Production of Carbitol
INTRODUCTION
Carbitol or Diethylene glycol ethyl ether (DEGEE) is colorless
liquid with faint, sweet, pleasant odor and bitter taste. Its
boiling point is relatively high and vapor pressure and evaporation
rate low. Like most glycol ethers, DEGEE has very good solubility
and mixes completely with water and with both polar and non-polar
solvents.
In 1993 DEGEE was registered as an ingredient in 178 Swedish
chemical products, and estimated annual use was just under 500 tons
of pure substance. The major area of use was as solvent, but the
substance was also used in paint, varnish, cleaners and binders. In
Sweden DEGEE is not used in pharmaceuticals or non-prescription
diet supplements, but does occur in cosmetics and skin care
products (personal communication, Cecilia Ulleryd, Swedish Medical
Products Agency, Nov. 15, 1996).
In the United States, DEGEE was reported to occur in 80 cosmetic
preparations in 1981. The substance, under the name Transcutol, is
used in skin medications and it has also been found in chemical air
fresheners for consumer use. The use of DEGEE in Sweden increased
rapidly from 1985 to 1992, and in the following year as well.
DEGEE, along with mono-(EGBE) and diethylene glycol butyl ether
(DEGBE), have been identified as the solvents most widely used in
water-based paints and varnishes . Global use in 1993 was estimated
to be 31,000 tons.
PROPERTIES AND USES OF CARBITOL
Chemical and physical characteristics of carbitol.
PropertiesValue
Chemical FormulaCH3CH2OCH2CH2OCH2CH2
OH
Molecular weight134.2
Density (20oC)0.99
Boiling point202 oC
Melting point- 76 oC
Vapor pressure (25oC)19 Pa (0.14 mm Hg)
Relative evaporation rate0.02 (n-butyl acetate = 1)
Saturation concentration (25oC)180 ppm
Vapor pressure (25oC)0.13 mm Hg
Vapor density4.62
Flash point96 C
Freezing point-76 C
Refractive index (20o C)1.4300
Latent heat of vaporization85 cal/g
Heat of combustion-6330 cal/g
FLAMMABILITY (FLASH POINT):
This chemical has a flash point of 96 C (205 F). It is
combustible. Fires involving this material can be controlled with a
dry chemical, carbon dioxide or Halon extinguisher. A water spray
may also be used. The auto ignition temperature of this compound is
204 C (400 F).
REACTIVITY:
This chemical can react with oxidizing materials. It is
incompatible with strong acids, acid chlorides and acid anhydrides
. It is also incompatible with alkalies. It may react with
peroxides, oxygen, nitric acid and sulfuric acid.
STABILITY:
This chemical is hygroscopic. ethanol or acetone should be
conditions.Solutions of this chemical in water, 95% stable for 24
hours under normal lab
USES:
This compound is used as a solvent for dyes, nitrocellulose and
resins.
It is used in non-aqueous stains for wood, for setting the twist
and conditioning yarns and cloth, in textile printing, textile
soaps, lacquers, cosmetics and quick-drying varnishes and enamels.
This compound is also used in brake fluid diluent and in organic
synthesis. It used to determine saponification values of oils and
as neutral solvent for mineral oil-soap and mineral oil-sulphated
oil mixtures (giving fine dispersions in water)
PRODUCTION OF CARBITOL.
Polyethylene glycols as high as the octaethylene glycol have
been prepared. The ethers of these polyethylene glycols have
attained commercial significance, and this is particularly true of
the diethylene glycol ethyl ether, which is sold commercially under
the name of carbitol. This product is especially useful in the
manufacture of laminated glass, wherein the celluloid interleaf is
misted over with a spray of carbitol, which increases the adhesion
to the glass. In the printing and dyeing of textile fabrics, it
makes possible more economical use of dyes. Carbitol is a solvent
with a mild odor, a low rate of evaporation, and a boiling point of
201.9oC. it enters into the manufacture of wood stains and
automobile polishes and is used as a lacquer solvent. However, it
is particularly valuable in the cosmetic field, where it is
employed for creams and hair tonics.Preparation of Glycol Ethers
(carbitol)
C2H5OH + CH2CH2OHO
CH2CH2 OC2H5
EthylEthyleneCellosolve
AlcoholOxide
CH2OC2H5CH2CH2OC2H5
+O
CH2
CH2OHCH2
CH2
CH2OH
CARBITOL
There is a large quantity of data available on the German
process for the manufacture of the glycol ethers. The monomethyl,
ethyl, and n-propyl ethers of ethylene glycol are manufactured by
the continuous reaction of ethylene glycol are manufactured by the
continuous reaction of ethylene oxide with the anhydrous alcohol at
about 200oC and at 25-45 atm pressure. One volume of ethylene oxide
and 6 volume of alcohol are fed to a pressure tower packed with
iron raschig rings. Excess of alcohol is used to give the required
high ratio of glycol ether: diglycol ether, to control the heat
librated in the reactor, and to avoid high concentrations of the
ethylene oxide and alcohol is exothermic, about 20-25 kg-cal per g
mole of ethylene oxide reacted. The reaction product emerges from
the base of the pressure tower and is distilled semi continuously.
The alcohol is recycled; the pure glycol and diglycol ethers are
isolated by batch fractionation. After removal of excess alcohol,
the crude product contains about 85 percent glycol ether, 10
percent diglycol ether, and 2-3 percent polygylcol ethers. The
yield of ethers is about 90-95 percent on ethylene oxide and
alcohol consumed. The reaction is controlled to give complete
conversion of ethylene oxide. Conditions in the pressure tower and
feed ratios for methyl, ethyl and n-proply alcohols are given in
table. The contact time has been calculated on the assumption that
the reactor capacity is 3.5 cubic m.
Either acids or bases may be used as catalysts in the reaction
between ethylene oxide and alcohol. However, acids are corrosive
and must be neutralized before treating the crude reaction product,
and alkalies lead to the formation of resins with the acetaldehyde
present in the ethylene oxide. Because of the above reasons a
noncatalytic process was developed. The Germans also developed a
process using aqueous ethyl alcohol, but as this involves a
difficult products separation, it is not preferred.
Conditions for Preparation of Glycol Ethers
AlcoholTemp, oCPressureRate, lb/hrContact
atmtime, hr
EthyleneAlcohol
oxide
20040120720About 4
Methyl
Ethyl200-21035100600About 5
220-2302580-90450-540About 6
n-propyl
MATERIAL BALANCE
Basis: 100 Kg moles of C2H4O The conversion is taken as 95%Of
the 100 Kg moles of ethylene oxide sent, 95% is reacting.
The crude stream a contain the products and 5 Kg moles of C2H4O
Neglecting the 5 Kg moles of C2H4O i.e. assuming that all the 5 Kg
moles of C2H4O are going out with the top product of the tower.
On a ethylene oxide free basis.The weight fraction (%) of crude
product Stream 'a' is:
85% - Monoglycol ether
10% - Diglycol ether
5% - Triglycol ether
The mole fraction of this stream 'a' on a ethylene oxide free
basis is as: 89.885% - Monoglycol ether
7.101% - Diglycol ether
3.011% - Triglycol ether
For a crude product of 100 Kg moles (C2H4O-free),
89.885 Kg moles of monoglycol ether = 89.885 Kg moles of C2H4O
7.101 Kg moles of carbitol = 14.202 Kg moles of C2H4O3.011 Kg moles
of triglycol ether = 9.033 Kg moles of C2H4O
Therefore, Total ethylene oxide that has reacted for 100 Kg
moles of (C2H4O-free) crude product =113.12 Kg moles of C2H4O.
For 100 Kg moles of C2H4O sent only 95 Kg moles reacts.
Therefore,for 95 Kg moles of C2H4O reacting,the total crude
product (C2H4O-free) formed is
(95x 100)/113.12 = 83.981 Kg moles of crude product
The crude product is (83.981+5)Kg moles for 100 Kg moles of
C2H4O feed. The total C2H5OH reacting = [(83.981x89.885)/100 +
(83.981x7.101)/100] = 81.4504 Kg moles
The recycle must be such that the ratio is 1:6
Total recycle is: (6x100) 81.450 = 518.5 Kg moles The
composition of the stream b is
518.55 Kg moles of C2H4O+(83.981+5)Kg moles of crude product The
composition of stream a on a C2H4O free basis is known.
For tower (1):
F = 83.981
F = D+W----------(1)
And FxF = DxD + WxW ----------(2)
Now fixing xD as 0.99 and xW =0.05, xF=0.89885 We have solving
(1) and (2):
83.981x 0.89885= Dx0.99+(83.981-D)0.05 Solving for D we have
D = 75.837 Kg moles And W = 8.143 Kg moles
Let us take the 0.01 mole fraction lost is distillate is all
carbitol(it is lighter then triglycol ether)
Therefore carbitol lost above is 0.75837 Kg moles
Now W= 8.143 Kg moles and has (8.143x0.05)=0.40715 Kg moles of
mono-glycol ether
The carbitol in crude stream a on a C2H4O free basis is =
(83.981x7.101)/100= 5.963 Kg moles
Of this 5.963 Kg moles of carbitol, 0.75837 Kg moles of carbitol
is lost in distillate.
Therefore carbitol present =5.963 0.75837 =5.205 Kg moles
The remaining in triglycol ether =8.143 5.205 0.40715 =
2.5307
For tower (2):
F =8.143 Kg moles
Fixing xD as 0.99 and this is the purity of the product and xW
=0.05 We have F =D + WFxF =DxD + WxW
And xF of carbitol =5.205/8.143 =0.639 8.143x0.6391 = Dx0.99 +
(8.143-D)x0.05 Solving for D we get
D=5.1040 Kg moles And W =3.038 Kg moles
For 100 Kg moles of feed we have 5.1040 Kg moles of carbitol of
0.99 purity 25 TPD = 186.567 Kg moles/day
The ethylene oxide feed is 3655.309 Kg mole/day
The C2H5OH feed is (81.450x186.567)/5.1040
= 2977.249 Kg mole/dayCrude product(C2H4O-free) is
(83.981x3655.309)/100
=3069.765 Kg mole/day Crude product with C2H4O is
(88.981x3655.309)/100
= 3252.530 Kg moles/day The C2H4O recycle is
(518.55x3655.309)/100
= 18954.558 Kg mole/day
Mono glycol ether,distillate tower (1) =75.837x36.55309
=2772.069 kg moles/dayF of tower (2) is (8.143x3655.309)/100
= 297.6510 Kg moles/day And D is (5.1040x3655.309)/100= 186.566
Kg moles/day
And w is (3.038x3655.309)/100 = 111. 048 Kg moles/day
ENERGY BALANCE
The datum temperature is taken as 30c.
1) ENERGY BALANCE AROUND THE REACTOR
Feed of ethylene oxide and ethyl alcohol can be assumed at room
temp (30c).
1 hour of operation is taken as the basis.
Heat input + generation rate heat output - heat supplied
externally b. Heat input = (mcpt)alcohol + (mcpt)EtO 0 + 0 ,(
treference = tfeed) 0
c. Generation rate
The heat of reaction is 23kcal/gmol of EtO reacted.
Generation rate = +Rfeed rateconversion 234180103152.300.95
13.9100159109J/hr c. Heat output (mcpt)alcohol+
(mcpt)mge+(mcpt)carbitol+(mcpt)tge+(mcpt)EtO
The product outlet temperature is taken as that of reaction
temp,200c. (mcpt)alcohol = 789.77175170kJ/hr
(mcpt)mge = 114.97210170 kJ/hr
(mcpt)EtO = 7.614115170 kJ/hr
(mcpt)carbitol = 9.0825481.4170 kJ/hr
(mcpt)tge = 439.6143.85125170 kJ/hr
Therefore heat output = 28.780106 kJ/hr
d. Heat supplied externally = ( 28.780- 13.910 )106 kJ/hr
14.8700109 J/hr 4130.5kW.
2) 2) ENERGY BALANCE AROUND THE ALCOHOL RECOVERY TOWER
Heat input + Reboiler load= heat output + heat load of
condenser
a. Heat input
output from the reactor is feed for the tower hence, heat input
= 28.780109J/hr.b. Heat output
The products of the tower are at 100c.The output of the tower
are distillate and residue.
heat out with distillate = (mcpt)alcohol= 789.2715070 kJ/hr.heat
out with residue = (mcpt)carbitol + (mcpt)mge + (mcpt)tge
9.0825343.2570 + 114.9710570 3.8512376.7570 1.16718kJ/hr
The total heat output = 9.453106kJ/hr
c. Condenser load- Reboiler load = (28.780-9.453)109
19.327109J/hr.
3) ENERGY BALANCE AROUND TOWER 1 ( MGE-SEPARATION)
Heat input + Reboiler load= heat output + heat load of
condenser
a. Heat input
output from the alcohol recovery tower is feed for the tower
hence, heat input = 1.1671109J/hr.b. Heat output
The products of the tower are at 150c.
The output of the tower are distillate and residue.heat out with
distillate = (mcpt)mge= 114.97205120 kJ/hr. heat out with residue =
(mcpt)carbitol + (mcpt)tge 9.0825418.99120 + 3.8512457.52120
0.667106 kJ/hr
The total heat output = 3.4952106 kJ/hr
c. Reboiler load - Condenser load = (3.4952-1.1671)109
2.3281109J/hr.
4) ENERGY BALANCE AROUND TOWER 2 ( PRODUCT TOWER). Heat input +
Reboiler load
heat output + heat load of condenser
b. Heat input
output from tower 1 is feed for the tower hence, heat input =
0.667109J/hr.b. Heat outputThe products of the tower are at
202c.The output of the tower are distillate and residue.
heat out with distillate = (mcpt)carbitol
= 9.0825481.6172 kJ/hr. heat out with residue = (mcpt)tge
3.8512505.6172 0.3349106 kJ/hr
The total heat output = 1.0872106 kJ/hr
c. Reboiler load - Condenser load = (1.0872-0.667)109
0.4202109J/hr
MAJOR EQUIPMENT DESIGN
DISTILLATION COLUMN
Process Design
The feed rate for the tower is 297.6510 kmols/day Therefore F =
297.6510 kmols/day
= 12.4021 kmols/hr The mole fraction of carbitol in the feed is
0.639. i.e Xf = 0.639
Neglecting the residual mono-glycol ether and other impurities,
thus the mixture can be treated as binary mixture of carbitol and
triglycol ether.
The distillate rate is 186.566 kmols/day
Therefore D = 186.566 kmols/day
= 7.7735 kmols/hr
The mole fraction of carbitol in the distillate is 0.99. i.e xd
= 0.99
The residue rate is 111.048 kmols/day
Therefore W = 111.048 kmols/day = 4.627 kmols/hr
The mole fraction of carbitol in the residue is 0.05 i.e xw =
0.05
The feed is taken as saturated liquid, since it is coming from a
Reboiler.
Therefore q = ( Hv - Hf )/ ( Hv Hl )
= 1
Therefore the slope of the q-line is q/(q-1) =
Vapor pressure data
TemperatureVapor pressure of MCBVapor pressure of DCB
( K )(mm Hg)(mm Hg)
405700190
408810210
411900240
4151000260
4181050300
4221150330
4261300370
4291400400
4331500450
4371700500
4411800575
4442000625
4482200680
4532400740
Vapor liquid equilibrium data
Temperaturexaya
(K)( mole fraction of( mole fraction of
MCB in liquid )MCB in gas )
4080.920.98
4110.790.94
4150.680.90
4180.610.85
4220.520.79
4260.420.72
4290.360.66
4330.300.59
4370.220.49
4410.150.36
4440.100.26
4480.050.15
4530.010.04
From graph:Minimum reflux ratio ( Rm ) Operating reflux
ratio
= 0.17857 = 1.5 x Rm
= 0.2678 Slope of operating line xd/( R+1) = 0.7808 From the
graph the number of ideal stages is eight. Therefore total no of
trays in the column is seven. Number of stages in enriching section
is four. Number of stages in stripping section is three.
Liq flow rate in enriching section L
d. = D x R
7.7735 x 0.2678
2.0817 kmols/hr
Vapor flow rate in the enriching section G
G = L + D
( 1 + R )D
1.2678 x 7.7735
9.855 kmols/hr
Liq flow rate in the stripping section L
d. = L + qF
2.0817 + 12.4021
14.4838 kmols/hr
Vapor flow rate in the stripping section G
e. = G + ( q - 1 )F
= 9.855 kmols/hr
EVALUATION OF AVERAGE PROPERTIES OF THE MIXTURE
DENSITY CALCULATION
Liq phase calculationsCarbitol density at various
temperatures
( ! )202c = 5.977kmol/m3. ( ! )213c = 5.857kmol/m3. ( ! )261c =
5.379kmol/m3.Triglycol ether ( TGE ) density at various
temperatures( ! )202c = 6.6209kmol/m3. ( ! )213c = 6.5141kmol/m3. (
! )261c = 6.0870kmol/m3.
Vapor phase calculations Assuming ideal gas
PV = nRT With P = 1 atm
R = 0.082 m3atm/kmol k T = temp in Kelvin
Carbitol density at various temperatures ( ! )204c =
0.025565kmol/m3.
( ! )218c = 0.024835kmol/m3. ( ! )282c = 0.21970kmol/m3.
Triglycol ether density at various temperatures ( ! )204c =
0.25566kmol/m3.
( ! )218c = 0.02483kmol/m3. ( ! )282c = 0.021972kmol/m3.
Mixture properties
( !l ) kg/m3 = avg molecular wt/( x1 !carbitol + x2 !TGE )
( !l )202c = 134.16/(0.99/5.977 + 0.01/6.6209)
= 802.65 kg/m3 ( !l )213c = 838.248 kg/m3( !l )261c = 885.248
kg/m3( !v )204c= 3.4298 kg/m3
( !v )218c= 3.3550 kg/m3
( !v )282c= 3.233 kg/m3
The average properties in the enriching section and stripping
section can be summarized into the table as below.
PROPERTYENRICHINGSTRIPPING
SECTIONSECTION
TOPBOTTOMTOPBOTTOM
Liq flow rate2.08172.081714.483814.4838
kmols/hr
Vap flow rate9.8559.8559.8559.855
kmols/hr
Liq mole0.990.70.70.13
fraction x
Vap mole0.990.930.930.18
fraction y
Avg ( liq )134.16138.80138.80147.92
molecular wt
g/mol
Avg ( vap )134.16135.12135.12147.12
molecular wt
g/mol
Temp liq (c)202213213261
Temp vap(c)214218218282
Liq flow rate279.28288.9392010.382142.44
kg/hr (L)
Vap flow rate1322.141331.601331.601449.86
kg/hr(G)
Liq density !l802.65838.32838.32885.248
kg/m3
9DS GHQVLW\ !v3.42983.355023.355023.233
kg/m3
L/G[ !v !l)0.50.013800.0137260.095500.08930
ENRICHING SECTION DESIGN
1.TOWER DIAMETER CALCULATIONThe maximum value of L/G[ !v !l)0.5
is 0.1380 at the top of the section
Choosing Intalox saddles, polypropylene Nominal size =
25mmSurface area = 206 m2/m3.
Fp = 105 per meter
From steam tables the density of water at 202c is
802.65kg/m3.
!water !liq = 862.28/802.65 = 1.0742
l1/3(mix) = x1 11/3 + x2 21/3
0.99(0.41/3) + 0.01(0.921/3)
0.7360
l (mix) = 0.4 cp
From graph (18-38)G2)S% 0.2 !g !l*g) = 0.22
Solving we obtain Gf = 7.954 kg/m2s
Choosing 65% flooding
We get G = 0.65Gf = 5.17027kg/m2s.
Therefore the cross sectional area of the tower
Ac = mass flow rate/G
1322.14/( 3600 x 5.17027)
0.07103 m2
The diameter of the column = {(4xAc)/}0.5
Dc = 0.30073 mTaking the corrected Dc = 0.300 m Therefore Ac =
0.07068m2= 0.76085ft2
tower dia /packing dia (Dc/dp) = 0.30/0.025 = 12 ( > 10 )
WETTING RATE
Lmin = 279.28/(802.65x60x4.546097x10-3)Gallons/min =
1.2756GPM
Degree of wetting Lmin/tower area in ft2
1.2756/0.760855
1.676536 GPM/ft2
Hence acceptable ( greater than the limits of 0.5 GPM/ft2 ).
PRESSURE DROP CALCULATIONS
Pressure drop can be calculated using the equationS &2 *10
(C3*Utl) !g*Utg2- equation 18.48
where
C2, C3 are constants to be evaluated from table 18.7 Utl, Utg
are superficial liq and vapor flow rate ( ft/sec ).!g isavg vapor
density ( lb/ft3 ).
SSUHVVXUH GUop in inch water/ft packing
)
Utl =L /$F !
l
=284.11/( 3600x0.7068x802.48x0.3048)
= 4.4647x10-3 ft/s.Utg = G $F !g)
= 1326.87/(3600x0.7068x3.39241x0.3048) =5.0431 ft/s.Density of
gas !g = 3.3924 Kg/m3 = 0.21177 lb/ft3
C2 = 0.31C3 = 0.0222S&2 *10 (C3*Utl) !g*Utg
0.31*10 (0.0222*0.004464)*0.21177* 5.04312
1.6700 in water/ft packing
139.06 mm water/m packing
STRIPPING SECTION DESIGN
1.TOWER DIAMETER CALCULATIONThe maximum value of L/G[ !v !l)0.5
is 0.09550 at the top of the section
Choosing Intalox saddles, polypropylene Nominal size =
25mmSurface area = 206 m2/m3.
Fp = 105 per meter
From steam tables the density of water at 213c is
848.662kg/m3.
!water !liq = 862.28/838.32 = 1.01233
l1/3(mix) = x1 11/3 + x2 21/3
0.7(0.41/3) + 0.3(0.941/3)
0.7663
l (mix) = 0.45 cpFrom graph (18-38)
G2)S% 0.2 !g !l*g) = 0.15
Solving we obtain Gf = 7.038 kg/m2s
Choosing 65% flooding
We get G = 0.65Gf = 4.574kg/m2s.
Therefore the cross sectional area of the tower
Ac = mass flow rate/G
1331.60/( 3600 x 4.574)
0.0808 m2
The diameter of the column = {(4xAc)/}0.5
Dc = 0.3205 m Taking the corrected Dc = 0.300 m Therefore Ac =
0.07068m2= 0.76085ft2
tower dia /packing dia (Dc/dp) = 0.30/0.025 = 12 ( > 10 )
WETTING RATE
Lmin = 2010.38/(838.32x60x4.546097x10-3)Gallons/min =
8.791GPM
Degree of wetting Lmin/tower area in ft2
8.791/0.760855
11.555 GPM/ft2
Hence acceptable ( greater than the limits of 0.5 GPM/ft2
).PRESSURE DROP CALCULATIONS
Pressure drop can be calculated using the equationS &2 *10
(C3*Utl) !g*Utg2- equation 18.48
where
C2, C3 are constants to be evaluated from table 18.7 Utl, Utg
are superficial liq and vapor flow rate ( ft/sec ).!g isavg vapor
density ( lb/ft3 ).
SSUHVVXUH GUop in inch water/ft packing
Utl = L / $F !l)
2076.41/( 3600x0.7068x861.78x0.3048)
0.031067 ft/s.
Utg = G $F !g)= 1390.73/(3600x0.7068x3.29401x0.3048)
=5.443 ft/s.'HQVLW\ RI JDV !g = 3.29401 Kg/m3 = 0.20565
lb/ft3
C2 = 0.31C3 = 0.0222S&2 *10 (C3*Utl) !g*Utg
0.31*10 (0.0222*0.0.31067)*0.20565* 5.4432
1.8917 in water/ft packing
157.60 mm water/m packing
AVERAGE CONDITIONS FOR ENRICHING AND
STRIPPING SECTION
PROPERTYENRICHINGSTRIPPING
SECTONSECTION
Liq flow rate kmols/hr2.081714.4838
Liq flow rate kgs/hr284.102076.41
Vapor flow rate9.8559.855
kmols/hr
Temp of liq c207.5247.5
Temp of vap c211250
'HQVLW\ RI OLT !liq820.485861.784
kg/m3
'HQVLW\ RI YDS !vap3.392413.2940
kg/m3
l cp0.4250.46
vap cp0.0100250.010756
liq dynes/cm12.62215.6123
DlAB cm2/sec16.408x10-617.58x10-6
DVAB cm2/sec0.052780.06082
Liq Schmidt number315.609303.627
Nsl
Vap Schmidt number0.5590.5368
Nsg
TOWER HEIGHT CALCULATION
ENRICHING SECTION
The height of the enriching section can be given by
Z = H og *NogWhere
Hog = Hg + m*Gm/Lm*Hl
Nog = NTln -1)is the stripping factor given by mGm/Lm
Hg% 'c1.11*Z0.33*Scg / (L*f1*f2*f3*f4)0.5
From fig 18.65
Dc = 0.3m
L = 284.10/(0.7068x3600)
= 1.11653 kg/s m2
f1l water )0.16
( 0.425/1.0 )0.16
0.872
f2 !water !l )1.25
(1000/820)1.25
1.0322
f3water/ l)0.8
= ( 72.8/12.622 )0.8= 4.0625
Hg =
0.029*50*(0.3)1.11*(Z)0.33*(0.559)0.5/(1.11653*0.872*1.3022*4.0625)0.5
= 0.5365*(Z)0.33
By Cornell method eqn 18-56Hl !& l !lDl)0.5(z/3.05)0.15
Liquid rate = 5.2147 Kg/m2 s
! P IURP ILJ -60)C = 0.8 (from fig 18-59)
Hl = 0.035*0.8/3.28 * (315.69)0.5 * (Z/3.05)0.15
= 0.1282*(Z)0.15
H og = 0.5365*(Z)0.33 + 0.1214*4.734*0.1282*(Z)0.15 =
0.5365*(Z)0.33 + 0.0737*(Z)0.15Nog = NTln -1)is the stripping
factor given by mGm/Lm
m is the slope of equilibrium line in enriching section. m (top)
= 0.1
m (bottom) = 0.1428 m (avg) = 0.1214 Gm/Lm = 9.855/2.0817
4.734 = 0.5750 NT = 4
Nog = 4ln0.5750/(0.5750-1)
= 5.2144
Z = H og *Nog {0.5365*(Z)0.33 + 0.0737*(Z)0.15}5.2144
2.796*(Z)0.33 + 0.3844*(Z)0.15 Solving the above equation by
trail and error, we get Z = 5.35m Therefore height of enriching
section is 5.35mSTRIPPING SECTION
The height of the stripping section can be given by
Z = H og *NogWhere
Hog = Hg + m*Gm/Lm*Hl
Nog = NTln -1)is the stripping factor given by mGm/Lm
Hg% 'c1.11*Z0.33*Scg / (L*f1*f2*f3*f4)0.5
From fig 18.65
Dc = 0.3m
L = 2076.41/(0.7068x3600)
= 8.160 kg/s m2
f1l water )0.16
( 0.46/1.0 )0.16
0.8831
f2 !water !l )1.25
(1000/861.78)1.25
1.204
f3water/ l)0.8
( 72.8/15.612 )0.8
3.427
Hg =
0.029*50*(0.3)1.11*(Z)0.33*(0.559)0.5/(6017.69*0.8831*1.204*3.427)0.5
= 0.052*(Z)0.33
By Cornell method eqn 18-56Hl !& l !lDl)0.5(z/3.05)0.15
Liquid rate = 8.160 Kg/m2 s
! P IURP ILJ -60)C = 0.8 (from fig 18-59)
Hl = 0.07*0.8/3.28 * (303.627)0.5 * (Z/3.05)0.15
= 0.2516*(Z)0.15
H og = 0.0527*(Z)0.33 + 0.86751*0.25167*(Z)0.15
= 0.0527*(Z)0.33 + 0.21826*(Z)0.15
Nog = NTln -1)is the stripping factor given by mGm/Lm
m is the slope of equilibrium line in stripping section. m (top)
= 0.3
m (bottom) = 2.25 m (avg) = 0.1.275 Gm/Lm = 0.6804= 0.86751
NT = 3
Nog = 3ln0.8675/(0.8675-1)
= 3.218
Z = H og *Nog {0.0527*(Z)0.33 + 0.21826*(Z)0.15}3.218
0.1673*(Z)0.33 + 0.7023*(Z)0.15
Solving the above equation by trail and error, we get Z =
0.84m
Therefore height of stripping section is 0.84m Total height of
the tower is
Z = (5.35+0.84) = 6.19m(B) MECHANICAL DESIGN OF DISTILLATION
COLUMN
Diameter of the tower Di = 0.3m
Working pressure = 1 atm=1.0329 kg/m2
Design pressure pd = 1.1362 kg/m2
Shell material Plain Carbon steel
Permissible tensile stress (ft)= 950kg/cm2
Insulation thickness = 50mm
Density of insulation = 770 kg/m3
Top disengaging space = 1m
Bottom separator space = 2m
Skirt height = 2m
Density of material column = 7700 kg/m3
Wind pressure = 130 kg/m2
1) Shell thicknessts= PDi/(2fJ-P) + C
P= design pressure in kg/cm3 f=allowable tensile stress
kg/cm3
C= corrosion allowance ( 2 mm )
J=joint factor
ts= (1.1362*300)/(2*950*0.85-1.1362) + C ts = 2.2112 mm
minimum thickness allowable is 6mm ts= 6mm
2) Head Design
Shallow dished and torispherical head
Thickness of head is given by
th= PRCW/2fJ
Rc=crown radius,300mm
W=stress intensification factor
W= 0.25(3 + (RC/RK)
Rk= knuckle radius ,6% of crown radius.
W= 1.7706 mm
th= 1.1362*300*1.7706/2*950*0.85 ts= 0.3737 mmminimum thickness
is ts= 6mm
3) Shell thickness at different heights
At a distance X m from the top of the shell the stress are;
Axial Stress: (tensile)fap = pi Di___4(ts C)
1.1362*300/4*(6-2)
142.025 kg/m3
4)Compressive stress due to weight of shell up to a distance X
fds = /4 * ( Do2 Di2 ) s X/4 * ( Do2 Di2 ) s X
0.77X kg/m3
5) Compressive stress due to weight of insulation fd(ins) = Dins
tin ins Dm (ts C )
fd(ins) = 412*50*770*X
306*(6-2) fd(ins) = 1.1015X
6) compressive Stress due to the weight of the liquid and
packing
fd = OLT DQG SDFNLQJ ZW XQLW KHLJKW ; Dm(ts-C)For the chosen
packing, 25mm plastic intalox saddles
= 0.91
approximate wt/m3 kg/m3 = 76 Number of elements per m3 =
55800
Total volume of the packing = *Dc2*h/4
= 0.4375m3
Total void volume = 0.91*0.4375 = 0.3981m3Total volume of the
actual packing = 0.039379m3
Average density of the liq in the column is 841.13kg/m3.
fd = (76+841.13*0.3981)/0.4375*X Dm(ts- c)
24.42X kg/cm2
7) Stress due to the weight of the attachments The total weight
of the attachments
The weight of the head is taken as 1020kgs Wa = (1020 +
140X)
Fd(att) = (1020 + 140X)/(*30.6*0.4)
26.525 + 3.640X kg/cm2
8) Total compressive dead weight stress at height X ( sum of
2-7) fds=29.9315X + 26.525 9) Stress due to wind load at distance X
fws = 1.4*PwX2/*Do(ts-c)
(1.4*130*X2)/(*31.2*.4)
4.642X2
10) Stress in upwind side
fmax= fws+fap-fdx0.8*950 = 4.642X2+142.025-29.9315X-26.525
4.642X2-29.9315X-644.5 = 0 solving for X
X = 15.2177 m 11) stress in down sidefmax=
fws+fap+fdx4.642X2-29.9315X-928.55 = 0 X = 17.730 m
From this, whole tower of 6mm thickness is enough.
12) Skirt design
The material of construction for skirt is carbon steel
IS:2062-1962
Minimum weight of vesselWmin= (Di+ts) ts(H-2)s + 2WH
H=11.29 (Total height of tower including skirt height)
s= 7700kg/m3 (specific weight of shell material)
WH= 1020 kg (weight of head)
Wmin= (0.3+0.006) (0.006)(11.29-2)7700 + 2*1020
Wmin= 2171.33 kg
Maximum weight of vesselWmax=Ws+Wi+Wl+WaWs=10800kg(weight of
shell during test) Wi= 4200kg (weight of insulation)
Wl= 656.67 kg (weight of water during test) Wa= 4400 kg (weight
of attachments)
Wmax=20056.67 kg
Wind loadPW= K1pwHD
For minimum weight of column,D = 0.3m
Pw(min)= 0.7*130*11.29*0.3 = 308.217 kg
Pw(max)= 0.7*130*11.29*0.312 = 320.54 kg
Minimum wind momentMW(min) = PW(min)*H/2 308.217*11.29/2 1739.88
kg m
Maximum wind moment
MW(min) = PW(min)*H/2 320.54*11.29/2 1809.44 kg m
Bending stresses
fb(min)= 4Mw(min) *D2*t
= 4*1739.88 *0.32*t= 2.461/tkg/cm2
fb(max)= 4Mw(max) *D2*t
4*1809.88 *0.32*t
2.5604/t kg/cm2
Minimum dead load stressFds(min)= Wmin/dt 2171.33 /*0.3*t
0.23038/t kg/cm2
Maximum dead load
Fds(max)= Wmax/dt
= 0.21280/t kg/cm2
Maximum tensile stress without any eccentric load fz=
fbs(max)-fbs(min)
980*0.8= 0.0994/t t= 0.1449mm
Maximum compressive stress without any eccentric load
fz= fbs(max)-fbs(min)fz= 0.125 E (t/Do) 0.125*2.04*106*t/0.3
850000t
850000t = 2.5604/t+2.461/t t=2.4305 m
Minimum skirt thickness is 7mm,by providing 1mm corrosion
allowancets=8mm
DESIGN OF SKIRT BEARING BOLTS
Maximum compressive stress between bearing plate and foundation
fc= Wmax/A + Mw/2
A=(Do-l)/2l=outer radius of bearing plate minus outer radius of
skrit Z=Rm2 lRm=(Do-l)/2
fc = 20056 /((0.3-l)l) + 1809.44/((0.3-l)2l
The allowable compressive stress of concrete foundation varies
from 5.5 to 9.5 MN/m20.55*106 = 20056 /((0.3-l)l) +
1809.44/((0.3-l)2l
l=57mm
As required width of bearing plate is very small a 100 mm width
is selected
l=0.1 m
therefore fc= 0.50253*106 kg/m2
thickness of bearing plate
tbp=l(3fc/f)
= 100(3*0.5026*106/96*106) = 12.53 mmBearing plate thickness of
12.53 mm is required
As the plate thickness required is less than 20mm, no
reinforcement is required.
fmin= Wmin/A Ms/Z
2171.33/(0.3-0.1)0.1 + 1739.88/(0.3-0.1)20.1 103897kg/m2
The modulus value is taken, less than zero implies that the
vessel must be anchored to the concrete foundation by means of
anchor bolts toprevent overturning owning to the bending moment
induced by the wind or seismic load.
Therefore anchor bolts are to be used Pbolt*n=fmin*AwherePbolt =
load on one anchor boltfmin = stress determined by eqn10.211
A = area of the contact between bearing plate and foundation
103897.5*3.14*(0.3-0.1)*0.1
6528.07 kg/m2
For hot rolled carbon steel f=5.73*106 kg/m2 (arn)f=nPbolt
arn=1139.2 mm2 For 12X1.5 ,ar=63mm2
Number of bolts=11392/63=18 bolts
Therefore 18 bolts are to be distributed equally.
MINOR EQUIPMENT DESIGN
CONDENSER
Process Design
The reflux condenser, condenses vapor of the column and send it
as the reflux.
The vapor flow rate is 9.855kmols/hr. Tvap = 202c. ( sat vap
).Molecular wt of vap = 134+HDW RI FRQGHQVDWLRQ 202c =
21000Btu/lbmol 364.52kJ/kg.
48846kJ/kmol.
Heat load Q = 9.85548846 481377.33kJ/hr. 133.71kW.
Let the overload be 10%.
Therefore Q = 1.1133.71kW= 147.0875kW.
Let water at a temp of 25c be used to condense the vapor. Fixing
the outlet temp of water as 35c.
The water flow rateWc = 147.0875/Cpwater(Tout- Tin)
147.0875/4.18710
3.512kg/s. Tvap = 202c,
7lmtd = {(202-35)-(202-25)}/ln{(202-35)/(202-25)} = 171.95c
Let us assume an overall heat transfer coefficient,( U ) of
567.83J/m2sc. As the heat load is very low, we shall use a
DPHE.
Required area for heat transfer is A Q/ 7lmtdU
147.08875/(171.95567.83)
1.369m2. N = A/( Ld02 ) Choosing 2 NPS, 40SCH and 1.25 as the
tubes. Inner dia of 2 NPS pipe= 5.25cm
Outer dia of 2NPS pipe= 6.032cm
Inner dia of 1.25 NPS pipe= 3.505cm
Outer dia of 1.25 NPS pipe= 4.216cm
Taking the length of the pipe as 6 ft ( 1.828m) Length available
for heat transfer = 1.528m Heat transfer area, outside area of
inner pipe = Ld02N where N, is the number of hairpins. Therefore ,
= 1.369m2/(1.5280.042162) = 3.382 Taking the number of hairpins N
as 4. Therefore the corrected heat transfer area = 1.61906m2
Corrected overall heat transfer coefficient = 480W/m2c Location of
the fluids , the vapor is taken in the annulus and water in the
tube. OVERALL HEAT TRANSFER COEFFICIENT CALCULATION (Ud) The
overall heat transfer coefficient is given by the equation 1/Ud =
1/ho+(Do/Di)(1/hi)+ {xwDo/(Dwk)}+ dirt factor where, ho, hi are
outside and inside heat transfer coefficients. xw,Dw are wall
thickness and mean wall diameter. k is wall material thermal
conductivity. 1)ANNULUS SIDE (carbitol vapors) The individual heat
transfer coefficient ho = 0.725{[K3!2g @ >' 7@`0.25where,
K is the thermal conductivity of condensate = 0.12461W/mK! LV
WKH GHQVLW\ RI FRQGHQVDWHNJ P3.
g is acceleration due to gravity= 9.81m/s2.
LV WKH KHDW RI FRQGHQVDWLRQN- NJ
D is the outside dia of the tube= 0.04216m
LV WKH YLVFRVLW\ RI WKH FRQGHQVDWHFS
7 LV WHPS GLIIHUHQFH RI WKH FRQGHQVDte and the wall =
202-116
= 86c.
Therefore,ho = 959.92W/m2K
2) Inside heat transfer coefficient ( tube side)
hidi/K = 0.023(Re)0.8(NPr)0.4Where,
Re is the Reynolds number. NPr is the prandtl number.
5H'9!
4Mass flow rate/(' 43.512/(0.035050.8510-3) 150092.0
NPr = CP . 4.181030.8510-3/0.16
5.82
Therefore
hidi/K = 0.023(150092)0.8(5.82)0.4 = 643.84
hi = 643.84K/di 643.840.61/0.03505
11205.3W/m2K 3) Wall Resistance can be expressed as
Mean temp of the wall is 116c.
{xwDo/(Dwk)} = 0.003550.04216/(0.0384945)= 8.6410910-5 m2K/W
4) Dirt factor of 0.0005 is assumed.
Therefore,1/Ud = 1/ho+(Do/Di)(1/hi)+ {xwDo/(Dwk)}+ dirt
factor
0.0889/(0.0736611205.3) + (1/959.92) + 8.6410910-5 + 0.0005
1.7358810-3. Ud = 576.076W/m2K
Since the Uassumed < Ud, the design is acceptable from heat
transfer point of view.
PRESSURE DROP CALCULATIONS
1)TUBE SIDEPT=(4fLvt2/2gDi)tg
Tube side Reynolds number=NRe= 150092.0
Friction factor f = 0.079(NRe)-1/4 =0.079(150092) -1/4
=4.01310-3
Tube side velocity vt = 3.656m/s
PT =
(44.01310-314.6243.6562/(29.80.03505)9969.8
44.54 kPa.
2) ANNULUS SIDE PA= (4fLva2/2gDH)g
Annulus side Reynolds number can be calculated as DH Mass flow
rate/( area of annulus)
0.010640.3668/(0.009710-37.68310-4)
508.91103 Friction factor f = 0.079(NRe)-1/4
=0.079(508910) -1/4 =2.95710-4
Annulus side velocity is 138.784m/s at one end negligible at
other end. Therefore the annulus side velocity = (138.78 + 0 )
0.5
= 69.392m/s.
PA =
(42.95710-414.62469.3922/(29.80.01064))3.349.8 13.0728 kPa.
Hence the pressure drops are acceptable.
Mechanical Design
Let the material of construction be 15C-8, carbon steel.
T( tensile strength ) = 410 N/mm2.
y( yield strength ) = 220N/mm2. The pressure in the annulus is
taken as 1 atm. The design pressure be taken as
P = 1.251 atm = 1.25atm
1)The total load of the bolt is given by
Fa = Pressure annular cross sectional area
(1.25-1)7.68310-3
19.461N
2)The total load capacity of the bolt is given by
= C(Ar)1.418
= 2.29(Ar)1.418
Therefore , stress area of the bolt Ar =
(19.461/4.52268)-1.418
= 4.52268mm2.
3)From table 9.8, for the obtained Ar , bolt dia d = 3mmpitch=
o.5mm.
4) Initial tension load in a bolt
Fi = 2805d
4815N
5) Effect of applied load on bolt stress
The final load on the bolt = K Fa + Fi , K for asbestos gasket =
0.6 F = 0.619.461 + 4815
= 4826.67N5) Number of bolts,
An empirical rule for the number of bolts in pipe joints is
given by
N = 0.024Dc+ 2 , Dc = dia of cylinder 60.32mm N = 3.44768
As a standard we can provide 6 bolts
6) The maximum spacing of the bolt in any fluid tight joint
s G
s = 6d
s = 18mm
7) The extension at one end is the same as that of the pipe. The
bolt circle diameter is given byD2 = D1 + 3.2d
D1 = 1.8D +20 = 128.576mmD2 = 128.57 + 3.23
138.176mm
8) The flange thickness is given by
t = 0.35D + 9 mm
0.3560.32 + 9
31.12mm
POLLUTION AND SAFETY
The raw material ethylene oxide used for the production of
carbitol will explode in the presence of common igniters. Pure EtO
vapor is difficult to ignite compared to oxide-air or
hydrocarbon-air mixtures, requiring spark energies about ten
thousand times larger.
The design of the plant should be such that the gas mixtures
handled are always outside the explosive limit. The actual safe
operating ranges are dependent on operating temperatures,
pressures, equipment configurations, gas composition, dynamics of
catalyst and instrumentation.
Health affects
Toxic effects
There are no reports on effects of occupational exposure. There
is one case report describing a man who drank about 300 ml of
DEGEE. He developed severe symptoms of poisoning: CNS effects,
breathing difficulty, thirst, acidosis and albuminuria .
An unpublished report (Kligman, 1972) cited by Opdyke describes
dermal application of 20% DEGEE in petroleum jelly, under
occlusion, to 25 volunteers for 48 hours. The application resulted
in no irritation or sensitization. In another sensitization study,
pure DEGEE was applied under occlusion to the backs of 98 young men
for 7 days, followed by a 3-day application 10 days later. No skin
sensitization or edema was observed, but 7 of the men had
pronounced skin reddening.
ACUTE/CHRONIC HAZARDS:
When heated to decomposition this compound emits acrid smoke,
irritating fumes and toxic fumes of carbon monoxide and carbon
dioxidePersonal Protection and Exposure Control
Eye: Wear goggles to avoid splash. Skin: Use protective
gloves.Inhalation: Use mask while spraying. Avoid inhalation.
Engineering Controls: Sufficient ventilation to keep within OSHA
PEL/TLV limit.The dried film of the product may contain all or some
of the following OSHA chemicals And may become a dust nuisance when
removed by sanding or grinding.OSHA recommends a PEL/TWA of 15
mg/m3 for the respirable fraction. ACGIH recommends TLV/TWA of 10
mg/m3 for total dust. Use approved mask, eye goggles and gloves
while sanding, or grinding. Skin absorption may contribute to the
overall exposure of the material.
HEALTH EFFECTS OF OVER EXPOSURE AND FIRST AID
Primary RoutesSymptomsFirst Aid
EyeMay cause burning andFlush with water. Seek
irritation uponmedical attention
direct contact.
SkinDirect skin contact mayWash with soap and
cause skinwater.
irritation and dermatitis.
IngestionSevere oral intoxicationDo not induce
will lead tovomiting. Seek medical
intense burning of theattention.
throat and may
result in drowsiness,
numbness and
headache followed by
weakness &
nausea.
InhalationAcute overexposure inRemove person to fresh
mist form mayair. Apply
result in irritation ofartificial respiration.
throat & lungs.
COST ESTIMATION AND ECONOMICS
Marshall and Shift index in 1992 for equipments is 943.1 Cost
estimation based on the equipment
Cost of the bare columns is 3*11000. Saddles cost 720/m3.
Total cost of the packing ,for the 3 columns 3*315. Cost of the
reactor is 16000.
There are five heat exchangers, reboilers and condensers, of
shell and tube type. The approximate cost of the each exchanger is
3000.
The cost of one DPHE is 1500.
Therefore the total cost of the equipment = 66445. Taking 1 = Rs
63.
Total cost of the equipment = Rs 66445*63*1048/943.1. = Rs
4.65115*106.
ESTIMATION OF DIRECT COST
COMPONENTSCOSTS
Purchased equipment cost ( E )Rs 4.65115*106
Purchased equipment installationRs 1.8139*106
( 39% of E )
Instrumentation (installed cost),Rs 1.3023*106
28%E
Piping installed, 31%ERs 1.4418*106
Electrical installation, 10%ERs 0.465115*106
Yard improvement, 10%ERs 0.465115*106
Service facility, 55%ERs 2.55813*106
Land , 6%ERs 0.279069*106
TOTAL DIRECT COST (D)Rs 13.999*106
ESTIMATION OF INDIRECT COST1.Engg and supervision ( 32% E )= Rs
1.48836*106
2.Construction + contractor fees ( 25% direct cost )
= Rs 3.4997*106
Therefore total indirect cost ( I )= Rs 4.988118*106
DIRECT AND INDIRECT COST (TOTAL) = Rs 18.9871*106
Contingence ( 10%D+I )= Rs 1.89871*106
Fixed capital investment ( FCI ) ,
contingence + D + I= Rs 20.88801*106
Working Capital: (10-20% of Fixed-capital investment)Consider
the Working Capital = 15% of Fixed-capital investment i.e., Working
capital = 15% of 20.88801*106= 0.15 20.88801*106= Rs
3.13320*106
Total Capital Investment (TCI):
Total capital investment
e. Fixed capital investment + Working capital
f. Rs 24.02121*106
i.e., Total capital investment = Rs 24.02121*106
Estimation of Total Product cost:
I. Manufacturing Cost = Direct production cost + Fixed charges +
Plant overhead cost.
A. Fixed Charges: (10-20% total product cost)
i. Depreciation: (depends on life period, salvage value and
method of calculation-about 13% of FCI for machinery and
equipment and 2-3% for Building Value for Buildings) Consider
depreciation = 13% of FCI for machinery and equipment and 3% for
Building Value for Buildings)i.e., Depreciation = Rs.
2.71541106
ii. Local Taxes: (1-4% of fixed capital investment) Consider the
local taxes = 3% of fixed capital investment i.e. Local Taxes =
0.0320.88801*106
Rs. 0.6266106
iii. Insurances: (0.4-1% of fixed capital investment) Consider
the Insurance = 0.7% of fixed capital investment i.e. Insurance =
0.007 Rs 20.88801*106
Rs. 0.14621607106
iv. Rent: (8-12% of value of rented land and buildings) Consider
rent = 10% of value of rented land and buildings Rent = Rs.
0.1302322x106
Thus, Fixed Charges = Rs. 3.61844106
B. Direct Production Cost: (about 60% of total product cost)
Now we have Fixed charges = 10-20% of total product charges
(given) Consider the Fixed charges = 15% of total product cost
Total product charge = fixed charges/15%
Total product charge = 3.61844106/15%
Total product charge = 3.61844106/0.15
Total product charge(TPC) = Rs. 24.1229106
Raw Materials: (10-50% of total product cost)
Consider the cost of raw materials = 25% of total product
cost
Raw material cost = 25% of 24.1229106
Raw material cost = Rs. 6.03073106
ii. Operating Labour (OL): (10-20% of total product cost)
Consider the cost of operating labour = 12% of total product cost
operating labour cost = 12% of 24.1229106
Operating labour cost = Rs 2.89478106
iii. Direct Supervisory and Clerical Labour (DS & CL): iv.
(10-25% of OL)Consider the cost for Direct supervisory and clerical
labour= 12% of OL
Direct supervisory and clerical labour cost
12% of 2.89478106
0.34736106
iv. Utilities: (10-20% of total product cost)
Consider the cost of Utilities = 12% of total product cost
Utilities cost = 12% of 24.1229106
= 0.12 24.1229106
Utilities cost = Rs. 2.89464106
v. Maintenance and repairs (M & R):
(2-10% of fixed capital investment)
Consider the maintenance and repair cost
= 5% of fixed capital investment i.e. Maintenance and repair
cost = 0.0520.88801106= Rs. 1.0444106
vi. Operating Supplies: (10-20% of M & R or 0.5-1% of FCI)
Consider the cost of Operating supplies = 15% of M & R
Operating supplies cost = 15% of 1.0444106Operating supplies cost =
Rs. 0.15666106
vii. Laboratory Charges: (10-20% of OL) Consider the Laboratory
charges = 15% of OL Laboratory charges = 15% of 2.89478106
Laboratory charges = Rs. 0.434212106
Patent and Royalties: (0-6% of total product cost)
Consider the cost of Patent and royalties = 4% of total product
cost
d. Patent and Royalties = 4% of 24.1229106
e. Patent and Royalties cost = Rs. 0.964919106
Thus, Direct Production Cost = Rs. 14.7676106
C. Plant overhead Costs (50-70% of Operating labour,
supervision, and maintenance or 5-15% of total product cost);
includes for the following: general plant upkeep and overhead,
payroll overhead, packaging, medical services, safety and
protection, restaurants, recreation, salvage, laboratories, and
storage facilities.
Consider the plant overhead cost = 60% of OL, DS & CL, and M
& R Plant overhead cost
= 60% of ((2.8947106) + (0.3473106) + (1.04441106)) Plant
overhead cost = Rs. 2.5719106
Thus, Manufacture cost = Direct production cost + Fixed charges
+ Plant overhead costs.Manufacture cost = Rs. 20.9580106
AI. General Expenses = Administrative costs + distribution
and
selling costs + research and development costs
Administrative costs:(2-6% of total product cost) Consider the
Administrative costs = 5% of total product cost Administrative
costs = Rs. 1.206145106
B. Distribution and Selling costs: (2-20% of total product
cost); includes costs for sales offices, salesmen, shipping, and
advertising. Consider the Distribution and selling costs = 15% of
total product cost Distribution and selling costs = 15% of
2.41229107
Distribution and Selling costs = Rs. 3.6184106
C. Research and Development costs: (about 5% of total product
cost) Consider the Research and development costs = 5% of total
productcost
Research and Development costs = 5% of 2.41229107
Research and Development costs = Rs. 1.2010106
D. Financing (interest): (0-10% of total capital investment)
Consider interest = 5% of total capital investment i.e. interest =
5% of 2.40212107
Interest = Rs. 1.20106106
Thus, General Expenses = Rs. 7.23178106
III. Total Product cost = Manufacture cost + General Expenses
Total product cost = Rs. 28.1897106
V. Gross Earnings/Income:
Wholesale Selling Price of carbitol per ton = 70Hence Wholesale
Selling Price of carbitol per ton. = 63 70 = Rs. 4400 Total Income
= Selling price Quantity of product manufactured
= 4400 (25 T/day) (330 days/year) Total Income = Rs. 36.3106
Gross income = Total Income Total Product Cost
= 36.3106 24.122106
Gross Income = Rs. 12.1771106
Let the Tax rate be 45% (common) Taxes = 45% of Gross income=
45% of 12.1771106
Taxes = Rs. 5.4796106
Net Profit = Gross income - Taxes = Gross income (1- Tax rate)
Net profit = (12.1771106 ) (5.4796106)= Rs. 6.6974106
Rate of Return:
Rate of return = Net profit100/Total Capital Investment Rate of
Return = 6.6974106100/ (24.02121106) Rate of Return = 27.881%
PLANT LOCATION AND layout
THE LOCATION OF THE PLANT CAN HAVE A CRUCIAL EFFECT ON THE
PROFITABILITY OF A PROJECT, AND THE SCOPE FOR FUTURE EXPANSION.
MANY FACTORS MUST BE CONSIDERED WHEN SELECTING A SUITABLE SITE, AND
ONLY A BRIEF REVIEW OF THE PRINCIPAL FACTORS WILL BE GIVEN IN THIS
SECTION. THE PRINCIPAL FACTORS TO BE CONSIDERED ARE:
g. LOCATION, WITH RESPECT TO THE MARKETING AREA.
h. RAW MATERIAL SUPPLY.
i. TRANSPORT FACILITIES.
j. AVAILABILITY OF LABOUR.
k. AVAILABILITY OF UTILITIES: WATER, FUEL, POWER.
l. AVAILABILITY OF SUITABLE LAND.
m. ENVIRONMENTAL IMPACT, AND EFFLUENT DISPOSAL.
n. LOCAL COMMUNITY CONSIDERATIONS.
o. CLIMATE.
p. POLITICAL STRATEGIC CONSIDERATIONS.
MARKETING AREA
FOR MATERIALS THAT ARE PRODUCED IN BULK QUANTITIES: SUCH AS
CEMENT, MINERAL ACIDS AND FERTILIZERS, WHERE THE COST OF THE
PRODUCT PER TON IS RELATIVELY LOW AND THE COST OF TRANSPORT A
SIGNIFICANT FRACTION OF THE SALES PRICE, THE PLANT SHOULD BE
LOCATED CLOSE TO THE PRIMARY MARKET. THIS CONSIDERATION WILL BE
LESS IMPORTANT FOR LOW VOLUME PRODUCTION, HIGH-PRICED PRODUCTS;
SUCH ASPHARMACEUTICALS. IN AN INTERNATIONAL MARKET, THERE MAY BE AN
ADVANTAGE TO BE GAINED BY LOCATING THE PLANT WITHIN AN AREA WITH
PREFERENTIAL TARIFF..
RAW MATERIALS
THE AVAILABILITY AND PRICE OF SUITABLE RAW MATERIALS WILL OFTEN
DETERMINE THE SITE LOCATION. PLANTS PRODUCING BULK CHEMICALS ARE
BEST LOCATED CLOSE TO THE SOURCE OF THE MAJOR RAW MATERIAL; WHERE
THIS IS ALSO CLOSE TO THE MARKETING AREA.
TRANSPORT
THE TRANSPORT OF MATERIALS AND PRODUCTS TO AND FROM PLANT WILL
BE AN OVERRIDING CONSIDERATION IN SITE SELECTION.
IF PRACTICABLE, A SITE SHOULD BE SELECTED THAT IS CLOSE AT LEAST
TWO MAJOR FORMS OF TRANSPORT: ROAD, RAIL, WATERWAY OR A SEAPORT.
ROAD TRANSPORT IS BEING INCREASINGLY USED, AND IS SUITABLE FOR
LOCAL DISTRIBUTION FROM A CENTRAL WAREHOUSE. RAIL TRANSPORT WILL BE
CHEAPER FOR THE LONG-DISTANCE TRANSPORT OF BULK CHEMICALS.
AIR TRANSPORT IS CONVENIENT AND EFFICIENT FOR THE MOVEMENT OF
PERSONNEL AND ESSENTIAL EQUIPMENT AND SUPPLIES, AND THE PROXIMITY
OF THE SITE TO A MAJOR AIRPORT SHOULD BE CONSIDERED.AVAILABILITY OF
LABOUR
LABOUR WILL BE NEEDED FOR CONSTRUCTION OF THE PLANT AND ITS
OPERATION. SKILLED CONSTRUCTION WORKERS WILL USUALLY BE BROUGHT IN
FROM OUTSIDE THE SITE, BUT THERE SHOULD BE AN ADEQUATE POOL OF
UNSKILLED LABOUR AVAILABLE LOCALLY; AND LABOUR SUITABLE FOR
TRAINING TO OPERATE THE PLANT. SKILLED TRADESMEN WILL BE NEEDED FOR
PLANT MAINTENANCE. LOCAL TRADE UNION CUSTOMS AND RESTRICTIVE
PRACTICES WILL HAVE TO BE CONSIDERED WHEN ASSESSING THE
AVAILABILITY AND SUITABILITY OF THE LABOUR FOR RECRUITMENT AND
TRAINING.
UTILITIES (SERVICES)
THE WORD UTILITIES IS NOW GENERALLY USED FOR THE ANCILLARY
SERVICES NEEDED IN THE OPERATION OF ANY PRODUCTION PROCESS. THESE
SERVICES WILL NORMALLY BE SUPPLIED FROM A CENTRAL FACILITY; AND
WILL INCLUDE:
I. ELECTRICITY: - POWER REQUIRED FOR ELECTROCHEMICAL PROCESSES,
MOTORS, LIGHTINGS, AND GENERAL USE J. STEAM FOR PROCESS HEATING: -
THE STEAMS REQUIRED FOR THE PROCESS ARE GENERATED IN THE TUBE
BOILERS USING MOST ECONOMIC FUEL.
B. COOLING WATER: - NATURAL AND FORCED DRAFT COOLING TOWERS ARE
GENERALLY USED TO PROVIDE THE COOLING WATER REQUIRED ON SITE.
C. WATER FOR GENERAL USE: - THE WATER REQUIRED FOR THE GENERAL
PURPOSE WILL BE TAKEN FROM LOCAL WATER SUPPLIES LIKE RIVERS, LAKES
AND SEAS. BECAUSE OF THIS REASON ALL THE PLANTS LOCATED ON THE
BANKS OF RIVER.
D. DEMATERIALIZED WATER: - DEMATERIALIZED WATER, FROM WHICH ALL
THE MINERALS HAVE BEEN REMOVED BY ION-EXCHANGE IS USED WHERE PURE
WATER IS NEEDED FOR THE PROCESS USE, IN BOILER FEED WATER.
E. REFRIGERATION: - REFRIGERATION IS NEEDED FOR THE PROCESSES,
WHICH REQUIRE TEMPERATURES BELOW THAT ARE PROVIDED BY THE COOLING
WATER.
F. INERT-GAS SUPPLIES.
G. COMPRESSED AIR: - IN AN ETHYLENE OXIDE PLANT COMPRESSED AIR
IS ONE OF THE RAW MATERIALS. IT IS ALSO NEEDED FOR PNEUMATIC
CONTROLLERS ETC.
EFFLUENT DISPOSAL FACILITIES: - FACILITIES MUST BE PROVIDED FOR
THE EFFECTIVE
DISPOSAL OF THE EFFLUENT WITHOUT ANY PUBLIC NUISANCE.
ENVIRONMENTAL IMPACT, AND EFFLUENT DISPOSAL
ALL INDUSTRIAL PROCESSES PRODUCE WASTE PRODUCTS, AND FULL
CONSIDERATION MUST BE GIVEN TO THE DIFFICULTIES AND COAT OF THEIR
DISPOSAL. THE DISPOSAL OF TOXIC AND HARMFUL EFFLUENTS WILL BE
COVERED BY LOCAL REGULATIONS, AND THE APPROPRIATE AUTHORITIES MUST
BE CONSULTED DURING THE INITIAL SITE SURVEY TO DETERMINE THE
STANDARDS THAT MUST BE MET.
LOCAL COMMUNITY CONSIDERATIONS
THE PROPOSED PLANT MUST FIT IN WITH AND BE ACCEPTABLE TO THE
LOCAL COMMUNITY. FULL CONSIDERATION MUST BE GIVEN TO THE SAFE
LOCATION OF THE PLANT SO THAT IT DOES NOT IMPOSE A SIGNIFICANT
ADDITIONAL RISK TO THE COMMUNITY.
LAND (SITE CONSIDERATIONS)
SUFFICIENT SUITABLE LAND MUST BE AVAILABLE FOR THE PROPOSED
PLANT AND FUTURE EXPANSION. THE LAND SHOULD BE IDEALLY FLAT, WELL
DRAINED AND HAVE LOAD-BEARING CHARACTERISTICS. A FULL SITE
EVALUATION SHOULD BE MADE TO DETERMINE THE NEED FOR PILING OR OTHER
FOUNDATIONS.CLIMATE
ADVERSE CLIMATIC CONDITIONS AT SITE WILL INCREASE COSTS.
ABNORMALLY LOW TEMPERATURES WILL REQUIRE THE PROVISION OF
ADDITIONAL INSULATION AND SPECIAL HEATING FOR EQUIPMENT AND PIPING.
STRONGER LOCATIONS WILL BE NEEDED AT LOCATIONS SUBJECT TO HIGH WIND
LOADS OR EARTHQUAKES.
POLITICAL AND STRATEGIC CONSIDERATIONS
CAPITAL GRANTS, TAX CONCESSIONS, AND OTHER INDUCEMENTS ARE OFTEN
GIVEN BY GOVERNMENTS TO DIRECT NEW INVESTMENT TO PREFERRED
LOCATIONS; SUCH AS AREAS OF HIGH UNEMPLOYMENT. THE AVAILABILITY OF
SUCH GRANTS CAN BE THE OVERRIDING CONSIDERATION IN SITE
SELECTION.
PLANT LAY OUT
The economic construction and efficient operation of a process
unit will depend on how well the plant and equipment specified on
the process flow sheet is laid out. The principal factors are
considered are:
Economic considerations: construction and operating costs.
The process requirements.
Convenience of operation.
Convenience of maintenance.
Safety.
Future expansion. Modular construction. Costs The cost of
construction can be minimized by adopting a layout that gives the
shortest run of connecting pipe between equipment, and at least
amount of structural steel work. However, this will not necessarily
be the best arrangement for operation and maintenance. Process
requirements An example of the need to take into account process
consideration is the need to elevate the base of columns to provide
the necessary net positive suction head to a pump or the operating
head for a thermosyphon reboiler. Operations Equipment that needs
to have frequent attention should be located convenient to the
control room. Valves, sample points, and instruments should be
located at convenient positions and heights. Sufficient working
space and headroom must be provided to allow easy access to
equipment. Maintenance Heat exchangers need to be sited so that the
tube bundles can be easily withdrawn for cleaning and tube
replacement. Vessels that require frequent replacement of catalyst
or packing should be located on the out side of buildings.
Equipment that requires dismantling formaintenance, such as
compressors and large pumps, should be places under cover.
Safety
Blast walls may be needed to isolate potentially hazardous
equipment, and confine the effects of an explosion.
At least two escape routes for operators must be provided from
each level in process buildings.
Plant expansion
Equipment should be located so that it can be conveniently tied
in with any future expansion of the process.
Space should be left on pipe alleys for future needs, and
service pipes over-sized to allow for future requirements.
Modular construction
In recent years there has been a move to assemble sections of
plant at the plant manufacturers site. These modules will include
the equipment, structural steel, piping and instrumentation. The
modules are then transported to the plant site, by road or sea. The
advantages of modular construction are:
f. Improved quality control.
g. Reduced construction cost.
h. Less need for skilled labour on site. Some of the
disadvantages are; Higher design costs & more structural steel
work.
More flanged constructions & Possible problems with
assembly, on site.
THE PLANT LAYOUT KEYWORDS
Raw material Storage
Product Storage
Process Site
Laboratories
Workshop
Canteen & Change house
Fire Brigade
Central Control Room
Security office 10.Administrative Building Site for Expansion
Project.
Effluent treatment plant 13.Power house
Emergency water storage 15.Plant utilities
A detailed plant layout is drawn and submitted with this thesis
report. This plant layout is just a reference plant layout. There
may be a lot of changes in actual plant layout.
BIBLIOGRAPHY
(1)
a. R. H. PERRY AND DON W. GREEN, PERRYS CHEMICAL ENGINEERS HAND
BOOK, 6TH ED. MC-GRAW HILL INTERNATIONAL EDITION,
b. H.SAWISTOWSKI &W.SMITH, MASS TRANSFER PROCESS
CALCULATIONS, INTERSCIENCE PUBLISHERS,
c. R. K. SINNOTT, COULSON AND RICHARDSONS CHEMICAL ENGINEERING
SERIES, VOLUME-6, CHEMICAL EQUIPMENT DESIGN 3RD ED., BUTTER
WORTH-HEINEMANN, PAGE NO: 828-855, 891-895
d. JOSHI M. V., PROCESS EQUIPMENT DESIGN, 2ND ED., MC-MILLAN
INDIA LTD, e. MAX S. PETERS AND KLAUS TIMMERHAUS, PROCESS PLANT
DESIGN AND ECONOMICS FOR CHEMICAL ENGINEERS, 3RD ED., MC-GRAW HILL
BOOK COMPANY, PAGE NO: 207-208, 484-485. f. (7) B.C BHATTACHARYA,
CHEMICAL EQUIPMENT DESIGN, CHEMICAL ENGINEERING EDUCATION
DEVELOPMENT CENTRE.g. h. (8) L.E. BROWNELL AND E.H. YOUNG, PROCESS
EQUIPMENT DESIGN, JOHN WILEY & SONS INC. NEW YORK,