Table of ContentsIntroduction2Process Description3Block Diagram4Process Design Basis5Overall Material Balance6Design of column, reboiler and condenser7Equipment List15PFD, Piping and Instrumentation Diagram and Plot plan16Line Sizing17Pump Hydraulic Calculations19

IntroductionThe instructions given in the following pages are for the design of 100 MT/day Liquid N2O4 plant.

Process DescriptionRaw liquid N2O4 containing a small amount of Nitric acid is branched off from the discharge of N2O4 pump P-1 and fed to N2O4 distillation column T-1 through a flow control valve.The distillation column is provides with stripping and rectification sections and has beds of intalox saddles.The heat required for the vaporization of N2O4 is supplied by indirect steam in the thermosyphon reboiler E-1. Saturated LP steam at 3 kg/cm2 g pressure will be utilized for this purpose. The temperature at the bottom of the column is maintained at approx. 25 degC.Pure N2O4 vapors leaving the top of the column at approx 21 degC are condensed and sub cooled to 0 degC in condenser E-2 using brine. A controlled quantity of condensed N2O4 is fed back to the column as reflux for proper rectification. The quantity of reflux required depends upon the purity of product obtained.The bottom product containing N2O4 and nitric acid is returned to tank D-2 by gravity.Excess N2O4 from the condenser is drawn off as product and is led to collecting tank D-3 from where it can be transferred to the main storage tanks using transfer pumps. If there is no draw-off from D-3 to main storage, the product overflows into D-1.

Block Diagram

Feed TPD95 % (w/w) N2O4Temperature 0 degCBottoms84 % (w/w) N2O4Temperature 25 degCDistillate100 TPD99.9 % (w/w) N2O4Temperature 0 degCDistillation

Process Design Basis

Plant CapacityThe plant has been designed to produce 100 MT/day of pure N2O4Product QualityLiquid N2O4 Concentration 99.9% (w/w)Temperature 0 degCFeedstock and UtilitiesRaw N2O4 Temperature 0 degCQuality 95% (w/w)LP SteamPressure 3 Kg/cm2 gTemperature SaturatedBrinePressure 4 Kg/cm2 gTemperature -13 degCInst AirPressure More than 4 Kg/cm2 gTemperature AmbientDew Point -40 degC

Overall Material Balance

Product Distillate flow rate = 100 TPDProduct concentration = 99.9% (w/w) N2O4Bottom concentration = 84% (w/w) N2O4Feed concentration = 95% (w/w) N2O4Let feed, distillate and bottom mass flow is identified by F, D and W respectively.Also let feed, distillate and bottom mass fraction of component N2O4 is identified by Xf, Xd and Xw respectively.Therefore at steady state taking material continuity balance,F = D + WXf.F = Xd.D + Xw.WTherefore,F = 100 + W.95F = .999*100 + .84WSolving above two linear equations in two unknowns,F = 144.5455TPD = 6022.7273 kg/hrW = 44.5455 TPD = 1856.0606 kg/hr

Design of column, reboiler and condenserTaking material balance across the distillation column

Distillate product rate = 100TPD

External reflux ratio = 4

Hence,

Reflux rate to column = 400TPD

The reflux is cooled to 0 deg C and top temp is 21 deg C.

Taking material balance across condenser,

Vapor in to condenser = Distillate product + Reflux flow

Therefore,

Vapor to condenser=400+100 =500TPD

Above vapor value is the vapor leaving the enriching section. But the vapor flow through the

enriching section will be more than above value because cold reflux is used.

Taking material and energy balance at top end of enriching section to find the vapor flow in

enriching section,

Vapor in enriching section - Vapor flow out of enriching section = Vapor condensed

Also,

Heat of vaporisation at 21 degC X Vapor condensed = Sensible heat rise of reflux (i.e 400 TPD)

from 0 to 21 degC

Heat of vaporisation of n2o4 and hno3 at 21 degC is 75.04828 and 148.4839 Kcal/Kg respectively

Taking weighted average,

Heat of vaporisation of vapor mixture at 21 degC is equal to 75.12172Kcal/Kg

Now heat capacity of distillate at 0 deg C is equal to0.198728Kcal/Kg.degK

Now heat capacity of distillate at 21 deg C is equal to0.200991Kcal/Kg.degK

Avg heat capacity is equal to 0.19986Kcal/Kg.degK

Putting above values in the equation for calculating condensed vapor amount

75.12172 X Vapor condensed = 400000 X 0.19986 X (21-0)

Therefore, Vapor condensed = 22348.01Kg per day

=22.34801TPD

Therfore liquid flow rate in enriching section is equal to =422.348TPD

Since vapor was condensed from vapor in enriching section

the vapor flow rate in enriching section is equal to =522.348TPD

Taking material balance at feed point,

From the observed value of q for feed it can be predicted that the increase in liquid flow rate

in stripping section due to subcooled feed is negligible. Hence it can be neglected safely.

Liquid flow in enriching section =422.348TPD

Feed flow = 144.5455TPD

Liquid flow in stripping section =566.8935TPD

Vapor flow in stripping section

is same as in enriching section =522.348TPD

Therfore, bottom flow rate,

= 566.8935-522.348 =44.54549TPD

The maximum L/G ratio is obtained at bottom of column. Also maximum gas and liquid flows

are at bottom of column

Vapor and Liquid densities at 25 degC and prevailing pressure are as follows

CompntLiquidVaporUnit

n2o489.647620.232422lb per cuft

hno394.258040.1614lb per cuft

Comp

n2o40.840.985498

hno30.160.014502

0.009374.24012

0.0016970.089852

Den mix90.354740.230948lb per cuft90.12379

Flow52074.4747982.55lb per hr1.002563

1.00128

=1.086669

For above value of function the corresponding value of ordinate from graph for a pressure drop

of .05 inH2O per ft of packing is .003

For 25mm ceramic intalox saddle packing, Cf =981.54E-08

J=1.502

gc=4.18E+08

Liquid viscosity=0.395977cP

3.141593

Gas flow square =194531

Gas flow=441.0567lb/sqrft.hr

Therfore column cross section =108.79sqft

=10.107sqmtrTherforeDia=1.772454meter

Plate dimensions

pi X diameter =5.568328meter

18.26879feet

Following plate dimensions are available at leading indian alluminium plate manfg Hindalco

Al AlloysTemperThicknessLengthWidth

mmmmmm

10506-2001000-7000500-1550

1060

1070

1080F and O

1100

1200

3003

3105

8011

Design of Vertical Thermosyphon

Vapor flow in stripping section = 522.348TPD

=47990.72Lb per hr

A circulation ratio of 4:1 shall be used because Alluminium alloy MOC is used and alluminium

nitrate salt will get deposited on tubes for smaller ratios.

3/4 in OD tubes 16 BWG on a triangular pitch of 1 inch shall be used.

Let initial guess of tube length as 6 feet.

Heat of vaporisation of n2o4 and hno3 at 25 degC is 75.04828 and 148.4839 Kcal/Kg respectively

Weigthed average Heat of vaporisation for wt fraction of n2o4 = .84 is

Heat of vaporisation = 84.0 Kcal/Kg

Therefore, Heat of Duty, Q = 84 X 522.348 =1828218Kcal/hr

The maxium allowable heat flux will be 12000 Btu/hr.sqft = 32571.3Kcal/hr.m2

Therefore, Heat transfer area = 56.12972sq.mtr = 603.9558sq.ft

Heat transfer area for 3/4 in OD tube per unit length is 0.1963 sq.ft per ft of length

Therefore, No of 6ft tube lengths =512.7829

Nearest tube count for 1-Pass, 3/4 in OD tube on 1 in triangular pitch is,

Shell ID, 27 inch; Total tube count 559

Check for pressure drop,

Static pressure of reboiler psi

Vapor density =0.230948lb per cuftSpecific volum = 4.329972

Liquid density =90.35474lb per cuftSpecific volum = 0.011067

Substuting above,

Static pressure of reboiler leg =0.057524psi

Frictional resistance through reboiler :

Flow are :

Total tube flow area =Tube flow area per tube X No of total tubes

= 0.302 X 559 =168.818sq.inch

=1.172347sq.ft

Mass velocity,G = Total mass flow rate/ Total tube flow area

= 5 X 47990.73 / 1.1723 =204686.2lb per hr per sq.ft

At 25 deg C,

viscos, nu =0.03lb/hr.ft

ID of tube =0.0517ft

Nre=DG/nu352742.6

friction fac0.00011sq.ft per sq.in

Avg.sp.volm2.17052

delta P =0.004721psi

Total resistance = 0.0575 + 0.0047 =0.062245psi

Driving force, Liquid leg X Density of liquid =3.764781psi

Since Driving force is much more than Total resistance a circulation ratio of 4:1 is assured.

At Nre =352742

Coulborn factor, jH =700

Thermal condu, k=0.162685Btu per hr per ft per degF

sp heat capacity, Cp=0.210588Btu per lb per degF

k(Cp.nu/k)^1/3 = 0.055091

hi = (jH.k/D)(Cp.nu/k)^1/3 = 745.9165

This exceeds the maximum. Use 300

hio = hi X ID/OD =300 X 0.62/0.75 = 248 Btu per hr per sq.ft per degF

ho for condensing steam = 1500Btu per hr per sq.ft per degF

Clean overall coefficient Uc:

Uc =hio.ho/(hio + ho) =1500 X 248/(1500+248) =213Btu per hr per sq.ft per degF

Assuming steam chest pressure to be just above atmospheric so as to keep temp driving force

below critical temp difference. Therefore steam temperature is 100 degC

Due to above steam control valve can be designed for enough pressure drop.

LMTD =75degC

UD = 40.36097Btu per hr per sq.ft per degF

Dirt factor Rd:

Rd = Uc -Ud/Uc.Ud =0.020082hr.sqft.degF/Btu

Design of Total Condenser with subcooling

3/4 inch OD, 16 BWG tubes on 1 inch triangular pitch shall be used.

The condensation shall be in tube side.

According to Nusselt's theory the transition from laminar to turbulent shall occur for Re=1400

The length of tube at which this transition shall take place from top is given by,

where,

lambda = heat of vaporisation, Btu/lb=135.0869

nu = viscosity of condensing film, lb/hr.perft=1.000964

rho = density of film, lb/cu.ft=90.02211

k = thermal conductivity of film, Btu/hr.ft.F=0.083496

g = acceleration due to gravity = 4.18E+8 ft/sq.hr=4.18E+08

Tv = Condesing temperature, degF=69.8

tw = Temperature of wall, degF=8.6

specific heat capacity,Btu/lb.degF=0.200991

Therefore,

xc = 4.703844feet

6 feet tube length shall be satisfactory.

Condensing load of condenser = 500 TPD =45930lb/hr

Condensing heat load = 45930 X 135.0869 = 6204541Btu/hr

Condesing film coefficient can be obtained from Colburn's semiempirical condensation curve,

Re = 4G'/nu

where,

G' = Loading per tube perimeter, lb/hr.ft

Inside perimeter for 3/4 inch OD, 16 BWG tubes = 0.162316feet

Assume 27 inch Shell ID, 1-P. Total number of tubes = 559

G' = 506.2025lb/hr.ft

Re = 4 X 506.2025/1.001 =2022.86

This is the lowest value of Re on Colburns curve. So any change in loading from above value will

only increase Re and shall increase condensing coefficient.

Ordinate from Colburns curve for above Re is 0.15

Therefor, condensing coefficient = 187.9769Btu/hr.sqft.F

Condesing area = Condensing heat duty/(Condensing coeff X (Tv-tw)) =539.3291sq.ft

Heat transfer area sqft per linear ft =0.1623sq.ft

Length of tubes required for to accommodate above heat load from above heat transfer area is

=5.944612feet

Therefore length remaining for subcooling is 6 - 5.9446 = 0.055388feet

Heat transfer area corresponding to this remaining tube length =5.025098sq.feet

Subcooling from 21 degC to 0 degC heating duty =193861.8Btu/hr

According to McAdams heat transfer coeff for subcooling is,

Therefore,

Subcooling film coefficient =4739.914Btu/hr.sqft.F

LMTD = 21.84289=71.31721degF

Therefore, area for subcooling =0.573492sq.ft

The availabe area for subcooling 5.02 sq.ft is much more than required value. Hence the heat

exchanger is suitable for condensation and subcooling.

Equipment ListEQUIPMENT LIST FOR PLANT SECTION

ITEMDESIGNATIONNUMBERMEDIUMTECHNICAL DATAMATERIALDRIVEREMARKS

ab

T-1N2O4 PRODUCTION TOWER1-N2O4,HNO3H = 5000 MMAL 99.5-I

D = 1780 MM

E-1CONDENSER1N2O4, BRINEA = 61.2 SQ.MTRSHELL : AL 99.5-I

Q = 1.6E+6 KCAL/HRTUBE : AL 99.5

E-2REBOILER1N2O4,HNO3A = 61.2 SQ.MTRSHELL : AL 99.8-I

STEAMQ = 1.8E+6 KCAL/HRTUBE : AL 99.8

D-1N2O4 FEED TANK1N2O4V = 210 CU.MTRAL 99.5-I

D = 4100 MM

L = 16000 MM

P-1 A,BN2O4 FEED PUMP11N2O4F = 5.1 CU.MTR/HRSS 304 LE.MOTORI

H = 21 mLC

P-2 A,BREFLUX PUMP11N2O4F = 14.7 CU.MTR/HRSS 304 LE.MOTORI

H = 8 mLC

PFD, Piping and Instrumentation Diagram and Plot plan

Line SizingLine sizing:The Feed pump suction:From pump rated flow equal to 5.1 cu.mtr per hrSelecting velocity as 0.6 m/s,The diameter is equal to 54.88 mm.Select DN 50 Sch 40 pipe.The Feed pump discharge:From pump rated flow equal to 5.1 cu.mtr per hrSelecting velocity as 1.6 m/s,The diameter is equal to 33.6 mm.Select DN 40 Sch 40 pipe.For vapor line from column to condenser:Vapor flow from material balance = 500 TPD = 20833 kg per hrAt top temperature vapor density = 3.6 kg per cu.mtrTherefore, volumetric flow rate = 5787 cu.mtr per hrSelecting vapor velocity = 15 m/sDiameter = 370 mmSelect DN 400 Sch 40 pipe.For condenser outlet line as reflux pump suction:Condensate mass flow = 20833 kg per hrCondensate density = 1470 kg per cu.mtrTherefore, volumetric flow rate = 14.7 cu.mtr per hrSelecting velocity = 0.6 m/sDiameter = 94 mmSelect DN 90 Sch 40 pipe.For reflux pump discharge:For volumetric flow rate = 14.7 cu.mtr per hrSelecting velocity = 1.6 m/sDiameter = 57 mmSelect DN 50 Sch 40 pipe.For column to reboiler line:Mass Flow = 108900 kg per hrVolumetric flow = 75 cu.mtr per hrSelecting velocity = 0.3 m/sDiameter = 296 mmSelect DN 300 Sch 40 pipeFor reboiler to column vapor line:For circulation ratio of 4:1 density of mixture does not change appreciably.Hence for the same velocity select a DN 350 Sch 40 pipe.

Pump Hydraulic CalculationsPump hydraulics:Feed pumps:Rated flow through pump = 5.1 cu.mtr per hrThe elevation of feed point from grade is 8.5 mtr. But for ease of maintenance the feed control valve will be located on second floor at elevation of 11 mtr from grade.The pump head will be calculated for 11 mtr discharge elevation.The available positive suction head shall be avoided for this calculation.Calculating total discharge head,Selected discharge velocity = 1.6 m/sThe fittings in discharge line can be estimated by following,Selecting straight length of pipe as 20 meters,Le/L = 1 + (0.347d^0.5 + 0.216) FcWhere d = pipe NB inches.Fc = Correction factor = 1.0Selected discharge pipe is DN 32 i.e NB 1.1/4Therefore,Le/L = 1.61Le = 32.2 metersHence, Total equivalent length = 52.2 metersAdding 30 % of above value for pressure drop across feed control valve,Therefore, Final total equivalent length = 1.3 X 60 = 67.9 mtrCalculation of friction factor,ID of pipe = 35.048 mmVelocity = 1.6 m/sDensity = 1470 kg/cu.mtrViscosity = 0.4 cP = 0.0004 Pa.sNre = 206083Roughness factor for pipe, episilon = 0.025Darcy Friction factor,f = 0.0198Pressure drop = f. (L/ID).(V2/2g) = 0.0198 X (68/0.03505) X (sqr1.6/2X9.8)=5.1 mLC=72278 Pa=0.74 Kg per sq.cmTherefore total discharge head = Static head + Discharge losses = 11 + 5.1 = 16.1 mLCCalculating Head loss through suction line:Distance between feed pump suction and tank outlet = 3.5 metersSelected suction velocity = 0.6 m/sThe fittings in suction line can be estimated by following,Selecting straight length of pipe as 3.5 meters,Le/L = 1 + (0.347d^0.5 + 0.216) FcWhere d = pipe NB inches.Fc = Correction factor = 1.0Selected suction pipe is DN 50 i.e NB 2Therefore,Le/L = 1.71Le = 5.985 metersHence, Total equivalent length = 9.485 metersCalculation of friction factor,ID of pipe = 52.506 mmVelocity = 0.6 m/sDensity = 1470 kg/cu.mtrViscosity = 0.4 cP = 0.0004 Pa.sNre = 115776Roughness factor for pipe, episilon = 0.025Darcy Friction factor,f = 0.0198Pressure drop = f. (L/ID).(V2/2g) = 0.0198 X (9.485/0.0525) X (sqr0.6/2X9.8)=0.066 mLCTotal Head = 16.1 + 0.066 = 16.2 mLCPump shutoff head can be estimated as 30% plus above value.Therefore, Pump shutoff head = 16.2 X 1.3 = 21.06 meters.Pump shutoff pressure = Pump Shutoff head X Density X g = 21.06 X 1470 X 9.8= 303390.4 Pa=3.09 Kg per sq.cm (abs)

Reflux pumps:Condensate mass flow = 20833 kg per hrCondensate density = 1470 kg per cu.mtrTherefore, volumetric flow rate = 14.7 cu.mtr per hrSelecting velocity = 1.6 m/sDiameter = 94 mmSelect DN 50 Sch 40 pipe.Rated flow through pump = 14.7 cu.mtr per hrFor ease of maintenance and operation reflux pumps shall be located on first floor at elevation of 5 mtr from grade.The elevation of reflux point from grade is 12 mtr.The pump head will be calculated for 12-5 = 7 mtr discharge elevation.The available positive suction head shall be avoided for this calculation.Calculating total discharge head,Selected discharge velocity = 1.6 m/sThe fittings in discharge line can be estimated by following,Selecting straight length of pipe as 7 meters,Le/L = 1 + (0.347d^0.5 + 0.216) FcWhere d = pipe NB inches.Fc = Correction factor = 1.0Selected discharge pipe is DN 50 i.e NB 2Therefore,Le/L = 1.71Le = 11.97 metersHence, Total equivalent length = 19 metersCalculation of friction factor,ID of pipe = 52.506 mmVelocity = 1.6 m/sDensity = 1470 kg/cu.mtrViscosity = 0.4 cP = 0.0004 Pa.sNre = 308736Roughness factor for pipe, episilon = 0.025Darcy Friction factor,f = 0.018Pressure drop = f. (L/ID).(V2/2g) = 0.018 X (19/0.0525) X (sqr1.6/2X9.8)=0.851 mLCTotal Head = 7+0.851 = 8 mLCNeglecting the losses in suction line and positive suction head,Pump shutoff head can be estimated as 30% plus above value.Therefore, Pump shutoff head = 8 X 1.3 = 10.4 meters.Pump shutoff pressure = Pump Shutoff head X Density X g = 10.4 X 1470 X 9.8= 149822.4 Pa=1.53 Kg per sq.cm (abs)