DPS Issue No

8/3/2019 DPS Issue No

1/18

DPSDESIGN OF PLUMBING SYSTEMS

Official Publication of the New Master

Plumbers / Plumbing Engineers

SUSTAINED PROFESSIONAL EDUCATION FOR RMPS AND PES

HYDRAULICCALCULATION

FORAUTOMATIC

WET SPRINKLER

SYSTEMS


2/18

INTRODUCTION

This publication is the result of inspiration from the previous publication of American Society of

Plumbing Engineers. No attempt however has been made to copy ASPEs publication but emulate

its effort to provide materials for continuing education to its members and non-members.

Design of Plumbing Systems (DPS) is a working pamphlet and the result of syndicated effort of

concerned RMPs (Registered Master Plumbers) to extend and sustain education to peers who

cannot attend formal CPE ( Continuing Professional Education) seminars or schooling. This is also

a preparatory material to RMPs as-would be Plumbing Engineers in the near future.

HOW TO USE THIS PAMPHLET


This pamphlet is a guide to system designers of wet sprinkler systems as outlined in the

following pages in simplified approach.

This guide is not in any way or manner intended to supplant any systems used by qualified fire

protection engineers to design an efficient automatic wet sprinkler system and any

pretense to provide a complete and sufficient guide is hereby explicitly disclaimed.

A questionnaire is provided at the end of this pamphlet. The user of this guide is suggested toanswer the questions related to discussion and return to the publishers the questionnaire

sheet with answers for validation purposes. The publishers are now in the process of seeking

accreditation from the Professional Regulation Commission (PRC) for CPE credit points.

The user will find in this Guide a straight-forward approach in the design which calls for

his/her creativity, judgment, and decisiveness which cannot be incorporated in any book or

instructional manuals.

The user should always refer to NFPA No. 13 during the process of calculations.

English system is used throughout in the calculation.

DPS PublishersIssue No. 001

December 2011


3/18

HYDRAULIC CALCULATION FOR WET SPRINKLER SYSTEM

(TREE- TYPE)

1. After creating the sprinkler piping layout determine the most remote area of operation.

2. Calculate the number of sprinkler heads and branches within the sprinkler area of operation.

3. Determine the fire hazard classification of the commodity being protected.

4. Select the applicable density from density/area curve of NFPA No. 13.

5. Select the applicable area of coverage for the remotest sprinkler head from NFPA No. 13

table.

6. Use the standard form for hydraulic calculation from NFPA No. 13.

7. Provide sizes on the sprinkler piping layout from branches, cross main, feed main, riser and

down to the riser pump discharge outlet.

CALCULATION PROCEDURES


There are three basic occupancy hazard classifications outlined in NFPA No. 13:

1. Light Hazard

2. Ordinary Hazard: Group 1 and Group 2.

3. Extra Hazard: Group 1 and Group 2.

OCCUPANCY HAZARD CLASSIFICATIONS

DPS-Issue No. 001- December 2011 Page 3


4/18


In this pamphlet we are going to present an example of calculation for an open space

office with no drop ceiling. In NFPA No. 13 offices are considered under light hazard

classification.

DENSITY AND REMOTE AREA

As can be seen from the above graph for light hazard curve and 1,500 sq. ft. most

demanding remote area of sprinkler operation, the density to use is 0.10 gpm/sq. ft . This

means that sprinklers will operate effectively within the 1,500 sq. ft. in any fire situation.

Density- the amount of water that must be delivered to the protected area every minute for

every square foot of floor space.

Remote Area- the farthest area of the floor space where minimum water should be

delivered. For example in a protected remote area of 1,500 sq. ft. the minimum amount of

water required should be 1,500 sq. ft. X 0.10 gpm/sq. ft. = 150 gpm.


REMOTE AREA AS INDICATED BY HATCHED LINES


5/18



CALCULATIONS

Based on the selected 1,500 sq. ft. (as shaded in the sample layout) area of sprinkler

operation, calculate now the number of effective sprinkler heads:

No. of Sprinkler Heads= Design Area/ Sprinkler Coverage Area = 1500/200= 7.5 say 81/2

No. of Sprinklers on branch Line= 1.2A/S where A= sprinkler operation area, sq. ft.

S= sprinkler head spacing, ft.1/2

Thus, No. of Sprinklers on Branch Line= 1.2(1,500)/ 10 = 4.65 calculate 4.

Since there are 8 sprinklers calculated two branches are considered for calculation in the

remote area.

NOTES:

As per table below the maximum sprinkler coverage for light hazard application is

225 sq.ft for hydraulically calculated with a maximum sprinkler head to head spacing

of 15 ft. In the calculation above 200 sq. ft. and 10 ft spacing are used.

It is a matter of judgment and consideration of engineering economy to assign lower

values without sacrificing the integrity of the system design.

Protection Areas and Maximum Spacing (Standard SprayUpright/Standard

Spray Pendent) for Light Hazard

Protecti

Construction Type System Typeft

2

n Area Spacing (

m

2 f

Noncombustible obstructed and Pipe schedule 200 18.6 15

unobstructed and combustible

unobstructedHydraulically

calculated

225 20.9 15

Combustible obstructed All 168 15.6 15

Combustible with members less than 3 All 130 12.1 15

ft on center


6/18


prinkler

ocationFlow (gpm) Pipe Size

Fittings &

DevicesPipe Equivalent Length

C-

factorpsi/ft/ Req. Pressure

K-

factor

q Nom

None

lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe


HYDRAULIC CALCULATION FORM


7/18



Using the hydraulic calculation form and referring to sample layout above calculate nowthe flow requirement of the farthest sprinkler head as follows:

Q = As x D where: As= Area of sprinkler coverage, sq. ft.

D = Density, gpm/sq. ft.

Q = 200 x 0.10 = 20 gpm

Calculate the operating pressure required for the 20 gpm flow from:

P1 = (Q1/k) where: Q1= 20 gpm

k = 5.6

P1 = (20/5.6) = 12.75 psi say 13 psi.

Per NFPA -13 minimum operating pressure required is 7 psi.

Now, calculate the pressure loss due to pipe and fittings from sprinkler head to sprinklerhead using Hazen- William formula:

STEP 1

Pf =4.52Q1.85 L

C1.85

D4.87

Where,

Pf= Total friction loss in piece of pipe (psi)

Q = Total flow in pipe (gpm)

L = Length of pipe (feet)C = C-factor of pipe (from NFPA-13 (2007)

D = Diameter of pipe (inches)

Per NFPA -13 minimum operating pressure required is 7 psi.

Q = 20 gpm

C = 120 from NFPA-13 table as shown below.

L = 10 ft. from sample layout.

D = 1.049 inch inside diameter from any pipe handbook.


8/18



The value of friction loss due to pipes and fittings after substitution, Pf= 0.13

Pressure required due to pipes and fittings shall be, Pf= 12 x 0.13 = 1.6 psi.

Assume the there is no change in elevation, pressure required due to elevation is Pe = 0.

The pressure required now from Sprinkler to Sprinkler shall be, Pt = 13 + 1.6 + 0 = 14.6psi.

Sprinkler

LocationFlow (gpm) Pipe Size

Fittings & DevicesPipe Equivalent Lengt

C-

factorpsi/ft/Req. Pressure

K-

factor

q Nom 1 1 elbow lgth 10 120 0.13 Pt 13. 5.6

Q 20 Act. 1.049ftg 2 Pf 1.6tot 12 P 0

q Nom lgth Pt 14.6

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf


9/18



Equivalent Schedule 40 Steel Pipe Length Chart

Fittings and Valves Fittings and Valves Expressed in E

1/2

in.

3/4

in.

1

in.11/4

in.

11/2

in.

2

in.21/2

in.

3 3

in. in

45 elbow 1 1 1 2

90 standard elbow 1 2 2 3 4

90 long-turn elbow 0.5 1 2 2 2

Tee or cross (flow turned 90) 3 4 5 6 8

2 3

5 6

3 4

10 12

3

7

5

15 1

Butterfly valve

Gate valve

6 7 10

1 1 1

Swing check* 5 7 9

For SI units, 1 in. = 25.4 mm; 1 ft = 0.3048 m.Notes:

1. This table applies to all types of pipe listed in Table 8-4.4.5.

11 14 16 1

2. Information on 1/2-in. pipe is included in this table only because it is allowed under 5-13.20.2 and 5-13.20.3.

* Due to the variations in design of swing check valves, the pipe equivalents indicated in this table are considered ave

Light Hazard Pipe SchedulesSteel Copper

1 in. 2 sprinklers 1 in. 2 sprinklers

11/4 in.

11/2 in.

2 in.

3 sprinklers

5 sprinklers

10 sprinklers

11/4 in.

11/2 in.

2 in.

3 sprinklers

5 sprinklers

12 sprinklers

21/2 in.

3 in.

30 sprinklers

60 sprinklers

21/2 in.

3 in.

40 sprinklers

65 sprinklers

31/2 in.

4 in.

100 sprinklers

See Section 5-2

31/2 in.

4 in.

115 sprinklers

See Section 5-2

For SI units, 1 in. = 25.4 mm.


10/18



Sprinkler


Fittings & DevicesPipe Equivalent Lengt

C-

factorpsi/ft Req. Pressure

K-

factor

q Nom 1 1 elbow lgth 10 120 0.13 Pt 13. 5.6

Q 20 Act. 1.049ftg 2 Pf 1.6tot 12 P 0

q 21 Nom 1 1 tee lgth 10 120 0.49 Pt 14.6 5.6to

Q 41 Act. 1.049ftg 5 Pf 7.3tot 15 P 0

q Nom lgth Pt 21.9 5.6

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

tot P

q Nom lgth Pt

Q Act.ftg Pf

STEP 2

Compute discharge capacity of sprinkler as follows:

Q = k p

Where,Q = Flow from individual sprinklerk = Sprinkler k-factor (See NFPA-13, 2007) = 5.6p = Pressure from an individual sprinkler. = 14.6 psi.

Substituting the values in the formula, Q2 = 21.4, say 21 gpm.

Now, add sprinkler and sprinkler = 20 + 21 = 41 gpm

Compute for pressure loss using the formula in page 7, Pf = 0.49 psi/ft

Pressure required due to pipes and fittings, Pt to = 0.49 x 15 = 7.35 say 7.30 psi.

Again, there is no change in elevation and we equate, Pe to = 0


11/18



Sprinkler


Fittings &


C-

factorpsi/ft/Req. Pressure

K-

factor

q Nom 1 1 elbow lgth 10 120 0.13 Pt 1

5.6

Q 20 Act.1.049 ftg 2 Pf 1.6

tot 12 Pe 0

q 21 Nom 1 1 tee lgth 10 120 0.49 Pt 14.6 5.6to

Q41

Act.1.049 ftg 5 Pf 7.3

tot 15 Pe 0

q 26 Nom 11/4 1 tee lgth 10 120 0.32 Pt 21.9 5.6 to Q 67 Act. 1.38 ftg 6 Pf 5.1

tot 16 Pe 0 q 29 Nom 11/2 1 tee lgth Pt 27.0

t o Q

96Act.

1.61 ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q

Nom

lgth

Pt

Q Act.ftg Pf

STEP 3

Compute the flow capacity of sprinkler using the formula in page 10, Q3 = 26.2 or 26 gpm.

Now, add sprinkler and sprinkler= 41 + 26 = 67 gpm

Pressure loss from equation in page 7 shall now be= 0.32 psi/ ft.

From equation in page10, q4 = 29 gpm

Then, add Q2-3 and q4 = 67 + 29 = 96 gpm. This is the total flow requirement from sprinklerup to sprinkler.


12/18



STEP 4

The total flow requirement of the 4 remote sprinklers along the remote branch at 96 gpm isnow known. Looking back at the sample layout in page 4, there is 1 more branch to consider

as covered by the 1,500 sq.ft operating and most demanding area ( See hatched area).

To determine the total flow requirement of the most demanding area we proceed as follows:

Calculate the pressure required from sprinkler to point at the cross main. (The crossmain is the sub-main pipe where the branches are connected. The branches are pipes where

the sprinkler heads are installed. And the feed main is the main pipe connected to the riser.)

From Hazen-Williams formula, p 3-4 = 0.29 psi/ft

The flow requirement does not change and still at 96 gpm. But the elevation of the cross main

is assumed at 1 foot below the branch pipe,

Thus, Pe5 = 1 ft x 1 psi/ 2.32 ft. = 0.433 psi. Use 0.4 psi.

From calculation sheet in page 14, the total pressure required at is 33.5 psi.

This time the flow coefficient , k= 5.6 does not apply anymore. Since the flow and pressurerequirements are known, 96 gpm and 33.5 psi, respectively, we can now compute for the

system flow coefficient, ks.1/2

From formula in page 10, ks = 96/ (33.5) = 96/ 5.79 = 16.58 say 16.6

The pressure loss along the cross main from to, P 5-6 = 0.013 psi/ft

Pressure required from to, Pf 5-6 = 25 ft x 0.013 psi/ft = 0.3 psi

Total pressure required at now, Pt = 33.5 + 0.3 + 0 = 33.8 psi1/2

Flow at, Q6 = 16.6 x (33.8) = 96.5 gpm which is the requirement of the second branchwithin the remote operating area.

The system flow requirement of the most demanding area calculated, Q = Q5-6+ Q6

Q = 96 + 96.5 = 192.5 gpm. The 0.5 gpm takes care of the possible loss from toand balances the flow requirements of the 2 branches within the area of operation.


13/18



Hose Stream Demand and Water Supply Duration Requirements forHydraulically Calculated Systems

Occupancy or Commodity Classification Inside Hose (gpm) Total Combined Inside

and Outside Hose (gpm

Light hazard 0, 50, or 100 100

Ordinary hazard 0, 50, or 100 250

Extra hazard 0, 50, or 100 500

Calculation does not end here. The hose requirement should be added to the calculated system

flow requirement. From the table below select 50 gpm to represent inside hose.

Thus, Qs = 192.5 + 50 = 242.5 gpm . Calculate 250 gpm standard capacity of fire pump perNFPA No. 20. ( This will be covered in future issues of DPS.)

Assuming that our sample layout is located at the fifth floor of a 5-storey office building 4

meter high per storey. Fire pump is located 4 meters horizontally from the riser(base of) andthe following valves and fittings are to be considered before the fire pump: 3 elbows, 2 gate

valves, 1 alarm check valve1 swing-type check valve. Size of the riser and discharge pipe is

4 (100mm).

Calculation continues: From to there is a required floor control valve in the feed mainbefore the connection to the riser. Using the system flow of 250 gpm compute the totalpressure required from to and from to the fire pump .

Finally, the fire water flow and pressure requirements of the sample layout are calculated as

shown in the calculation sheet below: 250 gpm by 72.2 or 72 psi.


14/18



CALCULATION SHEET

Sprinkler


Fittings &


C-


K-

factor

q Nom 1 1 elbow lgth 10 120 0.13 Pt 13

5.6

Q 20 Act. 1.049ftg 2 Pf 1.6tot 12 Pe 0

q 21 Nom 1 1 tee lgth 10 120 0.49 Pt 14.6 5.6 to

Q41

Act.1.049 ftg 5 Pf 7.3

tot 15 Pe 0

q 26 Nom 11/4 1 tee lgth 10 120 0.32 Pt 21.9 5.6 to Q 67 Act. 1.38 ftg 6 Pf 5.1

tot 16 Pe 0 q 29 Nom 11/2 2 tee lgth 5 120 0.29 Pt 27.0 16.6

t o Q

96Act.

1.61 ftg 16 Pf 6.1tot 21 Pe 0.4

q 0 Nom 3 1 tee lgth 10 120 0.013 Pt 33.5

Q 96 Act. 3.068ftg 15 Pf 0.3

tot 25 Pe 0 to q 96.5 Nom 3 1 gate valve

1 elbow

lgth 86.9 120 0.075 Pt 33.8

Q192.5

Act.3.068 ftg 8 Pf 7.1

tot 94.9 Pe 0

Inside Hose q 50 Nom 4 2 gate valve

3 elbow

lgth 65.6 120 0.02 Pt 40.9

Q242.5 or

250Act.

4.026 ftg 78 Pf 2.9tot 143.6 Pe 28.4

q Nom 1 check valve

1 ac. valve

lgth Pt 72.2

Q250

Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pftot Pe

q Nom lgth Pt


15/18



SPE Questions- Hydraulic Calculations for Wet Sprinkler System

About This Month IssueThis December 2011 issue deals withhydraulic calculations for a tree-type

wet sprinkler layout. This pamphlet

presents the calculations in a step by

step manner from determining the

number of sprinkler heads to fire

water and pressure requirements of

the system. Methodology is based on

NFPA No. 13.

Sustained Education from Design of Plumbing Systems Version

For those who find it difficult to attend a formal

continuing education course due to reasons oftime and distance availability this issuance of

DPS on- line is a good means to help RMPs

maintain their design technical know-how and

obtain CPE points. (CPE accreditation from PRC is in

process.)

Questionnaire and sample design problem is

provided below to test your understanding of the

article presented. An answer sheet is also includedfor your submission to [email protected]

for grading purposes. Corrected and graded answer

will be sent back to you.

1. A calculated sprinkler system in which

pipe sizes are selected on a pressure loss

basis to provide a prescribed water

density, flow and minimum discharge

pressure.

a. Hydraulically design system.

b. Pipe schedule system.

c. none of the above.

2. A sprinkler system in which the pipe

sizing is selected from a schedule that is

determined by the occupancy classifications.

a. Hydraulically design system.

b. Pipe schedule systemc. All of the above.

3. The pipes in which the sprinklers are installed

and usually connected to the cross main.

a. Riser.

b. Feed main.

c. None of the above.

4. The pipes supplying the branch lines

directly or through the riser.

a. Cross main.b. Feed main.

c. All of the above.

5. The pipes supplying cross mains.

a. Feed main.

b. Riser.

c. Branch line.

6. The vertical pipes supply pipes in a

sprinkler system.

a. Downspout.

b. Riser

c. Vertical feed main.

7. The above ground supply pipe directly

connected to the water supply.

a. System riser.

b. Main pipe.

c. None of the above.
mailto:[email protected]:[email protected]


16/18



8. In the sample problem below what is the hazard classification of the occupancy? Operating

area of the sprinklers? Design density?

9. How many effective sprinklers are there in the area of sprinkler operation? Branches?

Sprinklers per branch?

10. Provide a tree-type wet sprinkler layout in the tenth floor of a 10- storey residential building.

The tenth floor is a covered open area used by the residents for functions. The riser location is as

indicated in the floor layout. Each floor is 3.5 meters high and there is no drop ceiling. The fire

pump is located about 3 meters from the base of the riser. Calculate the system fire water and

the pressure required to effectively meet the sprinkler system requirement using the hydraulic

calculation sheet provided in the answer sheet.


17/18


Sprinkler


Fittings &Devices

Pipe Equivalent LengthC-


K-

factor

q Nom

None

lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act. ftg Pftot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe

q Nom lgth Pt

Q Act.ftg Pf

tot Pe


Answer Sheet- Email to [email protected]

1.0 ___________

2.0 ___________3.0 ___________

4.0 ___________

5.0 ___________

6.0 ___________

7.0 ___________8.0 ___________

9.0 ___________

10.0 See calculation sheet below.

HYDRAULIC CALCULATION FORM


18/18


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The Plumbing Professionals is the

official magazine of the New Master

Plumbers/ Plumbing Engineers and

a mouthpiece for all serious

plumbing practitioners whose

aspirations are to realize a true andprofessional organization . The

Plumbing Professionals will deal on

all issues related to the practice of

plumbing locally and internationally

DPS Publishers wish to extend their sincerest

gratitude to relentless and untiring support of

all contributors. No amount of water nor any

kind of sprinkler systems can extinguish the fire

in their hearts to serve and share their technical

know -how to all plumbing professionals.

Official Magazine of the New Master Plumbers / Plumbing Engineers

Vol.1 No.1

Dec.2011

DPS Issue No

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