water & drainage equipement design
TRANSCRIPT
PROJECT /
PROPOSED (...+...+...) FLOORS
…………... BUILDING.
LOCATION \
………... - ………. - QATAR
OWNER \
………………………………
CONSULTING \
GCGGULF CONSULTING GROUP
المياه تغذيه لعمال الهيدروليكيه الحسابات
PLUMPING Calculation Sheet
Estamited Water Requirement
Office buildingTotal Area of Floors = 37038Gallon Per Day Per Capita = 20 GallonHour Per Day Usage = 9 HoursPeak Flow Rate = 2.5 X Average flow Rate
No. of Person = 4001 PersonDaily Consumption = 80020 GallonAverage Flow Rate = 149 GpmPeak Flow Rate = 372.5 Gpm
Resturant buildingTotal Area of Floors = 37038Gallon Per Day Per Capita = 25 GallonHour Per Day Usage = 9 HoursPeak Flow Rate = 2.5 X Average flow Rate
No. of Person = 4001 PersonDaily Consumption = 100025 GallonAverage Flow Rate = 186 GpmPeak Flow Rate = 465 Gpm
Appartement HouseNo of Bed Rooms = 1750 RoomGallon Per Day Per Capita = 100 GallonHour Per Day Usage = 15 HoursPeak Flow Rate = 3.0 X Average flow Rate
No. of Person = 1000 PersonDaily Consumption = 100000 GallonAverage Flow Rate = 112 GpmPeak Flow Rate = 336 Gpm
HotelsNo of Suit = 1750 SuitGallon Per Day Per Capita = 130 GallonHour Per Day Usage = 11 HoursPeak Flow Rate = 3.0 X Average flow Rate
No. of Person = 1000 PersonDaily Consumption = 130000 GallonAverage Flow Rate = 197 GpmPeak Flow Rate = 591 Gpm
m2
m2
HospitalsNo. of Bed = 500 BedConsumption per year = 50000000 Gal/yearConsumption per month = 5000000 Gal/monthPeak Hourly Flow = 15000 Gal/hour Minimum hourly flow = 1500 Gal/hour Peak Installatin flow = 500 GpmConsumption per day = 150000 Gal/dayAverage Installatin flow = 250 Gpm
University
Dormitory Student (Resedential)
Area of Capita = 9650No. of Students = 1043 Student
Fixture Usage /Day Gal/ Usage Gal/ Usage Water closet 3 5 15Shower 1 20 20Washup 2 10 20Miscellaneous 20
Total 78225
Kitchen and ResturantArea of Capita = 9650No. of Students = 1043 StudentNO of Meals per Day = 3 MealAssume Gal/meal / Capita = 2 GallonTotal = 6258 Gallon
Non-Residential Student
Area of Capita = 9650No. of Students = 1043 Student
Fixture Usage /Day Gal/ Usage Gal/ Usage Water closet 2 5 10Washup 2 5 10Miscellaneous 5
Total 25
Faculty and Staf
Area of Capita = 9650No. of Staf = 1043 Staf
Fixture Usage /Day Gal/ Usage Gal/ Usage
m2
m2
m2
m2
Water closet 2 5 10Washup 2 5 10Miscellaneous 5
Total 26075
General Laboratory use
No. of Fixture Usage Hour /Day Gal/ Usage Gal/ Usage 10 1 10 100
General Kitchen
Area of Capita = 9650No. of Students = 1043 StudentNO of Meals per Day = 2 MealAssume Gal/meal / Capita = 2 GallonTotal = 4172 Gallon
Gym
No. of Showers Usage /Day Gal/ Usage Gal/ Usage 10 8 4 1600
Get All University Water Demand Per Day 123511.3 Gallon
Shopping Centers
Area of Capita = 46296Domestic Water Consumption = 49999.68 GallonAverage Flow Rate = 83.3328 GpmPeak Flow Rate = 166.6656 Gpm
Parking and Pouplation Methods
No. of Parking Space = 4000 SpaceApproximae Customer /day = 24000 CustomerWater Consumption per Coustmure = 1.25 Gallon/dayDomestic Water Consumption = 30000 Gallon
No. of Empolyee = 800 EmpolyeeWater Consumption per Coustmure = 20 Gallon/dayDomestic Water Consumption = 16000 Gallon
Total load = Empolyee load + Coustumer Load1.1*( 30000 + 16000 )= 50600
Average Flow Rate = 85 GpmPeak Flow Rate = 170 Gpm
m2
m2
Court Houses And Detention Jails
Occupancey Population GPDC Total GPDNo. of Staf 200 20 4000Office Personnel 2000 20 40000Short Term Detainess ( prisoners) 1000 10 10000Long Term Detainess ( prisoners) 1000 80 80000
TOTAL 134000
Flow Rate Staf 4000 / 480 min =Office 40000 / 480 min =Short Term 10000 / 360 min =Long Term 80000 / 900 min =
Average Flow Rate
Peak Flow Rate 208.5 X 2 = 417
Gal/ Usage 15202020
78225
Gal/ Usage 10105
25
Gal/ Usage
10105
26075
Gal/ Usage 100
Gal/ Usage 1600
Gallon
8.483.427.888.9
208.5
Gpm
Building Specification :Building Lowest point level (Ground Tank Place) 0 mBuilding Highest point level (Roof Tank Place) 116 mSo get total building height (static head) 116 m
No. of Basement Floor 2 BasementNon. Typical Floor Levels 15 FloorNo. of Typical Floor in Building 5 FloorLast Floor in Building 1 FloorBefor Last Floor in Building 1 Floor2 Floor Befor Last in Building 1 FloorGet Total Building Floor 25 FloorRoof Area 250Site Plan Area 800Total Land Area 1050
For residential :No. of bed room in the building : 100 Bed roomNo. of person in the room : 2 PersonNo of person 200 PersonAssume daily water consumption in L/s/person 180 L/dayThen the total water consumption of building 36
For commercial center:-daily water consumption = total area of all commercial floors / 10 m2 per person * daily water consumption per personAssumption daily consumption per person in liter/day 70 L/Day/PersonArea of total commercial floors 0Daily water consumption 0
For Office Building 8 Hour Shift :-NO. of Person 0 Personassume daily water consumption in L/s/person 60 L/dayDaily water consumption 0
For Factories 7 Hour Shift :-NO. of Person 0 Personassume daily water consumption in L/s/person 60 L/dayDaily water consumption 0
For Hotels up to 3 Stars :-NO. of room 0 roomAssume daily water consumption in L/room 170 L/roomDaily water consumption 0
For Hotels up to 4 Stars With All Services :-NO. of rooms 0 roomAssume daily water consumption in L/room 700 L/roomDaily water consumption 0
For Hotels up to 5 Stars With All Services :-NO. of rooms 0 room
m2
m2
m2
m3
m2
m3
m3
m3
m3
m3
Assume daily water consumption in L/room 1200 L/roomDaily water consumption 0
For Car Parking :-NO. of cars 0 carAssume daily water consumption in L/car 30 L/carDaily water consumption 0
For Restaurant and Cafes Per Meal :-NO. of Person 0 PersonAssumed no. of meal per person 1 mealAssume daily water consumption in L/s/meal 30 L/mealDaily water consumption 0
For Laundry in Hotels Per Bed :-NO. of Bed 0 BedAssume daily water consumption in L/bed 130 L/bedDaily water consumption 0
For Laundry in Hospital Per Bed :-NO. of Bed 0 BedAssume daily water consumption in L/bed 600 L/bedDaily water consumption 0
For School :-NO. of Student 0 StudentAssume daily water consumption in L/student 75 L/student/dayDaily water consumption 0
For Airport :-NO. of Passenger Per Day 0 PassengerAssume daily water consumption in L/passenger 20 L/passenger/dayDaily water consumption 0
For Meeting Area :-NO. of person 0 PersonAssume daily water consumption in L/person 10 L/person/dayDaily water consumption 0
For Public Building :-NO. of person 0 PersonAssume daily water consumption in L/person 50 L/person/dayDaily water consumption 0
For Camps :-NO. of person 0 PersonAssume daily water consumption in L/person 75 L/person/dayDaily water consumption 0
For Swimming Pools & Shore :-NO. of person 0 Person
m3
m3
m3
m3
m3
m3
m3
m3
m3
m3
Assume daily water consumption in L/person 40 L/person/dayDaily water consumption 0 m3
For Mosque :-NO. of person 0 PersonAssume daily water consumption in L/person 30 L/person/dayDaily water consumption 0
For Public Toilet Per Unit :-NO. of toilet Unit 0 UnitAssume daily water consumption in L/toilet 120 L/toiletDaily water consumption 0
For Public Urinal :-NO. of urinal 0 urinalAssume daily water consumption in L/urinal 40 L/urinalDaily water consumption 0
For Public Lavatory :-NO. of Lavatory 0 LavatoryAssume daily water consumption in L/Lavatory 60 L/lavatoryDaily water consumption 0
For Public Shower :-NO. of showers 0 LavatoryAssume daily water consumption in L/shower 560 L/showerDaily water consumption 0
For Public Slaughterhouse :-NO. of Caw 0 CawAssume daily water consumption in L/caw 400 L/cawDaily water consumption 0
For Commercial Laundry Per Kg of Clothes :-NO. of Kg 0 kgAssume daily water consumption in L/kg 35 L/kgDaily water consumption 0
For Farms :-No. of Caws 0 cawNo. of Calf 0 calfNo. of goats & sheep 0 goat & sheepNo. of horses 0 horsesNo. of egg Chicken 0 ChickenNo. of turkey Chicken 0 ChickenNo. of ducks 0 duckNo. of camel 0 CamelDaily water consumption 0
m3
m3
m3
m3
m3
m3
m3
m3
Irrigation water consumption calculations water consumption for irrigation changed according to kind of planters existsNo. of palm trees 1 UnitNo. of spreading 1 UnitNo of evergreen 1 UnitNo. of Shrubs 1 UnitNo. of Ordinary Trees 1 UnitArea of grass 1Area of Gr.Cover 1get daily water consumption for irrigation 0.33OR Insert Total Green Area for The Building 350
Daily water consumption 5.25
Total water consumption :-water consumption per day for the building 36
m2
m2
m3
m2
Assume 1.5 cm of water per 1 m2
m3
m3
Daily Domestic Water Consumption :-water consumption per day for the building 36
Daily Irrigation Water Consumption:-water consumption per day for the building 5.3
Total Daily Water Consumption For the Building :-total consumption of water per day 41.3muliplay the water consumption in 10% safety factor 45.5
Under Ground or Ground Water Tank the storage capacity for 2 days 91
Roof Water Tank Capacity the storage will be for 1 day 45.5
m3
m3
m3
m3
m3
m3
We have 2 Possible Way to Calculate Sizing of Septic Tank
Proposed No.1
Assumption total possible drain = daily water consumption
From water consumption get daily water consumption / day = 36Assumption diversity factor
Get Waste drain flow per day 36 X 0.7 = 25.21 = 6702 Gallon
= 25200 LitersAccording to Qatar Public Work Authority Drainage AfairsSeptic tank Type Type F
Proposed No.2
According to Uniform Plumbing Code Table K-3 Page 337
Type of Occupancy-01
Type of Occupancy-02
Type of Occupancy-03
Type of Occupancy-04
Total GPD
Dance Halls 5 100 0 100 0 100 0 100 500
Total GPD
Airport 15 100 5 100 0 100 0 100 2000
Total GPD
Airport 15 100 5 100 0 100 0 100 2000
Total GPD
Factories 25 50 35 100 5 100 0 300 5250
Get Total Capacity for Septic Tank From 4 Diferent Occupancy 9750 GPD
Recommended Design Criteria. Sewage disposal systems sized using the estimated waste/sewage flow rates should be calculated as follows:(1) Waste/sewage flow, up to 1500 gallons/day (5677.5 L/day)
Flow x 1.5 = septic tank size(2) Waste/sewage flow, over 1500 gallons/day (5677.5 L/day)
Flow x 0.75 + 1125 = septic tank size
Septic Tank Capacity for 1 day = 8437.5 GPD = 31944.375 Liters = 31.944375
According to Qatar Public Work Authority Drainage AfairsSeptic tank Type Type F
m3
m3
Occupancy Type
Gallons / person
No. OF OCCUP.
No. OF OCCUP.
No. OF OCCUP.
No. OF OCCUP.
Occupancy Type
GPD/ Employee
No. OF OCCUP.
GPD/ Passenger
No. OF OCCUP.
No. OF OCCUP.
No. OF OCCUP.
Occupancy Type
GPD/ Employee
No. OF OCCUP.
GPD/ Passenger
No. OF OCCUP.
No. OF OCCUP.
No. OF OCCUP.
Occupancy Type
No showers/ employee
No. OF OCCUP.
With showers/
employee
No. OF OCCUP.
Cafeteria, add /
employee
No. OF OCCUP.
No. OF OCCUP.
m3
Storm Water Drainage Calculation Sheet
150
1620
Diferent Rainfall Rate per HR (Sq. ft)
Horizontal Pipes Slop
According to National Plumbing CodeSize of Vertical Conductors and Leaders
Diameter of vertical pipe or leaders 2 Inch.Get Maximum GPM in Conductors 23 GPM
Diameter of Horizontal pipe 3 Inch.Get Maximum GPM in Conductors 68 GPM
Diameter of roof Gutters drain 8 InchMaximum GPM for Gutters 83 GPMNO. of Gutters Required 1 GUILTERS
In Case of Actual Design
Actual Gutter No. 2 GUILTERSActual Gutter Flow 41.5 GPMDiameter of Gutter Required 7 Inch
Actual Gutters Diameter 6 Inchmaximum Flow for Actual Gutters 40 GPMGutter No. 3 GUILTERS
Storm Water Holding Tank250
800
Total Plan Area 1050Rainfall Rate per HR (Sq. ft) 1InchGet total Volume of Rain Water Holding Tank = 26.7For tank volume equal to 2 daysGet total Volume of Rain Water Holding Tank = 53.4
Roof Floor Area in m2 m2
Get Roof Area in ft2 ft2
From Plan we have Roof Area in m2 m2
Site Plan Area in m2 m2
m2
m3/day
m3
Table of Total FixturesENTER NO. OF UNIT ACCORDING TO FIXTURE TYPE PRIVATE OR PUBLIC
According of our bulding Specification we have = 2 Basement
According of our bulding Specification we have = 15 Non. Typical Floor
According of our bulding Specification we have = 5 Typical Floor
According of our bulding Specification we have = 1 Last Floor
According of our bulding Specification we have = 1 Before Last Floor
According of our bulding Specification we have = 1 2 Level befor Last Floor
According of our bulding Specification we have = 25 Total Floor
Fixture type : private
FIXTURES / POINT NO. B3 B2 B1 G.F. F.F 2nd.F 3rd F 4th F. 5th F 6th F TYP. F Last Floor
Water closet (flush valve) 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Water closet (flush tank) 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Water Bidet 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Hose for W.C 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Lavatory (& H.B.) 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Kitchen sink 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Service sink office,ect 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Shower 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Bath tub 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Dishwashing machine 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Laundry 8 Ib 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Laundry 16 Ib 1 2 3 4 5 6 0 0 0 0 0 0 0 2 23
Ablution (Public) 1 2 3 4 5 6 0 0 0 0 0 0 0 1 22
Urinal (1" flush valve) 1 2 3 4 5 6 0 0 0 0 0 0 0 1 22
Urinal (3/4" flush valve) 1 2 3 4 5 6 0 0 0 0 0 0 0 1 22
Urinal (flush tank) 1 2 3 4 5 6 0 0 0 0 0 0 0 1 22
Drinking fountain 3/8 in 1 2 3 4 5 6 0 0 0 0 0 0 0 1 22
Sum of Total Fixture in All Building Floor 386
Fixture type : public
FIXTURES / POINT NO. B3 B2 B1 G.F. F.F 2nd.F 3rd F 4th F. 5th F 6th F TYP. F Last Floor
Water closet (flush valve) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Water closet (flush tank) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Water Bidet 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Hose for W.C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Lavatory (& H.B.) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Kitchen sink 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Service sink office,ect 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Shower 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Bath tub 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Dishwashing machine 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Laundry 8 Ib 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Laundry 16 Ib 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Ablution (Public) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Urinal (1" flush valve) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Urinal (3/4" flush valve) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Urinal (flush tank) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Drinking fountain 3/8 in 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Sum of Total Fixture in All Building Floor 0
2 level b.last
1 level b.last
Total Fixtures
2 level b.last
1 level b.last
Total Fixtures
Pump Calculation Sheet :- According to International Plumbing Code
Type of Occupancy
Type of Water Heater
Type Of Flush
Proposed Location Of Water Filter
1- Lift Pump Calculations Total length of water pipe for lift pump from ground to roof water tanksVertical Length from Pump To Filling Point In Roof Water Tank
Pump Head = 1.20*(static head + static/10 + total pipe length from pump to roof tank*4/100+5 m losses in filter if it`s befor tank + 5 m losses in float valve
Get Pump Head = 1.20 * ( 10 + 10 /10 + 100 *4/100 + 0 + 5 =
According to Fixture Unit Method From International Plumbing Code We have Total Building Fixture Unit =No .of Pumps Will Be Used In Lifting 2 Duty + 1 Standby
Pump Flow 98 GPMPump Head 35 PSIPump HP 3.3 HPNo. of Pumps 2 Duty + 1 Standby
2- Boster Pump Calculations
No .of Pumps Will Be Used for Boosting 2 Duty + 1 StandbyFrom National Plumping Code Get Total Fixture Unit load =
Pump Flow 17 GPMPump Head 30 PSIPump HP 0.5 HPNo. of Pumps 2 Duty + 1 Standby
Pressure Vessel
From above we get boster pump flow ==
& Estimated pressure for booster pump =it is usual to arrange for pumps to have an ON-OFF cycle of 3-4 min Pump Capacity should be such that peak water demand is not likely to exceed half the averagepump out put PVO = 0.5 * APO X time of pump run in seconds
Pump run time 240
PVO. ( pressure vessel output) = 0.5 * 0.66938 * 300 =
from boyles low V1 = V1-V2/(1-(P1/P2)where
P1 = Cut in pressure head = 14.0136 meter 20 P1 = Cut out pressure head = 24.5238 meter 35
Get V1
V1 = = 234.281For Working Volume Say 20%
V1 = = 282
Volume of Vessel 300 Liters
Notes:-- In Case of Individual Water Heater Pump Will Use To Deliver (Hot & Cold) Water for last
3 floors only and by gravity for the rest of the building- In Case of Centeral Water Heater Pump Will Use To Deliver (Hot & Cold) Water for Last
3 Floors & Hot Water to The Rest of Building
P1 . V1 = P2 . V2
m3
m3
100 m10 m
24 m
386 UnitStandby
Standby62.25
17 GPM1.071 L/S
21.122449 m
100.40625 Liters
PsiPsi
Pressure Pump Calculation Pressure pump will used in case on suppling the building from main network directly or from ground or underground water tank as there is no ablity to make booster pump and roof elevated tank
The Calculation based on fixture unit load from fixture unit load of the building get
Type of Water HeaterType Of Flush
No .of Pumps Will Be Used for Boosting 2 Duty 2From National Plumping Code Get Total Fixture Unit load = 823.5
Total Distance Between Pump to Farther Fixture Unit at Last FloorVertical Length from Pump To Farther Fixture Unit at Last Floor
Pump Head = 1.20*(static head + static/10 + total pipe length from pump to roof tank*4/100+5 m losses in filter + 5 m loss in water heater if centeral + 14 m residual pressure at last fixture unit
= 1.2 X ( 20 + 20 / 10 + 4 /100* 55 + 5 + 5= 57.84 meter head
Pump Flow 186.82 94 GPMPump Head 83 PSIPump HP 8 HPNo. of Pumps 2 DUTY 2 STANDBY
Note It is better to use 3 pumps 1 duty and 2 standby satisfy the full load of systemIn Case only for Governement network supply you can reduce pump head for pump by value of gov. network pressure (about 1.5 bar or 20 psi)
Calculation for Hydropneumatic Tank System
Assuming the maximum no. of pumping cycles be six per hour 5 min an & 5 min ofFrom Perevious Calculation for Pump Capacity Equal to = 941/2 of Pump Capacity = 47For 5 min working = 235
then 235 Gallons should equal 25 per cent withdrawal.100 percent then equal to 235 X 4 = 940
From Calculation get pump Head Equal to = 83Assuming Low working pressure = 80Then get maximum working pressure = 100
Water Capacity . Percent of Total Tank CapacityFrom Calculation Low & high Pressure Between
Water Capacity , Percent Of Total Tank Capacity = 40Get Tank Capacity = 940Get Water Capacity in the tank = 376Get Percent of Water Drawal To Start Pump = 16Get The Volume of Water Drawal To Start Pump = 150
Pump Flow 183.7 94Pump Head 103Pump HP 10No. of Pumps 2 DUTY 2Hydropneumatic Tank 940Water Capacity in the tank 376Volume of Water Drawal To Start Pump 150
Note :-it is required to increase pump head to high pressure of hydropneumatic Tank
The Calculation based on fixture unit load from fixture unit load of the building get
StandbyFixture unit
55 m20 m
+ 14
It is better to use 3 pumps 1 duty and 2 standby satisfy the full load of systemIn Case only for Governement network supply you can reduce pump head for pump
GPMGPM
Gallon
Gallon Tank
PsiPsiPsi
%GallonGallon
%Gallon
GPMPSIHP
STANDBYGallonGallonGallon
it is required to increase pump head to high pressure of hydropneumatic Tank
Submersible Pumps will be designed according to fixture unit method According to our fixture unit we got :-
Type Of Flush
The receiving basin will collect sanitary from
The receiving basin will collect sanitary from
The receiving basin will collect sanitary from
The receiving basin will collect sanitary from
The receiving basin will collect sanitary from
The receiving basin will collect sanitary from
3RD BASEMENT GROUND FLOOR Fixture unit Private Public Private Public Private Public Private Public Private Public PrivateWater closet (flush valve) 1 1 4 0 0 0 0 0 0 0 0Water closet (flush tank) 1 1 4 0 0 0 0 0 0 0 0Water Bidet 1 1 4 0 0 0 0 0 0 0 0Hose for W.C 1 1 4 0 0 0 0 0 0 0 0Lavatory (& H.B.) 1 1 4 0 0 0 0 0 0 0 0Kitchen sink 1 1 4 0 0 0 0 0 0 0 0Service sink office,ect 1 1 4 0 0 0 0 0 0 0 0Shower 1 1 4 0 0 0 0 0 0 0 0Bath tub 1 1 4 0 0 0 0 0 0 0 0Dishwashing machine 1 1 4 0 0 0 0 0 0 0 0Laundry 8 Ib 1 1 4 0 0 0 0 0 0 0 0Laundry 16 Ib 1 1 4 0 0 0 0 0 0 0 0Ablution (Public) 1 1 4 0 0 0 0 0 0 0 0Urinal (1" flush valve) 1 1 4 0 0 0 0 0 0 0 0Urinal (3/4" flush valve) 1 1 4 0 0 0 0 0 0 0 0Urinal (flush tank) 1 1 4 0 0 0 0 0 0 0 0Drinking fountain 3/8 in 1 1 4 0 0 0 0 0 0 0 0
From Fixture Unit Loads get Pump Head = 233.4 GPM
The distance between low water line in the receiver up to the highest point pumped 5 MeterTotal distance for lifting line from lowest level to pumped level + static head 15 Meter
Get required pump head = 1.20*(static head + static/10 + total pipe length *4/100)= 1.2 X ( 5 + 5 / 10 + 4 / 100 X 15 ) =
= 7.4 Meter
Pump Flow 233.4 GPMPump Head 11 PSIPump HP 2.5 HP
Note--
Ejector Basin Sizing Time Required to Empty the Sum pit Minimum Capacity of Storage = Pump Capacity X Time Required to empty sum pit
= 233.4 X 10 = 2334 GallonAssumed using of delux system the standard diameter of tank will be 4 ft or 1.20 m
= 2334 / ( 0.7854 X 4^2 X 7.5 ) = 24.7929 ft distance between high level & low level = 7.55686 m
Adding :-Invert level of the entering pipe 0.9 meter ( Below finish floor level )Low level of water must be a minimium of 0.15 cm take it = 0.2 meter
Getting receving basin depth = Distance between low & high level + invert level of entering pipe + suction level= 7.55686 + 0.9 + 0.2 = 8.65686 meter
Getting Capacity of Storage = 2334 GallonGetting Receving Basin Depth = = 8.7 MeterReceving Basin Diameter = 1.2 MeterReceiving Basin Area = 9.8Diameter of lifting pipe Assumed velocity 1 m/sec 136.909 150 mm
Sump Basin SizingTime Required to Empty the Sum pit
Area of Sandy soil = 200 Equal 2160
Get 43.2 GPM
Area of clay soil = 200 Equal 2160
Get 21.6 GPM
Area of Paved soil = 250 Equal 2700
Get 112.5 GPM As 4 in per hour rainfall
Get total Capacity = 177.3 GPM
Minimum Capacity of Storage = Pump Capacity X Time Required to empty sum pit = 177.3 X 10 = 1773 Gallon
Assumed using of delux system the standard diameter of tank will be 4 ft or 1.20 m
= 1773 / ( 0.7854 X 4^2 X 7.5 ) = 18.8337 ft distance between high level & low level = 5.7405 m
The distance between low water line in the receiver up to the highest point pumped 10 MeterTotal distance for lifting line from lowest level to pumped level + static head 25 Meter
Get required pump head = 1.20*(static head + static/10 + total pipe length *4/100)= 1.2 X ( 10 + 10 / 10 + 4 / 100 X 25 ) =
= 14.4 Meter
Pump Flow 177.3 GPMPump Head 21 PSIPump HP 3.6 HP
Invert level of the entering pipe 0.9 meter ( Below finish floor level )Low level of water must be a minimium of 0.15 cm take it = 0.2 meter
Getting receving basin depth = Distance between low & high level + invert level of entering pipe + suction level= 5.7405 + 0.9 + 0.2 = 6.8405 meter
Getting Capacity of Storage = 1773 GallonGetting Receving Basin Depth = = 6.9 MeterReceving Basin Diameter = 1.2 MeterReceiving Basin Area = 7.8Diameter of lifting pipe Assumed velocity 1 m/sec 119.326 150 mm
when the receiving basin collects the discharge sanitary wastes it is called an Ejector Pumpwhen the receiving basin collects storm water or other clear water it is called a Sump Pump
Getting the depth of the basin = Storage Capacity / ( 0.7854 * D2 * 7.5 )
M2
m2 ft2
Assumed 2 gpm for each 100 ft2
m2 ft2
Assumed 1 gpm for each 100 ft2
m2 ft2
Assumed 1 gpm for each 24 ft2
Getting the depth of the basin = Storage Capacity / ( 0.7854 * D2 * 7.5 )
M2
TotalPublic Private Public
0 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 10 5 1
distance between high level & low level
distance between high level & low level
Centeral Hot Water Calculation Sheet
Calculation Will be According to Type of The Building & Fixture Unit Methods:-
Choose the Type of Your Building
Fixture QTY Demand Total Gallon/Hr
private lav. 2 2 4public lav. 2 8 16bathtub 0 30 0dishwasher 1 0 0foot basin 2 12 24kitchen sink 0 0 0laundry 0 0 0pantry sink 0 0 0shower 0 225 0service sink 0 0 0hydrotherapentic shower 0 0 0hubbard bath 0 0 0leg bath 0 0 0arm bath 0 0 0sitz bath 0 0 0continuous flow bath 0 0 0circular wash sink 0 0 0semicircular wash sink 0 0 0
44DEMAND FACTOR 0.4Probable Maximum Demand 17.6Heater or coil Capacity 17.6STORAGE CAPACITY Factor 1Storage Tank Capacity GALLONS 17.6Storage Tank Capacity Liters 67Heater Power 1.7673513514Heater Storage Capacity in Liters 80Heater Power in Kw 1.75
Water Heater Selection :-Heater Storage Capacity in Liters =### 80 LitersHeater Power in Kw =### 1.75 Kw
Centeral Hot Water Calculation Sheet
Calculation Will be According to Type of The Building & Fixture Unit Methods:-
Choose the Type of Your Building
Type of Pipes will be used
Type of Pipes will be used
In Case of insulated pipe u don’t need to choise type of pipe used
Fixture QTY Total Gallon/Hr
private lav. 10 2 20public lav. 10 6 60bathtub 30 20 600dishwasher 10 100 1000foot basin 20 3 60kitchen sink 2 20 40laundry 15 28 420pantry sink 20 10 200shower 30 75 2250service sink 2 20 40hydrotherapentic shower 0 400 0hubbard bath 0 600 0leg bath 0 100 0arm bath 0 35 0sitz bath 0 30 0continuous flow bath 0 165 0circular wash sink 0 20 0semicircular wash sink 0 10 0
4690DEMAND FACTOR 0.25Probable Maximum Demand 1172.5Heater or coil Capacity 1172.5STORAGE CAPACITY Factor 0.6Storage Tank Capacity GALLONS 703.5
Heat loss in Hot Water Pipes :-
Nominal pipe size
1/2" 100 35 11480
3/4" 50 43 7052
1" 30 53 5215.2
1-1/4" 100 65 21320
1-/2" 100 73 23944
2" 50 91 14924
2-1/2" 70 108 24796.8
3" 0 129 0
4" 0 163 0
5" 0 199 0
6" 0 233 0
8" 0 299 0
Total Qloss for Hot Water Network 108732
Get Total heat loss in the system = 108732 Btu/hr
For each 10,000 BTU/HR Need 1 GPM Required GPM for heat loss make up = Total heat loss in the system / 10,000
= 10.9 GPM
Total Farther Pipe length Supply & Return = 100 mto get pump head the only loss we have in the line friction loss in all pipes and fitting get pump head = 1.5*(friction loss)
= 1.5 * ( 4/100 * Total pipe length )= 4.8 meter head
6.8 Psi
Pipe length in meter Q loss/Ft Total Qloss
BTU/HR
Conclusion for Centeral Water System
Heat or coil Capacity 1172.5 Gallon / Hour
Storage Tank Capacity 703.5 Gallon
Circulation Pump Selection : -
Pump Flow 10.9 GPMPump Head 6.8 PSIPump HP 0.1 HP
Note :- You do not need to chose type of pipe in case of insulated pipe
Calculation for Pressure Reducing Station
The Most Popular Way to Choose The No. of PRV. Station and locations :
Most of fixture unit installed need between 8 Psi to 25 Psi as residual pressure at it is inlet.Assuming that the last fixture unit need About =Maximum pressure to keep the velocity less than 1.2 m/sec, or 10 fps =So we have additional pressure in network about =
=
From Previous Data we have the building height equal to ==
Floor Level Height ==
Getting No. of PRV. Installed in Riser =So We Will Get That PRV. Sataion Will Fixed Every =
Note :-Minimum pressure required for flushing water closet varies from 12 Psi to 25 Psiso we can assume 20 Psi as residual pressure is most suitable.
Most of fixture unit installed need between 8 Psi to 25 Psi as residual pressure at it is inlet.20 Psi.70 Psi.50 Psi.
115.5 Feet
116 Meter380.5 Feet
3.65 Meter12 Feet
3 PRV. Unit10 Story
Minimum pressure required for flushing water closet varies from 12 Psi to 25 Psi
Chilled Drinking Water System :-
Type of Building Water Inlet TemperatureRoom Temperature Assumed Circulation Pump HP
No. of Outlet 60 OutletGallon per Outlet per Hour 5 Gal./Outlet/HRBTU Per HR. Per Gallon 167 BTU/GAL.
Get from Above Usage load = 50100 BTU/HR
Length of Pipe According to Pipe Size
Length of 1/2" Pipe Size in Meter = 0 m 0 ftLength of 3/4" Pipe Size in Meter = 0 m 0 ftLength of 1" Pipe Size in Meter = 182.9 m 600 ftLength of 1.25" Pipe Size in Meter = 60.98 m 200 ftLength of 1.5" Pipe Size in Meter = 30.49 m 100 ftLength of 2" Pipe Size in Meter = 0 m 0 ftLength of 2.5" Pipe Size in Meter = 0 m 0 ftLength of 3" Pipe Size in Meter = 0 m 0 ft
Circulation Losses in Pipes
1/2" Pipe Size in Meter = 0 X 390 = 0 BTU/HR3/4" Pipe Size in Meter = 0 X 420 = 0 BTU/HR1" Pipe Size in Meter = 600 X 490 = 2940 BTU/HR1.25" Pipe Size in Meter = 200 X 550 = 1100 BTU/HR1.5" Pipe Size in Meter = 100 X 610 = 610 BTU/HR2" Pipe Size in Meter = 0 X 700 = 0 BTU/HR2.5" Pipe Size in Meter = 0 X 800 = 0 BTU/HR3" Pipe Size in Meter = 0 X 940 = 0 BTU/HR
Circulation Pump Heat Input = 1908 BTU/HR
Total BTU/HR = 56658 BTU/HRAdding 15% as Factor of Safety = 65157 BTU/HR
Circulation Pump Calculation:-
1/2" Pipe Size in Meter = 0 X 11.1 = 0 GPH3/4" Pipe Size in Meter = 0 X 11.8 = 0 GPH1" Pipe Size in Meter = 600 X 12.8 = 77 GPH1.25" Pipe Size in Meter = 200 X 14.6 = 29 GPH1.5" Pipe Size in Meter = 100 X 15.7 = 16 GPH
Total GPH = 122 GPHAdding 20% as Factor of Safety = 146 GPH
= 2.4 GPM
No of. Circuit = 5 CircuitAssume 3 gpm per CircuitGet total GPM = 15 GPM
So get Circulation Pump Capacity = 15 GPM
Circulation Pump Head
Total Farther Pipe length Supply & Return = 100 Mto get pump head the only loss we have in the line friction loss in all pipes and fitting get pump head = 1.2*(friction loss)
= 1.5 * ( 4/100 * Total pipe length )= 4.8 meter head
6.8 Psi
BTU = GALLON TO BE COOLED X TEMPERATURE DIFFERENCE 65157 = GALLON TO BE COOLED X ( 65 - 45 ) X GALLON TO BE COOLED = 392.5 Gallon
Assume 50% of Water Will Return Back to Chiller And 50% Will Come From Make up Water Make Up Connection Should Be Between Pump and Cooling Unit
Storage Capacity Will be = 196.3 Gallon
Chiller Cooling load 183.7 65157 Btu/hrStorage Capacity 196 GallonNo. of Pumps 1 Duty 1 StandbyCirculation Pump Capacity 15 GpmCirculation Pump Head 6.8 PsiCirculation Pump HP 0.1 HP
X 8.38.3
Irrigation Pump Selection
Calculation of manual System
Insert No. of Irrigation Pump Will be Used 1 Duty 1Enter Total Time to Finish Manual Irrigation 2 HoursPump Capacity 2.65 m3/hrEnter Diference in Head Between Pump & Irrigation Line 0.3 mEnter Total Lengh Of Irrigation Pipe 150 m
Pump Head = 1.20*(static head + static/10 + total pipe length from pump to farther irrigation point *4/100+3 bar residual pressure at farther point
= 64 psi Irrigation Pump Selection :-
Pump Capacity 11.7 GPMPump Head 64 PsiPump HP 0.73 HPPumps 1 Duty 1 StandBy
Pressure Vessel
From above we get pump flow = 11.7 GPM= 0.7361112 L/Sec
& Estimated pressure for boster pump = 64 PSI
it is usual to arrange for pumps to have an ON-OFF cycle of 3-4 min Pump Capacity should be such that peak water demand is not likely to exceed half the averagepump out put PVO = 0.5 * APO X time of pump run in secondsPump run time
PVO. ( pressure vessel output) = 0.5 * 0.4600694813 X 300= 69 Liters
from boyles low V1 = V1-V2/(1-(P1/P2)where
P1 = Cut in pressure head = 35.1 m P1 = Cut out pressure head = 45.1 m
Get V1V1 = 312 Liters
For Working Volume Say 20% V1 = 375 Liters
Volume of Vessel 400 Liters
P1 . V1 = P2 . V2
StandBy
Pump Capacity should be such that peak water demand is not likely to exceed half the average
V1 = V1-V2/(1-(P1/P2)
Liters
Automatic Irrigation System Calculation
Irrigation Water Demand Requirement
Plant Description Area / Quantity Watering Rate LPDShrubs 431.47 16 liters/ day/m2 6903.52
Turf or Grass 1872 12 liters/ day/m2 22464Ground Cover 1160 12 liters/ day/m2 13920
Trees 115 No. 120 liters/ day/m2 13800Palms 8 No. 150 liters/ day/m2 1200
Total Watering Rate 58287.52Adding 2% as Factor Of Safety 59453.27
From Perivious Data Get Storage tank for 1 day = 59.5Storage tank for 2 days = 119
m2
m2
m2
m3
m3
Over Flow Swimming Pool Calculation Sheet
Surface Area of Swimming Pool = 38Pool average depth = 1.7 mGot Pool Volume = 64.6
According To Egyptian Swimming Pool Code for every 25 m2 surface area need 1 inletNO. Water Inlets = 1.52 inletsSo Get the No. Of inlets = 2 inlets
According to Type of Swimming Pool Turn Over Period Will be Selected
Swimming Pool Type
Enter No. of Turn Over Period = 0.25 Time / Day
Get Swimming Pool Pump Capacity = Pool Volume / ( Turn Over Period * 60 ) = 4.31
No. Of pump will used = 1 Pumpno. Of standby pumps = 1 Pump
Get pump flow rate = 4.31
Balance tank volume = 0.01 X pool surface area
= 45
Balance tank volume = 45
To calculate pipe sizing All Folwing Equation According to SPATA Swimming Pool Manule Design
Supply pipe maximum velocity will be 2 m/sNo. Of inlet = 2 InletPump flow rate = 258.6
Each inlet give flow rate = 129
Dia 1 = 151.23451 Taking it = 150 mm.Dia 2 = 213.8779 Taking it = 150 mm.Dia 3 = 261.94586 Taking it = 150 mm.Dia 4 = 302.46902 Taking it = 150 mm.Dia 5 = 338.17064 Taking it = 150 mm.Dia 6 = 370.44738 Taking it = 150 mm.Dia 7 = 400.1289 Taking it = 150 mm.Dia 8 = 427.75579 Taking it = 150 mm.Dia 9 = 453.70353 Taking it = 150 mm.Dia 10 = 478.24551 Taking it = 150 mm.Dia 11 = 501.58812 Taking it = 150 mm.Dia 12 = 523.89171 Taking it = 150 mm.
NOTE :-No. of pipe will be taken according to No. of inlet for 4 inlet in series take to Dia 4
m2
m3
m3/min
m3/min
40 L/m2 X Pool Surface area + 10 min X pump flow rate +
m3
m3
m3/hr
m3/hr
Suction pipe maximum velocity will be 1.2 m/sPump flow rate = 258.6No. Of outlet = 4 Outlet
Each outlet give flow rate = 64.65
Dia 1 = 136.68888 Taking it = 150 mm.Dia 2 = 193.30727 Taking it = 150 mm.Dia 3 = 236.75209 Taking it = 150 mm.Dia 4 = 273.37776 Taking it = 150 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Over flow pipe Grille velocity will be 0.3m/sNo. Of Grille = 10 GrillesPump Flow Rate = 258.6
Each outlet give flow rate = 25.86
Dia 1 = 94.077548 Taking it = 100 mm.Dia 2 = 133.04574 Taking it = 150 mm.Dia 3 = 162.94709 Taking it = 150 mm.Dia 4 = 188.1551 Taking it = 150 mm.Dia 5 = 210.36379 Taking it = 150 mm.Dia 6 = 230.44199 Taking it = 150 mm.Dia 7 = 248.9058 Taking it = 150 mm.Dia 8 = 266.09149 Taking it = 150 mm.Dia 9 = 282.23264 Taking it = 150 mm.Dia 10 = 297.49933 Taking it = 150 mm.Dia 11 = 312.01993 Taking it = 150 mm.Dia 12 = 325.89419 Taking it = 150 mm.
NOTE :-No. of pipe will be taken according to No. of over flow unit for 4 unit take to Dia 4
Getting the filter area to be used
Pump flow rate = 258.6
Filter area = 7.4Assume we have to use a filter with diameter = 130 cmNo.of filter required 7.2513474 7 Filter
Heating Calculation for swimming pools
Heater is required for heating purposes at winter seasons if required take inconsideration that it's not particle to heat the pool without covers at coldest winter monthsFor calculation purposes the following conditions to be considered:-The recommended Pool water temperature is around 28 º C deg.Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C) : 6/ 10 deg º C BD/WB -winter condition From SPATA p5, Sec3, the Average Daily heat loss (DH) in KWhr / m2 of pool surface area
m3/hr
m3/hr
m3/hr
m3/hr
m3/hr
m2 at 15 gpm / ft2
Correction Value Will be Changed According to type of Swimming Pool And it`s Location
Swimming Pool Type and Location
correction value to be used i = 1.25
hours per day available for heatingNo. of hours per day available for heating = 96 hours
= Heater required Pool Volume * 1000 * ( Tc2 - Tc1) / ( 860 * Avaliable Heating Hour) = 64.6* 1000 ( 28 - 10 ) / 860 * 96 = 17.7 Kw
Cooling Calculation for Swimming Pools
For cooling purpose in summer seasons chilled water is required to cooling the water temperaturein summer to recommended temperatures The recommended Pool water temperature is around 28 deg º COutdoor air condition : 46 / 30 deg º C DB/ WB - summer condition , Max 122 deg º F
: 10 / 06 deg º C DB/ WB - winter condition During summer time the outdoor air temperature will be around 45 deg º C and hence the watertemperature will be the sameSo to estimate the cooling load required for the chiller to reduce water temperature to 28 deg º Cwill be two part as following
hours per day available for coolingNo. of hours per day available = 72 hours
The first part is the cooling load required to reduce the water temperature to 28 deg º C as following
where :- V= volume of pool Cp = specific heat of water 4.18 (KJ/Kg.C)
64.6 * 4.186 * 1 * ( 40 - 28 ) / ( 3600 * 72 )
= 12.5 KwThe second part is the cooling load required to over come the heat lose from the water surfaces of swimming pool
where :- CF= Correction Factor DH = average daily loss from swimming pool surface
from SPATA tables there is no information about heat loss from pool areas the gulf area condition
Q1 = V * Cp * P * (d T ) / Time Required for cooling
P = specific heat of water = 1 Kg/Ld T = temperature diference
Q1 =
Q2 = CF *DH * Pool surface area / Time Required for cooling
the avaliable data only at 15deg º C out door temp which is = 3.48 Kwh/m2
Assuming linearity condition the excepeted heat loss at 45deg º C = (45/15)x3.48 = 10.44 Kwh/m
1.25 * 38 * 10.44 / 72
= 6.9 Kw
= 12.5 + 6.9= 19.4 Kw= 66251 Btu/hr= 5.5 TR
Pump head calculation for swimming poolsHeater pressure drop = 10 ftFilter pressure drop = 5 ftChlorination auto pilot unit = 2 ftPipes and fittings losses = 7 ftStatic head heat pump = 5 ft
Pump head = 34.8 ftPump HP = 20 HP
Swimming Pool Lighting :For lighting we need about 5.4 watt to 16.30 watt for each 1m2Take the watt required for each 1m2 = 10 WattTotal watt required = 380 Wattfor lamp have capacity of = 300 WattNo. of lamp required = 1 Lamp
Swimming Pool Conglusion
Pump Capacity 1139 GPMPump Head 15 PSIPump HP 20 HPBalance tank V. 45Filter 7 Filter with Diameter 130 cmHeater Capacity 17.7 KwCooling Capacity 5.5 TRLighting 1 Lamp, Each one 300 Watt
Q2 =
So the Total Ciller Capacity will be = Q1 + Q2
m3
Pool Volume / ( Turn Over Period * 60 )
No. of pipe will be taken according to No. of inlet for 4 inlet in series take to Dia 4
X Pool Surface area + 10 min X pump flow rate +
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
No. of pipe will be taken according to No. of over flow unit for 4 unit take to Dia 4
Heater is required for heating purposes at winter seasons if required take inconsideration that
Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C)
at 15 gpm / ft2
For cooling purpose in summer seasons chilled water is required to cooling the water temperature
During summer time the outdoor air temperature will be around 45 deg º C and hence the water
So to estimate the cooling load required for the chiller to reduce water temperature to 28 deg º C
The second part is the cooling load required to over come the heat lose from the water
from SPATA tables there is no information about heat loss from pool areas the gulf area condition
= CF *DH * Pool surface area / Time Required for cooling
Assuming linearity condition the excepeted heat loss at 45deg º C = (45/15)x3.48 = 10.44 Kwh/m2
Skimmer Swimming Pool Calculation Sheet
Surface Area of Swimming Pool = 130Pool average depth = 1.7 mGot Pool Volume = 221
According To Egyptian Swimming Pool Code for every 25 m2 surface area need 1 inletNO. Water Inlets = 5.2 inletsSo Get the No. Of inlets = 5 inlets
According to Type of Swimming Pool Turn Over Period Will be Selected
Swimming Pool Type
Enter No. of Turn Over Period = 6 Time / Day
Get Swimming Pool Pump Capacity = Pool Volume / ( Turn Over Period * 60 ) = 0.61
No. Of pump will used = 1 Pumpno. Of standby pumps = 1 Pump
Get pump flow rate = 0.61
surface water for swimming pool
No. of Skimmer = 3 Skimmer Unit
To calculate pipe sizing All Folwing Equation According to SPATA Swimming Pool Manule Design
Supply pipe maximum velocity will be 2 m/sNo. Of inlet = 5 InletPump flow rate = 36.6
Each inlet give flow rate = 7
Dia 1 = 35.983813 Taking it = 50 mm.Dia 2 = 50.888796 Taking it = 50 mm.Dia 3 = 62.325792 Taking it = 65 mm.Dia 4 = 71.967626 Taking it = 75 mm.Dia 5 = 80.462252 Taking it = 75 mm.Dia 6 = 88.141981 Taking it = 100 mm.Dia 7 = 95.20422 Taking it = 100 mm.Dia 8 = 101.77759 Taking it = 100 mm.Dia 9 = 107.95144 Taking it = 100 mm.Dia 10 = 113.79081 Taking it = 100 mm.Dia 11 = 119.34481 Taking it = 100 mm.Dia 12 = 124.65158 Taking it = 100 mm.
NOTE :-No. of pipe will be taken according to No. of inlet for 4 inlet in series take to Dia 4
m2
m3
m3/min
m3/min
Acoording to egyptian code for swimming pool 1 skimmer for each 46.5 m2 from
m3/hr
m3/hr
Suction pipe maximum velocity will be 1.2 m/sPump flow rate = 36.6No. Of outlet = 4 Outlet
Each outlet give flow rate = 9.15
Dia 1 = 51.423244 Taking it = 50 mm.Dia 2 = 72.723449 Taking it = 75 mm.Dia 3 = 89.067671 Taking it = 100 mm.Dia 4 = 102.84649 Taking it = 100 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Getting the filter area to be used
Pump flow rate = 36.6
Filter area = 1Assume we have to use a filter with diameter = 50 cmNo.of filter required 2.5477707 3 Filter
Heating Calculation for swimming pools
Heater is required for heating purposes at winter seasons if required take inconsideration that it's not particle to heat the pool without covers at coldest winter monthsFor calculation purposes the following conditions to be considered:-The recommended Pool water temperature is around 28 º C deg.Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C) : 6/ 10 deg º C BD/WB -winter condition From SPATA p5, Sec3, the Average Daily heat loss (DH) in KWhr / m2 of pool surface area
Correction Value Will be Changed According to type of Swimming Pool And it`s Location
Swimming Pool Type and Location
correction value to be used i = 1.4
hours per day available for heatingNo. of hours per day available for heating = 24 hours
= Heater required Pool Volume * 1000 * ( Tc2 - Tc1) / ( 860 * Avaliable Heating Hour) = 221* 1000 ( 28 - 10 ) / 860 * 24 = 269.9 Kw
Cooling Calculation for Swimming Pools
For cooling purpose in summer seasons chilled water is required to cooling the water temperaturein summer to recommended temperatures The recommended Pool water temperature is around 28 deg º COutdoor air condition : 46 / 30 deg º C DB/ WB - summer condition , Max 122 deg º F
: 10 / 06 deg º C DB/ WB - winter condition During summer time the outdoor air temperature will be around 45 deg º C and hence the watertemperature will be the sameSo to estimate the cooling load required for the chiller to reduce water temperature to 28 deg º C
m3/hr
m3/hr
m3/hr
m2 at 15 gpm / ft2
will be two part as following
hours per day available for coolingNo. of hours per day available = 72 hours
The first part is the cooling load required to reduce the water temperature to 28 deg º C as following
where :- V= volume of pool Cp = specific heat of water 4.18 (KJ/Kg.C)
221 * 4.186 * 1 * ( 40 - 28 ) / ( 3600 * 72 )
= 42.8 KwThe second part is the cooling load required to over come the heat lose from the water surfaces of swimming pool
where :- CF= Correction Factor DH = average daily loss from swimming pool surface
from SPATA tables there is no information about heat loss from pool areas the gulf area condition
1.4 * 130 * 10.44 / 72
= 26.4 Kw
= 42.8 + 26.4= 69.2 Kw= 236318 Btu/hr= 19.7 TR
Pump head calculation for swimming poolsHeater pressure drop = 10 ftFilter pressure drop = 5 ftChlorination auto pilot unit = 2 ftPipes and fittings losses = 7 ftStatic head heat pump = 5 ftPump head = 34.8 ftPump HP = 2.8 HP
Swimming Pool Lighting :For lighting we need about 5.4 watt to 16.30 watt for each 1m2Take the watt required for each 1m2 = 10 WattTotal watt required = 1300 Watt
Q1 = V * Cp * P * (d T ) / Time Required for cooling
P = specific heat of water = 1 Kg/Ld T = temperature diference
Q1 =
Q2 = CF *DH * Pool surface area / Time Required for cooling
the avaliable data only at 15deg º C out door temp which is = 3.48 Kwh/m2
Assuming linearity condition the excepeted heat loss at 45deg º C = (45/15)x3.48 = 10.44 Kwh/m
Q2 =
So the Total Ciller Capacity will be = Q1 + Q2
for lamp have capacity of = 50 WattNo. of lamp required = 26 Lamp
Swimming Pool Conglusion
Pump Capacity 161 GPMPump Head 15 PSIPump HP 2.8 HPNo. of Skimmer 3 UnitFilter 3 Filter with Diameter 50 cmHeater Capacity 269.9 KwCooling Capacity 19.7 TRLighting 26 Lamp, Each one 50 Watt
Pool Volume / ( Turn Over Period * 60 )
No. of pipe will be taken according to No. of inlet for 4 inlet in series take to Dia 4
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Heater is required for heating purposes at winter seasons if required take inconsideration that
Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C)
For cooling purpose in summer seasons chilled water is required to cooling the water temperature
During summer time the outdoor air temperature will be around 45 deg º C and hence the water
So to estimate the cooling load required for the chiller to reduce water temperature to 28 deg º C
at 15 gpm / ft2
The second part is the cooling load required to over come the heat lose from the water
from SPATA tables there is no information about heat loss from pool areas the gulf area condition
= CF *DH * Pool surface area / Time Required for cooling
Assuming linearity condition the excepeted heat loss at 45deg º C = (45/15)x3.48 = 10.44 Kwh/m2
Jacuzzi Calculation Sheet
The design of Jacuzzi is same as the design of swimming pool but with small change in it there is Air Blower, Domestic Booster Pump , Skimmer, Jet Nozzle
Surface Area of Jacuzzi = 11Jacuzzi Average Depth = 0.9 mGot Jacuzzi Volume = 9.9
1- Calculation for Drain outlet, Skimmer, Wall inlets, Pipes , Filter, & Filtration or Circulation pumps
Turn Over Period = 0.3 Time / DayJacuzzi Filtration Pump Capacity = Pool Volume / ( Turn Over Period * 60 )
= 0.55
No. Of pump will used = 1 PumpNo. Of standby pumps = 1 Pump
Get pump flow rate = 0.55
Surface Water for Swimming PoolGot No. of Skimmer = 1 Unit
Assuming 50% of water suck from skimmer & 50% will suck from floor drains outletGot Flow Rate will discharge in pipes = 0.275
Suction pipe maximum velocity will be 1.2 m/s from skimmer & drain outletPump flow rate = 16.5No. Of outlet = 2 Outlet
Each outlet give flow rate = 8.25
Dia 1 = 48.828782 Taking it = 50 mm.Dia 2 = 69.054326 Taking it = 65 mm.Dia 3 = 84.573932 Taking it = 75 mm.Dia 4 = 97.657565 Taking it = 100 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
m2
m3
m3/min
m3/min
According to Egyptian code for Swimming Pool 1 Skimmer for each 46.5 m2 from
m3/min
m3/hr
m3/hr
Supply pipe maximum velocity will be 2 m/sPump flow rate = 16.5No. Of inlets = 1 Inlets
Each Inlet give flow rate = 16.5
Dia 1 = 54 Taking it = 50 mm.Dia 2 = 76.4 Taking it = 75 mm.Dia 3 = 93.6 Taking it = 100 mm.Dia 4 = 108 Taking it = 100 mm.Dia 5 = 120.8 Taking it = 100 mm.Dia 6 = 132.3 Taking it = 150 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Getting the filter area to be used Pump flow rate = 33
Filter area = 0.9Assume we have to use a filter with diameter = 50 cmNo. of filter required = 2 Filter
Heating Calculation for Swimming Pools
Heater is required for heating purposes at winter seasons if required take inconsideration that it's not particle to heat the pool without covers at coldest winter monthsFor calculation purposes the following conditions to be considered:-The recommended Pool water temperature is around 28 º C deg.Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C) : 6/ 10 deg º C BD/WB -winter condition From SPATA p5, Sec3, the Average Daily heat loss (DH) in KWhr / m2 of pool surface area
Correction Value Will be Changed According to type of Swimming Pool And it`s Location
Jacuzzi Type and Location
correction value to be used in = 1.25
Hours per day available for heatingNo. of hours per day available for heating = 24 hours
= Heater required Pool Volume * 1000 * ( Tc2 - Tc1) / ( 860 * Available Heating Hour)
= 9.9 * 1000 * ( 28 - 10 ) / 860 * 24 = 10.8 Kw
Cooling Calculation for Swimming Pools
For cooling purpose in summer seasons chilled water is required to cooling the watertemperature in summer to recommended temperatures The recommended Pool water temperature is around 28 deg º COutdoor air condition : 46 / 30 deg º C DB/ WB - summer condition , Max 122 deg º F
: 10 / 06 deg º C DB/ WB - winter condition
m3/hr
m3/hr
m3/hr
m2 at 15 Gpm / ft2
During summer time the outdoor air temperature will be around 45 deg º C and hence thewater temperature will be the sameSo to estimate the cooling load required for the chiller to reduce water temperature to 28 deg º C will be two part as following
hours per day available for coolingNo. of hours per day available = 72 hours
The first part is the cooling load required to reduce the water temperature to 28 deg º C as following
where :- V= volume of pool Cp = specific heat of water 4.18 (KJ/Kg.C)
9.9 * 4.186 * 1000 * ( 40 - 28 ) / ( 3600 * 72 )
= 1.9 KwThe second part is the cooling load required to over come the heat lose from the water surfaces of swimming pool
where :- CF= Correction Factor DH = average daily loss from swimming pool surface
from SPATA tables there is no information about heat loss from pool areas the gulf area
Assuming linearity condition the excepted heat loss at 45deg º C = (45/15) x 3.48 = 10.44 KWhr/m2
1.25 * 11 * 10.44 / 72
= 2 Kw
= 1.9 + 2= 3.9 Kw= 13318.5 Btu/hr= 1.1 TR
Pump head calculation for swimming poolsHeater pressure drop = 10 ftFilter pressure drop = 5 ftChlorination auto pilot unit = 2 ftPipes and fittings losses = 7 ftStatic head heat pump = 5 ftPump head = 34.8 ftPump HP = 2.6 HP
Q1 = V * Cp * P * (d T ) / Time Required for cooling
P = specific heat of water = 1 Kg/Ld T = temperature diference
Q1 =
Q2 = CF *DH * Pool surface area / Time Required for cooling
condition the available data only at 15deg º C out door temp which is = 3.48 KWhr/m2
Q2 =
So the Total Ciller Capacity will be = Q1 + Q2
2- Calculation for Air Blower, Jet Nozzle, Air Nozzle , Pipes , & Booster Pump,
Assuming Booster pump Capacity equal to 80% from Filtration pump CapacityFiltration Pump Capacity = 0.55
No. Of pump will used = 1 Pumpno. Of standby pumps = 1 Pump
Booster Pump Capacity = 0.44
Suction pipe maximum velocity will be 1.2 m/s from skimmer & drain outletPump flow rate = 26.4No. Of outlet Floor Drains = 3 Outlet
Each outlet give flow rate = 8.8
Dia 1 = 50.43015 Taking it = 50 mm.Dia 2 = 71.319002 Taking it = 75 mm.Dia 3 = 87.347582 Taking it = 75 mm.Dia 4 = 100.8603 Taking it = 100 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Jet pipe maximum velocity will be 2 m/sPump flow rate = 26.4No. Of Jet Fitting = 12 InletsEach Jet give flow rate = 2.2
Dia 1 = 19.7 Taking it = 40 mm.Dia 2 = 27.9 Taking it = 40 mm.Dia 3 = 34.2 Taking it = 40 mm.Dia 4 = 39.5 Taking it = 40 mm.Dia 5 = 44.1 Taking it = 40 mm.Dia 6 = 48.3 Taking it = 50 mm.Dia 7 = 52.2 Taking it = 50 mm.Dia 8 = 55.8 Taking it = 50 mm.Dia 9 = 59.2 Taking it = 65 mm.Dia 10 62.4 Taking it = 65 mm.Dia 11 65.4 Taking it = 65 mm.Dia 12 68.3 Taking it = 65 mm.Dia 13 71.1 Taking it = 75 mm.Dia 14 = 73.8 Taking it = 75 mm.Dia 15 76.4 Taking it = 75 mm.Dia 16 78.9 Taking it = 75 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Booster Pump head calculation Pipes and fittings losses = 7 ftStatic head heat pump = 5 ftPump head = 15.6 ft
m3/min
m3/min
m3/hr
m3/hr
m3/hr
m3/hr
Pump HP = 0.9 HP
Air Blower & Air Jet Nozzle Calculation
No. of get nozzels fixed in Ground = 6 Air Nozzel
Assuming each Jet nozzel give flow equal to = 0.3
No. Of duty air blower = 1 BlowerNo. Of standby air blower = 1 Blower
Get Air Blower Capacity = 1.8
Air nozzel pipe sizing
Dia 1 = 10.6 = 1/2 InchDia 2 = 21.2 = 3/4 InchDia 3 = 31.8 = 1 1/4 InchDia 4 = 42.4 = 1 1/4 InchDia 5 = 53 = 1 1/4 InchDia 6 = 63.6 = 1 1/4 InchDia 7 = 74.2 = 1 1/4 InchDia 8 = 84.8 = 1 1/4 InchDia 9 = 95.3 = 1 1/4 InchDia 10 = 105.9 = 1 1/2 Inch
NOTE :-No. of pipe will be taken according to No. of jet nozzel for 2 nozzel in series take to Dia 2
Swimming Pool Lighting :For lighting we need about 5.4 watt to 16.30 watt for each 1m2Take the watt required for each 1m2 = 10 WattTotal watt required = 110 Wattfor lamp have capacity of = 50 WattNo. of lamp required = 2 Lamp
Jaccuzi Pump Room Equipemet
Filteration Pump 145 GPM & 15 PSI & 2.6 HPNo. of Pumps 1 Duty & 1 StandbyNo. of Skimmers 1 Skimmer with minimum flow rate 73 GPMNo. Of Filter 2 Filter with Diameter 50 cmHeater Capacity 10.8 Kw Cooling Capacity 1.1 TR Booster Pump 116 GPM & 7 PSI & 0.9 HPNo. of Pumps 1 Duty & 1 StandbyAir Blower 64 CFM No. of Blower 1 Duty & 1 StandbyLighting 2 Lamp, Each one 50 Watt
m3/min
m3/min
The design of Jacuzzi is same as the design of swimming pool but with small change
Pool Volume / ( Turn Over Period * 60 )
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Heater is required for heating purposes at winter seasons if required take inconsideration that
Outdoor air condition : 46 / 30 deg º C DB/WB –summer condition, max 122deg F (50 C)
9.9 * 1000 * ( 28 - 10 ) / 860 * 24
at 15 Gpm / ft2
The second part is the cooling load required to over come the heat lose from the water
= CF *DH * Pool surface area / Time Required for cooling
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
No. of pipe will be taken according to No. of jet nozzel for 2 nozzel in series take to Dia 2
1 Skimmer with minimum flow rate 73 GPM
Fountain Design
The basic point in design of fountain from catalogue according to the shape of fountain and water height and water through and degree of water
One of Companies working in this field is Fontana , made in Greece Check catalogue in following site
From Catalogue chose the type of Nozzle you want Insert Following Data
Nozzel Model = MC-125Nozzel Flow Rate = 72 L/minNozzel Tip Pressure = 9.8 mMaximum Flow Height = 1 mSize of Inlet Pipe = 1.25 InchQuantity = 1 Unit
No. Of pump will used = 1 PumpNo. Of standby pumps = 1 Pump
Get pump flow rate = 4.32
Assuming flow from drain outlet to vacum point = 50 %
Suction pipe maximum velocity will be 1.2 m/s from skimmer & drain outletPump flow rate = 2.16No. Of outlet = 2 OutletEach outlet give flow rate = 1.08
Dia 1 =17.666918 Taking it = 50 mm.Dia 2 =24.984795 Taking it = 50 mm.Dia 3 = 30.6 Taking it = 50 mm.Dia 4 =35.333836 Taking it = 50 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Vacum pipe maximum velocity will be 1.2 m/s from skimmer & drain outletPump flow rate = 2.16No. Of Vacum Point = 1 Outlet
Each outlet give flow rate = 2.16
Dia 1 =24.984795 Taking it = 40 mm.Dia 2 =35.333836 Taking it = 40 mm.Dia 3 =43.274935 Taking it = 40 mm.Dia 4 =49.969591 Taking it = 50 mm.
m3/hr
m3/hr
m3/hr
m3/hr
m3/hr
CatalogueCatalogue
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Supply pipe maximum velocity will be 2 m/s
Dia 1 = 27.6 Taking it = 40 mm.Dia 2 = 39.1 Taking it = 40 mm.Dia 3 = 47.9 Taking it = 50 mm.Dia 4 = 55.3 Taking it = 50 mm.
NOTE :-No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Getting the filter area to be used Pump flow rate = 4.32
Filter area = 0.1Assume we have to use a filter with diameter = 60 cmNo. of filter required = 1 Filter
Fountain Lighting :For lighting we need about 5.4 watt to 16.30 watt for each 1m1
Surface Area of Fountain Pool = 4Take the watt required for each 1m2 = 15 WattTotal watt required = 60 Wattfor lamp have capacity of = 50 WattNo. of lamp required = 2 Lamp
Pump head calculation for swimming poolsFilter pressure drop = 2 ftChlorination auto pilot unit = 2 ftMaximum distance from farest way to pump = 3 ftStatic head heat pump = 1 ft
Pump head = 44.7 ftPump HP = 6 HP
Filteration Pump 108 GPM & 19 PSI & 6 HPNo. of Pumps 1 Duty & 1 StandbyNo. Of Filter 1 Filter with Diameter 60 cm
NOTE :-you can use Submersible or filteration pump
m3/hr
m2
m2
The basic point in design of fountain from catalogue according to the shape of fountain
Suction pipe maximum velocity will be 1.2 m/s from skimmer & drain outlet
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
Vacum pipe maximum velocity will be 1.2 m/s from skimmer & drain outlet
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
No. of pipe will be taken according to No. of outlet for 2 outlet in series take to Dia 2
at 15 Gpm / ft2
Grease Interceptor Design
There is two method to Calculate the Required Grease Interceptor
Method No.1 :-To determine the correct size grease trap use the dimensions of the waste sinks. Grease traps are diferentiated by their pounds and gallons per minute. Grease Traps may be placed under the sink or they may be buried
Grease No.Dimension in Inch
Length Width Depth
1 12 10 5 1.95 1002 10 10 5 1.625 503 0 0 0 0 254 0 0 0 0 255 0 0 0 0 256 0 0 0 0 257 0 0 0 0 258 0 0 0 0 25
GPM OF ALL OTHER SOURCES Grease GPM Required
Required Grease Interceptor 9 = 9For one -minute sink drain 9 = 9For two -minute sink drain 5 = 5
For Selection From Zurn CalalogueFor one -minute sink drain = GT2700-10, GT2702-10For two -minute sink drain = GT2700-07, GT2702-07
Method No.2 :-This method depends on the no.of fixture unit will be connected to grease interceptor
No. of Fixture unit = 8For 3 GPM For 1 Fixture UnitRequired GPM = 24For one -minute sink drain = 24For two -minute sink drain = 12
For Selection From Zurn CalalogueFor one -minute sink drain = GT2700-25, GT2702-25For two -minute sink drain = GT2700-15, GT2702-15
Note:-After Cetting GPM & Gallon Required go to Catalouge to Select Model Must be connected to internet connection
Calculated GPM
Percentage %
WattsWatts ZurnZurn
To determine the correct size grease trap use the dimensions of the waste sinks.
1.950.8125
00000069
GPMGallonGallon
GT2700-10, GT2702-10GT2700-07, GT2702-07
This method depends on the no.of fixture unit will be connected to grease interceptor
Unit
GPMGallonGallon
GT2700-25, GT2702-25GT2700-15, GT2702-15
After Cetting GPM & Gallon Required go to Catalouge to Select Model
Selected GPM
ZurnZurn