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CEP ASSIGNEMENT final environmental engineering.docx

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The document outlines the Multan Development Authority's plan to enhance water resources by utilizing surface water from the River Ravi due to declining groundwater levels. It details the design and analysis required for a reservoir, treatment facility, and distribution network, including calculations for optimal storage, treatment processes for low-quality water, and hydraulic flow analysis in the distribution system. The document emphasizes the importance of maintaining public health, environmental protection, and the economic benefits of efficient water treatment facilities while addressing challenges like aging infrastructure and climate change.

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Title: Complex Urban Water Distribution Network Design Statement: Multan Development Authority (MDA) has been provided a task to enhance the city’s water resource from River Ravi, as groundwater levels are declining in the area. By tapping into river’s surface water, a more resilient and reliable water supply is anticipated for the residents of Multan City. For this purpose, design of reservoir volume considering river flows and demands, design of treatment system, and analysis and design of water distribution network are the key task to perform. The following information are provided to complete the task. 1. Direct Supply Reservoir The inflow to a reservoir during the design drought is as follow and in units of 106 m3 . Every month is assumed to be 30 days long for convenience. The reservoir is partially full to begin with. Compensation flow of 1.80 x 106 m3 must be allowed through the reservoir at all times. The following outputs are required to be determined. Calculations Outflow (10^6 m³): Sum of all the net flow divided by 12 months =171.3/12 =14.275 Total Demand per Month: Sum of Compensation Flow (10^6 m³) and Outflow (10^6 m³)
a. Maximum Uniform Yield: 14.275 million m³ per month. b. Initial Storage: To meet the demand in the initial months when inflow is less than the yield, we need to store water. The cumulative deficit is:  Jan: -10.775  Feb: -21.35  Mar: -31.525  Apr: -40.1  May: -44.375 The maximum deficit is -44.375 million m³. So, the initial storage required below the starting level is 44.375 million m³. c. Additional Storage: To store the excess water in high-flow months for later use, we need additional storage. The maximum cumulative surplus occurs in July: Jan to Jul: 31.725 million m³. So, the additional storage required above the starting level is 31.725 million m³. d. Total Storage: Total storage required = Initial storage + Additional storage + 10% dead storage = 44.375 + 31.725 + 3.1725 = 79.2725 million m³. e. Maximum Uniform Yield with Limited Initial Storage: If the initial storage is limited to 30 million m³, we can't maintain a uniform yield of 1.1 million m³ per month. We'll need to adjust the yield to match the available storage and inflow. A detailed analysis, possibly involving a simulation model, would be required to determine the optimal yield in this case.
Jan Feb Mar Apr May Jun Jul Aug 0 50 100 150 200 250 13.2 26.4 39.6 52.8 66 79.2 92.4 105. Chart Title Commulative Inflow (10^6 m³) Commulative Total Demand per Month Com Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 0 50 100 150 200 250 13.2 26.4 39.6 52.8 66 79.2 92.4 105.6 118.8 132 145.2 158.4 Chart Title Commulative Inflow (10^6 m³) Commulative Total Demand per Month #REF! Commulative New Total demand per Month
2. Water Treatment Facility Provide typical sequence of unit treatment processes for lowland river supply of moderate to poor quality. Water treatment facilities are plants or installations designed to remove contaminants and pollutants from raw water sources, such as rivers, lakes, and underground aquifers, to produce safe and clean drinking water for human consumption. Types of Water Treatment Facilities 1. Conventional Treatment Plants: These plants use physical and chemical processes to remove contaminants from water. 2. Biological Treatment Plants: These plants use biological processes, such as aerobic and anaerobic treatment, to remove organic matter and nutrients from water. 3. Desalination Plants: These plants use reverse osmosis or other technologies to remove salt and other minerals from seawater or brackish water. 4. Wastewater Treatment Plants: These plants treat wastewater from households, industries, and institutions to remove pollutants and contaminants before discharging the treated water into the environment. Components of a Water Treatment Facility 1. Intake: The point where raw water is drawn from the source. 2. Pretreatment: Physical processes, such as screening and grit removal, to remove large objects and debris. 3. Coagulation and Flocculation: Chemical processes to remove dirt and other suspended particles. 4. Sedimentation: Physical process to remove suspended solids by settling. 5. Filtration: Physical process to remove remaining suspended solids and contaminants. 6. Disinfection: Chemical or physical process to kill bacteria, viruses, and other microorganisms. 7. Storage and Distribution: Treated water is stored in tanks or reservoirs and distributed to consumers through a network of pipes.
Importance of Water Treatment Facilities 1. Public Health: Water treatment facilities play a critical role in preventing waterborne diseases and protecting public health. 2. Environmental Protection: Water treatment facilities help to protect the environment by removing pollutants and contaminants from wastewater. 3. Economic Benefits: Water treatment facilities support economic development by providing a reliable source of clean water for industries, agriculture, and other uses. Challenges Facing Water Treatment Facilities 1. Aging Infrastructure: Many water treatment facilities are aging and in need of upgrade or replacement. 2. Climate Change: Climate change is affecting water quality and quantity, making it more challenging for water treatment facilities to operate. 3. Emerging Contaminants: New contaminants, such as pharmaceuticals and personal care products, are being detected in water sources, requiring new treatment technologies. Future Directions for Water Treatment Facilities 1. Advanced Treatment Technologies: Development and implementation of new treatment technologies, such as membrane bioreactors and advanced oxidation processes. 2. Water Reuse and Recycling: Increased focus on water reuse and recycling to conserve water resources and reduce wastewater discharge. 3. Sustainable and Resilient Design: Designing water treatment facilities that are sustainable, resilient, and adaptable to changing environmental conditions. 3: Flow from Treatment Plant storage reservoir to different distribution reservoirs To calculate the design flows in each branching pipeline, we need to apply the Darcy-Weisbach equation for head loss due to friction in pipes. This equation is typically used in conjunction with the Bernoulli equation to determine the flow in a pipe system. The general procedure involves calculating the head loss due to friction in each of the pipes and determining the flow rate.
Step 1: Understanding the given data We are provided with the following data:  Elevations: o A = 480 m (above datum) o C = 390 m (above datum) o D = 310 m (above datum)  Pipe information: o Pipe 1: Length L1=39 km, Friction factor f1=0.02, Diameter D1=0.8m o Pipe 2: Length L2=11 km, Friction factor f2=0.03, Diameter D2=0.6m o Pipe 3: Length L3=6 km, Friction factor f3=0.04, Diameter D3=0.5 m  Assumptions: o Minor losses are negligible. o The system is a branched pipeline where the water flows from the treatment plant to three distribution reservoirs. Step 2: Determine the total head loss for the system The total head loss due to friction in the system can be written as the difference in elevation between the water surfaces at the different points: ΔhAC=hA−hC for Pipe 1 to Pipe 2 ΔhAD=hA−hD for Pipe 1 to Pipe 3 Where:  hA=480 m (elevation at the treatment plant)  hC=390 m (elevation at reservoir C)  hD=310 m (elevation at reservoir D) ΔhAC= 480 m - 390 m =90m ΔhAD=480 m - 310 m =170m Step 3: Apply the Darcy-Weisbach equation The Darcy-Weisbach equation for head loss due to friction in a pipe is: hf=fLV2 /2gD
Where:  hf= Head loss due to friction (m)  f = Friction factor (dimensionless)  L= Length of the pipe (m)  D= Diameter of the pipe (m)  V = Velocity of water (m/s)  g = Acceleration due to gravity (9.81 m/s²) ΔhAC=f1L1V1 2 /2gD1 + f2L2V2 2 /2gD2 90m=(0.02x39000xV1 2 /2x9.81x0.8) +0.03x11000x V2 2 /2x9.81x0.6 90m=49.694V1 2 +28.03V2 2 V2 2 =3.21-1.773 V1 2 ΔhAD=f1L1V1 2 /2gD1 + f3L3V3 2 /2gD3 170 m = (0.02x39000xV1 2 /2x9.81x0.8) +0.04x6000x V3 2 /2x9.81x0.5 170 m = 49.694V1 2 + 24.46V3 2 V3 2 = 6.95-2.032V1 2 Step 4: Velocity of water in each pipe The velocity of water V in the pipe can be determined from the continuity equation: Q=A×V Where:  Q = Flow rate (m³/s)  A= Cross-sectional area of the pipe A=πD2 /4  V = Velocity of water (m/s) A1V1= A2V2+ A3V3 πD1 2 /4x V1= πD2 2 /4x V2+πD3 2 /4x V3 put the values of V2 and V3 A1=0.503m, A2=0.283m, A3=0.196m
0.503V1= 0.283x (3.21-1.773V1 2 )1/2 +0.196 (6.95-2.032V1 2 )1/2 V1=1.2125m/s , V2=0.777m/s , V3=1.991m/s Q1=0.610m3 /s, Q2=0.219m3 /s, Q3=0.390m3 /s, Q1= Q2+ Q3 0.610=0.219+0.390 V1=1.2125m/s , V2=0.777m/s , V3=1.991m/s Q1=0.610m3 /s, Q2=0.219m3 /s, Q3=0.390m3 /s, Q1= Q2+ Q3 0.610=0.219+0.390 4: Flow in distribution system: One of the hydraulic networks contains single source from reservoir 3 above and having four loops as shown below in the diagram. Use your roll numbers to calculate diameters for pipes (see hint below to calculate it). Lengths (in m) for different pipes and nodal flows (m3 /sec) are provided in the figure below.
For Loop 1 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 100 0.134 5.61E-05 0.75 0.5873045 5011.8 7 2210 2946.6 1 to 4 100 0.134 5.61E-05 0.85 0.7403294 5011.8 7 - 2786 3277.4 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.8 7 2.956 14.778 4 to 5 125 0.234 0.000847 0.05 0.0039183 5011.8 7 -1.22 24.405 Total - 574.1 6263.3 Δ=-Σh/nΣh/Q Δ= 0.049545695
For Loop 2 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 2 to 3 125 0.234 0.000847 0.25 0.0769465 5011.87 23.9 6 95.853 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.87 - 2.95 6 14.778 3 to 6 125 0.234 0.000847 0.1 0.0141254 5011.87 4.39 9 43.99 5 to 6 75 0.344 0.005534 0.1 0.0141254 5011.87 - 0.40 4 4.0416 Total 25 158.66 Δ=-Σh/nΣh/Q Δ= -0.085 For Loop 3 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 4 to 5 125 0.234 0.000847 0.05 0.00392 5011.872 1.2203 24.41 4 to 7 75 0.344 0.005534 0.4 0.18357 5011.872 -5.252 13.13 5 to 8 100 0.134 5.61E-05 0.15 0.02991 5011.872 112.54 750.2 7 to 8 125 0.234 0.000847 0.2 0.05092 5011.872 -15.86 79.29 Total 92.646 867.1 Δ=-Σh/nΣh/Q Δ= -0.0578
For Loop 1 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 100 0.6096 0.089778 0.75 0.587305 5011.872 1.381087 1.84145 1 to 4 100 0.6096 0.089778 0.85 0.740329 5011.872 -1.74094 2.04816 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.872 2.955569 14.77784 4 to 5 125 0.234 0.000847 0.05 0.003918 5011.872 -1.22025 24.40507 Total 1.375467 43.07252 Δ=-Σh/nΣh/Q Δ= -0.017261484 Now I am taking the correction value in the loop as per commercial standard diameter of pipe: For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 5 to 6 75 0.34 4 0.00553 4 0.1 0.0141 3 5011.87 2 0.404 2 4.042 5 to 8 10 0 0.13 4 5.61E-05 0.15 0.0299 1 5011.87 2 -112.5 750.2 6 to 9 30 0 0.53 4 0.04711 1 0.1 0.0141 3 5011.87 2 0.189 9 1.899 8 to 9 20 0 0.34 1 0.00530 3 0.05 0.0039 2 5011.87 2 -0.312 6.24 Total -112.3 762.4 Δ=-Σh/nΣh/Q Δ= 0.0795 9
For Loop 3 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 4 to 5 12 5 0.234 0.000847 0.158599 0.033156 5011.872 10.32556 65.10465 4 to 7 75 0.344 0.005534 0.42231 0.202964 5011.872 -5.80719 13.75102 5 to 8 10 0 0.6096 0.089778 0.250388 0.077168 5011.872 0.181465 0.724735 7 to 8 12 5 0.234 0.000847 0.108662 0.016472 5011.872 -5.12979 47.20865 For Loop 2 Pipe L (m ) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 2 to 3 12 5 0.23 4 0.00084 7 0.25 0.07694 7 5011.87 2 23.9631 8 95.85273 2 to 5 15 0 0.34 3 0.00545 6 0.2 0.05092 2 5011.87 2 -2.95557 14.77784 3 to 6 12 5 0.23 4 0.00084 7 0.1 0.01412 5 5011.87 2 4.39901 6 43.99016 5 to 6 75 0.34 4 0.00553 4 0.1 0.01412 5 5011.87 2 -0.40416 4.041554 Total 25.0024 7 158.6623 Δ=-Σh/nΣh/Q Δ= - 0.08518 For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 5 to 6 75 0.344 0.005534 0.1 0.014125 5011.872 0.404155 4.041554 5 to 8 100 0.6096 0.089778 0.15 0.029907 5011.872 -0.07033 0.468852 6 to 9 300 0.534 0.047111 0.1 0.014125 5011.872 0.189899 1.898987 8 to 9 200 0.341 0.005303 0.05 0.003918 5011.872 -0.31199 6.239753 Total 0.211739 12.64915 Δ=-Σh/nΣh/Q Δ= -0.00905 For Loop 3 Pipe L (m ) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 4 to 5 12 5 0.234 0.00084 7 0.05 0.00391 8 5011.87 2 1.22025 3 24.4050 7 4 to 7 75 0.344 0.00553 4 0.4 0.18357 4 5011.87 2 - 5.25242 13.1310 5 5 to 8 10 0 0.6096 0.08977 8 0.15 0.02990 7 5011.87 2 0.07032 8 0.46885 2 7 to 8 12 5 0.234 0.00084 7 0.2 0.05092 2 5011.87 2 - 15.8585 79.2923 Total - 19.8203 117.297 3 Δ=-Σh/nΣh/Q Δ= 0.09133 8 For Loop 1 Pip e L (m ) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 10 0 0.6096 0.08977 8 0.73273 9 0.56254 3 5011.87 2 1.32285 9 1.805363 1 to 4 10 0 0.6096 0.08977 8 0.86726 1 0.76838 3 5011.87 2 -1.80691 2.083461 2 to 5 15 0 0.343 0.00545 6 0.26756 5 0.08724 6 5011.87 2 5.06384 4 18.92565 4 to 5 12 5 0.234 0.00084 7 0.15859 9 0.03315 6 5011.87 2 -10.3256 65.10465 Total -5.74577 87.91912 Δ=-Σh/nΣh/Q Δ= 0.035325874 For Loop 2 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 2 to 3 125 0.23 4 0.000847 0.16482 0.035602 5011.872 11.08725 67.26884 2 to 5 150 0.34 3 0.005456 0.267819 0.087399 5011.872 -5.07272 18.94088 3 to 6 125 0.23 4 0.000847 0.01482 0.000413 5011.872 0.128654 8.681134 5 to 6 75 0.34 4 0.005534 0.17613 0.040252 5011.872 -1.1517 6.53893 Total 4.99148 5 101.4298 Δ=-Σh/nΣh/Q Δ= -0.0266
Total -0.42995 126.7891 Δ=-Σh/nΣh/Q Δ= 0.001833 For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 5 to 6 75 0.344 0.005534 0.09095 0.011852 5011.872 0.339104 3.72847 5 to 8 100 0.6096 0.089778 0.250388 0.077168 5011.872 -0.18146 0.724735 6 to 9 300 0.534 0.047111 0.09095 0.011852 5011.872 0.159333 1.75188 8 to 9 200 0.341 0.005303 0.05905 0.00533 5011.872 -0.42442 7.187532 Total -0.10745 13.39262 Δ=-Σh/nΣh/Q Δ= 0.004337
5:Use EPANET software to validate the optimized design: o Import the modified network layout into EPANET. o Simulate flow distribution and pressure under various demand scenarios. o Compare results from the Hardy Cross method with EPANET outputs to ensure accuracy and compliance with all constraints. :Expected Outcomes  A comprehensive report detailing the analysis, design calculations, and summarized results with proper referencing.  Validation of the optimized design through EPANET simulations, ensuring that the network meets regulatory standards and operational requirements. 10 9 8 7 6 5 4 3 2 1 3 14 13 12 11 10 8 7 6 5 4 2 1
Elevation Base Demand Initial Quality Demand Node ID m LPS LPS Junc 1 0 1.6 0 1.60 Junc 2 0 .3 0 0.30 Junc 3 0 .15 0 0.15 Junc 4 0 .4 0 0.40 Junc 5 0 0 0 0.00 Junc 6 0 .1 0 0.10 Junc 7 0 .2 0 0.20 Junc 8 0 .3 0 0.30 Junc 9 0 .15 0 0.15 Tank 10 30 #N/A 0 -3.20 water distribution system Network Table - Nodes EPANET 2.2 water distribution system
Network Table - Nodes Head Pressure Quality Node ID m m Junc 1 324.30 324.30 0.00 Junc 2 324.30 324.30 0.00 Junc 3 324.30 324.30 0.00 Junc 4 324.30 324.30 0.00 Junc 5 324.30 324.30 0.00 Junc 6 324.30 324.30 0.00 Junc 7 324.30 324.30 0.00 Junc 8 324.30 324.30 0.00 Junc 9 324.30 324.30 0.00 Tank 10 324.30 294.30 0.00 EPANET 2.2 Page 2 Length Diameter Roughness Flow Link ID m mm LPS Pipe 1 100 609 100 0.74
Pipe 2 125 234 100 0.16 Pipe 4 200 341 100 0.12 Pipe 5 125 234 100 -0.14 Pipe 6 75 344 100 -0.34 Pipe 7 125 234 100 0.11 Pipe 8 75 344 100 0.37 Pipe 10 100 134 100 0.03 Pipe 11 100 609 100 0.86 Pipe 12 300 534 100 0.27 Pipe 13 150 343 100 0.29 Pipe 14 134 534 100 3.20 Pipe 3 1000 234 100 0.01 Velocity Unit Headloss Friction Factor Reaction Rate Link ID m/s m/km mg/L/d Pipe 1 0.00 0.00 0.000 0.00 Pipe 2 0.00 0.00 0.104 0.00
Pipe 4 0.00 0.00 0.000 0.00 Pipe 5 0.00 0.00 0.000 0.00 Pipe 6 0.00 0.00 0.000 0.00 Pipe 7 0.00 0.00 0.000 0.00 Pipe 8 0.00 0.00 0.000 0.00 Pipe 10 0.00 0.00 0.000 0.00 Pipe 11 0.00 0.00 0.515 0.00 Pipe 12 0.00 0.00 0.000 0.00 Pipe 13 0.00 0.00 0.172 0.00 Pipe 14 0.01 0.00 0.043 0.00 Pipe 3 0.00 0.00 0.000 0.00 Link ID Quality Status Pipe 1 0.00 Open Pipe 2 0.00 Open Pipe 4 0.00 Open Pipe 5 0.00 Open Pipe 6 0.00 Open Pipe 7 0.00 Open Pipe 8 0.00 Open Pipe 10 0.00 Open Pipe 11 0.00 Open Pipe 12 0.00 Open Pipe 13 0.00 Open Pipe 14 0.00 Open Pipe 3 0.00 Open EPANET 2.2 Page 1
CEP ASSIGNEMENT final environmental engineering.docx

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CEP ASSIGNEMENT final environmental engineering.docx

  • 1. Title: Complex Urban Water Distribution Network Design Statement: Multan Development Authority (MDA) has been provided a task to enhance the city’s water resource from River Ravi, as groundwater levels are declining in the area. By tapping into river’s surface water, a more resilient and reliable water supply is anticipated for the residents of Multan City. For this purpose, design of reservoir volume considering river flows and demands, design of treatment system, and analysis and design of water distribution network are the key task to perform. The following information are provided to complete the task. 1. Direct Supply Reservoir The inflow to a reservoir during the design drought is as follow and in units of 106 m3 . Every month is assumed to be 30 days long for convenience. The reservoir is partially full to begin with. Compensation flow of 1.80 x 106 m3 must be allowed through the reservoir at all times. The following outputs are required to be determined. Calculations Outflow (10^6 m³): Sum of all the net flow divided by 12 months =171.3/12 =14.275 Total Demand per Month: Sum of Compensation Flow (10^6 m³) and Outflow (10^6 m³)
  • 2. a. Maximum Uniform Yield: 14.275 million m³ per month. b. Initial Storage: To meet the demand in the initial months when inflow is less than the yield, we need to store water. The cumulative deficit is:  Jan: -10.775  Feb: -21.35  Mar: -31.525  Apr: -40.1  May: -44.375 The maximum deficit is -44.375 million m³. So, the initial storage required below the starting level is 44.375 million m³. c. Additional Storage: To store the excess water in high-flow months for later use, we need additional storage. The maximum cumulative surplus occurs in July: Jan to Jul: 31.725 million m³. So, the additional storage required above the starting level is 31.725 million m³. d. Total Storage: Total storage required = Initial storage + Additional storage + 10% dead storage = 44.375 + 31.725 + 3.1725 = 79.2725 million m³. e. Maximum Uniform Yield with Limited Initial Storage: If the initial storage is limited to 30 million m³, we can't maintain a uniform yield of 1.1 million m³ per month. We'll need to adjust the yield to match the available storage and inflow. A detailed analysis, possibly involving a simulation model, would be required to determine the optimal yield in this case.
  • 3. Jan Feb Mar Apr May Jun Jul Aug 0 50 100 150 200 250 13.2 26.4 39.6 52.8 66 79.2 92.4 105. Chart Title Commulative Inflow (10^6 m³) Commulative Total Demand per Month Com Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 0 50 100 150 200 250 13.2 26.4 39.6 52.8 66 79.2 92.4 105.6 118.8 132 145.2 158.4 Chart Title Commulative Inflow (10^6 m³) Commulative Total Demand per Month #REF! Commulative New Total demand per Month
  • 4. 2. Water Treatment Facility Provide typical sequence of unit treatment processes for lowland river supply of moderate to poor quality. Water treatment facilities are plants or installations designed to remove contaminants and pollutants from raw water sources, such as rivers, lakes, and underground aquifers, to produce safe and clean drinking water for human consumption. Types of Water Treatment Facilities 1. Conventional Treatment Plants: These plants use physical and chemical processes to remove contaminants from water. 2. Biological Treatment Plants: These plants use biological processes, such as aerobic and anaerobic treatment, to remove organic matter and nutrients from water. 3. Desalination Plants: These plants use reverse osmosis or other technologies to remove salt and other minerals from seawater or brackish water. 4. Wastewater Treatment Plants: These plants treat wastewater from households, industries, and institutions to remove pollutants and contaminants before discharging the treated water into the environment. Components of a Water Treatment Facility 1. Intake: The point where raw water is drawn from the source. 2. Pretreatment: Physical processes, such as screening and grit removal, to remove large objects and debris. 3. Coagulation and Flocculation: Chemical processes to remove dirt and other suspended particles. 4. Sedimentation: Physical process to remove suspended solids by settling. 5. Filtration: Physical process to remove remaining suspended solids and contaminants. 6. Disinfection: Chemical or physical process to kill bacteria, viruses, and other microorganisms. 7. Storage and Distribution: Treated water is stored in tanks or reservoirs and distributed to consumers through a network of pipes.
  • 5. Importance of Water Treatment Facilities 1. Public Health: Water treatment facilities play a critical role in preventing waterborne diseases and protecting public health. 2. Environmental Protection: Water treatment facilities help to protect the environment by removing pollutants and contaminants from wastewater. 3. Economic Benefits: Water treatment facilities support economic development by providing a reliable source of clean water for industries, agriculture, and other uses. Challenges Facing Water Treatment Facilities 1. Aging Infrastructure: Many water treatment facilities are aging and in need of upgrade or replacement. 2. Climate Change: Climate change is affecting water quality and quantity, making it more challenging for water treatment facilities to operate. 3. Emerging Contaminants: New contaminants, such as pharmaceuticals and personal care products, are being detected in water sources, requiring new treatment technologies. Future Directions for Water Treatment Facilities 1. Advanced Treatment Technologies: Development and implementation of new treatment technologies, such as membrane bioreactors and advanced oxidation processes. 2. Water Reuse and Recycling: Increased focus on water reuse and recycling to conserve water resources and reduce wastewater discharge. 3. Sustainable and Resilient Design: Designing water treatment facilities that are sustainable, resilient, and adaptable to changing environmental conditions. 3: Flow from Treatment Plant storage reservoir to different distribution reservoirs To calculate the design flows in each branching pipeline, we need to apply the Darcy-Weisbach equation for head loss due to friction in pipes. This equation is typically used in conjunction with the Bernoulli equation to determine the flow in a pipe system. The general procedure involves calculating the head loss due to friction in each of the pipes and determining the flow rate.
  • 6. Step 1: Understanding the given data We are provided with the following data:  Elevations: o A = 480 m (above datum) o C = 390 m (above datum) o D = 310 m (above datum)  Pipe information: o Pipe 1: Length L1=39 km, Friction factor f1=0.02, Diameter D1=0.8m o Pipe 2: Length L2=11 km, Friction factor f2=0.03, Diameter D2=0.6m o Pipe 3: Length L3=6 km, Friction factor f3=0.04, Diameter D3=0.5 m  Assumptions: o Minor losses are negligible. o The system is a branched pipeline where the water flows from the treatment plant to three distribution reservoirs. Step 2: Determine the total head loss for the system The total head loss due to friction in the system can be written as the difference in elevation between the water surfaces at the different points: ΔhAC=hA−hC for Pipe 1 to Pipe 2 ΔhAD=hA−hD for Pipe 1 to Pipe 3 Where:  hA=480 m (elevation at the treatment plant)  hC=390 m (elevation at reservoir C)  hD=310 m (elevation at reservoir D) ΔhAC= 480 m - 390 m =90m ΔhAD=480 m - 310 m =170m Step 3: Apply the Darcy-Weisbach equation The Darcy-Weisbach equation for head loss due to friction in a pipe is: hf=fLV2 /2gD
  • 7. Where:  hf= Head loss due to friction (m)  f = Friction factor (dimensionless)  L= Length of the pipe (m)  D= Diameter of the pipe (m)  V = Velocity of water (m/s)  g = Acceleration due to gravity (9.81 m/s²) ΔhAC=f1L1V1 2 /2gD1 + f2L2V2 2 /2gD2 90m=(0.02x39000xV1 2 /2x9.81x0.8) +0.03x11000x V2 2 /2x9.81x0.6 90m=49.694V1 2 +28.03V2 2 V2 2 =3.21-1.773 V1 2 ΔhAD=f1L1V1 2 /2gD1 + f3L3V3 2 /2gD3 170 m = (0.02x39000xV1 2 /2x9.81x0.8) +0.04x6000x V3 2 /2x9.81x0.5 170 m = 49.694V1 2 + 24.46V3 2 V3 2 = 6.95-2.032V1 2 Step 4: Velocity of water in each pipe The velocity of water V in the pipe can be determined from the continuity equation: Q=A×V Where:  Q = Flow rate (m³/s)  A= Cross-sectional area of the pipe A=πD2 /4  V = Velocity of water (m/s) A1V1= A2V2+ A3V3 πD1 2 /4x V1= πD2 2 /4x V2+πD3 2 /4x V3 put the values of V2 and V3 A1=0.503m, A2=0.283m, A3=0.196m
  • 8. 0.503V1= 0.283x (3.21-1.773V1 2 )1/2 +0.196 (6.95-2.032V1 2 )1/2 V1=1.2125m/s , V2=0.777m/s , V3=1.991m/s Q1=0.610m3 /s, Q2=0.219m3 /s, Q3=0.390m3 /s, Q1= Q2+ Q3 0.610=0.219+0.390 V1=1.2125m/s , V2=0.777m/s , V3=1.991m/s Q1=0.610m3 /s, Q2=0.219m3 /s, Q3=0.390m3 /s, Q1= Q2+ Q3 0.610=0.219+0.390 4: Flow in distribution system: One of the hydraulic networks contains single source from reservoir 3 above and having four loops as shown below in the diagram. Use your roll numbers to calculate diameters for pipes (see hint below to calculate it). Lengths (in m) for different pipes and nodal flows (m3 /sec) are provided in the figure below.
  • 9. For Loop 1 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 100 0.134 5.61E-05 0.75 0.5873045 5011.8 7 2210 2946.6 1 to 4 100 0.134 5.61E-05 0.85 0.7403294 5011.8 7 - 2786 3277.4 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.8 7 2.956 14.778 4 to 5 125 0.234 0.000847 0.05 0.0039183 5011.8 7 -1.22 24.405 Total - 574.1 6263.3 Δ=-Σh/nΣh/Q Δ= 0.049545695
  • 10. For Loop 2 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 2 to 3 125 0.234 0.000847 0.25 0.0769465 5011.87 23.9 6 95.853 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.87 - 2.95 6 14.778 3 to 6 125 0.234 0.000847 0.1 0.0141254 5011.87 4.39 9 43.99 5 to 6 75 0.344 0.005534 0.1 0.0141254 5011.87 - 0.40 4 4.0416 Total 25 158.66 Δ=-Σh/nΣh/Q Δ= -0.085 For Loop 3 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 4 to 5 125 0.234 0.000847 0.05 0.00392 5011.872 1.2203 24.41 4 to 7 75 0.344 0.005534 0.4 0.18357 5011.872 -5.252 13.13 5 to 8 100 0.134 5.61E-05 0.15 0.02991 5011.872 112.54 750.2 7 to 8 125 0.234 0.000847 0.2 0.05092 5011.872 -15.86 79.29 Total 92.646 867.1 Δ=-Σh/nΣh/Q Δ= -0.0578
  • 11. For Loop 1 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 100 0.6096 0.089778 0.75 0.587305 5011.872 1.381087 1.84145 1 to 4 100 0.6096 0.089778 0.85 0.740329 5011.872 -1.74094 2.04816 2 to 5 150 0.343 0.005456 0.2 0.050922 5011.872 2.955569 14.77784 4 to 5 125 0.234 0.000847 0.05 0.003918 5011.872 -1.22025 24.40507 Total 1.375467 43.07252 Δ=-Σh/nΣh/Q Δ= -0.017261484 Now I am taking the correction value in the loop as per commercial standard diameter of pipe: For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 5 to 6 75 0.34 4 0.00553 4 0.1 0.0141 3 5011.87 2 0.404 2 4.042 5 to 8 10 0 0.13 4 5.61E-05 0.15 0.0299 1 5011.87 2 -112.5 750.2 6 to 9 30 0 0.53 4 0.04711 1 0.1 0.0141 3 5011.87 2 0.189 9 1.899 8 to 9 20 0 0.34 1 0.00530 3 0.05 0.0039 2 5011.87 2 -0.312 6.24 Total -112.3 762.4 Δ=-Σh/nΣh/Q Δ= 0.0795 9
  • 12. For Loop 3 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 4 to 5 12 5 0.234 0.000847 0.158599 0.033156 5011.872 10.32556 65.10465 4 to 7 75 0.344 0.005534 0.42231 0.202964 5011.872 -5.80719 13.75102 5 to 8 10 0 0.6096 0.089778 0.250388 0.077168 5011.872 0.181465 0.724735 7 to 8 12 5 0.234 0.000847 0.108662 0.016472 5011.872 -5.12979 47.20865 For Loop 2 Pipe L (m ) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 2 to 3 12 5 0.23 4 0.00084 7 0.25 0.07694 7 5011.87 2 23.9631 8 95.85273 2 to 5 15 0 0.34 3 0.00545 6 0.2 0.05092 2 5011.87 2 -2.95557 14.77784 3 to 6 12 5 0.23 4 0.00084 7 0.1 0.01412 5 5011.87 2 4.39901 6 43.99016 5 to 6 75 0.34 4 0.00553 4 0.1 0.01412 5 5011.87 2 -0.40416 4.041554 Total 25.0024 7 158.6623 Δ=-Σh/nΣh/Q Δ= - 0.08518 For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 5 to 6 75 0.344 0.005534 0.1 0.014125 5011.872 0.404155 4.041554 5 to 8 100 0.6096 0.089778 0.15 0.029907 5011.872 -0.07033 0.468852 6 to 9 300 0.534 0.047111 0.1 0.014125 5011.872 0.189899 1.898987 8 to 9 200 0.341 0.005303 0.05 0.003918 5011.872 -0.31199 6.239753 Total 0.211739 12.64915 Δ=-Σh/nΣh/Q Δ= -0.00905 For Loop 3 Pipe L (m ) D(m) D4.87 Q(m3/ s) Q1.85 C1.85 h h/Q 4 to 5 12 5 0.234 0.00084 7 0.05 0.00391 8 5011.87 2 1.22025 3 24.4050 7 4 to 7 75 0.344 0.00553 4 0.4 0.18357 4 5011.87 2 - 5.25242 13.1310 5 5 to 8 10 0 0.6096 0.08977 8 0.15 0.02990 7 5011.87 2 0.07032 8 0.46885 2 7 to 8 12 5 0.234 0.00084 7 0.2 0.05092 2 5011.87 2 - 15.8585 79.2923 Total - 19.8203 117.297 3 Δ=-Σh/nΣh/Q Δ= 0.09133 8 For Loop 1 Pip e L (m ) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 1 to 2 10 0 0.6096 0.08977 8 0.73273 9 0.56254 3 5011.87 2 1.32285 9 1.805363 1 to 4 10 0 0.6096 0.08977 8 0.86726 1 0.76838 3 5011.87 2 -1.80691 2.083461 2 to 5 15 0 0.343 0.00545 6 0.26756 5 0.08724 6 5011.87 2 5.06384 4 18.92565 4 to 5 12 5 0.234 0.00084 7 0.15859 9 0.03315 6 5011.87 2 -10.3256 65.10465 Total -5.74577 87.91912 Δ=-Σh/nΣh/Q Δ= 0.035325874 For Loop 2 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 2 to 3 125 0.23 4 0.000847 0.16482 0.035602 5011.872 11.08725 67.26884 2 to 5 150 0.34 3 0.005456 0.267819 0.087399 5011.872 -5.07272 18.94088 3 to 6 125 0.23 4 0.000847 0.01482 0.000413 5011.872 0.128654 8.681134 5 to 6 75 0.34 4 0.005534 0.17613 0.040252 5011.872 -1.1517 6.53893 Total 4.99148 5 101.4298 Δ=-Σh/nΣh/Q Δ= -0.0266
  • 13. Total -0.42995 126.7891 Δ=-Σh/nΣh/Q Δ= 0.001833 For Loop 4 Pipe L (m) D(m) D4.87 Q(m3/s) Q1.85 C1.85 h h/Q 5 to 6 75 0.344 0.005534 0.09095 0.011852 5011.872 0.339104 3.72847 5 to 8 100 0.6096 0.089778 0.250388 0.077168 5011.872 -0.18146 0.724735 6 to 9 300 0.534 0.047111 0.09095 0.011852 5011.872 0.159333 1.75188 8 to 9 200 0.341 0.005303 0.05905 0.00533 5011.872 -0.42442 7.187532 Total -0.10745 13.39262 Δ=-Σh/nΣh/Q Δ= 0.004337
  • 14. 5:Use EPANET software to validate the optimized design: o Import the modified network layout into EPANET. o Simulate flow distribution and pressure under various demand scenarios. o Compare results from the Hardy Cross method with EPANET outputs to ensure accuracy and compliance with all constraints. :Expected Outcomes  A comprehensive report detailing the analysis, design calculations, and summarized results with proper referencing.  Validation of the optimized design through EPANET simulations, ensuring that the network meets regulatory standards and operational requirements. 10 9 8 7 6 5 4 3 2 1 3 14 13 12 11 10 8 7 6 5 4 2 1
  • 15. Elevation Base Demand Initial Quality Demand Node ID m LPS LPS Junc 1 0 1.6 0 1.60 Junc 2 0 .3 0 0.30 Junc 3 0 .15 0 0.15 Junc 4 0 .4 0 0.40 Junc 5 0 0 0 0.00 Junc 6 0 .1 0 0.10 Junc 7 0 .2 0 0.20 Junc 8 0 .3 0 0.30 Junc 9 0 .15 0 0.15 Tank 10 30 #N/A 0 -3.20 water distribution system Network Table - Nodes EPANET 2.2 water distribution system
  • 16. Network Table - Nodes Head Pressure Quality Node ID m m Junc 1 324.30 324.30 0.00 Junc 2 324.30 324.30 0.00 Junc 3 324.30 324.30 0.00 Junc 4 324.30 324.30 0.00 Junc 5 324.30 324.30 0.00 Junc 6 324.30 324.30 0.00 Junc 7 324.30 324.30 0.00 Junc 8 324.30 324.30 0.00 Junc 9 324.30 324.30 0.00 Tank 10 324.30 294.30 0.00 EPANET 2.2 Page 2 Length Diameter Roughness Flow Link ID m mm LPS Pipe 1 100 609 100 0.74
  • 17. Pipe 2 125 234 100 0.16 Pipe 4 200 341 100 0.12 Pipe 5 125 234 100 -0.14 Pipe 6 75 344 100 -0.34 Pipe 7 125 234 100 0.11 Pipe 8 75 344 100 0.37 Pipe 10 100 134 100 0.03 Pipe 11 100 609 100 0.86 Pipe 12 300 534 100 0.27 Pipe 13 150 343 100 0.29 Pipe 14 134 534 100 3.20 Pipe 3 1000 234 100 0.01 Velocity Unit Headloss Friction Factor Reaction Rate Link ID m/s m/km mg/L/d Pipe 1 0.00 0.00 0.000 0.00 Pipe 2 0.00 0.00 0.104 0.00
  • 18. Pipe 4 0.00 0.00 0.000 0.00 Pipe 5 0.00 0.00 0.000 0.00 Pipe 6 0.00 0.00 0.000 0.00 Pipe 7 0.00 0.00 0.000 0.00 Pipe 8 0.00 0.00 0.000 0.00 Pipe 10 0.00 0.00 0.000 0.00 Pipe 11 0.00 0.00 0.515 0.00 Pipe 12 0.00 0.00 0.000 0.00 Pipe 13 0.00 0.00 0.172 0.00 Pipe 14 0.01 0.00 0.043 0.00 Pipe 3 0.00 0.00 0.000 0.00 Link ID Quality Status Pipe 1 0.00 Open Pipe 2 0.00 Open Pipe 4 0.00 Open Pipe 5 0.00 Open Pipe 6 0.00 Open Pipe 7 0.00 Open Pipe 8 0.00 Open Pipe 10 0.00 Open Pipe 11 0.00 Open Pipe 12 0.00 Open Pipe 13 0.00 Open Pipe 14 0.00 Open Pipe 3 0.00 Open EPANET 2.2 Page 1