design of distillation column
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DESIGN OF DISTILLATION COLUMNSeparation of BIO DIESEL from METHANOL
DESIGN STEPS OF DISTILLATION COLUMN
Calculation of minimum reflux ratio rm Calculation of actual reflux ratio Calculation of theoretical number of
stages Calculation of actual number of stages Calculation of diameter of the column Calculation of flooding point Calculation of entrainment Calculation of pressure drop Calculation of the height of the column
Xf =0.2848
Xd = 0.99
Xw = 0.01Now consider binary mixture of methanol and biodiesel.Total pressure of system =10 mmhg Relative volatility of mixture = 8
Calculation of minimum reflux ratio:From Coulson vol # 02 equation 11.15
yaSolving for Yf:
Xf=0.2848
yaYf =0.76
Know for minimum reflux ratio:
Rm
Rm = xd - yf
yf –xf
Rm= 0.99-0.76 0.76-.2848 Rm=0.503Since,R= 1.3RmR= 1.3(0.503)R=0.6
According to mass balance F=2129.649 kg/hr D=606.524 kg/hr W= 606.524kg/hr since,R=Ln /DLn = 394.2
SO,
AND WE KNOW THAT VN = LN + D
VN = 1000 KG/HR
SIMILARLY FOR STRIPPING SECTION AS WE ENTER THE FEED AT ITS BOILING POINT SO, LM = LN + F
LM = 394.2+ 2129 LM = 2523.2 KG/HRFROM PLATE BALANCE WE KNOW
LM = VM + W
VM = 1916KG/HR
Theoratical Number Of Stages:
from equation 11.58 (page 524) Coulson vol#06
Nm = So, Nm =ln(0.99/0.01)d (0.01/0.99)b
we get, ln 75 Minimum no plates = Nm =2
Theoretical number of plates:
= 0.75 [ 1 – ( R - ]0.566 ] By putting values
= 0.75 [ 1 – ( ]0.566 ]
N = 7 PLATES
Location of feed platesWe use Kirkbride’s method for finding feed point location. (Eq 11.62 Coulson vol#06)
log[] = 0.206 log[()()F {}2]
B and D are bottom and distillate respectively
log[] = 0.206log[()() {}2] log[] = 0.6895So
= 0.6895 And we know that
+ = 7
So , = 3.8 & = 3.2 So feed will b at 3rd plate from the bottom
Calculation of actual number of stages :
For finding Column efficiency() we need viscosity so for finding viscosity.
Avg. temp. off column =
= = 80 ℃Feed viscosity at average column temperature = ave = 0.033 Ns/m²
Relative volatility = 28
So efficiency is calculated by O'Connell's correlation formula (eq 11.67 Coulson vol # 06)
Efficency =
= 51 – 32.5 [log(0.033)(28)] = 82%Number of actual tray =
= 9 PLATES
Calculation of column dia: Vapor load at bottom= Qv = (average molecular weight of vapour)
= maximum flow rate in stripping section = 1916Kg/hr v = density of vapour = 850 Kg/m³
Avg molecular weight of vapour = 60 kg/kmol
So, Qv =
and
Qv = 0.037 3 /s
Liquid load at bottom = Qv = (average molecular weight of vapour)
= 2523 kg/hr
= density of vapour = 850 Kg/m³
Avg molecular weight of vapour = 60 kg/kmol
So, QL =
and
QL = 0.0423 /s
Tray Dynamics:Lv = liquid vapors factor
Lv = ()()0.5 = ()(
= 0.0184Capacity ParameterAssumed tray spacing = 700mmSurface tension of system = σ = 50 dynes/cm
Csb = 0.1048
Net Vapor Velocity At Flooding: Vnf = Csb () 0.2 [ ( ) 0.5 Vnf = 5.87 m/s
Actual Vapor Velocity:
VA = 0.80 Vnf = 0.80 * 2.3865 VA = 4.7 m/sVolumetric flow rate of vapours= 2.12 m3 /sec
Net-Area: An = Volumetric flow rate of vapor Actual vapor velocity = 10/4.7
An = 2.12 m2
Assume that down comer occupies 15% of Cross-sectional Area; then Cross-Sectional Area: AC = 2.12/0.85
Ac = 2.49m2
Diameter of Column Since we have Ac = D2
D =
D =
D = 1.8m
Calculation Of Flooding Point Vn = =
= 0.0174 m/s
Net Area: Net area = An = 1.687 m2
Active area = A a = A c - 2A d
Down comer area = Ad = 0.15 Ac
Ad = 0.15 * 2.49 = 0.3735 m2 now, A a = A c - 2A d
= 2.49 – 2(0.3735) = 1.74 m2
Pressure Drop Over The Entire Column:Pressure drop per tray= o.5 kpa ∆PT = 0.64 * 7 ∆PT = 4.48 Kpa
Height of Distillation ColumnHeight of column= (NT -1)x tray spacing + bottom space + t op space =(6 x 0.7) +0.7+0.7 = 5.6 m= 6 m
THANKS...
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