Multi effect distillation desalination

8/17/2019 Multi effect distillation desalination

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DESIGN OF A PLANT TO PRODUCE 1000 m3 OF FRESH

WATER BY THERMAL DESALINATION USING MULTI

EFFECT DISTILLATION

A Project Report

Submitted by

PRASHANT SHARMA (1071210024)

SAURABH MISHRA (1071210048)

Under the guidance of

Mrs. E. Poonguzhali

(Assistant Professor, Department of Chemical

Engineering)

In partial fulfillment of the requirements for the award of the

degreeof

BACHELOR OF TECHNOLOGY

in

CHEMICAL ENGINEERING

Department of Chemical Engineering

Faculty of Engineering & Technology

SRM University, SRM Nagar, Kattankulathur-603203

May 2016


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BONAFIDE CERTIFICATE

Certified that this project report titled “DESIGN OF A PLANT

TO PRODUCE 1000 m3

OF FRESH WATER BY THERMALDESALINATION USING MULTI EFFECT DISTILLATION”

is the bonafide work of “PRASHANT SHARMA (1071210024),

SAURABH MISHRA (1071210048)”, who carried out the project

work under my supervision. Certified further, that to the best of my

knowledge the work reported here in does not form any other

project report or dissertation on the basis of which a degree or

award was conferred on an earlier occasion on this or any other

candidate.

SIGNATURE SIGNATURE

Mrs. E. Poonguzhali

M.Tech. Dr. M. P. Rajesh

GUIDE HEAD OF DEPARTMENT

Assistant Professor Department of Chemical Engineering

Department of Chemical Engineering

Signature of the Internal Examiner Signature of External Examiner


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ABSTRACT

Desalination plants separate sea and brackish water into two flows

consisting of a freshwater stream (permeate in reverse osmosis,

condensate in thermal processes) with a low salt content and a stream

with a high salt concentration (brine or concentrate). Every desalination

technology requires energy for this separation process, which is supplied

to the system by thermal or mechanical means (generally as electrical

power). The thermal desalination process is based on evaporation and the

subsequent condensation of the steam.

Multiple-effect distillation (MED) is a distillation process often used for

sea water desalination. It consists of multiple stages or "effects". In each

stage the feed water is heated by steam in tubes. Some of the water

evaporates, and this steam flows into the tubes of the next stage, heating

and evaporating more water. Each stage essentially reuses the energy

from the previous stage.

This project aims at designing a plant that can produce 1000 m3

of fresh

water using MED technology. Thus for fulfilling this process an

appropriate Material balance, Energy balance and design of equipment

like condenser and evaporator carried out. This project aims at using and

analyzing all these ideas and the available data to come up with a plant

design which can be used for production of fresh water from sea water


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ACKNOWLEDGMENT

We are extremely thankful to Dr. M.P. Rajesh, head of department

Chemical Engineering, for allowing us to work on this project and for all

the support and guidance he has provided us.

We take immense pleasure in expressing our deepest gratitude to our

project guide Ms. E. Poonguzhali, Assistant Professor (Sr.G.),

Department of Chemical Engineering on her invaluable guidance and

encouragement at every stage of our Project.

We also thank the staff members of Chemical Engineering Department

for their technical assistance and support.

We owe a huge gratitude to our parents who have been of immense moral

support.


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TABLE OF CONTENT

ABSTRACT

ACKNOWLEDGEMENT

LIST OF SYMBOLS

1. INDRODUCTION 10

2. OBJECTIVE 11

3. BASIC CONCEPTS 12

4. VARIOUS DESALINATION AND

COMPARISION BETWEEN THEM 14

5. WORK PLAN 17

6. BASICS ON MED 18

7. PROCESS FLOW DIAGRAM 20

8. PROCESS DESCRIPTION 22

9. MATERIAL BALANCE 27

10. ENERGY BALANCE 28

11. PROBLEM FOR EXPLANNATION 29

OF MASS & ENERGY BALANCE

12. FALLING FILM EVAPORATORS 39

13. DESIGN OF EQUIPMENTS 41

14. COST ESTIMATION 48

15. PLANT LAYOUT 54

16. INSTRUMENTATION & 59

PROCESS CONTROL


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17. ENVIRONMENT ASPECTS & 64

SAFETY

18. CONCLUSION 73

16. REFERENCES 74


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LIST OF SYMBOL ABBREVIATIONS

Ts – Temperature of motive steam

Tn = Temperature of vapour in last effect

Xn- Salt conc of brine in last effect

Xf- Salt conc of feed stream

Mf = Mass of feed

Md= Mass of distillate

Bn= Mass of brine

λ= Latent Heat

A1 = Heat transfer area in first effect

Ac = Downward condenser heat transfer area

Qc = Heat load of condenser

Uc = overall heat transfer co efficient of condenser

Tf = Temperature of feed

Tcw = Temperature of rejected cooling water

Tn = Temperature in last effect vapour

SMcw =specific cooling water flow rate

Brine flow rate =B1, B2 ……. Bn

Distillate flow rate =X1, X2 …. Xn-1, Xn

Effect temperature = T1, T2 ….. Tn-1, Tn

A=Area of evaporator

d=diameter of evaporator

G=mass velocity

U= overall heat transfer coefficient

∆V = Hvap


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Q=heat flow

R= Thermal resistance

T = Temperature

HTC= heat transfer coefficient

VF= Void fraction

K= Thermal conductivity

ư = dynamic viscosity

P = mass density

Nt = No of tubes

Pt=square pitch

Db= Tube bundle diameter

Nr= No. of tube in central row

U=heat coefficient

U= Viscosity

K= Thermal conductivity

Cp= Specific Heat

Uo=the overall coefficient

ho=outside fluid film coefficient

hi=inside fluid film coefficient

hod=outside dirt coefficient

hid=inside dirt coefficient

kw= thermal conductivity of wall

di=ID=tube outside diameter

do=OD=tube outside diameter


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As= Cross flow area

Gs= Mass flow rate

De= Equivalent diameter

Re=Reynolds’s Number

=shell side pressure drop

=tube side pressure drop


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INTRODUCTION

Desalination

Desalination or desalinization is a process that removes minerals from saline water.

Desalination is a process of separation of water from its solution.

Importance of desalination

Only 0.007% of the water available on earth’s surface is useful. The

remaining portion is either inaccessible or saline.

Similar to the Earth’s natural water cycle.

Sea is the inexhaustible source of water and Desalination can help in

providing fresh water from sea.

Typically sea water has a salinity of 55 parts per thousand

Major Stages

Evaporation

Condensation

Precipitation

Collection

Thermal desalination

Desalination plants separate sea and brackish water into two flows consisting of a

freshwater stream (permeate in reverse osmosis, condensate in thermal processes)

with a low salt content and a stream with a high salt concentration (brine or

concentrate). Every desalination technology requires energy for this separation

process, which is supplied to the system by thermal or mechanical means (generally

as electrical power). The thermal desalination process is based on evaporation and the

subsequent condensation of the steam.


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OBJECTIVE

To design a thermal desalination(multi effect desalination) plant that produces

1000 cubic meter fresh water per day

To determine the material balance, energy balance and economic balance for

the proposed plant.

To design the major equipment for producing the product.


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BASIC CONCEPTS

In this process energy, in the form of heat is used to evaporate water and subsequently

condense it again.

Thermal Energy Raw Water water vapour condense fresh water.

Vapour can be formed repeatedly by giving more thermal energy (boiling),which is

the method followed in MED OR

By reducing the pressure of the evaporation chamber (flashing) which is the method

followed in MSF

Note that we have to supply latent heat for converting liquid to vapour and the same is

recoverable to a reasonable extent while condensing.


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World desalination capacity by process

Advantages of MED over MSF

Lower capital cost and running cost.

Unit size is considerably lesser than MSF. Hence, saves space.

Lesser electricity consumption .

Advantages of MED over RO

Production of water with good quality (5 to 50 ppm) RO produces 10 to 500 ppm,

dissolved salt concentration.

Leading technology for large‐scale seawater distillation (in use for 40 years) RO is sensitive to feed water quality and requires extensive feed water treatment to

limit scaling and membrane fouling complicated feed water pre-treatment.

RO uses non-ecofriendly membrane materials.


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MULTI EFFECT DISTILLATION

CONSTRUCTION

Multiple stages

Tube bundles

Sprayers

Ejectors

Pump

Steam tube

WORKING

Following are the steps which take place :

• The steam enters the plant and is used to evaporate heated seawater.

• The secondary vapour produced is used to generate tertiary steam at a lower

pressure.

• This operation is repeated along the plant from stage to stage.

• Latent steam heat is transferred at each stage by steam condensation through

the heat transfer surfaces to the evaporated falling film of seawater.

The product water is the condensate that accumulates from stage to stage.


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Advantages of MED

The MED process operates at low temperatures, which results in:

• small to medium-sized plant sizes

• reduced scaling risk

• low thermal energy consumption

• reduced operating costs

Multi Stage Flashing

The typical process flow diagram of Multi stage flash distillation is given above.

It is usually rectangular in construction. Vertical baffles divide the stages. Tube

bundles are provided on the top portion. Demisters are provided on the vapor path

from flash chamber to the condensing tubes

.


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Product trays are located below the tube bundles but above the brine chamber.

Product water tray is generally sloped from hot end to cold end for natural drain from

one end to the other .

The vacuum in the plant is first created and then maintained by a steam ejector which

continually removes air and non-condensable gases from the plant.

Each stage is maintained at different pressure to enable flashing. Here the boiling

occurs in each stage because temperature of feed seawater in each stage is kept always

higher than the saturation temperature corresponding to the pressure maintained in the

chamber.


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WORK PLAN

To study Multi Effect Desalination process and make a process flow diagram for the

process.

Determine the mass balance, energy balance and economic balance.

To find the optimum number of effects and multiple effect calculation.

Select and design the equipment involved in process.

Indicate the plant layout.

Mention the environmental and safety aspects of the process.


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BASICS ON THE MED PROCESS

A MED evaporator consists of several consecutive cells (or effects) maintained at

decreasing levels of pressure (and temperature) from the first (hot) cell to the last one

(cold). Each cell mainly consists in a horizontal tubes bundle. The top of the bundle is

sprayed with sea water make-up that flows down from tube to tube by gravity

.Heating steam is introduced inside the tubes. Since tubes are cooled externally by

Make-up flow, steam condenses into distillate (freshwater) inside the tubes. At the

same time sea water warms up and partly evaporates by recovering the condensation

heat (latent heat). Due to Evaporation, sea water slightly concentrates when flowing

down the bundle and gives brine at the bottom of the cell. The vapour raised by sea

water evaporation is at a lower temperature than heating steam. However it can still

be used as heating media for the next effect where the process is repeated. The

decreasing pressure from one cell to the next one allows brine and distillate to be

drawn to the next cell where they will flash and release additional amounts of vapour

at the lower pressure. This additional vapour will condense into distillate inside the

next cell. This process is repeated in a series of effects (Multiple Effect Distillation).

In the last cell, the produced steam condenses on a conventional shell and tubes heat

exchanger. This exchanger, called “distillate condenser” is cooled by sea-water. At

the outlet of this condenser, part of the warmed sea-water is used as make-up of the

unit; the other part is rejected to the sea. Brine and distillate are collected from cell to

cell till the last one from where they are extracted by centrifugal pumps. The thermal

efficiency of such evaporator can be quantified as the number of kilos of distillate

produced per one kilo of steam introduced in the system. Such number is called the

Gain Output Ratio (GOR).

The GOR of the evaporator can be enhanced by addition of a thermo compressor

between one of the cells and the hot one. Using LP or MP steam this static

compressor will take part of the vapour raised in one of the cells and recycle it into

higher pressure vapour to be used as heating media for the first one.


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When no steam is available, it is still possible to use the MED process with a

Mechanical Vapour Compressor (MED-MVC). In such case the vapour is recycled

from the cold cell to the hot one by means of a centrifugal compressor driven by an

electrical engine. The electrical consumption of such system is in the range of 8 to 15

kWh/m3. Due to current limitation in compressors technology the maximum capacity

of MED-MVC units is 5000m3/day.


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PROCESS FLOW DIAGRAM


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PROCESS DESCRIPTION

The intake seawater flows into the condenser of the last effect at a flow rate of

Mcw+Mf. This stream absorbs the latent heat of vapors formed in the last effect and

flashing box. The seawater stream is heated from the intake temperature, Tcw, to a

higher temperature, Tf. The function of the cooling seawater, Mcw is to remove the

excess heat added to the system in the first effect by the motive steam. In the last

effect, this heat is equivalent to the latent heat of the boiled off vapors. On the other

hand, the feed seawater, Mf, is heated by the flashed off vapors formed in the last

effect and the associated water flash box. The cooling seawater, Mcw is rejected back

to the sea. The feed seawater, Mf, is chemically treated, deaerated, and pumped

through a series of preheaters. The temperature of the feed water increases from Tf to

t2 as it flows inside the tubes of the preheaters. Heating of the feed seawater is made

by condensing the flashed off vapours from the effects, dj, and the flash boxes, dj.

The feed water, Mf, leaves the last preheater (associated with the second effect) and is

sprayed inside the first effect. It is interesting to note that the preheater of the first

effect is integrated in the heat exchanger of the effect. This is because there is no flash

box in the first effect or flashed off vapors within the effect. The brine spray forms a

thin film around the succeeding rows of horizontal tubes. The brine temperature rises

to the boiling temperature, T, which corresponds to the pressure of the vapour space.

The saturation temperature of the formed vapour, Tvp is less than the brine boiling

temperature by the boiling point elevation, (BPE) i.

A small portion of vapor, Dj, is formed by boiling in the first effect. The remaining

brine, Mf - D, flows into the second effect, which operates at a lower temperature and

pressure. Vapor is formed in effects 2 to n by two different mechanisms, boiling and

flashing. The amount vapor formed by boiling is Dj and the amount formed by

flashing is dj. Flashing occurs in effects 2 to n because the brine temperature flowing

from the previous effect, Tj. is higher than the saturation temperature of the next

effect, Ty. Therefore, vapour flashing is dictated by the effect equilibrium. In effects 2

to n, the temperature of the vapour formed by flashing, T'y, is lower than the effect

boiling temperature, Tj, by the boiling point elevation (BPE) j and the non-


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equilibrium allowance (NEA) j. In the flash boxes, a small quantity of flashing

vapors, d;, is formed with a temperature equal to T'y-. This temperature is lower than

the vapour condensation temperature in effect j, T, by the non-equilibrium allowance

(NEA)'j. Motive steam, Mg, extracted from an external boiler drives vapour formation

in the first effect. The vapor formed by boiling in the first effect, D1, is used to drive

the second effect, which operates at a lower saturation temperature, T2. Reduction in

the vapor temperature is caused by boiling point elevation, non-equilibrium

allowance, and losses caused by depression in the vapour saturation pressure by

frictional losses in the demister, transmission lines, and during condensation. These

losses can be represented as an extra resistance to the flow of heat between

condensing vapor and boiling brine. Therefore, it is necessary to increase the heat

transfer area to account for these losses. The amount of vapour formed in effect j is

less than the amount formed in the previous effect. This is because of the increase in

the latent heat of vaporization with the decrease in the evaporation temperature.

The condenser and the brine heaters must be provided with good vents, first for

purging during start-up and then for removing non-condensable gases, which may

have been introduced with the feed or drawn in through leaks to the system. The presence of the non-condensable gases not only impedes the heat transfer process but

also reduces the temperature at which steam condenses at the given pressure. This

occurs partially because of the reduced partial pressure of vapour in a film of poorly

conducting gas at the interface. To help conserve steam economy venting is usually

cascaded from the steam chest of one preheater to the steam chest of the adjacent one.

The effects operate above atmospheric pressure are usually vented to the atmosphere.

The non-condensable gases are always saturated with vapour. The vent for the last

condenser must be connected to vacuum-producing equipment to compress the non-

condensable gases to atmosphere. This is usually a steam jet ejector if high-pressure

steam is available. Steam jet ejectors are relatively inexpensive but also quite

inefficient. Since the vacuum is maintained on the last effect, the unevaporated brine

flows by itself from effect to effect and only a blow down pump is required on the last

effect. Summary of different processes that takes place in each effect, the associated

flash box and feed preheater is shown in Fig. 4. As is shown the brine leaving the

effect decreases by the amount of vapor formed by boiling, Dj, and by flashing, dj.


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The distillate flow rate leaving the flash box increases by the amount of condensing

vapors from the previous ef fect, Dj’ and dj’. The brine concentration increases from

Xj’ to Xj upon vapor formation. The effect and flash box temperatures decrease from

Tj’ to Tj and from T'j-1 to T'j, respectively.

Comparison of the process layout for MSF and MEE, show that MSF is a special case

of the MEE process. This occurs when the entire vapor formed in the effects is used to

preheat the feed in the preheaters and non-is left for the evaporator tubes. In this case,

the first effect, the flashing boxes, and the bottom condenser of the MEE replace the

brine heater, the distillate collecting trays, and the heat rejection section of the MSF,

respectively.

This includes:

Mass balance

Heat balance

This also includes heat transfer coefficient, thermodynamic loss, ∆P and physical

properties.

The various results obtained from the above balances are:

Performance ratio

heat transfer area

Cooling water flow rate

Temperature

Pressure

Flow rate

Salinity

Assumptions are:

System is at steady state condition.

Distillate is salt free.

Features are:

Equal heat transfer area

(Heat transfer area = Area for brine + Area for vapour)


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MEE PARALLEL FLOW

The data which is given are:

distillate flow rate

Brine concentration ( initial and final concentration)

Temperature of steam

The various factors to be considered are:

Cp – Constant for different temperature and concentration

∆Tloss is constant

Constant heat transfer area

No vapour flashing

Equal thermal load

Driving force = Condensing – evaporating temperature

No energy loss

PERFORMANCE PARAMETERS

Performance ratio = Md/Ms

Ms = D1 λv1/ λ s

Surface area = ∑ (Ai + Ac) / Md

Ac = Qc/Uc(LMTD)c

(LMTD)c = (Tf – Tcw)/ln(Tn – Tf)

{A1 = Heat transfer area in first effect}

{Ac = Downward condenser heat transfer area}

{Qc = Heat load of condenser}

{Uc = overall heat transfer co efficient of condenser}

{Tf = Temperature of feed}

(Tcw = Temperature of rejected cooling water}


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{Tn = Temperature in last effect vapour}

Qc = Dn λ n

Specific cooling water flow rate is defined as:

SMcw = Md/Mcw

PROBLEM

No of effects , n = 6

Ts= 100 degree Celsius

Md= 11.57kg/s

Xf= 35000 ppm-

X6= 50000ppm

∆Tloss= 2 deg Celsius

Sea water temp leaving condenser Tf = 35 deg Celsius

Intake seawater Tcw = 25 deg Celsius

The various unknown quantitie30s are:

Brine flow rate (B1 , B2 ……. Bn)

Distillate flow rate (X1 , X2 …. Xn-1)

Effect temperature ( T1 , T2 ….. Tn-1)

Steam flow rate

Heat transfer area

The known values are:

Ts – Temperature of motive steam

Tn – Temperature of vapour in last effect

Xn- Salt conc of brine in last effect

Xf- Salt conc of feed stream

Md – Distillate flow rate


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MATERIAL BALANCE

Overall material balance :

• Mf = Md + Mb

Salt material balance :

• XfMf = Mb Xb ;

Mf Xf = Bn Xn ;

Bn = ((Xf)/(Xn-Xf)) x Md

•

Here , Md = 11.57 kg/s

Xf = 35000 ppm

Xn = 50000 ppm

• Bn =((35000)/(50000-35000)) x 11.57

• Bn = 27 kg/s

• Now , Md + Mb = Mf

• Mf = 11.57 + 27= 38.57 kg/s

Mf

Md

Mb


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ENERGY BALANCE

In the first effect, the latent heat of the condensing steam is used to increase the

temperature of feed seawater from Tf to the boiling temperature T1 and to provide the

heat required to evaporate a controlled mass of vapor, D1 at T1. This gives

Ms λs = Mf Cp (T1-Tf) + D1 λv1

• Qi = Ai Ui ∆Ti

• Heat transfer = Thermal load

•

Q1 = Ms λs or Qi =Di λvi

• ∆T = Ts – Tn

• Di λvi = Di-1 λv-1 ( from 2 to n)


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SOLUTION

The overall material balance for the process is :

Mf = Md + Bn , where

Mf = Mass of feed

Md= Mass of distillate

Bn= Mass of brine

The component material balance :

Xf . Mf = X6 . Bn

Bn = Xf/(Xn-Xf) * Md

Now , Md = 1000 cu m / day = 11.57 kg/s

Bn = 35000 /(50000) X 11.57

Hence , B6 = 26.997 kg/s = 27kg/s

Mf = 27 + 11.57 = 38.57 kg/s

∆Ttotal = Ts – Tn = 60 deg C

TEMPERATURE PROFILE

Q1 = Q2 = Q3 …. = Qn

Q1 = Ms λs

Qi = Di λvi , where

λs = latent heat of steam at Ts

λvi = latent heat of vapour at ( Ti -∆Tloss)

Qi = Ai . Ui. ∆Ti

Q1/A1 = Q2/A2= Q3/A3 = …. Qn/An

Also ,

U1∆T1 = U2∆T2 = … Un∆Tn


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Temperature drop

∆ T = Ts-Tn

Ts = Temperature of steam

Ts = 100 deg

Tn = Temperature at last effect

Tn= 35 deg

∆T = 100 – 40 = 60 deg

∆T = ∆T1 +∆T2 + … ∆Tn

∆T2 = (∆T1 U1)/ U2

∆T3 = (∆T2 U2)/ U3 = ∆T1U1 / U3

∆T4 = (∆T3 U3)/ U4

∆T5 = (∆T4 U4)/ U5

∆T6 = (∆T5 U5)/ U6

∆T = ∆T1U1 (1/U1 + 1/U2 + …. 1/U6)

T1 = Ts - ∆T1 (for first effect)

T2 = T1 - ∆T (U1/Ui)

Ti = Ti-1 - ∆T (U1/Ui)

Now,

Md = D1+D2 +D3+D4+D5+D6

D3 λv3 = D2 λv2, for 2 to n

D2 = D1 λv1 / λv2

D3 = D2 λv2 / λv3

D3 = D1 λv1 / λv3

Therefore, Di = D1 λv1 / λvi

Where i = 2 to n


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Therefore,

Md = D1 + D1 λv1 / λv2 + D1 λv1 / λv3 + ….. D1 λvi / λvn

Hence,

D1 = Md / λv1 (1/ λv1 + 1/ λv2 + … 1/ λ vn)

D2 = D1 λv1/ λv2

And so on,

Dn = D1 λv1/ λvn

BRINE FLOW RATE

B1 = Mf – D1

For 2 to n, Bi = Bi-1 – Di

SALT BALANCE

X1B1 = XfMf

X1 = XfMf / B1

Xi = Xi-1- Bi-1 / Bi

Heat transfer area is:

A1 = D1 λv1 / (U1 (Ts – T1)

For 2 to n

Ai = Di λi / (Ui (Ti-∆Tloss))

The given values are as follows:

Tcw = 25 deg

Tf = 35 deg

Tn = 40 deg

Xf = 35000 ppm


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Xb = 50000 ppm

Ŋo = 90 % (preheater)

R fi + R fo = 1.75 X 10 ^ (-4) sq m deg / W

∂0 = 31.75 mm

∂1 = 19.75 mm

Brine velocity = 1.55 m /s

Top brine temp in first effect = 60 – 110 deg

Range: 4-12 ; U2 = U1 X 0.95 (If not taken as constant)

SOLUTION

No of effects – 6

Md – 11.57 kg/s

Ts – 100 deg

Xf – 35000 ppm

∆T loss – 2 deg

Tcw = 25 deg

Tf – 35 deg

λs = 2499.5698 – 2.204864Ts – 2.304 X 10 ^ (-3) Ts^2

= 2499.5698 – 2.204864 (100) – 2.304 X 10 ^ -3 Ts^2

= 2256. 043 KJ/kg

λv6 = 2499.5698 – 2.204864 (40-2) – 2.304 X 10 ^ -3 (38) ^ 2

= 2412 .45 KJ kg

B6 = XfMd/ (X6 – Xf)

= 35000 / (50000) X 11.57

= 27 kg/s

Mf = Md + B6

Mf = 11.57 + 27 = 38.57 kg/s


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∆T total = 100 – 40

∆T = 60 deg

U1 = 2.4 kW/sq m deg C

If we take U as constant then

1 / U1 = 1/U1 + 1/U2 + …. 1/U6 1/Ui =2.5 kW/m2KTemp drop in first effect is :

∆T1 = ∆Tt/ (U1 (1/U1)) = 60/(2.4 (2.5)) = 10 deg

∆T2 = 10 deg

∆T3 = 10 deg

∆T4 = 10 deg

∆T5 = 10 deg

∆T6 = 10 deg

But if we take Ui+1 =0.95i then

∆T per effect increases as the temperature is reduced. Assumptions are:

Constant heat transfer area

Lower overall heat transfer coefficient at low temperature

Constant Q

T1 = Ts - ∆T1 = 100 – 10 = 90deg

T2 = T1 - ∆T2 = 80 deg

T3 = T2 - ∆ T3 = 70 deg

T4 = 60 deg

T5 = 50 deg

T6 = 40 deg

TS T1 T2 T3 T4 T5 T6


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100 90 80 70 60 50 40

Inlet temperature of process stream T = 40 deg

Outlet temp of process stream = 35 deg

Inlet temp of water t = 25 deg

Outlet temp of water t2 = 35 deg

To calculate latent heat of steam

v1 = 2499.5698 – 2.204864 Tv1 – 2.304 X 10^ -3 Tv1 ^2

= 2499.5698 – 2.204864 (91.2-2) – 2.304 X 10^ -3 ( 91.2-2)^2

= 2284.47Kj/kg

Similarly,

λv2 = 2308.4 KJ/kg

λv3 = 2308.4 KJ/kg

λv4 = 2333.17 KJ/kg

λv5 =2385.21 KJ/kg

λv6 = 2412.46 KJ/kg

D1 = Md/ ( 1 + λv1/ λv2 + ………. + λv1/ λv6) = 1.98 kg /s

Similarly we calculate the values of D2 to D6 which are as follows:

D2 = 1.96 kg/s

D3 = 1.93kg/s

D4 = 1.91 kg/s

D5 = 1.896 kg/s

D6 = 1.875 kg/s


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Brine flow rate

B1 = Mf – D1

= 38.57 – 1.98

= 36.59 kg/s

B2 = B1 – D2

= 36.59 – 1.96

= 34.63 kg/s

B3 = B2-D3

= 34.63 – 1.93

= 32.7 kg/s

B4= B3-D4

= 30.79 kg/s

B5 = B4-D5

= 28.894 kg/s

B6 = B5-D6

= 27.019 kg/s

This value checks with the initial mass balance.


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SALT BALANCE

Therefore,

X1B1 = XfMf

Hence X1 = XfMf / B1 = (35000 x 38.57)/ 36.59 = 36893.9 ppm

X2 = (X1B1)/B2 = 38982 ppm

X3 = 41282.87 ppm

X4 = 43843.7 ppm

X5 = 46720.7 ppm

X6 = X5B5/ B6

= 63174.8 (19.249) / 17.347 =50,000 ppm

Effect 1 2 3 4 5 6

D

(kg/s)

1.98 1.96 1.93 1.91 1.89 1.87

B

(kg/s)

36.59 34.63 32.7 30.79 28.894 27.019

X

(ppm)

36894 38982 41282 43843 46721 50000


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HEAT TRANSFER AREA

A1= D1 λv1/ (u1 (Ts-T1))

= 1.98 (2284.47) / 2.4 (100- 91.24)

= 215 .14 sq m

A2= D2 λ v2/U2(∆T2-∆Tloss) = 235.6 sq m

A3 = 234.5 sq m

A4 = 234.6 sq m

A5 = 232.9 sq m

A6 = 235.6 sq m

Hence, Am = SIGMA A/n = 231.3 sq m

Now, Ms λs = D1 λv1

Ms = D1 λv1 / λs

= 1.98 x 2280.7 / (2256.04)

= 2 kg/s

Performance ratio = Md/Ms = 11.57 /2 = 5.785

Performance ratio is equal to the number of effects

Qc = D6 x λV6

= 4523.3 KJ/s

(LMTD)c = (Tf-Tcw)/ln((T6-∆Tloss- Tcw)/(T6-∆Tloss-Tf))

= 35-25 / ln ((40-2-25)/(40-2-35))

= 6.819 deg K


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CONDENSER HEAT TRANSFER

Ac = Qc / Uc (LMTD) c

= 4523.3 / 1.75 (6.819)

= 380 sq m

Now calculating specific heat transfer area

SA = SIGMA(Ai + Ac) / Md

= (1769.6) / 11.57

= 152.9 sq m

Cooling water flow rate

D6 λv6 = (Mf + Mcw) Cp (Tf – Tcw)

1.875 x 2412.5 = (38.57+ Mcw) (4.2) (35-25)

4523.43 = (38.57 + x) 42

Mcw = 62.22 kg/s

Md = 11.57 kg/s

Therefore , total water coming in = Mf +Mcw = 38.57+ 62.22 = 100.7 kg/s


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HEAT TRANSFER IN FALLING FILM EVAPORATOR

MED is involved in 20 % of the world’s desalination operations.

The horizontal falling film evaporator is the key element of MED .

The most common technique is to disperse and partly evaporate seawater on

horizontal tube handle.

The heat is supplied by steam condensing in tubes.

The design of heat exchanger area ie - tube bundle is decisive for plant efficiency

and operation costs

Overall heat balance of the tube bundle

Q= Ah∆t

1/U0 = (1/ni)*(r0 x ri) + (Rfi) * (r0/ri) + (r0/n (r0/ri)) + Rf + 1/ho

This way Uo of 2 to 4 is predicted

A=Area of evaporator

d=diameter of evaporator

G=mass velocity

U= overall heat transfer coefficient

∆V = Hvap

Q=heat flow

R= Thermal resistance

T = Temperature

HTC= heat transfer coefficient

VF = Void fraction

K = Thermal conductivity

ư = dynamic viscosity

P = mass density

Heat transfer tube inside


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Steam enters tube and condenses and releases latent heat of vapour . We should know

operating conditions , tube properties and steam properties .

Flow Pattern

Two types are:

Annular

Stratified

Heat transfer tube outside

Falling film mode three types are:

Droplet

Jet

Sheet

Nucleate boiling or surface film evaporation leads to dry and superheated spot

formation on tube which leads to scaling. The reasons are

High heat flux

High temperature difference

Fouling Resistance

It is the deposition of unwanted material on equipment surface. It affects the tube

outside. The reason are salinity and temperature.

Non Condensable Gases

NCG accumulates inside tube surface. It forms due to de-gasing of sea water and air

leakage

Heat transfer decreases due to additional resistance.

Decrease in temperature at which steam condenses at given pressure .

1 wt % NCG can decrease the heat transfer coefficient by 10 %

We use correction factor.


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Reynolds number decreases downstream

Temp drop can be equal to 1 C inside and outside the tube surface due to

pressure loss inside the tube and boiling point elevation on the outside .

DESIGN OF EVAPORATOR

Conditions

Ts=100 C

Pr=0.3 bar

Xf=35000ppm

Xn=36939.6 ppm

Ms=11.57 kg/s

SOLUTION

Energy balance for effect I:

MfHf+Mss=(Mf-Ms1)H1+Ms1Hs1

Mf=38.75 kg/s ,

Ms=2 kg/s , Ms1=D1=1.98 kg/s,

s=2257 kJ/kg

Calculated value of Hf and Hs1 is put. H1 is taken as reference.


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Effect 1 2 3 4 5 6

D

(kg/s)

1.98 1.96 1.93 1.91 1.89 1.87

B

(kg/s)

36.59 34.63 32.7 30.79 28.894 27.019

X

(ppm)

36894 38982 41282 43843 46721 50000

HEAT TRANSFER AREA

A1= D1 λv1/ (u1 (Ts-T1))

= 1.98 (2284.47) / 2.4 (100- 91.24)

= 215 .14 sq m

We take the evaporator area as 215 m2.

Tube side

Let us select 1¼ inch nominal diameter, 80 schedule, brass tubes of 12 ft in length.

Outer tube diameter (do) = 42.16 mm

Inner tube diameter (di) = 32.46 mm

Tube length (L) = 12 ft = 3.6576 m

Surface area of each tube (a) = π × do × L = π × 42.164×10-3 × 3.6576 = 0.4845 m 2

Number of tubes required providing 10% overdesign (Nt) = A /a = (215/0.4845)

443 Tube pitch (triangular), PT = 1.25 × do = 1.25 × 42.164 = 52.71 53 mm

Total area occupied by tubes = Nt (1/2) ×PT × PT × sinθ (where θ = 60°) = 443 × 0.5

×(53×10-3 )2x 0.866= 0.538 m 2

This area is generally divided by a factor which varies from 0.8 to 1 to find out the

actual area. This allows for position adjustment of peripheral tubes as those can’t be

too close to tube sheet edge.


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Actual area required = 0.538/ 0.9 (0.9 is selected) = 0.6 m2

The central downcomer area is generally taken as 40 to 70% of the total cross

sectional area of tubes. Consider 50% of the total tube cross sectional area.

Therefore, downcomer area = 0.5 × [Nt × (π/4) × do2

] = 0.5 × [443 × (π/4) ×(0.04216)2 ] = 0.309 m2

Downcomer diameter = √[(4 ×0.1661)/π] = 0.627 m

Total area of tube sheet in evaporator = downcomer area + area occupied by tubes =

0.538+ 0.6 m2 = 1.138 m 2

Tube sheet diameter = √[(4 ×0.4877)/ π] = 1.33 m


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DESIGN OF CONDENSER

As from the previous slide

• Heat removed from the vapor Q=Msƛs=D6ƛv6=1.875x2412.46=4523.33kJ/s

AMOUNT OF WATER CIRCULATED

Mcw Cp ∆T=Q .Therefore Mcw=108.2 kg/s

• (LMTD)=(40-35)-(35-25)/ln(40-35)/(35-25)

LMTD=7.2 deg C.

• Area of Condenser

Assume U=1750 w/m2K

• Therefore Ac=Qc/Uc(LMTD)c= 4523.33/1.75x7.21=337sq m

SPECIFICATIONS

• We take OD=20 mm and ID=16.8mm pipe of length 4.88m.

• Surface area of one tube=20x10^-3x3.14x4.88=0.305 sq m

• No of tubes Nt=337/0.305=1105

• Use square pitch, Pt.=1.25x20=25mm

• Tube bundle diameter Db.=20(1105/0.305)^(1/2.263)

Db. =1000mm

• No. of tube in central row Nr=1000/25=40

SHELL SIDE HEAT TRANSFER COEFFICIENT

Temp of vapour coming in =40 C

Avg temp of water=25+35/2=30 C

Wall temp=30+40/2=35 C

Film temp=40+35/2=37.5 C

PHYSICAL PROPERTIES OF WATER

• So the property of water is taken at 37.5 C

• Viscosity u=0.000692 kg/ms


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• Thermal conductivity K =628x10^-3 W/mK

• Density = 993 kg/m^3

• Specific Heat Cp=4.178 KJ/kgK

• Mass flow rate = Wc/(Nt) x L

MFR= 3.4X10^-3 Kg/m deg C

• hc=* +

• hc=518.9 kW/sq m Tube side coefficient

Tube CSA =3.14/4(16.8X10^-3)^2x1105/4

Tube CSA=0.06 sqm

Density of water at 37.5 Celsius=993 kg/m^3

Tube velocity=108.2/993x(1/0.06)=1.7 m/s=ut

Ut=1.7 m/s

hi= 7268.9 W/m^2 Fluid factors: neglected in calculation

Kw=50 W/m OVERALL HEAT TRANSFER COEFFICIENT

•

()

• 1/U=1/518+1/6000+{20X10^3ln(20/16.8)/2x50}+20/16.8x1/6000+

20/16.8x1/7268

HENCE we get U=1735 W/m^2. Close enough to estimate firm up design.


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SHELL SIDE PRESSURE DROP

Use pull through floating head no need for close clearance

Select baffle spacing=shell diameter ,45 % cut

Shell ID=1035+95=1130mm

• Cross flow area As= (25-20)/25 x1130x1130x10^-6

=0.255 sq m

Mass flow rate Gs=1.876/0.255=7.35 kg/s

Equivalent diameter, de=1.27/20(25x25-0.785x20x20)

=19.8mm

Vapour viscosity =0.00001 mNs/m2

Re=7.35x19.8x10^-3/0.00001 (Reynolds no)

Re=14553

Jf=1.5x10^-1 (from graph Jf vs. Re)

Shell side velocity Us=Gs/density of vapour

Us=7.35/8=0.9 m/s

• Take Pressure drop =50% of inlet flow rate

•

• =1.16 kPa ( within the limits)


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TUBE PRESSURE DROP

Viscosity=0.6 mNs/

Re=

Re=1.7x993x16.8x10^-3/0.6x10-3

Re=47266.6

Jf at this value=4X10^-3

[ { }]

=61 kPa

It is within the limit of 70 kPa.


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COST ESTIMATION

1. Estimation of total cost investment

The total capital investment “TCI” involves

a. The total fixed capital investment in the process area (TFCI)

b. Auxiliary Investment or Auxiliary Cost (AC)

c. The investment in working capital (WC)

Total Capital investment = TFCI + AC + WC

EQUIPMENT COST RUPEE QUANTITY TOTAL COST

EVAPORATOR 1500000 6 9000000

CONDENSOR 264000 1 264000

BOILER 1650000 1 1650000

PUMP 12000 4 48000

JET EJECTOR 30000 1 30000

TOTAL 10992000

Components Range of

FCI,%

Nominalized

percentage

Estimated

cost (Rupees)

Rounded

cost(Rupees)

Purchased

equipment

20 15.748 10992000 10992000

Equipment

installation

10 7.874 5496000 5496000


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Instrumentation

and controls

6 4.724 3297320.08 3297000

Piping 11 8.661 6045320.08 6045000

Electrical systems 6 4.724 3297320.08 3297000

Buildings(including

services)

10 7.874 5496000 5496000

Yard improvements 4 3.150 2198670.9 2198000

Service facilities 20 15.748 10992000 10992000

Land 2 1.575 1099330.9 1099000

Components Range of FCI,% Nominalized

percentage

Estimated cost

(Rupees)

Rounded

cost(Rupees)

Engineering and

Supervision

12 9.449 6595330.9 6595000

Construction

Expenses

11 8.661 6045320.1 6045000

Legal Expenses 2 1.575 1099339.6 1100000

Contractor's Fee 3 2.362 1648660 1648000

Contingency 10 7.874 5496000 5496000

TOTAL 127 100 69796000


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Total fixed capital investment = Direct Cost factor + Indirect Cost Factor=69796000

ITEM COST FACTOR

Auxiliary Building 0.05

Water Supply 0.02

Process Waste System 0.01

Electric Substation 0.015

RM Storage 0.01

Fire Protection 0.007

Roads 0.005

Sanitary & Waste Disposal 0.002

Communication 0.002

Yard lighting 0.002

Total cost factor =0.123+1(fixed capital investment)=1.123

Auxiliary cost of plant =1.123x69796000=78380908

TOTAL INSTALLED COST=TFCI+AC

Total installed cost=148176908

Working Capital (WC)

This is the capital tied up in the interest of the system in the form of ready cash to

meet

Operating expenses, inventories of raw material and product.

The working capital may conveniently be taken as 15 % of total investment made inPlant


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Therefore,

WC=0.15X148176908=22226536

TOTAL CAPITAL INVESTMENT= TOTAL INSTALLED COST+WC

= 148176908+22226536

=170403444

2. ESTIMATION OF MANUFACTURING COST (MC):

The manufacturing cost i.e. the cost of the day to day operation of the process can be

divided into three items as follows

2.1 Estimation of cost proportional to total investment

2.2 Estimation of cost proportional to production rate

2.3 Estimation of cost proportional to labor requirement.

2.1 Estimation of Cost Proportional to Investment

This includes the factors which are independent of the production rate and

proportional to the fixed investment such as:

• Maintenance e.g. Labor and material

• Property taxes

• Insurance

• Safety

• General services e.g. Laboratory, roads, etc.

For this process we shall charge 15% per year of the Total Installed Cost.

0.15X78380908=11757136

2.2 Estimation of Cost Proportional to Product Rate

• Factors proportional to product rate

• Raw material cost

• Utilities cost: power, fuel, water, steam, etc.

• Maintenance cost


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• Chemical ware house shipping

For this process we shall charge 20% of total capital investment

0.20x170403444=34080688

2.3 Estimation of Cost Proportional to Labor (L.R)

The manufacturing cost proportional to labor might take as an amount to 10% of

manufacturing cost

I.e. 10% of investment production

= 0.10(34080688+11757136)

= Rs. 4583782.4

2.4 Raw Material

Steam=487355

Chemical=221451

Hence total cost of raw material=708206

1 Cost proportional to

Investment

15% of (AC + TFC) Rs. 11757136


product rate

20% of the Total

Capital Investment+

Raw Material Cost

Rs.34788894


labor requirement

10% of Total

Manufacturing Cost

Rs.458378251

Therefore the total manufacturing cost=51129812

3. SALES PRICE FIXATION

The market price of desalinized water = ₹ 200/cubic meter

No. of days of production=330

Production per day =1000 cubic meter


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Total production=330x1000=330000

Total income=330000x200=66000000

4. DEPRECIATION

Using sinking fund method for calculating depreciation

R= (V-L)*i/(1+i)n

R = uniform annual payment made at the end of each year

V = installed cost of plant

L = salvage value of plant (consider 0)

N = life period (25 years)

I = annual interest rate (6%)

R= 1428626

5. GROSS PROFIT

Gross profit = net income from sales - annual manufacturing cost

=66000000-51129812=14870188

6. THE NET PROFIT RATE

It is defined as the expected annual returns on investment after deducting depreciation

and taxes tax period is assumed to be 40%

Net profit = gross profit – depreciation – (gross profit *tax rate)

=14870188-1428626-(14870188*0.40) = 7493486.8

7. ANNUAL RATE OF RETURNS

Annual percent return on the total investment after income taxes

= 100*(Net Profit / (TCI))

=100*(7493486.8/(170403444))

=4.397

Pay-out time = Fixed capital investment/ (Depreciation + Net profit)

=69796000/(1428626+7493486.8)=7.8 years


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PLANT LAYOUT

Properly planned plant building and reactors will help save the cost of construction

and also the operational cost. So, it is important to come up with optimized layout,

which will also include safety considerations. Following factors are needed to be

considered while during the plant layout.

1. The main reactors and its auxiliaries are sited first.

2. Storage tanks and utility generation systems should be apart from the main reactor

system for safety reasons.

3. Administrative buildings are the one where maximum outside persons are present.

These buildings hence should be in the front part of layout and the reactors should be

in the back part.

The plant layout plays and important role in the efficient functioning as well as the

safety of a plant. A good plant layout ensures easy accessibility to various services,

efficient movement of materials and a safe operation, thereby reducing constructional

as well as operating costs.

The features of good layout are:

Sufficient interplant/inter equipment space.

Ensuring high level of security and safety

Minimized material handling

Maximum proximity

Ease in supervision/coordination of activity

Built in flexibility

The plant layout is made keeping in mind safety regulations, quick and easy transfer

of materials between the equipment; operational convenience, future expansion,

economics.


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The equipment and buildings should be placed considering the direction of wind and

seismic variations. Fire stations should be constructed such that if there is fire in any

part of the pant the fire brigade should reach there without any obstacles within few

minutes. The various buildings are services required on the site in addition to the main

plant are

Administrative building

Storage for raw material and product

R & D Building

Maintenance workshops

Canteen and parking

Fire station and other emergency services

Utilities and Generation Unit

Process control room

The storage vessels and tank farms are located such that there is easy access

by road from the entry point for ease of loading and buildings. The administrative section, R & D, canteen building are segregated from the

production units and situated near the entry gate, so that access to

administration building is not through the plant area. For easy access the

transport, maintenance and firefighting, roadways are laid surroundings the

pant section storage tanks, utilities, and administration building. The storage

tanks and utilities are placed near the plant area for easy accessibility for all

the equipment in the plant. A site layout foe the plant is shown on the

following page.

SITE SELCTION

Plant location plays a critical role in the economic viability of the process. Hence

it is desirable to select a plant location with safer working condition, cheap and

skilled labour, and easy availability of raw material and probable effect of waste

generated. The choice of location of chemical plant depends on the number of

factors and their effects.


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Raw Materials

Marine processing waste, Lactic acid

Transportation and handling are the major contributors to the cost of the raw

materials.

Land:

Cost of the land should be as low as possible

Product market:

Most of the industries, which use chitin, should be situated near the site.

Environmental factors:

The product must not cause any irritation to skin or be detrimental to one’s health.

Tax rates:

The tax rate should be as low as possible

Support from state government:

This factor is also every important especially in a country like India where the

administrative delay in approval and licensing can transform a possible fortunemaking project into hapless one.

Fabrication facilities:

The fabrication facilities in the site and nearby areas are also important to speed

up the plant construction. Its significance is not much in this case.

Energy consideration

Continuous supply of power and fuel is critical parameter in running a continuous plant. To enhance productivity continuous supply of energy is essential.

Labour:

Hostile labour can affect the continuous profile of the plant.

Storage:

Timely delivery of raw materials should be ensured to minimize inventory.

Transportation:


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It is advantageous to have all kinds of transport nearby for quick transfer.

Rivers:

Water is the basic utility required in each and every chemical industry. Hence

supply of water must be continuous.

Infrastructure:

High quality infrastructure should be available with housing facility.

Drainage system should be as good as possible. Considering all the above factors

we have selected our site at Vishakhapatnam, Andhra Pradesh.


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INSTRUMENTATION AND PROCESS CONTROL

In any chemical plant, all processes are subject to disturbances that tend to change

operating conditions, compositions and physical properties of the streams. In order to

minimize the effects that could result from such disturbances, chemical plants are

equipped with adequate and proper instrumentation and control equipment. These

instruments monitor the important process variables during plant operation and ensure

smooth functioning of the plant. The control loops in the chemical process compare

the value of process variable from the system with set point and then give signal to

controlling element for the final controlling action. In critical cases and in especiallylarge plants the instrumentation is computer monitored for convenience, safety and

optimization. Before going for designing process control system for the plant we need

to first understand the process and find out the controlled, measured and manipulated

variables and then decide upon a proper controlling strategy for the process. Selection

of control loops is done considering following aspects:

Safety

Product specification

Operational constraints

Economics

Feed Back Control System Feedback control involves the detection of the controlled

variable and counteracting of charges its value relative to set point, by adjustment of a

manipulated variable. This mode of control necessities that the disturbance variable

must affect the controlled variable itself before correction can take place. Hence the

term 'feedback' can imply a correction 'back' in terms of time, a correction that should

have taken place earlier when the disturbance occurred. Thus a set point is set and

given to controller so that when the variable fluctuates (temperature or pressure) the

controller adjusts the flow rate by controlling the valve.


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Proportional Control: (a) Accelerates the response of a controlled process. (b)

Produces an offset (i.e.) non-zero steady state error for all process except those with

terms 1/s in their transfer function

.

Integral Control: (a) Eliminates any offset. (b) The elimination of offset usually

comes at the expense of higher maximum deviations. (c) Produces sluggish, long

oscillating responses.

Derivative Control: (a) Anticipates future errors and introduces appropriate action.

(b) Introduces a stabilizing effect on the closed loop response of the process.

Conventionally used controllers are Proportional (P), Proportional-Integral (PI) or

Proportional-Integral-Derivative (PID) type controls.

Temperature control: The temperature control is achieved by manipulating the flow

of water through the jacket of the reactor. Between the manipulated variable and the

measured temperature, we have two rather slow processes: (1) heat transfer between

the reacting mixture and the temperature sensor.

(2) Heat transfer from the mixture to the cooling water.

Any offset cannot be tolerated for temperature controller and a P controller is

therefore out of question. We expect, therefore, that the overall response would be

rather sluggish and a PI controller will make it even more sluggish. So, we use a PID

controller. The temperature control is achieved by manipulating the flow of cooling

water. This is done by using a PID controller as well.

Temperature control in reactor

Temperature Indicator (TI): Used to indicate the temperature inside the reactor when

reaction is going on.

Temperature Transmitter (TT): Transmitter reads the temperature of the reacting

mixture and transmits it to temperature recorder and controller.

Temperature recorder and controller (TRC): This records the temperature input from

the transmitter and send the signals to the control valve accordingly That is if


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temperature is above desired limit (4 C) then flow rate of cooling water is increased

and vice versa. A PID controller is used .

FULL PROCESS CONTROL

FEEDBACK CONTROL

A typical feedback control system (Figure 8.23m) consists of measuring the productconcentration with a density sensor and controlling the amount of steam to the first

effect by a three mode controller. The internal material balance is maintained by level

control on each effect.


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CASCADE CONTROL

A typical cascade control system is illustrated in Figure 8.23n. This control system,

like the feedback loop in Figure 8.23m, measures the product density and adjusts the

heat input. The adjustment in this instance, however, is through a flow loop that is

being set in cascade from the final density controller, an arrangement that is

particularly effective when steam flow variations (outside of the evaporator) are

frequent. It should be noted that with this arrangement the valve positioner is not

required and can actually degrade the performance of the flow control loop


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PROCESS CONTROL OF EVAPORATOR:


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ENVIRONMENTAL ASPECTS AND SAFETY

The outfall of the discharge flow inevitably poses the risk of changing the quality of

the seawater in the near vicinity of the discharge point. However, there is limited

information regarding the effect on aquatic life as a result of discharge from

desalination plants. With this potential impact but also limited information a design

incorporating a maximization of dispersion and dilution around the outfall was

considered necessary for the desalination plant. The effect considered most likely is a

change in the marine assemblages within the mixing zone. Aquatic ecology more

suited to highly saline environments will tend to migrate to this area whereas the more

sensitive ecology will tend to avoid the area.

With the majority of desalination plants extracting water directly through open water

intakes in the ocean, there is a direct impact on marine life. Fish and other marine

organisms are killed on the intake screens (impingement); organisms small enough to

pass through, such as plankton, fish eggs, and larvae, are killed during processing of

the salt water (entrainment). The impacts on the marine environment, even for a

single desalination plant, may be subject to daily, seasonal, annual, and even decadal

variation, and are likely to be species- and site-specific.

Another major environmental challenge of desalination is the disposal of the highly

concentrated salt brine that contains other chemicals used throughout the

process. Because all large coastal seawater desalination plants discharge brine into

oceans and estuaries, steps must be taken to ensure its safe disposal; at this stage, we

know very little about the long-term impacts of brine disposal on the marine

environment. Twice as saline as the ocean, the brine is denser than the waters into

which it is discharged and tends to sink and slowly spread along the ocean floor,

where there is typically little wave energy to mix it. There are several proven methods

to disperse concentrated brine, such as multi-port diffusers placed on the discharge

pipe to promote mixing. Brine can also be diluted with effluent from a wastewater

treatment plant or with cooling water from a power plant or other industrial user,

although these approaches have their own drawbacks that must be addressed.


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A major issue to consider is land use in the proximity of a proposed desalination

plant site . If planners place a desalination plant in densely populated areas, it may

impact the residential environment. Some desalination plants generate noise and gas

emissions. For example, reverse osmosis plants generate noise because of the use of

high-pressure pumps. If located near population centers or other public facilities,

plans should include steps to mitigate the noise pollution such as using canopies or

acoustical planning .

Desalination plants can have an indirect impact on the environment because many

plants receive energy from the local grid instead of producing their own.

The burning of fossil fuels and increased energy consumption allows more air

pollution and gas emissions to occur. Gaseous emissions from desalination stacks

include carbon monoxide (CO), nitric oxide (NO), nitrogen dioxide (NO2), and

sulphur dioxide (SO2). These air pollutants can have a harmful impact on public

health (Al-Mutaz 1991). There is also concern regarding the large amounts of

chemicals stored at the plants. Chemical spill risks require storing chemicals away

from residential areas.

Project Health and Safety

As with any plant construction and operation, a multitude of health and safety aspects

need to be considered. These are in place not only to secure the health and safety of

workers on the site but to ensure that the design comes in on budget or below.

Task specific hazards:

Working around heavy plant (piling rigs)

Working with cranes

Deep excavations

Falling objects (sheet piling, pile cages)

Working with concrete (injuries caused by burns)

Exposed steel reinforcement


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Vibration and noise

Slips, trips and falls due to uneven surfaces

Hazard mitigation measures:

All piling rigs and cranes to be accompanied by a slinger/signaler at all times. All

operatives to wear personal protective equipment (PPE) to improve visibility to

vehicle/plant operators. All pedestrians to be directed by signalers to avoid moving

plant.

All deep excavations to be screened off with high visibility fencing to prevent falls.

No operatives to work directly under crane lifting zones. Slinger/signaler to ensure

lifting area is clear of all operatives before any lifts take place.

All concrete placement operatives to wear overalls covering all skin as well as

wearing

waterproof gloves and wellington boots to prevent concrete coming into contact with

exposed skin and causing burns.

All exposed ends of steel reinforcement to be covered with yellow or green

mushroom

caps.

All operatives working near plant producing noise above 85dB should wear ear

protection.

Operatives using vibrating tools should restrict time using the equipment and wear

anti

vibration gloves.

Pedestrians must adhere to marked safe routes to prevent walking across unsafe

surfaces


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Mass Excavations

Task Specific Hazards

Working around heavy plant (excavators)

Deep excavations

Contact with buried services

Noise

Unstable slopes

Working at height (unguarded edges)

Slips, trips and falls due to uneven surfaces

Hazard mitigation measures

All excavators to be accompanied by a slinger/signaler at all times. All operatives to

wear PPE to improve visibility to vehicle/plant operators. All pedestrians to be

directed by signalers to avoid moving plant.

All deep excavations to be screened off with high visibility fencing to prevent falls.

Permits to dig must be issued before any excavation is to take place. A map detailing

known

services and their depths must be included. Excavation to cease if any services that

are not on the map are encountered, with a full cat scan of the area to follow to locate

the extent of the services.

All operatives working near plant producing noise above 85dB should wear ear

protection.

Slopes to be cut back to a shallow batter to prevent unforeseen collapse. For sand,

30%

would be an acceptable batter for the mass excavation.


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No operatives to work near unguarded edges of excavations. All unguarded edges

must be fenced off as soon as possible.

Pedestrians must adhere to marked safe routes to prevent walking across unsafe

surfaces.

Concrete Slabs & Pile Caps

Task Specific Hazards

Contact with concrete

Manual Handling

Use of cutting abrasive wheels to cut steel

Vehicle Movements (concrete wagons)

Working with cranes (to move steel reinforcement)

Use of compressed air tools

Hazard mitigation measures

All concrete placement operatives to wear overalls covering all skin as well as

wearing

waterproof gloves and wellington boots to prevent concrete coming into contact with

exposed skin and causing burns.

All operatives to be briefed on correct manual handling techniques. Equipment to be

provided it objects heavier than 30kg need to be carried by one person.

Hot works permits to be issued to any operatives using abrasive wheels to cut steel.

Fire

extinguishers to be provided in work area, with a safe cutting area cordoned off. Area

to

be inspected 30 minutes after work has ceased to ensure no embers can cause a fire.

Signalers to control traffic flow and escort every vehicle around work areas. No

vehicle

Is to reverse without a signaler present.


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No operatives to work directly under crane lifting zones. Slinger/signaler to ensure

lifting area is clear of all operatives before any lifts take place.

Operatives using compressed air tools must wear goggles at all times and must also

wear

ear protection due to the high level of noise. Care must be taken to ensure that air

nozzles

do not come into contact with the skin whilst operating.

STEPS TO BE FOLLOWED IN CASE OF EMERGENCIES

In the event that a plant must be shut down immediately due to an emergency

situation, the following steps must be followed

o Stop pumps if running.

o Close drains if open.

o Stop agitators if running.

o Close the main process valve and the cooling water, steam line valves.

o Leave the plant building checking that nobody is left inside.

In every establishment, wherein fifty or more workers are ordinarily employed, the

contractor appoints safety officers with qualifications and experience. (Regulation of

employment and conditions of service Act, 1996)

Safety Training:

The contractor provides safety training to all the workers as well as those appointed

by his sub-contractors, at least quarterly, through a faculty which possesses the

minimum qualification of safety officer.

Safety Inspections:


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The contractor will schedule regular inspection of various job sites and activities by

developing a checklist appropriate to the task and hazards involved therein and

implement the findings of the inspection.

Emergency Action Plan:

The contractor will ensure that an EAP is prepared to deal with emergencies arising

out of:

a. Fire and explosion

B. Collapse of buildings sheds etc.

c. Gas leakage or spillage of dangerous goods or chemicals

d. Landslides getting workers buried; floods, storms and other natural calamities.

Safety in work site:

i. Housekeeping: The contractor will be responsible for maintaining good

housekeeping and safety standards in the workplace.

a. Loose materials that are not required for use will not be placed or left behind so

dangerously as to obstruct workplaces or passage ways.

b. All projecting nails shall be removed or bent to prevent injury.

c. Workplaces and passageways that become slippery owing to spillage of oil or other

causes shall be cleaned up or strewn with sand, ash or the like.

ii. Lighting and ventilation:

a. All practical measures shall be taken to prevent smoke, fumes etc. from obscuring

any workplace or equipment at which any worker is engaged. b. Adequate and

suitable artificial lighting is to be provided where natural light is not sufficient. The


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artificial lighting so provided shall not cause any danger, including that of glare or

disturbing shadows

iii. Dangerous and harmful environment:

a. No worker shall be allowed to enter any such space unless a responsible person has

certified it to be safe and fit for the entry.

B. Gas test shall be carried out ensures that the confined space is completely free from

combustible gases and vapors.

c. Workers shall take necessary precautions to prevent unburned gases from escaping

inside a tank or vessel or other confined space.

iv. Excessive noise:

Noise level in no case shall exceed as in exposure in excess of 115 dBA over the

period of a quarter of an hour cannot be permitted.

a. Use of earplugs/muffs and anti-vibration gloves shall be ensured to protect the

workers from the impact of exposure to such dangers.

v. Corrosive substances:

a. While protection of the body could be ensured by the use of corrosion resistant

apparel/overalls, suitable goggles, gloves, apron, gum boots etc. shall be made

available to all concerned personnel.

b. To deal with an accidental spillage of a corrosive substance on a body of the

worker, the facility of eyewash fountain or water shower, as the case may be, shall be

installed, within the easy reach of the workplace.

vi. Eye protection:

a. Suitable personnel protective equipment for the protection of eyes shall be provided

and used by the building worker engaging in operations like welding, cutting,

chipping, grinding or similar operations which may cause hazard to his eyes.

vii. Head protection and other protection apparel:


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a. Workers are provided with safety helmets of the type approved and tested in

accordance with the national standards

b. Work in rain or in similar wet condition, shall be provided with waterproof coat

with hat

OTHER SAFETY INSTRUCTIONS

Hazard identification

Potential Acute Health Effects: Very hazardous in case of ingestion. Hazardous in

case of skin contact (irritant), of eye contact (irritant), of inhalation. Noncorrosive for

skin. Non-sensitizer for skin. Non-permeate or by skin. Potential Chronic Health

Effects: Very hazardous in case of ingestion. Hazardous in case of skin contact

(irritant), of eye contact (irritant), of inhalation. Noncorrosive for skin. Non-sensitizer

for skin. Non-permeate or by skin. The substance is toxic to lungs, the nervous

system. Repeated or prolonged exposure the substance can produce target organs

damage.

First Aid Measures

Eye Contact: Check for and remove any contact lenses. Immediately flush eyes with

running water for at least 15 minutes, keeping eyelids open. Cold water may be used.

Do not use an eye ointment. Seek medical attention. Skin Contact: After contact with

skin, wash immediately with plenty of water. Gently and thoroughly wash the

contaminated skin with running water and non-abrasive soap. Be particularly carefulto clean folds, crevices, creases and groin. Cold water may be used. Cover the irritated

skin with an emollient. If irritation persists, seek medical attention. Wash

contaminated clothing before reusing. Serious Skin Contact: Wash with a disinfectant

soap and cover the contaminated skin with an anti-bacterial cream. Seek immediate

medical attention. Inhalation: Allow the victim to rest in a well-ventilated area. Seek

immediate medical attention.


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CONCLUSION

The MED process of desalination described above has the following advantages:

• Efficient water distribution and tube wetting

• High heat-transfer rates

• Absence of dry patches

• Low scale formation and tube damage

• Efficient disengagement of vapors and non-condensable gases

• Proper venting of the non-condensable gases, and

• Simple monitoring of scaling and fouling.

The following results were obtained:

Flow rate of incoming water = 100.7 kg/s

Area of condenser = 337 m2

Total capital investment = Rs 17,04,03,444

Total pay-out time = 7.8 years


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REFERENCES

• Amjad Z. (2008). Scale inhibition in desalination applications: an overview,

Corrosion 230. [Book on scale prevention with anti-scalants]

• Awerbuch L. (1997). Dual purpose power desalination/hybrid systems/energy

& economics. IDA

• Desalination Seminar , Cairo, Egypt, September. [Energy and economy of dual

purpose thermal desalination].

• Awerbuch L. (2004). Hybridization and dual purpose plant cost

considerations. Proceedings of the International Conference on Desalination

Costing, 204-221, Limassol, Cyprus. [Costing of thermal desalination

processes

• Sourcebook of alternative technologies for freshwater augmentation in small

island developing states , United Nations Environment Programme

• www.pua.edu.eg/pua/site/uploads/file/engineering

• Fundamentals of desalination by H.T El Dessouky and Ettouney ,Pg 80-142,

1st

edition 2002• Modeling and thermodynamic analysis of a multi effect desalination plant for

sea water desalination by Artin Hatzikioseyian, Roza Vidali, Pavlina Kousi

National Technical University of Athens (NTUA), School of Mining and

Metallurgical Engineering , March 2002

• Design and simulation of a multi effect evaporator using vapour bleeding,

project done by Monalisha Nayak in final year B Tech National Institute of

Technology, Durgapur , March 2007 .

• nptel .ac.in/courses/103103027/pdf/mod7.pdf

• Desalination technologies , VA Tech WABAG , http://www.wabag.com/wp-

content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdf

• http://personal.ee.surrey.ac.uk/Personal/R.Webb/MDDP/2010/Desalination_3.

pdf

• http://pacinst.org/publication/desal-marine-impacts/

• http://www.scientificamerican.com/article/the-impacts-of-relying-on-

desalination/

http://www.pua.edu.eg/pua/site/uploads/file/engineeringhttp://www.pua.edu.eg/pua/site/uploads/file/engineeringhttp://www.wabag.com/wp-content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdfhttp://www.wabag.com/wp-content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdfhttp://www.wabag.com/wp-content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdfhttp://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.scientificamerican.com/article/the-impacts-of-relying-on-desalination/http://www.wabag.com/wp-content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdfhttp://www.wabag.com/wp-content/uploads/2012/04/WABAG_Desalination_englisch-20081.pdfhttp://www.pua.edu.eg/pua/site/uploads/file/engineering


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• Environmental Issues of Desalination ,Tamim Younos ,Virginia Polytectnic

Institute and State University (UNIVERSITIES COUNCIL ON WATER

RESOURCES JOURNAL OF CONTEMPORARY WATER RESEARCH &

EDUCATION ISSUE 132, PAGES11-18, DECEMBER 2005)

• Max S. Peters, Kaus D. Timmerhaus, Ronald E. West. (2004) Plant Design

and Economics for Chemical Engineers, 5th Ed., Mc Graw Hill

• Max Kurtz. (1920) Engineering Economics for Professional Engineers’

Examinations, 3rd Ed., Mc Graw Hill

• Chemical Process Control An Introduction to Theory and Practice George

Stephanopoulos

• http://www.alibaba.com/

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Multi effect distillation desalination

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