‫ﺟﺎﻣﻌﮫ اﻟﻘﺎدﺳﯾﺔ‬

‫ﻛﻠﯾﺔ اﻟﮭﻧدﺳﺔ‬
‫ﻗﺳم اﻟﮭﻧدﺳﺔ اﻟﻛﯾﻣﯾﺎوﯾﺔ‬
‫‪٢٠١٨-٢٠١٧‬‬

‫)‪(Production of glyoxylic acid‬‬

‫إﴍاف‬

‫ﻋﺒﺎر‬ ‫أ‪.‬م‪ .‬د‪ .‬ﻋﲇ ﺣﺴ‬

‫اﻋداد اﻟطﻠﺑﺔ‬
‫زﯾﻧب ﺣﺎﻛم ﻣﺎﺟد‬
‫ھﺑﮫ زﯾد ﻓﯾﺻل‬
‫ﻧور اﻟﺣﺳﯾن ﺣﻣﯾد ﻋﺑود‬
 Table of content

Chapter one: introduction

Chapter Two: material balance

Chapter Three: energy balance

Chapter four: equipment design

Chapter five: cost estimation

References
Objective
1.Design plant for production glyoxylic acid from glyoxal with a production
capacity

2.Determine the materials requirements by using the principal of material

balance

3.Determine the total energy requirements

4.Design two equipment based on the principal of chemical engineering design

5.Calculate the total cost of the project

Recommendation
1.Design a process for produc on glyoxylic acid by electrochemical
method

2. Production of glyoxylic acid from glycolic acid

3.Design a process for produc on of tartaric acid

4.Design a process for produc on glyoxal

 Chapter one

Introduction

Glyoxylic acid [298-12-4], oxoacetic acid, glyoxalic

acid,OHC COOH,Mr 74.04, is the simplest
α-oxocarboxylic acid. It is found in plants
and is involved in the metabolic cycle of animals.
Glyoxylic acid was discovered by Debus,
who established its formula in 1856[1]. This
formula was described more precisely as that of
the monohydrate by Perkin in 1868 [2].
In 1957, Kronberg and Krebs established
the basis of the metabolic cycle of glyoxylic acid
[3], in which isocitrate is converted to malate
anaerobically. This cycle acts as a relay in the
tricarboxylic acid cycle under specific conditions
of plant life, such as germination or ineffective
photosynthesisGlyoxylic acid is strongly hydrophilic; it is
soluble in alcohols, with which it reacts, and in
water-miscible solvents; it has a very low solubility
in ether and other organic solvents.
Glyoxylic acid forms complexes with ions of
alkali and alkaline-earth metals; it has a chelating
power of 30 mg of calcium carbonate per
gram of 50% aqueous solution

Glyoxylic acid Quick Details

Chemical Name: Glyoxylic acid

CAS No.: 298-12-4
EINECS No.: 206-058-5
Molecular Formula: C2H2O3

Chemical Structure:
Formula Weight: 74.04
Assay: 50%MIN
Other Names: Oxalaldehydic acid,2-oxoethanoic acid ,GA,oxo-
acetic acid,Glyoxalate, Glyoxylic acid 50% in water;glyoxylic
acid free acid;aldehydoformic acid;alpha-Ketoacetic acid.
Capacity: 5500MT per month.
Sample: available
Applications Summary : hair dye,hair care product,skin care
product

Glyoxylic acid Typical Properties

Item Specification Test Result

Appearance Yellowish transparent liquid Conform
Glyoxylic Acid 50.0%±0.5% 50.18%
Oxalic Acid ≤1.0% 0.83%
Nitric Acid ≤0.20% None Detected
Glyoxal ≤0.25% 0.21%
Conclusion Qualified

Glyoxylic acid Packaging and Shipping

 plastic drum, 25 kg
 plastic drum, 250kg
 IBC drum, 1250 kg
 ISO-TANK, 20-22tons
 Physical Properties

 Glyoxylic acid crystallizes as the monohydrate
 upon controlled concentration of its aqueous
 solution. When dry, the monohydrate has mp
 52 – 53 C. The dissociation constant in aqueous
 solution is 4.7×10 4, and the specific heat
 is 1.80 kJ kg 1 K 1. The density of a 50%aqueous
 solution of the acid, the only commercial
 form, is 1.34 g/cm3 at 20 C; the refractive index
 (20 C) is 1.416.
 The use of NMR spectroscopy (1H [6] and
 13C[7]) has confirmed the existence of glyoxylic
  acid in aqueous solution as the
dihydroxy acid (1) together
with a small Proportion of the
linear dimer (2)
'

Chemical Properties

Glyoxylic acid contains two functional groups:

the carbonyl group, which undergoes reactions
characteristic of aldehydes (→Aldehydes,
Aliphatic and Araliphatic, Chap. 2.2.), and the
carboxylic acid group (→Carboxylic Acids,
Aliphatic, Chap. 3.). The aldehyde group reacts
readily with nucleophilic reagents; the hydrated
aldehyde and the hemiacetal react similarly.
With ambident nucleophiles, the carboxylic
group may also react, leading to intramolecular
ring formation. Various heterocyclic compounds
can be obtained by coupling polynucleophiles
with glyoxylic acid: o-phenylenediamine
[95-54-5] gives 2-hydroxyquinoxaline
[1196-57-2] (3); with urea [57-13-6] and an acid
catalyst, allantoin [97-59-6] (4) is obtained in 60% yield [8]

Glyoxylic acid Applications

1.Used as material for methyl vanillin, ethyl vanillin in flavor
industry.
3.Used as intermediate of varnish material, dyes, plastic,
agrochemical, allantoin and daily-use chemical etc.It is popular
in the cosmetic industry,for hair dye;hair care product;skin
care product ect.
4.Glyoxylic acid is the material for water
purificants,pesticides.It is used as an intermediate of varnish
material and dyes.
5.Glyoxylic acid also can be used in the preservation of food,as
a crosslinking agent of polymerization and as a plating 2.used
as intermediate for D-hydroxybenzeneglycin, broadspectrum
antibiotic, ,acetophenone ,amino acid etc.

Glyoxylic acid storage

1.keep sample bottle tightly sealed.
2.Store in cool ,dry place in tightly closed containers. Ensure go
od ventilation at the workplace.
3.store away from oxidizing agents.

Production of glyoxylic acid from glycolic acid

A process for the production of glyoxylic acid involving the enzymatic
oxidation of glycolic acid. The process provides a commercially practical
method involving the reaction of glycolic acid in an aqueous solution at a
starting concentration range of 200 mM to 2,500 mM in the presence of
oxygen, glycolate oxidase and catalase at a pH of 7 to 10 and in the
presence of an amine such as ethylenediamine, or tris(hydroxymethyl)-
methylamine
A method for glyoxylic acid production using cells
of Alcaligenes sp. GOX373
Abstract
A microorganism capable of producing glyoxylic acid from glycolic acid
was obtained by enrichment culture with medium containing 1,2-
propanediol. It was identified as a member of the genus Alcaligenes.
The maximal activity of the enzyme responsible for glyoxylic acid
production by this strain was obtained by culturing at pH 5–6 for 3 days.
The optimum conditions of cell-reaction for accumulating high
concentration of

glyoxylic acid were investigated using lyophilized cells of 3 day

cultivates: Approximately 0.65 M glyoxylic acid was produced from 1.0 M
glycolic acid by incubating at 20°C for 15 days in 1.0 M TES-NaOH
buffer, pH 7.0. In addition, the production of glyoxylic acid was improved
to 0.95 M by keeping the reaction pH at neutral region. The method
proposed here is more advantageous for glyoxylic acid production than
enzymatic methods.
crystallization. The solution is then
purified by passage through an anion-exchange
resin or by electrodialysis, which removes the
residual nitric acid.
Glyoxal can also be oxidized at the anode of
a two-compartment electrolytic cell in the presence
of chloride ion [20]. Glyoxylic acid may be
 Chapter two

Oxidation

F2
31.6% HNO3

68.4 H2O

Oxidation
Glyoxal 28.3%
F1 F3Glyoxlic acid 15.329%
H2O 71.7% Glyoxal acid 1.387%
F4

NO2 1% HNO3 3.974%

50000 Tan/year Oxalic acid 0.7877%

50000*1000/300*24=6944 kg/hr H2O 78.521%

6944/74.64=93.03 kmol

Yelid=84%

Conv=90%

0.84=6944/amount of glyoxal=8267.2 kg/hr

F1=glyoxal+H2O

Glyoxal=8267.2/0.283=29212.7Kg/hr

Amount of 8267.2-29212.7=20945.5 Kg/hr

Mole of glyoxal =29212.7/58.04=142.4 Kmol/hr

Reacted from glyoxal =128.2 Kmol/hr

Themean reaction

CHO-CHO+HNO3 CHO-COOH+NO+H2O

128.2 128.2 128.2 128.2

Mole ra o = 1/1.2 = 142.4/HNO3 =glyoxal/HNO3

HNO3 =170.88Kmol/hr =10765.44 kg/hr=10765.44//0.316 =34067.83 kg/hr

Amount of H2O =34067.83 - 10765.44 =23302.41

Amount of glyoxlic acid in output =128.2*0.96 =123

128.2 -123 = 5.2

CHO-CHO+HNO3 CHO-COOH+NO+H2O

5.2 5.2 5.2 5.2

HNO3reacted =128+5.2=133.4 Kmol/hr =10765.44 – 133.4 =2361.24

Glyoxal in F3 =142.4 -128.2 = 824.2

H2O = 1163.6 + 1294.58 + 6 + 128.2 = 46662.8 Kg/hr

Glyoxlic acid = 123* 74.04 =9106.92 Kg/hr

Oxlic acid = 5.2* 90=468 Kg/hr

NO2 = (128.2+ 5.2) * 30 =4002 Kg/hr

materal F1 Wt% F2 Wt% F3 Wt% F4 Wt%

glyoxal 8267.2 0.283 824.2 0.01387
glyoxlic 9106.42 0.15329
oxlic 468 0.007877
HNO3 10765.44 0.31054 2361.24 0.03974
H2O 20945.5 0.71699 23302.41 0.67217 46649.11 0.78521
NO2 4002 1
Exchange resin HNO3 1%
Glyoxalic acid 15.329%
F5 F5 F7
Glyoxal 1.387% Exchange Glyoxalic 50%
F3
HNO3 3.974% resin H2O 50%

Oxalic acid 0.7877% F6

H2O 78.521% Oxalic acid 0.7576%

H2O 152.192%

Glyoxal acid 1.3342%

Glyoxalic acid 18.5656%

 Material F3 Wt% F5 Wt% F6 Wt% F7 Wt%
Glyoxlic 9106.92 0.15329 11468.16 0.185656 2361.24 0.5
glyoxal 824.2 0.01387 824.2 0.013342
HNO3 2361.24 0.03974 2361.24 1
Oxlic 468 0.007877 468 0.007576
H2O 46649.11 0.78521 94010.35 1.52192 2361.24 0.5

Exchange resin

Oxalic acid 1%
OxalicFacid F6 F8
0.7576%
f
Glyoxal
1.3342% EXChange2
H2O 79.948%
Glyoxlic
Glyoxal 1.3444%
18.5656%
Glyoxlic 18.707% F9
material F6 Wt% F8 Wt% F9 Wt%
Oxlic acid 468 0.007576 468 1
glyoxal 824.2 0.013342 824.2 0.013444
Glyoxlic 11468.16 0.185656 11468.16 0.18707
H2O 49010.35 1.52192 49010.35 0.79948
 Evapora on[1]

Glyoxal 1.344% F9
Glyoxlic acid 18.707% H2O 1%
Evaporation
H2O 79.948%
d

F13 F12

Glyoxal 4.6714%

Glyoxlic acid 65%

H2O 30.33%

Glyoxalic acid in F13 =11468.16 =11468.16/0.65 = 1764.32

=glyoxalic acid + glyoxal + H2O

H2O = glyoxlic acid + glyoxal

= 1764.32 –( 11468.16 + 824.2 ) =5350.96

The amonut of H2O evaport = 49010.35 – 4350.96 = 43659.39

material F9 Wt% F12 Wt% F13 Wt%

H2O 49010.35 0.79948 43659.39 1 5350.96 0.3033
Glyoxlic 11468.16 0.18707 11468.16 0.65
glyoxal 824.2 0.013444 824.2 0.04671
 COOLER[1]

Glyoxal 4.6714%
Glyoxal 4.6714% F14 H2o 30.33%
glyoxlic 65%
Glyoxlic 65%
F13
H2O 30.33% COOLER[1]

50% of glyoxlic acid is solid = 11468.16 * 0.5 = 5734.08 Kg/hr

material F13 Wt% F14 Wt%

Glyoxlic acid 11468.16 0.65 5734.08 0.65
Glyoxal 824.2 0.046714 824.2 0.046714
H2O 5350.96 0.3033 5350.96 0.3033
 Filter

Glyoxal 7.3118%
F14
Glyoxlic diss 50.869%
Filter
F15
Glyoxalic 32.5% H2o 41.818%

Glyoxal 4.671%

H2o 30.32%
F15 F16

Soild 90%

H2O 10%

Total = 5734.08/0.9 = 6371.2

= 6371.2 – 5734.08 = 637.12

= 5350.26 – 637.12 = 4713.84

Material F14 Wt% F15 Wt% F16 Wt%
Glyoxlic diss 5734.08 0.325 5734.08 0.50869
Glyoxlic soild 5734.08 0.325 5734.08 0.9
glyoxal 824.2 0.04671 824.2 0.073118
H2O 5350.26 0.3032 637.12 0.1 4713.84 0.41818
 Evapor on[2]

glyoxal 9.342%

glyoxlic 65%
H2o 41.818%
Evapor on[2]
Glyoxal 7.3118% F16
H2o 25.65%

Glyoxlic 50.869% F18

F17 H2O 1%

Total = 5734.08 / 0.65 = 8821.66

= 8821.66 –( 5734.08 + 824.2 ) =2263.38

H2O evaporat = 4713.84 – 2263.38 = 2450.46

material F16 Wt% F17 Wt% F18 Wt%

Glyoxlic 5734.08 0.50869 5734.08 0.65
H2O 4713.84 0.41818 2450.46 1 2263.38 0.2565
Glyoxal 824.2 0.073188 824.2 0.09342
 COOLER

Glyoxal 9.342%
Glyoxal 9.342%
H2o 25.65%
COOLER[20]
Glyoxlic 65% C Glyoxlic diss 32.5%
F19
H2O 25.65%l
Glyoxlic soild 32.5%
F18

50% glyoxlic acid soild

= 5734.08 * 0.5

= 2867.04

Material F18 Wt% F19 Wt%

Glyoxlic soild 2867.04 0.325
Glyoxlic diss 5734.08 0.65 2867.04 0.325
H2O 2263.38 0.2565 2263.38 0.2565
Glyoxal 824.2 0.09342 824.2 0.09342

Filter

F19 F21

Glyoxlic soild
32.5% filter Glyoxal 14.62%

Glyoxlic diss 32.5% Glyolic 50.86%

F20
Glyoxa l 9.342% H2o 34.499%

H2O 25.65% Soild 90%

H2o 9.99%

2867.04/0.9= 3185.6

H2O = Total - glyoxlic acid soild

 = 3185.6 – 2867.04 = 318.96

H2O out =2263.38 – 318.96 = 1944.42

Material F19 Wt% F20 Wt% F21 Wt%

Glyoxlic soild 2867.04 0.325 2867.04 0.9
Glyoxlic diss 2867.04 0.325 2867.04 0.5086
H2O 2263.38 0.2565 318.52 0.0999 1944.42 0.34499
glyoxal 824.2 0.09342 824.2 0.1462

Dryer

F22

F20

Dryer Total
Glyoxlic 90%
9533.64
H2O 9.99%

H2O

955.78
F23
 Chapter three

Energy Balance

1-Oxidation
Cp=1.80 kj/kg.kglyoxlic acid=48.56

Cp=1.044 j/g.k =28.16 j/mg.k

H2o A B C

92.053 -3.9953*10^-2 -2.1103*10^-4

HNO3 214.478 -7.6762*10^-1 1.4970*10^-3

NO 33.227 -2.3628*10^-2 5.3156*10^-5

Oxalic 0.206 5.0483*10^-1 -4.6577*10^-4

Q= H

H=H out- H in

H in=H1-H2

H1=0 ,H2=0

H out=H3+H4

H4=m cp T

Cp=33.227+(-2.3628*10^-2)(353)+5.3156*10^-5

=24.886

H4=174*24.886*(353-298)

=238159.02 J

H5=H glyoxalic acid +H glyoxal +H H2O+H HNO3+H oxalic

H glyoxalic =m cp T

= 122.9*48.56(353-298)
 =328241.32 J

H glyoxal =14.2*28.16*(353-298)

=21998.8J

H H2O =m cp T

Cp = a+bT+cT^2

=92.053+(-3.9953*10^-2)(353)+(-2.1103*10^-4*353)^2

=77.95J/mole.k

HH2O =2591.6*77.95*(353-298)

=11110910.94J

H Oxalic =m cp T

Cp =0.206+5.0483*10^-1*353+(-4.6577*10^-4*353)^2

=178.4

H Oxalic=178.4*5.2*(353-298)

=51022.4J

H HNO3 =m cp T

Cp =214.478+(-7.6762*10^-1*353)+(1.4970*10^-3*353)^2

= -56.212

=42.16*-56.212*(353-298)

=-130359.8

H out =51022.4+11110910.94+328241.32+21998.8-130359.8

=11381813.66 J

H =H out-H in

H=11381813.66J for physical system

 H+ Hr=for chemical system

CHO-CHO+HNO3-CHO-COOH+NO+H2O

Hr= Hf (product)- Hf(react)

Hf H2O =-285.48KJ/mole

Hfglyoxal =-193.86 J/mole

Hfglyoxlic = -458.75KJ/mole

H NO=90.37KJ/mole

Hf HNO3 = -173.23KJ/mole

Hr = (90.37*128.2-458.75*128.2-285.84-(-193.86*128.2-173.23*128.2)

= -36810.066

Q = H + Hr

= 11381813.66*10^-3 -36810.066

= -25428.25

2-Evaporation
H = H out – H in

H in =Hf9

H out = Hf12 + Hf13

H in = H glyoxalic + Hglyoxal + H H2O

H glyoxalic =m cp T

= 154.84*48.56*(313 – 298)

=112821.87

H glyoxal =14.2*28.16*(313-298)
 =5998.08

Cp H2O = 92.053 – 3.9953 *10^-3*313+(-2.1103*10^-4*313)^2

=79.55

H H2O|=2722.74*79.55*(313-298)

=3248977.73

H in =3248977.73 + 5998.08 + 112821.87

= 3367797.73

F12 H H2O = m cp T + ƛm

=2425.52*79.55*(313-298)+2425.52*2406.0

=8730052.86

F13 = H H2O + H glyoxal + H glyoxalic

H H2O = 297.27 * 79.55 * (313-298)

= 354724.05

H glyoxalic =154.89*48.56*(313-298)

=112822.90

H glyoxal =14.2*28.16*(313-298)

=5998.08

H F13=5998.08+112822.90+354724.05

=473545.03

H out=Hf12+Hf13

=2894251.74+473545.03

=3367796.77

H=H out-H in

=3367796.77-3367797.73
 =+5835800.16

3-Cooler(1)
H in =Hf13

H out=Hf14

H13=H H2O+H glyoxal+Hglyoxalic

=354724.05+112822.90+5998.08

=473545.03

H14=glyoxalic(s)+glyoxal(l)+glyoxal+H2O

H glyoxal=14.2*28.16*(293-298)

=-1999.36

H H2O=m cp T

Cp=92.053-3.9953*10^-2*293+(-2.1103*10^-4)

=80.35

H H2O=297.27*80.35*(293-298)

=-119428.22

H glyoxalic=m Cp T

=154.89*48.56*(293-298)

=-37607.29

H out =-1999.36-119428.22-37607.29

=-159034.87

H=H out –H in

H=-159034.87-473545.03

=632579.9
4-Filter
H in= H f14

H out =H f15+Hf16

H in=-159034.87

H 15=h glyoxalic (s)+H H2O

H glyoxalic=m cp T

=77.44*48.56*(293-298)

=-18802.43

H H2O=m Cp T

=35.39*80.35*(293-298)

=-14217.93

H 15=-33020.36

Hf10 =H glyoxalic +H glyoxal+H H2O

H glyoxalic=-18802.43

H Glyoxal=-1999.36

H H2O=261.88*80.35(293-298)

=-105210.29

H16=-18802.43-1999.36-105210.29

= -126012.08

H out =Hf15+Hf16

= -33020.36 – 126012.08

= -159032.44

H = H out – H in

=-159032.44-(-159034.87)
 H = 2.43

5-Evaporater
H in = Hf16

H out = Hf17 + Hf18

H in =H glyoxalic +H glyoxal + H H2O

H glyoxalic = 77.44*48.56*(313-298)

= 56407.29

H H2O = 261.1*79.55*(313-298)

=311557.57

H in = f16 =373962.94

H17 H2O = m cp T +mƛ

=136.13*79.55*(313-298)

=489965.9

Hf18 = H glyoxalic + H glyoxal + H H2O

H glyoxalic = 77.44*48.56*(313-298)

= 56407.29

H glyoxal = 5998.08

H H2O = 125.74*79.55*(313-298)

= 150039.255

Hf18= 212444.62

H out =Hf17 + Hf18

= 489965.9 + 212444.62

= 702416.52

H = H out – H in
 = 702410.52 -373962.94

= 328447.58

6-Cooler(2)
Hf18 = 212444.62

F19 = H glyoxalic + H glyoxal + H H2O

H glyoxal = 14.2*28.16*(293-298)

= -1999.36

H glyoxalic = m cp T

= 77.44*48.56*(293-298)

= -18802.43

H H2O = m cp T

= 125 .74*80.35*(293-298)

= -50516.045

H out = -1999.36 – 18802.43 – 50516.045

= -71317.83

H = H out – H in

= -71317.83 – 212444.62

= - 283762.45
 Chapter four

1- evaporator :
we select forced circulation evaporator it consist of heat exchanger &condenser
& drum as shown in fig.6.1

Fig.6.1.circulation magma evaporator

Qsteam=Δh.msteam
H150=2752.26 ,H143=597.41
ΔH=2154.85
Q=2154.85 ,msteam=2708.21 Kg/hr
Ts=143 C T=80
Material Property of material Inlet stream at Outlet stream at 80 C
40 C
Specific heat KJ/kg.K 79.55 79.55
H2O Density Kg/m3 1000 1000
Thermal conductivity W/m.K 0.65 0.65
Specific heat KJ/kg.K 28.16 28.16
glyoxal Density Kg/m3 1270 1270
Thermal conductivity W/m.K 0.155 0.155
Specific heat KJ/kg.K 48.56 48.56
Density Kg/m3 1490 1490
glyoxylic
Thermal conductivity W/m2.K 0.168 0.168
Specific heat KJ/kg.K
Density Kg/m3
Thermal conductivity W/m.K
N/m2.S viscosity ………

Density (mean)=∑Xiρi
Density(ρmean)=XρH2O+Xρglyoxal+Xρglyoxylic
ρmean=(0.79948*1000)+(0.01344*1270)+(0.18707*1490)
ρmean=1095.2831kg/m3
viscosity (µmean)=∑Xiµi
viscosity(µmean)=7.9948*10^-4 m pa s
Specific heat Cpmean=∑Xicpi
Cpmean =X H2O *Cp H2O + X glyoxal*Cpglyoxal + X glyoxylic*Cpglyoxylic
Cpmean= 0.79948*79.55 + 0.01344*28.16 + 0.18707*48.56
= 73.0612236 KJ/Kg.K
Thermal conductivity (Kmean)=∑Xiki
kmean= X H2O *K H2O+ X glyoxal*K glyoxal + X glyoxylic*K glyoxylic
 kmean = 0.79948*0.196+ 0.01344*0.155 + 0.18707*0.168
kmean= 0.19020904 w/m. K
At T=80 C
Density (mean)=∑Xiρi
ρ mean =1095.2831 Kg/m3

Liquid phase
Density (ρmean)L= X H2O *ρ H2O
(ρmean)L=0.79948*1000
(ρmean)L=799.48 kg/m3

we will use the shell & tube heat exchanger type in this process
Calculate temperature mean
T1=temperature of hot fluid (steam) inlet
t1=temperature of cold fluid (stream 22)
T2=temperature of hot fluid (steam) outlet
t2=temperature of cold fluid (stream 27)
Tlm= [(T1-t2)-(T2-t1)]/ln[(T1-t2)/ (T2-t1)]

Tlm= [(150-80)-(143-50)]/ln[(150-80)/ (143-50)] = 80.9 C

We don’t need correction factor because the heat exchanger is vertical &the
pipes &the cover is contain one pass & we could not use more one pass
because we have a solution of a mixture(slurry).
From table 12.1 volume 6 the overall coefficient for (vaporizer)Steam&light
organics solution(1000-1500) so we take U=1250W/m2. C
U =1250w/m2. C
QH=UA Tm
A =(5835800.16*103J/3600s)/(1250W/m2. C*80.9 C)
A =55.506m2
Using a split-ring floating head exchanger for efficiency and ease of cleaning.
fluid is corrosive, and the operating pressure is not high, so a plain stainless
steel
can be used for the shell and tubes.
The mixture is contain solid material &corrosive so we put it in the tubes and
the
steam in the shell.
Use 25 mm outside diameter, 21 mm inside diameter, 2 m Long tubes
(a popular size) on a triangular pitch
pitch/outside tube diameter= 1.25
pitch=1.25*0.025=31.25m
Area of one tube (neglecting thickness of tube sheets)= *d*L
Area of one tube = *0.025*2
Area of one tube =0.1571 m2
Number of tubes = 15.25m2/0.1571 m2=97.07tubes

For 4 pass Number of tubes=97.07/4= 24.28 tubes per pass

Check the tube-side velocity at this stage to see if it looks reasonable.
Tube cross-sectional area(Ac)=( /4)*d2
(Ac)=( /4)*(0.021)2
(Ac) = 0.0003461 m2
area per pass = 24.28* 0.0003461 =0.01m2
Volumetric flow(Vt) =maverag/(3600*ρaverage)
Volumetric flow(Vt) =61302.71kg/(3600s*3760kg/m3)
(Vt) =4.5288*10^-3 m3/s
Tube-side velocity, ut= Vt / area per pass
 ut= 4.5288*10^-3m3/s /0.01m2
ut=0.45288m/s
The velocity is satisfactory
From Table 12.4, for 4 tube passes, K1 =0.175, n1 = 2.285,
Bundle diameter Db= do(Nt/k1) (1/n1)
Db=0.025(97.07/0.175)(1/2.285)
Db =0.4m

Figure6.2.Shell bundle clearance

For a split-ring floating head exchanger the typical shell clearance from Figure
6.2=57mm=0.057m
the shell inside .diameter Ds=Db +shell clearance
Ds=0.4+0.057=0.47m

Tube-side heat transfer coefficient(in our case we have two phase)

Re=ρ*ut*di/µ
Re=1000*0.45288*0.021/ 0.001
Re= 9510.48
Pr=µ*cp/k
Pr= 0.001*4200/0.65
Pr=6.4615
L/di=2000mm/0.021mm=95.24
 Figure 6.3.tube side heat transfer factor

From Figure ( 6.3 )jh=0.003

Nu =hi*di/kfL=jh*ReL*prL0.33*(µ/µw)0.14
µ=fluid viscosity at the bulk fluid temperature Ns/m
µw=fluid viscosity at the wall temperature Ns/m
hfc=the forced convection coefficient
ReL=ρL*ut*di/µL
Re=1000*0.45288*0.021/0.001
Re=9510.48
Prl=µL*cpL/kL
Pr=0.001*4200/0.65
Pr=6.4615
(µ/µw)0.14this term is neglect because it is small
hi=0.003*4325.7*(32.45)0.33*0.5332/0.021
hi=hfc =1038.8W/m2. K

1/Xtt=[0.815/(1-0.815)]0.9*[880.1157/0.425]0.5*[0.386*10-3/0.0047]0.1
1/Xtt= 134.6206
 Figure 6.4.convective boiling enhancement factor

From figure (6.4 )fc=70

h′fc=the convection boiling coefficient
h′nb=the nucleate boiling coefficient
hcb=the effective heat transfer coefficient
fc=the convective boiling coefficient factor
h′fc=hfc*fc
h′fc=1038.8 *70= 72716W/m2. K
from volume 1 the the nucleate boiling coefficient h′nbis 20 percent from the
effective heat transfer coefficient hcb
The contribution from nucleate boiling is lower than that from convection
through the liquid film .It may be assumed to be 20% of the total heat
transfer.(12)

hcb= 72716+0.2hcb
0.8hcb = 72716
hcb = 72716/0.8=90895w/m2. K

Shell-side heat transfer coefficient

 In our case we have condensation of steam

As a first trial take the baffle spacing= Ds/2=0.47/2=0.235m

This spacing should give good heat transfer without too high a pressure drop.
As=(pt-do)DsLB/pt
As=(31.25-25)*0.47*0.235*/31.25
As =0.02209m2

de=4.5288/25*(31.25^2-0.917*25^2)
de=17.75mm
Volumetric flow of shell side (Vs) =msteam/(3600s*ρ)
(Vs) = 2708.21kg /(3600s*1000)
(Vs)= 0.4762m3/s
Shell side velocity ut=(Vs) /As = 0.4762/0.03=15.8733m/s
Re=ρ*ut*de/µ
Re=1000* 4.5288 *0.01775/0.001
Re =268.9173
Pr=µ*cp/k
Pr=0.001*4200/0.65
Pr =0.488

Ӷv=4368.3/(3600*97.07* *0.025)
Ӷv = 0.1336kg/m.s
Rec=4* 0.1336/0.00011 = 4858.2
Pr=µL*cpL/kL
3
Pr=0.00011*4.32*10 /0.685=0.6937
 Figure 6.5Condensation coefficient for vertical tubes

from fig.( 6.5)

0.18

hc=0.18*0.685[(0.00011)2/{918*(918-2.548)*9.81}]-1/3
hc=10849.6824w/m2.K
from table 12.2 fouling factor hod=7500w/m2.C ,hid=5000w/m2.C
thermal conductivity for stainless steel=16.269w/m. C

the overall coefficient based on the outside area of the tube W/m2.C=Uo
outside fluid film coefficient W/m2.C=ho
inside fluid film coefficient W/m2.C=hi
outside dirt coefficient (fouling factor) W/m2.C=hod
inside dirt coefficient W/m2.C=hid
thermal conductivity of the tube wall material W/m2.C=kw
tube inside diameter m=di
tube outside diameter m.=do
 1/Uo= (1/10849.6824)+ (1/7500)+[0.025*ln(25/21)/(2*16.269)]
+[25/(5000*21)] +[25/(21*1038.8W)]
Uo=1275.3W/m2. C
Tube side pressure drop

(µ/µw=neglect this term is small)

Figure6.6. tube side friction factor

From fig.(6.6) jf=0.006

ΔPt=4*[8*0.006*(2000/21)+2.5]*690.4838*[(1.1)2 /2]
ΔPt=11816N/m2
ΔPtotal for tubes=ΔPt*fc= 11816*70=827120N/m2

shell side pressure drop

Δp=Np[8jf(Ds/De)(ρus2/2)*µ/µw]
Np=4 ,Re=268.9173 ,us=15.8733m /s ,ρ=2.548kg/m3 ,Ds=0.49m ,de=0.01775m
(µ/µw=neglect this term is small)
Figure 6.7 shell side friction factor,segmental baffles

From fig.( 6.7) jf=0.07

ΔPs=4[8*0.07*(0.49/0.01775)*2.548]*[(15.8733m)2/2]
ΔPs=19850N/m2
2-cooler
Tm=(T1-T2)-(T2-T1)/Ln T1-T2/T2-T1

=(40-25)-(20-10)/Ln 40-25/20-10 =12.33

S=T2-T1/T1-T2=25-10/40-10=0.5

R =T1-T2 /T2-T1 =40-20/ 25 -10 =1.3

From gure(12-19) page(567)

F=0.8

Tm =F* Tm

= 0.8*12.33 =9.864

A = Q/U Tm = 632579.9/ 500*9.864 =128.26 m^2

Assume OD = 20mm

ID=16 mm L=4.83 m

Area of one Tube =4.83*20*10^-3*π =0.303 m^2

NO of Tube = 128.26/0.303 = 423

D b = do (N1/K1)^1/N1

D b = dandle diameter mm

Do =Tube outside diameter mm

K1and n1 can found table (12-4) page (649)

K= 0.249 N=2.207

D b =20(423/0.249)^1/2.207 =581.5 mm

Use a split – ring o ng head type from gure(12-10) vol(6) dandle diameter
clearance = 60

Shell diameter Ds = D b+ D b

581.5+60 = 641.5 mm
Mean Water Temperature = ( 25+ 10) /2 = 17.5

Tube cross sectional Area = (π/4)* 16 ^2 = 201 mm

Tube per pass = 423/2 = 211.5

Total ow Area = 211.5 * 201 * 10 ^-6 =0.0425

Q = m*Cp* T

Total ow Area = 211.3 * 201* 16 ^ -6 = 0.0425

Q = m*Cp * T

M = Q/ Cp T = 632579.9 /(25 -10) *(91.63) = 460.24

Water velocity = 460.24 /0.0425 = 10828.2

Density of water = 995 Kg /m^3

Water liner velocity (4) = 10828.2 /995 =10.88

Hi = 4200(1.35 + 0.02t) ut ^0.8/ di^ 0.2

Hi = inside coe cient for water W/m^ 2.c

T= water temperature

V= water velocity

D = tube inside diameter

= 4200(1.35 (0.02 *17.5)* 10.9 ^ 0.8 ) /16 ^ 0.2 = 60330.06

Viscosity of water = 0.8 m Ns / m^ 2

Thermal conduc vity = 0.59 w/m .c

Re = p*u*d/ µ = 995* 10.9* 16 *10^ -3 / 0.8 *10 ^ -3 = 216910

Pr =cpµ /K1

= 91.63 *0.8 * 10 ^ -3 / 0.59 = 0.1242

L/d =4.83 * 10^3 / 16 = 302

From figure (12-23) , vol (6)

Jh = 3*10^ -3

Hi =(0.59 /16 *10^-3 ) * 3*10 ^ -3 * 216910 = 23995.6 w/ m^2 .c

Shell side coefficient

Choose baffle spacing = Ds /5 = 641.8 /5 = 128.3

Tube pitch = 1.25 * 20 = 25mm

Cross flow area = (25 -20 )/25)* 641.5 * 128.3 * 10 ^ -6 =0.0164

Mass density = (17643.32)/3600*(1/0.0164)

=298.83

Equivalent Diameter = (1.1/do)*(pt^2 – 0.917 * do^2 )

= (1.1/20)*(25^2 – 0.917 * 20^2 )

= 14.4

Mean shell side

Temperature = (40+20) / 2

= 30C

Physical properties of Mixture

X glyoxal = 4.6714

X glyoxlic = 65

X H2O = 30.33

Pm = X glyoxal P + X glyoxlic P + X H2O P H2O

= 4.67*1270 + 65*1490 + 30.33*1000

= 102780.9

CP = X glyoxlicCp + XglyoxalCp + X H2O Cp H2O

= 4.67 * 28.16 + 65*48.36 + 30.33*91.64

 =6227.53

Km = 4.67*0.155 + 65*0.168 + 30.33*0.6

Km = 29.841

Re = (Gs*de)/µ

=(298.83*14.4)/1*10^-3

= 4.303

Pr = µ*(cp/kf)

= 1*10^-3 *(4200/0.6)

= 6.5

From figure below calculate Jn = 8*10^-1

Overall Coefficient

1/v = 1/no + 1/nod +( do ln (do/d1) )/2KW – do/d1* 1/nod + do/d1*1/ni

hs = -0.196/14.2*10^-3 *0.8*4.303*(- 0.196)^1/3

= 27.6

1/v = 1/27.6 + 1/27.6*16 +( 20ln 20/16) /2*50 – 20/16 *1/60.33*16 +

20/16 + 1/60.33

V = 600

Error = (600 – 500 /600 )

Pressure drop

pt = NP (Bjf(L/d)(µ/µw)^m + 2.5 ) * pui^2/2

= 2*(8*0.8*(4.83*10^-3/16)^1*10^-3 +2.5)*(995*2.49^2/2)

= 7499.21 N/m^2
-Mechanical design
1- thickness of shell

P = 12 bar

Design pressure take 10% above Opera ng pressure

P = (12-1)*1.1

= 12 bar

P = 1.21 N/mm^2

Design temperature = 413C

From table 13.2 typical design stress F =89 N/mm^2

e = P* Di/2*f - P

= 1.21 *16 /2*89 -1.21

= 0.109

Add Corrosion allowance 2mm

e = 0.109 +2

= 2.109 mm

2- demand head

Flat heat

Use Cp = 0.4

De = diameter let = 2.109 *10^-3 m

e = Cp*De* √p/√f

= 0.4* 2.109*10^-3 *√1.21/√89

= 9.83*10^-5 mm

Add Corrosion allowance = 2 mm

e = 9.83*10^-3 + 2 =11.83 *10^-3

3- Weight load

For the steel

Wc = 240 *Cv * Dm *(H + 0.8 +Dm) *t

Wc = total weight

Cv = 1.08

H=2m

Dm = (Di + t *10^-3)

Dm = 0.584 + 5.103*10^-3

= 0.5891m

Wc = 240 * 1.08 *0.5891 (2+ 0.8 *0.5891 ) *8.53*10^-3

= 3.143337 KN

= 314.337N

4- Skirt support

For straight cylinder G = 90 made from steel

Wind Loading

Take cylinder wind pressure = 1280

Deff = Ds +Ds * (t+e)

Assume t = 5.103 mm

e = 50 mm (insula on)

De = 0.397 + 0.397 (5.105 + 50 )*10^-3

= 0.419
Fw = Pw + Deff

= 1280 + 0.419 = 536.32 N/m^2

Bending moment at bottom

Mx = fw *h^2/2

Mx = 536.32*4.83^2/2

= 6255.877N

Mx = 6.255877KN

Bending moment at base of skirt

Take height of skirt = 1m

Height of heat exchange = 4.83 m

Bending = Mx *h^2 /2

= 6.255877*5.83^2/2

= 106.315 KN/m

bs = 4MS/[π(Ds + ts )*ts*Ds ]

Ds = inside diameter let = 0.5 m

bs = 4*106.31512 *1000 *1000/[π(500+5.103)*5.103 *500]

= 105.0872 N/mm^2

ws = W/[π(Ds+ts)*(ts) ]

ts = thickness = 5.23

Ds =0.5

Approximate weight = [π/4 *(0.5^2 *4.83)*7800*9.81]

= 72567.24N

= 79472.24N

Ws = 72567.24/[π(500+5.103)*5.103]
 = 8.96N/mm^2

Ws(Operate) = 79472.08/[π(500+5.103)*5.103]

= 9.814N/mm^2

Criteria desing

Tuking joint factor (J) = 0.85

Desing stress (f) = 100

θ= 90

s less than (f) * sin θ )

(1.94 less than (100*0.85 * sin 90)

9. 814 less than 85

So ,the criteria is satisfied .

 Chapter Five

COST ESTIMATION

Fixed capital investment for cost index ( 2010 )

Cost index for ( 2017 )

Present cost = original cost ( index value at present time/ index value
at time
original cost was
obtained)
=200488335*(567.5/550.8)
=206567048.1

Estimation of total investment cost:

1- Direct cost:
a- Purchased equipment cost:
(15 - 40% of FCI ) Assume 20 % of FCI
= 206567048.1*0.2=41313409.63

b- Installation cost:
(35 - 45% of PEC) Assume 40 % ,where PEC , Purchased
equipment cost
= 206567048.1*0.4=82626819.24

c- Instrument and control installed:(6 -30% of PEC) Assume 11 % of

PEC
= 206567048.1*0.11=22722375.29

d- Piping installation cost:(10 -80% of PEC) Assume 40 %

= 206567048.1*0.4=82626819.24
e- Electrical installation cost:(10 - 40% of PEC) Assume 20 % of
PEC
= 206567048.1*0.2=413409.62

f- Building process and auxiliary (10-70% of PEC) Assume 35 %

= 206567048.1 * 0.35=72298466.84

g- Service facilities:(30-80% 0f PEC) Assume 55 %

= 206567048.1* 0.55=113611876.5

h- Yard improvement:(10-15% of PEC) Assume 12 %

= 206567048.1* 0.12=24788045.77

i- Land:( 4-8% of PEC) Assume 6 %

= 206567048.1*0.06=12394022.89

Therefore direct cost = a + b + c + d + e + f + g + h + i

=41313409.63+82626819.24+22722375.29+82626819.24+41313409.
62+72298466.84+113611876.5+24788045.77+12394022.89=493695
245.02

Indirect cost:
Expenses which are not directly involved with material and labour
of actual installation or complete facility

a- Engineering and supervision(5-30% of DC) Assume 15 %

= 493695245.02 * 0.15=74054286.75
b- Construction expenses: (10% of DC)

= 493695245.02 *0.10=49369524.5

c- Contractors fee(2-7% 0f DC) Assume 4 %

= 493695245.02 * 0.04=19747809.8

d- Contingency: (8-20% of DC) Assume 16 %

= 493695245.02 * 0.16=78991239.2
Therefore total indirect cost = a + b + c + d
=222162860.3

Fixed capital investment

Fixed capital investment(FCI) = DC+IC

=493695245.02+222162860.3=715858105.3
Working capital investment: 10 -20% of FCI Assume 15%
= 715858105.3* 0.15=107378715.8

2- Total capital investment

Estimation of total product cost(TPC):
Fixed charges:
a- Depreciation: (10% of FCI for machinery)

= 715858105.3*0.10=71585810.53
b-Local taxes: (3-4% of TPC= FCI) Assume 3.5 %
= 715858105.3 * 0.035=25055033.69
c- Insurances(0.4-1% of FCI) Assume 0.7 %
= 715858105.3 * 0.007=5011006.737
d-Rent: (8-12% of FCI) Assume 10 %
= 715858105.3*0.10=71585810.53
Therefore total fixed charges = a + b + c + d
=173237661.5
But, Fixed charges = (10-20% of TPC) Assume 20%
Therefore Total product cost = total fixed charges / 0.2 or * 100/20
=173237661.5/0.2
=866188307.5

Direct production:
a- Raw material: (10-50% 0f TPC) Assume 25%
=866188307.5*0.25=216547076.9
b- Operating labor(OL): (10-20% of TPC) Assume 15 %
=866188307.5*0.15=129928246.1
c- Direct supervisory and electric labor (10-25% of OL) Assume 20 %
=129928246.1*0.2=25985649.23
 d- Utilities (10-20% of TPC) Assume 15 %
=866188307.5*0.15=129928246.1
e- Maintenance (2-10% of FCI) Assume 6 %
=715858105.3*0.06=42951486.32
f- Operating supplies (OS): (10-20% of maintenance) Assume 15 %
=42951486.32*0.15=6442722.948
g- Laboratory charges (10-20% of OL) Assume 12 %
=129928246.1*0.12=15591389.53
h- Patent and royalties (2-6% of TPC) Assume 4 %
=866188307.5*0.04=34647532.3

Plant overhead cost: 50-70% of (OL+OS+M) Assume 65 %

=(129928246.1+25985649.23+42951486.32)*0.65=129262498.1

General expenses:
a- Administration cost: (40-60% of OL) ASSUME 55 %
=129928246.1*0.55=71460535.36
b- Distribution and selling price (2-30% of TPC) Assume 20 %
=866188307.5*0.2=173237661.5
c- Research and development cost: (3% of TPC)
=866188307.5*0.03=25985649.23
Therefore general expenses(GE) = A + B + C
=270683846.1
Therefore manufacturing cost(MC) = Product cost +fixed charges
+Plant overhead expenses
=866188307.5+173237661.5+129262498.1=1168688467
Total production cost:
Total production cost= MC + GE
=1168688467+270683846.1=1439372313

Gross earnings and rate of return:

The plant is working for say 300 days a
year Selling price =2925000
Total income =produce day rate * number production days(in
year) *price per unit
=166.666*300
2925000=1.46249415*10^10
Gross income =Total income - total product
1.46249415*10^10-1439372313=722099524
Tax =50%
Net profit= Gross income – (Gross income * 50%)
=722099524-(722099524*0.5)=361049762
Rate of return =net profit/total capital investment
=7134528.11
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(1981) 1 – 26.
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