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Chapter - 9

Energy Conservation in Chemical Sector: Utilization of Non-Conventional

Energy Resource

9.1.0 Introduction: This chapter deals with a novel approach of “Effective Utilization” of Non-Conventional Energy Resource – Hidden Energy (Heat of Reaction) via Energy Conservation Technique (ECT). Thermal Engineering Aspects and Energy Conservation are in separable. To consider Thermal Engineering Aspects, the knowledge of Thermodynamic data on molal heat capacities, latent heat of vaporization & heat of reaction etc is essential. If sufficient thermodynamic data is not available, there is a general tendency in Thermal Engineering Studies, to keep such case studies in abeyance. Further, for the production of different chemicals in Industrial Sector, Thermal Engineering Aspects (TEA) are of utmost importance because of the fact that all such production units are generally highly energy intensive. Utilization of non-conventional energy resource – Hidden Energy (Heat of Reaction) in particular appears to be an area which has been generally neglected – because of non-availability of thermodynamic data for various parameters for different chemicals like raw materials, intermediate products and final product inclusive of co-products. Hence, in this investigation – probably for the first time – an attempt has been made to show how this “Hidden Energy” having a lot of potential can play the most important positive role via energy conservation in manufacturing of different selected chemicals in Industrial sector. The required various thermodynamic properties if not readily available, can be calculated predicted by use of “Group Contribution Technique”. The various statistical techniques can be utilized conveniently while predicting the required thermodynamic data on different properties. Once data regarding all the thermo chemical properties is, thus, available, exhaustive Material and Energy Balances around the different units of a manufacturing plant can be performed. Based on “exhaustive literature survey” and also based on “industrial visits” to different manufacturing units, various technological alternatives available for the manufacturing of a particular chemical has to be critically surveyed initially. Then, using principles of MEB, detailed Material Balance Flow Sheet can be prepared. Simultaneously, detailed Energy Balance Flow Sheet can also be prepared for various technological alternatives. These MEB flow sheets can be analyzed critically from Thermal Engineering Point view. Though such a study is time consuming and tedious, it is the most essential requirement in this novel approach of “Effective Utilization of non-conventional Energy Resources”. Keeping in mind the above mentioned background, the novel approach – “Utilization of non-conventional energy resource – Hidden Energy (Heat of Reaction) via Energy Conservation” has been attempted in this investigation.

9.2.0 Case Study: Effective Utilization of Non-Conventional Energy Resource in Chemical Sector by Conservation” has been achieved in a plant manufacturing a petrochemical product: Formaldehyde. This novel approach wherein energy conservation has been achieved in a plant manufacturing Petrochemical Product: Formaldehyde. It is known that any manufacturing process is generally either exothermic or endothermic in nature. To keep temperature of reaction constant in a reactor, heat energy has to be supplied to the reactor for an endothermic reaction and heat energy has to be removed by using cooling media in the case of a reaction which is exothermic in nature. Based on “exhaustive literature survey” and also based on “industrial visits” to different Petrochemical Manufacturing units wherein Formaldehyde is being manufactured, it is observed that utilization of non-conventional energy resource – Hidden Energy (Heat of Reaction) can be utilized conveniently via Energy Conservation Technique. Thus, a manufacturing process which is either exothermic or endothermic in nature and which is highly energy intensive process can be converted to a “zero energy requirement process by carrying out both the manufacturing reactions simultaneously in a shell and tube Heat exchanger.

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Indepth literature survey was performed and industrial units visited for the technological study were the following:

1. Simalin Chemicals Industries Ltd, Nandesari, Vadodara; 2. Balaji Formalin, Motibhoyan, Taluka: Kalol, Dist.: Gandhinagar; 3. Omkar Chemicals, Nandesari, Vadodara 4. Meru Chem Pvt. Ltd. Mulund (West), Mumbai

9.3.0 Routes Available for Production of Formaldehyde:

Being an important petrochemical, formaldehyde can be manufactured by two different processes namely – dehydrogenation and oxidation. Here in this discussion, the process (oxidation and dehydrogenation) for manufacturing the product is considered separately as two different routes. The detailed discussion for the production of formaldehyde via these two routes is given as under:

9.3.1 Manufacturing Processes:

a) Route – I Unit Process – Dehydrogenation

Formaldehyde can be manufactured from methanol by dehydrogenation and oxidation process. In dehydrogenation process, the temperature maintained is 600°C. the reaction is carried out in presence of oxides of ferrous & chromium. The reaction is highly endothermic in nature. Conversion is 70 – 80 % and the yield of the product is 99%. (HCHO + H2) vapors absorbed in water & H2 gets separated, hence 37% solution of HCHO so obtained called as “Formalin”. Being highly energy intensive process, it requires large quantity of heating media. Unreacted methanol separated from formalin by distillation b) Route – II Unit Process – Oxidation

In oxidation process, here the process is maintained at a temperature of 650°C. the reaction is carried out in presence of oxides of silver & molybdenum. The reaction is highly exothermic in nature. Conversion is 80 – 90 % and the yield of the product is 90%. Being highly energy intensive process, it requires large quantity of cooling water. Non – purified air, compressed to about 0.2 atm. gauge is preheated by heat exchange with reacting gases and then it is fed to methanol evaporator, where at high temperature condition methanol is evaporated. The ratio of CH3OH to O2 is maintained in the range of 30 – 50%, so that the required conditions are achieved. The mixed gases from evaporator are then heated in preheater and sent to a reactor where oxides of silver or molybdenum act as catalysts. The activity of catalyst is controlled to maintain a balance between endothermic dehydrogenation and exothermic oxidation reactions at the reaction conditions of 450 – 650°C. Some complete combustion reaction takes place. The product gases are absorbed in a water scrubber which is cooled by external circulation and sent back to light end stripper. The bottom of stripper is then fed to alcohol fractionator where approximately 15% of unreacted methanol is recovered from top and is recycled. The bottom of fractionator contains product which is 37% solution of Formaldehyde called as “Formalin”. Flow Sheet For Route –II: Unit Process Oxidation For Production of Formaldehyde is as attached. c) Route – III Combined Process: This route is based on the principles of carrying out dehydrogenation and oxidation reactions simultaneously using methanol as a raw material using shell and tube heat exchanger (Combined Route) as a reactor, then heat evolved on tube side wherein oxidation reaction is being carried out can be utilized conveniently to heat the entire mass from shell side where dehydrogenation reaction is being carried out as is seen in figure 9.1.

Methanol Formaldehyde

CH3OH HCHO + H2

Methanol Formaldehyde

CH3OH + ½ O2 HCHO + H2O

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Methanol

Heated Air

Air

Vent Gases Water

Recycle Methanol Vapor

Scr

ub

ber

Me

tha

no

l Eva

po

rato

r

Ca

taly

tic R

eac

tor

Lig

ht E

nd

Str

ipp

er

Alc

oh

ol S

trip

pe

r

Flow Sheet For Route –II: Unit Process Oxidation For Production of Formaldehyde

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Figure 9.1 Straight tube heat exchanger 9.4.0 Thermodynamic Considerations: Estimation of Thermodynamic data: (Yaws182, and Poling et al127) ∆HR, ∆G and lnK for Formaldehyde production: Route – I • Scheme of reaction: Unit Process – Dehydrogenation

Route – II • Scheme of reaction: Unit Process – Oxidation

9.4.1 Basic Thermodynamic Data: The required data (Reid et al145 and Yaws182)is tabulated in the table 9.1

Components Molecular

weight A B x 101 C x 104 D x 108 ∆H°

f, 298 ∆G°298

CH3OH 32 21.15 0.7092 0.2587 -2.852 -201300 -162600 HCHO 30 23.48 0.3157 0.2985 -2.3 -116000 -110000 H2 2 27.14 0.0927 -0.1381 0.7645 -- -- O2 32 28.11 0.00001 0.1746 -1.065 -- -- H2O 18 32.24 0.01924 0.1055 -0.3591 -242000 -228800

Table 9.1 Basic Thermodynamic Data • Equation for CP

°, CP° = A + BT + CT2 + DT3 in J/mol K

• ∆H°R, 298 = Σ(∆H°

f, 298)Products - Σ(∆H°f, 298)Reactants in J/mol

• ∆G°, 298 = Σ(∆G°

, 298)Products - Σ(∆G°, 298)Reactants in J/mol

Methanol Formaldehyde

CH3OH HCHO + H2

Methanol

CH3OH + ½ O2

Formaldehyde

HCHO + H2O

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(∵equation – (1))

9.4.2 Route – I – Unit Process: Dehydrogenation: (Smith et al161 and Ried et al144) a) Calculation of ∆H°R, 298:

∆H°R, 298 = -116000 – [-201300]

= 85300 J/mol

b) Calculation of ∆A, ∆B, ∆C, ∆D: The values of ∆A, ∆B, ∆C, ∆D are as in table 9.2

∆A ∆B ∆C ∆D

29.47 -0.0301 -0.0983 x 10-4 1.32 x 10-8

Table 9.2 Values of ∆A, ∆B, ∆C, ∆D

c) Calculation of ∆H°R:

( )∫ ∆+∆+∆+∆+∆=∆ °°T

T

RR dtTDTCTBAHH0

32298 ***

( ) ( )

( ) ( )448

334

22

2984

10*32.1298

3

10*0983.0

2982

0301.029847.29[85300

−+−

−−−−+=∆

−−

°

TT

TTH R

49362 *10*3.3*10*27.3*015.047.297.77910 TTTTH R−−° +−++=∆

∆H°R by equation at 298 K = 85300 J/mol

d) Calculation of ∆G°:

∆G° 298 = -110000 – [-162600] = 52600 J/mol

( ) ∫∫ ∆−∆+∆−∆−∆=∆ °°°°T

T

P

T

T

PRR T

dTCTdTCGH

T

THG

00

298298,0

298,

410372 *10*6.6*10*1.8*005.09581358 TTTTG −−° +−+−=∆

∆G° by equation at 298 K = 52589 J/mol

e) Calculation of ln k

∆G° 298 = -RTlnK , 23.21

298*314.8

)52600(ln −=−=K

311284 *10*9.7*10*7.9*10*66.9785

43.11ln TTTT

k −−− −++−=

ln k by equation at 298K = -21.22

Values of ∆Hr, ∆G and lnK are as in table 9.3

T (K) ∆Hr (J/mol) ∆G (J/mol) 1/T lnK K

298 85300 52588 0.0034 -21.22 6.07E-10

300 85340 52392 0.0033 -21 7.56E-10

350 86297 47471 0.0029 -16.31 8.25E-08

400 87175 42523 0.0025 -12.78 2.80E-06

450 87974 37549 0.0022 -10.03 4.39E-05

500 88697 32548 0.002 -7.827 0.0004

550 89345 27522 0.0018 -6.016 0.0024

600 89922 22469 0.0017 -4.501 0.011

(1)

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Table 9.3 Values of ∆Hr, ∆G and lnK Graph of ∆HR Vs T

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 200 400 600 800 1000 1200

Temp (K)

Del

ta H

r (J

/mo

l)

Figure 9.2 Graph of ∆HR Vs T

-25

-20

-15

-10

-5

0

5

0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004

1/T

lnK

650 90431 17391 0.0015 -3.215 0.040

700 90876 12289 0.0014 -2.109 0.121

750 91262 7163 0.0013 -1.146 0.317

800 91592 2014 0.0013 -0.3 0.740

850 91874 -3157 0.0012 0.4491 1.57

900 92111 -8349.1 0.0011 1.1181 3.06

950 92312 -13561 0.0011 1.7192 5.58

1000 92482 -18792 0.001 2.2624 9.61

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(∵equation – (2))

Figure 9.3 Graph of lnK Vs 1/T

9.4.3 Route – II – Unit Process: Oxidation(Smith et al161) a) Calculation of ∆H°

R, 298:

∆H°R, 298 = [-242000 – 116600] – [-201300] = -156700 J/mol

b) Calculation of ∆A, ∆B, ∆C, ∆D: Values of ∆A, ∆B, ∆C, ∆D are as in table 9.4

∆A ∆B ∆C ∆D

20.51 -0.03741 0.058 x 10-4 0.7254 10-8

Table 9.4 Values of ∆A, ∆B, ∆C, ∆D

c) Calculation of ∆H°R:

( )∫ ∆+∆+∆+∆+∆=∆ °°T

T

RR dtTDTCTBAHH0

32298 ***

( ) ( )

( ) ( )448

334

22

2984

10*7254.0298

3

10*058.0

2982

03741.029851.20[156700

−+−

+−−−+−=∆

−−

°

TT

TTH R

49362 *10*813.1*10*833.1*0187.051.205.161216 TTTTH R−−° ++−+−=∆

∆H°R by equation at 298 K = -156700 J/mol

d) Calculation of ∆G°:

∆G° 298 = [-228800 – 110000 ] – [-162600] = -176200 J/mol

( ) ∫∫ ∆−∆+∆−∆−∆=∆ °°°°T

T

P

T

T

PRR T

dTCTdTCGH

T

THG

00

298298,0

298,

410372 *10*63.3*10*83.4*0063.017.55159217 TTTTG −−° ++−−−=∆

∆G° by equation at 298 K = -176210 J/mol

e) Calculation of ln K

∆G° 298 = -RTlnK, 12.71

298*314.8

)176200(ln =−−=K

311284 *10*37.4*10*82.5*10*1.95.19150

63.6ln TTTT

k −−− −−+−=

ln k by equation at 298K = 71.2

Values of ∆Hr, ∆G and lnK are as in table 9.5

T (K) ∆Hr (J/mol) ∆G (J/mol) 1/T Ln k K

298 -156700 -176210 0.00336 71.2 8.36E+30

500 -155282 -188309 0.002 46.06 1.01E+20

525 -155185 -189836 0.0019 44.35 1.81E+19

550 -155105 -191369 0.00182 42.8 3.87E+18

575 -155040 -192908 0.00174 41.41 9.60E+17

600 -154990 -194454 0.00167 40.14 2.7E+17

(2)

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Table 9.5 Values of ∆Hr, ∆G and lnK

-160000

-150000

-140000

-130000

-120000

-110000

-100000

0 100 200 300 400 500 600 700 800 900

Temp (K)

Del

ta H

r (J

/Mo

l)

Figure 9.4 Graph of ∆HR Vs t

0

5

10

15

20

25

30

35

40

45

50

0 0.0005 0.001 0.0015 0.002 0.0025

1/T

lnK

Figure 9.5 Graph of lnK Vs 1/T

625 -154954 -196005 0.0016 38.99 8.61E+16

650 -154931 -197561 0.00154 37.95 3.03E+16

675 -154922 -199124 0.00148 37 1.17E+16

700 -154924 -200691 0.00143 36.14 4.96E+15

725 -154938 -202264 0.00138 35.36 2.26E+15

750 -154964 -203842 0.00133 34.64 1.11E+15

775 -154999 -205425 0.00129 34 5.80E+14

800 -155044 -207013 0.00125 33.41 3.22E+14

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9.5.0 Critical Thermodynamic Analysis: Critical Thermodynamic Analysis of Formaldehyde for three routes under considerations.

9.5.1 For Route – I: Unit Process – Dehydrogenation: Dehydrogenation reaction is highly endothermic in nature having heat of reaction value in the range of 85 to 93 KJ / mol of Methanol. The values of “lnK” ranges from -21.0 to 2.0. It indicates that reaction is highly reversible, so stringent control measures are required. The values of lnK are very low; practically zero up to 700 K. Being highly energy intensive process, it requires large quantity of heating media. Conversion is 80% and yield is above 99%. The value of conversion level can be increased at high temperature i.e. beyond 900 K. This happens to be an environmentally friendly route as O2 (from air) is not required & has higher yield.

9.5.2 For Route – II: Unit Process – Oxidation: Oxidation reaction is very highly exothermic in nature having the heat of reaction value in the range of -155 to -156 KJ / mol of Methanol. The value of “lnK” ranges from 33.0 to 71.0. It indicates that reaction is highly irreversible, highly feasible and conversion level obtained can be of the order of 90%. The yield of the product is also 90%. Thus there is a full scope to increase the actual conversion levels and raw material requirements can be reduced. Stringent air pollution control measures are required. Being highly energy intensive process, large quantity of cooling media is required. The reaction is irreversible and it is a most favorable reaction for cleaner production implementation.

9.5.3 For Route – III: Combined Route: This route is based on the principles of carrying out dehydrogenation and oxidation reactions simultaneously using methanol as a raw material. In this investigation, as the new technological option, Route – I being endothermic and Route – II being exothermic, combined reactor will give maximum benefits carrying both reactions (dehydrogenation and oxidation) simultaneously; the route can be converted to environmentally friendly route. Route – II is not an environmentally friendly, not a green reaction and not a green technology. However, if it couples with Route – I becomes a green reaction scheme. Conversion in dehydrogenation reaction and oxidation reaction is 80% and 90% respectively. This combined route will provide the most suitable conditions for eco – efficient environment. Route – III seems to be more environmentally friendly and more energy efficient route. Thus, it seems to be a most favorable route for energy conservation implementation. 9.5.4 Conclusion Based on Thermodynamics: Thus, all the two reactions under considerations, first reaction being highly exothermic in nature, the removal of heat have to be done very effectively, where as the second reaction is endothermic in nature, addition of energy has to be done very effectively. The relevant material balance and energy balance flow sheet can be prepared based on actual plant capacity data. The data tabulated in above tables is expected to be very useful while implementing energy conservation in plant manufacturing formaldehyde. Dehydrogenation being endothermic in nature, it requires large amount of energy in terms of heating media and oxidation reaction is highly exothermic in nature, it requires large amount of energy in terms of cooling media. In other words, both the routes are highly energy intensive. If both Routes – (I) & (II) are combined i. e. oxidation process as well as dehydrogenation process occur simultaneously using shell and tube heat exchanger as a reactor, then heat evolved on tube side wherein oxidation reaction is being carried out can be utilized conveniently to heat the entire mass from shell side where dehydrogenation reaction is being carried out. Thus full energy conservation can be achieved. Energy requirements – cooling media for Route – (II) as well as heating media for Route – (I) reduces to zero. According to energy balance flow data, energy requirement from external source for this combined route is zero. Thus for the implementation of energy conservation principles, thermodynamics aspects are of utmost importance. Finally it can be summarized that Formaldehyde production by combined route i. e Route – III is a good example of energy conservation principles.

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9.6.0 Analysis of Technologies from Energy Conservation Point of View: (Rao and Sittig140, Othimer 117 &118) 9.6.1 Analysis for Route I: In Route – I, formaldehyde is produced by the dehydrogenation of methanol which would produce anhydrous or highly concentrated formaldehyde solutions. In dehydrogenation process, the temperature maintained is 600°C and the reaction is carried out in presence of oxides of ferrous & chromium. Fresh methanol with recycled quantity of methanol is fed to the reactor operated at 600°C and 1 atmospheric pressure to produce formaldehyde. The reaction is highly endothermic in nature. Conversion is 70 – 80 % and the yield of the product is 99%. (HCHO + H2) vapors absorbed in water & H2 gets separated, hence 37% solution of HCHO so obtained called as “Formalin”. Being highly energy intensive process, it requires large quantity of heating media, (Lefferts et al96 and Carcia24) Unreacted methanol separated from formalin by distillation. Route-I happens to be environmentally friendly route as O2 (from air) not required and has higher yield than Route-II. 9.6.2 Analysis for Route II: In Route – II, formaldehyde is produced by the oxidation of methanol which would produce anhydrous or highly concentrated formaldehyde solutions. Methanol was oxidized over a copper catalyst, but this has been almost completely replaced with silver. The silver-catalyzed reactions occur at essentially atmospheric pressure and 600 to 650°C. The reaction is highly exothermic (-156 KJ /mol) in nature. Conversion is 80-90% and yield is 90%. The mixture passes through a super heater to a catalyst bed of silver crystals or layers of silver gauze. The product is then rapidly cooled in a steam generator and then in water cooled heat exchanger and fed to the bottom of an absorption tower. Absorber bottoms go to a distillation tower where methanol is recovered for recycle to the reactor. The base stream from distillation, an aqueous solution of formaldehyde, is usually sent to an anion exchange unit which reduces the formic acid to specification level. HCHO vapors absorbed in H2O and one obtains 37% solution of HCHO called as Formalin. The product contains up to 55% formaldehyde and less than 1.5% methanol. The reaction occurs at essentially adiabatic conditions. Recycled methanol required for a 50–55% product is 0.25–0.50 parts per part of fresh methanol. With increasing energy costs, maximum methanol conversion is desirable, eliminating the need for the energy-intensive distillation for methanol recovery. In another process, tail gas from the absorber is recycled to the reactor. This process can produce 50% formaldehyde with about 1.0% methanol without a distillation tower. Methanol recovery can be obviated in two-stage oxidation systems where, for example, part of the methanol is converted with a silver catalyst, the product is cooled, excess air is added, and the remaining methanol is converted over a metal oxide catalyst such as that described below. In another two-stage process, both first and second stages use silver catalysts. Formaldehyde–methanol solutions can be made directly from methanol oxidation product by absorption in methanol. Aqueous formaldehyde is corrosive to carbon steel, but formaldehyde in the vapor phase is not. All parts of the manufacturing equipment exposed to hot formaldehyde solutions must be a corrosion-resistant alloy such as type-316 stainless steel. Theoretically, equipment can be of carbon steel, but in practice alloys are required in this part of the plant to protect the sensitive silver catalyst from metal contamination. Water is also generated along with the product. Being highly energy intensive process, it requires large amount of cooling media (water) (Lefferts et al96 and Carcia24), This route produces air emissions such as Nox, CO etc. Hence stringent air pollution control measures required. 9.6.3 Analysis for Route III: Route – III suggested in this investigation appears to be more attractive and more environmentally than Route – I and Route - II. This new technological option is considered as a combined route, in which both the reactions are carried out simultaneously. If both is processes are combined i. e. oxidation reaction as well as dehydrogenation reaction occur simultaneously using shell and tube heat exchanger as a reactor, then heat evolved on tube side wherein oxidation reaction is being carried out can be utilized conveniently to heat the entire mass from shell side wherein dehydrogenation reaction is being carried out. Thus, full energy conservation can be achieved. Energy requirements in terms of heating media for route-I as well as cooling media for route-II reduces to zero. Energy evolved in oxidation process due to high exothermic reaction is fully utilized in dehydrogenation as if the reaction is highly endothermic in nature. By this option sufficient amount of energy can be saved and energy costing by thus can be reduced. Hence, this combined route can be considered to be very cost effective. Based on

- 240 -

material balances and energy balances data, energy requirement from external source for the combined route is, thus, zero.

9.7.0 Material Balance and Energy Balance for Formaldehyde Production: 9.7.1 Material Balance for Formaldehyde Production: a) Route – (I): Dehydrogenation Process: CH3OH HCHO + H2 Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kgmoles/day

b) Route – (II): Oxidation Process: CH3OH + ½ O2 HCHO + H2O Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kgmoles/day Figure 9.7 Material Balance diagram for Route – II (Formaldehyde

Production)

Figure 9.6 Material Balance diagram for Route – I (Formaldehyde Production)

O2: 1111.1

(30 TPD)

N2: 4179.9 (117.0 TPD)

CH3OH: 185.2 (5.9 TPD)Recycle

CH3OH: 1851.9 (59.3

HCHO: 1666.7 (50 TPD)

N2: 4179.9 (117.0 TPD)

O2: 185.2 (5.9 TPD)

Oxidation T = 600°C P = 1 atm X = 90%

Separator

CH3OH: 1666.7 (53.4 TPD)

CH3OH: 185.2 (5.9 TPD) Recycle

(35.6 TPD)

H2O: 1666.7

CH3OH: 1666.6 (53.33 TPD)

CH3OH: 2083.3 (66.66 TPD)

HCHO: 1666.6 (50 TPD)

CH3OH: 4167 (133.3 TPD) Recycle

H2: 1666.7 (3.33 TPD)

CH3OH: 416.7 (13.33 TPD) Recycle

Dehydrogenation T = 600°C P = 1 atm X = 80%

Separator

- 241 -

c) Route – (III): Combined Route: Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kgmoles/day

Figure 9.8 Material Balance diagram for Combined Route (Formaldehyde Production)

CH3OH: 271.6 (8.7 TPD) Recycle - I

CH3OH: 1086.6 (34.76 TPD)

N2: 1309.1 (36.65 TPD)

O2: 58 (1.8 TPD)

H2: 1086.6

HCHO: 1086.6 (32.6 TPD)

Separator

CH3OH: 271.6 (8.7 TPD) Recycle – I

(2.16 TPD)

H2O: 580 (10.4 TPD)

CH3OH: 64.4

CH3OH: 580 (18.5 TPD)

CH3OH: 64.4 (2.1 TPD) Recycle – II

COMBINED ROUTE

T = 600°C P = 1 atm

CH3OH: 644.4 (20.6 TPD)

O2: 348

CH3OH: 1358.2 (43.46 TPD)

N2: 1309.1 (36.65 TPD)

Separator (11.1 TPD) (2.1 TPD)

Recycle – II

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9.7.2 Material Balance Calculations for Formaldehyde Production: a) Route – I: Unit Process – Dehydrogenation:

Basis: 50 TPD (1666.6 kgmoles / day) of Formaldehyde produced From reaction, 1 kmol of HCHO = 1 kmol of CH3OH ∴ Fresh CH3OH required = 1666.67 kgmoles / day Data available from literature, the conversion is 80%

∴Total CH3OH fed to reactor = TPDdaykgmoles 7.66/33.20838.067.1666 ==

So, Methanol unreacted = Methanol supplied – Methanol reacted = 2083.33 – 1666.67

= 416.7 kgmoles / day = 13.3 TPD

This quantity is recycled with fresh CH3OH Now, 1 kmol of HCHO = 1 kmol of H2

∴ H2 produced = 1666.67 kgmoles / day = 3.33 TPD b) Route – II: Unit Process – Oxidation:

Basis: 50 TPD (1666.67 kgmoles / day) of Formaldehyde produced From reaction, 1 kmol of HCHO = 1 kmol of CH3OH ∴ CH3OH required = 1666.67 kgmoles / day Data available from literature, the conversion is 90%

∴ CH3OH fed to reactor = TPDdaykgmoles 3.59/9.18519.0

67.1666 ==

So, Methanol unreacted = Methanol supplied – Methanol reacted = 1851.9 – 1666.67 = 185.23 kgmoles / day = 5.9 TPD Now, 1 kmol of CH3OH= 0.5 kmol of O2

∴ Theoretical requirement of O2 = 925.9 kgmoles / day O2 supplied = 925.9 x 1.2 = 1111.1 kgmoles / day = 35.6 TPD Air contains 21% O2 and 79% N2.

So, moles of air supplied = TPDdaykgmolesx 1.148/52911.111121

100 ==

∴Moles of N2 = 4179.9 kgmoles / day = 117 TPD ∴ O2 unreacted = O2 supplied - O2 reacted = 1111.1 – 925.9 = 185.2 kgmoles / day = 5.9 TPD Moles of N2 in exit stream = 4179.9 kgmoles / day = 110 TPD Now, 1 kmol of HCHO = 1 kmol of H2O ∴ H2O produced = 1666.67 kgmoles / day = 30 TPD

CH3OH HCHO + H2

CH3OH + ½ O2 HCHO + H2O

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c) Route – III: Combined Route: Unit Process – Dehydrogenation Unit Process – Oxidation In combined route, oxidation as well as dehydrogenation reaction is combined and carried out in single reactor. Material balance calculation for this route is as under. Let, formaldehyde produced in reaction – 2 is 580 kgmoles / day, so formaldehyde produced in reaction – 1 is 1086.6 kgmoles / day. From reaction, 1 kmol of HCHO = 1 kmol of CH3OH ∴ Fresh CH3OH required = 1086.6 kgmoles / day Data available from literature, the conversion is 80%

∴ Total CH3OH fed to reactor = TPDdaykgmoles 5.43/2.13588.0

6.1086 ==

So, Methanol unreacted = Methanol supplied – Methanol reacted = 1358.2 – 1086.6 = 271.6 kgmoles / day = 8.7 TPD This quantity is recycled with fresh CH3OH Now, 1 kmol of HCHO = 1 kmol of H2

∴ H2 produced = 1086.6 kgmoles / day = 2.17 TPD Now, formaldehyde produced in reaction – 2 is 580 kgmoles / day From reaction, 1 kmol of HCHO = 1 kmol of CH3OH ∴ Fresh CH3OH required = 580 kgmoles / day Data available from literature, the conversion is 90%

∴ Total CH3OH fed to reactor = TPDdaykgmoles 6.20/4.6449.0

580 ==

So, Methanol unreacted = Methanol supplied – Methanol reacted = 644.4 – 580 = 64.4 kgmoles / day = 2.1 TPD This quantity is recycled with fresh CH3OH Now, 1 kmol of HCHO = 0.5 kmol of O2 = 580 x 0.5 ∴ Theoretical requirement of O2 = 290 kgmoles / day = 9.28 TPD O2 supplied = 2900 x 1.2 = 348 kgmoles / day = 11.1 TPD Air contains 21% O2 and 79% N2.

So, moles of air supplied = TPDdaykgmolesx 8.47/1.165734821

100 ==

∴Moles of N2 = 1309.1 kgmoles / day = 36.65 TPD ∴ O2 unreacted = O2 supplied - O2 reacted = 348 – 290 = 58 kgmoles / day = 1.85 TPD Moles of N2 in exit stream = 1309.1 kgmoles / day = 36.65 TPD Now, 1 kmol of HCHO = 1 kmol of H2O ∴ H2O produced = 580 kgmoles / day = 10.4 TPD

CH3OH HCHO + H2

CH3OH + ½ O2 HCHO + H2O

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9.7.3 Energy Balance for Formaldehyde Production: a) Route – (I): Dehydrogenation Process: CH3OH HCHO + H2 Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kJ/day

b) Route – (II): Oxidation Process: CH3OH + ½ O2 HCHO + H2O Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kJ/day

Figure 9.10 Energy Balance Diagram for Route – II (Formaldehyde Production)

Figure 9.9 Energy Balance Diagram for Route – I (Formaldehyde Production)

Dehydrogenation

T = 600°C P = 1 atm

∆HR = 14.21 x 107

CH3OH:

HCHO: 0

Separator ∆HP = 0.0

Q = 11.8 x 107

Preheater T = 65°C

Q = 11.8 x 107

CH3OH: 0 Recycle

H2: 0

Enthalpy

Out:

Q1 = 14.21 x 107

Q2 = 0

CH3OH: 0

CH3OH: 0 Recycle

11.8 x 107 11.8 x 107

HCHO: 0

H2O: 0

Separator ∆HP = 5.06 x 107 Q = 20.33 x 107

CH3OH: 0

O2: 0 N2: 0 Q1 = 0

Q2 = 26.1 x 107

Oxidation T = 600°C P = 1 atm

∆HR = -26.1 x 107

Enthalpy

Out:

O2: 0

Preheater T = 65°C

Q = 15.27 x 107

N2: 0

CH3OH: 0

CH3OH: 0 Recycle

CH3OH:

N2:

O2: 10.51 x 107

0.586 x

4.17 x 107

15.27 x 107 Recycle

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c) Route – (III): Combined Route: Basis: 50 TPD (1666.7 kgmoles/day) of Formaldehyde produced Unit: kJ/day

Figure 9.11 Energy Balance Diagram for Combined Route (Formaldehyde Production)

N2:

Preheater T = 600°C

Q = 7.7 x 107

CH3OH: 0

O2: 0

CH3OH: 0 Recycle – I

CH3OH: 0

CH3OH: 0 Recycle – II

Separator

∆HP = 1.82 x 107 Q = 6.92 x 107

Separator ∆HP = 0.0

Q = 7.7 x 107

ST

OR

AG

E

O2: 0 N2: 0

H2: 0

HCHO: 0

H2O: 0

CH3OH: 0

Q2 = 0

Recycle - I

Recycle - II

Dehydrogenation Reaction T = 600°C P = 1 atm

∆HR = 9.1 x 107

Oxidation Reaction

T = 600°C P = 1 atm

∆HR = -9.1 x 107

Enthalpy Out:

Enthalpy Out:

Q1 = 0

5.10 x 107

7.7 x 107

Preheater T = 600°C

Q = 5.1 x 108

CH3OH: 3.61 x 107

O2: 0.18 x 107

N2: 1.31 x 107

CH3OH:

7.7 x 107

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9.7.4 Energy Balance Calculations for Formaldehyde Production: a) Route – I: Unit Process – Dehydrogenation (T = 600°C = 873 K) (CH3OH)in = (mCp∆T + mλ) = 2083.3[37.461(873 – 298) + 35254] = 2083.3[21540 + 35254] =56794 x 2083.3 = 11.8 x 107 kJ / day ∴Total enthalpy in: 11.8 x 107 kJ / day Heat of reaction = 85300 x 1666.67 = 14.21 x 107 kJ / day ∴ Heat to be added during reaction: 14.21 x 107 kJ / day If products are heated to 873K, the energy requirements are as under. (CH3OH)out = (mCp∆T + mλ) = 416.7[37.461(873 – 298) + 35254] = 416.7[21540 + 35254] = 416.7 x 56794 = 2.36 x 108 kJ / day (HCHO)out = (mCp∆T + mλ) = 1666.67[27.19(873 – 298) + 21550] = 1666.67[15634 + 21550] = 1666.67 x 37184 = 6.20 x 107 kJ / day (H2)out = mCp∆T) = 1666.67 x 16.88(873 – 298) = 1666.67 x 9706 = 1.60x 107 kJ / day

∴ Energy required to preheat: 10.16 x 107 kJ / day However, enthalpy of product comes out of dehydrogenation reactor is 11.8 x 107 kJ / day Hence, energy required to be removed, Q = 11.8 x 107 kJ / day. b) Route –II: Unit Process – Oxidation (T = 600°C = 873 K) (CH3OH)in = (mCp∆T + mλ) = 1851.9[37.461(873 – 298) + 35254] = 1851.9 [21540 + 35254] =56794 x 1851.9 = 10.51 x 107 kJ / day (O2)in = ½ (mCp∆T) = ½[1111.1 x 18.36 (873 – 298)] = ½ [1111.1 x 10557] = 0.586 x 107 kJ / day (N2)in = (mCp∆T) = 4179.9 x 17.35 (873 – 298) = 4179.9 x 9976 = 4.17 x 107 kJ / day ∴Energy required to preheat: 15.27 x 107 kJ / day

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Heat of reaction = (-156700 x 1666.67) = -26.1 x 107 kJ / day ∴ Heat to be removed during reaction: -26.1 x 107 kJ / day If products are heated to 873K, the energy requirements are as under. (CH3OH)out = (mCp∆T + mλ) = 185.2[37.461(873 – 298) + 35254] = 185.2 [21540 + 35254] = 185.2 x 56794 = 1.05 x 107 kJ / day (HCHO)out = (mCp∆T + mλ) = 1666.67[27.19(873 – 298) + 21550] = 1666.67[15634 + 21550] = 1666.67 x 37184 = 6.20 x 107 kJ / day (O2)out = (mCp∆T) = 185.2 x 18.36 (873 – 298)] = [185.2 x 10557] = 0.196 x 107 kJ / day (N2)out = (mCp∆T) = 4179.9 x 17.35 (873 – 298) = 4179.9 x 9976 = 4.17 x 107 kJ / day (H2O)out = (mCp∆T + mλ) = 1666.67[20.19(873 – 298) + 40626] = 1666.67[11609 + 40626] = 1666.67 x 52235 = 8.71 x 107 kJ / day ∴ Energy required to preheat: 20.33 x 107 kJ / day Energy to be required in terms of cooling for phase change

∆HP = 20.33 x 107 – 15.27 x 107

= 5.06 x 107 kJ / day C) Route –III: Combined Route (T = 600°C = 873 K) Unit Process - Dehydrogenation (CH3OH)in = (mCp∆T + mλ) = 1358.2[37.461(873 – 298) + 35254] = 1358.2[21540 + 35254] =56794 x 1358.2 = 7.7 x 107 kJ / day ∴Total enthalpy in: 7.7 x 107 kJ / day Heat of reaction = 85300 x 1086.6 = 9.1 x 107 kJ / day ∴ Heat to be added during reaction: 9.1 x 107 kJ / day If products are heated to 873K, the energy requirements are as under. (CH3OH)out = (mCp∆T + mλ) = 271.6[37.461(873 – 298) + 35254]

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= 271.6 [21540 + 35254] = 271.6 x 56794 = 1.53 x 107 kJ / day (HCHO)out = (mCp∆T + mλ) = 1086.6[27.19(873 – 298) + 21550] = 1086.6 [15634 + 21550] = 1086.6 x 37184 = 4.04 x 107 kJ / day (H2)out = mCp∆T = 1086.6 x 16.88(873 – 298) = 1086.6 x 9706 = 1.05x 107 kJ / day

∴ Energy required to preheat: 6.62 x 107 kJ / day However, enthalpy of product comes out of dehydrogenation reactor is 7.7 x 107 kJ / day Hence, energy required to be removed, Q = 7.7 x 107 kJ / day. Unit Process - Oxidation (CH3OH)in = (mCp∆T + mλ) = 644.4[37.461(873 – 298) + 35254] = 644.4[21540 + 35254] =56794 x 644.4 = 3.61 x 107 kJ / day (O2)in = ½ (mCp∆T) = ½[348 x 18.36 (873 – 298)] = ½ [348 x 10557] = 0.18 x 107 kJ / day (N2)in = (mCp∆T) = 1309.1 x 17.35 (873 – 298) = 1309.1 x 9976 = 1.31 x 107 kJ / day ∴Total enthalpy in: 5.1 x 107 kJ / day Heat of reaction = (-15670 x 5800) = -9.1 x 107 kJ / day ∴ Heat to be removed during reaction: -9.1 x 107 kJ / day If products are heated to 873K, the energy requirements are as under. (CH3OH)out = (mCp∆T + mλ) = 64.4[37.461(873 – 298) + 35254] = 64.4[21540 + 35254] = 64.4 x 56794 = 0.36 x 107 kJ / day (HCHO)out = (mCp∆T + mλ) = 580[27.19(873 – 298) + 21550] = 580[15634 + 21550] = 580 x 37184 = 2.12 x 107 kJ / day (O2)out = (mCp∆T) = 58 x 18.36 (873 – 298)]

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= [58 x 10557] = 0.07 x 107 kJ / day (N2)out = (mCp∆T) = 1309.1 x 17.35 (873 – 298) = 1309.1 x 9976 = 1.31 x 107 kJ / day (H2O)out = (mCp∆T + mλ) = 580[20.19(873 – 298) + 40626] = 580[11609 + 40626] = 580 x 52235 = 3.03 x 107 kJ / day ∴ Energy required to preheat: 6.92 x 107 kJ / day Energy to be required in terms of cooling for phase change

∆HP = 6.92 x 107 – 5.10 x 107

=1.82 x 107 kJ / day

9.8.0 Critical Analysis of Material and Energy Balance for Formaldehyde Production:

9.8.1 Route: 1 Unit Process – Dehydrogenation:

Formaldehyde is manufactured by the dehydrogenation reaction using methanol as a raw material. To produce 50 TPD (1666.7 kgmoles / day) of formaldehyde, the total quantity of methanol required to feed into reactor is 66.66 TPD (2083.3 kgmoles / day). Since the conversion is 80%, the quantity of fresh methanol required is 53.33 TPD (1666.7 kgmoles / day). This methanol is heated in preheater. Energy required to preheat the raw material is 11.8 x 107 kJ / day. Since the dehydrogenation reaction is highly endothermic in nature, the amount of energy that must be supplied in terms of heating media is 14.21 x 107 kJ / day. The product stream coming out of the reactor contain equimolar mixture of main product formaldehyde 50 TPD (1666.7 kgmoles / day), co – product hydrogen 3.33 TPD (1666.7 kgmoles / day). This product stream also contains unreacted amount methanol 13.33 TPD (416.7 kgmoles / day), that is separated from the product and co – product and sent back to recycle with the fresh methanol stream to make the total quantity as 66.66 TPD (2083.3kgmoles / day). Hence the requirement of raw material can thus be reduced. The product stream comes out form the reactor is allowed to cool and co – product as well as unreacted amount is separated from the product. The amount of energy that must be supplied in terms of cooling is 13.4 x 107 kJ / day.

9.8.2 Route –II: Unit Process – Oxidation: In this route, formaldehyde is manufactured by the oxidation reaction using methanol and oxygen from air as a raw material. To produce 50 TPD (1666.7 kgmoles / day) of formaldehyde, the total quantity of methanol required to feed into reactor is 59.3 TPD (1851.9 kgmoles / day). The quantity of air used is 152.6 TPD (5291 kgmoles / day) containing 35.6 TPD (1111.1 kgmoles / day) of oxygen as an oxidizing agent and nitrogen 117 TPD (4179.9 kgmoles / day) as an inert. Since the conversion is 90%, the quantity of fresh methanol required is 53.4 TPD (1666.7 kgmoles / day). The raw material (methanol and air) is heated in preheater. Energy required to preheat the raw material is 15.27 x 107 kJ / day. Since the oxidation reaction is highly exothermic in nature, the amount of energy that must be supplied in terms of cooling media to remove the heat is 26.1 x 107 kJ / day. The product stream coming out of the reactor contains mixture of main product formaldehyde 50 TPD (1666.7 kgmoles / day); water 30 TPD (1666.7 kgmoles / day); 5.9 TPD (185.2 kgmoles / day) of unreacted oxygen and nitrogen 1170 TPD (41799 kgmoles / day). This product stream also contains unreacted amount methanol 5.9 TPD (185.2 kgmoles / day), that is separated from the product and by – product and sent back to recycle with the fresh methanol stream so as to make the total quantity 59.3 TPD (1851.9 kgmoles / day). Hence the requirement of raw material can thus be reduced. The product stream comes out form the reactor is allowed to cool and by – product as well as unreacted amount is separated from the product. The amount of energy that must be supplied in terms of cooling is 20.33 x 107 kJ / day.

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9.8.3 Route –III: Combined Route: As a new technological option, this route is considered and treated as combined route. In this route total 50 TPD of formaldehyde is produced by carrying out oxidation and dehydrogenation reaction in a common shell and tube type reactor. So, a common material and energy balance calculation is done for entire reactor. In tube side oxidation reaction is taking place. As both the reactions are competitive reaction, as a result of oxidation reaction 17.4 TPD (580 kgmoles / day) of formaldehyde as a product produced. The total quantity of methanol required to feed into reactor is 20.6 TPD (644.4 kgmoles / day). The quantity of air used is 47.75 TPD (1657.1 kgmoles / day) containing 11.1 TPD (348 kgmoles / day) of oxygen as an oxidizing agent and nitrogen 36.65 TPD (1309.1 kgmoles / day) as an inert. Since the conversion is 90%, the quantity of fresh methanol required is 18.5 TPD (580 kgmoles / day). The raw material (methanol and air) is heated in preheater. Energy required to preheat the raw material is 5.1 x 107 kJ / day. Since the oxidation reaction is highly exothermic in nature, the amount of energy that must be supplied in terms of cooling media to remove the heat is 9.1 x 107 kJ / day. The product stream coming out of the reactor contains mixture of main product formaldehyde 17.4 TPD (580 kgmoles / day); water 10.4 TPD (580 kgmoles / day); 1.8 TPD (58 kgmoles / day) of unreacted oxygen and nitrogen 36.65 TPD (1309.1 kgmoles / day). This product stream also contains unreacted amount methanol 2.1 TPD (64.4 kgmoles / day), that is separated from the product and by – product and sent back to recycle with the fresh methanol stream so as to make the total quantity 17.4 TPD (580 kgmoles / day). Hence the requirement of raw material can thus be reduced. The product stream comes out form the reactor is allowed to cool and by – product as well as unreacted amount is separated from the product. The amount of energy that must be supplied in terms of cooling and for separation is 6.92 x 107 kJ / day. In dehydrogenation route, to produce 32.6 TPD (1086.6 kgmoles / day) of formaldehyde, the total quantity of methanol required to feed into reactor is 43.46 TPD (1358.2 kgmoles / day) in shell side. Since the conversion is 80%, the quantity of fresh methanol required is 34.76 TPD (1086.6 kgmoles / day). This methanol is heated in preheater. Energy required to preheat the raw material is 7.7 x 107 kJ / day. Since the dehydrogenation reaction is highly endothermic in nature, the amount of energy that must be supplied in terms of heating media is 9.1 x 107 kJ / day. This energy is supplied from the tube side where oxidation reaction is taking place. So, energy requirement from external source is zero. The product stream coming out of the reactor contain equimolar mixture of main product formaldehyde 32.6 TPD (1086.6 kgmoles / day), co – product hydrogen 2.16 TPD (10866 kgmoles / day). This product stream also contains unreacted amount methanol 8.7 TPD (271.6 kgmoles / day), that is separated from the product and co – product and sent back to recycle with the fresh methanol stream to make the total quantity as 43..46 TPD (1358.2 kgmoles / day). Hence the requirement of raw material can thus be reduced. The product stream comes out form the reactor is allowed to cool and co – product as well as unreacted amount is separated from the product. The amount of energy that must be supplied in terms of cooling is 7.7 x 107 kJ / day. This is the most favorable route in terms of energy conservation. 9.8.4 Comparison of Three routes based on Material Balance Principles:

Finally we can summarize the entire three routes in common table by material balance principles for the production of 50 TPD of formaldehyde as under:

Dehydrogenation route Oxidation route Combined route CH3OH required 66.66 TPD 59.3 TPD 64.06 TPD Oxygen required -- 35.6 TPD 11.1 TPD Air required -- 152.6 TPD 47.7 TPD Energy required 1.645 KJ/s 3.021 KJ/s 0.0 KJ/s Table No 9.6 Material Balance From the above table 9.6, it could be seen that formaldehyde production by combined route is a good example of in terms of waste minimization and energy Conservation and also it is taken as an eco – friendly route.

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9.8.5 Comparison of Three routes based on Energy Balance Principles: The Energy requirement for entire three routes for the production of formaldehyde can be summarized with respect to energy balance in a common table as under:

Preheating (kJ / day)

Heat to be absorbed / removed (kJ / day)

Separation (kJ / day)

TOTAL (kJ / day)

Route – I Dehydrogenation 11.8 x 107 14.21 x 107 11.8 x 107 37.81 x 107

Route – II Oxidation 15.27 x 107 26.1 x 107 20.33 x 107 61.7 x 107

Route – III Combined Route 12.8 x 107 0.0 14.62 x 107 27.42 x 107

Table No 9.7 Energy Balance From the above comparison, it is seen that among three routes total energy requirement (kJ / day) in terms of preheating the raw materials and separation of products from other chemical spices in combined route (Route – III) is very low in comparison with the other two routes. In other words, Dehydrogenation route (since it is not so commonly used) requires total energy 37.81 x 107 kJ / day, whereas combined route requires total energy of 27.42 x 107 kJ / day. Hence in this combined route (Route – III) approximately 27 – 28% of total energy can be saved. This is the good example of energy conservation. In comparison to Route – II (most commonly used process) and Route – III, it is seen from the data tabulated above, the total energy required in Route – II is 61.7 x 107 kJ / day, whereas Route - III requires total energy of 27.42 x 107 kJ / day. Hence in this combined route (Route – III) approximately 45% of total energy can be saved. So, this combined route can also become the best example of energy conservation. As a result, this route is considered as the new technological option, in case of good energy conservation and waste minimization, for the formaldehyde production. Furthermore it requires no (zero) energy (as the energy evolved during oxidation reaction is fully absorbed by dehydrogenation reaction) from external source in either terms of heating or cooling media. Thus, good energy conservation can be achieved by this option. So, combined route can be considered as most energy efficient and more eco – friendly route. 9.9.0 Conclusions: As discussed earlier, this petrochemical can be manufactured by two different processes namely – dehydrogenation and oxidation. Route – I refers to dehydrogenation reaction, basically endothermic in nature. So this process requires large amount of heating media. Thus the process becomes highly intensive in nature. Critical analysis of material and energy balance states that, for the production of 50 TPD of formaldehyde, the total quantity of methanol required to feed into reactor is 66.66 TPD. Also co – product hydrogen 3.33 TPD is coming out with the main product formaldehyde. (HCHO + H2) vapors absorbed in water & H2 gets separated, hence 37% solution of HCHO so obtained called as “Formalin”. Route-I happens to be environmentally friendly route as it does not require O2 (from air) and has higher yield than Route-II. The total amount of energy required in this route for the production of formaldehyde is 37.81 x 107 kJ / day. Thus Route –I happens to be a highly energy intensive route. Route – II, formaldehyde is manufactured from methanol and air either by a process using a silver catalyst or one using a metal oxide catalyst refers to oxidation reaction, basically exothermic in nature. So this process requires large amount of cooling media. Thus the process becomes highly intensive in nature. Critical analysis of material and energy balance states that, for the production of 50 TPD of formaldehyde, the total quantity of methanol required to feed into reactor is 59.3 TPD. The quantity of air used is 152.6 TPD containing 35.6 TPD of oxygen as an oxidizing agent and nitrogen 117 TPD as an inert. The total amount of energy required in this route for the production of formaldehyde is 61.7 x 107 kJ / day. Generation of water is 30 TPD for this route. Compared this situation with Route – I, 50% more energy is required in this route. Thus Route –II also happens to be a highly energy intensive route. Route – III, proposed in this investigation, can be considered as more environmentally friendly route and least energy intensive route. As the new technological option, both the routes are carried out in a combined reactor. The total amount of energy require in this route for the production of formaldehyde is 27.42 x 107 kJ / day. Energy required from the external source for either removal or absorption of energy is zero. It is seen that among three routes, total energy requirement in terms of preheating the raw materials and separation of products in combined route is very low in comparison

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with the other two routes. Hence, compared with Route – II approximately 50% of total energy can be saved. Also raw material requirement in this route can also be reduced significantly. Thus full energy conservation can be achieved. Energy requirements, cooling media for Route – (I) as well as heating media for Route – (II) reduces to zero. Thus, quantity of gaseous stream coming out of plant containing pollutants gets reduced considerably by 65%. Thus, intensity of air pollution gets reduced by a factor of three. Finally it can be summarized that formaldehyde production by combined route is a good example in terms of energy conservation for the implementation of energy conservation principles. The energy requirements for the three routes under consideration can be summarized as under: Among three routes total energy requirement (kJ / day) in terms of preheating the raw materials and separation of products from other chemical spices in combined route (Route – III) is very low in comparison with the other two routes.

• In other words, Dehydrogenation route requires total energy 37.81 ×107 kJ / day, whereas combined route requires total energy of 27.42 ×107 kJ / day.

• Hence in this combined route (Route – III) approximately 27 – 28% of total energy can be saved. This is the good example of energy conservation.

• In comparison to Route – II and Route – III, the total energy required in Route – II is 61.7 ×107 kJ / day, whereas Route - III requires total energy of 27.42 ×107 kJ / day.

• Hence in this combined route (Route – III) approximately 45% of total energy can be saved. • So, combined route can also become the good example of Energy conservation. • Hidden Energy (Heat of Reaction) can be utilized effectively.

Thus, in a plant of Formaldehyde manufacture having production capacity of 50 Tons/Day by implementing Process-Design Modification – i.e. Conventional Tubular bed rector to be replaced by Shell & Tube Heat Exchanger as a reactor – energy saving to the extent of 14.2×107 kJ / day minimum / 26.1 × 107 kJ / day maximum can be achieved depending on the route utilised for the production of Petrochemical – Formaldehyde. This is the most important unique feature of the Ph.D. Research Project. Production of some other Petrochemicals like Vinyl chloride, Acrylonitrile etc can also be under taken using this novel concept of Utilization of non conventional energy resource – Hidden Energy (Heat of Reaction) via Energy Conservation Technique.