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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
2 Cost proportional to
product rate
20% of the Total
Capital Investment+
Raw Material Cost
Rs.34788894
3 Cost proportional to
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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