670.6071 0.0235 42.3594 1802.5315

Ethanol 5.9026 9589.1975 1624.5718 0.0118 11.8954 1008.0918

Water 24.5266 27740.4651 1131.0359 0.0491 24.8027 505.1472

Ethylene 0.0001 0.18918 1891.8261 0.1476 98.5085 667.4023

Sum 42.1608 68660.0792 7318.0409 0.232 177.5662 3983.1728

| P a g e 52
 3.6 ENERGY BALANCE AROUND OF DEE COLUMN (T-1201)

66

99

‫ا‬

| P a g e 53
 Component Α
DEE 2.2544 0.2782 0.8723
Ethanol 0.5800 0.14000 0.1092
Water 0.2534 0.5817 0.0184
Ethylene 0.8426 3.5048* 6.7803*

q at boiling point =1

by trial and error

Rmin + 1 =

Rmin = 1.268667-1= 0.268667

R = 3*Rmin =0.806001

| P a g e 54
From volume 6 , chapter 11 ,figure (11.11)

N=6.176

Ln= R*D = 0.806001*13.3138 = 10.7309

Vn= D*(R+1) = 13.3138*(0.806001+1) = 24.044

Lm= Ln+F =10.7309 + 42.1610 = 52.8919

Vm= Lm-W = 52.8919-28.8471 = 24.0448

| P a g e 55
 3.7 ENERGY BALANCE AROUND OF DEE CONDENSER
(E-1203)

E-1203

14 15

Saturated vapor saturated liquid

Cw

Component Α

DEE 2.2544 1 0.8723

Ethanol 0.5800 0.2572 0.429

Water 0.2534 0.11240 0.1637

Ethylene 0.8426 0.3737 1.814*

| P a g e 56
Composition and Amount (TOP) :-

Component Composition (X) Amount (D)

DEE 0.8723 11.6142
Ethylene 6.7803 9.0272
Ethanol 0.1092 1.4544
Water 0.0184 0.2452

 T bubble

̅ ∑

193.9165 kpa 1454.58 mmhg

Log =A -

Log (1454.58)=6.96559 -

=53.9959 C֩

| P a g e 57
  T dew

* +

= 255.503 kpa

255.503 kpa 1916.44 mmhg

Log (1916.44) = 6.96559 -

= 63.166 C֩

Q=V ( )

Component Cp
Dee 1031.9465
Ethanol 704.4715
Water 817.0099
Ethylene 450.9329

∫ =A+BT+C +D

CP= ( ) + /2 ( ) + /3 ( ) + D/4 ( )

Q = 121138.7441 J/sec

| P a g e 58
3.8 ENERGY BALANCE AROUND OF DEE REBOILER (E-1204)

E-1204

10 11

Saturated liquid saturated vapor

Component α

DEE 2.2544 8.896 1.805*

Ethanol 0.5800 2.2888 0.2656

Water 0.2534 1 3.3216

Ethylene 0.8426 3.3257 2.365*

| P a g e 59
Composition and Amount (Bottom) :-

Component Composition (X) Amount (D)

DEE 4.0697* 0.1174
Ethylene 1.9931* 5.7498
Ethanol 0.1541 4.4482
Water 0.8417 24.2815

 T bubble

̅ ∑

142.207 kpa 1066.90 mmhg

Ln =A -

Ln (1066.90)=18.3036 -

=109.94 C֩

| P a g e 61
  T dew

* +

=628.0762 kpa

628.0762 kpa 4710.9595 mmhg

Ln =A -

= 113.32 C֩

Q=V ( )

Component Cp
Dee 228.4193
Ethanol 137.1547
Water 112.6949
Ethylene 112.4969

∫ =A+BT+C +D +E

CP= ( ) + /2 ( ) + /3 ( ) + D/4 ( ) + E/5 (

)

Q =11401.8068 J/sec

| P a g e 60
3.9 ENERGY BALANCE AROUND OF ETHANOL COLUMN
(T-1202)

19

| P a g e 62
 Component Α
DEE 2.2544 4.0697* 0.0246
Ethanol 0.5800 0.1541 0.9243
Water 0.2534 0.8417 0.0509
Ethylene 0.8426 1.9931* 1.2069*

q at boiling point =1

by trial and error

Rmin + 1 =

Rmin+1 = 5.70912

Rmin= 4.70912

R = 3*Rmin =14.12736

| P a g e 63
From volume 6 ,chapter 11 ,figure (11.11)

N=14.9855

Ln= R*D = 14.12736*4.7635 = 67.2956

Vn= D*(R+1) = 4.7635*(14.12736+1) = 72.05917

Lm= Ln+F =67.2956 + 28.8471 = 96.1427

Vm= Lm-W = 96.1427-24.0833 = 72.0594

| P a g e 64
3.10 ENERGY BALANCE AROUND OF ETHANOL
CONDENSER (E-1206)

E-1206

24 25

Saturated vapor saturated liquid

Component α

DEE 3.476 3.0593 8.04105*

0.9243
Ethanol 1.13620 1

Water 0.5070 0.4462 0.11407

Ethylene 46 2.98104*
40.4858

| P a g e 65
Composition and Amount (TOP) :-

Component Composition (X) Amount (D)

DEE 0.0246 0.1173
Ethylene 1.2069 5.7498
Ethanol 0.9243 4.4037
Water 0.0509 0.2428

 T bubble

̅ ∑

171.1324 kpa 1283.9153 mmhg

Log =A -

Log (1283.9153)=7.68117 -

=92.11 C֩

| P a g e 66
  T dew

* +

= 183.1227 kpa

183.1227kpa 1373.8721 mmhg

Log (1373.8721)=7.68117 -

= 93.9942 C֩

Q=V ( )

Component Cp
Dee 239.8349
Ethanol 158.4497
Water 163.8101
Ethylene 106.2838

∫ =A+BT+C +D

CP= ( ) + /2 ( ) + /3 ( ) + D/4 ( )

Q = 123357.6881 J/sec

| P a g e 67
3.11 ENERGY BALANCE AROUND OF ETHANOL REBOILER
(E-1207)

E-1207

20 21

Saturated liquid saturated vapor

Component Α

DEE 3.476 6.8560 6.0563*

Ethanol 1.13620 2.2410 8.2449*

Water 0.5070 1 0.9981

Ethylene 46 90.729 2.6313*

| P a g e 68
Composition and Amount (Bottom) :-

Component Composition (X) Amount (D)

DEE 4.1522* 0.0001
Ethylene 0 0
Ethanol 1.8477* 0.0445
Water 0.9981 24.0387

 T bubble

̅ ∑

174.5810 kpa 1309.7883 mmhg

Ln =A -

Ln (1309.7883)=18.3036 -

= 116.149 C֩

| P a g e 69
  T dew

* +

=174.81178 kpa

174.81178 kpa 1311.5197 mmhg

Ln =A -

= 116.043 C֩

Q=V ( )

Component Cp
Dee 729.61
Ethanol 420.018
Water 353.31
Ethylene 362.63

∫ =A+BT+C +D +E

CP= ( ) + /2 ( ) + /3 ( ) + D/4 ( ) + E/5 (

)

Q = 16523.8364 J/sec

| P a g e 71
 CHAPTER IV
CHEMICAL DESIGN

| P a g e 70
4.1 CHEMICAL Design of Heat Exchanger (E_-1201)

The temperature inlet in the shell side is 63. 5°C and the outlet temperature is
177°C.
The temperature of the heat exchanger tube inlet is 200°C and the outlet
temperature is 150°C.
The pressure equals to 10bar.
The flow is counter current.
( ) ( )
( )
( )

( ) ( )
=50°C

( )
( )

( )
( )

( ) ( )
( ) ( )

| P a g e 72
Figure 4-1 Correction-factor plot for exchanger with one shell pass and two, four,
or any multiple of tube passes.(Reference: Holman , Heat Transfer 10th edition

By using the values of P and R through the above figure the value of correction
factor is obtained.

| P a g e 73
Assuming the overall heat transfer coefficient equals 500 by using Table
12.1

U=500 for organic solvents

℃

Where:

Q : heat transferred per unit time , W .

U : overall heat transfer coefficient , w/m.c .

A : heat transfer area , .

mean temperature difference , ℃

F : temperature correction factor.

Q=2487723.4104 from energy balance

Select :
Tube length 16 ft 4.83 m
Tube out side diameter 20 mm 0.02 m
Tube thickness 16 mm 0.016 m

Surface area of one tube :

o
A=3.14*0.02*4.83

| P a g e 74
_The recommend tube pitch distance between

tube center is 1.25 times the tube out side diameter.

Use triangular pitch ,

( )

K1=0.319,n1=2.142

( )

Where:

Clearnace=61mm
Ds=527+61=588mm

| P a g e 75
Tube_side coefficient:
Mean water temperature = ℃

Tube per pass=353tube pass

Where:
∶ ℎ 𝑢𝑏
∶ 𝑓

Steam mass velocity =

⁄

203428.57

( )

From figure 12.23

| P a g e 76
 ( )

( )

Shell_ side cofficient

pt = 1.25 do = 1.25 20 = 25mm

Where

( ) )

( )

Equivalent dimeter (de)= ( )

Mean shell Temperture =

| P a g e 77
 ρ( ) Cp(J/kg °C( μ (pa.s) K(w/m. °C(

Water 967 4251 0.000229 0.684

Ethanol 690 3482 0.0002 0.123
Mole fraction of ethanol =70%
Mole fraction of water=30%

Total density =967*0.3+690*0.7=773.1

Cptotal=4251*0.3+3482*0.7=3712.7J/kg. °C
μ total=0.000229*0.3+0.0002*0.7=2.087
Ktotal =0.684*0.3+0.123*0.7=0.29 w/m. °c

Re=

Re= 60827.98

From figure (12-29 ) choose 25percent baffle cut at Re 60827.98

Jh=2.7*

Hs=ho= ( )

ho=3303.7(w/m.°C(
Checking:
Hod=5000,Hid=3000 from table (12.2) volume6
From table (12.6)for steel kw=35w/ . °C

| P a g e 78
overall heat transfer coefficient based on the outside area of the tube.
( )

℃
℃
∶ ℃
∶ ( ) ℃ )
∶ ℃
∶ ℃
∶
∶
( ) (
)

Uo=500.638

Pressure drop of tube side

[ ( )( ) ]

Where

| P a g e 79
 ( )

Re = 203428.57 from figure 12.24

[( ( ) ) ]

| P a g e 81
Pressure drop of shell side

𝑢
( )( ) ( )
𝑏

Where

L = tube length=4.83mm ,

De=14.2mm

= baffle spacing .

For Re = 60827.98

from figure 12.30

Number of baffles ( )

| P a g e 80
Summary
Tube length 4.83 m
Tube thickness 0.016 m
Shell inlet temperature 63.5C
Shell outlet temperature 177C
Tube inlet temperature 200C
Tube outlet temperature 150C
Number of tubes 353
Bundle diameter 527 mm
Number of baffles 196mm
Overall coefficient 500
Pressure drop of tube side 17.66 psi
Pressure drop of shell side 25.6 psi
Shell diameter 14.2
mm

| P a g e 82
4.2 CHEMICAL DESIGN OF REACTOR :

| P a g e 83
Conditions Value
Inlet Temperature (K) 450
Inlet Pressure (kpa) 1200
Outlet Temperature (k) 410.6
Outlet pressure (kpa) 1200
Molar flow rate (A) 29.5725
(mole/sec)
Other information
Type of Reactor Packed bed rector
operating condition Adiabatic reactor
Density of catalyst 3950
(Kg/m3)
E 0.5

( )
ethanol diethyl ether
A 1/2 C + 1/2 D

θA = 1 , θc = 0 , 𝜃d =

Fc = 11.829 mole/sec
FC=FA°(θc + 1/2 X)

FA°= = 29.5725 mole/sec

FA= FA° (1-X) = 29.5725 (1-0.8) = 5.9145 mole/sec

FD= FA° (𝜃d+ X) =29.5725( +( ))

Ftotal = FA°( 𝜃 𝜃d + 1/2 x +1/2 x)

Ft = FA°( )
Ft = FA°(1.428)

| P a g e 84
PA = Pt * y
°( )
PA = Pt *
( )

PA = 840 (1-x)
( )
Pc = Pt * ( )

Pc = 420.168 x
°( )
PD = Pt * ( )

( )
PD = 1200 *

PD = 359.663 + 420.168x

Hf(ethanol) = -234.81 KJ/mole , Hf(DEE) = -252.7 KJ/mole ,

Hf(water) = -241.8 KJ/mole

( )
Ethanol diethyl ether
ΔHr = ∑(n ΔHf)product - (n ΔHf)react
ΔHr(298) = (1 * -241.8) + (1 * -252.7)⦌ - (2 * -234.81)⦌
ΔHr(298) = - 24.88 KJ/mole = - 24880 J/mole

| P a g e 85
Substance Cpi (J/Kmol . K)
A 1 27.091+1.1055*10-1T + 1.0957*
10-4T2 - 1.5046 *10-7T3 + 4.6601 * 10^-11
T4
C 0 35.979 + 2.8444 * 10^-1 T- 1.2673 * 10^-6
T^2 – 1.0128 * 10^-7 T^3 + 3.4529*10^-11
T^4
D 0.428 33.933 – 8.4186 * 10^-3T + 2.9906 * 10^-5
T^2 – 1.7825 * 10^-8T^3 + 3.6934 *10^-12
T^4

At T = 450 K CPA = 87.2266 , CPc = 164.2133 , CPD = 34.7277

At T = 410.6 K CPA = 81.8646 , CPc = 146.5268 , CPD = 34.3893

CPmix = yA CPA + yc CPc + yD CPD

°( ) ( ) °( )
CPmix = CPA *
( )
+ CPD * ( )
( )

( ) ( ) ( )
CPmix = CPA * ( )
+ CPD * ( )
( )

At x = 0 and T = 410.6 K CPmix1 = 67.6352

At x = 0.8 and T = 450 K CPmix2 = 78.3509

CPmix(average) = =

CPmix(average) = 72.993 J/mole.k

| P a g e 86
 ∫

Solve by simpson’s rule

∫ 𝑓( ) (𝑓 𝑓 𝑓 𝑓 )

∫ 𝑓( )

Weight of catalyst
W=V*d
W = 6.196 * 3950
W = 24474.2 kg

Volume of reactor =
( )

Vr =
( )

Vr = 12.392

| P a g e 87
12.392 = * 4D

12.392 = 3.14

L = 4 * 1.580
L = 6.32 m

Summary of Reactor :
Catalyst volume 6.196 m3
Catalyst weight 24474.2 kg
Reactor volume 12.392 m3
Reactor diameter 2.451 m
Reactor Length 1.634 m

| P a g e 88
4.3_ CHEMICAL DESIGN OF HEAT EXCHANGER (E_1202)

Hot in: T = 410 K Cold in: T = 298 K

E_1202

Cold out: T = 313 K Hot out: T = 313 K

Tube side ( water )

Shell side (gases : D.E.E , ethanol , ethylene and water )

-Type of heat exchanger is one shell , two tube pass

-The flow is counter current

| P a g e 89
Where

Q : heat transferred per unit time , W .

U : overall heat transfer coefficient , w/m.c .

A : heat transfer area , .

: log mean temperature difference , ℃ .

F : temperature correction factor.

From energy balance Q = 250525.7903 J/S

( ) ( )
( )
( )
( ) ( )
℃

= 0.8667

= 0.15366

| P a g e 91
* From figure 4.1 : F = 0.91

fig. 4.1 . temperature correction factor; one shell pass; two or more tube pass

* let overall heat transfer coefficient : U = 20

℃

Select :

Tube length 16 ft 4.83 m

Tube outside diameter 20 mm 0.02 m
Tube thickness 16 mm 0.016 m

| P a g e 90
 𝑢 𝑓 𝑓 𝑢𝑏 ( )

𝑢 𝑏 𝑓 𝑢𝑏 ( )

_The recommend tube pitch distance between

tube center is 1.25 times the tube out side diameter.

Use triangular pitch ,

𝑏𝑢 ( ) ( )

( )

| P a g e 92
* Bundle diameter clearance = 77.9

𝑏 𝑓𝑓 ( )

| P a g e 93
Tube side coefficient ( )

Mean tube temperature =

P = 10 bar = 1000 Kpa

ρ ⁄

𝑢𝑏

𝑢𝑏

𝑢𝑏 𝑢𝑏

𝑢 ⁄

| P a g e 94
𝑢

From figure 12.23,

| P a g e 95
 ( )

Where

𝑓 𝑓

𝑢 𝑓 𝑢𝑏 𝑓𝑢

𝑓 𝑓𝑢 𝑢

[ ]

( )

( )

Shell side coefficient (ho)

P=10 bar = 1000Kpa

= 1.5905 kg

Cp = 1835.7118 J/Kg.K

| P a g e 96
 K = 0.02286 w/m.℃

Where De is the equivalent diameter

( )

( )

Where As is the shell cross flow area

Tube pitch = Pt = 1.25 *do

( )

| P a g e 97
Overall coefficient U

( )

Where

U = the overall coefficient based on the out side area of the tube ℃

ho = outside fluid film coefficient , ℃

hi = inside fluid film coefficient , ℃

hod = out side dirt coefficient ( fouling factor ) , ℃

hid = in side dirt coefficient , ℃

= thermal conductivity of the tube wall material , ℃

di = tube inside diameter , m

do = tube out side diameter , m

* From table 12.2 volume 6

℃ ℃

| P a g e 98
Pressure drop of tube side

𝑢
[ ( )( ) ]

Where

𝑢𝑏 𝑢 ( )

𝑢 𝑏 𝑓 𝑢𝑏

𝑢 𝑢𝑏

𝑓 𝑢𝑏

Re = 811.7433 from figure 12.24

| P a g e 99
 [( ( ) ) ]

Pressure drop of shell side

𝑢
( )( ) ( )
𝑏

Where

L = tube length ,

= baffle spacing .

For Re = 50525.95078

from figure 12.30

| P a g e 011
Number of baffles ( )

| P a g e 010
Summary

Tube outside diameter 0.20 m

Tube length 4.83 m
Tube thickness 0.016 m
Cold inlet temperature 25 C
Cold outlet temperature 40 C
Hot inlet temperature 137.6 C
Hot outlet temperature 40 C
Number of tubes 1028.454
Bundle diameter 869.1034 m
Number of baffles 24.5555
Overall coefficient 21.10
Pressure drop of tube side 5.52 * psi
Pressure drop of shell side 2.5039 psi
Shell diameter 947.0034 mm

| P a g e 012
4.4 CHEMICAL DESIGN OF ETHANOL COLUMN (T-1202):

From energy balance:

4.70912
11.389
Ø 0.953
R 14.12736
72.05917
67.2956
72.0594
96.1427
0.8248

0.9338

0.76

N 14.9855

| P a g e 013
Cloumn diameter :

Top section

Liquid vapour flow parameter ,FLV 

FLV= √

Where: FLV = liquid vapor flow factor

Ln = liquid mass flow rate; Kg/s

Vn = vapor mass flow rate; Kg/s

V = vapor density; Kg/m3

L = liquid density; Kg/m3

Vapor density: =

=(74.12*0.0246)+(46.07*0.9243)+(18*0.0509)+(28*1.2069* )
=45.25g/mol

=2.61 kg/

| P a g e 014
 ( )
ρl=A*

constant of liquid density .

Comp A B N Tc(K)
Diethyl ether 0.27267 0.27608 0.29358 466.70
H2O 0.34710 0.27400 0.28571 647.13
Ethylene 0.21428 0.28061 0.28571 282.36
Ethanol 0.26570 0.26395 0.23670 516.25

density of component in the top

comp Ρl(kg/m^3) Xd
Diethyl ether 618 0.0246
H2O 961 0.0509
ethanol 717.2 0.9243
ethylene 959.78 1.2069*10^-5

SO

ρl= =967 kg/

FLV= √ =0.048 

Taking spacing plate (L) =0.5m 

( -0.047) *√

=0.87 m/s

| P a g e 015
 From figure 11.27 find value of K1.

K1 = 0.08

Uf= K1 √ =1.537 m/s

Ufmax = 0.85 *1.537 = 1.306 m/s

√ =0.855 m

Bottom section

FLV= √ 

Vapor density: =

=(74.12*4.1522* )+(46.07*1.8477* )+(18*0.9981)+(28*0)

=18.05g/mol

| P a g e 016
 ℃

=2.11 kg/

( )
ρl=A*

density of comonent in the bottom

comp Ρl(kg/m^3) Xw
Diethyl ether 583 4.1522*
H2O 939.257 0.9981
ethanol 691.10 1.8477*
ethylene 259.4 0

SO

ρl= =947 kg/

FLV= √ =0.063 

( -0.047) *√

| P a g e 017
 From figure 11.27 find value of K1

K1 =0.088

Uf= K1 √ =1.862 m/s

Ufmax = 0.85 *1.862 = 1.5827 m/s

√ =0.998m

Plates efficiency :

(∑

𝑓 ∶

component
DEE 0.126 4.0697*10^-3
ethylene 0.0057 1.9931*10^-6
Ethylene 0.36 0.1541
Water 0.29 0.8417

( )=0.50 

| P a g e 018
 Column High
𝑓
𝑓 𝑢

0.5=

Actual No. of stage N=30 

Safety factor =0.1*33=3.3 

Total stage =30+3=33 plates 

Column high =0.5*33=16.5m 

Provisional plate design

Column diameter

Column Area ( )=

= ( )

=0.77

At 12 percent of area 

Down comer Area

=0.0924

Net area (An) = (Ac-Ad)=0.77 – 0.0924= 0.6776 m2

Active area (Aa)=Ac– 2× Ad== 0.77 - 2×0.0924 = 0.5852 m2

Hole area take 10 percent from Aa as first trial

Ah= 0.1 ×0.5852 =0.05852 m2

| P a g e 019
 Weir length:

(Ad/Ac) *100%= 12

Lw/Dc = from figure 11.31 = 0.76

Lw= Dc* 0.76

Lw=0.99*0.76=0.7524m

Hole Diameter =2.5 to 12.5mm 

Take Hd=5mm 

Plate thickness=55mm 

The weir high= 40 to 90 mm 

| P a g e 001
Take weir hight =50mm=hw 

Cheek weeping :

- Maximum liquid rate (Lw)=

=0.482 kg/s

Manimum liquid rate (Lw)=70 percent maximum liquid rate

=0.7 *0.482 =0.3374 kg/s

Maximum how = 750( ) 

=750( )

Minumum how=750( ) =4.5mm 

| P a g e 000
K2 = constant = 32 figure 11.30

Minimum design vapor velocity (m/s) : 

( ( ))
Ûh =

( )
Ûh = = 9.39m/s

Actual min vapor velocity = min vapor velocity/Ah= 0.7*0.81/0.05852 = 9.688

m/s

* So minimum operating rate will be well above weep point

| P a g e 002
plate pressure drop :

dry plate drop

Max. vapor velocity through holes(𝑢 )=0.081/0.05852= 13.8 m/s

From figure 11.34,

| P a g e 003
The pressure drop through the dry plate can be estimated using an expression ;

Hd= 30.67 mm liquid

Residual head (hr)= = 12.67 mm liquid

Total plate pressure drop ht= hd +(hw+how) +hr

=30.67+(50+5.78)+12.67

=30.67+55.78+12.67

=99.12mm liquid

 Down comer liquid backup:

hap=50 – 10 = 40 mm

area under apron Aap= 0.04 × 0.7524 = 0.0301 m2

= 0.0474mm

 Backup in down comer

Hb= (hw+how) + ht + hdc = 55.78+99.12+0.0474=154.9474mm

=0.155m

So:
( plate spacing + weir height )

=1/2(0.5+(50*10^-3))

| P a g e 004
=0.275

-So tray spacing is acceptable

 Check Residence time :

Tr=

Tr= = 14.1 sec > 3 sec so the design is satisfactory;

 Check entrainment:

Percentage flooding =85%

For FLV=0.063

= 0.06 well below 0.1

 Trail Layout:

Use cartridge type construction allow 50 mm un performed strip round edge , 50

mm wide clawing zones .

50 mm

0.7524 0.99 m
Өc
m

50 mm
| P a g e 005
perforated area

From fig (11.32) Lw/Dc = θc =99

Angle subtended at plate edge by unperforated = 180-99 = 81

-Mean length , un perforated edge strips

= ( 0.99 – 50 × )×π× = 1.33 m

-area of unperforated edge strips = 0.05× 1.33 = 0.0665 m2

-mean length of calming zone= (0.99- 0.05) ×sin (99/2) = 0.714 m

| P a g e 006
-area of calming zones= 2( 0.714×0.05) =0.0714 m2

 total area for perforations

Ap = 0.5852 – 0.066– 0.0714 = 0.4478

Ah/Ap = 0.05852/0.4478 = 0.13

Lp/dh = 2.7 , Satisfactory , Whithin ( 2.5 to 4 )

-Area of one hole = =π*(5*10^-3) 1.964 ×

 Number of holes:

Number of holes =

| P a g e 007
 CHAPTER V
MECHANICAL DESIGN

| P a g e 008
5.1 MECHANICAL DESIGN OF HEAT EXCHANGER (E_1201)

Condition
Operating pressure 10 bar
Material Stainless steel
Density of material 7800 ⁄
Shell temperature ℃
Tube temperature 200_150 ℃
Corrosion allowance 2 mm
Shell diameter 588 mm
Number of tubes 353
Insulation thickness 75 mm

Pi=1.2N/m

with corrosion allowance as follows

2.361+2=4.361mm

Determining ℎ
Choosing Tori spherical

( )

| P a g e 009
 ( √ )

.771

( )

Determining Dead weight

t=4.361 hv=4.83m

(H )
( )
3.549kN

Determining Weight of pipes

= 3.05kN

| P a g e 021
Determining Weight of insulation

=2.2

= 2.9000kN
Double=2.9000*2=5.8KN

Total weight
Shell 3.549
Insulation 5.8
Pipes 3.05
Total 12.399

Determining Wind loading M

(H )
( )

| P a g e 020
 (H )

Where
ℎ 𝑏 𝑏

Analysis of stresses
.

( ) ( )

( )

( )

| P a g e 022
 ( )

( )

80.898 - (-40.438) =121.336N/m

Check elastic stability (buckling)

Critical buckling stress:

( ) ( ) ⁄

The maximum compressive stress will occur when the vessel is not under
pressure .

well below the critical buckling stress. So design is satisfactory

| P a g e 023
Support design

Choose saddle support must be designed to withstand the load imposed by the
weight of the vessel and content . they are constructed of bricks or concrete, or
are fabricated from steel plate . the contact angle should not be less than 120° ,
and will not normally be greater than 150° . wear plates are often welded to the
shell wall to reinforce the wall over the area of contact with the saddle .

The dimensions of typical "standard" saddle designs are given in figure 13.26
chemical engineering Vol.6 page 847.

Choose the design that fit that diameter frim fig 13.26.

| P a g e 024
Flange design

For stainless steel

Where:
𝑓 𝑏

1-For shell outlet:

℃

| P a g e 025
2-For shell inlet:
℃.
( )

3-For tube inlet:

At T=200

( )

| P a g e 026
4-For tube outlet:
T=150

( )

Summary
Shell thickness 4.361 mm
Head thickness 6.139mm
Total weight 12.399kN
Loading 854.222 N/m
Pressure stress 40.449 ⁄
Pressure stress 80.898 ⁄
Dead weight stress 0.00152 ⁄
Bending stress 0.0083 ⁄
(𝑢 ) 40.456 ⁄
( ) -40.438 ⁄

| P a g e 027
5.2_ MECHANICAL DESIGN OF REACTOR :
Condition
Operating Pressure (bar) 12
Material Carbon Steel
Catalyst weight, kg 24474.2
Operating Temperature c 137-177
Diameter, m 1.58
Height, m 6.32
Insulation thickness, mm 75

( )
ethanol diethyl ether

Design pressure, take as 10 per cent above operating pressure

= (12−1) 1.1= 12.1 𝑏
=1.21 /
Design temperature 150 c.
From Table 13.2 Chemical Engineering vol. 6 page 812, typical design
stress = 115 N/mm2

add corrosion allowance 8 + 2 = 10 mm

Elliposoidal head

| P a g e 028
Dead weight
Take =1.08
= +e
Dm =1.58 + 1(0 * 10^-3) = 1.59
𝑊v=240 ( +0.8 )
( )
Wv =36288.7577 N = 36.2887 KN

Weight of insulation:
mineral wool density = 130 kg/m3
𝑊 𝑓 𝑢 = 𝑢 9.81
𝑢 of insulation =
𝑢 of insulation = 3.14 * 1.58 * 6.32 * 75 * 10^-3 = 2.35 m^3
𝑊e 𝑓 𝑢 = 130 * 2.35 * 9.81
𝑊 𝑓 𝑢 = 2996.955 N = 2.996 KN
Double this to allow for fitting = 2.996 * 2 = 5.992 KN

Wt = 𝑊 +𝑊

Wt = 36.2887+ 5.992 = 42.2807

Wind loading
Take dynamic wind pressure as 1280 N/m2.
Mean diameter, including insulation ( + 75) 0.001= 1 75

Loading (per linear meter) =1280* 1.75 =2240 N/m

Bending moment at bottom tangent line

( )

| P a g e 029
Analysis of stresses

At bottom, tangent line

Pressure stresses:

Dead weight stress:

( ) ( )

Bending stresses:
Do = Di + 2t
0=1.58 *
( ) ( )

( )
𝑏 ( )

The resultant longitudinal stress is:

= + ∓ 𝑏

is compressive and therefore negative

(𝑢 )= – 1.74 + 2.2794= 48.3344
( )= 47.795 – 1.74 – 2.2794 = 43.7556

| P a g e 031
The greatest difference between the principal stresses will be on the
down-wind side
95.59 – 43.7556 = 51.8344 N/
well below the maximum allowable design stress

Check elastic stability (buckling)

( )
( )
= + 𝑏 = 1.74 + 2.2794 = 4.0194

Vessel support design :

Try a straight cylindrical skirt (Өs = 90˚)

Material of construction = carbon steel

F = 115 N/mm^2

Youngˊs models = 200000 N/mm

Skirt high = hs = 3 m

Skirt thickness = ts = 10 mm

Join factor = 0.85

Weigt of skirt

Approximate weight = H

Approximate weight = = 121559.8459 N

Approximate weight = 121.559 KN.

Weight of vessel = 42.2807 kN

SO total weight = 42.2807 + 121.559 = 163.8397 KN

| P a g e 030
Analysis of stresses of skirt .

1-Bending moment of skirt, Ms

Wind loading = 2.24 kN

Ms =0.5 * Fw(Hv + Hs)^2

=0.5 * 2.24( 6.32+3)^2 = 97.285 KNm

2-bending stress of skirt ,

( )
= ( )

3- deal weight sress skirt ,

( ) ( )

operation = ( ) ( )

4- resultant stress in skirt ,

Maximum , ( )

Maximum , ( )

Maximum ( )

Maximum ( )

| P a g e 032
5 - Criteria for design

J = 0.85 where Ө˚ =90

(tensile) < fsJ sin Ө˚

6.67 < 115*0.85*sin 90

6.67 < 97.75

s (compressive) < 0.125 E (ts/Ds) sinӨ˚

11.84 < 0.125 * 200000 (10/1580) sin 90

11.84 < 158.2278

Both criteria are satisfied, add 2 mm for corrosion, gives a design

thickness of 12 mm.

Base ring and anchor bolts:

1. Bolts smaller than 25 mm diameter should not be used.

2. Minimum number of bolts = 8 .

3. Use multiple of 4 bolts .

4. Bolts pitch should not be less than 600 mm.

 Approximate Pitch circle diameter, Db = 2.2 m

 Circumference of bolt circle = 2200π
 Closest multiple of 4, Nb = 12 bolts .
 Take Bolt design stress fb = 125 N/mm^2.
 Fw =2240 N/m.
 Ms =97.285 KNm .

Number of bolts required, at minimum recommended bolt spacing

= Circumference of bolt circle / 600 = 2200π / 600 = 11.5

Take W = operating value = 42.2807 kN

| P a g e 033
 , 𝑏= ⦌

, 𝑏= ⦌

Ab = 90 mm^2

Use M24 bolts (BS 4190:1967) root area = 353 mm^2 (from figure 11.30
volume 6) (page – 849)

Bolt root diameter =( ) =( )^0.5 = 11 mm

Total compressive load on the base ring per unit length:

𝑏= ⦌

Where, Fb = the compressive load on the base ring, Newtons per linear
metre Ds = skirt diameter, m

Fb = ⦌ = 58165.7339 N/m = 58.165 kN/m

The minimum width of the base ring:

𝑏 = ( b / 𝑓 ) * (1/ 10^3) .------ ref 7

Where,

 Lb = base ring width, mm

 fc = the maximum allowable bearing pressure in the concrete
foundation pad, which will depend on the mix sed, and will
typically range from 3.5 to 7 N/mm2 (500 to 1000 psi)

Taking the bearing pressure as 5 N/mm^2

Lb= (58.165*10^3) / 5*10^3 = 12 mm

This is the minimum width required; actual width will depend on the
=chair design. Actual width required ( figure 13.30 v6 ) .

Lr + ts + 50 mm

= 76+12+50= 138 mm

| P a g e 034
  Actual bearing pressure on concrete foundation:

The minimum thickness is given by.

√ ---------- ref 7

 Lr = the distance from the edge of the skirt to the outer edge of the
ring, mm.
 tb = base ring thickness, mm.
 f’c = actual bearing pressure on base, N/mm^2 .
 fr = allowable design stress in the ring material, typically 140
N/mm^2.

tb = 76 √ = 7.8 mm
round off to 8 mm

The chair dimensions from figure 13.30 for bolt size M24.

Skirt is to be welded flush with outer diameter of column shell.

| P a g e 035
 Flange Design
For input steam at T = 450 k
From material balance :
= ƒ = 1.59 /s

Calculate =

Material
Ethanol 1.25 0.857 46.07 1.07
Watwr 0.488 0.143 18.02 0.0698
Total =
1.14

For carbon steel

= 282 × (G)0.52 × (ρ) −0.37
= 282 × (1.59)0.52 × (1.14) 0.37

= 296

300 323.9 440 22 44 365 4 M20 12 22 395 355

99

| P a g e 036
 d1=323.9

d2=22

12
h=44
b=22
3
d3=355
d4=365
K=395
D=440

For output steam at T = 410.6 k

From material balance :
= ƒ = 1.59 /s

Calculate =

Material
DEE 2.20 0.548 74.12 1.21
ethanol 1.37 0.171 46.07 0.234
water 0.535 0.278 18.02 0.149
ethylene 0.833 0.00261 28.05 2.17 * 10^-3
total ∑x=1 = 1.60

| P a g e 037
 For carbon steel
= 282 × (G)0.52 × (ρ) −0.37
= 282 × (1.59)0.52 × (1.60) −0.37
= 268

250 273 375 22 44 312 3 M16 12 18 335 295

d1=273

d2=18

12
h=44

b=22
3
d3=295
d4=312
K=335
D=375

| P a g e 038
Summary

Shell thickness 10 mm
Total weight 42.2807 KN
Loading fw 2240 N/m
Pressure stress L 47.795
Pressure stress H 95.59
Dead weight stress w 1.74
Bending stresses b 2.2749
z (upwind) 48.3344
z (downwind) 43.7556

| P a g e 039
5.3 MECHANICAL DESIGN OF HEAT EXCHANGER
(E_1202)
Condition
Operating pressure 10 bar
Material Stainless steel
Density of material 7800 ⁄
Shell temperature 40 - 137.6 ℃
Tube temperature 25 – 40 ℃
Corrosion allowance 2 mm
Shell diameter 947.0034 mm
Number of tubes 1028.454
Insulation thickness 75 mm

𝑏 ⁄

 Design stress from table 13.2 vol.6

𝑓 ⁄

 Temperature design = 137.6 ℃

Shell thickness ( e ) ( )

Head thickness
 Elliposoidal

| P a g e 041
Dead weight of vessel

 Take Cv = 1.08

𝑊 ( )

𝑊 ( )

Weight of pipes

𝑊 𝑓

Weight of insulation

 Mineral wool density = 130 ⁄

Approximate volume of insulation

Approximate volume of insulation

Approximate volume of insulation

Weight of insulation 𝑓 𝑢

Double this to allow for fitting = 2748.7816 N = 2.7487 KN

| P a g e 040
Total weight

Shell 75.2352

Pipe 8.9002

Insulation 2.7487

86.8841 KN

Wind loading

 : dynamic wind pressure = 1280 ⁄

Mean diameter ( )

Mean diameter ( )

Loading ( per linear meter )

Bending moment at bottom tangent line ( Mx )

Analysis of stress

| P a g e 042
Dead weight stress

( )

⁄
( )

Bending stress

( )

( )

( )

( ) ⁄

The resultant longitudinal stress is

(𝑢 ) ⁄

( ) ⁄

 The greatest difference between the principal stress will be on the

down wind side

Well below the maximum allowable design stress

| P a g e 043
Check elastic stability (buckling)
Critical buckling stress:

( ) ( ) ⁄

The maximum compressive stress will occur when the vessel is not under
pressure .

well below the critical buckling stress. So design is satisfactory.

| P a g e 044
Support design
Choose saddle support must be designed to withstand the load imposed
by the weight of the vessel and content . they are constructed of bricks or
concrete, or are fabricated from steel plate . the contact angle should not
be less than 120° , and will not normally be greater than 150° . wear
plates are often welded to the shell wall to reinforce the wall over the area
of contact with the saddle .

The dimensions of typical "standard" saddle designs are given in figure

13.26 chemical engineering Vol.6 page 847.

 Choose the design that fit that diameter from fig 13.26.

| P a g e 045
Flange design
 For stainless steel

Where

𝑓 ⁄

 for input gases to shell at :

℃ ⁄

| P a g e 046
 for out put gases from shell at :
℃ ⁄

 for in put water to tube at :

℃ ⁄

 For out put from tube at :

℃ ⁄

| P a g e 047
Summary
Shell thickness 6.00480 mm
Head thickness 6 mm
Total weight 86.8841KN
Loading 1315. 84 N/m
Pressure stress 47.35017 ⁄
Pressure stress 94.70034 ⁄
Dead weight stress 0.8112 ⁄
Bending stress 3.6094 ⁄
(𝑢 ) 50.1483 ⁄
( ) 42.9295 ⁄

| P a g e 048
5.4_ MECHANICAL DESIGN OF ETHANOL COLUMN (T-
1202) :

Cylindrical shell thickness calculate by

Where: e = thickness of shell; mm

Pi = design pressure; N/mm2 = 0.1925N/mm2

Di = shell inside diameter = 0.998m =998 mm

f = design stress; N/mm2=126n/mm2

C= corrosion allowance = 2 mm for carbon steel material

Head thickness
Choose elipsoidal head :

3 mm

Supports:
Weight loads:

Weight of vessel (shell):

Wv = 240 CV Dm (HV +0.8 Dm)t ------( ref 7)

Where Wv = total weight of vessel excluding internal fitting N

CV = constant = 1.15

| P a g e 049
HV = height or length of shell m = 7.8 m

t = e= 3 mm shell thickness

Dm = Di + (t × 10-3) --------(ref 7)

Dm = 0.998 + (3 × 10-3) = 1.001 m.

Hv=16.5

Wv = 240*1.15*1.001(16.5+ 0.8* 1.001)*3 = 14324.4 N = 14.3244 KN

Plates

Weight of plate including liquid loading:

WP = 1.2 Ac N

Where: WP = total weight of plates KN

Ac = cross sectional area of column = 0.77 m2

N = number of plate =30

Wp = 1.2* 0.77* 30 = 27.72kN = 27720 N

Weight of insulation:

W = v*g*ρ

where : mineral wool density D 130 kg/m3

approximate volume of insulation =

= * 0.998* 16.5 * 0.075 = 3.879 m^3

W = 4946.8887N = 4.946 KN .

Double this to allow for fitting = 2* 4.946 = 9.892 KN

 So Total weight = Wv+Wp+Wi

14.3244+27.72+ 9.892 = 51.9364 KN .
Added 20% above total weight

= 51.9364 * 0.2 = 10.38728 KN = 10387.3 N

| P a g e 051
Wind load

Take dynamic wind pressure as 1280 N/m2.

Mean diameter, including insulation = Di +2(t+ t insulated). ----- (ref 7)

t for insulation take 75 mm.

Dm = 998+2(3+75) = 1154mm=1.154m

Loading (per linear metre) Fw = Pw * Dm = 1280*0.157 = 1477.12 N/m

Bending moment at bottom tangent line Mx =

Analysis of stresses

At bottom tangent line

Pressure stresses:

 Thickness is taken as 9mm as maxmimum

, ref (7)

1-Longitudial stresses due to pressure are given by :

 Circumferential stress due to pressure are given by :

Dead weight stress:

( ) ------------ ref 7

( )
( compressive) .

| P a g e 050
Bending stresses

∓ ( ) ------------(ref 7)

where ; Mx = is the total bending moment at the plane being considered

Iv = second moment of area of the vessel about the plane of

bending .

Iv ( )= ----------- (ref 7 ).

Where Do outside diameter = Di+2t = 998 +2* 9= 1016mm.

Iv= ( ) *10^9 mm^4

∓ ( ) ∓

The resultant longitudinal stress is:

∓ 𝑏

(dwonwind) = 5.336-1.824-0.0283 = 3.4837 N/mm^2

(Up wind) = 5.336-1.824+0.0283=3.54 N/mm^2

The greatest difference between the principal stresses.

( )

= 10.673-3.4837= 7.1893N/mm^2

Check elastic stability (buckling)

Critical buckling stress:

( ) ( )

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The maximum compressive stress will occur when the vessel is not under
pressure .

well below the critical buckling stress. So design is satisfactory.

Vessel support design :

Try a straight cylindrical skirt (Өs = 90˚)

Material of construction = carbon steel

F 126 N/mm^2

Youngˊs models = 200000 N/mm

Skirt high = hs = 4 m

Skirt thickness = ts = 9 mm

Join factor = 0.85

Weigt of skirt
Approximate weight = l

= = 119909 N
Wapprox = 119.9 KN.

SO total weight = 51.9364+119.9 = 171.8364KN

Analysis of stresses of skirt .

1-Bending moment of skirt, Ms
Ms =0.5 *Fw(Hv + Hs)^2
=0.5 * 1.447( 16.5+4)^2 = 304.05 KNm

| P a g e 053
2-bending stress of skirt ,

( )
= ( )

3- deal weight sress skirt ,

( ) ( )

operation = ( ) ( )

4- resultant stress in skirt ,

Maximum , ( )

Maximum ( )

5 - Criteria for design

Ө˚
( ) Ө˚
47.011 < 126*0.85*sin 90
47.011 < 107.1

| P a g e 054
 ( ) ( ) Ө˚

1.748 < 0.125 * 2*10^5 (9/998) sin 90

40.9759< 225.451 N/mm^2

Both criteria are satisfied, add 2 mm for corrosion, gives a design

thickness of 11 mm.

Base ring and anchor bolts:

1. Bolts smaller than 25 mm diameter should not be used.

2. Minimum number of bolts = 8 .

3. Use multiple of 4 bolts .

4. Bolts pitch should not be less than 600 mm.

 Pitch circle diameter, Db = 2.5 m


 Closest multiple of 4, Nb = 12 bolts .
 Blots design stress fb = 125 N/mm^2.
 Fw =1477.12N/m.
 Ms =304.05KN .
Number of bolts required, at minimum recommended bolt spacing

, = 1 / 𝑏𝑓𝑏 × ((4 * )/ 𝑏 – 𝑊)

= (1/16*125)*((4*304050))/3.2-51936.4) = 164.063 mm^2

Use M24 bolts (BS 4190:1967) root area = 353 mm2 ( from figure
11.30 ref 7.

( ) ( * )

= 14.453 mm

| P a g e 055
Total compressive load on the base ring per unit length:

𝑏 = (( 4 / ^2) +( 𝑊 / ))--------ref 7

Where, Fb = the compressive load on the base ring, Newtons per

linear metre

Ds = skirt diameter, m

(( * * ) ( * ))

=48.5248 KN/m

The minimum width of the base ring:

𝑏 = ( b / 𝑓 ) * (1/ 10^3) .------ ref 7

Where,

 Lb = base ring width, mm

 fc = the maximum allowable bearing pressure in the concrete
foundation pad, which will depend on the mix sed, and will
typically range from 3.5 to 7 N/mm2 (500 to 1000 psi)
Lb= (48.5248*10^3) / 5*10^3 = 9705 mm

This is the minimum width required; actual width will depend on the
=chair design. Actual width required ( fihure 13 .30 ref 7 ) .

Lr + ts + 50 mm

= 76+16+50= 142 mm

 Actual bearing pressure on concrete foundation:

𝑓𝑏
𝑓

| P a g e 056
The minimum thickness is given by.

𝑏 √ ---------- ref 7

 Lr = the distance from the edge of the skirt to the outer edge of
the ring, mm.
 tb = base ring thickness, mm.
 ’
 fr = allowable design stress in the ring material, typically 140
N/mm.

tb = 76 √ = 6.487 mm .
The chair dimensions from figure 13.30 for bolt size M24. Skirt
is to be welded flush with outer diameter of column shell.

| P a g e 057
 Design of Nozzles:

Optimum duct diameter, dopt = 293

Where:-

G=flowrate, kg/s
𝞺=density, kg/

 Feed Stream:
G=6.509 kg/s

𝞺=920.46 kg/

 dopt=293( ) ( ) = 62.3mm

 Nozzle thickness , t=

= ( )

So, the thickness of nozzle =corrosion +0.0048

=2+0.0048=2.0048mm

 Top Stream:
G=0.0216 kg/s

𝞺=2.61 kg/

 dopt=293( ) ( ) =47mm

| P a g e 058
  Nozzle thickness , t=

= ( )

So, the thickness of nozzle =corrosion +0.0036

=2+0.0036=2.0036mm

 Bottom Stream:
G=4.35kg/s

𝞺=947 kg/

 dopt=293( ) ( ) = 70mm

 Nozzle thickness , t=

= ( )

So, the thickness of nozzle =corrosion +0.0053

=2+0.0053=2.0053mm

| P a g e 059
 Flange design:-

The summary of flange design

Feed stream:

50 60.3 140 14 28 90 3 M12 4 14 110 80

d1=60.3

d2=14

12
h=28

b=14
3
d3=80
d4=90
K=110
D=140

| P a g e 061
 Top stream:

40 48.3 130 14 26 80 3 M12 4 14 100 70

d1=48.3

d2=14

12
h=26

b=14
3
d3=70
d4=80
K=100
D=130

| P a g e 060
 Bottom stream:

65 76.1 160 14 32 110 3 M12 4 14 130 100

d1=76.1

d2=14

12
h=32

b=14
3
d3=100
d4=110
K=130
D=160

| P a g e 062
REFERENCES :
1- Chemical - Reaction - Engineering Third Editionn - Octave Levenspiel

Department of Chemical Engineering Oregon State University.

2- CARL. YAWS – PROFESSOR OF CHEMICAL ENGINEERING –

LAMAR UNIVERSITY BEAUMONT, TEXAS.

3- Coulson & Richardson's - CHEMICAL ENGINEERING –(VOLUME

6) THIRD EDITION - Chemical Engineering Design - R. K. SINNOTT -
Department of Chemical and Biological Process Engineering University
of Wales Swansea.

4- Coulson and Richardson’s - CHEMICAL ENGINEERING -

(VOLUME 2) FIFTH EDITION - Particle Technology and Separation
Processes.

5- CHEMICAL ENGINEERING THERMODYNAMICS -EIGHTH

EDITION.

6- Heat Transfer - Tenth Edition - J. P. Holman - Department of

Mechanical Engineering Southern Methodist University.

7- https://diethyl-ether.weebly.com/weebly

8- https://webwiser.nlm.nih.gov/substance?substanceId=
11&identifier=Diethyl%20Ether&identifierType=name&menuItemId=62
&catId=83

9- https://www.slideshare.net/100003690100228/diethyl-ether-dee-
literature-review

10- https://byjus.com/chemistry/diethyl-ether/Byjus

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