Integrated FO-RO waste water treatment system for the production of potable water

1. 1

2. Design of Integrated Forward Osmosis- Reverse Osmosis (FO-RO)
Wastewater Treatment System for the Production of Potable Water
GHULAM ISHAQ KHAN INSTITUTE OF ENGINEERING SCIENCES & TECHNOLOGY
Advisor Name: Dr. Muhammad Usman Farooq
Co-Advisor Name: Dr. Javaid Rabbani Khan
DEPARTMENT OF CHEMICAL ENGINEERING
Muhammad Ammar Saeed, 2017205
Muhammad Rehan Anjum, 2017305
Sohnal Rameez, 2017441
Zain Ul Hassan, 2017508

3. LITERATURE REVIEW
• INITIAL WORK
 1748 by Jean-Antoine Nollet ,osmotic process by using a pig’s
bladder as a membrane
 Functional synthetic RO membrane in 1964
 First RO plant Coalinga, California in 1965 by Sydney Loeb
• ENVIRONMENTAL IMPACT
 Disposal of brine to the water sources poses a potentially
serious threat to marine ecosystems
 Environmental damage to fragile ecosystems like
corals(Siegmann & Shezad, 2006)
• INTEGRATED TECHNIQUES
 Nano filtration-Multi stage flash (NF-MSF) thermal hybrid.
 Forward osmosis-Multi stage flash (MSF) hybrid system.
 Membrane Bio reactor-reverse osmosis-advanced oxidation
process
 Forward Osmosis-Reverse Osmosis (FO-RO)

4. IMPORTANCE OF WORK/MOTIVATION
2.4-3.2 Billion
population around the
world in facing water
scarcity as expected in
2025
Deteriorating Quality of
Water in Pakistan
To reduce fouling
phenomena
Explore the
performance of a
hybrid FO-RO process
and unexplored
commercial avenues
Substantial change in
discharged
concentrated seawater
volume can be
4

5. GAP ANALYSIS AND DEFINED OBJECTIVES
Current State Desired Future
State
Gaps Remedies
Pumping Energy High consumption Low consumption Comparatively high
power required for
motor
Once through
circulation system
Foundation and extensive
civil work
Complex Simplistic Two stage process
multiple intakes
one common feed
water header and
pretreatment unit
Common seawater
intake(Rastogi, 2016)
2016)
Feed Water Temperature 50-60°C 28-35°C Increased RO Flux Utilizing the heat
rejection of MSF
Permeation Flux and Life of
Membrane
1-2 years 3-5 years Absence of Physical
Modelling
Single stage RO
process
Assessment of
membrane material
Calculation of
membrane area,
module and Vessels.
Selection of pump
Calculation of pump
energy consumption.
Identification of the
components causing
fouling in the feed
water
5

6. 6
COMMERLISATION POSSIBLITIES AND SOCIAL IMPACT
Implementation
• Technologically and economically viable
• Water reclamation is accomplished
Further
improvements
• Pretreatment can enhance the process
• Improving feed channel hydraulics (Loeb, 1976) in FO-membrane
Economic
impact
• Less operational cost then conventional method
• Total water cost per cubic meter produced water has lower cost
reduction(Ren & McCutcheon, 2014)

7. PROCESS DESCRIPTION
PROCESS FLOW DIAGRAM
MATERIAL BALANCE
ENERGY BALANCE
PROCESS DESIGN
COST ESTIMATION
7 MAIN POINTS TO BE FOCUSED

8. PROCESS DESCRIPTION
Forward
Osmosis
• Forward osmosis (FO) uses a saline stream to extract water from a source of impaired water.
• FO uses an osmotic pressure differential as the driving force drawing water through a
semipermeable membrane
Seawater
Dilution
• Seawater is diluted by an impaired water stream through an FO membrane in the first osmotic
dilution stage
• The diluted seawater is processed through an RO desalination system which rejects salts and
dissolved contaminants.
Second Stage
Osmosis
• A second-stage osmotic dilution process can be implemented to dilute seawater before discharge and to
further concentrate.
• The saline water is diluted during the first-stage osmotic dilution process the energy required is reduced
• Thus the energy demand of the desalination plant is decreased, and two significant barriers are in place to
reject contaminants in the impaired stream.
8

9. PROCESS FLOW DIAGRAM
9

10. MATERIAL BALANCE
10

11. MATERIAL BALANCE
 Material balances are the basis of process design.
 A material balance taken over the complete process will determine the quantities of
raw materials required and products produced.
 Balances over individual process units set the process stream flows and compositions.
11

12. ASSUMPTION
S
Basis: 100
m3/d of
Permeate
Mostly
theoretical
Based
Fluxes and
recoveries
were
assumed
after proper
literature
review.
Generation
and
consumption
=0
No loses in
Process
Continuous
flow
Ideal
condition
supposed
12

13. PROPERTIES
SEWAGE WASTERWATER CHARACTERISTICS SEAWATER CHARACTERISTICS
PARAMETERS CONCENTRATIONS(mg/l) PARAMETERS CONCENTRATIONS(mg/l)
PH 7 TDS 35000
TSS 220 PH 7.5-8.5
TDS 500 Sodium 11200
COD 250 Calcium 400
BOD 500 Magnesium 1400
Ammonia 2.5 Chloride 19750
Nitrate 10.1 Sulphate 2650
Bicarbonate 140
Potassium 370
13

14. MATERIAL BALANCE ON RO:
 𝑚5
0
= 𝑚6
0
+ 𝑚7
0
 𝑚7
0
= 102300
𝑘𝑔
𝑑
𝑜𝑟
100𝑚3
𝑑
 Recovery =
𝑝𝑒𝑟𝑚𝑒𝑎𝑡𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑚7
0
𝑓𝑒𝑒𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑚5
0 .
 =
102300
𝑓𝑒𝑒𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
= 0.50
Feed flow rate (𝑚5) = 204600 kg/d
Now using main equation we get 𝑚0= 102300 kg/d
Here m_3^0= m_5^0= 204600 kg/d (because mass flow rate is same before and after pump)
14

15. MATERIAL BALANCE ON FO
 For Municipal wastewater feed:

Recovery =
𝑝𝑒𝑟𝑚𝑒𝑎𝑡𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
𝐹𝑒𝑒𝑑 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
 =
𝑝𝑒𝑟𝑚𝑒𝑎𝑡𝑒 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒
168750
= 0.64
 Municipal wastewater Permeate flow rate= 108000 kg/d
 Now 204600-108000= 96600 kg/d= 𝑚𝑜1= Seawater feed
 𝑚𝑜2 = 𝑝𝑒𝑟𝑚𝑒𝑎𝑡𝑒 + 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡e
168750=108000+concentrate
concentrate of municipal wastewater= 60750 kg/d.
15

16. REJECTION:
Rejection = 99% in both membranes. only water is permissible.
RECOVERIES:
FO = 64%
RO = 50%
16

17. ENERGY BALANCE
17

18. ENERGY BALANCE
 Energy balance on any equipment plays an important role in any industry to get an
idea about the complete energy utilization in plant.
 In case if design is to be made more energy efficient, we can do changes in energy
balances to make plant more energy efficient.
 Energy balance
Main energy equation is as follows
Z1
𝑔
𝑔𝑐
− 1
2
𝑣𝑑𝑝 +
𝑣12
2𝑎𝑔𝑐
+ 𝑊 = 𝑧2
𝑔
𝑔𝑐
+
𝑣22
2𝑎𝑔𝑐
+ 𝐹
W= 1
2
𝑣𝑑𝑝
Power = W = 𝑣(𝑝2 − 𝑝1)
18

19. RO PUMP ENERGY CONSUMPTION
 Operational pressure 19 bar (278.73 psi)
 In Flow rate 200 m3/d(36.690 GPM)pump
efficiency 85%
 Hp (Horse power) = ∆P(psi)*Q(GPM)/1714(conversion factor)*η(efficiency)
 Hp = 7.01=5.25 KWh [This much energy is required for pumping 8.3 m3 /hr. i-e (200
m3 /d)]
19

20. FO PUMP ENERGY CONSUMPTION
 Sea water Flow rate = 128m3/d(23.48 GPM)
 Wastewater Flow rate=112.5m3/d(20.63 GPM)
 Efficiency of pump 85%
 Pressure 2 bar (29.34 psi)
 Hp(Horsepower)=∆P(psi)*Q(GPM)/1714(conversionfactor)*η(efficiency)
Hp(for seawater Pump)=0.472=0.354 KWh
Hp(for wastewater pump)=0.414=0.311 KWh.
20

21. ENERGY BALANCE ON MEMBRANE
Q0-W0= ΔH0+ ΔK.E0+ ΔP.E0
ΔK.E0 = 0;
ΔP.E0= 0;
W0 = 0 ;
Q0 = 0
Therefore,
ΔH0 = 0.
21

22. PROCESS DESIGN
22

23. STEPS IN
DESIGNING
MEMBRANE
PROCESS
MEMBRANE
AREA
PUMP
SE;ECTION
PIPE
SELESCTION
NUMBER
OF
ELEMENTS
NUMBER
OF VESSELS
ELEMENTS/
VESSELS
23

24. MEMBRANE AREA
 Membrane area has great importance in membrane processes because the desired product that
needs to be achieved majorly depends on membrane area.
 If large products are required than larger area will be required.
RO SECTION
We have a flux=16.85 l/m2hr
[Standard conditions PH-8, T= 25 0C, 32 g/lNacl]
Permeate flow rate= 4166.6 l/ hr
So the Area of membrane = Permeate flow rate/Flux
= 4166.6 l /hr ÷ 16.85l/m2.hr
SO,
Membrane Area = 247.3 m2
24

25. FO SECTION
 We have a Flux = 16.85 l/m2.hr [Standard condition 1M NaCl, T= 20 0C]
 Permeate Flow rate = 8333.3 l/hr
Hence the membrane area = Permeate flow rate/Flux
=8333.33 l /hr ÷ 16.85l/m2.hr
SO,
Membrane Area =495 m2
25

26. PUMP SELECTION
 In order to flow the liquid, we need pumps so it’s very important to design the pump.
 Here we will simply choose the pump according to our required pressure and flow rate.
 For larger feeds larger pumps will be required which can provide both high flow rate and required pressure to get the
product.
 Operational pressure 19 bar (278.73 psi)
 In Flow rate 200 m3/ d (36.690 GPM) Pump efficiency is 85%
 Hp (Horse power) = ∆P (psi)*Q (GPM)/1714(conversion factor)*η(efficiency)
 Hp = 7.01=5.25 KWh [This much energy is required for pumping 8.3 m3 /hr. i-e (200 m3 /d)]
 We will use multistage Centrifugal pump [from Pump selection Chart using Pressure and Flow]
26

27. PIPE SELECTION
 After designing pumps another important thing to design is pipe for providing
channel to the liquid.
 Without pipes process designing is incomplete. So we need to know about the pipe
dimensions and its material so that it can bear the pressure exerted by pump.
 Material = PVC
 Operating Pressure = 19 bar = 275 psi
 Schedule No = 40
 Nominal Diameter = 1/2” [for 200 m3/d]
 [From ASTM D2466-06 Standard Specification for PVC]
27

28. NUMBER OF ELEMENTS
 For membrane process we can’t install a single large module to hold the membrane because
in this case we need a very large pump to provide required transmembrane pressure.
 Ultimately it would not be economically viable, so we need to calculate number of
elements/modules for an efficient process.
 FOR RO SECTION
Number of elements = Membrane area/ Membrane area per module
= 247.3/2.5 [DOW Company]
= 99
 FOR FO SECTION
Per module area = 2.5 m2
Number of elements = Membrane area/ Membrane area per module
= 495/2.5
= 198
For commercial plant standard for elements per vessel = 7
28

29. NUMBER OF VESSELS
Following the designing of modules, we also need to know about the vessels in which
membrane are housed. Without proper housing process cannot be run properly.
 FOR RO SECTION
Number of vessel = total elements/ no of element in a vessel
= 99/7
14 vessels for RO
 FOR FO SECTION
Number of vessel = total elements/ no of element in a vessel
= 198/7
28 vessels for FO
29

30. COST ESTIMATION
30

31. COST ESTIMATION
• MEMBRANE COST:
𝐶𝑚 = 𝐶𝑚𝑒𝑚 ∗ 𝐴𝑚⁄𝐴𝑚𝑜𝑑
𝐶𝑚 = $259.99 ∗ 247.3⁄2.5
𝐶𝑚 = $25718.21
• HOUSING COST
For Fiber glass housing = $430
No of Housings= 14
𝐶ℎ = 14 ∗ $430 = $6020
• PUMP COST:
Ce= CB * (
𝑄
𝑄𝑏
)𝑀
𝐶𝑒 = 1.97 ∗ 103 ∗ (
5.25
1
)35
Ce = $3519.83
𝐶1
𝐶2
=
𝑖𝑛𝑑𝑒𝑥1
𝑖𝑛𝑑𝑒𝑥 2
=
558.3
435.8
= 1.28
31

32. 𝐶𝑒 = 1.28 ∗ $3519.83 = $4509.22
Where Ce= Equipment Cost with capacity Q
CB= Known base cost for equipment with capacity QB
M= Constant depending upon equipment
Capital Cost = Membrane Cost + Housing Cost + Pump Cost
Capital Cost = $25718.21 + $6020+$4509.22
Capital Cost= $36247.43
Operational Cost = Labor Cost + Maintenance Cost + D energy
Energy Index 2016-17 = 117.65KWh/m3
= 117.65*73000m3/year
=8588450 KWh/year
Unit Energy Cost = $0.09/KWh
Denergy= 0.09*8588450
Denergy=$772960.5/year
32

33.  Maintenance Cost= 0.02* Capital Cost
Maintenance Cost= 0.02*$36247.43
Maintenance Cost= $724.94
 Labor Cost = 0.048*Capital Cost
Labor Cost= 0.048*$36247.43
Labor Cost = $1739.87 So,
Operational Cost= $775425.316
 Installation Cost = 0.4*Capital Cost
Installation Cost = 0.4*$36247.43
Installation Cost = $14498.97
33

34. Installation Cost = 0.4* CapitalCost
Installation Cost = 0.4*$36247.43
Installation Cost = $14498.97
Pretreatment Cost=1417*𝑸𝟎.8=1417*(94.4 +112.5)0.8
Pretreatment Cost =$100920
Total Cost=Installation Cost +Capital Cost+Operational Cost+ Pretreatment Cost
Total Cost = $14498.97 + $36247.43 + $ 775425.316+$100920
Total Cost =$927091.716
Cost of potable water/m3 = $5
Yearly production= 320*100 m3/year =32000m3/year
Yearly revenue= 5*32000 = $160000
Pay back period=Total cost/yearly revenue
Payback period= $826171.7/160000=5.7 year
34

35. INSTRUMENTATION
AND
PROCESS CONTROL

36.  Instrumentation is the basic process control in the industry and control number of
variables like temperature, flow, level, pressure and distance.
 These variables can be interdependent variables in a single process, requiring a complex
microsystem for total control.
 The main objective of specifying and using of instruments and control are:
 To save the plant operation by keeping the variables within the
limits and detects the dangerous situation when developed.
 To control the product rate and quality within the specified quality.
 To process at the lowest production cost.
36

37. SENSORS AND TRANSMITTORS
 Sensor is an object whose purpose is to detect events or changes in its environment, and then
provide a corresponding output.
 A sensor is a type of transducer; sensors may provide various type of output, but typically use
electrical or optical signals.
 They are interfaces between the process and its control systems, and its function is to convert the
sensor signal into control signals.
 Set points: It is a desired process output that an automatic control system will aim to reach.
37

38. PROCESS VARIABLES
 The process variable is a dynamic feature of the process which may change rapidly.
 There are four commonly measured variables which affect chemical and physical process like:
PRESSURE
CONTROL
TEMPERATURE
CONTROL
LEVEL
CONTROL
FLOW
CONTROL
38

39. PRESSURE AND TEMPERATURE CONTROL:
PRESSURE
Pressure control is a process in which change of pressure of an equipment is measured
and the pressure into or out of the system is adjusted to achieve a desired average
pressure.
TEMPERATURE
CONTROL:
Temperature control is a process in which change of temperature of a space and objects
is measured or otherwise detected, and the passage of heat energy into or out of the
space to achieve a desired average temperature.
WORKING:
• If the desired temperature inside a boiler increases or decreases as compared to its limit,
then temperature controller gives signal to the system and then action is taken.
39

40. LEVEL AND FLOW CONTROL:
LEVEL
CONTROL:
Fluids and fluidized solids flow to become essentially level in their containers because of gravity whereas most bulk solids
pile at an angle of response to peak the substance to be measured.
FLOW
CONTROL:
Level control system is installed for the control of liquid level inside the boiler up to (30-40) %. When the level exceeds
limit a level controller gives a signal to the system.
A flow controller is designed and calibrated to control a specific type of fluid or gas at a particular range of flow rates.
WORKING:
The flow controller can be given a set point from 0 to 100 of its full-scale range but is typically operated in range of (10-
of full scale where the best accuracy is achieved.
A digital flow controller is used and is able to control more than one type of fluid or gas whereas an analog controller is
limited to the fluid for which it was calibrated.
40

41. FINAL CONTROL ELEMENT AND CONTROLLER
 In a majority of systems, the final control system is an automatic control valve which throttles the
flow of manipulated variables.
 There are three basic type of controllers which are:
.
Proportional action which moves the control valve indirectly proportional
proportional to the magnitude of the error
Integral action which moves the control valve based on the time
integral of the error and the purpose of integral action is to drive the
process back to its set point when it has been disturbed
Ideal deviation action and its purpose is to anticipate where the process is
heading by looking at the time and rate of change of error.
41

42. FEEDBACK CONTROL SYSTEM
Controller
Final control element
Flow meter
Set point
+
-
Membrane Process
42
Permeate flow rate

43. FEEDBACK CONTROL SYSTEM
 The purpose of instrumentation is to maintain the power of the pump at its desired value in spite
of disturbances.
 The controller will compare the measurement signal of the controlled variable to the set point (the
desired value of the controlled variable) and the difference is called the error.
 Depending upon the magnitude and sign of the error, the controller takes the appropriate action
by sending a signal to the final control element which provides an input to the process to return
the power to the set point.
 Here our system is a feedback control system because information about the deviation of the
system is fed back to the controller which utilizes this information to change the system in some
way.
43

44.  It’s a closed loop control system because the controller automatically acts to retain the controlled variable at
its desired variable.
 It’s a negative type signal feedback because the error signal is computed from the difference between the set
point and the measured signal.
 The type of controller used is the PID controller. It will eliminate the error completely as well as reduce or
vanish the oscillatory action.
 The flow can be measured using any power measuring device depending upon the process economics and
safety issues.
FEEDBACK CONTROL SYSTEM(CONTINUED)
44

45. HAZOP STUDY

46. INTRODUCTION AND BASIC METHOD OF
HAZOP:
A HAZOP is a qualitative technique based on guide words and is
carried out by a multi-disciplinary team.
A hazard and operability (HAZOP) team is used to identify and evaluate
problems that may represent risks to personnel or equipment or prevent
efficient
Basic method of HAZOP includes a piping and instrumentation diagram,
which is examined is small sections such as individual items of equipment
or pipes between them.
46

47. HAZOP PROCESS
The HAZOP process is
a systematic examination.
The team approach to
a HAZOP makes it a
multidisciplinary study.
The HAZOP team
utilizes operational
experience.
The process covers
safety as well as
operational aspects.
Solutions to the
problems identified
may be indicated.
HAZOP consider
operational
procedures.
HAZOP cover
human errors.
47

48. HAZOP PROCESS(continued):
The HAZOP
studies led by
independent person.
HAZOP studies
results are recorded.
A HAZOP does not
require considerable
technical expertise
for technique
formulation.
As a systematic
process it provides
rigor for focusing on
system elements and
hazards.
The HAZOP process
is a team effort with
many viewpoints.
48

49. 49
HAZOP STUDY OF A PLANT

50. CONCLUSION AND ADVANTAGES
HAZOP studies focuses on single events rather than combinations of possible events.
The HAZOP studies focus on guide-words allows it to overlook some hazards not related to a guide-
word.
HAZOP technique is very effective for identifying plant operability problems, threats to the environment,
product quality, plant throughput and for highlighting critical maintenance requirements
HAZOP studies are typically very time consuming and thus expensive.
Training is essential for optimum results, especially for the facilitator
50

51. EXECUTION PLAN AND EXPECTED OUTCOMES
Execution Plan Expected Outcomes
Overall cost reduction of
16% could be achieved
85% Removal of BODs and
TSS
Selection of High recovery
ratio
51
1-Oct 20-Nov 9-Jan 28-Feb 19-Apr
literature Survey
Review of different projects
Selection of Title
Capacity Selection
Process Comparison and Selection
Prepare preliminary investigation report
Process flow Diagram
Material And Energy Balance
instrumentation and control
HAZOP study & Cost estimation
Simulation and final report

52. 52