drainage systems lecture (i)
TRANSCRIPT
1
1
Sanitary Drainage systemsSanitary Drainage systems
Lecture NotesLecture Notes
2
Table of ContentsTable of Contents
Chapter Chapter --1 Sanitary Drainage Fixture Units 1 Sanitary Drainage Fixture Units Page 1Page 1--2121Chapter Chapter --2 Vent system 2 Vent system Page 22Page 22-- 39 39 Chapter Chapter --3 Storm water & drainage systems 3 Storm water & drainage systems Page 40Page 40--5151Chapter Chapter --4 Sizing the Underground Sewage Network for Buildings 4 Sizing the Underground Sewage Network for Buildings Page 52Page 52--7474Chapter Chapter --5 Septic tank capacity 5 Septic tank capacity Page 75Page 75--8787Chapter Chapter --6 General example problem 6 General example problem Page 88Page 88--9696Chapter Chapter --7 Sanitary Appliances & Arrangements 7 Sanitary Appliances & Arrangements Page 97Page 97--103103Chapter Chapter --8 Applications 8 Applications Page 104Page 104References References Page 122Page 122
2
3
Sanitary Drainage Fixture UnitsSanitary Drainage Fixture UnitsThe suggested values of DFU ( table 1 & 2) were designed for The suggested values of DFU ( table 1 & 2) were designed for application in conjunction with the probability of simultaneous application in conjunction with the probability of simultaneous use of fixtures so as to establish maximum permissible use of fixtures so as to establish maximum permissible drainage loads, in terms of fixture units rather than in drainage loads, in terms of fixture units rather than in numbers of specific types of fixtures or gallons per minute of numbers of specific types of fixtures or gallons per minute of drainage flow, for each of the various parts of sanitary drainage flow, for each of the various parts of sanitary drainage systems. drainage systems. In general, the sanitary drainage fixture unit value assigned toIn general, the sanitary drainage fixture unit value assigned toa particular fixture is based on the average volume discharged a particular fixture is based on the average volume discharged and the average rate of discharge for the fixture. This value and the average rate of discharge for the fixture. This value is determined from the fixture’s total discharge flow, in is determined from the fixture’s total discharge flow, in gallons per minute, divided by gallons per minute, divided by 7.5,7.5, or, in other words, its total or, in other words, its total discharge flow in cubic feet per minute.discharge flow in cubic feet per minute.
ChapChap--11
4Table 1Table 1
3
5Table 1Table 1
Ref [2]Ref [2]
6
Table 2Table 2
Ref [2]Ref [2]
4
7
By Size of trapBy Size of trap
Is used for fixture not listed in the previous table. For exampIs used for fixture not listed in the previous table. For example le the the floor drain floor drain with 2” pipe diameter ,the corresponding DFU is 3with 2” pipe diameter ,the corresponding DFU is 3
Ref [1]Ref [1]
8Junction Box SystemJunction Box System
W.C.W.C.
BathBath
BidetBidet BidetBidet
4”UT
8 F.UTotal DFU
4”4W.C.s (flash Tank),
1
1.1/2”-2”
2Bathtub1
3”3Floor drain *1
1.1/2”-2”
1Bidet1
1.1/2”-2”
1Lavatory 1
Diameter
DFUFixturesNumber
5
9
8 F.UTotal DFU
4”4W.C.s (flash Tank), 11.1/2”- 2”2Bathtub1
3”3Floor drain *11.1/2”- 2”1Bidet11.1/2”- 2”1Lavatory 1DiameterDFUFixturesNumber
* Some references does not include F.D. in the calculation.A shower head over a bathtub does not increase the F.U.
10
10 F.U
Total DFU
4”4W.C.s (flash Tank),
1
1.1/2”-2”
2*2Bathtub2
3”3Floor drain *1
1.1/2”-2”
1Bidet1
1.1/2”-2”
1Lavatory 1
Diameter
DFUFixturesNumber
ShowerShower
6
11
7 F.U
Total DFU
4”4W.C.s(flash Tank),
1
1.1/2”- 2”2Bathtub2
3”3Floor drain *
1
1.1/2”- 2”1Lavatory 1
DiameterFUFixturesNumber
Clean out SystemClean out System
12Drainage Stacks and BranchesDrainage Stacks and BranchesBased on the computed drainage stack flow capacity for Based on the computed drainage stack flow capacity for stacks flowing stacks flowing 7/24 full at terminal velocity7/24 full at terminal velocity, the , the corresponding number of fixture units may be determined corresponding number of fixture units may be determined from design load charts or tables (1,2 &3) so as to establish from design load charts or tables (1,2 &3) so as to establish the total load which may be placed on a tall drainage stack. the total load which may be placed on a tall drainage stack. For example, the computed flow capacity for a 4For example, the computed flow capacity for a 4--in (10 cm) in (10 cm) stack flowing at stack flowing at 7/247/24 full is 143 gpm (9.02 L/s). From design full is 143 gpm (9.02 L/s). From design load charts or tables, it may be found that this rate of flow isload charts or tables, it may be found that this rate of flow isequivalent to equivalent to 500 fixture units. 500 fixture units. This is the total load that may This is the total load that may be received from all branches on a 4be received from all branches on a 4--in (10 cm) tall stack. in (10 cm) tall stack. However, to avoid excessive interference between flow However, to avoid excessive interference between flow entering the stack and that coming down the stack, it is entering the stack and that coming down the stack, it is necessary to limit the amount of flow, which may be, allowed necessary to limit the amount of flow, which may be, allowed to enter the stack at each of the branches.to enter the stack at each of the branches. Thus, in a building Thus, in a building of just a few stories in height, the amount of flow entering of just a few stories in height, the amount of flow entering the stack through a branch may be greater than what would be the stack through a branch may be greater than what would be permissible in a building of many stories.permissible in a building of many stories.
7
13Table 3Table 3 for sizing for sizing drainage stacks drainage stacks provides different provides different permissible loading for permissible loading for stack of more than stack of more than 3 3 stories in heightstories in height. . Included in the table Included in the table ,the maximum loads ,the maximum loads permitted on any permitted on any horizontalhorizontal fixture fixture branch of a branch of a short stackshort stack..
Table 3Table 3
14
Vertical for Vertical for each flooreach floorHorizontal per Horizontal per
floorfloor
8
15
As a sample exampleAs a sample example : Calculate the total number of DFU , and : Calculate the total number of DFU , and size the horizontal branch connecting the two adjacent size the horizontal branch connecting the two adjacent bathrooms , knowing that, The total fixture unit of each bathrooms , knowing that, The total fixture unit of each bathroom consists of (water closet, bidet, lavatory and bathroom consists of (water closet, bidet, lavatory and bathtub or shower) = bathtub or shower) = 8 FU8 FU’’ss
Total fixture unit of two adjacent bath rooms connected to Total fixture unit of two adjacent bath rooms connected to the same horizontal branch pipe is : the same horizontal branch pipe is : 8 x 2 = 16 FU8 x 2 = 16 FU’’s.s.
As can be seen from table (3 ) for any horizontal branches , As can be seen from table (3 ) for any horizontal branches , the 3the 3”” can handle up to 20DFU but , due to the presence of can handle up to 20DFU but , due to the presence of the the W.C.W.C.’’ss the the 44” ” pipe diameter is selected which can handle pipe diameter is selected which can handle up to 160 DFU.up to 160 DFU.
16
Table 4Table 4
Table 4Table 4 for sizing for sizing drainage stacks provides drainage stacks provides different permissible different permissible loading for stack of loading for stack of 3 3 stories or less in height stories or less in height and for stacks more than and for stacks more than 3 stories in height3 stories in height. . Included in the table are Included in the table are the maximum loads the maximum loads permitted on any permitted on any horizontal fixture branch horizontal fixture branch of a short stack and at of a short stack and at any any 1 story of stack more 1 story of stack more than 3 stories in heightthan 3 stories in height..
Ref [2]Ref [2]
9
17
Slopes for horizontal drains are shown inSlopes for horizontal drains are shown in (Table 5) ,(Table 5) ,
Which are applicable for building underground sewers Which are applicable for building underground sewers and drains as well as those running at the level of the ceiling and drains as well as those running at the level of the ceiling of basements, service tunnels, etc. Readersof basements, service tunnels, etc. Readersshould note that the carrying capacity of horizontal drains isshould note that the carrying capacity of horizontal drains issubstantially lower than that for vertical pipes. Diameter of a substantially lower than that for vertical pipes. Diameter of a vertical stack may have to be increased when it runs vertical stack may have to be increased when it runs horizontally due to its reduced capacity in that position.horizontally due to its reduced capacity in that position.
18Table 5Table 5
For 4” Pipe diameter , having a slope of 1.04% , the Max. DFU is 180 , However if the slope is 4.2% , the DFU becomes 250
10
19
Sanitary building drains are designed to flow Sanitary building drains are designed to flow half full at peak loadhalf full at peak load. To . To avoid backup of flow from the building drain into branches, eachavoid backup of flow from the building drain into branches, eachbranch connection to the building drain should be made to its upbranch connection to the building drain should be made to its upper half per half or its airor its air--space portion. This may be achieved for 90 degrees branch space portion. This may be achieved for 90 degrees branch connections by means of a oneconnections by means of a one--sixth bend and a 45 degrees Y branch or sixth bend and a 45 degrees Y branch or a longa long--sweep onesweep one--quarter bend and a Y branch. The Yquarter bend and a Y branch. The Y--branch fitting branch fitting may be rotated so that the branch is at 45 degrees angle above tmay be rotated so that the branch is at 45 degrees angle above the he horizontal when the onehorizontal when the one--sixth bend is to be used and at a vertical angle sixth bend is to be used and at a vertical angle when the longwhen the long--sweep onesweep one--quarter bend is to be used. Less invert quarter bend is to be used. Less invert elevation is lost with the oneelevation is lost with the one--sixth bend and Y combination (see Fig ).sixth bend and Y combination (see Fig ).
Connections to Sanitary Connections to Sanitary Building DrainsBuilding Drains
20
Figure 4Figure 4
Two pipe system S.S.Two pipe system S.S.
Vent pipeVent pipe
Vent pipeVent pipeVent pipeVent pipe
11
21
Vent pipeVent pipe
One pipe system S.S.One pipe system S.S.
(Most popular )(Most popular )
Ref [2]Ref [2]
22
VentVent SystemSystem
ChapChap--22
12
23
Sanitary drainage system of a building should be provided Sanitary drainage system of a building should be provided with an attendant system of vent piping designed so as to with an attendant system of vent piping designed so as to permit gases and odors in all parts of the drainage piping to permit gases and odors in all parts of the drainage piping to circulate up through the system and escape into the circulate up through the system and escape into the atmosphere above the building and to permit the admission atmosphere above the building and to permit the admission and emission of air in all parts of the system so that and emission of air in all parts of the system so that siphonage, aspiration, or backsiphonage, aspiration, or back--pressure conditions will not pressure conditions will not cause an excessive loss of trap seal uncause an excessive loss of trap seal under ordinary der ordinary conditions of use. The sizing, arrangement, and installation conditions of use. The sizing, arrangement, and installation of attendant vent piping should be designed so as to limit of attendant vent piping should be designed so as to limit airair--pressure variations in all fixture drains to a differential pressure variations in all fixture drains to a differential not exceeding 1 in (2.5 cm) of water column above or below not exceeding 1 in (2.5 cm) of water column above or below atmospheric pressure. atmospheric pressure.
IntroductionIntroduction
24
A vent system is a pipe in a drainage system used :
1. To provide a flow of air to and from a drainage system so as to ventilate it.
2. To provide a circulation of air within such a system to eliminate trap siphonage and reduce back pressure and vacuum surge .
3. To insure the rapid and silent flow of waste
13
25Table 5Table 5 is used in sizing vents in is used in sizing vents in
accordance with drainage accordance with drainage capacity loadscapacity loads. Permissible . Permissible lengths of vents are less than lengths of vents are less than those computed by formulas (in those computed by formulas (in which additional allowance need which additional allowance need to be made for the equivalent to be made for the equivalent length of pipe fittings) that the length of pipe fittings) that the stated length may be applied stated length may be applied directly as permissible developed directly as permissible developed length of pipe . This table is length of pipe . This table is applied for applied for vent stacks and vent stacks and branch vent sizing.branch vent sizing.
Developed length of pipe =straight length of pipe + equivalent
length of fittings
26
Ref [1]Ref [1]
14
27
Ref [1]Ref [1]
28
15
29
Ref [1]Ref [1]
30
16
31
32
Traps. Traps. A fixture trap, illustrated in A fixture trap, illustrated in Fig. , is a UFig. , is a U--shaped section of pipe shaped section of pipe of the necessary depth to retain of the necessary depth to retain sufficient liquid required by code. All sufficient liquid required by code. All fixtures and equipment directly fixtures and equipment directly connected to the sanitary drainage connected to the sanitary drainage system are required to have traps. system are required to have traps. All traps must be All traps must be ventedvented in an in an approved manner, except for specific approved manner, except for specific conditions waived by local code conditions waived by local code requirements or authorities.requirements or authorities.
Ref [3]Ref [3]
17
33
Ref [3]Ref [3]
34
Ref [3]Ref [3]
18
35
Ref [2]Ref [2]
36
Ref [2]Ref [2]
19
37
Ref [2]Ref [2]
38
Ref [2]Ref [2]
20
39
Ref [1]Ref [1]
40
Storm waterStorm water drainage system drainage system & &
Rain WaterRain Water pipespipes
ChapChap--33
21
41
Roof drainage systems Roof drainage systems
A roof drainage system is composed of stormA roof drainage system is composed of storm--water collection devices water collection devices located in the roof and piping , connected to the collection devlocated in the roof and piping , connected to the collection devices, ices, which transforms the runoff out of the building to the ground. which transforms the runoff out of the building to the ground. Spacing Spacing and location of the roof drains are dependent on a number of locand location of the roof drains are dependent on a number of local al conditions and building characteristics. Consideration should beconditions and building characteristics. Consideration should be given given to such criteria as the local climatic conditions, type of roof,to such criteria as the local climatic conditions, type of roof, slope of slope of roof, location of pipe chases, and available ceiling space to inroof, location of pipe chases, and available ceiling space to install stall piping.piping.
42It has been found that a storm producing a rainfall intensity It has been found that a storm producing a rainfall intensity of 75 mm/hr may occur for 5 minutes once in 4 years, Can of 75 mm/hr may occur for 5 minutes once in 4 years, Can cause a serious damage .cause a serious damage .The rate of runThe rate of run--from roof +balconies from roof +balconies is calculated as follows:is calculated as follows:
Where Q = The rate of runWhere Q = The rate of run--off from roof and balconies.off from roof and balconies.A = effective area m2.A = effective area m2.P = impermeability factor which is 0.9 (concrete)P = impermeability factor which is 0.9 (concrete)
For asphalt in good order is (0.875).For asphalt in good order is (0.875).R = Rainfall intensity mm/hr, ( R = Rainfall intensity mm/hr, ( 7575--100 mm/h100 mm/h ))
For example: For example: Calculate the flow rate from a concrete roof having an Calculate the flow rate from a concrete roof having an
effective area of 50 m when the rainfall intensity is 75 effective area of 50 m when the rainfall intensity is 75 mm/hr.mm/hr.
]/[10003600
RPAQ 3 sm×××
=
.sec/1]/[001.010003600
579.005Q 3 literisthatsm=×××
=
22
43
The following procedure should be used in designing a roof drainThe following procedure should be used in designing a roof drainage age system:system:((11)) Lay out the position of the roof drains, deck drains and rainwaLay out the position of the roof drains, deck drains and rainwater ter leaders. Consideration should be given to placing an overflow drleaders. Consideration should be given to placing an overflow drain ain adjacent to each roof drain.adjacent to each roof drain.((22)) Determine the tributary area to each roof drain, deck drain, scDetermine the tributary area to each roof drain, deck drain, scupper, upper, gutter, or rainwater leader. The tributary area is the surface gutter, or rainwater leader. The tributary area is the surface area of roof area of roof that drains towards a specific drain. This tributary area shouldthat drains towards a specific drain. This tributary area should include include the effects of runoff from adjacent walls which drain onto the wthe effects of runoff from adjacent walls which drain onto the walls, alls, fig fig (R(R--1)1) indicates the wall area that should be added to roof area to indicates the wall area that should be added to roof area to determine the total tributary area for each drain.determine the total tributary area for each drain.((33)) Determine the routing and slope of the stormDetermine the routing and slope of the storm--water conductors. First, water conductors. First, determine the points from which, and to which, the conductors mudetermine the points from which, and to which, the conductors must be st be installed. Then determine the space available for installing theinstalled. Then determine the space available for installing the stormstorm--water conductors. Finally, the routing and slope of the stormwater conductors. Finally, the routing and slope of the storm--water water conductors.conductors.
Roof Drainage Design Procedure
44
Fig (RFig (R--1)1)
Ref [2]Ref [2]
23
45((44) ) Determine the rainfall rate to be used in sizing of the roof draDetermine the rainfall rate to be used in sizing of the roof drainage inage system. The rainfall rate (also known as the rainfall intensity)system. The rainfall rate (also known as the rainfall intensity) is a term is a term that relates the quantity of rainfall to a unit of time. Such rathat relates the quantity of rainfall to a unit of time. Such rainfall rates infall rates are usually expressed in inches per hour or centimeters per hourare usually expressed in inches per hour or centimeters per hour..((55) ) Determine the flow rate (volume per unit time) of equipment suchDetermine the flow rate (volume per unit time) of equipment such as as pumps, ejectors, airpumps, ejectors, air--conditioning equipment, and similar equipment conditioning equipment, and similar equipment which discharge into the roof drainage piping. Then convert theswhich discharge into the roof drainage piping. Then convert these flow e flow rates into equivalent roof area. Flow rate is a term expressing rates into equivalent roof area. Flow rate is a term expressing a volume a volume of water over a period of time such as cubic feet per second (cuof water over a period of time such as cubic feet per second (cubic bic meters per hour), and gallons per minute (liters per second). Thmeters per hour), and gallons per minute (liters per second). The e following equations determine the roof area which will produce rfollowing equations determine the roof area which will produce runoff at unoff at a flow rate equal to the flow rate of the equipment:a flow rate equal to the flow rate of the equipment:
Equivalent roof area = 96/R * flow rate of the equipmenEquivalent roof area = 96/R * flow rate of the equipment ft² t ft² Equivalent roof area = 359/R * flow rate of the equipmeEquivalent roof area = 359/R * flow rate of the equipment m²nt m²
where R is the rainfall rate used in the design of the roof drwhere R is the rainfall rate used in the design of the roof drainage ainage system in inches per hour (centimeters per hour). The flow rate system in inches per hour (centimeters per hour). The flow rate of the of the equipment is expressed in gallons per minute (liters per second)equipment is expressed in gallons per minute (liters per second)..
46
((66) ) Calculate the total roof area drained by each segment of the rooCalculate the total roof area drained by each segment of the roof f drainage system. This calculation should include all roof areas drainage system. This calculation should include all roof areas calculated in step (2) and the equivalent roof area calculated icalculated in step (2) and the equivalent roof area calculated in step (5). n step (5). Express the total area in square feet (square meters).Express the total area in square feet (square meters).((77) ) Determine the size of the roof drains and stormDetermine the size of the roof drains and storm--water conductors or water conductors or the gutters and rainwater leaders. Sizes can be determined usingthe gutters and rainwater leaders. Sizes can be determined using table table 11through table through table 22. . These tables list the maximum roof These tables list the maximum roof area in square feet (square meters) which can be handled by area in square feet (square meters) which can be handled by stormstorm--water drainage piping of different sizes and slopeswater drainage piping of different sizes and slopesfor various rainfall rates. for various rainfall rates. An example of Roof rain water distribution is shown in figure An example of Roof rain water distribution is shown in figure (R(R--2) 2) Area supplied by a drain pipe =
= (Area of the balcony) +(area of the adjacent wall)+ Part of the roof area.+ Part of the roof area.
24
47
900090005,7655,7653,0703,0701,4701,47048048066
10,80010,8006,9206,9203,6803,6801,7601,76057557555
13,50013,5008,6508,6504,6004,6002,2002,200720720( ( 4 )4 )
17,99517,99511,53011,5306,1306,1302,9302,93096096033
27,00027,00017,30017,3009,2009,2004,4004,4001,4401,44022
54,00054,00034,60034,60018,40018,4008,8008,8002,8802,88011
Maximum tributary area (ftMaximum tributary area (ft²² ))Rain fall rate (inch/h)
2 3 4 5 2 3 4 5 6 6
Size of drain pipe or leader (inch)
Table (R-1), is used to size, roof drains, vertical rainwater leaders or storm water conductors.
Ref [2]Ref [2]
48
39,65039,65047,60047,60059,50059,50072,80072,800109,000109,0001515
22,20022,20026,65026,65033,30033,30044,40044,40066,60066,6001212
13,80013,80016,58016,58020,70020,70027,60027,60041,40041,40010107,6007,6009,2009,20011,50011,50015,33015,33023,00023,000883,5663,5664,2804,2805,3505,3507,1337,13310,70010,700662,2272,2272,6752,6753,3403,3404,4534,4536,6806,680551,2561,2561,5041,5041,8801,8802,5062,5063,7603,760445485486576578228221,0961,0961,6441,64433
Maximum tributary area (ft² )Pipe sizing (inch)
2 3 (4 ) 5 6
Rainfall rate (in/hr)Rainfall rate (in/hr)
Table (R-2),is used to size conductors or rain water leader installed at a slope 1/8 in/ft (1cm/m)
25
49
Roof Rain water Drain Roof Rain water Drain
Figure ( RFigure ( R--2)2)
50Example : “Sizing Rain water pipe”Example : “Sizing Rain water pipe”
Suppose we decide to size the rain water pipe Suppose we decide to size the rain water pipe ( shown in figure ( shown in figure RR--3)3) for a 5 floors building having the following data :for a 5 floors building having the following data :
11-- One pipe is used to collect the rain water from two One pipe is used to collect the rain water from two adjacent balconies and part of the roof . This part of roof adjacent balconies and part of the roof . This part of roof has a 65 mhas a 65 m22 area ( refer to figure Rarea ( refer to figure R--3)3)
22-- The balcony area is 10 mThe balcony area is 10 m22 each.each.
33-- The adjacent balcony wall area is 15 mThe adjacent balcony wall area is 15 m22 each ( refer figure each ( refer figure RR--1)1)Solution: Solution: Area supplied by the drain pipeArea supplied by the drain pipe = = = (Area of the balcony) +(area of the adjacent wall) + Part of the roof area+ Part of the roof area= [(2 = [(2 xx 10) 10) xx 5] + [(155] + [(15xx 2)/ 2)/ 22 x 5) +65 = 195 mx 5) +65 = 195 m22 , that is (2166.6 ft, that is (2166.6 ft22))
From Table (RFrom Table (R--2) a D= 4 in at 4 in/ hr Rain water intensity can handle flow fr2) a D= 4 in at 4 in/ hr Rain water intensity can handle flow from om 2500 ft2500 ft2 2 areare . The 4 inch pipe is selected for this example. . The 4 inch pipe is selected for this example.
26
51
AREA OF ROOF PART = 65 M2Roof Drain
BALCON OF AREA = 10 m2
WALL AREA = 15 M2
BALCON OF AREA = 10 m2
WALL AREA = 15 M2
Figure ( RFigure ( R--3)3)
52
Sizing the Underground Sizing the Underground Sewage Network for BuildingsSewage Network for Buildings
ChapChap--44
27
53
Type of undergroundType of undergroundDrainage For Drainage For
buildingsbuildings
Separated Sewer Separated Sewer & rain water& rain water
system system Fig (UFig (U--1) 1)
Combined Combined Sewer + Rain waterSewer + Rain water
Fig( UFig( U--2)2)
54
Separate System of drainageSeparate System of drainage
Drainage below ground connectionDrainage below ground connection
Fig ( UFig ( U--1)1)
Ref [3]Ref [3]
28
55
Combined Rain + Sewer drainCombined Rain + Sewer drainFig ( UFig ( U--2)2)Ref [3]Ref [3]
56
In the case of combined system In the case of combined system ( ( Sewer +Rain waterSewer +Rain water), rainwater ), rainwater must be connected to the foul must be connected to the foul water drain through a back inlet water drain through a back inlet gully, to prevent the smellgully, to prevent the smellas shown in Fig. (Uas shown in Fig. (U--3).3). In the case In the case of separate system of separate system ( ( Rain water onlyRain water only), it is not ), it is not necessary to provide a trap before necessary to provide a trap before the rainwater pipe .It is connected the rainwater pipe .It is connected to the surface water drain, and to the surface water drain, and therefore a rainwater shoe, as therefore a rainwater shoe, as shown in Fig. (Ushown in Fig. (U--4), may be used.4), may be used.
Connections of the rain water DrainConnections of the rain water Drain
Fig. (UFig. (U--3).3).
Fig. (UFig. (U--4).4).
29
57
Garage Gully trapGarage Gully trap
The public Health Act 1936 The public Health Act 1936 section 34 defines certain section 34 defines certain prohibited discharges into drains prohibited discharges into drains or sewers asor sewers as1. anything that may injure a drain 1. anything that may injure a drain or sewer or interfere with the free or sewer or interfere with the free flow or treatment and disposal flow or treatment and disposal processes,processes,2. hot liquids with a temperature 2. hot liquids with a temperature exceeding 43.3 C,exceeding 43.3 C,3. petroleum spirit and calcium 3. petroleum spirit and calcium carbide.carbide.This means that the floor washings This means that the floor washings of large garages, petrol stations of large garages, petrol stations and indeed small garages should be and indeed small garages should be provided with some means of provided with some means of intercepting petrol before it intercepting petrol before it enters the drain or sewer. For the enters the drain or sewer. For the floor washings of a small garage, it floor washings of a small garage, it is sufficient to provide a garage is sufficient to provide a garage gully as shown in Fig. (Ugully as shown in Fig. (U--5). 5).
Garage DrainageGarage Drainage
Fig ( UFig ( U--5)5)
Ref [3]Ref [3]
58Grease Traps (Fig,UGrease Traps (Fig,U--6)6)Special gullies for the collection of Special gullies for the collection of grease are not required for houses, grease are not required for houses, but for canteen kitchens where the but for canteen kitchens where the waste water from the sinks and waste water from the sinks and dishwashers contains a considerabledishwashers contains a considerableamount of grease they are essential.amount of grease they are essential.When grease is hot or contained in When grease is hot or contained in hot water, it is in the form of an hot water, it is in the form of an emulsion, and if it is allowed to flow emulsion, and if it is allowed to flow into the drain it will cool and adhere into the drain it will cool and adhere to the sides of the pipes. The to the sides of the pipes. The principle of operation of the grease principle of operation of the grease trap is that of cooling down the trap is that of cooling down the grease in a large volume of water, grease in a large volume of water, which will generally be cool,which will generally be cool,so that the grease is solidified and so that the grease is solidified and floats on the surface. At periodic floats on the surface. At periodic intervals, the tray is lifted out of theintervals, the tray is lifted out of thetrap, which at the same timetrap, which at the same timecollects the grease.collects the grease.
Fig ( UFig ( U--6)6)Grease TrapsGrease Traps
Ref [3]Ref [3]
30
59Flow under gravity conditions ( Manning Formula )Flow under gravity conditions ( Manning Formula )
Manning , after carrying out a series of experiments , Manning , after carrying out a series of experiments , deduced the following equation which is the most commonly deduced the following equation which is the most commonly used for open channel flow and for water, sewer flows used for open channel flow and for water, sewer flows freely in pipes and conduits when both ends are open to freely in pipes and conduits when both ends are open to atmospheric pressure .atmospheric pressure .Calculations :Calculations :
).(.486.121
32
unitsSUSRn
V =
2/13/2486.1 SRn
AQ ×××=Where Where Q= flow rate ft3/sec.Q= flow rate ft3/sec.A = Wetted area ft2, (half pipe cross sectional area)A = Wetted area ft2, (half pipe cross sectional area)N= roughness of surface from table( ).N= roughness of surface from table( ).R = Hydraulics radius (Area/wetted perimeter).R = Hydraulics radius (Area/wetted perimeter).S= Slope 0.5 S= Slope 0.5 --1 % from 1 % from ChezyChezy formula formula
60The The determination of the hydraulics radiusdetermination of the hydraulics radius R for flow not running full R for flow not running full was explained before (chapwas explained before (chap--10 Dr. Hammoud lecture notes). 10 Dr. Hammoud lecture notes). In an open channel , the slope S can be determined as follows :In an open channel , the slope S can be determined as follows :Since the flow velocity is the same and the depth pressure does Since the flow velocity is the same and the depth pressure does not not change , the general energy equation becomes :change , the general energy equation becomes :
We can express this equation on a unit of length basis by dividiWe can express this equation on a unit of length basis by dividing both ng both sides by the length of the channel under consideration . Changesides by the length of the channel under consideration . Change in in elevation divided by change in distance yields the slope :elevation divided by change in distance yields the slope :
( ft / ft ) or dimensionl( ft / ft ) or dimensionless ess
From the above formula , it is clear that the flow down is causeFrom the above formula , it is clear that the flow down is caused by d by the difference in potential energy or gravity . On the other hanthe difference in potential energy or gravity . On the other hand the d the variable n known as Manning s , is a measure of the roughness ofvariable n known as Manning s , is a measure of the roughness of the the channel . Table (Uchannel . Table (U--1) lists the values of n for some of the more common1) lists the values of n for some of the more commonmaterials . materials .
P Vg
Z h P Vg
ZL1 1
2
12 2
2
22 2γ γ+ + − = + +
. . Z Z hL1 2− =
S Z Z L h LL= − =( ) / ( / )1 2
31
61
The following procedure should be used in designing a the The following procedure should be used in designing a the underground sewer pipe system:underground sewer pipe system:
(1)(1) Lay out should be drawnLay out should be drawn(2)(2) The total DFU connected to the sewer pipe should be The total DFU connected to the sewer pipe should be
calculate. calculate. (3)(3) From load tablesFrom load tables cconvert the DFU to gpm or L/s,onvert the DFU to gpm or L/s,(4)(4) Select the value “Select the value “nn” based on the pipe material.” based on the pipe material.(5)(5) Select a value of “Select a value of “S”S” , recommended underground slope , recommended underground slope
S=0.5S=0.5--1 % .1 % .(6)(6) Use Manning formula to determine the pipe diameter.Use Manning formula to determine the pipe diameter.
Note :Note :
PVC pipe is used where n = 0.01 , flow Running half full & PVC pipe is used where n = 0.01 , flow Running half full & recommended slope is 1% . recommended slope is 1% .
62
Values of Manning’s nValues of Manning’s n
Table (UTable (U--1)1)
32
63Example Example
S.I. unitS.I. unit
Water at the rate of 0.1 mWater at the rate of 0.1 m33 /s flows through a 1 m pipe diameter vitrified /s flows through a 1 m pipe diameter vitrified sewer when the sewer pipe is sewer when the sewer pipe is halfhalf-- fullfull . Find the slope of the water , if . Find the slope of the water , if Manning’s n is 0.013 .Manning’s n is 0.013 .
Solution :Solution :Given discharge ,Given discharge ,
Q = 0.1 mQ = 0.1 m33 /s/sDiameter of pipe D = 1 mDiameter of pipe D = 1 mArea of flow , A = ( 3.14/8) (0.5)Area of flow , A = ( 3.14/8) (0.5)22 =0.2777 m2=0.2777 m2
Wet Perimeter P = 3.14Wet Perimeter P = 3.14×× DD/2 = 3.14/2= 1.57m/2 = 3.14/2= 1.57m
Hydraulic radius Hydraulic radius
Manning s constant n = 0.013 Manning s constant n = 0.013
Find the slope S:Find the slope S:
S = ( 0.1 / 8.477 )S = ( 0.1 / 8.477 )22 = 1 /7186= 1 /7186
R AP
D mH = = = =0 3931 57 4
0 25..
.
1.0.)25.0(013.0
2777.0..121
32
21
32
=== SSRAn
Q
64
33
65
66
34
67
ManholesManholes
Usually constructed of Usually constructed of brickwork, brickwork, precastprecast concrete or concrete or plastic. Shallow manholes, plastic. Shallow manholes, which sometimes called which sometimes called inspection chamber built in 113 inspection chamber built in 113 mm of brickwork, providing that mm of brickwork, providing that they are not in a road or they are not in a road or waterlogged ground. waterlogged ground. Fig. (UFig. (U--7) shows a detail of brick 7) shows a detail of brick manhole whereas Fig. (Umanhole whereas Fig. (U--8) 8) shows A detail of a shows A detail of a precastprecastconcrete manhole.concrete manhole.
Fig. (UFig. (U--7)7) Ref [3]Ref [3]
68
Fig. (UFig. (U--8)8)
35
69
Dimensions of Brick ManholesDimensions of Brick Manholes
Cover sizes for depths up to 2.7 m are 600 mm. x 600 Cover sizes for depths up to 2.7 m are 600 mm. x 600 mm, and for depths up to 3.3 m aremm, and for depths up to 3.3 m are900 mm x 600 mm. For depths above 3.3 man access 900 mm x 600 mm. For depths above 3.3 man access shaft may be constructed above the main chamber.shaft may be constructed above the main chamber.
70
Precast Concrete Precast Concrete Manhole Manhole
36
71
The Building Regulations 1992 require accessThe Building Regulations 1992 require accessto drains at the following points:to drains at the following points:1. at a bend or change of direction;1. at a bend or change of direction;2. at a junction, unless each run can be2. at a junction, unless each run can be
cleared from an access point.cleared from an access point.3. On or near the head of each drain run;3. On or near the head of each drain run;4. on long runs;4. on long runs;5. at a change of pipe size. 5. at a change of pipe size. Figs (UFigs (U--9), and (U9), and (U--11) show the positions of access points. 11) show the positions of access points. The distances marked The distances marked ““AA”” depend on the type of access, depend on the type of access, see Table Rsee Table R--..
Sitting of Access PointsSitting of Access Points
72
Fig ( UFig ( U--9)9)
Ref [3]Ref [3]
37
73
Figure (R ) Junctions between drains and sewers. Note: 1,2,3 and 4 are alternative positions of the inspection chambers.
Fig ( UFig ( U--10)10)
74
Ref [1]Ref [1]
38
75
Septic Tank Septic Tank calculationcalculation
ChapChap--55
76The Septic tank capacity is calculated as follows:The Septic tank capacity is calculated as follows:
The type of building & the number of persons is first The type of building & the number of persons is first calculated and then multiply by the average wastecalculated and then multiply by the average waste--water( table water( table SS--1& S1& S--2) per person a day . {2) per person a day . {Rain water is not included}Rain water is not included}
For example :For example :
Suppose we decide to determine the septic tank capacity for a Suppose we decide to determine the septic tank capacity for a luxury home having 10 persons . From table (Sluxury home having 10 persons . From table (S--1) the daily 1) the daily waste water per person is between 75waste water per person is between 75--150 gpm /person/day . 150 gpm /person/day . If we select 110 gpm as an average value Then the daily waste If we select 110 gpm as an average value Then the daily waste water flow is: 110 gpm x 10= 1100 gpm /day . The volume of the water flow is: 110 gpm x 10= 1100 gpm /day . The volume of the septic tank should be sized for at least 10septic tank should be sized for at least 10--15 days (if no city 15 days (if no city sewer net work is available ) & for 2 days [if a city sewer netsewer net work is available ) & for 2 days [if a city sewer network is available + pump (electricity cut work is available + pump (electricity cut --off)]. off)].
The vent pipe size for the septic tank is shown in table (SThe vent pipe size for the septic tank is shown in table (S--3)3)
Practically for ordinary buildings a value of 200LPractically for ordinary buildings a value of 200L--250 250 L/Person/day is satisfactory. L/Person/day is satisfactory.
39
77SS--11
78
SS--22
40
79SS--33
80
1/3 L1/3 L 2/3 L2/3 L
SS--44
Length and structure of a septic tankLength and structure of a septic tank
Ref [1]Ref [1]
41
81Septic Tank CapacitySeptic Tank CapacitySS--55
Ref [1]Ref [1]
82
SS--66
Ref [1]Ref [1]
42
83
84Water Water --Drainage Pumping (Fig.)Drainage Pumping (Fig.)
Wherever possible, drains should be laid soWherever possible, drains should be laid sothat the liquid flows by gravity to the that the liquid flows by gravity to the sewer, or other point of disposal. In some sewer, or other point of disposal. In some cases, however, the water pipe or point of cases, however, the water pipe or point of disposal is above the drain, and pumping is disposal is above the drain, and pumping is therefore required. For the pumping of therefore required. For the pumping of surface water, a pumping installation as surface water, a pumping installation as shown in Fig. ( Sshown in Fig. ( S--7) may be used. 7) may be used. For larger installations, two pumps should For larger installations, two pumps should be installed, so that one of the pumps may be installed, so that one of the pumps may be used for Stand be used for Stand ––by purposes.by purposes.This type of installation is used for This type of installation is used for basements and boiler rooms tobasements and boiler rooms todeal with seepage of water, floor washing deal with seepage of water, floor washing or the draining down of the boilers and or the draining down of the boilers and heating pipe work,heating pipe work,
Fig. ( SFig. ( S--7)7)
Ref [3]Ref [3]
43
85Sump pumps (For waste water Sump pumps (For waste water drainage):drainage):
The sewer pipes are located below the The sewer pipes are located below the city network; in this case, a city network; in this case, a submersible pump will be used where submersible pump will be used where the motor and the pump section are the motor and the pump section are submersed in the liquid. Usually, two submersed in the liquid. Usually, two parallel sump pumps accompanied with parallel sump pumps accompanied with automatic switches are used.automatic switches are used.
Figure (SFigure (S--8) shows the operation 8) shows the operation principle of the pumps set .When the principle of the pumps set .When the liquid reaches a certain level, pump (No liquid reaches a certain level, pump (No 1) will start first, next to the second 1) will start first, next to the second level, pump (No 2) starts according to level, pump (No 2) starts according to the position of the level switches. For the position of the level switches. For further safety, the system is further safety, the system is accompanied with an alarm signalaccompanied with an alarm signal.. Fig. ( SFig. ( S--8)8)
86
Fig. ( SFig. ( S--9)9)
44
87Example:Example:Estimate the sump pump power required to evacuate a tank of Estimate the sump pump power required to evacuate a tank of 10 m10 m33 in 30 minutes to the city network pipe. The height is 8 in 30 minutes to the city network pipe. The height is 8 m and the total effective length L = 20 m. the material is m and the total effective length L = 20 m. the material is smooth pipe type L. Take the unit head loss for 6 ft/100ft. smooth pipe type L. Take the unit head loss for 6 ft/100ft. Assuming an overall pump efficiency Assuming an overall pump efficiency ηη =52%=52%Solution:Solution:10 m10 m33 /30 min.= 333 L/min= 88 gpm/30 min.= 333 L/min= 88 gpmSelect the pipe size that can transfer 88 gpm at the Select the pipe size that can transfer 88 gpm at the recommended pressure drop. From the pipe flow chart of recommended pressure drop. From the pipe flow chart of smooth pipe, the diameter is about 2.5smooth pipe, the diameter is about 2.5”” and corresponding and corresponding flow velocity is about 6.2 ft/s.flow velocity is about 6.2 ft/s.From the general energy equation we get:From the general energy equation we get:hhLL = h= h11x L = (6/100) x20 = 1.2 mx L = (6/100) x20 = 1.2 mhhAA = Z= Z11 –– ZZ22 + + hhLLhhAA = 8 + 1.2 = 8 + 1.2 ≅≅ 9.2 m9.2 mPPoutout = = γγ xx QQVV xx hhAA = 9.2 = 9.2 x x 9.819.81x x 0.00555 = 0.5 Kw 0.00555 = 0.5 Kw ≅≅ 0.68 hp0.68 hpPPelecelec = 0.68/= 0.68/ηη = 0.68/0.52 = 1.3 hp.= 0.68/0.52 = 1.3 hp.
88
General Example General Example problemproblem
ChapChap--66
45
89
Example 3. Determine the diameter of the main waste and soil stack for a five-storey Motel, having 6 W.C.s (flash valve), 8 bathtub, 3 urinals –wall lip and 2 Lavatories (1.1/2”)on each floor connected to one single S.S. riser.From Table 1 ,2 & 3Each floor 6 W.c.s, × 6 = 36 DFU , 8 Bathtub × 33 = 24 DFU3 urinals × 44= 12 DFU , 2 Lav × 2 = 4 DFUTotal = 76 DFU in each floor .
From table (4) horizontal fixture branch for the 76 DFU ,the 4” is selected because it can handle up to 160 DFU. The same table shows that the vertical S.S diameter can be 4” since it can handle up to 90DFU per floor which is sufficient for the 76 DFU that connected in at each T-Y connection.
90
Per floor76 DFU
4”366W.C.s(flash valve),
6
1.1/2”42Lavatory2
2.1/2”124urinals –wall lip
3
2”243Bathtub8
Diameter
Total DFU
DFUFixturesNumber
The Total for five floors =76The Total for five floors =76xx 5= 5= 380380 DFUDFU
46
91
Table 4Table 4
Vertical for Vertical for each flooreach floor
Horizontal Horizontal per floorper floor At Basement floor At Basement floor
connectionconnection
92The total DFU for the whole Motel is The total DFU for the whole Motel is 380380 DFU:DFU:30 30 W.c.sW.c.s, , ×× 66 = = 180180 DFU , 40 Bathtub DFU , 40 Bathtub ×× 33 = = 120120 DFUDFU15 urinals 15 urinals ×× 44= = 6060 DFU , 10 sinks DFU , 10 sinks ×× 2 = 2 = 2020 DFUDFUThe Total for five floors = The Total for five floors = 380380 DFUDFU
According to table (4) the horizontal branch connection at ¼ According to table (4) the horizontal branch connection at ¼ in 1 inch ft ( basement connection at high level) should be 5” in 1 inch ft ( basement connection at high level) should be 5” .Since the 4” branch pipe can only handle .Since the 4” branch pipe can only handle 216216 @ ¼ in per 1 ft @ ¼ in per 1 ft whereas our requirement is whereas our requirement is 380 DFU 380 DFU @ ¼ in per 1 ft slope@ ¼ in per 1 ft slope. The . The 5”5” branch pipe can handle branch pipe can handle 480480 DFU @ ¼ in per ft which is DFU @ ¼ in per ft which is enough.enough.
As a Summary:As a Summary:
The horizontal branch in each floor is 4 inch The horizontal branch in each floor is 4 inch
The vertical riser for the whole Motel pipe is 4 inch.The vertical riser for the whole Motel pipe is 4 inch.
The horizontal connection at the ground floor or basement is The horizontal connection at the ground floor or basement is 5 inch. 5 inch.
47
93
Table 5Table 5
From table(5) four values of From table(5) four values of DFU is available for the 4DFU is available for the 4”” S.S S.S that is, 43, 140, 320 & 540 that is, 43, 140, 320 & 540 DFU . Our values is DFU . Our values is 380 which380 whichis between 320 & 540 DFU is between 320 & 540 DFU The higher value is selected The higher value is selected (540 DFU ).(540 DFU ).The pipe diameter of the vent The pipe diameter of the vent pipe handling 540 DFU at a 50pipe handling 540 DFU at a 50--150 ft effective height is 150 ft effective height is between 2.1/2between 2.1/2”” & 3& 3””. The . The higher value is selected (3inch)higher value is selected (3inch)
Refer to the following Refer to the following schematic drawing schematic drawing
Size the vent pipeSize the vent pipe
94
Building drain or sewer connection pipeFor 5" pipe ( maximum) 480 DFU @1/4 in per ft
5" SS.
For 4" pipe ( maximum) 160 DFU Any Horizontal short fixture branch
4" SS.
2" V.pipe
Total @ 1 story or 1 branch intervalFor 4" pipe ( maximum) 90 DFU
3" V.S.
Roof
2" V.pipe
76 DFU each floor 5 x76 =380 DFU less than 500 D
4" SS.
Vertical 4" S.S. is enough
For 4” S.S. For 4” S.S. pipe the max. pipe the max.
FU is 500 FU is 500
48
95Now it is required to size the underground pipe diameter, Now it is required to size the underground pipe diameter, S=1%S=1% ,flow half full, ,flow half full, L= 100 mL= 100 m . .
As mentioned previously the total DFU = 380 As mentioned previously the total DFU = 380 ,the corresponding flow rate is 105 gpm = ,the corresponding flow rate is 105 gpm = 6.63 L/s6.63 L/s (from load table for flash tank) = (from load table for flash tank) = 0.01 m0.01 m33/s ./s . The value of n =0.01The value of n =0.01
01.0..121
32
== SRAn
Q
001.0)01.0(
01.001.05.02/1
32
=×
=×
=×S
QnRA
D= 0.15 m (6”) D= 0.15 m (6”) →→ This is the minimum This is the minimum diameter for the out flow of the building. diameter for the out flow of the building.
96
49
97
Sanitary Appliances & Sanitary Appliances & ArrangementsArrangements
ChapChap--77
98Types of Sanitary Types of Sanitary
Appliance WCAppliance WC
TwoTwo--trap Siphonic WC trap Siphonic WC panpan
Single Siphonic WC (most Single Siphonic WC (most popular)popular)
Ref [3]Ref [3]
50
99
UrinalsUrinals--types types
Ref [3]Ref [3]
100
51
101
BathsBaths
There is a large variety of bath shapes There is a large variety of bath shapes Ref [3]Ref [3]
102
There is a large variety of kitchen shapes There is a large variety of kitchen shapes
Kitchen sink Kitchen sink
Ref [3]Ref [3]
52
103Kitchen sink Kitchen sink
Ref [3]Ref [3]
104
Fixture ConnectionFixture Connection
& & Pipe sizingPipe sizing
ApplicationsApplications
ChapChap--88
From Reference [4]From Reference [4]
53
105
106
54
107
108
55
109
110
56
111
Ref [4]Ref [4]
112
Ref [4]Ref [4]
57
113
114
Ref [4]Ref [4]
58
115
116
59
117
Ref [4]Ref [4]
118
2”2”
2”2”
60
119
120
washer
25
fridge
dish
875
washer
dish fridge
1425
575
338
1175
338 500
950
n
H.W. H.W.
Draw & size the drain pipesDraw & size the drain pipesThe location of the Sewer Stack are shown The location of the Sewer Stack are shown
61
121
Example 1Example 1. . Find the internal diameters of the soil stack for Find the internal diameters of the soil stack for an eightan eight--storey office, having five storey office, having five WC.sWC.s, and five basins on , and five basins on each floor, assuming public use of fittings.each floor, assuming public use of fittings.
Example 2Example 2 Find the internal diameter (If the soil and waste Find the internal diameter (If the soil and waste stack for a four storey office having four stack for a four storey office having four W.c.sW.c.s, and four , and four basins on each floor, assuming public use of fittings.basins on each floor, assuming public use of fittings.
H.WH.W..
122
Example 3Example 3. . Find the internal diameters of the Rain water Find the internal diameters of the Rain water riser pipe serving eightriser pipe serving eight--balconies 10 mbalconies 10 m22 each . each .
Example 4Example 4. . Find the septic tank capacity for Motel serving Find the septic tank capacity for Motel serving 100 persons ( no sewer net work). Size the pump , and the 100 persons ( no sewer net work). Size the pump , and the corresponding vent pipe. Knowing that the septic tank must corresponding vent pipe. Knowing that the septic tank must be recovered weekly.be recovered weekly.
ProjectProject.. The drawingThe drawing entitled entitled ““tahertaher”” consist of 7 floors consist of 7 floors building . It required to:building . It required to:
1)1) Draw & size sewerage layout for each bathroom include the Draw & size sewerage layout for each bathroom include the location and the size of the vent pipe.location and the size of the vent pipe.
2)2) Draw & size the drainage riserDraw & size the drainage riser ..3)3) Draw & size the rain water pipesDraw & size the rain water pipes4)4) Draw and size the underground septic tank Draw and size the underground septic tank
62
123
ReferencesReferences11-- Mechanical & electrical equipment for buildings Mechanical & electrical equipment for buildings ––by by Stein/Reynolds, Ninth edition ,John Wiley, 2000.Stein/Reynolds, Ninth edition ,John Wiley, 2000.
22--Practical Plumbing Engineering , Cyril Practical Plumbing Engineering , Cyril M.Harris,ASPE,1998.M.Harris,ASPE,1998.
33-- Building Services & equipment , F. Hall, Third Building Services & equipment , F. Hall, Third edition, 1994.edition, 1994.
44-- Upland engineering , Mechanical consulting office, Upland engineering , Mechanical consulting office, Dr. Ali hammoud.Dr. Ali hammoud.