NUMBER OF PLATES
Fenske Equation
=3.069 plates with Reboiler
q = 1 (feed at its bubble point)By hit and Trial
θ = 1.57654
Rm = 0.019698 R = 0.21
N= 7
FEED PLATE LOCATION
By, using kirkbridge Equation
So feed is entering at 5 plates from bottom.
Column Efficiency
For Mono Ethylene Glycol & Water System
So, Actual no. of plates = 7 / 0.60 = 12 plates
NB=4 / 0.60=7
ND=3 / 0 .60=5
COLUMN DIAMETER
The plate diameter is calculated based on the flooding considerations
Maximum liquid flow rate in rectifying section
Ln = D Rm = (0.21 ´0.19326) = 0.0405846 kmol/sec
Maximum vapor flow rate in rectifying section
Vn = Ln + D = (0.0405846 + 0.19326) =0.2338446 kmol/sec
Maximum liquid flow rate in stripping section
Lm = Ln + F = ( 0.0405846+ 0.3186682) = 0.3592528 kmol/sec
Maximum vapor flow rate in stripping section
Vm = Lm – B = (0.3592528– 0.125412456)) =0.2338446 kmol/sec
Density of liquid in the stripping section
ρLm = 1052kg/ m3
Density of the vapour in the stripping section
ρVn = 0.557 kg/ m3
Density of the liquid in the rectifying section
ρVm = 976 kg/ m3
Density of the vapour in the rectifying section
ρLn = 0.190694 kg/ m3
Average Molecular weight at the top=18.10 kg/kg moles
Average molecular weight at the bottom= 55.46 kg /kg moles
Maximum volumetric flow rate of Vapour
Top m3/sec
Bottom m3/sec
Maximum Volumetirc flow rate of liquid
Top m3/sec
Bottom m3/sec
Calculate FLV = liquid vapor flow factor
LW = liquid flow rate kgmoles/s
VW = vapor flow rate, kgmoles/s
=vapour density (kg / m3)
=liquid density (kg/ m3)
= 0.002489427
BOTTOM CALCULATION
= 0.035389951
and for a tray spacing of Ts=609 mm.
Csb=0.0105+(8.127/10000)*(Ts^0.755)*EXP(-1.463*FLG^0.842)
FLOODING PARAMETER
Csb, flood = 0.104801008 m/s.
Now,
Unf = Csb, flood × (σ / 20) ^0..2 [(r l - r g) / r g] 0.5
where
Unf = gas velocity through the net area at flood, m/s (ft/s)
Csb, flood = capacity parameter, m/s (ft/s, as in fig.18.10)
σ = liquid surface tension, mN/m (dyne/cm.)
r l = liquid density, kg/m3 (lb/ft3)
Unf=Csb,flood×(42.2/20)^0.2[(1052 - 0.557)/0.557] 0.5
= 5.283950816 m/sec
Flooding = 80 %
Let
Actual velocity, Un= 0.8×Unf
i.e., Un = 4.227 m/s
Net area available for gas flow (An)
Net area = (Column cross sectional area)-(Down comer area.)
An = Ac - Ad
An = (0.235133*55.46 / (0.557))/4.4227 = 5.53 m2
Let Ad = 15 % of the Ac (cross sectional are) Ac= An / 0.85
Ac = 5.53/0.85 Ac = 6.51m2
DIAMETER OF THE COLUMN AT THE BOTTOM
D = √(4 * Ac/pi) = √(4 * 6.51/(22/7)
= 2.88 m
CALCULATION AT THE TOP
= 0.002489427
and for a tray spacing of Ts=609 mm.
Csb=0.0105+(8.127/10000)*(Ts^0.755)*EXP(-1.463*FLG^0.842)
FLOODING PARAMETER
Csb, flood = 0.112490634m/s
Now,
Unf = Csb, flood × (σ / 20) ^0..2 [(r l - r g) / r g] 0.5
Where
Unf = gas velocity through the net area at flood, m/s (ft/s)
Csb, flood = capacity parameter, m/s (ft/s)
σ = liquid surface tension, mN/m (dyne/cm.)
r l = liquid density, kg/m3 (lb/ft3)
Unf=(0.11249063)×(42.2/20)^0.2[(1052 - 0.557)/0.557] 0.5
= 10.14278254 m/sec
Flooding = 50 %
Let
Actual velocity, Un= 50 ×Unf
i.e., Un = 4.227 m/s
Net area available for gas flow (An)
Net area = (Column cross sectional area) - (Down comer area.)
An = Ac - Ad
An = (0.235133*18 / (0.19069))/ (5.07139)
= 4.376 m2
Let Ad = 15 % of the Ac (cross sectional are)
Ac= An / 0.85
Ac = 4.376 / 0.85
Ac = 5.148 m2
DIAMETER OF THE COLUMN AT BOTTOM
D = √(4 * 5.148/pi)
=√(4 * 6.51/(22/7) = 2.5598 m
PROVISIONAL PLATE DESIGN
Column diameter (base) = 2.88
Column Area Ac
Ac = 6.51 m2
Downcomer area Ad = 0.9765 m2
Net area An = Ac – Ad
= 6.51 – 0.9765 = 5.53 m2
Active area Aa =Ac – 2Ad = 6.51 – 2(0.9765) =4.557 m2
Hole area Ah take 10% Aa as first trial = 0.1 × 4.557 = 0.4557 m2
WEIR LENGTH
Ad / Ac = 0.9765/ 6.51 = 0.15
Lw / dc = 0.81 Lw =2.33 m
Take weir height , hw = 30mm
Hole diameter, dh = 3.175 mm
Plate thickness= 5mm
CHECK WEEPING
Maximum liquid rate=12.72 kg/sec
Minimum liquid rate at 70% turn down = 8.90 kg/sec
how = weir crust
for segmental downcomer
Maximum
= 40.14 mm liquid
Minimum = 31.63 mm liquid
at minimum hw + how =30 + 31.63 = 61.63 mm liquid
K2 = 30.3
= 13.79 m/s
Actual minimum vapour velocity
= 35.67 m/s
So minimum vapor rate will be well above the weep point.
PLATE PRESSURE DROP
Maximum vapour velocity through holes
Uh = Max vapour vol flow Rate/Hole Area= 23.22/0.4557 = 50.95 m/sec
Plate thickness / hole dia =5 / 3.175 =1.57
And Ah/Ap = Ah/Aa =0.1
Co=0.93
= 41.046 mm liquid
RESIDUAL HEAD
mm liquid
TOTAL PRESSURE DROP
ht = hd + (hw + how) + hr
Total pressure drop = 41.046 + (30 + 31.63) + 11.88
ht = 114.556 mm liquid
Down comer Liquid Backup
Take hap = hw – 10 = 20mm
Area under apron = hap* Lw
= 0.02 *2.33 = 0.0466 m2
= 11.17 mm liq.
BACKUP IN DOWN COMER
hb= (hw + how) + ht + hdc =(30+ 31.63) + 114 +11.17
= 186.8 mm liq. = 0.186 liq. m
½ (Tray spacing + weir height) = ½(0.609+0.03)=0.3195m
0.186 < ½ (Tray spacing + weir height)
So tray spacing is acceptable
CHECK RESIDENCE TIME
tr = (Ad* hbc* ℓL)/Lwd
= 21.84 sec
Ø 3 sec. so, result is satisfactory
CHECK ENTRAINMENT
Uv = Maximum Volumetric Flow Rate of vapors/Net Area
UV = 23.22 /5.53 = 4.199 m/s
Percent flooding =4.199/5.28 = 0.79 = 79 %
FLV(base) = 0.035
Fractional Entrainment y = 0.072 well below 0. 1 Satisfactory
NO OF HOLES
Diameter of one hole = 3.175 mm =0.003175 m
Area of one hole = 22/7*(0.003175/2)2 = 7.92 *10-6
Total Hole Area =0.4557 m2
No of Holes =0.4557 m2 / 7.92 * 10-6 = 57534.31
HEIGHT OF COLUMN
# Of plates = 18
Spacing between each plate = 0.609 m
Space for disengagement of vapor and liquid on top = 0.609 m
Space for disengagement of vapor and liquid in bottom = 0.609 m
Height of column = (#of plates - 1) × space between each plate) + (space for disengagement on top and bottom)
= ((12-1)×0.609) + (0.609+0.609) = 7.91 m
So height of column = 7.91 m
No comments:
Post a Comment