Abstract

Liquefaction of air and subsequent rectification of the liquid air is the basis of the Linde- frankls processon which this project is based . Prof.Linde is credited today with the first man producing liquid oxygen on a commercial scale. This thesis report is a design of an oxygen plant of capacity 205 tonnes /day, along with oxygen this plant also produces nitrogen of very high purity . In it a complete design of a distiallation column and a 1-2 shell and tube condenser has been done.The Linde Frankls process is the most important process for the commercial production of oxygen in this process in which the air stream is divided in two , one is cooled by J-T effect and the other by incoming pure nitrogen which is the most striking thing about this process and which makes it more economical .Results have been obtained with mechanical calculation and then verified by chemical enggineering software chemsep ,kamlex and chem office. Fianlly a safety and preventions have been delat with along with breif overview of plant layout and locations of the plant.

ContentsChapter page no.1.Introduction 12. Selection of process 83.Mass & Energy balance 194. Equipment Design 305. Conclusion and discussion 51

List of Tables Table page no.1. Analysis of air 202. Conditions and enthalpy of streams 243. Enthalpy chart 274. The equilibrium data for lower column 285. Excell sheet 37

List of figures Fig. Page no. 1. T-S diagram for cla 102. Flowsheet of the linde frankls process 123. T-S for linde frankls process 134. Typical distillation column 315. Condenser flow pattern 436. LMTD calculation 447. Designed condenser 498. Schematic diagram 50

List of graphs Graph page no.1. Equilibrium data 342. McCabe Thiele rough 353. McCabe Thiele by chemsep 384. Relative volatility vs stage 395. Entropy Vs stage 41

Chapter 1.

Introduction

INTRODUCTION, PROPERTIES AND USES

Oxygen has been known to the scientists from as early as the 3rd and 4th centuries. The Chinese were the first to detect its presence as an essential element in the atmosphere without which life and combustion is not possible.In 1770, Henry Cavendish, Joseph priestly and Karl withelm Scheele investigated the properties of oxygen. Later on priestly carried out some experiments and produced it in small quantities.It combines with hydrogen forming water covering around 70% of earth crust combine with various metals and nonmetals to be present in around 99% of rocks. Overall oxygen constitutes 49.2%of the mass of the earth and 23.2%of the atmosphere is oxygen.Oxygen can be produced in the laboratory by heating oxides, peroxides, permanganates and bicarbonates in the presence of sulphuric acid. One of the first industrial manufactures of oxygen was Brin process which was based on the property possessed by the monoxide of barium of absorbing oxygen readily from the air at a temperature of about 540C forming dioxide, this dioxide at 8700C gives up the oxygen absorbed, barium being restored to monoxide.Another process is the electrolysis of water using caustic soda solution in distilled water as electrolyte.

Now coming to the modern process, which is the liquefaction of air and subsequent rectification of the liquid air. Which will be the main concern in this report; few pioneers who were responsible in the gradual development are Noothmore, who was the first to produce chlorine in its liquid form in 1806,

Faraday who succeeded in liquefying many gases by the application of pressure. But still gases such as H2, O2, and N2 for many years resisted all attempts to liquefy them and were therefore called permanent gases.Investigations in 1863by Andrew revealed that gases could not be liquefied however great the pressure applied, until the temperature of the gas is below a certain limit, this temperature is called critical temperature. Further in this field two French scientists Pictat and Caillete in 1877 actually succeeded in liquefying first oxygen and then many other gases.Pictat merely elaborated the Faraday tube experiment using high pressure and low temp gas already liquefied as the refrigerant, while Caillete on the other hand obtained this refrigerating effect by suddenly expanding a gas under high pressureRest was for Deval, Linde and Hampson in 1844 to adopt Cailletets method to a practical continuous process and prof.Linde is credited today with the first man producing liquid oxygen on a commercial scale.

PROPERTIES AND USES

OXYGEN IN SOLID STATE: It is a hard, pale blue, doubly refracting crystalline solid. Melting point: -218.810C Density at -252.50C: 1.4256 gm/cc Specific heat at -2560C: 0.078 calHeat of fusion at -2190C : 313 cal/gmOXYGEN IN LIQUID STATE : It is a pale steel blue, transparent and very mobile liquid

Boiling point : -182.020C Density at boiling point: 1.14gm/cc Surface tension at B.P. : 13074 dynes/cmIt is a non conductor of electricity and strongly magnetic when compared to iron.OXYGEN IN GASEOUS STATE: It is a colorless, odorless, tasteless, diatomic gas, a volume of it slightly heavier than equal volume of air. One Litre of oxygen under standard condition weighs 1.42901gm and the corresponding weight of air is 1.2929gm. The oxygen is only slightly soluble in water at ordinary temperature and pressuresThe value of Ostwald coefficient=conc. in liquid phase/conc. in gas phase =0.308 at 250 C and 1atm.Physiochemical properties: Atomic mass: 15.9994 Atomic number: 8Isotopes: O16 99.759% O17 0.037% O18 0.204%Thermo physical properties:Gas at 1 atm: Density (0): 1.42908gm/litDensity (21.11): 1.327gm/litDensity at B.P: 4.467gm/litDensity at triple point: 0.0108Viscosity, at 250: 206.39 mili poise

CHEMICAL PROPERTIES: Action of oxygen with hydrogen: oxygen and hydrogen at an ambient temperature when activated by a catalyst such as platinum metal, forms water. 2H2 +O2 = 2H2OAction of oxygen on metals: all metals react with oxygen, although not with same rate or the same energy. Action of metals on non metals: oxygen reacts directly with all the non metals although direct action often is not the most effective way of producing oxide. Even fluorine forms a compound with oxygen, OF2

USESIron and steel applications: Around 98%pure oxygen is needed in steel industry for blast furnace to enrich the air from 21%normal oxygen to 26%.It is required for cutting and welding and for scarfing billets, for this purpose oxygen should be 99.5% pureApplication in chemical industry: In chemical industry, oxygen is used in the manufacturing of synthetic materials such as acetylene, ethyline oxide, methanol, acrolein, hydrogen peroxide, titaniumdioxide etc.Also find its wide application in waste water treatment.Oxygen for cutting purposes:oxygen cutting of ferrous metals into shapes for fabrication is practiced on a large scale, in this technique the metal is heated first by means of a fuel oxygen torch to the point of combustion &then a narrow, high velocity(1500ft/sec or more)stream of oxygen is introduced through the center of the flame. It is used that oxygen should be of high purity of 99.55 or better.Aerospace uses: Aerospace activities have placed an expanding demand on oxygen production. Oxygen continues to be the preferred oxidant for large scale& first stage firing because it is readily prepared at nominal cost and unlike fluorine or the oxides of nitrogen, use does not make an area uninhabitable or require extensive decontamination.

\Oxygen in life support systems: The development of high fly commercial aircraft, recent projects to explore both inner and outer space require oxygen in the life support systems.Moreover under the sea as well as well above the sea level in the mountains oxygen cylinders are needed. In the medical applications, oxygen cylinders are employed almost routinely for patients suffering respiratory function &represent the most frequent contact of the patient with medical oxygen. Use of oxygen in pediatrics incubator has been an important factor in increasing the survival rate of premature infants.Modern anesthesia routinely uses oxygen as a component of the gaseous mixture, herby overcoming the diluting effects of gaseous mixture, herby overcoming the diluting effects of gaseous anesthetics & ensuring an adequate supply for life support.

Chapter 2

Selection of process

DIFFERENT PROCESSES AND SELECTION OF PROCESS

DIFFERENT MANUFACTURING PROCESS:There are four different processes for the manufacture of oxygen from air,they are:1. Low temp. rectification of liquid air2. Electrolysis of water.3. Cyclic chemical absorption& desorption from air by barium oxide at elevated temp. This is called Brin process.4. A process similar to Brin process but using cobalt compounds.Out of these only low temp. Rectification process was found to be the most economical & feasible.This process mainly comprises of two steps. One is the liquefaction & then the separation of liquid air into oxygen & nitrogen. Since the process mainly requires low temp., refrigeration is necessary. Different cycles are in use to have required refrigerating effect.For the separation also different types of columns are in use like simple, compund & double. Taking into consideration the two stages various combinations were proposed.1. Heylandt liquid oxygen process2.Le rought process3. Kellog low pressure process4. Claude process5. Linde frankl processOut of the above processes the Claude process and lindey frankl process are really important processes for the manufacture of oxygen.Briefly giving an overview of these two processes.CLAUDE PROCESS:This process is characterized by the double expansion engine. After passing through the preliminary heat exchanger, at 60 atm pressure, part of the air traverses the liquefier from top to bottom & is admitted to the base of the dephlagmetar after expanded to 4 atm.The reset is expanded like wise to 4 atm,in the first stage of the expansion engine, after which it is again separated into two portions. One is added to the partially liquid air behind the valve, the other is warmed by passing up the upper part of the liquefier,& is then expanded to 1 atm.In the second stage of the engine, the expanded air there upon mixes with the cold nitrogen vapor emerging from the top of the column & returns through the lower part of the liquefier & through the preliminary heat exchanger. The oxygen in this air is wasted. The refrigerant between the tubes of the dephlagmator is liquid oxygen from the column, part of which is withdrawn through tube.

fig 1 t-s diagram for claud process.

LINDE-FRANKL PROCESS:This is the most important process for the commercial production of oxygen in this process at first air is filtered & compressed to 6.8 atm in turbo compressor. During the compression cooling is done to maintain the temp to 35 -400C.After compression the air is divided into two streams. One is 65% stream & the other is 35%, now the larger stream is then passed through after cooler and heat exchanger where it is cooled to -1500C to -1700C by the incoming pure nitrogen & waste nitrogen streams produced from rectification columns. The smaller stream is passed through reciprocating compressor to increase the pressure to about 200atm.Here the air temp is maintained at 4-80C by intermediate cooling between stages using cold water obtained by ammonia refrigeration. Then the air goes through high pressure heat exchanger where the temp of air is brought down to about-120 -1400C. Now the air undergoes expansion to about 6.5 atm in the expansion engine .The temperature of air is brought down to -170 to-1740Cby joule Thompson effect. Now the air will be in liquid state &mixes with the larger stream & changes the whole air stream into saturated liquid state. This saturation liquid is fed to Linde rectification column. This column may be single, double or compound depending on requirement. the liquid product coming out will have a purity of about 99.4 -99.99%.This liquid is partially vaporized in condenser, to liquefy the nitrogen vapor &the rest may be taken as liquid product or it may be obtained in gaseous state if it is used for cooling of incoming air, the other products that obtained are pure nitrogen of purity above 98% & waste nitrogen product of purity of about 92-96%.These cold streams are utilized for cooling air, this process is most economical for tonnage oxygen plants &most widely used in the world.

Fig 2. Flow sheet of linde frankl process

Fig 3.T-S diagram for the process

PROCESS VARIABLES:

1. Initial temperature: The initial or the entrance temp for air is an independent variable. Although it varies from 10 to 500C depending on the weather conditions and the type of after cooler, for most cases, it is assumed as 27-300C.

2. Initial pressure: This is an important variable; it is independent in the case of liquid oxygen process. The lowest allowable value of pressure is controlled by the saturation temp. Of air or nitrogen, this must be higher than the boiling point of oxygen at column pressure. In a double column, the air pressure must be such that the nitrogen will condense at the temp of boiling point of oxygen in order to produce reflux. This will be in the 4-6.5atm range, depending on the pressure in the lower pressure column, and on the necessary temperature difference in the condenser boiler.

3. Temperature approaches: An analysis of the process shows that it is necessary to establish the minimum temp approaches for heat exchangers. For the exchange of sensible heat, the minimum approach has to be 3-50C

4. Temperature level of refrigerants: vapor liquid refrigerants may be used at almost any temp level down to liquid air temperature provided the right material is chosen. The following is a list of common refrigerants with the temp to which they can cool air.

Refrigerant temperature (0K)

Sulphur dioxide 263Methyl chloride 249F_12 244Ammonia 240Propane 231Ethane 184Ethylene 169Methane 111Nitrogen 77

5. Intake temperature to expander: this variable has the flexibility of being at almost any value between the room temp and one close to that of air liquefaction, without appreciable effect on the efficiency of expander. It is so chosen that the exhaust is saturated vapor.

6. Purity of product gases: this is fixed at some value less than 100%,with due regard to certain limitation that may set a definite upper limit with a single column, the product nitrogen purity cannot exceed 93%.in case of producing liquid oxygen, the available refrigeration may set a still lower limit.

7. Heat leak: this is one of the most difficult variables to be selected since it depends on factors which cannot readily be evaluated in advance. For e.g. the size of the plant is of at most importance. Although the leak on an hourly basis increases when expressed the basis increases as plant size increases when expressed the basis of a unit of air treated or oxygen produced, it decreases and in large plants, is almost negligible.

8. Energy requirement: once the various quantities, temp, pressure are established the work done by the compressors and expanders, and hence the energy requirements can be easily calculated.

9. Sizes of equipment units: The thermodynamic analysis plays an important role in this variable, since it deals with the driving forces in the heat exchangers and in distillation columns.

The type of column chosen also plays a big role in the operation of the unit. There are 3 types of columns that can be used for rectification namely simple, double and compound. Simple consists of only an exhausting column and liquid feed to it consists only the reflux. It has an adv of simplicity and economy of distribution and construction, but the disadvantage is that it suits only moderate yield of oxygen since nitrogen leaving the column can never be enriched beyond 93%.

The compound column has both exhausting and enriching sections, but it must be provided with a source of refrigeration at a very low temp, ie below the B.P.of nitrogen at the operating pressure of the column.

The double column has two rectification columns operated at two different pressures, so chosen that the nitrogen at high pressure column condenses at a temperature above the boiling point of oxygen of low pressure column. Usually the pressure in the high pressure column is 6-7 atm and in the other is 1.5-2 atm. This has the adv of high yield without auxiliary refrigeration, but it is expensive and complicated to manufacture. But on the tonnage scale, double column is preferred of high yield and high purity.

THEORETICAL MINIMUM ENERGY REQUIREMENT

The computation of absolute minimum work required for air separation and oxygen liquefaction with the case is possible. It known that this work is given as:W=H-T0.SIn air separation is zero if one neglects the enthalpy of mixing. The valve of S may be shown that the enthalpy change accompanying the mixing of pure oxygen and pure nitrogen to yield one mole of a mixture of x mole fraction of O2 is

S = RX lnX+R(1-X)ln(1-X)A similar expr may be written for a mix of mole fraction YSo the entropy of M1 moles of air relative to pure O2 andN2 is 0.21Rln0.21+0.79ln0.79 Accordingly if M1 moles of air to be separated into M2 moles of O2 rich gas of composition X2 mole fraction O2 and M3 moles of oxygen lean gas of X3 mole fraction of O2 ,the overall entropy change will be, S = R [ M2 { (2X2 lnX2 )+(1-X2)ln(1-X2) } + M3{ (X3 lnX3 )+(1- X3 )ln(1-X3 ) }] -RM1 {0.21ln0.21+0.79ln0.79}This expr when multiplied by ambient temp., usually taken as 303K yields the theoretical minimum energy requirement.Now the material balance equations are:

M1=M2+M3

0.21M1=X3 M3+X3 M3 Defining M2 and M3 for chosen values of M1,X2 and X2. SELECTION OF THE PROCESS

Prior to the selection of the process it should be emphasized that there are so many adjustable variables involved, it is very difficult to put all on a comparable basis. Usually on the smallest scale, the most important economic factors are capital and labor charges. Thermodynamic efficiency and hence power charges for compression are of less consequence. Charges for material such as compressor oil, chemicals for air purification are small in relation to the other costs. The labor charges also decreases when the scale increases and in this cases the capital costs and power costs dominate. When the plant produce a liquid product, as in this case the power requirement to provide necessary refrigeration is considerably increased and thermodynamic efficiency is of much importance. The essential requirement on a general scale is a cheap and simple plant, easy to operate for which a thermodynamic efficiency is not needed. But if the liquid is produced on a large scale, the thermodynamic efficiency becomes important. As the purpose of this project is to design large scale oxygen product plant the only process that seems economical is the LINDE FRANKL PROCESS.As the most important adv here is the high purity of oxygen (99.5) %, although the work of liquefaction of air is about 3.5kwh/gallon which is relatively higher than other process. Also the outgoing streams from the rectification column are used effectively in supplying the required refrigeration for cooling the incoming air.In view of all the above cited advantages the process For this project is linde-frankl.

Chapter 3

Mass & Energy balance

MASS AND ENERGY BALANCE CAPACITY of the unit is, 205 T/day of gaseous oxygen of 99.95 purity.And 10 t/day of liquid oxygen of 99.97% purity. So total oxygen production = 215 T/day = (215*1000)/(24*32) = 281.25 kg moles/hr

At standard temperature and pressure,1kg.mole occupies 22.4m3Oxygen produced in volumetric units =281.25*22.4 =6300Nm3/hr

Standard analysis of air:Component volume%Nitrogen 78.03Oxygen 20.99Argon 0.94Hydrogen 0.01Helium 0.0003Krypton 0.00011Xenon 0.00009Carbon dioxide 0.03-0.06Moisture 0.02-0.05

Table 1.Quantity of intake air:

Capacity of the unit = 6300m3/hr of oxygen Volume % of oxygen in the air =21% Quantity of air needed =6300/0.21 =30000m3/hr Assuming about 15% less of air due to removal of moisture,CO2 and from possible leaks.The quantity needed = 30000+30000 *.15 =34500m3/hrAtmospheric conditions :Temperature =300C,relative humidity =60%

Now the air passes through elements of the system

Enthalpy of air at 1 atm and 300C=kcal/kgmol. Enthalpy of air at 6.8 atm and380C=3766.23 kcal/kgmol Moles of air entering compressor=34500Nm3=1540.18 kgmol/hr Enthalpy of air entering compressor=1540.1883701kcal/hr=5701515.3kcal/hr Enthalpy of air leaving the compressor=1540.18*3766.23=5800672.1kcal/hr. So change in enthalpy of air = 5800672.1-5701515.3 99156.8kcal/hrNow the air is split into two streams one is 65% and the other is 35%. Larger stream passes through refrigeration cycle and the smaller stream passes through compression cycle and then expanded in expansion engine.

Mass & energy balance for the refrigeration cycle 65% streamAmount of air passing = 1540.18*0.65 = 1001.117 kg moles /hrFirst air passes through water coolerWater cooler Inlet 6.8 atm and 380 COutlet 6.5 atm and 80CRelative humidity = 60%So partial pressure of water at 380 C=.6*49.692=29.82cm of HgTotal pressure is 760*6.8 = 5168 mmHg Now

P = (H1 *Pt )/(Mw/Ma+H1) from Perrys handbook eq 15.6Where P = partial pressure in mm HgPt = total pressure in mm HgMw = molecular weight of water = 18 Ma = molecular weight of air =29H1 = molal absolute humidity in kg moles of water vapor / kg moles of dry air Substituting the values in the formula,We get H1 = 3.6*10-3 Let us assume air coming out from the cooler is saturated So final humidity = 100%Vapor pressure of water at 80 0C =8.045 mm Hg.Final partial pressure of water = 8.045 mm of HgTotal pressure = 6.5 *760 =4940 mm of HgUsing the same eqWe get H2 = 9.68*10-4 kg moles of water vapor /kg moles of dry air Let the amount of dry air be M kg molesNow M +.0036M=1001.117So M=997.52So amount of water contained =997.52(.0036-.000968)=2.625kg moles /hr Moles of air coming out of cooler 1001.117 - 2.625 =998.492 kg moles/hrComing to energy balance, from enthalpy chart of airEnthalpy of entering air = 3766.23 kcal/kgmol Enthalpy of outgoing air = 3573.09 kcal/kgmole Enthalpy of entering air =1001.117*3766.23 = 37770436.8 kcal/hrEnthalpy of leaving air = 998.492*3573.09=3567701.7 kcal/hrChange in enthalpy of air = 202735.1 kcal/hr Heat exchanger Air entering = 99.2492 kg moles /hrVolume % of CO2 in air = 0.045%Moles of CO2 present in air entering= 998.492*0.045/100= 0.4493 kg moles/hr Amount of dry air leaving heat exchanger = 997.52-0.4493 = 997.07kg moles /hrThis is cooled to -1680 C by circulating pure nitrogen and waste nitrogen streams coming from distillation column. Total Amount of pure nitrogen is 361.61 kg moles /hr It splits into 2 streams one passes through this heat exchanger and other through high pressure heat exchanger in the compression cycleLet the amount of pure nitrogen stream passing this exchanger is M kg. Amount of waste nitrogen 892.69 kg moles /hr

Conditions and enthalpy of streams: from enthalpy chart.

Stream Flow rate Kg moles/hr temp0 C Pressure (atm) Enthalpy(kcal/kg)

Dry air In-997.07 Out -997.07 8 -168 6.5 6.5 3573.09 2092.35

Pure N2 In- M Out M -178 4 1.5 1.5 1500 2750

Waste N2 In 892.69 Out 892.69 -172 6 1.5 1.5 1510 2800

CO2 In- 0.4493 Out- nil 8 solidifies at -56.6 Calculated below

Moisture In-.972 Out-nil 8 Solidifies at - 4 Calculated below.

table 2Heat given up by dry air = 997.07(3573.09-2092.35)=1476401.4 kcal/hrHeat given up by CO2 :CO2 at first cools from 8 to -56.6 0 C and then solidifiesSpecific heat Cp is 10.34 +0.0027T-195500/T in kcal/kg mol kT is absolute temp in kHeat lost by CO2 on cooling to -56.6 = T2 M CpdT T1 = 411.95 kcal/hr Heat of sublimation of CO2 = 1993.83 kcal/kg mole Heat lost by CO2 on freezing = 0.4498*1993.83 kcal/hr = 896.83 kcal/hrTotal heat lost by CO2 411.95 +896.83 = 1308.78 kcal/hrHeat lost by water molecule by cooling to -4= .972 *18(8-(-4))= 209.952 kcal /hrHeat lost by water molecule on freezing 0.972 *80*18 = 1399.6 kcal/hr Latent heat of condensation = 80 kcal/hr So total heat lost by water = 209.52+1399.6=1609.632 kcal /hr Heat joined by water nitrogen stream 892.67 (2800-1500) = 1160497 kcal /hr Heat gained by pure nitrogen = M(2750-1510) = 1240M 11160497+1240M = 1476401.4+1308.78+1609.632Or 1204M=318822.81M = 257.115 kg moles/hr The amount of waste nitrogen passing through heat exchanger = 257 kg moles /hr

Amount of air passing = 34500*.35=12075 1207/22.4=539.063 kg moles/hr This air passes through compressor Compressor Before compression, air is at 6.5 atm and 38 0C And after it is at 800 atm & 6 0C Enthalpy of inlet air = 3766.23 kcal/kgmol Enthalpy of outlet air = 3251.119 kcal/kgmol

So net enthalpy of entrance = 3766.23*539.063 kcal/hr = 2030235.2 kcal/hr And for exit air 539.063*3251.119 kcal/hrAssuming no condensation of water vapor The net change in enthalpy of air =27677.3 kcal/hr CO2 absorber: volume % ofCO2 in air =.045$% Moles of V present = .045*539.063/100=.2426 kg moles/hr From the earlier calculations, molar humidity of air + 3.6*10-3 If G is the amount of dry air in kg mole/hr then G+.0036G=539.063So G=538.478 kg moles/hrMoisture present 539.063-538.478 = 0.535kgmoles/hrIn the adsorber most of the moisture and CO2 is removed High pressure heat exchangerNo change in mass. Now heat balance, from enthalpy chart in Perrys handbook Stream Flow rate Kg moles/hr Temp. 0C Pressure atm Enthalpy Kcal/kg

Pure nitrogen In- 104.495 Out- 104.495 -178 4 1.5 1.5 1500 2750

Oxygen In -267.86 -178.2 2 30 30 2161.6 3328.64

Air In 538.478 Out-538.478 6 ? 200 200 3251.119 ?

Table 3. Heat taken by pure nitrogen stream= 104.495(2750-1500)= 130618.75 kcal/hr Heat taken up by oxygen = 267.86 (3328.64-2161.6) = 312603.33 kcal/hr Heat lost by air = 538.478(3251.119-H)Now H=2428.91 kcal/kg mol Now the corresponding temp of air is -1500 CSo the air entering the expansion engine is at 200 atm and -1500 C

Here air is expanded to 6.5 atm the temp. Of air after compression is -1740 C(liquid state) Now both the air streams are mixed. And finally the temp. Is at -1720 C and will be in saturated liquid state and fed to lower distillation column.

The set up of oxygen plant consists of two linde column:Upper column (1.5 atm)Lower column (6.5atm) The equilibrium data is as follows For lower column.

Temp. 0 C Liquid composition X% Vapor composition Y%

-160.9 0 0

-163 2.5 6.5

-165 17 32.5

-167 27 48.5

-168 32.5 50

-170 45 68

-172 60 79.5

-174 79 90

-175 89 95

-176 100 100

Table 4.

Chapter 4

Equipment design

Equipment design

Distillation column

Fig.4. Typical distillation column with reflux and reboiler

For mass balance, total air entering the column=997.07+538.478=1535.55 kg moles/hrFrom an overall balance of linde column kg moles of oxygen coming out = 6300/22.4=281.25kgmoles/hrSo amount of waste nitrogen issuing out = 1535.55-(361.61+281.25)=892.69 kg moles/hrLet the liquid to vapor ratio at the top of both columns to be .58So L1/V1=L2/V2=.58Suffix 1 for upper column

Now V2=361.61L2=.58*361.61=209.734 kg moles /hrThis L2 comes out of condenser in vapor state and fed at the top of upper column as liquid after passing through expansion valve.58V1+209.734=V1.42V1=209.734 so V1=499.366 kg moles/hr The amount L1 refluxed to the lower column after condensation occurs in the condenser =.58*499.366= 289.632 kg moles/hrLet F=amount of feed =1535.55kgmoles/hrW=bottom product from lower which is refluxed back to the top column ( rich liq)S=side stream from lower column, which is refluxed back to top column (impure liquid) Xf=mole fraction of N2 in feed = .79Xw=mole fraction of N2 in rich liquid =.65 Xs= mole fraction of N2 in IPL=.96X1= mole fraction of N2 in top product = .9999

Now, from overall& component balance F=W+S+L2 (1)FXF =WXw +SXs + X1L2 (2)From (1) 1535.55 = W + S +209.734

W +S = 1325.816From (2) (1535.55 *.079) = (W*.65)+(S*.96)+(209.734*0.9999) (4)1003.3714 = .65W + .96S Multiplying (3) by .650.65W +0.65S = 0.65 * 1325.816 (5) subtracting 5 from 4 141.591= s (0.96-0.65) S=456.746 kg moles /hrFrom 3 W = 1325.816-456.746 = 869.07 kg moles/hr Also L1/V1 = .58L1/V1 = (456.746-289.633)/499.366= .335The stream S is taken out as liquidV1 = V1=V1L!/V1 = (1535.55+167.112)/499.366=3.4

Using the equilibrium composition data at 6.5 atm, the curve is drawn from XL1=.9999 the operating line is drawn with L1/V1=0.58The side stream being taken out is impure liquid of 96% of N2 For this point from which the operating line of slope L1/V1 = 0.335 is drawn will be Xs1 = [(456.746*.96) + (209.734*0.9999)]/456.746+209.734 = 0.978These 2 operating lines are seen to be intersecting at X =0.96 Now the operating line for exhausting section is drawn from Xw =0.65 with slope L1/V1 =3.4 Hence from the graph no. of theoretical plates = 17.3

Graph 1.

Graph 2.

Plate efficiency

Where is the average molal viscosity of the feed in cp Of liquid O2 = 0.19 Cp Of liquid N2 =0.24 CpAvg molal of feed = .79*.24 +.21*.19= .2295 Cp = .6127 So actual no. of plates = 17.3/.6127= 27.249 =28 So no. of plates =28From graph, feed enters the 15th plate.Actual plate at which feed enters = 15/.6127 = 23.952 = 24Feed enter 24th plateFrom graph, side stream is taken out from 5th plateActual plate from which side stream is taken out is 5/0.6127=8.16 = 9th plate Height of column : No. of plates =28 Plate spacing = 12 Spacing at top & bottom = 12each Spacing for feed = 12 Total height =( 28+2)*12 =360 = 9.146

Computer aided design of column The software CHEMSEP from chem. Office was used, data were fed and the following excel sheet obtained.Stream Feed1 Top Bottom

Stage 14128

Pressure (N/m2) 650000200000650000

Vapour fraction (-) 110

Temperature (K) 10583.686104.821

Enthalpy (J/kmol) 5624000624800011000000

Entropy (J/kmol/K) 41261.142611.392166.4

Mole flows (kmol/s)

Oxygen 1200.390691119.609

Nitrogen 379300.72978.2712

Total molar flow 499301.12197.88

Mole fractions (-)

Oxygen 0.2404810.0012970.604452

Nitrogen 0.7595190.9987030.395548

Mass flows (kg/s)

Oxygen 3839.8812.50173827.38

Nitrogen 10617.38424.622192.69

Total mass flow 14457.28437.136020.06

Mass fractions (-)

Oxygen 0.2656040.0014820.63577

Nitrogen 0.7343960.9985180.36423

Vapour:

Mole weight (kg/kmol) 28.972328.0192

Density (kg/m3) 21.57238.0542

Viscosity (N/m2.s) 7.7E-065.67E-06

Heat capacity (J/kmol/K) 29204.529254.8

Thermal cond. (J/s/m/K) 0.0108640.008368

Liquid:

Mole weight (kg/kmol) 30.4227

Density (kg/m3) 889.88

Viscosity (N/m2.s) 0.0001

Heat capacity (J/kmol/K) 60360.7

Thermal cond. (J/s/m/K) 0.112383

Table 6.With the help of software the mcabe thiele diagram was obtained .

Graph 3. With chem. sep

Chem. Sep data sheet

Graph 4. The above graph gives the variation of relative volatility with stage no.

Following the similar pattern design of upper column can be done in the same fashion.

Graph 5.This is entropy vs stage graph.as obtained by the chemsep,from chemoffice

CONDENSER DESIGN

Basic Theory

The fundamental equations for heat transfer across a surface are given by:

Where w is tube side flow and W is shell side flow and CPt is specific heat fr tube side and CPs is for shell side In design, a correction factor is applied to the LMTD to allow for the departure from true countercurrent flowto determine the true temperature difference.Tm = Ft TmThe correction factor is a function of the fluid temperatures and the number of tube and shell passes and is correlated as a function of two dimensionless temperature ratios

The overall heat transfer coefficient U is the sum of several individual resistances as follows

Continuing with our problem .Nitrogen vapors enter counter currently through the tubes of the condenser, where it meets the liquid oxygen coming out on the bottom product of the top column. for this purpose, a 1-2 shell and tube condenser is chosen. It is a vertical type condenser.

fig 5.diagram indicating the flow pattern

Enthalpy of N2 vapor at 96.5K is 238.35 J/gm=1595.16 cal/kgmol=1595.16 Kcal/kgmol. Enthalpy of N2 vapor at 94.8 K is1592.05 cal/kgmolEnthalpy of N2 liquid at 94.8K is 443.52 cal/kg molSo total enthalpy change of stream= (499.367*1595.16) (209.734*1592.02+289.633*443.52) = 1399009.42KJ/hrLMTD calculations:fig.6Tha is inlet temp of nitrogen vapor = -176.50 CThb is outlet temp of nitrogen = -178.20 CTca is inlet oxygen liquid =-179.50 CTcb is outlet oxygen =-178.20 CTl = 1.491Taking overall heat transfer coefficient as 100 Btu/hrft2F0567W/m2 0Cnow Q = UATl From this A comes around 460 m2 Let N is the no. of tubes so total heat transfer area =N*DLWhere D is tube OD And L is the lengthWe are choosing 1 inch OD. and 14 BWG cu tubes of 8 inch length and calculating the no. of tubes obtained is 2360 Also taking a triangular pitch of 1.25 inch that ,nearest no of tubes found from TEMA regulations is 2362 and shell dia as 1.67 metersNow the corrected area becomes 460.5m2 So using this value of heat transfer area the U becomes 560.6W/m2 0C Inside and outside coefficient calculations:a. Nitrogen in tube side Flow area for the given tube is 3.52cm2Gas mass velocity = 499.367*28/(3.52*2362/10000*2)=33634.52kg/m2hrDensity of nitrogen =802.6kg/m3 So velocity = .01155 m/sec So Nre = 822 Condensate loading per linear foot is G = W/DN = 88.955 kg /m hrFrom graph in kern we get the value of hi = 240 btu/hr ft2 0 FAlso ho = hi*id/od= 22.16 btu/hr ft2 0F

Oxygen on shell side Density = 1140 kg/m3Thermal conductivity =k= 0.07 btu/ft.hr.0 F Here flow area = As = ID*C*B/144ptC is clearance taken as 0.25 inch B = baffle spacing = 1 inch i.d. =66 inch Pt =1.25 inch so As = 1.1 ft2 =11.84m2 Gas mass velocity Gs = W/As = 1535.55*32/11.84=4150.135kg/m2 hrFrom this NRE can be calculated by DeGs/ We get it as 72824Now from the graph of jH Vs Nre jH =175now ho = jHk/De*(c/k)^-1/3solving ho= 223.61 btu/hr ft2F=1267.8w/m2 0C

Clean overall coefficientUc=hio*ho/(hio+ho)=598.865 w/m2 0CDirt factor :Rd = Uc-Ud/(Uc*Ud)= 0.0000952m2 0C/w

Pressure drop:Shell sideNo. of crosses = N+1 =8 so no. of baffles =8-1 = 7For Nre 72824 Friction factor =0001.4 (graph)Sp gravity of oxygen = .00624Ds= inside diameter in ft and n is the no. of tube passesSo Ps=fGs 2Ds(N+1)/(5.22*10^10De*S) So Ps =1.522psi

For tube side:Pt= fGt 2Ln/(5.22*10^10*DS) ,psi = 0.155*10^-3 psiReturn losses: Pr = 4nV2/2Sg = 2.2 *10 ^-4 psi So PT = Pt+Pr = 3.75*10 ^-4

Design summary: no. of tubes =2362Shell ID = 66 inchClean overall H.T.C. = Uc =598.865 w/m2 0C Design overall = 560.6W/m2 0C Dirt factor Rd=) 0.0000952m2 0C/wPressure drop for tube side = 0.000375 psiPressure drop for shell side = 1.522 psi

Computer aided design of condenser Used KAMLEX heat exchanger software to design and test my results obtained from my designResults and diagram.

Fig 7.Condenser with geometry

Schematic diagram

Fig 8.

Results from kamlex.:No of tubes = 2362Overall heat transfer coefficient: Heat exchanged = 563.2 W/m2 0C Heat exchanged =1406504.1KJ/hrPressure drop for shell side: 11574.35 PPressure drop for tube side: 30.2Pascal Chapter 5

Conclusion and discussion

CONCLUSION AND DISCUSSIONOn the basis of above calculations, I propose an oxygen plant with a capacity of 205 T/day with the above design specifications. This plant can be actually called as an oxygen cum nitrogen plant as it is producing pure nitrogen along side the oxygen.Apart from equipment design there are certain other factors which have to kept in mind before designing and finally establishing a plant., like location analysis and a plant layout .Plant lay out and locationBut before making such a choice, he has to go through the detailed locational analysis considering various factors, which influence his decision. It is a long-term strategic decision, which cannot be changed once taken. An optimum location can reduce the cost of production anddistribution to a great extent. Thus great care and appropriate planning is required to select the most appropriate location.

The efficiency of production depends on how well the various machines;production facilities and amenities are located in a plant. An ideal plant layoutshould provide the optimum relationship among the output, floor area andmanufacturing process. An efficient plant layout is one that aims at achieving various objectives like efficient utilization of available floor space, minimizes cost, allows flexibility of operation, provides for employees convenience, improves productivity etc. The entrepreneurs must possess the expertise to lay down a proper layout for new orexisting plants. It differs from one plant to another. But basic principles to be followed are more or less same.Plant layout is applicable to all types of industries or plants. At the end, the layoutshould be conducive to health and safety of employees. It should ensure free andefficient flow of men and materials. Future expansion and diversification mayalso be considered while planning factory layout Safety precautionAll personnel being employed for work in connection with oxygen/rich air should be cautioned concerning the hazards involved and precautions to be observed. Oil grease or similar substances must not be allowed to come into contact with compressed oxygen or liquid oxygen. Contact of this substance with oxygen may result in an explosion. Personnel working in an area of possible oxygen concentration, such as near an oxygen vent or a liquid oxygen spillage, or in a trench where oxygen seepage and concentration might occur, must ensure that their clothing is free from contaminations of oxygen before lighting a cigarette or approaching a naked flames. It is essential that the clothes be dries for at least 15 minutes before approaching a flame after any such contamination.

The following precautions must be strictly observed at all times:

1. Thoroughly wash all oxygen fittings, valves and parts with clean Tricolor Ethylene / carbon tetra chloride (CTC) before installation. Never use petrol, kerosene or other hydrocarbon solvents for this purpose. All tubing, lines valves etc. to be used in oxygen service, must be of an approved type and must be thoroughly degreased and blown out with clean oil-free compressed air or Nitrogen before being placed in service.

2. Do not permit the release of Acetylene or other flammable gases in the vicinity of the plant air intake. A concentration of Acetylene exceeding 5 parts per million in liquid oxygen may explode with extreme violence. Strict supervision is essential to minimize the possibility of contamination.

3. The plant and the plant vicinity must be kept clean and free from abstractions at all times. Any oil leak within the plant surrounding must be rectified without delay. Oil spillage must be cleaned up immediately using rag and carbon Tetra Chloride.

4. Do not lubricate oxygen valves, regulators, gauges or fitting with oil or any other substance.

5. Ensure that insulation removed from the Air Separator jacket is not contaminated with oil or other inflammable materials. Personnel carrying out maintenance on the Air Separation Plant equipment must wear clean overalls and their hands and tools must be free of oil. This ensures that the insulation and equipment within the jacket is not contaminated with oil. Should contamination take place the affected materials must be discarded and replaced by clean new material?

6. Do not fasten electric conduits to the plant or its pipelines.

7. Do not use oxygen as a substitute for compressed air, spark present in an atmosphere of oxygen will immediately burst into flame.

8. Do not fill any container or pipe line with oxygen unless it has been thoroughly degreased with clean CTC or TCE.

9. When discharging liquid oxygen or rich liquid from drains, valves or pipe lines, open valves slowly to avoid the possibility of being splashed. In particular ensure that liquid does not run into shoes or gloves. Contact with liquid oxygen rich liquid will cause frostbite evidenced by whiteness and numbness of the skin. The affected parts must be batched at once in cold (not box) water and seek medical attention immediately.

10. Do not breathe cold oxygen vapor. The temperature of the vapor rising from liquid oxygen is approximately - 181 Deg C. A deep breath of vapor at this temperature can result in frost-bitten lungs with resultant serious illness and permanent disability or death.

References

Basic Theoretical Physics ,Springer new York ,part 4 ,pages 313-325 Cryogenic Engineering, Springer new York ,part 1,pages 3-27 and 146-160 US Patent 4072023 - Air-rectification process and apparatus Chemical and Petroleum Engineering journal, Air separation in plants with an external source of refrigeration ,,volume 4 ,pages 825-829 American Chemical society journal , some aspects of gas separation at low temperatures by W. H. Granville

air liquefaction: distillationencyclopedia of separation science, 2007, pages 1895-1910,science directR. agrawal, D.M. herron Cryogenic Process EngineeringEncyclopedia of Physical Science and Technology, 2004, Pages 13-36Klaus D. Timmerhaus

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