plumbing 2008 includes fu calculation fire fighting

8/10/2019 Plumbing 2008 Includes FU Calculation Fire Fighting..

1/203

1 1

Plumbin S stemsPlumbin S stemsPart IPart I

Lecture NotesLecture Notes

y r. . .y r. . .

LIULIU --20082008


2/203

DescriptionDescription of Architectureof Architecture

Design of Risers

Daily W. Requirement

Cold water distribution systemCold water distribution system

Load Values W.F.U.

Pipe sizing

CalculationCalculation

Hot water distribution systemHot water distribution system Circulating Pump

Types of pumps

CalculationCalculation Pipe sizing

Electrical W. heater

inside the flatsinside the flats Water storage heater

Instantaneous orsemi-inst. heaters

1 2

uest ons


3/203

Symbols & legendsSymbols & legends

SOIL STACK

WASTE STACK

SS

WS

STACK VENT

RAIN WATER

SV

RW

V VENT

RAIN WATER STACKRWS

SOFT COLD WATER

COLD WATER

POTABLE WATER

SW

CW

PW

TANK SUPPLY

DOMESTIC HOT WATER

DOMESTIC HOT WATER RETURNHWR

WTR

TS

HW

WATER

FIRE FIGHTING

G GAS

DR

F.F

DRAINAGE

1 3FUEL OIL SUPPLY

COMPRESSED AIR

FOS

V

A

VACUUM


4/203

BLACK STEEL PIPE ( SEAMLESS )

GALVANIZED STEEL PIPE ( SEAMLESS & WELDED )

POLYVINYLCHLORIDE PIPE

CAST IRON PIPE

PVC

BS

GSCI

CHLORINATED POLYVINYLCHLORIDE PIPE

POLYPROPYLENE RANDOM PIPE ( WATER )

POLYPROPYLENE PIPE ( DRAINAGE )

UNPLASTICIZED POLYVINYLCHLORIDE PIPE

P.P.R

C-PVC

PVC-U

P.P

POLYETHYLENE PIPE

COPPER PIPE

PE-X , ALUMINUM , PE-X ( TRIPLE LAYER ) PIPE

-

P.E

CU

-

PE-X / AL / PE-X

ASIATIC WATER CLOSET

AWC

. . .

EUROPEAN WATER CLOSET

S

B

LAV

EWC

SINK

BIDET

LAVATORY

KITCHEN SINK

DRINKING FOUNTAINDF

KS

BT BATHTUB

1 4FLASH VALVEFLASH TANK

FV

FT


5/203

JUNCTION BOX

FLOOR CLEANOUT

CEILING CLEANOUT

J.B

CO

FCO

CCO

CLEANOUT

FLOOR CLEANOUT

CEILING CLEANOUT

CO

FCO

CCO

CLEANOUT

WATER SOFTNER

ROOF VENT CAP

FIRE HOSE CABINET

WS

RVC

FHC

MH MANHOLE

WATER SOFTNER

ROOF VENT CAP

FIRE HOSE CABINET

.

WS

RVC

FHC

MH MANHOLE

WATER HEATERWHWATER HEATER

WH

FROM ABOVE

UNDER TILE

IW

UT

TB

FA

IN WALL

TO BELOW

FROM ABOVE

UNDER TILE

IW

UT

TB

FA

IN WALL

TO BELOW

BELOW FLOOR SLAB

IN FLOOR SLAB

UNDER CEILING LEVEL

UNDER GROUND

B.F.S

I.F.S

UCL

UG

BELOW FLOOR SLAB

IN FLOOR SLAB

UNDER CEILING LEVEL

UNDER GROUND

B.F.S

I.F.S

UCL

UG

HIGH LEVEL

LOW LEVELLL

UP

DN

HL

UP

DOWN

HIGH LEVEL

LOW LEVELLL

UP

DN

HL

UP

DOWN

1 5NOT TO SCALENTS

FM FROM

NOT TO SCALENTS

FM FROM


6/203

1 6


7/203

1 7


8/203

Block A 18 floors

Block B 17 floors

1 8


9/203

CarsCars

1 9


10/203

Waterstorage tanks

1 10


11/203

HOW TO READ AND DRAW THEHOW TO READ AND DRAW THE

. .

1 11


12/203

Example of some pipeExample of some pipe

distribution systemdistribution system

1 12


13/203

EXAMPLE OF WATER DISTRIBUTION SYSTEM INSIDEBATHROOM GALV. STEEL PIPES

1 13


14/203

1 14


15/203

DETAIL OF WATER DISTRIBUTION SYSTEM INSIDE

BATHROOM P.P.R PIPES

1 15


16/203

DETAIL OF WATER DISTRIBUTION SYSTEM INSIDE

BATHROOM PEX OR PEX AL-PEX PIPES

1 16


17/203

Madam Cury project water distribution system

1 17


18/203

Schematic water risers diagram for Madam Cury project

1 18


19/203

Before starting the detail calculation of

t e p um ing project . Stu ent s ou eable to read and understand all theArchitecture drawings of the project

.

1 19


20/203

1 20

Cha .2

o o wa ero o wa er


21/203

Calculation Of

. . Systems

. .Systems

Daily Water requirement

Load Values

Pressure requirement

Pipe sizing

Slide 211 Max Zornada (2002) 21Pump selection


22/203

Water Distribution Systems Up to 10 floors Bldg

n recn rec

1 22


23/203

Buildings above 20 floors

Pressure vessel

Pressure Reducer Break -Pressure reservoires Break- pressure ( Branch water supply )Direct supply ( Booster )

or frequency inverter

1 23


24/203

1 24

MultiMulti--pipes system is always preferablepipes system is always preferable

Muli-pipes system

Underground Tank

Each flat has its own inlet flow pipeEach flat has its own inlet flow pipe


25/203

1 25

a er s orage n u ngsa er s orage n u ngs

DomesticDomestic Fire fightingFire fighting

IrrigationIrrigation


26/203

1 26

omes c wa er s orage n u ngsomes c wa er s orage n u ngs


27/203

1 27

Water is stored in buildings due to the irregular supplyWater is stored in buildings due to the irregular supply

supply of city water .Normally water is stored insupply of city water .Normally water is stored in ..Roof tanks could one single tank for the whole building orRoof tanks could one single tank for the whole building orse arate tanks or each latse arate tanks or each lat

As shown in the following pages ,water tanks are providedAs shown in the following pages ,water tanks are provided, , ,, , ,

overflow and vent pipe.overflow and vent pipe.


28/203

1 28

Under round water stora e Pum sUnder round water stora e Pum s

Tanks ConnectionsTanks Connections

29


29/203

1 29

Roof tanks should be elevated enough above roof levelRoof tanks should be elevated enough above roof level

to have enough pressure for the upper apartment ,to have enough pressure for the upper apartment ,. .

--

--

--

1 30


30/203

1 30

Concrete Roof tanksConcrete Roof tanks

1 31


31/203

1 31

a van ze oo an sa van ze oo an s

1 32


32/203

1 32

. . .. . .

1 33Riser diagramRiser diagram


33/203

1 33Riser diagramRiser diagram

1 34

Riser diagram of the present projectRiser diagram of the present project


34/203

1 34Riser diagram of the present projectRiser diagram of the present project

1 35Chap 3


35/203

1 35Chap. 3

Desi n recommendationsDesi n recommendations&&

1 36


36/203

Dail Water Re uirementDail Water Re uirement

-- ai y water requirementai y water requirement an san scapacities. ( two methods are used tocapacities. ( two methods are used toe er ne e a y wa er requ re en , ee er ne e a y wa er requ re en , e

first is base on the number of occupants ,first is base on the number of occupants ,..

22-- Load value (W.f.u.)Load value (W.f.u.)

1 37


37/203

Type of Establishment GallonsGallonser da er ersoner da er erson

Table WTable W--11

Schools (toilets & lavatories only) 1515

Schools (with above plus cafeteria) 2525

Schools (with above plus cafeteria plus 3535

Day workers at schools and offices 1515

Residences 3535--5050

Hotels (with connecting baths) 5050

Hotels (with private baths, 2 persons per

room)

100100

1 38


38/203

Daily Water RequirementDaily Water Requirement for Storagefor Storage

ase on t e num er o occupantsase on t e num er o occupantsxamp e ca cu at on o a y omest c water requ rementxamp e ca cu at on o a y omest c water requ rement

Suppose we haveSuppose we have 2424 floors & each floor consists offloors & each floor consists of 44 flats,flats,

22 of them havinof them havin 33 bedroomsbedrooms22 of them havingof them having 22 bedrooms.bedrooms.++11 Mad each flat.Mad each flat.

s a ru e o um we a es a ru e o um we a e persons e room.persons e room.

Total number/floor =Total number/floor = 223322++222222++44 == 2424 Persons/floor.Persons/floor.Total number of occupants=Total number of occupants= 2424 2424 ++ 55 (labors+ concierges(labors+ concierges

etc) =etc) = 581581 Persons.Persons.From table WFrom table W--11 the daily water requirement is betweenthe daily water requirement is between 3535--5050

al/ da Residential Buildinal/ da Residential BuildinThe daily water requirement for the whole building is:The daily water requirement for the whole building is:=>=> 5050581581 == 2900029000 gallons /daygallons /day 110110 mm33/day/day

1 39


39/203

,,between the roof & underground tanks as follows:between the roof & underground tanks as follows:

''22 days on the ground floor ( standard ).days on the ground floor ( standard ).

men one e ore e o a amoun o wa er nee e or emen one e ore e o a amoun o wa er nee e or e

floors building isfloors building is 110110 mm33 ,this equivalent to,this equivalent to 110110 tones additionaltones additionalweight on the roof. On the other handweight on the roof. On the other hand 22 xx 110110 == 220220 mm33 must bemust bestored in the basement loor this ma a ect thestored in the basement loor this ma a ect the number onumber ocarscars in the basement.in the basement.

As a general rules (As a general rules ( one day water storageone day water storage on the roof &on the roof &

basement may be satisfactory ,basement may be satisfactory ,if water flowif water flow from well pump isfrom well pump isguarant eguarant e ).).

N.B. Drinking water tank capacity is calculated based onN.B. Drinking water tank capacity is calculated based on 1010--1212 L / person / dayL / person / day


40/203

1 41

Hose reel installation should be designed so that no part of the floor ish f h l h h h f ll d d h


41/203

more than 6 m from the nozzle when the hose is fully extended. The

gpm through the nozzle and also designed to allow not less than threehose reels to be used simultaneously at the total flow of 100 gpm for

.

The minimum required water pressure at the nozzle is 2 bar where the

axi u allowable ressure is 66 99 barbar Ade uate s ste ressuresAde uate s ste ressures isisaboutabout 44..55 barsbars .Booster pump is used for top roof flats..Booster pump is used for top roof flats.

The rubber hose reel length is 32 m & could be 1 or diameter(British standard), or 1.1/2(US standard), and the jet should have ahorizontal distance of 8 m and a height of about 5 m.

Riser main pipe diameter D= 2.1/2ranc p pe ame er= 1.1/2

Rubber hose reel diameter = 1 .

1 42


42/203

1 43


43/203

Siamese connectionSiamese connection

Located next to fire escapeLocated next to fire escape

1 44


44/203

Water storage or irrigationWater storage or irrigation

Irrigation systems could be by hose or automatically

usin um electrical valves timers & s rinklers.As a rule of thumb ,the water consumption forirrigation is estimated as follows:

The green area x 0.02 m /dayFor example :

uppose we ave a m green area aroun ehouse) to be irrigated. Estimate the water storage &

the u in rate er hour500 x 0.02 = 10 m3. & the pumping rate is 10 m3/h.

1 45


45/203

Determine the number of FUsDetermine the number of FUsFrom Table WFrom Table W--11

Determine the probable flow rate gpmDetermine the probable flow rate gpm

From ChartFrom Chart--11 or Table Wor Table W--22

Determine the Pipe sizeDetermine the Pipe sizePipe flow ChartPipe flow Chart--22

N.B. Pipe material should be known in orderN.B. Pipe material should be known in order

o use e correspon ng p pe ow c ar .o use e correspon ng p pe ow c ar .

1 46


46/203

. .. .

FixtureFixture TypeType UseUse FF..UsUs

National Plumbing Code of USA

.

Tab e W-2

WaterWater closetcloset -- FlushFlush tanktank (Private)(Private) 33

WaterWater closetcloset -- FlushFlush valvevalve ((PublicPublic)) 1010

BidetBidet (Private)(Private) 22

The value for separateThe value for separatehot and cold waterhot and cold water

BathBath tubtub (Private)(Private) 22

LavatoryLavatoryLavatoryLavatory

(Private)(Private) 11

((PublicPublic)) 22

taken astaken as of the total of the total

valuevalue

ShowerShower

ShowerShower

(Private)(Private) 22

((PublicPublic)) 33

UrinalUrinal -- FlushFlush tanktank ((PublicPublic)) 55

KitchenKitchen sinksink ---- 22

RestaurantRestaurant sinksink ---- 44

MopMop sinksink ---- 33

DrinkingDrinking fountainfountain ---- 11//22

DishDish washer,washer, washingwashing machmach.. (Private)(Private) 22


47/203

1 48

For theFor the

whole bldg.whole bldg.


48/203

Water Hammer ArrestorWater Hammer Arrestor

ChartChart --11 For eachFor eachflatflat


49/203

1 50


50/203

Ta e W-3

1 51

Volume Flow Rate (Cold+Hot) at The Inlet of FlatVolume Flow Rate (Cold+Hot) at The Inlet of Flat


51/203

Volume Flow Rate (Cold+Hot) at The Inlet of FlatVolume Flow Rate (Cold+Hot) at The Inlet of Flat..

FUs. For example suppose it is require to determineFUs. For example suppose it is require to determine

the inlet flow rate (gpm) of an apartment having thethe inlet flow rate (gpm) of an apartment having the33 W.C( flash tank) +W.C( flash tank) + 22 bidet +bidet + 33 lavatory +lavatory + 11 shower +shower +22 ++ 11 ++ 11

From table WFrom table W--11 we get :we get :((3333 F.U +F.U + 2222 F.U +F.U + 3311 F.U +F.U + 2211 F.U +F.U +2222 F.U +F.U + . +. + .. ..

From GraphFrom Graph--11 or tableor table--22 we select the probable waterwe select the probable water

demand for each identical flat : isdemand for each identical flat : is 2020 mm 11..2424L/s).L/s).

1 52

V l Fl R t C ld H t f th h l b ildiV l Fl R t C ld H t f th h l b ildi


52/203

Volume Flow Rate Cold+Hot for the whole buildinVolume Flow Rate Cold+Hot for the whole buildin ..

I two risers i e are used to su l water or the wholeI two risers i e are used to su l water or the wholebuildingbuilding The probable flow rate is determined asThe probable flow rate is determined asfollows:follows:

ssum ngssum ng oors eac oor asoors eac oor as en ca apar men sen ca apar men sAs calculated before the probable water demand for eachAs calculated before the probable water demand for each

. ,. ,

F.US let sayF.US let say 25002500 FUs.FUs.Inter Gra hInter Gra h--11 with a value ofwith a value of 25002500 FU and read theFU and read the

corresponding probable water demand for whole buildingcorresponding probable water demand for whole buildingwhich iswhich is 30003000 gpm . Since we have four risers thegpm . Since we have four risers the

,, ..Each riser will be sized based on this value i.e.Each riser will be sized based on this value i.e. 750750 gpm.gpm.

Without question the plumbing fixture will never operate simultaneouslyWithout question the plumbing fixture will never operate simultaneously, the diversity factor is included in Chart, the diversity factor is included in Chart --11

1 53

Pipe sizingPipe sizing


53/203

p gp g

Pipe flow charts are available which shows the relationPipe flow charts are available which shows the relationbetween the water flow in gpm or L/s , pressure drop in Psibetween the water flow in gpm or L/s , pressure drop in Psior ft /or ft / 100100 ft , pipe diameter in mm or inches and theft , pipe diameter in mm or inches and thecorresponding flow velocity in m/s or ft/s.corresponding flow velocity in m/s or ft/s.

To avoid erosion , water hammer & noise, the recommendedTo avoid erosion , water hammer & noise, the recommendedflow velocity is betweenflow velocity is between 11..22 andand 22..44 m/sm/s ( less than( less than 33 m/s).m/s).

e op mum ow ve oc y se op mum ow ve oc y s .. m s.m s.

The acceptable pressure drop perThe acceptable pressure drop per 100100 ft is aroundft is around 55Psi/Psi/100100ft ,that, in order to avoid excessive pressure loss andft ,that, in order to avoid excessive pressure loss andthe need for higher pressure to maintain the flow rate.the need for higher pressure to maintain the flow rate.

Low velocity pipe less thanLow velocity pipe less than 00..55 m/s can cause precipitation ofm/s can cause precipitation ofsand and others in the pipe .sand and others in the pipe .

Velocity limitation method recognized as good engineeringVelocity limitation method recognized as good engineeringpractice.practice.

1 54

11..3535 m/sm/s V=V=22 m/sm/s


54/203

.. 55 m/sm/s

DD

1 55How to use the pipe flowHow to use the pipe flow--chartchartThe use of the pipe flow chart is best presented by the


55/203

The use of the pipe flow chart is best presented by theo ow ng examp eo ow ng examp e : a r y roug stee p pe s use to

deliver 20 gpm of water at ordinary temperature with a.

recommended pipe size that can be used ?recommended pipe size that can be used ?

Solution: Enter the Fi ure alon the abscissa with the valueof 5 Psi/100 ft , move upward to the ordinate where QV is 20gpm .From the intersection ; read the values of ( D )and thecorrespon ng ow ve oc y .

Now it is clear that the intersection lies between 1.1/4 and . ,

be 15 Psi/100 ft which is greater than the given value . This s

is unacce table. If the 1.1/4 i e is used , the ressuredrop will be 4 Psi/100 ft which is less than the maximumallowed pressure drop .I would recommend D=1.1/4 with a flowve oc y ess an m s. e ow ve oc y s a ou . m s

.

1 56

Sizing the riser diagramSizing the riser diagram


56/203

g gg g

DD11 ??DD66 ??

4 Pressure relief valveHot water

.

Inlet water flow ?

Electrical water heater

Cold water 1"1"

D ?D ?

DD44 ??4 o t e tota xture un ts are use or co water

DD55 ??H.WH.W..

1 57


57/203

Open systemOpen system

n

loss

n

loss

fricti

fricti

Equa

Equa

1 58

Sizing the various pipes of the net workSizing the various pipes of the net work


58/203

g p pg p p

3/4 of the total fixture units are used for cold water

? "? " ? "? "? "

? "? " ? " ? " ? "

Determine the pipe sizes of the present drawing

H.WH.W..

1 59


59/203


60/203

1 61


61/203

--pipe feeding this apartment. Every bathroom should have twopipe feeding this apartment. Every bathroom should have twovalves one for cold and the second for hot water pipe.valves one for cold and the second for hot water pipe.

22-- Each plumbing Fixture should have and angle valve forEach plumbing Fixture should have and angle valve formaintenance reason.maintenance reason.

-- xposing pipes are insta e approximate yxposing pipes are insta e approximate y cm rom wa witcm rom wa wit

hangers and supports.hangers and supports.-- ..

55-- Pipe under tiles or in walls are PPR if however steel pipes arePipe under tiles or in walls are PPR if however steel pipes are

used the i e are wra ed with ute and as halt .used the i e are wra ed with ute and as halt .66-- Pipes crossing walls should be through pipe sleevesPipes crossing walls should be through pipe sleeves

1 62


62/203

11-- Pressure required during flow for differentPressure required during flow for differentfixtures.fixtures.

22-- Pressure required at the inlet of the flat.Pressure required at the inlet of the flat.

33-- The hydrostatic pressure available at each shutThe hydrostatic pressure available at each shut--o va ve.o va ve.

-- ressure re ucer va veressure re ucer va ve

1 63


63/203

ressure equ re ur ng ow orressure equ re ur ng ow orDifferent FixturesDifferent Fixtures

N.P.Code USAN.P.Code USA

1 64


64/203

As it well known the Hydrostatic pressure @ shutAs it well known the Hydrostatic pressure @ shut--off valve is given byoff valve is given by ::

WhereWhere is the specific weight kN/mis the specific weight kN/m33 & h is the pressure head in m& h is the pressure head in m

The maximum pressure at the inlet of the flat is Limited toThe maximum pressure at the inlet of the flat is Limited to 3030 m which is aboutm which is about.. ar , t at , avo excess ve pressuresar , t at , avo excess ve pressures

If the pressure is more thanIf the pressure is more than 22..99 BarBar ::

-- ..

The available pressure at the inlet of the flat, has to overcome the pressure lossThe available pressure at the inlet of the flat, has to overcome the pressure lossdue to i e friction and fittin s of the lon est branch and have a sur lus ressuredue to i e friction and fittin s of the lon est branch and have a sur lus ressureto operates the most critical fixture ( for example Dish washer or shower).to operates the most critical fixture ( for example Dish washer or shower).

Pressure Drop, P=Pressure Drop, P= x hx hLL + Surplus pressure ( h+ Surplus pressure ( hLL is the head loss due to pipeis the head loss due to pipe

Allowing additional pressure drop aroundAllowing additional pressure drop around 2525--3030% for fittings on straight pipe% for fittings on straight pipeor calculate the effective length for minor losses as described in Fluid Mechanicsor calculate the effective length for minor losses as described in Fluid MechanicsLecture notes. It is alwa s recommended to use the K value for the calculation ofLecture notes. It is alwa s recommended to use the K value for the calculation oft e pressure rop.t e pressure rop.

1 65

Example of high riserBuilding


65/203

floorsfloors

ABBRAGE BEIRUT

1 66

The hydrostatic pressure available at each shutThe hydrostatic pressure available at each shut--off valve.off valve.


66/203


67/203

1 68

1 " C.W.P 1 " C.W.P 1 " C.W.P1 " C.W.P


68/203

FLOATVALVE

1 1/2" C.W.P

UPPERDOMESTICWATERTANK3 * 10000 litres ( P.ETANKS)

1 1/4" C.W.P 1 1/4" C.W.P

FLOATVALVE

1 1/2" C.W.P

19TH. FLOOR

1 " C.W.P

1 1/4" C.W.P

"3"

p.r p.r 1 " C.W.P

1 1/4" C.W.PDrain pipe

1 " C.W.P 1 " C.W.P 1 " C.W.P 1 " C.W.P

3"18TH. FLOOR

1 1/2" C.W.P 1 1/2" C.W.P

1 1/2" C.W.P3" C.W.P

17TH. FLOOR

1 1/2" C.W.P

1 1/4" C.W.P 1 1/4" C.W.P 1 1/4" C.W.P 1 1/4" C.W.P

1 1/4" C.W.P 1 1/4" C.W.P 1 1/4" C.W.P 1 1/4" C.W.P

.

15TH. FLOOR


69/203

1 70


70/203

1 71


71/203

PRV

1 72

Pressure Reducer Valve PRVPressure Reducer Valve PRV


72/203

1 73


73/203

1 74


74/203

The head loss due to pipe frictionThe head loss due to pipe friction

. . , . ,

McGraw-Hill,2008.

1 75

Now !!

:


75/203

1- Calculate the daily water requirement for the given project & thecapacity of the overhead & underground tanks.

2- Recognize the drawing of water distribution system inside the flat.

3- Selectin the t e of the riser dia ram i.e. Direct or indirectwater supply. Sizing the riser diagram. Sizing the pipes inside thebathrooms etc..

- usti ie i t e y rostatic pressure at t e in et o t e at is

enough to overcome losses + the surplus pressure to operates the+ the surplus pressure to operates themost critical fixture .most critical fixture .

55-- Do we need a booster pump for top roof?Do we need a booster pump for top roof?

66--Do we need a breakDo we need a break -- ressure tank or ressure reducin valve ?ressure tank or ressure reducin valve ?

Now move on to the next part

Pump selection


76/203


77/203

the pipe size and pumps duty

11)) Prepare the drawing of the piping /pumping system, measure thePrepare the drawing of the piping /pumping system, measure thelength of the pipe connecting thelength of the pipe connecting the underground tankunderground tank to theto the overhead (overhead (

e very ane very an an coun a ngsan coun a ngs a ong e way .a ong e way .22)) Find the required volume flow rate for each flat. Then, add themFind the required volume flow rate for each flat. Then, add themup to obtain the total flow rate at the peak demand . The probableup to obtain the total flow rate at the peak demand . The probable

occupants or based on the total fixture units. ( It is not always easy tooccupants or based on the total fixture units. ( It is not always easy toknow the number of occupants in the early stage , so the secondknow the number of occupants in the early stage , so the secondmethod usin the T.F.Us becomes more reliable )method usin the T.F.Us becomes more reliable ) ..33)) Since the equal friction loss method is used , choose a value of frictionSince the equal friction loss method is used , choose a value of frictionloss rate for theloss rate for the main riser pipe based on the following limits :main riser pipe based on the following limits :

a )a ) The recommended friction loss rate is betweenThe recommended friction loss rate is between 11--55 m /m /100100 mm pipepipelength or (length or ( 11-- 55 ft/ft/100100ft , aboutft , about 22 Psi perPsi per 100100 ft ).ft ).b ) The velocity in the main should not exceedb ) The velocity in the main should not exceed 11..22--11..88 m / s ( saym / s ( say 11..55 m/sm/s)) in small systems , orin small systems , or22..44-- 33 m / sm / s in larger systems . The velocity inin larger systems . The velocity in

1 77

occup e areas s ou no exceeoccup e areas s ou no excee .. m sm s,, so as o preven no se.so as o preven no se.


78/203

building ( cont)

44)) Select aSelect apipe sizepipe sizefrom the pipe flowchartsfrom the pipe flowchartsbased on the above limits . We could also preparebased on the above limits . We could also prepare

tables which resent the i e diameters frictiontables which resent the i e diameters frictionfactor and flow rate . The tables are regarded asfactor and flow rate . The tables are regarded asmore accurate but the pipe flowcharts are moremore accurate but the pipe flowcharts are moreconvenientconvenient55)) Continuing along the circuit chosen , select theContinuing along the circuit chosen , select thesucceeding pipe sizes . This should be donesucceeding pipe sizes . This should be done

Determine by inspection which branch will be theDetermine by inspection which branch will be thelongest, or have the greatest equivalent length .longest, or have the greatest equivalent length .

a cu a e e pressure ropa cu a e e pressure rop n e onges c rcu .n e onges c rcu .

1 78


79/203

building ( cont)

66--Calculate :Calculate :

.. The actual pipe length + Equivalent length (due toThe actual pipe length + Equivalent length (due tofittings and valves etc.).fittings and valves etc.).

L L Leff e.=

e o a ea o or pre ure rope o a ea o or pre ure ropThe head loss per unit of length is about (The head loss per unit of length is about (55 ftftw./w./100100 ftft multi lied bmulti lied b the effective len th .the effective len th .

h h LL eff 1 .

1 79


80/203

building ( cont)

77) The approximated) The approximated pump s power is then calculatedpump s power is then calculatedas follows :as follows :

e ea e vere y e pump or e o a ea o ee ea e vere y e pump or e o a ea o epump: which is equal to the static head + the totalpump: which is equal to the static head + the totalhead loss ( case of o en tanks ).head loss ( case of o en tanks ).

The theoretical power requirement (Water power) isThe theoretical power requirement (Water power) is

LtsA

== xx AAxx VV ..(Where(Where

is the specific weight of water,is the specific weight of water, hhAA is theis theum head in m andum head in m and is the o eratin dischar eis the o eratin dischar e

mm33/s ). The operating discharge is taken from the/s ). The operating discharge is taken from theintersection of the pump characteristic curve with theintersection of the pump characteristic curve with thei e s stem curve.i e s stem curve.

1 80


81/203

Safety Margin

To avoid any miscalculation during pump selection, itTo avoid any miscalculation during pump selection, it

is recommended to a l a safet mar in ofis recommended to a l a safet mar in ofaroundaround 55%% for the estimated flow rate &for the estimated flow rate & 1010 % for% forthe estimated head.the estimated head.

For example :For example :

The recommended flow & head will be :The recommended flow & head will be :

,,

1 81


82/203

building ( cont)

88-- The shaft power of the pump can be determinedThe shaft power of the pump can be determinedby dividing water power by the pump efficiency.by dividing water power by the pump efficiency.

VA QhPowerPump

=

The motor power of the pump can be determined

efficiency.

VA QhPowerMotorPump =

1 82


83/203

The most popular types ofThe most popular types ofcentr uga pump use or cocentr uga pump use or co

water su l s stems in buildin swater su l s stems in buildin s

are:are:

Review !Review !

1 83

Vertical Multista e Pum sVertical Multista e Pum s


84/203

1 84

Horizontal multistage pump


85/203

1 85

VerticalVertical Line shaft submergedLine shaft submerged--pumpspumps


86/203

The usual pumping depth isThe usual pumping depth isa oua ou mm. owa ays, a. owa ays, adepth ofdepth of 250250 mm can becan beobtained with multistageobtained with multistageturbinesturbines..This kind of pumps is usedThis kind of pumps is usedfor clean water, sewa efor clean water, sewa eirrigation and fire fittings,irrigation and fire fittings,

etc.etc.A broad selection o dri erA broad selection o dri erheads is available to drive theheads is available to drive thepumps by most common primepumps by most common prime

.. .. ..High performance and lowHigh performance and lowmaintenance.maintenance.

1 86

TURBINE, VERTICAL TYPE, MULTISTAGE,,


87/203

These pumps develop high headThese pumps develop high heady us ng a ser es o smay us ng a ser es o smaimpellers rather than a largeimpellers rather than a largesingle one. The characteristicsingle one. The characteristiccurves for such pumps dependcurves for such pumps dependupon the number of stages orupon the number of stages orim ellers. Each im eller has theim ellers. Each im eller has thesame characteristic curve andsame characteristic curve and

the final curve is obtained bythe final curve is obtained by..

at a given discharge is the sumat a given discharge is the sumof individual heads (case ofof individual heads (case of

..pumps may deliver the liquid uppumps may deliver the liquid upfromfrom 400400 toto 500500 mm depth.depth.

1 87

ese pumps are common y useese pumps are common y usein tube wells, deep open wells,in tube wells, deep open wells,etc.etc.

SUBMERSIBLE PUMP


88/203

For high heads and low flow.

1 88


89/203

1 89


90/203

i

b ll t


91/203

balloon type

1 91


92/203

Example


93/203

1 93


94/203

Is used for boosting theIs used for boosting thewater to top floors, whenwater to top floors, when

e y ros a c pressuree y ros a c pressureat the inlet of the flat isat the inlet of the flat is

pressure requirement .pressure requirement .Location : In the attic orLocation : In the attic oron the roof.on the roof.

As a rule of thumb the vesselAs a rule of thumb the vesselcapacity is aboutcapacity is about 22 minuteminutess thetheactual um dischar e.actual um dischar e.

1 94

Domino pumps to replace theDomino pumps to replace the

pneumatic systempneumatic system


95/203

pneumatic systempneumatic system

Domino is a device used to replaceDomino is a device used to replace

the traditional booster system (the traditional booster system (

a oon ype . cons s s o ana oon ype . cons s s o an

electronic circuit, and diaphragm ,electronic circuit, and diaphragm ,

retaining spring and pressureretaining spring and pressure..

running.running.

Used for small and mediumUsed for small and medium

pressure system , up topressure system , up to 55..22 bar andbar and66..55 mm33/h it is a single phase pump ./h it is a single phase pump .

From LaworaFrom Lawora-- um s catalo ueum s catalo ue

1 95

1 96

Discharge & pressure headDischarge & pressure head

valvevalve


96/203

valvevalve

Estimated pumps

D ?D ?isc arge Gpm or m / s ma es ma ePump sPump sHead mHead m

Static (hs)

Each pump drawing should have the value of H & Q .

Review of the PerformanceReview of the Performance

Characteristics curves of aCharacteristics curves of a

water centrifugal pumpwater centrifugal pump


97/203

water centrifugal pumpwater centrifugal pump

--Efficiency curveEfficiency curve

Shaft power curveShaft power curve

NPSHNPSH

ReviewReview

1 97

11-- Head capacity curveHead capacity curve

The available head produced by the pumpThe available head produced by the pump

decreases as the discharge increases.decreases as the discharge increases.


98/203

ggAt Q=At Q= 00, the corresponding head is called, the corresponding head is called

shut off head point (shut off head point (11))

Point (Point (22) is called run out point below) is called run out point below

which the pump cannot operatewhich the pump cannot operate..&should be shut down&should be shut down

endend--of curveof curve

1 98

22-- Efficiency curveEfficiency curve

The efficiency of a centrifugal pump is the ratio of waterpower to brake power


99/203

y g p ppower to brake power.

powerWaterP = TheThe highest efficiency of a pumphighest efficiency of a pump

occurs at the flow where theoccurs at the flow where theincidence angle of the fluid enteringincidence angle of the fluid entering

the h draulic assa es bestthe h draulic assa es best

matches with the blade angle. Thematches with the blade angle. The

operating condition where a pumpoperating condition where a pump

referred to as the best efficiencyreferred to as the best efficiency

pointpoint B.E.PB.E.P..

1 99

33-- Power curvePower curve

..

The shaft power can be determined directly from the manufacturersThe shaft power can be determined directly from the manufacturers

catalogue plot or calculated from the following formulacatalogue plot or calculated from the following formula ::


100/203

QHPowershaft =

From the e uation, it is clear that the mainFrom the e uation, it is clear that the main

parameter affecting the shaft power is theparameter affecting the shaft power is the

discharge and not the headdischarge and not the head.. This is becauThis is becau

of the increase in the discharge for the samof the increase in the discharge for the sam

pipe diameter leading to additional lossespipe diameter leading to additional losses

which need more power to drive the pumpwhich need more power to drive the pump..

1 100


101/203

How to draw the pipe systemres s ance curves


102/203

1 102

1 103

& the Pumping Rate& the Pumping Rate


103/203

& ump ng& ump ng

In order to size the discharge pipe which feed the roof tanks , thefollowing data are needed:

-2- The pumping rate.N.B. To avoid disturbance & noise the Pumping time is limited to 4 hours

/day ( CIBSE B4).If for example , the Pump has to refill the empty overhead tank in 4

hours ,the pumping rate becomes 40 m3 / 4 h = 10 m3 /h.If however ,the Pump has to refill the empty overhead tank in 2 hours

.Decision has to be made by the consultant engineer to determine the

pumping time ,for example one or two hours .The um in rate is not the o eratin oint or dut oint of the um .

It is an estimated value used to estimate the flow rate in the pipe. Theactual pump discharge is obtained from =>Intersection of the pipesystem curve and pump performance curve.

e er o our ec ure no es e .

1 104


104/203

As it is known that , the role of the pump is toAs it is known that , the role of the pump is toovercome loss + elevation difference + dynamicovercome loss + elevation difference + dynamichead.head. V

ZZhh LA.2

2

2

12

The elevation difference represents the total staticThe elevation difference represents the total static

head whead which is the vertical distance between the watersurface level of the suction and discharge tanks. The dynamic head is too small, practically it canThe dynamic head is too small, practically it can

..

Operating point or duty point Operating point or duty point

corresponding to the intersection of its headcorresponding to the intersection of its head--capacity curve with the pipecapacity curve with the pipesystem curve. The intersection point is called system curve. The intersection point is called Duty point or operating pointDuty point or operating point


105/203

y py p y p p g py p p g p

required from the pumprequired from the pump

= the head given by the= the head given by the

um .Also At this ointum .Also At this oint

the pump would deliverthe pump would deliverthe maximum dischargethe maximum discharge

QQmaxmax ..

1 105


106/203

1515 L/sL/s

1313

1 106

Pump selection Pump selection

Pump is selected based on the B.E.P.Pump is selected based on the B.E.P. or nearly so . Howeveror nearly so . However tthe acceptable drop inhe acceptable drop in

efficiency is limited to onlyefficiency is limited to only 77 %% from its maximum efficiencyfrom its maximum efficiency .. As far as theAs far as the


107/203

1 107


108/203

1 108

ump s power


109/203

ump s power MonoMono--blockblock

1 109

The h draulic ower or water ower is iven b :The h draulic ower or water ower is iven b :

mV hQVAPVFpowerwater ===


110/203

powerwater

=P.

efficiencMotorefficiencyonTransmissiefficiencyPumppowerInput

=

Pump efficiency & motor power is selected from thePump efficiency & motor power is selected from the manufacturer cataloguesmanufacturer catalogues..

For Example ; The Transmition efficiencyFor Example ; The Transmition efficiency is taken as follows:is taken as follows:

11-- Case of shaft coupling =Case of shaft coupling = 11 ,,

22-- Case of flat belt Transmition =Case of flat belt Transmition = 00..99 toto 00..9393

1 110

33-- Case of VCase of V--belt Transmition =belt Transmition = 00..9393-- 00..9595..

Motor Power selectionMotor Power selection

ere s no s mp e ru e o um n mo or se ec on.. acere s no s mp e ru e o um n mo or se ec on.. ac

manufacturer suggest a safety margent for their motormanufacturer suggest a safety margent for their motor..


111/203

..

Example:Example: KSB pump catalogueKSB pump catalogue presents the followspresents the follows

Example:Example:

UP toUP to 77..55 kW addkW add 2020%%

FromFrom 77..55 -- 4040 kW add a roximatelkW add a roximatel 1515%%FromFrom 4040 kW and above add approximatelykW and above add approximately

..

1 111

Pumps powerPumps power


112/203

RequiredRequiredPumps ShaftPumps Shaft

anu ac urer anu ac urer Pumps powerPumps power

End curveEnd curve

MonoblockMonoblock-- PumpPump

1112


113/203


114/203

m3/hr

1 114

A centrifugal pump is used to supply water to the overhead tank- .

overhead tank is 30 m3.


115/203

1- Estimate the size of the rising main to overhead tank.2- Select the most suitable pump from the Lawora- pump

cata ogues.3- Estimate the power required which fits the water pipe system.4- Discuss the results.Assuming that:

The total length of the pipe is 50 m.The elevation difference is 31 m. from minimum water level of theunderground level up to the top Float switch of the overhead tank)2 gate valves full open and 6 (90 standard elbows) and one check

. .

The maximum running time of the pump is about 2 hours /day.The pumping of water is controlled automatically using automatic

1 115

.

Class exercise


116/203

A centrifugal pump is used to supply water to aA centrifugal pump is used to supply water to a1010-- floor buildin , which consists offloor buildin , which consists of 3535 flats.flats.Each flat is occupied byEach flat is occupied by 66 persons.persons.

11--Work out the daily water requirement, theWork out the daily water requirement, theun ergroun an over ea an capac y.un ergroun an over ea an capac y.Assuming that, each person requiresAssuming that, each person requires 3535 gal ofgal ofwater / er da .water / er da .

22-- Estimate the pumping rate of the pump.Estimate the pumping rate of the pump.The pumping of water is controlled automaticallyThe pumping of water is controlled automatically

using automatic water level switches.using automatic water level switches.

1 116


117/203

Variable Speed PumpsDriven by Frequency

.Direct supply system . UsedIn Hotels villas Hos italetc..

1 117


118/203

reductionreduction

PumpsPumps

1 118

powerpower


119/203


120/203

curve)


121/203

H

pressure transmitter, thepressure transmitter, the

pump ispump is balancingbalancing thethe

friction losses of systemfriction losses of system

curve.curve.

Q

6060 %% versus a full speedversus a full speed

um .um .

pressure increases topressure increases tocompensate for the addedcompensate for the added

1 121

. .

a n a n ng a cons an ow ra e

H


122/203

It can guarantee aIt can guarantee a

constant flowconstant flow atat

H

variable headvariable head

It can avoid toIt can avoid to run outrun out ofof

the curve when thethe curve when thesys em nee s ow easys em nee s ow ea Q

As the discharge changes .TheAs the discharge changes .The

cancan save energysave energy ncrease t e rpm .e. t encrease t e rpm .e. t epressure to maintain apressure to maintain aconstant discharge.constant discharge.

1 122

What happens to Flow, Head and Power with Speed?


123/203

~ RPM~ RPM

H ~ RPMH ~ RPM22SP ~ RPMSP ~ RPM33

1 123


124/203

1 124

Affinity laws (For the same pump)


125/203

1 125

D ublin th pump r t ti n lD ublin th pump r t ti n l


126/203

Doubling the pump rotationalDoubling the pump rotational

speespee ea s o:ea s o:-- . .

22-- Increase the total headIncrease the total headvalue by a factor ofvalue by a factor of 44..

--factor offactor of 88..1 126

Class ExerciseA d li 2000 L / i f t i t


127/203

A pump delivers 2000 L /min. of water against

running at shaft rotational speed of 3000 rpm.

Estimate the new pump characteristics if the

rotational s eed of the shaft is chan ed to4000 rpm. Assume the pump efficiency is

.

1 127

Summary of Exercise :


128/203

1 128

1 129Chap. 4


129/203

Domestic hot water systemDomestic hot water system

1 130

Hot water circulatin um (Hot water circulatin um (inside the apartment)inside the apartment)

f h l d f ff h l d f f


130/203

If water heater is located far away fromIf water heater is located far away from

p um ng x ures more anp um ng x ures more an m o wa erm o wa ercirculator should be provided in order to havecirculator should be provided in order to haveo wa er n e p p ng sys em a e me ano wa er n e p p ng sys em a e me an

hot to wait for a long time to have hot water.hot to wait for a long time to have hot water.o s ze e o wa er c rcu a oro s ze e o wa er c rcu a or s ca cu a es ca cu a ethe total hot water fixture units (as mentionedthe total hot water fixture units (as mentioned

or co wa er . orma y o wa er x ureor co wa er . orma y o wa er x ureunit isunit is 00..7575 of total fixture unit.of total fixture unit.

1 131Suppose we have a large flat having the followingSuppose we have a large flat having the followinglumbin fixtures estimate the circulated umlumbin fixtures estimate the circulated um

==discharge in gpm. :discharge in gpm. :

..44 lavatories xlavatories x 11 xx 00 7575 == 33


131/203

44 lavatories xlavatories x 11 xx 00..7575 == 33

..44 bidet xbidet x 22 xx 00..7575 == 66 sin s xsin s x xx .. ==

__________________Total =Total = 2121 FUFU

circulationcirculation 2121 FU/FU/ 2020 FU =FU = 11..0505 gpm .gpm .

Pump head is calculated by multiplying pipe effective length by thePump head is calculated by multiplying pipe effective length by the

pressure drop perpressure drop per 100100 ft as discussed earlier including pipe fittings.ft as discussed earlier including pipe fittings.

1 132

Supply Hot Water Pipe Sizing :Supply Hot Water Pipe Sizing :


132/203

pipe , the hot water Pipe is sized .pipe , the hot water Pipe is sized .rs o a ca cu a e e o a x urers o a ca cu a e e o a x ure

units (units (00..7575 of total FU) then find theof total FU) then find thecorresponding flow in gpm and thencorresponding flow in gpm and thene e o i e lo cha a d selec hee e o i e lo cha a d selec he

corresponding pipe size.corresponding pipe size.

1 133

Domestic Hot Water-Return Pipe Sizing.

, . .hot water return will be satisfactory.


133/203

f yHowever for lar e installation, the heat loss

from the return line becomes a majorconsideration. The following method is usedto s ze t e return p pe.

A. Determine the approximate total length ofa o wa er supp y + re urn p p ng.B. Multiply this total length by 30 Btu/Ft

. m , or nsu a e p pe anBtu/Ft (57.6 W/m) for un-insulated pipe to.

1 134

C. Divide the total heat loss by 10,000 to obtain the

pump capacity in L/s.[ 1Kg water /liter x 3600 sec/hr x 11 C=40 000][ 1lb water /gal x 60 min/hr x 20 F=10 000]

h 11 C i th ll bl t t d


134/203

where 11

C is the allowable temperature drop.

.GPM and obtain from pump curves the head created at

E. Multiply the head required by 100 (30.5) and divide

by the total length of the longest run of the hot waterreturn piping to determine the allowable friction loss per100 feet of pipe.

.circulating loop and size the hot water return pipe based

above step E.

1 135

Pipe insulation :Pipe insulation :

Hot water i es areHot water i es areinsulated in order toinsulated in order to

l f hl f h


135/203

prevent losses of heat toprevent losses of heat to

atmosp ere. nsu at onatmosp ere. nsu at onmaterial could bematerial could be

insulation like armaflexinsulation like armaflex

which comes in the form ofwhich comes in the form oftube and the pipe is pulledtube and the pipe is pulledinside it .inside it .

1 136

Pipe insulation:

reasons:

.

R d i


136/203

Reduce noise.

Control surface condensation. .

The insulating material is the same used in duct insulationThe insulating material is the same used in duct insulation

1 137

Insulation thicknessInsulation thickness

c nesses or nsu at on to e supp e an nsta e or t e var ousc nesses or nsu at on to e supp e an nsta e or t e var ous

systems shall conform to the following table:systems shall conform to the following table:


137/203

systems shall conform to the following table:systems shall conform to the following table:

a. Conditioned air supply and return (ductwork)a. Conditioned air supply and return (ductwork) 1 11 1//22""

b. Refrigerant suction and liquid lines:b. Refrigerant suction and liquid lines: 11//22""

c. Condensate drain pipesc. Condensate drain pipes 11//22""

d. Acoustic duct liner.d. Acoustic duct liner. 11e: ea ng wa er p pes up oe: ea ng wa er p pes up o ame er:ame er:

f. Heating water pipes abovef. Heating water pipes above 11 :: 1 11 1//22""

""..

k. Boiler, water heaterk. Boiler, water heater 22L. Boiler breeching & steel chimneyL. Boiler breeching & steel chimney 22..11//22

1 138

xpans on o pe ma er a


138/203

LL ==Amount of change in pipe length, mmAmount of change in pipe length, mm

== == ,, .. ..

TT== Temperature differenceTemperature difference, C, C

LL== is the original length of pipe mis the original length of pipe m

For example : AFor example : A 66 m copper pipe is subjected to temperature difference ofm copper pipe is subjected to temperature difference of 5050CCCalculate the pipe expansionCalculate the pipe expansion LL::

== .. xx xx == mmmm

1 139


139/203

1 140


140/203

Calculate the length of the Arm Ls ,assuming that D= 25 mm == mm.mm.

cm10452530Ls =


141/203


142/203

1 143

B


143/203

B= 200 +2 x LL ==200200++ 22xx4545 == 290290 mm.mm.


144/203

1 145


145/203

1 146


146/203

1 147

Domestic hot water systemDomestic hot water system


147/203

Water to water storage HeaterWater to water storage Heater

Boiler +Water StorageBoiler +Water Storage

--

types of heaterstypes of heaters

1 148

Electrical Water heater PowerElectrical Water heater Power

4 barHot water

1.25 "

Inlet water flow ?Pressure Releif valve

i


148/203

DrainThermostatThermostat

Temp. Electrical power :Electrical power :

11..55,, 22..44,, 44--55, &, & 99 KwKw

Cold water 1"1"

ze o :ze o :

1515,, 2020 ,, 3030,, 4040,, 5050 ,,

6666,, 8080 &&1 201 20 gallonsgallons3/4 of the total fixture units are used forc old water

KwEfficiencytimeHeating

TKghcSPower 2.475.036003

)1065(1502.4..

=

=

=

Where Shc is the specific heat capacity of water (Where Shc is the specific heat capacity of water ( 44..22 Kj/kg. K), Heating time or recovery period T =Kj/kg. K), Heating time or recovery period T = 33

hrs, The mass of waterhrs, The mass of water 150150 Liters =Liters = 150150 Kg., Temperature rise fromKg., Temperature rise from 1010 toto 6565 C.C.

Insulation efficiency isInsulation efficiency is 7575--8080 %.%.

1 149

In general, electric water heaters are fully automatic and haveIn general, electric water heaters are fully automatic and havea storage tank, one or more electric elements, and operatinga storage tank, one or more electric elements, and operatingan sa ety contro s. e eat ng e ements are ava a e n aan sa ety contro s. e eat ng e ements are ava a e n a

variety of standard voltages and wattages to meet thevariety of standard voltages and wattages to meet the. .

Electrical water heater are fitted with electrical resistanceElectrical water heater are fitted with electrical resistance


149/203

.. ,, .. ,, -- ,, ..horizontal or vertical type . EWH are insulated andhorizontal or vertical type . EWH are insulated and

EWHs should have pressure /temperature relief valveEWHs should have pressure /temperature relief valve..

EWH the use of noneEWH the use of none --return is not recommended. In thisreturn is not recommended. In thiscase ,the ressure relief valve is connected to nearestcase ,the ressure relief valve is connected to nearest

floor drain.floor drain.

electrical heater automatically.electrical heater automatically.

Available EWHs size areAvailable EWHs size are 1515,, 2020 ,, 3030,, 4040,, 5050 ,, 6666,, 8080 &&1 201 20 gallonsgallons

1 150

Resistance heating elementResistance heating element

Usually the electrical water heaters have a primaryUsually the electrical water heaters have a primaryres s ance ea ng e emen near e o om , an poss eres s ance ea ng e emen near e o om , an poss ea secondary element located within the upper quarter ora secondary element located within the upper quarter or


150/203

. .

Minimum wattage ratings of two element heaters are basedMinimum wattage ratings of two element heaters are based

For the upper unit (For the upper unit (88 watt / liter)watt / liter) of the tank capacity.of the tank capacity.For the lower unit (For the lower unit (55 watt/literwatt/liter) of the tank capacity.) of the tank capacity.

-- -- ,,

watt/litterwatt/litter) for each unit .) for each unit .,,

1 151

HotHot water de andwater de and

(definitions)(definitions)Demand of water supplyDemand of water supply ::is the rate of flow in pgm furnished by a wateris the rate of flow in pgm furnished by a water


151/203

is the rate of flow in pgm furnished by a wateris the rate of flow in pgm furnished by a water

supply system to various types of plumbing fixturessupply system to various types of plumbing fixturesand water outlets under normal conditions.and water outlets under normal conditions.

Maximum Demand:Maximum Demand:

is the eak value of the demand. The values Hot ofis the eak value of the demand. The values Hot ofwater demandwater demand areare shown previously in table (Wshown previously in table (W--22 ).).

Demand factorDemand factor ::is the ratio of the maximum demand of the hot wateris the ratio of the maximum demand of the hot water

ea ng sys em o e o a connec e oa or eea ng sys em o e o a connec e oa or etotal of the individual requirements of all the fixturestotal of the individual requirements of all the fixturesof the system.of the system.

1 152

Water to water storage heaterWater to water storage heater

oo --wa er s orage an mus mee co e requ remen s a epenwa er s orage an mus mee co e requ remen s a epen

on its size and pressure and the authority having jurisdiction. Itson its size and pressure and the authority having jurisdiction. Its

water in the tank may be drawn off before the temperature dropwater in the tank may be drawn off before the temperature drop

(caused by the incoming cold water) becomes unacceptable. A value(caused by the incoming cold water) becomes unacceptable. A value


152/203

( y g ) p( y g ) p

ofof 7070 percent usually is used in design calculations.percent usually is used in design calculations.

Heated water from boiler enter the coil of the water stora e vesselHeated water from boiler enter the coil of the water stora e vessel

where it will heat the water , and then returns to boiler at lowerwhere it will heat the water , and then returns to boiler at lower

temperature . Whereas the domestic Hot water leaves the top of thetemperature . Whereas the domestic Hot water leaves the top of thewater tank to the supply distribution piping upon demand for thewater tank to the supply distribution piping upon demand for the

various fixtures and apparatus. When recalculation of the hot watervarious fixtures and apparatus. When recalculation of the hot water

,,

water vessel or tank by a circulate pump.water vessel or tank by a circulate pump.

orma y ea ng wa er rom o er en ers e wa er an aorma y ea ng wa er rom o er en ers e wa er an a

8282 C ) and return to boiler atC ) and return to boiler at 160160 F (F ( 7070 C),C), T =T = 1212--1515 C.C.

1 153

Hot water re uirement for stora eHot water re uirement for stora e

TheThe hothot waterwater storagestorage forfor BuildingsBuildings && HotelsHotels willwill bebe calculatedcalculated basedbased onon unitunit hourlyhourly

demanddemand ratesrates asas followsfollows ::


153/203

galongalon perper hourhour

FixtureFixture typetype BuildingBuilding HotelsHotels

ava oryava ory gpgp gpgp

bathtubbathtub 2020 gphgph 2020 gphgph

ShowerShower 3030 gphgph 7575 gphgph

LaundryLaundry 2020 gphgph 2828 gphgph

DishwashersDishwashers 1515 gphgph 5050 gphgph

.. ..

StorageStorage factorfactor 11..2525 00..88DomesticDomestic hothot waterwater temperaturetemperature willwill bebe 6060 CC..

1 154

able Hwable Hw--11


154/203

1 155

Table HwTable Hw--22


155/203

1 156

Procedure for estimating the heating capacityrecover ca acit of a hot-water heatin s stemhaving a storage tank.

Step(Step(11)) TabulateTabulate thethe numbernumber ofof fixturesfixtures ofof eacheach typetype inin thethe buildingbuilding..

Step(Step(22)) ThenThen multiplymultiply thethe numbernumber ofof fixturesfixtures ofof eacheach typetype byby thethe probableprobablett tt


156/203

emaneman oror eaceac typetype oo xturexture..

Step(Step(33)) ObtainObtain thethe maximummaximum demanddemand byby takingtaking thethe sumsum ofof productsproducts ofof

Step(Step(44)) ThenThen obtainobtain thethe hourlyhourly heatingheating capacitycapacity byby multiplyingmultiplying thethe

maximummaximum demanddemand inin steste 33 bb thethe demanddemand factorfactor obtainedobtained fromfrom tabletable 22..

Step(Step(55)) multiplymultiply thethe hourlyhourly heatingheating capacitycapacity ofof stepstep 44 byby thethe storagestoragecapacitycapacity factorfactor givengiven inin tabletable HWHW--11 forfor thethe appropriateappropriate typetype ofof buildingbuilding totoobtainobtain t et e requirerequire capacitycapacity oo t et e storagestorage tanktank..

NN..BB.. NotNot toto bebe usedused forfor instantaneousinstantaneous oror semisemi--instantaneousinstantaneous typestypes ofof

1 157

Acce table Tem erature of domestic Hot waterAcce table Tem erature of domestic Hot water


157/203

.. ,, .. --

1 158

ExampleExample

,capacity , the boiler power of an apartment house

buildin havin the followin data:

60 Lavatories , 40 kitchen sinks ,and 10 laundry 60, .


158/203

,

Step1& 2. The probable water demand are as follows:

From table HW-1;

Lavatories = 60 x 2 h = 120 hKitchen sink = 40 x 10 gph = 400 gph .

aun ry = x gp = gp .

Shower = 60 x 30 gph = 1800 gph.

Dishwasher = 40 x 15 gph = 600 gph .

1 159

StepStep 33 ::

=((=((606022gph)+ (gph)+ (60603030gph)+ (gph)+ (40401010gph) + (gph) + (4040 xx 1515 gph)gph)

+(+(1010xx 2020 gph ))=gph ))= 31203120 gph.gph.StepStep 44Building demand factor =Building demand factor = 00..33 (from table HW(from table HW--11 ).).

== == ==


159/203

== .. == == ..gpm.=gpm.= 00..985985 liter /sec orliter /sec or 35463546 L/hr.L/hr.

The required capacity of the storage tank is:The required capacity of the storage tank is:

Usable capacity =Usable capacity = 936936 xx 11..2525 == 11701170 gal i.e.gal i.e. ((44294429 liters)liters)[[11..2525 is the storage factor from table Hwis the storage factor from table Hw--11]]

a e y s orage:a e y s orage:Since onlySince only 7070% of the tank is usable% of the tank is usable,, so theso the actualactual

..((63276327 liters).liters).

1 160

Boiler power CalculationBoiler power Calculation

From the basic equation Q = m C T

un s. .n

.


160/203

;)rate(flowgpmindemandwatercalculatedtheisgpmWhereTgpm500BTU/h)in(boilerQ 1

=

F]101[Fianal&initialbetweendifferenceetemperaturtheisT o=

rom e prev ous examp e, we averom e prev ous examp e, we ave gp =gp = .. gpmgpmof water to be heated , temperature riseof water to be heated , temperature rise T=T=101101 F.F.

QQTT== 500500 xx 1515..66 xx 101101 == 787800787800 BTU/hr =BTU/hr =230230 Kw.Kw.I I :I I :

KwCCKgKjKgKWPower 5.227)55(./2.4.sec/985.0 =


161/203

1 162

==


162/203

T=T=7272 CC

T=T=1010 CC


163/203

1 164For hot water systems in which piping from theFor hot water systems in which piping from theheater to the fixture or a liance is shortheater to the fixture or a liance is short 3030m), or less], circulating systems are not generallym), or less], circulating systems are not generally

used. But it is common ractice to rovideused. But it is common ractice to rovidecirculating pump in all hot water supply systems incirculating pump in all hot water supply systems inwhich it is desirable to have hot water availablewhich it is desirable to have hot water available


164/203

continuously at the fixtures.continuously at the fixtures.

Sizing of hot water circulating pump is simplifiedSizing of hot water circulating pump is simplified

bb ..11 gpm for everygpm for every 2020 fixtures units in the systemfixtures units in the system..Or:Or:

00..55 gpm (gpm (00..0316103161/s) for each/s) for each 00..7575-- oror 11 riser; riser;11 m (m (00..0631106311/s) for each/s) for each 11..2525 -- oror 11..55 riser; riser;22 gpm for eachgpm for each 22 riser. riser.

1 165

and high temperature and high temperature,and high temperature and high temperature,

circulatincirculatin u su s should be ade oshould be ade o bronzebronze oror

other material designed to withstand theseother material designed to withstand thesecondition.condition.


165/203

When the hot water piping exceed (30m ) , awa er c rcu a e pump s o en ns a e . scontrolled by an immersion thermostatthermostat (in there urn ne se o s ar an s op e pump over a11 C). However for continuous hot water supply

e ermos a s e m na e .

1 166

Calculation of circulatin umCalculation of circulatin um --11 ca acitca acit


166/203

Circulatingpump 1

R.H.W.H.W. S

H.W. SH.W.Storage tank

R.H.W.

Circulatin um

C.W.S

1 167

Suppose we have a building containing theSuppose we have a building containing thefollowin lumbin fixtures estimate the circulatefollowin lumbin fixtures estimate the circulatepumppump--11 flow rate in gpmflow rate in gpm. :. :

Lavatories = 60 x 1 x 0.75 = 45 Fus.

. .


167/203

Laundry = 10 x 2 x 0.75 = 15 Fus .

Shower = 60 x 2 x 0.75 = 90 Fus.

Dishwasher = 40 x 1 x 0.75 = 30 Fus .Total =Total = 240240 FUFU

FU =FU = 1212 gpmgpm

This is the discharge of the circulated pump , whichThis is the discharge of the circulated pump , whichcirculate the water from (boiler tank) to building.circulate the water from (boiler tank) to building.

1 168

Calculation of the circulating pumpCalculation of the circulating pump--22 capacitycapacity

PumpPump 11

H.W.Storage tankTP P.R.V. H.W. S


168/203

R.H.W.

Circulating pumpC.W.S

Expansion VesselFUEL Supply

PumPum 22

1 169

Estimating circulating pump 2 capacity

unitsU.S.In

1 lb/gal)8.3(T)60min/hour(GPM(BTU/h)Q


169/203

1

T/500BTU/hinboiler)60min/hourT(8.3/BTU/h)in(boilerQ(gpm)Q

In SI units:In SI units:

227KwPower

. K Kg.Kg sec.KWPower

=

.114.2114.2

=

=

=

,,circulate the water from boilercirculate the water from boiler--storage tankstorage tank-- Boiler.Boiler.

1 170

ea o e rcu a e pump

As it is known that , the role of the circulated pumpAs it is known that , the role of the circulated pumpis to overcome loss due to pipe friction & fittings.is to overcome loss due to pipe friction & fittings.


170/203

The elevation difference is not included .The elevation difference is not included .LA hh=The head loss due is determined from Darcy equationThe head loss due is determined from Darcy equation

as mentioned in chap.as mentioned in chap. 99

1 171

Suppose we have to estimate the head required of aSuppose we have to estimate the head required of ac rcu ate pump , assum ng t e o ow ng : e p pe engt sc rcu ate pump , assum ng t e o ow ng : e p pe engt s600600 ft.ft. and an allowance for fittings on straight pipe ofand an allowance for fittings on straight pipe of 2525 %%--

. .

1- Determine,Determine, thethe totaltotal effectiveeffective len thlen th EE..LL thatthat isis::


171/203

The actual pipe length + Equivalent length (due to fittings andThe actual pipe length + Equivalent length (due to fittings andvalves etc.)valves etc.)=

22-- TheThe totaltotal headhead lossloss oror pressurepressure dropdrop hLhL isis determineddetermined asas ::

--

e e.

11 ..multipliedmultiplied byby thethe effectiveeffective lengthlength ..

]11[5.3705.0750100/5. misthatftftftLh

e

effL

ff

=

=

Usually the Pump is oversized byUsually the Pump is oversized by 1010 % of head &% of head & 55% flow .% flow .

1 172

Instantaneous or semiInstantaneous or semi--instantaneousinstantaneous

n an aneou y en an aneou y e

The instantaneous indirect water heater is used to meet a demand for aThe instantaneous indirect water heater is used to meet a demand for astea y, continuous supp y o ot water. In t is type o unit,stea y, continuous supp y o ot water. In t is type o unit, t e water ist e water ish d i l i fl h h h b f h h i ilh d i l i fl h h h b f h h i il


172/203

heated instantaneously as it flows through the tubes of the heating coilheated instantaneously as it flows through the tubes of the heating coil..The heating medium (steam or hot boiler water) flows through the steelThe heating medium (steam or hot boiler water) flows through the steelpipe shell yielding a small ratio of hot water volume to heating mediumpipe shell yielding a small ratio of hot water volume to heating mediumvolume. Instantaneous water heaters are designed to provide sufficientvolume. Instantaneous water heaters are designed to provide sufficient

ca acit to heat the re uired uantit of water usuall ex ressed in mca acit to heat the re uired uantit of water usuall ex ressed in m(l/s)) at the time the hot water draw occurs. Storage tanks are not usually(l/s)) at the time the hot water draw occurs. Storage tanks are not usuallypart of an instantaneous water heater, although a separate storage tankpart of an instantaneous water heater, although a separate storage tank

..

high demand type, a circulating pump should be installed in both the boilerhigh demand type, a circulating pump should be installed in both the boilerwater and domestic water piping circuits.water and domestic water piping circuits.


173/203

1 174

Procedure for estimating the heating capacity forinstantaneous and semiinstantaneous waterheaters.

Step(Step(11)) TabulateTabulate thethe numbernumber ofof plumbingplumbing fixturesfixtures ofof eacheach typetype thatthat useusehothot waterwater..

teptep u t p yu t p y t et e num ernum er oo xturesxtures oo eaceac typetype yy t et e num ernum er oofi tfi t itit fi tfi t (( bt i dbt i d ff t blt bl HWHW 33 )) tt bt ibt i thth t t lt t l


174/203

fixturefixture unitsunits perper fixturefixture (( obtainedobtained fromfrom tabletable HWHW--33 )) toto obtainobtain thethe totaltotalnumbernumber ofof fixturefixture unitsunits..

Step(Step(33)) UsingUsing thethe totaltotal numbernumber ofof fixturesfixtures unitsunits obtainedobtained fromfrom stepstep 22 ,,determinedetermine thethe maximummaximum demanddemand inin gpmgpm usingusing thethe appropriateappropriate curvecurve givengiven ininc arc ar -- ..

Step(Step(44)) ToTo thethe demanddemand ofof stepstep 33 ,, addadd thethe demanddemand forfor hothot--waterwater fixturesfixtures..

Step(Step(55)) SelectSelect aa heaterheater thatthat willwill provideprovide thethe requiredrequired riserise inin temperaturetemperatureTT == 101101 FF forfor thethe totaltotal demanddemand ofof steste 33 && 44..

1 175

Example on the calculation of water demand usingsemisemi--instantaneous type of heaters .instantaneous type of heaters .

Determine the required capacity in gpm of a semi-

there are 6 wash fountains , 10 showers, 2 service sinks,1 pantry sink , and 4 private lavatory basins.

St 1& 2 T b l t th mb f pl mbi fixt & M ltiplM ltipl


175/203

Step-1& 2: Tabulate the number of plumbing fixture & MultiplyMultiplythese numbers by the number of fixture units per fixture ( obtainedthese numbers by the number of fixture units per fixture ( obtainedfrom table HWfrom table HW--33 ) to obtain the total number of fixture units.) to obtain the total number of fixture units.

. .

Service sink = 2 x 2.5 = 5 Fus .

an ry s n = x . = . us .

Showers = 30 x 1.5 = 45 Fus.= =. ,

Total Fixture units = 70.5 FUs

1 176Step-3 & 4 : Using the total number of fixtures unitsUsing the total number of fixtures units,,

using the appropriate curve given in chart HWusing the appropriate curve given in chart HW--44 ..

,,read the corresponding value forread the corresponding value for 7070..55 FUs , which isFUs , which is 1515 gpm.gpm.This is the hot water demand for fixtures that o erateThis is the hot water demand for fixtures that o erate

int mitt ntl N ss m t l st n fixt p t sint mitt ntl N ss m t l st n fixt p t s


176/203

intermittently . Now assume at least one fixture operatesintermittently . Now assume at least one fixture operatescontinuously , and it needs a demand ofcontinuously , and it needs a demand of 11 gpm.gpm.

The total water flow rate becomesThe total water flow rate becomes 1616 gpm ,this is the capacitygpm ,this is the capacity

of the semiof the semi--instantaneous Boiler.instantaneous Boiler.Select the desired temperature of the water leave & theSelect the desired temperature of the water leave & thetem erature of cold water enters the boiler in order totem erature of cold water enters the boiler in order to

calculate the boiler power.calculate the boiler power. Tgpm500BTU/h)in(boilerQ Kw236.6BTU/h000808F10116500)(boilerQ =


177/203

1 178nother way tonother way to

determine gpmetermine gpm

HWHW--44


178/203

1 179

(b)(b)HWHW--44


179/203

1 180

rawin o aterrawin o ater

Di t ib tiDi t ib ti


180/203

DistributionDistributionSystemsSystems

1 181


181/203

1 182


182/203

1 183


183/203

1 184BOILER SPECIFICATIONS

11--It should be capable ofIt should be capable of quick startquick start--upup..22--Should meet largeShould meet large load fluctuationsload fluctuations..33--OccupyOccupy less floor spaceless floor space..

-- ..

55--Should essentially possess the capacity of producingShould essentially possess the capacity of producing


184/203

55 Should essentially possess the capacity of producingShould essentially possess the capacity of producing. . .. . .

thermal efficiencythermal efficiency

66-- Simple in construction.Simple in construction.77--Tubes should be sufficiently strongTubes should be sufficiently strong to resist wear andto resist wear andcorrosion.corrosion.

--

platesplates..--The e ocit o water and that o ue as shou dThe e ocit o water and that o ue as shou d

minimumminimum..

1 185Selection of a boiler

The selection criteria of a boiler depends very much on theThe selection criteria of a boiler depends very much on the

purpose of the boiler i.e.purpose of the boiler i.e. the load requirementthe load requirement. Boiler may be. Boiler may bee er use o pro ucee er use o pro uce s eam o a s eams eam o a s eam ur ne, or orur ne, or orheating processheating process. If steam is required for power Generation. If steam is required for power Generation

turbine is essential.turbine is essential.h h h d f h b l dh h h d f h b l d f hf h


185/203

turbine is essential.turbine is essential.On the other hand, if the boiler is re uiredOn the other hand, if the boiler is re uired for a heatinfor a heatinprocess like an industrial load other applications likeprocess like an industrial load other applications like hospitals,hospitals,hotels, kitchens steam or hot water boilers must behotels, kitchens steam or hot water boilers must becons ere .cons ere .FoFor power generation we need essentially ar power generation we need essentially a water tube boilerwater tube boiler..

,,

possible. Hot water are usually produced aroundpossible. Hot water are usually produced around 100100

CC AndAndressure fromressure from 22 ~~ 88 atmatm, and hi h, and hi h--tem hot watertem hot water HTHWHTHWfromfrom 121121C~C~260260CC and pressure overand pressure over 1010..88 atmatm

1 186Hi

plumbing 2008 includes fu calculation fire fighting

Documents