CHAPTER 1
INTRODUCTION & DESCRIPTIONOF THE PROJECT
Introduction
CE-432 is an Environmental Engineering Sectional course. The project given in this course is to design the water supply network of an industrial village.
This industrial village includes a residential area, a commercial area, an industrial area and a portion for future increment.
Our objective was,
- To prepare an Organ gram for the whole area
- To find out the water requirement
- To design the well and pointing its position
- To design pumping requirement
- To design water supply network
- To design the plumbing network
- To design the sanitary network
- To design the sewer network
Description of the Project:
Our project is a medium sized industrial village. This includes a Textile Industry and some commercial & residential block. The whole project is located at the bank of a river which is at the North side of the industry. The main entrance is at East which connects the whole area with the main high way. An exit is also located at the west portion of the industrial village.
The power is supplied by the Power Plant which is located at the North-East point of the village. There is also a waste water treatment plant located near the river at the north portion of the industrial sector.
There is also a portion for providing the room for future extension. A play ground is located near this area.
Our Industrial Village includes 4 major areas, these are
- Ø Residential Block
- Ø Commercial Block
- Ø Industrial Block and
- Ø An open area for future extension
The Residential Block:
It includes the residential buildings like
- house of the General Manager
- house of Managing Director
- house of Second Class Officers
- the quarter for the AGM
- a guest house
- And also a Club for Officers.
The commercial block:
It includes
- workers housing facility
- parking facility
- office complex
- a daycare center
- a shopping mall
- a prayer hall/ mosque
- a hospital
- a bank &
- a fire station
The industrial block:
It includes,
- the textile industry,
- a workers restroom
- a canteen
- a waste inventory
- a finished goods store
- a truck stand
- a finished goods inventory
- & an effluent treatment plant.
In doing this project, we took the liberty of selecting the various components of any area. Also we had the liberty on selecting the number of officials and other population of the industrial village. Though most of the activity in selecting the components of various areas and determining the number of population came from our imagination, it was perfected and completed by the help of our respectable course teachers.
CHAPTER 2
ACTIVITY THROUGOUT THE PROJECT
Various activity done in the project are briefly described below,
1. Sketching the layout:
A sketch was provided to showing all areas of the industrial village. As we had the liberty to apply our own imagination to create an industrial village, many photos were collected from the internet on this purpose and were analyzed. After that, a rough sketch was presented which was corrected and perfected later. After the calculation of water requirement, the final sketch was provided. The final sketch is shown at the Drawing section of this project report (Figure 2.2.)
2. Organogram:
Finishing the Organogram was a difficult task. The Organogram helped us to understand the various classes of the population. This class is required to calculate the water consumption. A rough estimation of the total population was computed which would not be possible without the Organogram. The theme of the Organogram was also selected by the group members. The Organogram of BUET, some banks and industries were analyzed In order to create our Organogram. This Organogram was then corrected by our course teacher. The Organogram is added at the Drawing section of the Project report.
3. Determination of total population:
The Organogram was the base line in determining the total population of the industrial village. Our total population is shown in a table below.
The population showing in the above table is the total working population of the industry. Most of them will not stay in the village after the working hours. Thus, it is required to determine the resident population of the industrial village. After making some appropriate assumptions, which was later on corrected by the help of our teachers, a list of resident population was determined. The list is shown in a table below.
From the above table, we have a total count of resident population of 606.
4. Water requirement for entire area:
Determination of water requirement is the determination of total discharge of the area. Now, there are four types of discharge,
I. Residential discharge
II. Commercial or Facility discharge
- Industrial discharge
- & Fire discharge.
Their sample calculation is described below;
I. Residential discharge: The total residential population was counted. Then from BNBC (the chart is attached in appendix portion), their class of occupancy and water consumption in LPCD was determined. Then the flow was multiplied by peaking factor and peak flow was determined. As for example, the water requirement calculation for G.M quarter is shown below;
Total number of G.M = 4
Family Size = 6
Total population = 4*6
= 24 persons
Class of Occupancy = A2
Water Requirement (LPCD)= 135 LPCD
Total flow =3240 Lpd
Peaking Factor = 2.1
Peak Flow = 6804 Lpd
Detailed calculation is shown in the appendix (Table 2.3)
II. Commercial or Facility discharge: At first the population in various facilities was determined. Then, the class of occupancy and the water requirement (in LPCD) was determined. Then total flow was calculated and then it was multiplied by the peaking factor to calculate the peak flow. As for example, the sample calculation for office is shown below;
Facility: Office
Population= 74 person
Class of Occupancy= F1
Water Requirement (LPCD) = 45 LPCD
Total Flow= 3330
Peaking Factor= 1.8
Peak Flow= 5994 m3/d
Detailed calculation is shown in the appendix (Table 2.4)
III. Industrial discharge: The working population in the industry was calculated and the water requirement was determined taking Occupancy G1 and 40 LPCD. Total population was found 294. The water requirement for working population was found 11760 l/d. It was assumed that, 5 ton fabric is produced per day requiring .138 million liters of water. So, total discharge was found to be, 150092 l/d. The peaking factor 2.1 was selected. And the peak flow was found to be 1295.196 m3/d
Detailed calculation is shown in appendix (table 2.5)
IV. Fire discharge:
Diameter of Nozzle= 3 inch
Pressure of water 3ft/sec Where,
Flow Q= A*V =πD2/4 = 0.049008 ft3 Q= Flow rate
= 360329 l/d A= Cross section of Nozzle
This flow of water must be supplied for at least 30 min.
So, the total Discharge was found to be
= Q Residential + Q Industrial + Q Facility + Q Fire = 108850+102809+150092
+360329 l = 722161 Lpd
5. Design of well:
There are certain steps on designing the water well. They are stated below;
I. Grain size distribution for different soil layers
II. Locating the aquifer and the water bearing strata
III. Design of gravel pack material
IV. Determination of strainer length and position
V. Selection of strainer size
VI. Yield of well
These points are discussed briefly below;
I. Grain size distribution for different soil layers: to locate the expected depth of water, drilling is done and sieve analysis of various layers at 10’ interval is found out. The layer with high FM is the expected layer for water extraction. The sieve analysis of different layers are shown in the appendix( table 2.6-2.13 & figure 2.7-2.14)
The sample calculation of layer at 400’ is shown below;
DEPTH 400′
| Sieve No. | Sieve Size | Material Retained | Percent of Material Retained | Cumulative of Percent Retained | Percent Finer | F.M. |
| (mm) | (gm) | % | % | % | ||
| No.16 | 1.18 | 0.00 | 0.00 | 0.00 | 100.00 | |
| No.30 | 0.6 | 12.50 | 12.15 | 12.15 | 87.85 | |
| No.40 | 0.425 | 16.80 | 16.33 | 28.47 | 71.53 | |
| No.50 | 0.3 | 29.20 | 28.38 | 56.85 | 43.15 | 1.60 |
| No.100 | 0.15 | 34.90 | 33.92 | 90.77 | 9.23 | |
| No.200 | 0.075 | 8.50 | 8.26 | 99.03 | 0.97 | |
| Pan | 1.00 | 0.97 | 100.00 | 0.00 | ||
| Total | 102.90 |
Table 2.9: Sieve Analysis at depth 400
| Sample Depth | D10 | D30 | D60 | U= D60/D10 | % of Coarse Sand | % of Medium Sand | % of Fine Sand | FM |
| (ft) | (mm) | (mm) | (mm) | % | % | % | ||
| 340′ | 0.17 | 0.25 | 0.35 | 1.40 | 0.5 | 89.5 | 20 | 1.50 |
| 360′ | 0.18 | 0.24 | 0.35 | 1.46 | 0.5 | 89.5 | 20 | 1.49 |
| 380′ | 0.2 | 0.3 | 0.39 | 1.30 | 4 | 86 | 10 | 1.68 |
| 400′ | 0.15 | 0.24 | 0.38 | 1.58 | 12 | 68 | 20 | 1.60 |
| 420′ | 0.18 | 0.25 | 0.38 | 1.52 | 2 | 82 | 16 | 1.56 |
| 440′ | 0.18 | 0.27 | 0.3 | 1.11 | 10 | 75 | 15 | 1.67 |
| 490′ | 0.15 | 0.22 | 0.34 | 1.55 | 1 | 76 | 23 | 1.38 |
| 510′ | 0.16 | 0.21 | 0.29 | 1.38 | 0.5 | 72.5 | 27 | 1.30 |
Table 2.14: Summery of grain size distribution
II. Locating the aquifer and the water bearing strata
The summery of all sieve analysis are shown in tabular form below;
Higher FM indicates coarser particles & coarser particle can hold water. So, location of water bearing soil layer is 380’-440’.
Length of the casing pipe:
Casing pipe must be sufficient enough so that submersible pump always remains below water.
Static water level at present = 400’ (assumed)
Drawdown of 15′ is assumed while pumping each time.
Average rate of water level declination (per year) × design life = 2’×20 = 40′
Safety distance of 15′ as a factor of safety
Therefore, length of the casing pipe = 400+15’+40’+15’=470’
III. Design of gravel pack material:
Gravel pack size distribution is obtained from the sieve analysis curve of comparatively finest sand stratum among the water bearing soil layer. As the strata is finer and uniform, 30% size or percent finer is multiplied by 4 and a reference point is obtained. Through this point, a parallel smooth curve is drawn which has uniformity coefficient 2.5 or less. In the sieve analysis of gravel pack material, range of % retained is assumed to be ±8%.
From Graph, D10 = 0.7 mm, D30= 1.0 mm, D60 = 1.5 mm
So, uniformity co-efficient, U=D60/D10 =1.5/0.7 =2.14<2.5. So OK
| Sieve no. | Size (mm) | %finer from graph | Cumulative % retained | % retained | Range of % retained |
| 4 | 4.75 | 99 | 1 | 1 | 0-9 |
| 8 | 2.36 | 92 | 8 | 7 | 0-15 |
| 16 | 1.18 | 40 | 60 | 52 | 44-60 |
| 30 | 0.6 | 4 | 96 | 36 | 28-44 |
| 40 | 0.425 | 2 | 98 | 2 | 0-10 |
| 50 | 0.3 | 0 | 100 | 2 | 0-10 |
| 100 | 0.15 | 0 | 100 | 0 | – |
| 200 | 0.075 | 0 | 100 | 0 | – |
Table 2.15: Grain size of the Gravel Pack
IV. Determination of strainer length and position:
Strainer length depends on the length of the aquifer. There is a thumb rule that,
| Aquifer thickness | Screening |
| <25′ | 70% |
| 25′-50′ | 75% |
| >50′ | 80% |
Table 2.16: Thumb Rule of Selecting Screening (in %)
Here, the thickness of the aquifer is = 440-380 = 60′
So, length of the strainer will be = 60 × 0.8 = 48′.
V. Selection of strainer size:
Slot size of the strainer is obtained from D10 size of gravel pack material. From gravel pack size distribution curve,
D10 = 0.7 mm
D30= 1.0 mm
D60 = 1.5 mm
To retain 90% of gravel pack materials, slot = D10/25.4×1000 = 0.7/25.4*1000=27.55
4-6 inch diameter 30 slot strainer is selected having each opening area of 30/1000.
VI. Yield of well
Separate slot sizes have different opening area
| Slot size | Assumed opening(steel screen) |
| 40 slot30 slot20 slot | 20%15%10% |
| For PVC screen, opening area is considered to be half of the above mentioned areas. | |
Table 2.17: Different Slot Sizes & their Opening
Yield = (area of strainer × flow velocity)/ factor of safety
= (0.15*3.14*6/12*48*0.1)/2.5
= 0.45216 ft3/s
=0.45216*(0.3048)3*3600*1000 l/h
=46093 lph
Here, screen entrance velocity is assumed as 0.1 ft/s so that,
- Friction losses in screen openings will be negligible.
- Rate of incrustation will be minimum
- Rate of corrosion will be minimum.
Design consideration,
i. For one pump
ii. In one day
iii. 8hr continuous pumping
Therefore, yield = 46093×8 = 368744 l/d
No. of Required Tubewell:
Water demand at present is = 722161 l/d
Water demand after 20 years = 770211 l/d
No. of tubewell= demand/yield
At present; 722161/368744 = 1.95
At future; 770211/368744= 2.08
So, 2 deep tubewell will serve upto 20 years.
| SUMMERY SHEET OF WELL DESIGN | |||||||||||||||||||||||||||||||||||||
| 1 | Strainer Position: Between 400′-448′ i. e. 48′ long strainer | ||||||||||||||||||||||||||||||||||||
| 2 | Slot Size: maximum 30 slots with approximately 15% opening area of steel strainer | ||||||||||||||||||||||||||||||||||||
| 3 | Yield/Transmission Capacity: 46,093 l/h | ||||||||||||||||||||||||||||||||||||
| 4 | Gravel Pack Material:
| ||||||||||||||||||||||||||||||||||||
| 5 | Length of Casing: 470 ft | ||||||||||||||||||||||||||||||||||||
Table 2.18: SUMMERY SHEET OF WELL DESIGN
6. Design of pumping requirement:
Distribution System is the part of Water work which receives water from the pumping station or from conduits by the gravity flow and delivers it to the individual consumers in the required quality and under sufficient pressure.
Sample Calculation:
Total lift or head of the pump = H
Volume of water to be Pumped = Q (gpm)
Working Horse Power, WHP= HQ/3960
Breaking Horse Power, BHP = WHP/η
Pump Efficiency= η=75%
Now, H= hs+hd+hf+hv
Where, hs= Suction Head= 0 (Since submersible pump)
hd= Delivery Head
= Length of Casing Pipe+ Maximum Building Height + Roof thank Height
hf= Velocity Head
hf= Friction Head
Delivery Head:
Casing Pipe Length = 400’
Additional 5’ is added since the pipe generally is not exactly at ground level.
Assume, maximum building height= 60’ (6 story @ 10’)
Providing 25’ static head from roof Tank so that Minm pressure at top fixture will be 15 psi and additional 10’ horizontal pipe.
So, Delivery Head hd = 400’+5’+ 60’+ 35’+ 500’
Velocity Head:
Assuming, Velocity of water supply, v= 3 fps
Velocity Head, hs= v2/2g= (3)2/2*32.2= .167 ft
Frictional Head:
Frictional loss is assumed to be 10% of maximum length.
For residential and administrative area, Maximum pipe length = 1650’
For industrial area, Maximum pipe length= 1280’
hf (residential) = 2055* 0.10 = 205.5’
hf (industrial) = 460* 0.10 =46’
Pump Capacity:
Residential Area:
H= 500’+ .167’+ 205.5’ = 705.667’= 706’
Q= 366984 l/d = 67.27 gpm (yield of the pump)
WHP = QH/3960 =67.27*706/3960 =12 HP
BHP = WHP/0.75 = 12/0.75 = 16 HP
Indusrial Area:
H= 500’+ .167’+ 46’ = 546.17’
Q= 366984 l/d = 67.27 gpm (yield of the pump)
WHP = QH/3960 =67.27*546.17/3960 =9 HP
BHP = WHP/0.75 = 9/0.75 = 12 HP
7. Water supply network design:
to supply the extracted water to various consumers is a great task. In order to to so, the whole area is divided into many sub areas. In our project, we divided the total area in three sub areas and provided 3 type of network system. Two of them are Loop network and one of them are branched network. Our water supply network in the project area is shown below;
We used Hardy Cross method in solving the Looped networks. We assumed K=1 and n=2 for our calculation. We also assumed that velocity of water is 3 fps and water flow for Fire Fighting is 3 gpm. The sample calculation for Looped Network 1 is shown below
loop Network 1
| LOOP NETWORK-1 | ||||||||
| Trial 01 | ||||||||
| Pipe | Length (ft) | Q (gpm) | k | n | ∑KQⁿ | ∑KQⁿ-1 | ||
| AP | 800 | 30 | 1 | 2 | 900 | 60 | ||
| SP | 200 | 26 | 676.00 | 52.00 | ||||
| RS | 940 | 27.5 | 756.25 | 55 | ||||
| QR | 200 | 32 | 1024.00 | 64 | ||||
| AQ | 140 | 39 | 1521.00 | 78 | ||||
| ∑ | 3077.25 | 309.00 | ||||||
| ∆Q= | ∑KQⁿ | = 9.96 | ||||||
| ∑KQⁿ-1 | ||||||||
| The convention is to take clock wise flow as +ve and anti clock wise flow as – ve. Now, as the value of ∆Q is -9.96, it is required to add -9.96 with every flow. As for example in AP, the flow is anti clock wis, and from the table , we can see that it is -30 gpm. So, in next trial, the flow in AP would be, -30+ (-9.96) = 39.96.And in this method we have to correct our flows in the loop.Trial 02 | ||||||||
| Pipe | Length (ft) | Q (gpm) | k | n | ∑KQⁿ | ∑KQⁿ-1 | ||
| AP | 800 | 39.96 | 1 | 2 | 1596.70 | 79.92 | ||
| SP | 200 | 16.04 | 257.32 | 32.08 | ||||
| RS | 940 | 17.54 | 307.70 | 35.08 | ||||
| QR | 200 | 22.04 | 485.82 | 44.08 | ||||
| AQ | 140 | 29.04 | 843.39 | 58.08 | ||||
| ∑ | 297.53 | 249.25 | ||||||
| ∆Q= | -1.19 | |||||||
| Trial 03 | ||||||||
| Pipe | Length (ft) | Q (gpm) | k | n | ∑KQⁿ | ∑KQⁿ-1 | ||
| AP | 800 | 41.15 | 1 | 2 | 1693.52 | 82.30 | ||
| SP | 200 | 14.85 | 220.45 | 29.70 | ||||
| RS | 940 | 16.35 | 267.24 | 32.70 | ||||
| QR | 200 | 20.85 | 434.62 | 41.70 | ||||
| AQ | 140 | 27.85 | 775.49 | 55.70 | ||||
| ∑ | 4.27 | 242.09 | ||||||
| ∆Q= | -0.02 | |||||||
At trial 3, the value of ∆Q is -.02 which is negligible. So this is our final design for Loop network 1
| Original Pipe | Original Pipe Length (ft) | Branch Pipe | Q original(gpm) | Q (gpm) @ Branch Pipe | |
| AB | 150 | BB’ | 41.15 | 1 | |
| BC | 200 | CC’ | 40.15 | 2 | |
| CD | 38.15 | 1 | |||
| CE | 140 | EE’ | 37.15 | 1 | |
| EF | 30 | FF’ | 36.15 | 30 | |
| FG | 350 | GG’ | 6.15 | 25 | |
| GH | 370 | HH’ | -18.85 | 1.5 | |
| HI | 350 | II’ | -20.35 | 4.5 | |
| IA | 690 | -24.85 | |||
| FIRE FIGHTING | 3 | ||||
| FLOW IN AQ | 27.85 | ||||
| Sample calculation for pipe diameter:For pipe AB,Discharge= 41.15 gpm= 41.15/1.833×10-4=224495.4 lpd =(224495.4×3.283)/(24x60x60x1000) =0.092 ft3/s We know, Discharge, Q=A x V Where, cross sectional area, A= /4 x d2 and d=pipe diameter Velocity, V= 3 ft/s Q= /4 x d2 x V Or, 0.092= /4 x d2 x 3 Or, d=0.198 inch But, we will provide minimum 4 inch dia pipe
| |||||
| PIPE DIAMETER OF LOOP NETWORK-1 | |||||
| Original Pipe | Pipe Length (ft) | Demand Q (gpm) | pipe diameter (inch) | Pressure loss (ft) | |
| AB | 150 | 41.15 | 4 | 2.53 | |
| BC | 200 | 40.15 | 4 | 3.38 | |
| CE | 140 | 37.15 | 4 | 2.36 | |
| EF | 30 | 36.15 | 4 | 0.51 | |
| FG | 350 | 6.15 | 4 | 5.91 | |
| GH | 370 | 18.85 | 4 | 6.24 | |
| HI | 350 | 20.35 | 4 | 5.91 | |
| IA | 690 | 24.85 | 4 | 11.64 | |
| Table 2.19- Pipe diameter of Loop Network 1 | |||||
| Branch Pipe | Pipe Length (ft) | Q (gpm) | pipe diameter (inch) | Pressure loss (ft) | |
| BB’ | 40 | 1 | 4 | 0.68 | |
| CC’ | 40 | 2 | 4 | 0.68 | |
| CD | 35 | 1 | 4 | 0.59 | |
| EE’ | 40 | 1 | 4 | 0.68 | |
| FF’ | 350 | 30 | 4 | 5.91 | |
| GG’ | 380 | 25 | 4 | 6.41 | |
| HH’ | 40 | 1.5 | 4 | 0.68 | |
| II’ | 45 | 4.5 | 4 | 0.76 | |
Table 2.20 Pipe diameter of Branch of Loop network 1
Design of Brunch Network: Discharge at the end of the pipe line was measured. Then required pipe diameter was provided assuming the velocity 3 fps. After that, discharge of the next consumer is found out and this discharge is added with the previous discharge. As the pipe in this region is required to contain both of the discharge. So, the highest discharge is found near the pump. The diameter of the branches are shown below.
| Branch Pipe | Pipe Length (ft) | Q (gpm) | pipe diameter (inch) | Pressure loss (ft) |
| jkk’ | 340 | 1.27 | 4 | 5.74 |
| jj’ | 20 | 2.26 | 4 | 0.34 |
| ij | 180 | 3.54 | 4 | 3.04 |
| ii’ | 20 | 0.08 | 4 | 0.34 |
| ghi | 270 | 3.62 | 4 | 4.56 |
| gg’ | 20 | 5.44 | 4 | 0.34 |
| gg” | 30 | 0.45 | 4 | 0.51 |
| fg | 170 | 9.5 | 4 | 2.87 |
| ff’ | 25 | 0.19 | 4 | 0.42 |
| ff” | 25 | 6.19 | 4 | 0.42 |
| fc | 300 | 15.88 | 4 | 5.06 |
| dee’ | 190 | 0.85 | 4 | 3.21 |
| dd’ | 40 | 15.4 | 4 | 0.68 |
| cd | 200 | 16.25 | 4 | 3.38 |
| bc | 80 | 32.13 | 4 | 1.35 |
| bb’ | 40 | 4.4 | 4 | 0.68 |
| oab | 210 | 36.53 | 4 | 3.54 |
Table 2.24: Pipe diameter of Branch Section
8. Plumbing & Sanitary network design:
In plumbing, a Fixture Unit (FU) is equal to one cubic foot of water per minute. A Fixture Unit is not a flow rate unit but a design factor. One cubic foot of water is roughly 7.48 gallons. A Fixture Unit is used in plumbing design for both water supply and waste water.
Different fixtures have different flow requirements. In order to determine the required size of pipe, an arbitrary unit is used for pipe sizing which takes into account the likelihood that all the fixtures will not be used at the same time. This is called “fixture unit” (FU)
To determine the Plumbing Network of the Building, at first, a plan of one apartment was provided. In this plan, the fixture units were indicated. The plan of the apartment is attached in the Drawing section (Figure 2.20). Then the number of F.U. is calculated.
To determine the F.U for the Building is the primary calculation to design the Plumbing Network.
At first, the number of one F.U. of one apartment was calculated. Then the total F.U. of the building was calculated. A sample calculation is shown below,
| Fixture Units: | ||||||
| For One Toilet: | ||||||
| Name of t he Fixure | Occupancy | No | Type | F.U Per Fixure | Total F.U | |
| 1 | Shower Head | Private | 2 | Mixing Vulve | 2 | 4 |
| 2 | Lavatory | Private | 1 | Fuccet | 1 | 1 |
| 3 | Water Closet | Private | 1 | Flash Tank | 3 | 3 |
| Total= | 8 | |||||
| For Kitchen | ||||||
| Name of t he Fixure | Occupancy | No | Type | F.U Per Fixure | Total F.U | |
| 1 | Sink | Private | 1 | Fuccet | 2 | 2 |
| Total= | 2 | |||||
| For Dining | ||||||
| Name of t he Fixure | Occupancy | No | Type | F.U Per Fixure | Total F.U | |
| 1 | Lavatory | Private | 1 | Fuccet | 1 | 1 |
| Total= | 1 | |||||
| For Servant’s Toilet | ||||||
| Name of t he Fixure | Occupancy | No | Type | F.U Per Fixure | Total F.U | |
| 1 | Shower Head | Private | 2 | Mixing Vulve | 2 | 4 |
| 2 | Water Closet | Private | 1 | Flash Tank | 3 | 3 |
| Total= | 7 |
*All the F.U are provided According to BNBC code. which is attached in the appendix
Table 2.27 Fixture Unit calculation of one floor
In layout, there are 2 toilets with an additional servant’s toilet in one apartment.
Therefore, total F.U. for one apartment= 2*8+2+1+7= 26
There are two apartments in our floor. Fixture units per floor = 52 F.U.
Total load on a service pipe= 6*52= 312 F.U.
Height of the tank:
The minimum pressure for a shower in the top floor is 12 psi. But, if we consider this pressure, the height of the overhead tank will be = 12/.434= 27.65 ft, which is not usual. So, the pressure at top floor is considered as 8 psi according to BNBC. Considering this pressure, the height of the overhead tank is found= 8/.434= 18.4 ft. Now, the calculation for determination of pipe size, is shown below.
Sample Calculation for Pipe Size Determination of each Floor:
At fifth floor, an additional 10’ horizontal pipe length was considered.
Permissible pressure drop in the pipe from the tank to the top of the floor is the static gain. The fixture pressure is 8 psi and the branch loss is 1 psi. therefore,
p= 24*(.434)-8-1=1.4 psi
Assuming the fitting friction to be equal to 50% increase in actual length, the drop per 100’ is
= 2.8 psi/100’
At each other floor the gain is 10(.434)= 4.34 psi
= 28.9 psi per 100’
Building designated as flash tank type fixture with fixture pressure
of 8 psi
| Floor | F.U. | Accumulated F.U. | Demand Flow (gpm) | Actua Pipe Length | Equivalent Length | Total Equivalent Length | Pressure Drop (psi/100 ft) | pipe size (inch) |
| 5f | 52 | 312 | 112 | 34 | 17 | 51 | 2.8 | 3 |
| 4f | 52 | 260 | 102 | 10 | 5 | 15 | 28.9 | 2 |
| 3f | 52 | 208 | 93 | 10 | 5 | 15 | 28.9 | 2 |
| 2f | 52 | 156 | 81 | 10 | 5 | 15 | 28.9 | 2 |
| 1f | 52 | 104 | 69 | 10 | 5 | 15 | 28.9 | 1.5 |
| GF | 52 | 52 | 53 | 10 | 5 | 15 | 28.9 | 1.5 |
Table 2.28: Determination of pipe size of each floor
Sanitary design:
For sanitary design, a chart was followed which is attached in the appendix section (Table 2.29). The design of sanitary network is attached in the drawing section (Figure 2.21).
9. Sewer network design:
We designed interceptive sewer in our project area. This type of sewer carries the load coming from various house sewers. The industrial sewer contains different compounds than the house sewers, thus a separate sewer line is designed for the industrial sewage that takes the industrial load the Effluent Treatment Plant. Other loads coming from the house and other residential or commercial buildings are transferred to the Sewage Treatment Plant.
At first, flow from various zones was calculated. In case of residential and commercial zones, the waste water flow was considered as 75% of the total water consumed. And in case of industrial zone, the generated waste water is considered as the 95% of total flow consumed. The P.F that we used in the calculation of water supply is also used in this calculation. The calculation is shown below;
| FLOW CALCULATION FOR SEWER: | |||||||||
Point | Facilities | Flow (l/d) | future contribution(l/d) | Waste Water Flow (l/d) | Waste Water Flow (75%) (m3/d) | Type | Peaking Factor | Peak Flow (m3/d) | Total (m3/d) |
1 | Office Complex | 3330 | 30 | 3360 | 2.52 | Commercial | 1.8 | 4.536 | 4.536 |
2 | Club | 230 | 20 | 250 | 0.1875 | Commercial | 1.8 | 0.3375 | 31.308 |
Third Class Officers | 12180 | 1400 | 13580 | 10.185 | Residential | 2.5 | 25.4625 | ||
Community Center | 2400 | 20 | 2420 | 1.815 | commercial | 1.8 | 3.267 | ||
Mosque | 1600 | 60 | 1660 | 1.245 | commercial | 1.8 | 2.241 | ||
3 | Second Class Officers Quarter | 11880 | 1050 | 12930 | 9.6975 | Residential | 2.5 | 24.24375 | 137.655 |
MD | 2400 | 0 | 2400 | 1.8 | Residential | 2.5 | 4.5 | ||
GM | 9600 | 0 | 9600 | 7.2 | Residential | 2.5 | 18 | ||
Rest House | 1350 | 0 | 1350 | 1.0125 | Residential | 1.8 | 1.8225 | ||
AGM Quarters | 13500 | 350 | 13850 | 10.3875 | Residential | 2.5 | 25.96875 | ||
School and College | 21000 | 40 | 21040 | 15.78 | Commercial | 4 | 63.12 | ||
4 | Shopping Mall | 230 | 10 | 240 | 0.18 | Commercial | 1.8 | 0.324 | 46.73385 |
Bank | 675 | 20 | 695 | 0.52125 | Commercial | 1.8 | 0.93825 | ||
Hospital | 3000 | 0 | 3000 | 2.25 | Commercial | 1.8 | 4.05 | ||
Daycare Center | 1516 | 0 | 1516 | 1.137 | Residential | 1.8 | 2.0466 | ||
Worker Housing Facility | 17500 | 3500 | 21000 | 15.75 | Residential | 2.5 | 39.375 | ||
5 | Truck Stand | 17536 | 0 | 17536 | 13.152 | Industrial | 1.4 | 18.4128 | 18.4128 |
6 | Workers Restroom | 3000 | 0 | 3000 | 2.25 | Residential | 1.8 | 4.05 | 4.05 |
Canteen | 4546 | 0 | 4546 | 3.4095 | Residential | 1.8 | 6.1371 | ||
Table 2.30: Waste water Generated from Various Zones
After the flow calculation, the cumulative flow in each pipe was calculated. The cumulative flow in each pipe is shown in a chart below;
Line | From | To | Length of Sewer (ft) | Area (ft2) | Area (ha) | Peak Flow (m3/d) | Cumilitive Peak Flow (m³/d) | infiltration (m³/d) | Cumilitve Infiltration * (m3/d) | Total Cumilitive Peak Flow (m3/d) | Total Cumilitive Peak Flow (m3/s) |
1 | 1 | 2 | 300 | 224000 | 2.08 | 4.536 | 4.5360 | 29.15 | 29.15 | 33.69 | 0.00039 |
2 | 3 | 2 | 300 | 240000 | 2.23 | 137.66 | 137.66 | 31.23 | 31.23 | 168.89 | 0.00195 |
3 | 2 | 4 | 100 | 142500 | 1.32 | 0.00 | 142.19 | 18.54 | 78.92 | 221.11 | 0.00256 |
4 | 4 | 5 | 400 | 140000 | 1.30 | 31.31 | 173.50 | 18.22 | 97.14 | 270.64 | 0.00313 |
5 | 5 | 6 | 150 | 52500 | 0.49 | 46.73 | 220.2329 | 6.83 | 103.97 | 324.21 | 0.00375 |
6 | 6 | 7 | 350 | 126000 | 1.17 | 18.41 | 238.65 | 16.40 | 120.37 | 359.02 | 0.00416 |
7 | 7 | 8 | 150 | 54000 | 0.50 | 4.05 | 242.6957 | 7.03 | 127.40 | 370.09 | 0.00428 |
Table 2.31: Cumulative Flow in each Pipe in Residential & Commercial zone
At first, we assumed that the ground slope is 1 in 500. It is customary slope of the sewer is selected according to the ground slope. There are charts from which the slope is selected. But in our case, the discharge is so little that we could not find any value from the chart (the chart is attached in the appendix section). So, we selected the self cleansing velocity (0.75 m/s) as our prime consideration and determined the slope from the chart.
The calculation is shown below;
The upper invert level is calculated as =Ground surface – depth of the cover- the thickness of the pipe – diameter of the pipe.
For example, in case of pipe 1,
the upper IR = 30.3- 2-0.05-.1016=28.1484 m.
And the Lower invert elevation= Upper invert elevation- slope of sewer* length of sewer.
So the lower IR will be= 28.1484-.017*300/3.28=26.59 m
The calculation for Residential area and Commercial area is shown below;
Sewer Design for Residential Area:
| ground surface | sewer pipe invert | |||||||||
| elevation | elevation | |||||||||
Line | From | To | Sewer Diameter (mm) | Slope (m/m) | Capacity when full (m3/s) | Velocity when full (m/s) | at upper manhole | at lower manhole | upper end | lower end |
1 | 1 | 2 | 101.6 | 0.017 | 0.006 | 0.75 | 30.3 | 30.25 | 28.1484 | 26.59 |
2 | 3 | 2 | 101.6 | 0.017 | 0.006 | 0.75 | 30.33 | 30.25 | 26.59 | 25.04 |
3 | 2 | 4 | 101.6 | 0.022 | 0.007 | 0.85 | 30.25 | 30.21 | 25.04 | 24.37 |
4 | 4 | 5 | 101.6 | 0.017 | 0.006 | 0.75 | 30.21 | 30 | 24.37 | 22.29 |
5 | 5 | 6 | 101.6 | 0.022 | 0.007 | 0.85 | 30 | 29.98 | 22.29 | 21.29 |
6 | 6 | 7 | 101.6 | 0.017 | 0.006 | 0.75 | 29.98 | 29.92 | 21.29 | 19.47 |
7 | 7 | 8 | 101.6 | 0.017 | 0.006 | 0.75 | 29.92 | 29.87 | 19.47 | 18.70 |
Table 2.34: Calculation for determination of slope and Invert Level for Residential Area
The calculation for Residential area and Commercial area is shown below;
INDUSTRIAL SEWER DESIGN:
Line | From | To | Length of Sewer (ft) | Area (ft²) | Area (ha) | Peak Flow (m3/d) | Cumilitive Peak Flow (m³/d) | infiltration (m³/d) | Cumilitve Infiltration * (m3/d) | Total Cumilitive Peak Flow (m3/d) | Total Cumilitive Peak Flow (m3/s) |
1 | 1 | 2 | 300 | 80000 | 0.74 | 129.6 | 129.6 | 10.41 | 10.41 | 140.01 | 0.0016 |
2 | 2 | 3 | 150 | 80000 | 0.74 | 155.5 | 285.1 | 10.41 | 20.82 | 305.92 | 0.0035 |
Table 2.35: Cumulative Flow in each Pipe in Industrial Zone
| ground surface | sewer pipe invert | |||||||||
| elevation | elevation | |||||||||
Line | From | To | Sewer Diameter (mm) | Slope (m/m) | Capacity when full (m3/s) | Velocity when full (m/s) | at upper manhole | at lower manhole | upper end | lower end |
1 | 1 | 2 | 101.6 | 0.017 | 0.006 | 0.75 | 30.11 | 29.94 | 27.96 | 26.40 |
2 | 2 | 3 | 101.6 | 0.022 | 0.007 | 0.85 | 29.94 | 29.88 | 26.40 | 25.40 |
Table 2.36: Calculation for determination of slope and Invert Level for Residential Area
Storm Water calculation:
We calculated the amount of storm water that may generate in our project area. But we did not design any sewer for this purpose. The calculation is shown below;
zone | factor of propotionality F | mainstream length L | catchment area A | slope S | time of concentration, tc | rainfall intensity I | factor of propotionality F | run off coefficient C | peak discharge Q |
km | km² | m/km | min | mm/hr |
|
| m³/s | ||
A | 0.37 | 0.068 | 0.005 | 81.72 | 72 | 0.6 | 0.817 | ||
B | 0.39 | 0.075 | 0.005 | 85.29 | 68 | 0.75 | 1.063 | ||
C | 58.5 | 0.28 | 0.029 | 0.005 | 67.34 | 80 | 0.278 | 0.8 | 0.516 |
D | 0.17 | 0.014 | 0.005 | 43.97 | 95 | 0.6 | 0.222 | ||
E | 0.2 | 0.019 | 0.005 | 50.18 | 90 | 0.2 | 0.095 |
Table 2.37: Calculation for Storm Water
Position of Manhole:
Manholes are used to interconnect two or more sewers and to provide entry of sewer cleaning. For the sewers that are 1200 mm and smaller, manholes should be located at changes in size, slope or direction. In larger sewers these changes can be made without using a manhole. The positions of manhole are shown in the Figure 2.24 in drawing section.
Recommendation
We can see from our calculation of sewer network that, the flow in the sewer pipe is very low. To attain the self cleansing velocity, we had to provide a large slope. As a result, it was found that, the sewer pipe near the WWTP was at a lower elevation then the elevation of the river.
In order to avoid this, we should not design any Intercepting Sewer for the Industrial Village. Rather, we should design sanitary sewers (or House Connection). These sewers will be connected with the Intercepting Sewer of the Town.
The sanitary sewage, generated in the commercial or residential buildings will be transported to a Septic Tank. Then this sewage will be transported to the main sewers.
