hydraulic calculation chap-3

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Fire Fighting

ChapChap--33

Design RecommendationsDesign Recommendations& Hydraulic& Hydraulic ––CalculationCalculation

By Dr. Ali HammoudBy Dr. Ali HammoudBAUBAU--20052005

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DEVELOPING A SPRINKLER PLAN FOR APPROVALDEVELOPING A SPRINKLER PLAN FOR APPROVAL

TTo prepare a sprinkler plan for approval, it is necessaryo prepare a sprinkler plan for approval, it is necessarytoto::

1.1. Collect and review data.Collect and review data.2.2. Trace or scan building outline.Trace or scan building outline.3.3. Draw necessary building details.Draw necessary building details.4.4. Determine branch line logic.Determine branch line logic.5.5. Determine system type and configuration.Determine system type and configuration.

6.6. Determine the hazard class of the occupancyDetermine the hazard class of the occupancy..7.7.Determine the area protected by each sprinkler.Determine the area protected by each sprinkler.8.8.Determine the number of branch lines.Determine the number of branch lines.9.9.Determine the distance between branch lines.Determine the distance between branch lines.10.10.Determine the maximum allowable distance betweenDetermine the maximum allowable distance between

sprinklers.sprinklers.

Ref Ref --33



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11- Pipe sizing tables.12- Friction Loss FormulaFriction Loss Formula ”” HazenHazen--WilliamsWilliams””13- Hydraulic Calculation for sprinkler systemsHydraulic Calculation for sprinkler systems1414--exanple problemexanple problem

4

Installation Requirements Based onInstallation Requirements Based onNFPANFPA--1313

System Protection Area Limitations.System Protection Area Limitations.

1.1. Light hazardLight hazard —— 52,000 ft52,000 ft22 (4831 m(4831 m22))

2.2. Ordinary hazardOrdinary hazard —— 52,000 ft52,000 ft22

(4831 m(4831 m22

))3.3. Extra hazardExtra hazard ——Hydraulically calculatedHydraulically calculated40,000 ft40,000 ft22 (3716 m(3716 m22))

4.4. StorageStorage —— HighHigh--piled storagepiled storage —— 40,00040,000ftft22 (3716 m(3716 m22))

Ref Ref --11



5

77--Area of CoverageArea of Coverage

•• Determination of the Protection Area of CoverageDetermination of the Protection Area of Coverage

1.1. Along Branch Lines.Along Branch Lines. Is the distance betweenIs the distance betweensprinklers, defined assprinklers, defined as ““SS””

2.2. Between Branch Lines.Between Branch Lines. Is the perpendicular distanceIs the perpendicular distanceto the sprinkler on the adjacent branch line, definedto the sprinkler on the adjacent branch line, definedasas ““LL””

AsAs == S X LS X L

•• The maximum allowable protection area of coverageThe maximum allowable protection area of coveragefor a sprinkler (As)for a sprinkler (As) shall beshall be in accordance with thein accordance with the

value indicated in each type or style of sprinkler. Thevalue indicated in each type or style of sprinkler. Themaximum area of coverage of any sprinklermaximum area of coverage of any sprinkler shall notshall notexceed 400 ftexceed 400 ft22 (36 m(36 m22).).

6

LL

SS

AS AS

AreaArea

protectedprotected

by oneby one

sprinkler sprinkler

AsAs == S X LS X L



7

Area of Coverage (Area of Coverage (for Lightfor Light

HazardHazard))

• Protection Areas and Maximum Spacing (Standard SprayUpright/Standard Spray Pendent) for Light Hazardfor Light Hazard

4.64.615.615.6CombustibleCombustible

4.64.620.920.9NoncombustibleNoncombustible

Max SpacingMax SpacingSS (m)(m)

Max AreaMax AreaAAss (m(m22))

Construction typeConstruction type

Dr. Ali Hammoud BAUDr. Ali Hammoud BAU-- 20052005

8

Area of Coverage ( for Ordinary Hazardfor Ordinary Hazard)

• Protection Areas and Maximum Spacing (Standard SprayUpright/Standard Spray Pendent) for Ordinary Hazard

4.612.1 All

Max Spacing

SS (m)

MaxMax Area

AAss (m2)

Construction type



9

Area of Coverage (for High Hazard)

• Protection Areas and Maximum Spacing (Standard Spray

Upright/Standard Spray Pendent) for High Hazard

4.612.1 All, ρ < 0.25

3.79.3 All, ρ >= 0.25

Max Spacing

“SS”” (m)MaxMax Area

AAss (m2)

Construction type

10

Practical Spacing “S” of up-rightand pendent spray sprinkler

The maximum distance between sprinklers, either on branchThe maximum distance between sprinklers, either on branchlines or between branch lines, shall be as follows:(as perlines or between branch lines, shall be as follows:(as perNFPANFPA--13.item 413.item 4--4) :4) :

Light HazardLight Hazard

15ft (4.5 m)15ft (4.5 m)

Ordinary HazardOrdinary Hazard 15ft15ftExtraExtra –– HazardHazard 12ft (3.6 m)12ft (3.6 m)HighHigh –– Piled storage 12ftPiled storage 12ftThe distance from sprinklersThe distance from sprinklers to wallto wall shall exceedshall exceed oneone--half ofhalf ofallowable distance between sprinklers.allowable distance between sprinklers.



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88-- Determine the Number of branch Lines.Determine the Number of branch Lines.

The number of branch lines per bay is determined by dividingThe number of branch lines per bay is determined by dividingthe width of the bay, or the distance between column lines inthe width of the bay, or the distance between column lines infeet, by the maximum allowable distance between sprinklers,feet, by the maximum allowable distance between sprinklers,((LLmaxmax) . If the result is not a whole number,) . If the result is not a whole number, round up to theround up to thenearest whole number.nearest whole number.

The formula below can be used for this calculation:The formula below can be used for this calculation:

Number of branch lines = (Width of bay)/ Number of branch lines = (Width of bay)/ ( ( LLmax max ) )

The space between the beams along the column lines in exposed construction is

called a bay.

For example, an ordinary hazard system installed in a bay that iFor example, an ordinary hazard system installed in a bay that iss 35 '35 '--00" wide would" wide would

require three branch lines, since NFPA 13 mandates a maximum ofrequire three branch lines, since NFPA 13 mandates a maximum of 15 '15 '--0" between0" between

branch lines:branch lines: 35' /1535' /15’’ = 2.3 rounded to 3 branch lines= 2.3 rounded to 3 branch lines

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35' /1535' /15’’ = 2.3 rounded to 3 branch lines= 2.3 rounded to 3 branch lines

Number of branch LinesNumber of branch Lines

Ref Ref --33



13

99-- Determine the distance between branchDetermine the distance between branchLines (L).Lines (L).

linesbranchof Number

bayof WidthTotal L =

The acceptable distanceThe acceptable distance ““LL”” between the lines in the bay is obtained bybetween the lines in the bay is obtained by

dividing the width of the bay by the number of branch linesdividing the width of the bay by the number of branch lines

For exampleFor example, continuing with the building in the previous example and calc, continuing with the building in the previous example and calculate theulate the

distance L:distance L:

L= 35' /3 branches = 11.667 or 11L= 35' /3 branches = 11.667 or 11’’--88”” ..

However the distance from the end lines to the walls is L/2 = 5However the distance from the end lines to the walls is L/2 = 5 ’’--1010””

14

1010-- Determine the maximum allowableDetermine the maximum allowabledistance between sprinklers (S).distance between sprinklers (S).

NFPA 13 allows Smax to be 15 ft for light hazardlight hazard and ordinary and 1212ftft for extra hazard ,S is also obtained by the same formula ;

SS == AASS / L/ L

For example, for anFor example, for an ordinary hazardordinary hazard the NFPA 13 would requirethe NFPA 13 would require As =130 ftAs =130 ft22 (( 12.1 m12.1 m22

) per sprinkler . However for the) per sprinkler . However for the previous exampleprevious example L =11L =11’’--88”” the maximum spacingthe maximum spacing ““SS ““,,

becomes:becomes:

S=S= 130 /11130 /11’’--88”” = 11.14 ft= 11.14 ft 1111’’ ft. ( 1ft =12ft. ( 1ft =12””))



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Ordinary hazard spacingOrdinary hazard spacingexample in one bay ofexample in one bay ofbuilding .building .

Refer to drawing and checkRefer to drawing and checkto be certain that sprinklerto be certain that sprinklerspacingspacing ““SS ““ is acceptable :is acceptable :

As=As= SxSx LL

1111’’ x 11x 11’’--88”” =128.3 ft=128.3 ft22

From NFPAFrom NFPA--1313

AAmaxmax =130 ft=130 ft2.2.

Layout is AcceptableLayout is Acceptable

Ref Ref --33

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Maximum Distance from Walls& ceiling

• The distance from sprinklers to wallsshall not exceed one-half of theallowable distance ( L/2).

• Sprinklers shall be located a minimumof 4 in. (102 mm) from a wall.

Spacing (Standard Spray Upright/Standard Spray Pendent)Spacing (Standard Spray Upright/Standard Spray Pendent)

Ref Ref --11



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Deflector PositionDeflector Position

• Distance Below Ceilings: deflector and the ceilingshall be shall be a minimum of 1 in. (25.4 mm) and a maximum of12 in. (305 mm)

•• Under obstructed constructionUnder obstructed construction, the sprinklerdeflector shall be located within the horizontal planesof 1 in1 in. to 6 in6 in. (25.4 mm to 152 mm) below thestructural members and a maximum distance of 22 in22 in.(559 mm)

Standard Pendent and Upright Spray Sprinklers

The deflector distance is the dimension from of the ceiling to t The deflector distance is the dimension from of the ceiling to t he top of thehe top of the

sprinkler deflector.sprinkler deflector.

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11- Pipe sizing tables

•• Schedule size forSchedule size for light Hazardlight Hazard OccupanciesOccupanciessystems,systems,

•• Schedule size forSchedule size for Ordinary HazardOrdinary HazardOccupancies systems.Occupancies systems.

•• Schedule size forSchedule size for Extra Hazard occupanciesExtra Hazard occupancies



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Schedule size forSchedule size for light Hazardlight HazardOccupancies systemsOccupancies systems

Light Hazard Pipe SchedulesLight Hazard Pipe Schedules

(as per NFPA (as per NFPA --13.table613.table6--5.2.2)5.2.2)

SteelSteel CopperCopper

1 in.1 in.………………………… 2 sprinklers 1 in.2 sprinklers 1 in.………………………… 2 sprinklers2 sprinklers

11¼¼ in.in.…………………… 3 sprinklers 13 sprinklers 1¼¼ in.in.………………………… 3 sprinklers3 sprinklers

11 ½ ½ in.in.…………………… 5 sprinklers 15 sprinklers 1 ½ ½ in.in.………………………… 5 sprinklers5 sprinklers

2 in.2 in.………………………… 10 sprinklers 2 in.10 sprinklers 2 in.……………………………… 12 sprinklers12 sprinklers

22 ½ ½ in.in.……………… 30 sprinklers 230 sprinklers 2 ½ ½ in.in.………………………… 40 sprinklers40 sprinklers

3 in.3 in.…………………… 60 sprinklers 3 in.60 sprinklers 3 in.………………………… 65 sprinklers65 sprinklers

33 ½ ½ in.in.……………… 100 sprinklers 3100 sprinklers 3 ½ ½ in.in.………………………… 115 sprinklers115 sprinklers

Ref Ref --11

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Schedule size forSchedule size for Ordinary HazardOrdinary HazardOccupancies systemsOccupancies systems

Ordinary Hazard Pipe ScheduleOrdinary Hazard Pipe Schedule

(as per NFPA (as per NFPA --13.table613.table6--5.3.2(a))5.3.2(a))

SteelSteel CopperCopper

1 in.1 in. ………………………… 2 sprinklers 1 in.2 sprinklers 1 in. ……………………………… 2 sprinklers2 sprinklers

11¼¼ in.in.………………………… 3 sprinklers 13 sprinklers 1¼¼ in.in. ………………………… 3 sprinklers3 sprinklers

11 ½ ½ in.in.………………………… 5 sprinklers 15 sprinklers 1 ½ ½ in.in. ………………………… 5 sprinklers5 sprinklers

2 in.2 in.……………………………… 10 sprinklers 2 in.10 sprinklers 2 in. ……………………………… 12 sprinklers12 sprinklers

22 ½ ½ in.in.………………………… 20 sprinklers 220 sprinklers 2 ½ ½ in.in. ………………………… 25 sprinklers25 sprinklers

3 in.3 in. ………………………………40 sprinklers 3 in.40 sprinklers 3 in. ……………………………… 45 sprinklers45 sprinklers

33 ½ ½ in.in. …………………………65 sprinklers 365 sprinklers 3 ½ ½ in.in. ………………………… 75 sprinklers75 sprinklers




Ref Ref --11



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Schedule size forSchedule size for Extra HazardExtra Hazardoccupanciesoccupancies

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RemarkRemark ““ Sprinklers outletSprinklers outlet ””

••Minimum 1Minimum 1--in. (25.4in. (25.4--mm) outletsmm) outlets shall beshall be providedprovided

The branch lines and the sprinklers spacing andThe branch lines and the sprinklers spacing and

distance are now be dimensioned on thedistance are now be dimensioned on the

drawing with respect to the bays . Pipe sizing isdrawing with respect to the bays . Pipe sizing isthan be completed from the pipethan be completed from the pipe ScheduleSchedule

Ref Ref --11



23

H.WH.W--11

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1212--Friction Loss FormulaFriction Loss Formula ””HazenHazen--WilliamsWilliams””



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Friction Loss FormulaFriction Loss Formula ”” HazenHazen--WilliamsWilliams””

Pipe friction lossesPipe friction losses shall beshall be determined on the basis of the Hazendetermined on the basis of the Hazen--WilliamsWilliamsformula, as follows:formula, as follows:

WhereWherep = frictional loss factor inp = frictional loss factor in psi per foot of pipepsi per foot of pipeQ = flow inQ = flow in gpmgpmC = friction loss coefficient for new black steel pipe CC = friction loss coefficient for new black steel pipe C=120=120

d = actuald = actual internalinternal diameter of pipe indiameter of pipe in inchesinchesFor SI units, the following equation shall be usedFor SI units, the following equation shall be used::

WhereWherepm = frictional resistance inpm = frictional resistance in bar per meter of pipebar per meter of pipeQmQm = flow in L/min= flow in L/minC = friction loss coefficientC = friction loss coefficientdm = actual internal diameter in mmdm = actual internal diameter in mm

5

87.485.1

85.1

10)(05.6 ××

=d C

Q P f

87.485.1

85.152.4

d C

Q P f

×

×=

psipsi / foot / foot

Bar / mBar / m

26



27

Actual Actual internalinternal diameter diameter

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N.B. Choosing the Hydraulically most demandingN.B. Choosing the Hydraulically most demandingareaarea

The hydraulically most demanding area may be thegeographically most remote area, which is the area whoselinear distance from the sprinkler system riser is the longest,in feet .

Sometimes , the hydraulically most demanding area is not themost remote area. However “ When in doubt , calculate itWhen in doubt , calculate it

outout “ which means that if you are unsure which is thehydraulically most demanding area , perform calculations forother potential hydraulically most demanding areas until alluncertainty is removed.

Ref Ref --33



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13-Hydraulic -Calculation Procedures

The procedure consists of the following steps:The procedure consists of the following steps:1.1. Selecting occupancy,Selecting occupancy,

2.2. Selecting a hydraulic density,Selecting a hydraulic density,3.3. Determining the length of the hydraulically mostDetermining the length of the hydraulically most

demanding area,demanding area,4.4. Determining the number of sprinklers flowing along theDetermining the number of sprinklers flowing along the

length of the design area,length of the design area,5.5. Determining the configuration of sprinklers in theDetermining the configuration of sprinklers in the

hydraulically most demanding area,hydraulically most demanding area,6.6. Determining the minimum flow at the hydraulically mostDetermining the minimum flow at the hydraulically most

demanding sprinkler,demanding sprinkler,

7.7. Determining the minimum pressure at the hydraulicallyDetermining the minimum pressure at the hydraulicallymost demanding sprinkler,most demanding sprinkler,8.8. Determining friction loss in each pipe segment.Determining friction loss in each pipe segment.

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The correct selection of the occupancyThe correct selection of the occupancyclassification of a building or a portion of aclassification of a building or a portion of abuilding is the foundation for meaningful andbuilding is the foundation for meaningful andreliable hydraulic calculations.reliable hydraulic calculations.

Careful selection of the occupancy is theCareful selection of the occupancy is themost critical decisionmost critical decision that a fire protectionthat a fire protectionsystems designer makes during the hydraulicsystems designer makes during the hydrauliccalculation process.calculation process.A list of occupancies are summarized below;A list of occupancies are summarized below;

11-- Selection of occupancy ( refer to Chp.1)Selection of occupancy ( refer to Chp.1)

Hydraulic calculationsHydraulic calculations concon’’tt



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22-- Selecting a Hydraulic densitySelecting a Hydraulic density

Hydraulic calculationsHydraulic calculations concon’’tt

The area density curves shown in the accompanied figure 1 may be used asthe basis for your calculation in selecting the density.These curves are a function of a design density and the total design areaof sprinkler operation. The design density is the quantity of water perThe design density is the quantity of water persquare foot of remote area that has been found by experience tosquare foot of remote area that has been found by experience to bebeeffective in controlling a fire ofeffective in controlling a fire of given occupancygiven occupancy.. The design areaThe design area (A )(A ) isisan area whose size is related to the occupancy, where in all spran area whose size is related to the occupancy, where in all sprinklers ininklers inthethe area isarea is expected to actuateexpected to actuate at the same timeat the same time..

Ref Ref --33

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Area-Density Method

The area under which themaximum number of sprinklersshall operate among one risergrid, is selected from thearea-density curve

Example: From the extra

hazard occupancy group 1(curve 4) we choose the pointhaving the maximum densityand the lower protected areathat is: ( A = A = 2500ft²,0.3gpm/ft²)

Fig. 1Fig. 1



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Area/density curves. “Remarks”

• If the areas of sprinkler operation is lessthan 1500 ft2 (139 m2) for light and ordinaryhazard occupancies, the density for 1500 ft2

(139 m2) shall be selected that is ρ=0.1 and0.15 respectively.

• However for extra hazard occupancies if thecalculated areas of sprinkler operation is lessthan 2500 ft2 (232 m2), the density for 2500

ft2 (232 m2) shall be used.

Ref Ref --11

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Area-Density Method

The area under which themaximum number ofsprinklers shall operateamong one riser grid, isselected from the area-density curve

Example for ordinary hazardgroup 1 (curve 2) we choosethe point having the maximumdensity and the lowerprotected area (1500ft²,0.15gpm/ft²)

Fig. 1Fig. 1



35

Example of hydraulically most demanding area

22

11

33 44

55 Ref Ref --11

36

33-- Determining the Length of the Design AreaDetermining the Length of the Design AreaHydraulic calculationsHydraulic calculations ConCon’’tt

NFPA 13 provides a formula with which the hydraulically mostNFPA 13 provides a formula with which the hydraulically mostdemanding area is sized:demanding area is sized:The minimum length of the rectangle isThe minimum length of the rectangle is 1.21.2 times the square root of thetimes the square root of thedesign area (design area (AA):):

A designer has selected a design area ofA designer has selected a design area of 1.5001.500 feetfeet22 from thefrom thearea/density curves in Figure 1. Determine the minimum length ofarea/density curves in Figure 1. Determine the minimum length of thethedesign area.design area. ““LL”” shown in Figure 2.shown in Figure 2.

ThisThis ““ L =46.5 ftL =46.5 ft““ length is a minimum, because the length of the designlength is a minimum, because the length of the designarea must be increased soarea must be increased so thatthat the edge of the design falls at the pointthe edge of the design falls at the pointbetween two sprinklers. We will make this adjustment in the nextbetween two sprinklers. We will make this adjustment in the next step.step.

ft A L 5.4615002.12.1min ===



37

A L 2.1min =

Figure 2.Figure 2.

Ref Ref --33

38

Hydraulic calculationsHydraulic calculations ConCon’’tt44-- Number of sprinklers flowing along the lengthNumber of sprinklers flowing along the lengthof the design area of the Design Areaof the design area of the Design Area

S

L Ns min=

The number of sprinkler flowing along the length of the design aThe number of sprinkler flowing along the length of the design area isrea isdetermined by dividingdetermined by dividing LLminmin by the spacing between sprinklers ( S).by the spacing between sprinklers ( S).

For exampleFor example ; From previous we get; From previous we get LLminmin= 46.5 ft and it is known from NFPA= 46.5 ft and it is known from NFPA-- 1313the sprinkler spacing for Extrathe sprinkler spacing for Extra-- hazard group I , S= 12hazard group I , S= 12’’. Therefore:. Therefore:

.488.3/12

5.46 spisthat

sprinkler feet

feet Ns ==

The actual length of the design areaThe actual length of the design area LLactact, is determined by multiplying the, is determined by multiplying theactual number of sprinklersactual number of sprinklers , Ns ,, Ns , by the spacing between sprinklers ( S).by the spacing between sprinklers ( S).

ft S Ns Lact 48'124 =×=×=



39

44-- Configuration of sprinklers in the design AreaConfiguration of sprinklers in the design Area

Hydraulic calculationsHydraulic calculations ConCon’’tt

Once we have determined the actual length of the design area [mostremote area] , we must determine its width. As follows:

ft ft

ft

L

AW

act

25.3148

1500 2

===

act L

AW =

From Fig 3 the area A = 48' x 30' = 1440 square feet, containing 12 sprinklers.

This falls slightly short of the required 1.500 square foot design area. Since eachsprinkler covers a 12' x 10' area or 120 square feet we know that adding onesprinkler to 1440 sq ft results in an area of 1 440 sq ft plus 120 sq ft, or 1560 sqft. which meets the desired minimum criteria. The question is where is thisthirteenth sprinkler added? This fina1 sprinkler is always added closest to thecrossmain to assure the hydraulically most demanding area.

40

Given : A wet pipe sprinkler system , Occupancy ordinary hazard group I

Area of operation =1500 Sq. Ft , Density = 0.15 gpm per sq. foot

Lmin= 46.5 feet , Ns = 4 sprinklers along the length of the design area

Lact = 48 feet , W= 31.25 feet , Width covered by 3 branch lines is 30’

48’ x 30’ = 1440 sq.ft. [ 12 sprinklers]

14440 sq.ft is less 1500 sq.ft. must add one more sprinkler [13]

Fig.3Fig.3

closest to the crossmain



41

In short ,the Number of sprinklers in each branchIn short ,the Number of sprinklers in each branchline is determine as follows:line is determine as follows:

Where the design is based on area/density method, theWhere the design is based on area/density method, thedesign area shall be a rectangular area having adesign area shall be a rectangular area having adimension parallel to the branch linesdimension parallel to the branch lines at least 1.2at least 1.2timestimes the square root of the area of sprinklerthe square root of the area of sprinkleroperation (A)operation (A) used, whichused, which shallshall permit the inclusion ofpermit the inclusion ofsprinklers on both sides of the cross main.sprinklers on both sides of the cross main.

)(ftcurveDensityAreafromareaeRemontA

(ft)sprinklers betweenSpacingSWhere

2.1

line branchoneonSprinklersof Number

2−=

=

×=

S

A N

Light Hazard 15ft (4.5 m)

Ordinary Hazard 15ft

Extra – Hazard 12ft (3.6 m)

High – Piled storage 12ft

42

Example of estimating the total number ofExample of estimating the total number ofsprinklers operating at the same time & thesprinklers operating at the same time & thenumber of sprinklers in each branchesnumber of sprinklers in each branches

Ref Ref --11



43

55-- Minimum flow at the hydraulically most DemandingMinimum flow at the hydraulically most Demandingsprinkler (sprinkler ( one sprinklerone sprinkler ))

Hydraulic calculationsHydraulic calculations ConCon’’tt

gpm

ft sq ft sq gpm Asd Q

18

..120./15.0)(

=

×=×=

)( Asd Q ×=

This is the minimum required flow at the hydraulically most demaThis is the minimum required flow at the hydraulically most demandingndingsprinkler. For ordinary Hazard group Isprinkler. For ordinary Hazard group I ““AsAs”” is taken as 130 ftis taken as 130 ft22..

The minimum required flow at the hydraulically most demanding spThe minimum required flow at the hydraulically most demanding sprinkler isrinkler isobtained by multiplying the design density by the area covered bobtained by multiplying the design density by the area covered b y one y onesprinkler .sprinkler .

44

66-- Determining the minimum pressure at the hydraulicallyDetermining the minimum pressure at the hydraulicallymost demanding sprinklermost demanding sprinkler

Hydraulic calculationsHydraulic calculations

The flow at a sprinkler head is determined by the formula:

The flow at a sprinkler (Q) is equal to the sprinkler discharge coefficient(K) times the square root of the pressure (P). K is commonly referred to asthe K-factor ( coefficient of discharge) . Each sprinkler that is tested andlisted for use on a sprinkler system has a unique Ka unique K--factorfactor, or orificecoefficient, that is calculated for that sprinkler. Sprinklers havingnominal orifices of 1/2" generally have K-factors ranging from 5.3 to 5.8.

P K Q =



45

Flow Characteristic & Water spreading S.I

Units

Q=KQ=K √√PP

Where Q= flow rate inWhere Q= flow rate in L/minL/minK= K factor K= K factor

P= pressure inP= pressure in BarBar

46

Sprinkler Discharge Characteristics“K”

Yes¾ in NPT25013.5-14.519.0¾

Yes¾ in. NPT

Yes½ in. NPT

or

200

11.0-11.5

15.95/8

NO¾ in. NPT

or

1407.4-8.2

13.517/32

NO½ in. NPT1005.3-5.812.7½

Yes½ in. NPT754.0-4.411.07/16

Yes½ in. NPT502.6-2.99.53/8

Yes½ in. NPT33.31.8-2.08.05/16

Yes½ in. NPT251.3-1.56.4¼

Nominal Orifice

Size

Marked On

Frame

Thread

Type

Percent of Nominal l½ in.

Discharge

K

Factor1

(mm)

Nominal

Orifice

Size

(in)

K= 5.6



47

7- Friction Loss Formula “Hazen-Williams formula”

1)- Calculations shallshall begin at the hydraulically most remotesprinkler. Discharge at each sprinkler shall be based on thecalculated pressure at that sprinkler.

2)- Pipe friction lossPipe friction loss shall beshall be calculated in accordance withcalculated in accordance withthe Hazenthe Hazen--Williams formula.Williams formula.

87.485.1

85.152.4

d C

Q P f ×

×=

HazenHazen--Williams formula is one of the most popular friction lossWilliams formula is one of the most popular friction lossformula, recognized by NFPA 13 and considered as a standardformula, recognized by NFPA 13 and considered as a standard

formula for the calculation of the pressure drop.formula for the calculation of the pressure drop.

48

1414--Example ProblemExample ProblemThe next figure present a plan of wet pipe system layoutThe next figure present a plan of wet pipe system layout

Assuming the following data:Assuming the following data:

DensityDensity,,0.15 gpm/ft0.15 gpm/ft22 calculated over the most remote 1500calculated over the most remote 1500sq.ftsq.ft. In this example take [As= 120 ft. In this example take [As= 120 ft22],S=12 ft & L= 10ft],S=12 ft & L= 10ftandand thethe total number of Sp. is 13 sprinklers in most remotetotal number of Sp. is 13 sprinklers in most remotearea.area.

1.1. The minimum flow at the most remote sprinkler =The minimum flow at the most remote sprinkler =18 gpm18 gpm2.2. The minimum required pressure at the most remoteThe minimum required pressure at the most remote

sprinkler=10.33sprinkler=10.33 PsiPsi..Size the piping system & determine the pump duty.Size the piping system & determine the pump duty.Knowing that , the branch lines are schedule 40 black steel.Knowing that , the branch lines are schedule 40 black steel.(C=120 ) Crossmain lines are schedule 10 black steel.(C=120 ) Crossmain lines are schedule 10 black steel.



49

50

Sectional view of the buildingSectional view of the building

Ref Ref --3.3.



51

Solution in stepsSolution in steps

1. Number of sprinklers riser needed

2. Most remote area3. Area of coverage and sprinklers spacing

4. Expected total number of sprinklers per floor

5. Pipe sizing

6. Determination of the most remote area

7. Number of sprinklers in the most remote area

8. Number of sprinklers along one branch

9. Flowrate of the most remote sprinkler

10. Residual pressure of the most remote sprinkler

11. Total effective length between two adjacent sprinklers

12. Friction loss formula13. Calculating pump total pressure

14. Calculating pump total flowrate

52

Number of sprinklers riserneeded

Based on NFPA 13Based on NFPA 13--44--2.12.1

Since the parking area is less than 4831mSince the parking area is less than 4831m22, one, onesprinkler riser is sufficientsprinkler riser is sufficient



53

Area-Density Method

AAdd is theis the

mostmost

remote arearemote area

and is equaland is equal

to 1500 ftto 1500 ft22

for ordinary

Hazard 1

54

Area of Coverage/sprinkler spacing

Protection Areas and Maximum Spacing (StandardProtection Areas and Maximum Spacing (StandardSpray Upright/Standard Spray Pendent)Spray Upright/Standard Spray Pendent) forforOrdinary Hazard. NFPAOrdinary Hazard. NFPA--13 4.4.1.113 4.4.1.1

4.61512.1130AllAll

mftm2ft2System

type

Construction

type

Maximum

spacing

(s)

Protection

area (As

)



55

4. Expected total number of sprinklers per4. Expected total number of sprinklers perfloorfloor

The expected number of sprinklers is determined asThe expected number of sprinklers is determined asfollows :follows :

Number of sprinklers =Total area of the car parking / AsNumber of sprinklers =Total area of the car parking / As

Area of the car parking 13200 ftArea of the car parking 13200 ft22 ( 110 ft x 120 ft)( 110 ft x 120 ft),divided by the sprinkler coverage area As =120 ft,divided by the sprinkler coverage area As =120 ft22 we getwe get110 sprinklers.110 sprinklers.

56

1 in.1 in. ………………………… 2 sprinklers2 sprinklers

11¼¼ in.in.………………………… 3 sprinklers3 sprinklers

11½½ in.in.………………………… 5 sprinklers5 sprinklers

2 in.2 in.………………………………10 sprinklers10 sprinklers

22½½ in.in.………………………… 20 sprinklers20 sprinklers

Pipe sizingPipe sizing



57

Determining the flowrate of the most remoteDetermining the flowrate of the most remote

sprinklersprinkler(gpm(gpm)

s Adensityq ×=

2

=k

q p

Step 5- Determining the residual pressure of theDetermining the residual pressure of themost remote sprinklermost remote sprinkler

(psi)(psi)

58

Determining the total effective length betweenDetermining the total effective length betweentwo adjacent sprinklerstwo adjacent sprinklers

F LT +=T =total effective length

L= Pipe length

F= Equivalent pipe length



59

Friction Loss FormulaFriction Loss FormulaPipe friction losses shall be determined on the basisPipe friction losses shall be determined on the basis

of the Hazenof the Hazen--Williams formula, as follows:Williams formula, as follows:

Where,Where,p = frictional resistance in psi per foot of pipep = frictional resistance in psi per foot of pipeQ = flow in gpmQ = flow in gpmC = friction loss coefficient black steel C=120C = friction loss coefficient black steel C=120

d = actual internal diameter of pipe in inchesd = actual internal diameter of pipe in inches

87.485.1

85.1

52.4d C

Q P

×=

Dr. Ali Hammoud

60

Where,Where,PPff =friction loss in psi between two adjacent=friction loss in psi between two adjacentsprinklerssprinklers

P T P f

×=

12012 =+=

+= F LT



61

Calculate Total pressure drop PCalculate Total pressure drop PT T

Where,Where,PPTT = total pressure in psi= total pressure in psi

PPe = elevation pressure in psi= elevation pressure in psi

e previousT f T P P P P ++=

62

Calculate flow q of the second sprinklerCalculate flow q of the second sprinkler ““22””

T P K q =

The flowof bothsprinklers



63

1st Branch

Remember !Remember !

4 Sprinklers flow4 Sprinklers flow

q= 79.56 gpmq= 79.56 gpm

Whereas 4 x 18Whereas 4 x 18gpm =72 gpmgpm =72 gpm

12+6 =18’

64

18 gpm

19.08 gpm

37.08 gpm

20.16 gpm

22.32 gpm

79.56 gpm

79.56 gpm

57.24 gpm

Flow distributionFlow distributionin Branch 1in Branch 1



65

At = Top of riser nipple & AAt = Top of riser nipple & ABB = Bottom of riser nipple= Bottom of riser nipple

AB

AT

Pressure drop due to elevation 1.5 ‘Pe= Z x 0.433 Psi/ft

=1.5 x0.433 =0.65 Psi

F = Pressure drop due Tee = 10 ft ‘

BB

66

At = Top of riser nipple & AAt = Top of riser nipple & ABB = Bottom of riser nipple= Bottom of riser nipplePressure drop due to elevation 1.5 ‘

Pe= Z x 0.433 Psi/ft

=1.5 x0.433 = 0.65 Psi

Pressure drop due to [ 2” Tee ] = 10 ft

Friction loss Psi/100 ft from Hazen William equation with Q= 79.56 gpm &D=2” we get Pf=0.064 Psi.

The total equivalent length T = (1.5 +10) = 11.5 ft

The pressure drop Pf= 11.5x 0.064= 0.74 ft

The total pressure drop Pt = 21.35+ 0.65+ 0.74 =22.74 ft.



67

AABB = Bottom of riser nipple= Bottom of riser nipple

Branch LineBranch Line ““KK”” calculation: Imagine a huge sprinklercalculation: Imagine a huge sprinklerinstalled at nod Binstalled at nod BBB having K= 16.68.This K represents thehaving K= 16.68.This K represents thetotal flow of sprinklers 5 thru 8 @ node Btotal flow of sprinklers 5 thru 8 @ node BBB with an orificewith an orificecapable of discharging the total flow for 4 sprinklers atcapable of discharging the total flow for 4 sprinklers at

nodes 5 thru 8.nodes 5 thru 8.

Q=KQ=K xx√√PP

K= 79.56/K= 79.56/√√22.74 =16.6822.74 =16.68

Pressure drop in line A-B=10 x 0.007 = 0.07 where D= 3 ,

q=79.56 gpm and ,L= 10 ft

68

BBBB = Bottom of riser nipple & C= Bottom of riser nipple & CBB

For Pt = 22.81For Pt = 22.81 PsiPsi & K= 16.68 the flow in branch B& K= 16.68 the flow in branch B--CCq= 79.99 gpmq= 79.99 gpm

79.66 gpm



69

Sprinkler 13 flow rate Node DB

gpm

low P

high P

lowQ Adjusted Q 62.2537.11

05.2318 =×=×=

To adjust the pressure for the pressure differential betweenthe demand of sprinkler 13 at DB and sprinklers 1-12 at DB we

use the balancing equation.

It is important to note that at node 13 ,we use the same minimum flow that wecalculated for the most remote sprinkler (1 ) i.e 18 gpm as low flow for sprinkler 13.

70Dr. Ali Hammoud



71

79.66 gpm

239.3 gpm

79.56 gpm

80.08 gpm

159.22 gpm

264.3 gpm

Flow distribution in theFlow distribution in themain pipesmain pipes

25 gpm

72

Pressure drop From D-E

L=5+10+10+10+10+10=55ft.

F= 15 ft one Tee 3”

Leffective =70 ft

Pressure drop =

70x 0.062 ft==4.32 ft.

Node DB



73

Sectional view of the buildingSectional view of the building

90 Elbow =10 ft

Pressure drop From E-F

L=58’-6”. F=10’ [4” Elbow ],Leffective =68.5 ft

Pressure drop = 68.5x 0.018 = 1.32 ft.

Total Pressure is the summation of all Pt = 37.94 ft

E

74



75

Fire Hose streams:

•Add 100 gpm for light hazard occupancies

•250 gpm for ordinary hazard occupancies

•500 gpm for extra hazard occupancies

76



77

78

Equivalent Pipe Lengths of Valves and FittingsEquivalent Pipe Lengths of Valves and Fittings

(shall be used with a Hazen(shall be used with a Hazen--Williams C factor of 120 only)Williams C factor of 120 only)

Dr. Ali Hammoud BAUDr. Ali Hammoud BAU-- 20052005



79

Example problem IIExample problem II

80

Example of determining the number ofsprinklers to be calculated



81

77

99

66

88

44 55

Example from the NFPA 13Example from the NFPA 13 Car parkingCar parking

82

L=70 ftL=70 ft

77

66

99

44

22

11

33

4 DN4 DN 55



83

84

Hydraulic calculation example



85

86

Obstructions to sprinkler discharge pattern development



87

Obstruction

Wall

Obstruction

Sprinklers Location

C e i l i n g

O p e n w e b s t e e l o r

w o o d t r u s s

Dor C A 33f

[ ] D +− 8f

Ventilation ducts

88

Sprinklers under pitched roofs with sprinkler directlyunder peak; branch lines run up the slope.



89

Sprinklers at pitched roofs; branch lines run up the slope

90

Drainage

• All sprinkler pipe and fittings shall be soinstalled that the system can be drained

Riser or Main SizeRiser or Main Size SizeSize

1.1. Up to 2 in.Up to 2 in. 3/4 in.3/4 in.2.2. 2121//22 in., 3 in., 31/2 in.in., 3 in., 31/2 in. 11 ¼¼ in.in.3.3. 44in. and largerin. and larger 2 in.2 in.



91

System test connection on wetSystem test connection on wetpipe systempipe system

92

Aboveground Pipe and TubeAboveground Pipe and Tube

• Ferrous Piping (Welded and Seamless)Specification for Black and Hot-Dipped Zinc-Coated (Galvanized) Welded and Seamless SteelPipe for Fire Protection Use ASTM A 795

• Specification for Welded and Seamless Steel PipeANSI/ASTM A 53• Copper Tube (Drawn, Seamless)

Specification for Seamless Copper Tube ASTM B75Specification for Seamless Copper Water Tube

ASTM B 88



93

Underground Pipe

• Piping shall be listed for fire protection service and complywith the AWWA standards

• Cement Mortar Lining for Ductile Iron Pipe and Fittings forWater

• Polyethylene Encasement for Ductile Iron Pipe Systems• Ductile Iron and Gray Iron Fittings, 3-in. Through 48-in.,

for Water and Other Liquids• Rubber-Gasket Joints for Ductile Iron Pressure Pipe and

Fittings• Flanged Ductile Iron Pipe with Ductile Iron or Gray Iron

Threaded Flanges• Ductile Iron Pipe, Centrifugally Cast for Water• Steel Water Pipe 6 in. and Larger

• Coal-Tar Protective Coatings and Linings for Steel WaterPipelines Enamel and Tape — Hot Applied• Cement-Mortar Protective Lining and Coating for Steel

Water Pipe 4 in. and Larger — Shop Applied• Field Welding of Steel Water Pipe

94



95

References

1.1. NFPANFPA--200420042.2. F.FallF.Fall ““ Building services & equipmentBuilding services & equipment””

VolVol 11--22--333.3. R.R. gagnongagnon Fire protection systems.Fire protection systems.4.4. The Design Project & AutoCAD drawingThe Design Project & AutoCAD drawing

by Uplandby Upland ––engineeringengineering ““Dr. HammoudDr. Hammoud””

5.5. Photos From WebsPhotos From Webs……..

hydraulic calculation chap-3

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