Sunday 10 April 2011

COLUMN DESIGN


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,       d= 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* L)/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

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