DESIGN PROJECT

“GLYCEROL”

SRINIVAS REDDY CHERUKULA

(09bch020)

SANATH KUMAR VELLANKI

(09bch066)

MARKET ANALYSIS :

The global market for glycerin is projected to reach 4.4 billion

pounds in consumption by the year 2015 and 5.4 billion pounds by

the year 2017, propelled by the emerging use of glycerin in various

novel applications such as propylene glycol and growing demand

from oral care, personal care, pharmaceuticals and food & beverage

sectors.

There are over 2,000 established uses for glycerine in the drug, food,

beverage, chemicals and synthetic material industries

Demand-supply scenario in different

countries :

2011 Demand(ktons) Supply(ktons)

US 377 276

Europe 620 750

China 507 113

ASEAN 91 320

Japan 55 42

India 75 40

Latam 50 210

Africa 30 30

Middle East 40 0

Total 1845 1781

For every 1 tonne of biodiesel that is manufactured,

100 kg of glycerol is produced.( glycerin is a

byproduct in biodiesel manufacturing)

Usually this crude glycerol has to be purified, typically

by performing vacuum distillation. This is rather

energy intensive

Conclusion:

1)Demand for Glycerin in chemical industry is growing leaps and

bounds! But the supply is limited!

2) It is evident that in the very near future the demand-supply

gap increases enormously

PHYSICAL PROPERTIES :

1) Molecular weight = 92.09

2) Melting point =18.17ºC

3) Boiling point(760mm Hg) = 290ºC

4) Density (20ºC) =1.261 g/cm3

5) Vapor pressure = 0.0025 mm Hg at 50ºC

= O.195 mm Hg at 100ºC

= 4.3 mm Hg at 150ºC

= 46 mm Hg at 200ºC

6) Refractive index = 1.474

7) Surface tension = 63.4 dyne/cm at 20ºC(100% glycerol)

8) Compressibility (28.5ºC) = 2.1×10 MPa

9) Viscosity = 1499 c.p. at 20ºC (100% glycerol)

10) Specific heat = 0.5779 cal/gm at 26ºC(99.94%glycerol)

11) Heat of vaporization = 21060 cal/mole at 55ºC

= 18170 cal/mole at 195ºC

12) Heat of formation = 159.6 Kcal/gm mole

13) Heat of combustion = 1662 KJ/mole

14) Heat of fusion = 18.3 KJ/mole

Glycerol is completely soluble in water and alcohol. It is slightly soluble

in ether, ethyl acetate, and dioxane and insoluble in hydrocarbons.

Glycerol has useful solvent properties similar to those of water and

simple aliphatic alcohol's because of its three-hydroxyl groups.

Glycerol is a useful solvent for many solids, both organic and inorganic

which is particularly important for the preparation of

pharmaceuticals.The solubility of gases in glycerol,like other liquids is

temperature and pressure dependent.

CHEMICAL PROPERTIES:

1)Glycerol is a reactive molecule that undergoes all the usual reactions of

alcohols.

2)The two terminal primary hydroxyl groups are more reactive than the

internal secondary hydroxyl group.

3)Under neutral or alkaline conditions, glycerol can be heated to 250ºC

without formation of acrolein.

4)Reactions with glycerol are therefore best carried out under alkaline or

neutral conditions at 180ºC, alkaline glycerol begins to dehydrate

forming ether-linked polyglycerols.

5)At room temperature glycerol rapidly absorbs water. When dilute it is

attacked by microorganism

6) On oxidation, glycerol yields variety of product depending upon the

reaction conditions.

7)By the use of mild oxidizing agent it is possible to oxidize only one

hydroxyl group to yield Glyceraldehyde.

8) Nitric acid converts glycerol to glyceric acid CH2CHCHOHCOOH

melting at 134-135ºC

USES:

1. Glycerin has a role in virtually every industry.

2. it is present in combined form in all vegetable and animal fats

3. Glycerin as a food is easily digested and non-toxic

4. In candies and icings glycerin prevents crystallization of sugar

5. In drugs and the medicine glycerin is an ingredient of many

tinctures and elixirs and as a glyceride of starch is used in jellies

and ointments

6. In cosmetics glycerin is used in many creams and lotion to keep the

skin soft and replace skin moisture

7. In processing tobacco, glycerin is important part of casing solution

sprayed on tobacco before leaves are shredded and packed

8. Glycerin can be used as a lubricant in places where oil can fail.

MANUFACTURING PROCESSES :

Glycerol is the simplest trihydric alcohol which was originally

obtained merely as a byproduct of fat saponification

After 1940’s glycerin began to dominate the market and the

glycerol produced was no longer sufficient to meet the rising

demand

1)FROM FATS AND OILS :

In fats and oils glycerol exists as triglycerides

this varies between 8 to 14%

To obtain glycerol fats must be split

TRANSESTERIFICATION :

glycerol can be obtained from the continuous transesterification of

oils and fats to their methyl esters

CH2OCOR CH2OH

Lipase

CHOCOR + 3MeOH 3 R.COOMe + CHOH

CH2OCOR CH2OH

SAPONIFICATION:

Splitting of fats is done by the addition of alkali or alkali carbonate

Until 1949 all glycerin was obtained from the glycerides in the form of

fats and oils, but from that time on production of synthetic glycerin

increased until, in 1965 it accounted for about 60% of the market. The

first glycerol process put on stream in 1948,followed the discovery that

propylene could be converted to glycerol by several routes, the synthesis

of glycerol from propylene became possible.

2) FROM ALLYLCHLORIDE :

Hypochlorination:

Dehydrochlorination:

Hydrolysis of epichlorohydrin to glycerol:

3)HYDROGENATION OF CARBOHYDRATES :

Hydrogenation of natural polyalcohols such as cellulose, starch, or

sugar followed by distillation gives glycerol

Catalyst used in high temperature includes nickel, cobalt, copper,

chromium and tungsten as well as oxides of some of the lanthanides

The crude glycerin produced is poor in quality and requires drastic,

expensive refining methods.

4)AS A BYPRODUCT IN BIODIESEL PRODUCTION:

1)For every 1 tonne of biodiesel that is manufactured, 100 kg of

glycerol is

produced.( glycerin is a byproduct in biodiesel manufacturing)

2)Usually this crude glycerol has to be purified, typically by

performing

vacuum distillation. This is rather energy intensive

SELECTION OF PROCESS:

1)Out of these processes production of glycerin from allyl chloride

is the

better one

2)yield is high >90%

3)Majority of the glycerol producers i.e more than 80% of the total

4)glycerol capacity worldwide use allyl chloride as the feedstock

5)Yields from other processes are very low.

6)They are economically unimportant

7)The glycerol content of fats and oils varies between 8 and 14%. So

production from them doesn’t meet the needs

FLOSHEET:

REACTIONS:

1)Epichlorohydrin is hydrolyzed at 80-200°C in the precence of sodium carbonate

or sodium hydroxide at atmospheric pressure

2)The yield of dilute (10-25%) glycerol solution is > 98 %

3)This aqueous glycerol solution containing sodium chloride is

evaporated in a multistage evaporation plant under vacuum to a glycerol

concentration of > 75 %

4)Centrifuge is used to saperate the sodium chloride precipitated sodium chloride is

separated at the same tim

Enthalpy balance:

DATA:

SPECIFIC HEATS:

1) H2O = 4.18 KJ/Kg K (at 25 C)

2) Epichlorohydrin = 1.422 KJ/Kg K (at 25 C)

3) Sodium Hydroxide = 3.93 KJ/Kg K (at 25 C)

4) Sodium chloride = 3.31 KJ/Kg K ( at 31 C)

5) Glycerol = 2.4234 KJ/Kg K ( at 31 C)

STANDARD HEAT OF FORMATION :

H2O = -285.830kcal/mol

Epichlorohydrin = -35.6 kcal/mol

Sodium hydroxide= -210kcal/mol

Glycerol= -259.428 kcal/mol

Sodium chloride = -385.92 kcal/mol

CSTR:

H2O stream :

temperature: 298 K

Molefraction: H2O =1

H (in)=MCPdT

= 9972.93*4.18*(25)

=10421.7 KJ/kg

Epichlorohydrin stream:

Temperature :298 K

Mole fraction: epichlorohydrin = 0.69

sodium hydroxide= 0.31

H(in) = (0.69*1.422)+(0.31*3.93) *25

= 40350.56KJ/kg

Glycerol stream:

Temperature: 304

Mole fraction: glycerol= 0.611

sodium chloride= 0.389; H= 51479 KJ/kg

Heat Generated:

(Heat of formation of products) – (Heat of formation of reactants)

= -359.428-385.92-(-269.15-35.6-285.830) cal/mol

= 39455 KJ

Centrifuge:

Enthalpy of NaCl stream: H= 32361*3.31* 27 = 28921KJ

Enthalpy from the centrifuge: 10000KJ

Calculations of evaporator’s enthalpy

Are as follows:

Glycerol solution is to be concentrated from 7Wt% to a final concentration of 80Wt% The feed is

available at 50993 Kg/Hr and 27C. Steam is supplied at 103.66Kpa (Abs.) to the first effect and a

vacuum of 74.2(Abs.) mm of Hg is maintained in the last effect. Heat losses by radiation and by

entertainment are neglected. Condensate are assumed to enter each effect at its saturation

temperature. Glycerol in feed = 50993*0.07

= 3569.51 Kg/Hr

Feed rate WF = 50993/3600

= 14.16 Kg/Sec Bottom outlet from third effect = 3569.51/(0.83600)

= 1.24 Kg/Sec Total evaporation =14.16 – 1.24

=12.9 Kg/Sec

Assuming equal evaporation in all the three effects.

W1– Evaporation rate in first effect

W2- Evaporation rate in second effect

W3– Evaporation rate in the third effect

W1 = W2 = W3 = (12.9/3) = 4.3 Kg/sec

Outlet from first effect = WF - W1

= 14.16 – 4.3

= 9.86 Kg/sec

Outlet from second effect = WF – W1- W2

= 14.16 – 4.3 – 4.3

= 5.56 Kg/sec

Outlet from third effect = WF – W1 – W2 – W3 = 14.16 – 4.3 – 4.3 – 4.3 == 1.26 Kg/sec

Concentration of glycerol in outlet from first effect = (14.16 0.07 )/9.86

= 10 wt%

Concentration of glycerol in outlet from second effect =

(14.16.07)/5.5

6 = 0.178

= 17.8 wt%

Concentration of glycerol in outlet from third effect = (14.16.07)/1.26

= 0.79

= 79 wt%

Saturation temperature of inlet steam TS = 100C Boiling point of water in the third effect = 12C Temperature difference =100 – 12

= 88C

BOILING POINT RISE FOR GLYCEROL SOULTION 1) BPR for the third effect at 79 Wt% glycerol = 10.4C 2) BPR for the second effect at 17.8 Wt% glycerol = 1.5C 3) BPR for the first effect at 10 Wt% glycerol = 0.5C Sum

of the BPR of the glycerol solution = 10.4 + 1.5 + 0.5

= 12.4C

Effective temperature difference = 88 – 12.4

= 75.6C Now we assume that the amount of heat transfer from steam to the solution is the same in all

the three effects i.e. Q (heat transfer in the first effect) = Q (heat transfer in the second effect) =

Q (heat transfer in the third effect)

OVERALL HEAT TRANSFER COEFFICIENT Since it is a vertical effect evaporator, the overall heat transfer coefficient can safely be assumed as follows:

In the first effect U1 = 2325 W/mC

In the second effect U2 = 1275 W/mC

In the third effect U3 = 1031 W/mC

Now as we have assumed heat transfer rate’s to be equal , we have

Q1 = Q2 = Q3

(U1 A1 ∆t1 ) = (U2 A2 ∆t2 ) = (U3 A3 ∆t3 )

We design the triple effect evaporator such that the heating area in all the three effect is the same

A1 = A2 = A3

U1 ∆t1 = U2 ∆t 2 = U3 ∆t3

∆t2 / ∆t1 = U1 / U2

= 2352 /

1275 = 1.8

∆t3 / ∆t 2 = U2 / U3

=1275 /

1031 = 0.6

∆t1 + ∆t2 + ∆t3 = 75.6C

0.6 ∆t2 + ∆t2 + 0.6 ∆t2 =

75.6C ∆t2 = 32.3C

∆t1 = 21.54C

∆t3 = 21.64C

ACTUAL BOILING POINTS IN EACH EFFECT First effect:

T1 = TS - ∆t1

= 100 - 21.54

= 78.54C Second effect:

T2 = T1 - (BPR)1 -∆t2

= 78.5 - 0.5 - 32.3

= 45.7C Third effect:

T3 = T2 - (BPR)2 -∆t3

= 45.7 - 1.5 - 21.64

= 22.56C

Effect 1 ( ºC ) Effect 2 ( ºC ) Effect 3 ( ºC )

TS = 100 T1 = 78 T2 = 44.2

T1 = 78.5 T2 = 45.7 T3 =22.56

HEAT BALANCE FIRST EFFECT:

WS λS + WFHF = W1H1 + ( WF - W1 ) h1

latent heat of steam λS = 2257.86 KJ/ kg

HF – Enthalpy of feed at inlet temperature ( 27ºC) = Cpf × ( Tf - 0)

= ( 0.576 × 4.18) × 27

= 65 KJ / Kg

H1- Enthalpy of vapor leaving the first effect = H2S + (Cp)steam × (BPR1)superhea

= 2640 + (1.884 × 0.5)

= 2487 KJ/ Kg

H2s- Enthalpy of steam at 78ºC = 2640 KJ / Kg

(Cp)steam at 78ºC = 1.884 KJ/ Kg

h1 - enthalpy of outlet from first effect at 78.5ºC = Cp1 × ( t1 – 0 )

= 0.65 × 4.18 × 78.5

= 213.28 KJ/Kg

WS × (2257.86) + (14.16 × 65) = ( W1 × 2487 ) + ( 14.16 – W1 ) × 213.28

WS × (2257.86) = 2273.7 × W1 + 2099.6

WS = W1 + 0.93 ------------------------(1)

SECOND EFFECT:

W1 λ1 + ( WF – W1 ) h1 = W2 H2 + ( WF – W1 – W2 ) h2

H3S – Enthalpy of steam vapor at 44.2ºC = 2580 KJ /Kg

H2 - Enthalpy of vapor leaving the second effect = H3S + ( Cp)steam ×(BPR2)supreheat

= 2580 + ( 1.884 × 1.5 )

= 2583 KJ/Kg

h2-Enthalpy of outlet from the second effect at 45.7ºC = Cp2 × ( t2 - 0 )

= 0.6 × 4.18 × 45.7

= 114.62 KJ /Kg

W1 × 2162 + ( 14.16 – W1 ) × 213.28 = (W2 × 2583 ) + ( 14.16 – W1 – W2 )×114.62

2063.34 × W1 = 1397 + 2468.38 × W2

W1 + 0.667 = 1.196 × W2 -------------------(2) THIRD EFFECT:

W2 λ2+ (WF – W1 –W2 ) × h2 = W3 H3 + (WF – W1 - W2 - W3 ) ×h3

H4S -Enthalpy of steam vapor at 12 .16ºC = 2523 KJ/Kg

H3 -Enthalpy of vapor leaving the third effect = H4S + (Cp)steam × (BPR3)ssuperheat

= 2523 + (1.884 ×10.4 )

= 2543 KJ/ Kg

h3 – Enthalpy of outlet from third effect at 22.56ºC = Cp3 ×( t3 - 0)

= 0.57 ×4.18 ×22.56

= 53.75 KJ /Kg

W2 ×2393 + (14.16 – W1 – W2 ) × 114.62 = W3 × 2543 + ( 14.16 – W1 – W2 – W3 ) × 53.75

2332.13 ×W2 - 60.87 = 2489.25 × W3 + 861.92

W2 + 0.370 = 1.067 × W3 + 0.026 × W1 --------------------(3)

W1 + W2 + W3 = 12.9 Kg/ Sec -------------------------(4) Solving equations' (1),(2),(3) and (4), we

get: WS = 5.153 Kg/sec

W1 = 4.223 Kg/sec

W2 = 4.089 Kg/sec

W4 = 4.588 Kg/sec

EVAPORATOR DESIGN:

Now, Q1 = WSλS

=11634.75 KJ/sec

But Q1=U1A1∆t1

Therefore A1= (11634.75×103)/(21.54×2325)

= 232.32 m2

Q2 = W1λ1

= 9130.13 KJ/sec

But Q2 = U2A2∆t2

Therefore A2 = (9130.13×103)/(32.3×1275)

= 228 m2

Q3 = W2λ2

= 9784.98 KJ/sec

But Q3 = U3A3∆t3

Therefore A3 = (9784.98×103)/(21.64×1031)

= 235 m2

Thus the obtained areas are within the acceptable range of 5% difference

Therefore the average area per effect of the evaporator is 232 m2.

Tube details: Most generally used diameters today ranges from 1.25 to 2.00 in. outer diameter and most

generally used lengths of tubes ranges from 4 to 15 ft. Let us choose 5/4-in. nominal diameter, 80 schedule, brass tubes of 10-ft

length. Therefore Outer diameter do = 42.164 mm

Inner diameter di = 32.46

mm Length L = 10 ft

= 3.048 m

Tube pitch (∆)PT = 1.25 × do

= 1.25 × 42.164

= 52.705 mm

Surface area of each tube a = πdoL

= π × 52.705×10-3

×

3.048 = 0.4037 m2

Number of tubes required Nt = A /a

= 619

Area occupied by tubes = Nt × (1/2) ×PT × PT × sin∝

= 619 × 0.5 ×(52.705 ×10-3

)2 ×

0.866 = 0.7445 m2

Where ∝ = 60o

But actual area is more than this. Hence this area is to be divided by factor which varies from 0.8 to

1.0. Let us choose this factor as 0.9. Therefore actual area required = 0.7445/ 0.9

= 0.827 m2

The central downcomer area is taken as 40 to 70% of the total cross sectional area of tubes. Let us

take it as 50%.

Therefore Downcomer area = 0.5 × [Nt × (π/4) × do2]

= 0.5 × [619 × (π/4) ×

(0.04216)2] = 0.432 m

2

Downcomer diameter = √(4 ×0.432) / π

= 0.742 m Total area of tube sheet in evaporator = downcomer area + area occupied by tubes

= 0.432+ 0.827

= 1.259 m2

Thus tube sheet diameter = √(4 × 7.1025)/ π

= 1.27 m

MECHANICAL DESIGN OF EVAPORATOR: Amount of water to be evaporated = 15480 kg/hr

Heating surface required A = 232 m2

Design pressure = 5% extra of maximum working pressure

= 1.05 × 1.03

= 1.082 bar

= 1.103 kgf/cm2

Evaporator – low carbon steel

Tubes – brass

Permissible stress for low carbon steel = 980 kg/cm2

Modulus of elasticity for low carbon steel = 19 × 105 kg/cm2

Modulus of elasticity for brass = 9.5 ×105 kg/cm2

Tube sheet thickness:

To find tube sheet thickness

K = [Es×ts×(Do – ts)] / [Et×Nt×tt ×(do – tt)]

Es = elastic modulus of shell

Et = elastic modulus of tube

Do - outside diameter of shell = 3 m

do - outside diameter of tube = 60.325 mm

ts - shell thickness = 12 mm

ts - shell thickness = 12 mm

tt - tube wall thickness = 5.5 mm

Nt - number of tubes in shell = 2108

Therefore K = [19 × 105 ×12 × (1270– 10)] / [9.5 × 105 × 619 ×4.85 × (42.164 –

4.85)] = 0.027

F = √[(2 + K) / (2 + 3 × K)]

= √[(2 + 0.027) / (2 + 3 ×0.027)]

=0.99

The effective tube sheet thickness is given by

tts = FDo √[(0.25 × P) / f]

= 0.99×1270×√[(0.25 ×1.103) / 980]

= 21 mm

with corrosion allowance the thickness may be taken as 25 mm.

CHECK FOR TUBE THICKNESS: The tube thickness is given by

tt = Pdi / (2fJ – P)

The permissible stress for brass = 538 kg/cm2 and J = 1

Therefore tt = (1.103 ×32.46) / [(2 ×538 ×1) – 1.103]

= 0.033 mm

But provided thickness is 4.85 mm. Therefore chosen tubes have enough strength

to withstand within operating conditions

EVAPORATOR DRUM DIAMETER: The following equations help to determine the drum diameter..

Rd = (V/A) / [0.0172 × √{(ρl - ρv) / ρv}]

Where V – volumetric flow rate of vapour in m3/sec

A – cross sectional area of drum

For drums having wire mesh as entrainment separator device, Rd may be taken as

1.3.

A = V / [Rd × 0.0172 × √{(ρl - ρv) / ρv}]

= [15480 /(3600 × 0.258)] / [1.3 × 0.0172 × √{(1019 – 0.258)/0.258}]

= 11.87 m2

Therefore drum diameter = √{(4 ×11.87) /3.14} =3.89 m

Drum height can be taken as 2 to 5 times of tube sheet diameter.

Thus drum height = 2×1.27 = 3.81 m

ESTIMATION OF BOLT LOADS : Load due to design pressure H = (π/4)G^2 P = (π/4) × (1.3)^2×0.11 = 0.15 MN

Load to keep joint tight under operation

Hp = πG(2b)mP

= π ×1.3×(3.25×10-3 )× 2× 0.11

= 2.92 ×10–3 MN

Total operating load Wo = H + Hp = 0.15+2.92 ×10–3 = 0.153 MN

To find weight of vessel with contents: Weight of vapour drum = πdLtρ = π×1.27 × 3.81× 0.015 ×7820

W1= 1783 kg

Weight of tubes W2 = (π/4)Nt(do^2 – di^2) Lρ

= (π/4) × 619× (0.0422 2 – 0.032462) ×3.048× 8450

= 9150 kg

weight of tube sheet W3 = 2 (π/4)Ds^2 tρ

= (π/4) ×1.272× 0.025 ×7820 × 2

= 495 kg

Therefore total weight W = W1 + W2 + W3 = 10728 kg

ENTHALPY BALANCE

COST ESTIMATION: Evaporator: Evaporator type: falling film Area= 232m^2 = 2497 sq ft. C= 1.218 exp[2.2362-0.0126ln(2497)+0.0244(ln(2497))^2) Cost =Rs.632663$ No.=2 Total cost= 2* 632663 = 1265326$ Centrifuge: 1.68m^2 cost= 50,0000$ No.=2 Total cost= 2*500000= 1000000$ Pumps: number of pumps: 4 Cost= 22,000*4 =88000$ Reactor: volume= 14.8 cubic meter From perry’s hand book, for 0.38 m3 and N=0.53, cost= 465000$ C2 = C1 (Q2/Q1)n

C2= 465000*(14.8/0.38)^0.53

C2= 3238965$

Total cost of equipment= 5592291 $

Cost index in 2003 = 402

Cost index in 2012 = 588.8

Therefore the total cost of equipment = 5592291* 588.8/402

= 8* 10^6 $ ESTIMATION OF FIXED CAPITAL INVESTMENT:

DIRECT COST: Purchased equipment E= 8.190 * 10^6 $

Purchased equipment installation = 39% E =3194450$

Piping (installed), 31%E = 2539178 $

Electrical installed,10% E = 819089.7$

Land, 6% E =491453.8$

Total direct Cost =13.01* 10^6 $

INDIRECT COST : a) Engineering and supervision:(5-30% of DC)

Assume 30%

=3.90 *10^6 $

a) Construction expenses:(10% of DC)

=1.31* 10^6 $

b) Contractors fee:(2-7% 0f DC)

Assume 7%

=9.17*10^5$

c) Contingency:(8-20% of DC)

Assume 10%

=1.31*10^6 $

Therefore total indirect cost = 7.437 *10^6 $

Fixed capital investment: Fixed capital investment(FCI) = DC+IC

= Rs 20.447 * 10^6 $

Working capital investment: 10 –20% of FCI

Assume 15%

=3.06*10^6 $

Total capital investment: = FCI + WC

=23.514* 10^6 $

Estimation of total product cost(TPC):

Fixed charges: a) Depreciation:(10% of FCI for machinery)

=2.044*10^6 $

b) Local taxes:(3-4% of FCI)

Assume 3%

=6.13*10^5 $

c) Insurances:(0.4-1% of FCI)

Assume 1%

=2.044*10^5 $

d) Rent:(8-12% of FCI)

Assume 9%

=1.840*10^6 $

Therefore total fixed charges =4.70 *10^6 $

But, Fixed charges = (10-20% of TPC)

Assume 10%

Therefore Total product cost =155506507/0.1

=4.70*10^7 $

Direct production: a) Raw material:(10-50% 0f TPC)

Assume 30%

=1.41*10^7 $

b) Operating labour(OL):(10-20% of TPC)

Assume 15%

=7.05*10^6 $

c) Direct supervisory and electric labour:(10-25% of OL)

Assume 15%

=1.05*10^6 $

d) Utilities:(10-20% of TPC)

Assume 15%

=7.05*10^6 $

e) Maintainence:(2-10% of FCI)

Assume 6%

=1.22*10^6 $

e) Operating supplies (OS):(10-20% of maintainence)

Assume 15%

=1.84*10^5 $

f) Laboratory charges:(10-20% of OL)

Assume 15%

=1.05*10^6$

g) Patent and royalties:(2-6% of TPC)

Assume 4%

=1.88*10^6 $

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

Assume 60%

=5.077*10^6$

General expenses: a) Administration cost:(40-60% of OL)

Assume 55%

=3.87*10^6$

b) Distribution and selling price:(2-30% of TPC)

Assume 15%

=7.05*10^6$

c) Research and development cost:(3% of TPC)

=1.41*10^6$

Therefore general expenses(GE) =12.337*10^6

Therefore manufaacturing cost(MC)= Product cost+fixed

charges+Plant overhead expenses

= 56.777*10^6$

Total production cost: Total production cost =MC + GE

=69.114*10^6$

Gross earnings and rate of return: The plant is working for say 300 days a year

Selling price =2.71$/kg

Total income =100×300×1000×2.71

=81.30*10^6 $

Gross income =Total income – total product cost

=12.186*10^6$

Tax =40%

Net profit = Gross income - Taxes = Gross income× (1- Tax rate)

7.311*10^6$

Rate of return = net profit/total capital investment

=7.311/23.514 *100 =31.09%

SAFETY AND HAZARDS IN GLYCEROL

MANUFACTURING

Effects, Acute Exposure (GLYCEROL)

Skin Contact little to no effect (may soften

skin)

Skin Absorption slight; no toxic effects possible

by this route

Eye Contact may be slightly irritating

Inhalation mist may be slightly irritating

Gaseous emission:

Epichlorohydrin and accompanying organic materials in

exhaust gases from the hydrolysis of epichlorohydrin can

be removed, e.g. by adsorption on activated carbon.

Waste water:

Contaminated with epichlorohydrin can be treated with,

thereby converting the epichlorohydrin to

glycerol.Regulations exists in several countries concerning

exhaust gas and waste water purification's well as

measures for the protection of plant personnel.

Residues:

Residues from evaporation and distillations column should

be treated.

STABILITY / REACTIVITY:

Dangerously Reactive With strong oxidising agents may

cause explosion

Decomposes in Presence of heat above 290

Decomposition Products toxic acrolein fumes on thermal

decomposition

Storage and transportation:

Storage:

Glycerol is stable when stored below 100ºC:it is non

corrosive and presents little risk of ignition because of its

high flash point.Anhydrous glycerol does not corrode

steel,but storage tank of carbon steel must be protected

by surface coating to prevent rusting by residual

moisture.Glycerol is therefore stored in tanks of stainless

steel of aluminum.

Transportation:

Glycerol is shipped in tank trucks, containers and drums.

The tank trucks and containers are usually made of

stainless steel, galvanized or resin coated steel is used for

drums: for small drums, plastic is also employed.

Toxicology and occupational health:

Glycerol is not harmful to health. Ingestion even of large

amounts causes no harm to humans. The use of glycerol as

a food additive is permitted in most countries in particular

those off the EEC and in the United States. Slight irrition

of the skin or mucous is possible on contact with undiluted

glycerol because the strongly hygroscopic glycerol draws

water from the skin.

FIRST AID

SKIN: Wash with soap and plenty of water. Remove

contaminated clothing and do not reuse until thoroughly

laundered.

EYES: Wash eyes with plenty of water, holding eyelids

open. Seek medical assistance promptly if there is

irritation.

INHALATION: Remove from contaminated area promptly.

CAUTION: Rescuer must not endanger himself! If

breathing stops, administer artificial respiration and seek

medical aid promptly.

INGESTION: Give plenty of water to dilute product. Do not

induce vomiting (NOTE below). Keep victim quiet. If

vomiting occurs, lower victim’s head below hips to prevent

inhalation of vomited material. Seek medical help

promptly.