Plumbing System: Water Supply, Water Consumption and Water Demand

Plumbing Definitions

  • Plumbing: Includes water supply, distribution, and drainage within a building.

Water Consumption and Demand

Water Consumption

  • Definition: The amount of water consumed by residents and institutions when provided with water service facilities.

  • Units: Measured in liters/capita/day (lpd).

  • Typical Consumption:

    • Rural areas: 60 lpd (faucets)

    • Individual household connection: 100 lpd

    • Combined public: 80 lpd

    • Suburban areas: 120 lpd

    • Urban areas: 150 lpd

  • Breakdown of Usage (120 lpd):

    • Kitchen: 20% (24 liters)

    • Drinking: 10% (12 liters)

    • Shower: 40% (48 liters)

    • Others (carwash, water gardens): 10% (12 liters)

    • Toilet: 20% (24 liters)

  • Breakdown of Usage (150 lpd):

    • Kitchen: 20% (30 liters)

    • Drinking: 5% (7.50 liters)

    • Shower: 40% (60 liters)

    • Others: 8% (12 liters)

    • Toilet: 27% (40.50 liters)

Other Water Usage by Building Type (liters/capita/day)

  • Airports (per passenger): 11-19

  • Apartments (per resident): 151

  • Boarding Houses (per boarder): 150-189

  • Factories (per person/shift): 57-132

  • Hotels with T&B: 227

  • Hotels w/o T&B (per person): 189

  • Hospitals (per bed): 946-1514

  • Institutional Building (per person): 284-473

  • Large Residential Houses (per resident): 189-284

  • Restaurants:

    • with toilets (per patron): 26-38

    • without toilets (per patron): 9-11

    • with bar/cocktail (add): 8

  • Schools (per student):

    • with gym, canteen, shower: 95

    • w/o gym, canteen, shower: 57

  • Auditorium (per seat): 19

  • Stores (per toilet room): 151

  • Office (per person shift): 57-75

Average Daily Livestock Consumption (liters/day)

  • Each Milk Cow: 133

  • Each Steer or Dry Cow: 45

  • Each Horse: 45

  • Each Hog: 15

  • Each 100 Chickens: 8

Water Demand:

  • Definition: The sum of water consumption and unaccounted for water.

  • WaterDemand=WaterConsumption+UnaccountedforWaterWater Demand = Water Consumption + Unaccounted for Water

  • Unaccounted for Water: Water losses through leakages and pilferages.

  • Ideal percentage of unaccounted water: ≤ 15% of the total water sent to the distribution system.

Computing Water Demand

Data required to Compute Water Demand:

  1. Design population

  2. Design period

  3. Demand factors:

    • Average Day Demand (ADD)

    • Maximum Day Demand (MDD)

    • Maximum Hour (Peak Hour) Demand (MHD)

Design Population

  • Definition: Number of individuals within the area to be served during the design period.

  • Formula: Pp=1.15×PP_p = 1.15 \times P

    • Where:

      • PpP_p = Projected Population at the end of the design period

      • PP = Present Population

      • 1.15 = 3% annual increase x design period of 5 years.

Demand Factors

Average Day Demand (ADD)

  • Definition: Sum of the daily water demands divided by the number of days in that year.

  • Formula: ADD=DesignPopulation×WaterConsumptionRateADD = Design Population \times Water Consumption Rate

  • Use: Used to design the reservoir or storage tank capacity.

Maximum Day Demand (MDD)

  • Definition: The highest water demand occurring during a day, typically during the dry season on a Monday.

  • Formula: MDD=1.30×AverageDayDemandMDD = 1.30 \times Average Day Demand

  • Use: Used to determine the minimum pump capacity (except in hydropneumatic pressure systems).

Maximum Hour Demand (MHD)

  • Definition: The hour of the day when water demand is at its maximum, usually early morning (7-8 AM).

  • Formula:

    • If Design Population < 600 persons: MHD=(3×AverageDayDemand)+TimeMHD = (3 \times Average Day Demand) + Time

    • If Design Population > 600 persons: MHD=(2.5×AverageDayDemand)+TimeMHD = (2.5 \times Average Day Demand) + Time

    • Time = Duration of Usage

  • Uses:

    1. Estimating the diameter of transmission and distribution mains.

    2. Estimating the minimum pump capacity in Hydropneumatic Pressure Systems.

Example Problem

  • Proposed 15-Storey Office Building

    • Total Useable Floor Area: 15,000 square meters

    • Add 3% annual increase for a design period of 15 years (Not clear in the example. Assuming it is 5 years as per the formula for Projected Population).

    • Use 9.30 square meters per person

    • Daily Water Consumption: 75 liters per person/shift/day

    • Duration of Usage: 9 hours or 540 minutes

  1. Solving for the Design Population (PpP_p):

    • Pp=1.15×(15,000/9.30)=2,097P_p = 1.15 \times (15,000 / 9.30) = 2,097 persons

  2. Solving for the Average Day Demand (AVD):

    • AVD=2,097×75AVD = 2,097 \times 75 liters per person per day (lpd) = 157,275 liters

  3. Solving for Maximum Day Demand (MDD):

    • MDD=1.30×157,275=204,457.50MDD = 1.30 \times 157,275 = 204,457.50 liters

  4. Solving for the Maximum Hour Demand (MHD):

    • MHD=(2.5×157,275)/540MHD = (2.5 \times 157,275) / 540 min = 728.125 liters/min

    • MHD=12.13MHD = 12.13 liters/sec.

Seatwork

  • Proposed 10-Storey Office Building

    • Total Useable Floor Area: 10,000 square meters with 2 shifts

    • Add 3% annual increase for the design period of 5 years

    • Use 9.30 square meters per person

    • Daily Water Consumption: 60 liters per person/shift/day

    • Duration of Usage: 12 hours usage per shift.

Calculations:

  • Design Population: [(10,000/9.30)×2shifts]×1.15=2,474[(10,000 / 9.30) \times 2 shifts] \times 1.15 = 2,474

  • Average Day Demand: 2,474×60=148,4402,474 \times 60 = 148,440

  • Maximum Day Demand: 1.30×148,440=192,9721.30 \times 148,440 = 192,972

  • Maximum Hour Demand: (2.5×148,440)/(12×60×60)=8.59(2.5 \times 148,440) / (12 \times 60 \times 60) = 8.59 lps

Recap:

  • Average Day Demand: To design the Reservoir or Storage Tank capacity.

  • Maximum Day Demand: To find the minimum pump capacity (except in Hydropneumatic Pressure System).

  • Maximum Hour Demand:

    1. Estimating the diameter of transmission and distribution mains.

    2. Estimating the minimum pump capacity in Hydropneumatic Pressure System.

Water Tanks

Importance for Architects:

  • Pre-determine the size and dimensions for space allocation.

  • Pre-determine the weight for structural integrity (if overhead).

  • To verify if the water storage is the appropriate amount for the project.

Method 1: Average Water Demand / Average Day Demand

  • Formula: CWT = Average Day Demand \times # of Days

    • CWT = Capacity of Water Tank

Example: If Average Day Demand: 78,450 liters/day and # of Days: 1, then

  • CWT=78,450 liters/day ×1=78,450 litersCWT = 78,450 \text{ liters/day } \times 1 = 78,450 \text{ liters}

Sample Problems:

  1. Household of 6 people needs to store water for 2 days at most. Use 190 liters per capita per day as the water demand.

  • 6 people ×190 liters/capita/day ×2 days =2,280 liters6 \text{ people } \times 190 \text{ liters/capita/day } \times 2 \text{ days } = 2,280 \text{ liters}

  • If considering leakage and pilferage, multiply it by 1.03 (additional 3 percent). If not asked, don't include the add-on.

  1. A diner that serves only breakfast and lunch has a max seating capacity of 60 people per meal. Determine the capacity of the water tank for a day's worth of operation. Use 26L as the Average Water Demand (AWD).

  • 60 people x 2 meals x 26L Average Water Demand = 3,120 LITERS

Method 2: WSFU (Water Supply Fixture Units)

  • Formula:

    • CWT=LPS×SCWT = LPS \times S

    • LPS = water demand in liters per second

    • SS = seconds of operation

Example:

  • 2 WC: 2x3 = 6

  • 2 LAV: 2X1 = 2

  • 2 SHO: 2X2 = 4

  • 1 KS: 1X2 = 2

  • 2 HB: 2X3=6

  • TOTAL: 20 WSFU

  • CLT = LPS x S (seconds of operation) = 1.24 x 4 hours x 60 x 60 = 1.24 x 14,400 =17, 850 LITERS

Sample Problems:

  1. Compute the required capacity of the water tank in a residence with the following plumbing fixtures: 3WC, 3LAV, 3SHO, 2KS.
    WC:3X3=9
    LAV: 3X1=3
    SHOW: 3X2 = 6
    KS: 2X2=4
    TOTAL: 22 WSFU
    Based on the table, 22 wsfu is closer to 25 wsfu, we get 1.36lps
    = 1.36lps x 3hours x 60 x 60
    = 14,688 L

  2. Compute the required capacity of the water tank for a private office space with the following plumbing fixtures; 6WC, 4LAV, 2UR, ISS, 1KS.
    WC:6X3=18
    LAV: 4X1 = 4
    UR: 2X2 = 4
    SS: 1X2 = 2
    KS: 1 X 2 = 2
    TOTAL: 36 WSFU
    Since no 36 wsfu on the table, we will use 40 with a value of 1.66
    Duration of Operation: 4 hours
    = 1.66lps x 4x60x60
    = 23,904 Liters

Sizing of Storage Tanks or Reservoir

  • As a "Rule of Thumb", the storage tank volume (except in a hydropneumatic pressure system) should be at least equal to one-fourth (1/4) of the water demand of the population.

    • Maximum

      • Cst = Ave. Day Demand x 1.03*

    • Minimum

      • Cst = (0.25 x Ave. Day Demand)1.03*

      • *Accounts to losses due to evaporation

Calculate the Pump Capacity

A. Solving for Rate of Flow
Q=Maximum Day Demand
Q = 101,985 liters/day x 1 day /24hours x 1 hour/3600 seconds
Q = 3.15 liters/second

B. Calculate the Pump Total Dynamic Head (TDH).
TDH = Pump Setting + Tank Elevation + Friction Loss
For this example, the values are: C
Pump Setting: 45.00 meters
Tank Elevation: 0.00 meters
Friction Loss: 2.00 meters
Therefore: TDH = 45.00 meters

C. Calculate the Water Horsepower (WHP).
WHP = (Qx TDH) / 75
=(3.15 x 45.00)÷ 75
WHP = 1.89 Hp

D. Calculate the Brake Horsepower (BHP).
BHP = WHP ÷ pump efficiency assume at 40% to 60%
BHP = 1.89 ÷ 0.60
BHP = 3.15 Hp
use 4 Hp
Pump Specifications:
Q = 3.15 Ips
TDH = 45 meters
BHP = 4 Hp