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[247]

ANALYTICAL CALCULATION OF AUTOMATIC SPRINKLER FIRE

EXTINGUISHING SYSTEM (SFES)

Mr. Ketan Gurunath Shirsat*1

*1Project Engineer, Oriental Fire Tech System, Kurla- (East), Mumbai, India.

ABSTRACT

Fire Fighting Sprinkler System is a first aid means of controlling fire. This study is based on “How to calculate

pressure & flow rate of sprinkler fire extinguishing system (SFES)”. SFES has been proven for the safety of any

areas like commercial, residential Buildings, industries, etc. Fire can be control by eliminating 3 of any 1 factor.

i.e., fuel, oxygen & heat. SFES Controls the fire by means of removing heat by throwing water over the area

which is covered with fire. For Installing SFES, the installer must know the requirement of pressure & flow rate

to be set on the pump parameters. While Designing SFES for a particular area in India, the Designer must follow

the National Fire Protection Standard - 13(NFPA-13), IS- 15105 Standard & NBC Codes. Once the installation

has been completed according to the standard, the Designer must do the analysis of appropriate Design Density,

Flow Rate & Working Pressure requirement by finding Assume Maximum Area of Operation (AMAO).

Keywords: Hydraulic Calculation Of Branched Type Fire Sprinkler Protection System, Pressure, Flow Rate,

Design Density, Assume Maximum Area Of Operation (AMAO), Fitting & Valve Equivalent.

I. INTRODUCTION

SFES is of Four types. Those are Wet Pipe Sprinkler System, Dry Pipe Sprinkler System, Deluge Sprinkler

System, Dry Pipe Pre- Action System. It further classified based on Piping Layout i.e., Branched Piping with

Dead End, Grid Pattern, Grid Pattern with Loops. We will focus on Branched Piping with Dead End Calculations

only.

AMAO – It is an area which is at largest extremity from the pump. i.e., for buildings uppermost floor with

farthest extremity from Main Riser tapping. Which means if that area sprinklers get required pressure to

control fire of respective area, then rest of the area sprinklers will surely get required pressure to operate

during fire.

As we know SFES is First Aid means system. So, it will design only for the few numbers of sprinklers to be

activated (so that at start only the sprinklers will extinguish the fire)

For Calculating Pressure to be set for pump, we start from AMAO location to the pump location., from farthest

sprinkler of uppermost floor to the pump.

II. METHODOLOGY

Let’s take an example of floor layout of particular Light Hazard building structure. It is a 19-floor building

with 1 Podium & 2 Basements.

Firstly, check the AMAO location. How AMAO area shall be taken? It has been taken as per IS-15105 Standard.

From the table 1, for light hazard, we have taken AMAO = 904 ft2 (84 m2) & Minimum Design Density = 0.05522

gpm/ft2 (2.25 lpm/m2). As per IS-15105, Minimum sprinkler Discharge Pressure at any sprinkler nozzle shall

be 10.1526psi(0.70bar).

Note: Above details are for the Light hazard building structure installation only.

Table 1





[248]

Fig 1

(See the above Fig. 1)

A. At point ① Sprinkler

We take pressure as 10.1526 psi. Let us assume sprinkler is of K = 5.6

Relation between Flow Rate & Pressure for Sprinkler orifice,

Q = K P

Q = Design Density or Flow Rate through Sprinkler Orifice (gpm)

P = Pressure at the entry of Sprinkler Orifice (psi)

K = Sprinkler Constant (Given by Manufacturer as per orifice size)

So,

Q₁ = 5.6 x 10.1526 = 17.842 gpm Area Covered by Sprinkler at ① = 96.87 ft2 (9m2)

From Nodal point ① to Nodal point ③, frictional pressure loss in pipe line shall be calculated from Hazen -

Williams empirical formula,





[249]

p = L x 4.52 x Q1.85

C1.85x d4.87

p = Frictional Pressure Loss in Pipe (psi) Q = Flow Rate from Pipe (gpm)

L = Length of Pipe (feet)

C = Frictional Coefficient of Pipe = 120 (from NFPA-13 :- for schedule 40 Steel Pipe)

d = Internal Diameter of Pipe (inch)

Here are few simplified formulae of Frictional Pressure Loss through Pipe for each different internal diameter.

d = 1.049 inch (25mm)

p = 5.0989 x 10-4 x Q1.85 x L

d = 1.380 inch (32mm)

p = 1.3410 x 10-4 X Q1.85 x L

d = 1.610 inch (40mm)

p = 6.3301 x 10-5 x Q1.85 x L

d = 2.067 inch (50mm)

p = 1.8747 x 10-5 x Q1.85 x L

d = 2.469 inch (65mm)

p = 7.8899 x 10-6 x Q1.85 x L

d = 3.068 inch (80mm)

p = 2.7394 x 10-6 x Q1.85 x L

d = 4.026 inch (100mm)

p = 7.2931 x 10-7 x Q1.85 x L

d = 6.065 inch (150mm)

p = 9.9144 x 10-8 x Q1.85 x L

For ① to Nodal ③point,

Nominal length of pipe (l)= 0.9842 ft (300mm)

For finding equivalent length of fitting refer

Table 2

Table 2 [2]

Here we have

1T section (d = 25mm) = 5 ft --- (Table 2)

So total equivalent Length

L = length of pipe + fitting equivalent





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= 0.9842 + 5

= 5.9842 ft

Pressure loss through Pipe becomes

p = 5.0989 X 10-4 X (17.842)1.85 X 5.9842

= 0.6304 psi

Total Pressure required at Junction,

(P) = 0.6304 + 10.1526 = 10.7824 psi

B. For ② to Nodal Point ③

Length of Pipe (l) = 0.164 ft

1L Section (d = 25mm) = 2 ft ---- (Table 2)

Total equivalent length (L) = 0.164 +2

= 2.164 ft

Note: At junction (Nodal) point, there should be zero hydraulic pressure difference from all branches.

So, Total Pressure at junction ③ node,

(P) = 10.7824 psi

Area Covered by Sprinkler at ② = 96.87 ft2 (9m2)

10.7824 = Pressure at Sprinkler② + frictional loss through pipe between ② to ③

10.7824= (Q₂/K)² + 5.0989 x 10-4 x Q1.85 x L

10.7824= (Q₂/5.6)² + 5.0989 x 10-4 x Q1.85 x 2.164

Q₂= 18.1859 gpm

C. For Nodal Point ③ to Nodal Point ⑤

Total Flow Rate through Pipe (Q₃) = Q₁ +Q₂

= 17.842 + 18.1859

= 36.0279 gpm

For Diameter = 25mm


1 T section (d = 25mm) = 5 ft ---- (Table 2)

2 L Section (d = 25mm) = 2 x 2 ft =4 ft ---(Table 2)

Total Equivalent Length (L) (d =25) = 22.1234 ft

For Diameter = 32mm,

Total Length of Pipe (l) = 0.5249 ft

No Fitting of pipes used.

Total Equivalent Length (L) = 0.5249 ft

Pressure loss through Pipe between ③ to ⑤,

p = Pressure loss due to (d = 25mm) + Pressure loss due to (d = 32mm)

p = 5.0989 x 10-4 x Q₃1.85 x L + 1.3410 x 10-4 x Q₃1.85 x L

p = 5.0989 x 10-4 x (36.0279)1.85 x 22.1234 + 1.3410 x 10-4 x (36.0279)1.85 x 0.5249

p = 8.6061psi

Total Pressure required at Nodal Point ⑤,

(P) = 8.6061 + Pressure required at Nodal Point ③

= 8.6061 + 10.7824

= 19.3885 psi

D. For Point ④ to Nodal Point ⑤





[251]

So,

Total Pressure at junction ⑤ node,

(P) = 19.3885 psi

Area Covered by Sprinkler= 72.65 ft2 (6.74 m2)

19.3885 = Pressure at Sprinkler④ + Pressure loss through pipe between ④ to ⑤

19.3885 = (Q₄/K)² + 5.0989 x 10-4 x Q₄1.85 x L



So,

Total Equivalent Length of Pipe (L) = 4.4291 ft

19.388= (Q₄/5.6)² + 5.098 x 10-4 x Q₄1.85 x 4.4291

Q₄ = 24.1337 gpm

E. For Nodal Point ⑤ to Point ⑥

Total Flow Rate (Q₅) = Q₃ + Q₄

= 36.0279 + 24.1337

= 60.1616 gpm

For Diameter = 32mm


1 L Section (d =32) = 1 x 3 ft = 3 ft --- (Table 2)


For Diameter = 40mm

Total Length of Pipe (l) = 0.2296 ft



Pressure loss through Pipe between Nodal Point ⑤ to Point ⑥,

(p) = Pressure loss due to (d = 32mm) + Pressure loss due to (d = 40mm)

P = 1.3410 x 10-4 x Q₅1.85 x L + 6.3301 x 10-5 x Q₅1.85 x L

P = 1.3410 x 10-4 x (60.1616)1.85 x 8.4133 + 6.3301 X 10-5 x (60.1616 )1.85 x 0.2296

P = 2.237 psi

Total Pressure Required at Point ⑥,

P = 21.6255 psi

F. At point ⑥ I Sprinkler

Area Covered by Sprinkler at ⑥ = 90.4 ft2 (8.4 m2)

21.6255 = [(Q₆)I / 5.6]2

(Q₆)I = 26.041 gpm

Total Flow Rate required between point ⑧ to Nodal Point ⑥,

Q₆= (Q₆)I + Q₅ = 86.2034 gpm

G. For point ⑥ to Nodal Point ⑧

d = 40mm


1 T section (d=40mm) = 8 ft ---- (Table 2)

1 L Section (d=40mm) = 4 ft ---- (Table 2)

Total Length of Pipe (L) = 29.6837 ft

Pressure loss through Pipe,





[252]

p = 6.3301 x 10-5 x (Q₆)1.85 x L

p = 6.3301 x 10-5 x (86.2034)1.85 x 29.6837

p = 7.1556 psi

Total Pressure require at Nodal Point ⑧

p = 7.1556 + Pressure Require at Point⑥

p = 7.1556 + 21.6255

p = 28.7811 psi

H. For Nodal point ⑦ to Nodal Point ⑧

So,

Total Pressure at junction Point ⑧,

(P) = 28.7811 psi

Q₇ = (Q₇) I + (Q₇) II

Do trial & Error Method to find exact flow rate,

*As per NFPA = Pressure at each Sprinkler Should be greater than 10.1526 psi (0.70 bar)

1) Assuming pressure at ⑦I Point sprinkler

P = 15 psi

Area Covered by ⑦I Sprinkler =96.87 ft2(8.4 m2)

P = (Q /K)2

15 = [(Q₇) I /5.6]²

(Q₇) I = 21.688 gpm

Nominal Length of Pipe ⑦I to ⑦ = 11.4829 ft

1 L Section (d = 25mm) = 2 ft ---- (Table 2)

Total equivalent length of pipe ⑦I to ⑦,

L =13.4829 ft

Frictional loss through Pipe ⑦I to ⑦,

p = 5.0989 x 10-4 x (Qᴬ)1.85 x 13.4829

p = 2.038 psi

Pressure required at ⑦,

P = Pressure at ⑦I Point sprinkler + Frictional loss through Pipe ⑦I to ⑦

P = 15 + 2.038

P = 17.038 psi

Calculating flow rate through (Q₇) II sprinkler,

Area Covered by ⑦II Sprinkler =96.87 ft2 (8.4 m2)

Nominal Length of Pipe ⑦II to ⑦ = 5.0853 ft


Total equivalent length of pipe ⑦II to ⑦

L = 7.0853 ft

P at ⑦ = Pressure at Sprinkler ⑦II + Frictional loss through Pipe ⑦II to ⑦

17.038 = 5.0989 x 10-4 x [(Q₇) II]1.85 x 7.0853 + [(Q₇) II /5.6]²

(Q₇) II = 22.334gpm

(Q₇) = (Q₇) I + (Q₇) II

(Q₇) = 44.022 gpm

Nominal Length of Pipe ⑦ to ⑧,

l (d = 25mm) = 10.8268 ft





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l (d = 50mm) = 0.6561ft

1T Section (d = 25mm) = 5 ft ---- (Table 2)

No Fitting of pipes (d= 50mm) used.

Total equivalent length of pipe ⑦ to ⑧,

L (d = 25mm) = 15.8268 ft

L (d = 50mm) = 0.6561ft

Pressure required at Nodal point ⑧,

P = Frictional Loss through ⑦ to ⑧ (d =25) + Frictional loss through ⑦ to ⑧ (d =50) + Pressure at ⑦

P = 5.0989 x 10-4 x (Q₇)1.85 x 15.8268 + 1.8747 x 10-5 x (Q₇)1.85 x 0.6561 + 17.038

P = 25.9161 psi

Total pressure at junction ⑧ = 25.9161 psi (need pressure equal to 28.7811 psi)

Neglected


P = 16 psi


P = (Q /K)2

16 = [(Q₇) I /5.6]²

(Q₇) I = 22.4 gpm



Total equivalent length of pipe ⑦I to ⑦II,

L =13.4829 ft


p = 5.0989 x 10-4 x (Qᴬ)1.85 x 13.4829

p = 2.163 psi



P = 16 + 2.163

P = 18.163 psi


Area Covered by ⑦II Sprinkler = 96.87 ft2(8.4 m2)



Total equivalent length of pipe ⑦II to ⑦,

L = 7.0853 ft


18.163 = 5.0989 x 10-4 x [(Q₇) I I]1.85 x 7.0853 + [(Q₇) II /5.6]²

(Q₇) II = 23.064gpm

(Q₇) = (Q₇) I + (Q₇) II

(Q₇) = 45.464 gpm


l (d = 25mm) = 10.8268 ft

l (d = 50mm) = 0.6561ft






[254]



L (d = 25mm) = 15.8268 ft

L (d = 50mm) = 0.6561ft


P = Frictional Loss through ⑦ to ⑧ (d =25) + Frictional loss through ⑦ to ⑧ (d =50) + Pressure at ⑦

P = 5.0989 x 10-4 x (Q₇)1.85 x 15.8268 + 1.8747 x 10-5 x (Q₇)1.85 x 0.6561 + 18.163

P = 27.5865 psi

Total pressure at junction ⑧ = 27.5865 psi (need pressure equal to 28.7811 psi)

Neglected

* Note: - if pressure at sprinkler ⑦I = 15 psi, then total pressure at junction ⑧= 25.9161 psi

& If pressure at sprinkler ⑦I = 16 psi, then total pressure at junction ⑧= 27.5865 psi

So,

Rate of increase per psi = pressure due to 16 psi – pressure due to 15 psi

= 27.5865 – 25.9161

= 1.6704 psi

Required increase in psi = Pressure at ⑧ – pressure due to (p = 16 psi)

= 28.7811 – 27.5865

= 1.1946 psi

So, pressure at sprinkler ⑦I = 16 psi + (Required increase in psi / Rate of increase per psi)

= 16 + (1.1946/ 1.6704)

P = 16.686 psi


P = 16.686 psi


P = (Q /K)2

16.68 = [(Q₇) I /5.6]²

(Q₇) I = 22.8753 gpm



Total equivalent length of pipe ⑦I to ⑦,

L =13.4829 ft


p = 5.0989 x 10-4 x (Qᴬ)1.85 x 13.4829

p = 2.249 psi



P = 16.686 + 2.249

P = 18.9353 psi


Area Covered by ⑦II Sprinkler = 96.87 ft2(8.4 m2)



Total equivalent length of ⑦II to ⑦,





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L = 7.0853 ft


18.9353 = 5.0989 x 10-4 x [(Q₇) I I]1.85 x 7.0853 + [(Q₇) II /5.6]²

(Q₇) II = 23.5517 gpm

(Q₇) = (Q₇) I + (Q₇) II

(Q₇) = 46.427 gpm


l (d = 25mm) = 10.8268 ft

l (d = 50mm) = 0.6561ft




L (d = 25mm) = 15.8268 ft

L (d = 50mm) = 0.6561ft


P = Frictional Loss through ⑦ to ⑧ (d =25) +Frictional loss through ⑦ to ⑧ (d =50) + Pressure at ⑦

P = 5.0989 x 10-4 x (Q₇)1.85 x 15.8268 + 1.8747 x 10-5 x (Q₇)1.85 x 0.6561 + 18.9353

P = 28.7314 psi

Total pressure at junction ⑧ = 28.7314 psi (need pressure 28.7811 psi)

Accepted

I. For Nodal point ⑧ to ⑩

Q₈ = Q₇ + Q₆

= 86.2034 + 46.427

Q₈ = 132.6304 gpm

Nominal Length of Pipe = 5.9055 ft

1T Section (d = 50) = 10 ft ---- (Table 2)

Total equivalent length of pipe ⑧ to ⑩,

L = 15.9055 ft

Frictional loss through Pipe ⑧ to ⑩,

p = 1.8747 x 10-5 x Q₈1.85 x L

= 1.8747 x 10-5 x (132.6304)1.85 x 15.9055

p = 2.5198 psi

Total required pressure at junction ⑩,

P = Pressure at ⑧ + Frictional loss through Pipe ⑧ to ⑩

= 28.7314 + 2.5198

P = 31.3009 psi

For Nodal point ⑨ to ⑩,

Q₁₀ = Q₈ + Q₉

Q₉ = (Q₉)I + (Q₉)III + (Q₉)IV

Do Trial & Error Method to find exact flow rate,

*As per NFPA = Pressure at each Sprinkler Should be greater than 10.1526 psi (0.70 bar)

1) Assuming pressure at ⑨I Node sprinkler,

P = 20 psi

Area Covered by Sprinkler ⑨I = 79.43 ft2(7.37m2)





[256]

P = (Q /K)2

20 = [(Q₉)I / 5.6]²

(Q₉)I = 25 gpm

Nominal Length of Pipe ⑨I to ⑨III

l (d =25) = 200 + 1050 + 50 =1300mm = 4.2649 ft

l (d =32) = 200 + 1000= 1200mm = 3.9369 ft

l (d =40) = 20 mm = 0.0656 ft

2 L Section (d = 25mm) = 2 ft x 2 = 4 ft

1 T Section (d = 32mm) = 6 ft x 1 = 6 ft


Total equivalent length of pipe ⑨I to ⑨III,

L (d = 25mm) = 8.2649 ft

L (d = 32mm) = 9.9369 ft

L (d = 40mm) = 0.0656 ft

Frictional loss through pipe ⑨I to ⑨III,

p = pressure loss through (d=25) pipe + pressure loss through (d=32) pipe + pressure loss through (d=40)

pipe

= 5.0989 x 10-4 x [(Q₉)I]1.85 x 8.2649 +1.3410 x 10-4 x [(Q₉)I]1.85 x 9.9369 + 6.3301 x 10-5 x [(Q₉)I]1.85 x 0.0656

= 5.0989 x 10-4 x (25)1.85 x 8.2649 + 1.3410 x 10-4 x (25)1.85 x 9.9369 + 6.3301 x 10-5 x (25)1.85 x 0.0656

p = 2.1406 psi

Pressure at Nodal point (Q₉)III,

P = Pressure at sprinkler (Q₉)I + Frictional loss through pipe ⑨I to ⑨III

= 20 + 2.1406

P = 22.1406 psi

Flow rate though Sprinkler ⑨II,

Area Covered by Sprinkler ⑨II= 85.57 ft2 (7.97m2)

P = (Q /K)2

22.1406 = [(Q₉)II /5.6]²

(Q₉)II = 26.35 gpm

(Q₉)III = (Q₉)I + (Q₉)II

= 25 + 26.35

(Q₉)III = 51.35 gpm

Nominal Length of Pipe ⑨ to ⑨III,

l = 5380 + 1000= 6380 mm = 20.9317 ft

1T Section (d = 40mm) =8 ft x 1= 8 ft --(Table 2)

1L Section (d = 40mm) =4 ft x 1= 4 ft --(Table 2)

Total equivalent length of pipe ⑨ to ⑨III

L= 32.9369 ft

Frictional loss through pipe ⑨to ⑨III,

p = 6.3301 x 10-5 x (51.35)1.85 x 32.9369

= 3.0450 psi

Total pressure at ⑨,

P = Pressure loss through pipe ⑨ to ⑨III + Pressure at Nodal point ⑨III

= 3.0450 + 22.1406





[257]

P = 25.1856 psi

For calculating flow rate through sprinkler ⑨IV,

Area Covered by Sprinkler ⑨IV= 96.87 ft2 (9 m2)

Nominal Length of Pipe ⑨ to ⑨IV,

l (d =25mm) = 2410mm = 7.9068 ft

l (d =40mm) =170mm= 0.5577 ft

1 L Section (d = 25mm) =2 ft x 1 =2 ft --(Table 2)


Total equivalent length of pipe ⑨ to ⑨IV,

L (d= 25) = 9.9068 ft

L (d= 40) = 0.5577 ft


P = pressure at ⑨IV Sprinkler + Frictional loss through (d=25mm) + Frictional loss through (d=40mm)

25.1856 = [(Q₉)IV /5.6]² + 5.0989 x 10-4 x [(Q₉)IV]1.85 x 9.9068 + 6.3301 x 10-5 x [(Q₉)IV]1.85 x 0.5577

(Q₉)IV = 26.8276 gpm

Flow Rate at ⑨,

Q₉ = (Q₉) III + (Q₉) IV

= 51.35 + 26.8276

Q₉ = 78.1776 gpm

To find pressure at ⑩,

Nominal Length of Pipe ⑨ to ⑩,

l = 2000mm = 6.5616 ft

1T Section (d = 40mm) =8 ft x 1= 8 ft --(Table 2)

Total equivalent length of pipe ⑨ to ⑩

L = 14.5616 ft

Pressure Loss through pipe ⑨ to ⑩,

p = 6.3301 x 10-5 x (Q₉)1.85 x 14.5616

= 6.3301 x 10-5 x (78.1776)1.85 x 14.5616

p = 2.9296 psi

Total Pressure at ⑩,

P = 2.9296 + 25.1856

P = 28.1152 psi (required pressure = 31.30 psi)

Neglected

2) Assuming pressure at ⑨I Node sprinkler,

P = 22.27 psi

Area Covered by Sprinkler ⑨I = 79.43 ft2(7.37m2)

P = (Q /K)2

22.27 = [(Q₉)I / 5.6]²

(Q₉)I = 26.427 gpm

Nominal Length of Pipe ⑨I to ⑨III

l (d= 25) =200 + 1050 + 50 =1300mm =4.2649 ft

l (d= 32) = 200 + 1000= 1200mm = 3.9369 ft

l (d= 40) = 20 mm = 0.0656 ft

2L Section (d= 25mm) =2 ft x 2= 4 ft --(Table 2)





[258]

1T Section (d= 32mm) =6 ft x 1= 6 ft --(Table 2)


Total equivalent length of pipe ⑨I to ⑨III,

L (d= 25mm) = 8.2649 ft

L (d= 32mm) = 9.9369 ft

L (d= 40mm) = 0.0656 ft

Frictional loss through pipe ⑨I to ⑨III,

p = pressure loss through (d=25mm) pipe + pressure loss through (d=32mm) pipe + pressure loss through

(d=40) pipe

= 5.0989 x 10-4 x [(Q₉)I]1.85 x 8.2649 +1.3410 x 10-4 x [(Q₉)I]1.85 x 9.9369 + 6.3301 x 10-5 x [(Q₉)I]1.85 x 0.0656

= 5.0989 x 10-4 x (26.427)1.85 x 8.2649 + 1.3410 x 10-4 x (26.427)1.85 x 9.9369 + 6.3301 x 10-5 x (26.427)1.85 x

0.0656

p = 2.3721 psi

Pressure at Nodal point (Q₉)III,

P = Pressure at sprinkler (Q₉)I + Frictional loss through pipe ⑨I to ⑨III

= 22.27 + 2.3721

P = 24.6421 psi

Flow rate though Sprinkler ⑨II,

Area Covered by Sprinkler ⑨II= 85.57 ft2 (7.97m2)

P = (Q /K)2

24.6421= [(Q₉)II /5.6]²

(Q₉)II = 27.798 gpm

(Q₉)III = (Q₉)I + (Q₉)II

= 26.427 + 27.798

(Q₉)III = 54.225 gpm

Nominal Length of Pipe ⑨ to ⑨III,

l = 5380 + 1000= 6380 mm = 20.9317 ft

1 T Section (d = 40mm) = 8 ft x 1 = 8 ft

1 L Section (d = 40mm) = 4 ft x 1 = 4 ft

Total equivalent length of pipe ⑨ to ⑨III

L= 32.9369 ft

Frictional loss through pipe ⑨to ⑨III,

p = 6.3301 x 10-5 x (54.225)1.85 x 32.9369

= 3.3674 psi


P = Pressure loss through pipe ⑨ to ⑨III + Pressure at Nodal point ⑨III

= 3.3674 + 24.6421

P = 28.0095 psi

For calculating flow rate through sprinkler ⑨IV,

Area Covered by Sprinkler ⑨IV= 96.87 ft2 (9 m2)

Nominal Length of Pipe ⑨ to ⑨IV,

l (d =25mm) = 2410mm = 7.9068 ft

l (d =40mm) =170mm= 0.5577 ft

1L Section (d =25mm) =2 ft x 1 =2 ft --(Table 2)





[259]


Total equivalent length of pipe ⑨ to ⑨IV,

L (d= 25) = 9.9068 ft

L (d= 40) = 0.5577 ft


P = pressure at ⑨IV Sprinkler + Frictional loss through (d=25mm) + Frictional loss through (d=40mm)

28.0095 = [(Q₉)IV /5.6]² + 5.0989 x 10-4 x [(Q₉)IV]1.85 x 9.9068 + 6.3301 x 10-5 x [(Q₉)IV]1.85 x 0.5577

(Q₉)IV = 28.3017 gpm

Flow Rate at ⑨,

Q₉ = (Q₉) III + (Q₉) IV

= 54.225 + 28.3017

Q₉ = 82.5276 gpm

To find pressure at ⑩,

Nominal Length of Pipe ⑨ to ⑩,

l = 2000mm = 6.5616 ft

1T Section (d= 40mm) =8 ft x 1 =8 ft --(Table 2)

Total equivalent length of pipe ⑨ to ⑩

L = 14.5616 ft

Pressure Loss through pipe ⑨ to ⑩,

p = 6.3301 x 10-5 x (Q₉)1.85 x 14.5616

= 6.3301 x 10-5 x (82.5276)1.85 x 14.5616

p = 3.2383 psi

Total Pressure at ⑩,

P = 3.2383 + 28.0095

P = 31.25 psi (Required pressure = 31.30 psi)

Accepted

J. Flow rate at Nodal Point ⑩

Q₁₀ = Q₈ + Q₉

= 132.6304 + 82.5276

= 215.158 gpm

Pressure & Flow Rate required at ⑫,

Between ⑩ to ⑫,

Nominal length of pipe = 600mm =1.9685 ft

1Cross Section (d= 65mm)=12 ft ----(Table 2)

Total Equivalent length of pipe = 13.9685 ft

Pressure loss through pipe ⑩ to ⑫,

p = 7.8899 x 10-6 x (Q₁₀)1.85 x L

= 7.8899 x 10-6 x (215.158)1.85 x 13.9685

p = 2.2794 psi

K. For Nodal Point at ⑫

P = Pressure loss through pipe ⑩ to ⑫ + Pressure at Node ⑩

= 2.2794 + 31.30

P = 33.5803 psi

Flow rate required at ⑫,





[260]

Q₁₂ = Q₁₀ + Q₁₁

For finding Q₁₁,

Between ⑪ to ⑫

Nominal length of pipe (l)= 50mm = 0.164 ft

1 L Section (d = 25mm) = 2 ft ----(Table 2)

(For sprinkler point 1 L bow is attached at the end.)

Total Equivalent length of pipe (L) = 2.164 ft

Area Covered by sprinkler at ⑪= 96.87ft²(8.40m²)

Pressure at Nodal Point ⑫ = Pressure Loss through pipe between ⑪ to ⑫ + pressure at sprinkler ⑪

33.5803 psi = 5.0989 X 10-4 x (Q₁₁)1.85 x 2.164 + (Q₁₁ /5.6)²

Q₁₁ = 32.1225 gpm

So,

Q₁₂ = Q₁₀ + Q₁₁

= 215.158 + 32.1225

Q₁₂ = 247.2805 gpm

L. For Nodal Point at ⑬,

Nominal length of pipe ⑫ to ⑬,

l = 3000 + 1400 = 4400mm = 14.4356 ft

1T Section (d = 65mm) = 12 ft --(Table 2)

1Butter Fly Valve (d=65mm) =7 ft --(Table 2)

(For Floor Maintenance)

Total Equivalent Length ⑫ to ⑬,

L = 33.4356 ft

Frictional loss through pipe between ⑫ & ⑬,

p = 7.8899 x 10-6 x (Q₁₂)1.85 x L

= 7.8899 x 10-6 x (247.2805)1.85 x 33.4365

p = 7.058 psi

Total Required pressure at ⑬,

P = Pressure loss through pipe between ⑫ to ⑬ + Pressure at Nodal Point ⑫

= 7.058 + 33.5803

= 40.6389 psi

Total Required pressure at ⑬= 40.6389 psi

Flow Rate required at ⑬,

Q₁₃ = Q₁₂= 247.2805gpm





[261]

Fig. 2





[262]

(See the Fig. 2)

M. For Nodal Point at ⑭

From 19th floor to 2nd floor,

Each floor height = 3200mm

So, Nominal length of pipe from ⑬ to ⑭

L = 3200 x 18 floor = 57600mm = 188.9764 ft

1 T Section (d=150mm) = 30 ft = 30ft --(Table 2)

Total equivalent length of pipe = 218.9764 ft

Note: - Elevation Pressure required,

Pe = 0.433 x height of elevation

= 0.433 x h

So, Total pressure required at ⑭,

P = Frictional loss through Pipe ⑬ to ⑭ + Elevation loss through pipe ⑬ to ⑭ +Total Required pressure at ⑬

= 9.9144 x 10-8 x (Q₁₃)1.85 x 218.9764 + 0.433 x 188.9764 + 40.6389

= 0.5808 + 81.8267 + 40.6389

P = 123.0464 psi

Total pressure required at ⑭ (P) = 123.0464 psi

Fig. 3


N. For Nodal Point at ⑮,

For 1st floor,

Pressure loss in pipe between ⑭ & ⑮

Nominal length of pipe ⑭ to ⑮,

l = 600 + 3000 + 600 = 4200mm = 13.7795 ft

4 L Section (d=150mm) = 14 ft x 4 =56ft-(Table 2)

Total equivalent length,

L = 13.7795+ 56 = 69.7795 ft

Frictional loss in pipe between ⑭ & ⑮

p =9.9144 x 10-8 x (Q₁₃)1.85 x 69.7795

= 9.9144 x 10-8 x (247.2805)1.85 x 69.7795

= 0.185 psi

Total Pressure Required at ⑮,





[263]

P = Pressure loss in pipe between ⑭ to ⑮ + Total pressure required at ⑭

= 0.185 + 123.0464

= 123.2314 psi

Pressure required at Nodal Point⑯,

From 1st floor to ground floor

Nominal pipe length,

l = 3200 + 4500 + 5000 + 150

= 12850mm = 42.1787 ft

Total equivalent length of pipe (L)= 42.1787 ft

So,

Total pressure required at Nodal Point ⑯,

P = Frictional loss through Pipe ⑮ to ⑯+Elevation loss through pipe ⑮ to ⑯ +Total Required pressure at ⑮

= 9.9144 x 10-8 x (Q₁₃)1.85 x 42.1787 + 0.433 x 42.1787

= 0.1118 + 18.2633 + 123.2314

= 141.6065 psi

Total pressure required at ⑯= 141.6065 psi

Fig. 4


O. For Nodal Point at ⑰

For 1st Basement

Pressure loss in pipe between ⑯ to ⑰

Nominal length of pipe ⑯ to ⑰

l = 124800mm = 409.448 ft

26 L Section (d=150mm) =14ft x 26 ----(Table 2)





[264]

= 364 ft

2Butter Fly Valves (d=150mm) =10 ft x 2-(Table2)

= 20 ft

(At gong valve connection on the ground floor)

Total equivalent length ⑯ to ⑰,

L = 793.448 ft

Elevation Height at Gong valve Connection = 7.5459 ft

Pressure loss between ⑯ & ⑰

p = Frictional loss through Pipe ⑯ & ⑰ + Elevation loss through pipe at gong valve

= 9.9144 x 10-8 x (247.2805)1.85 x 793.448 + 0.433 X 7.5459

= 2.1046 + 3.2673

p = 5.3719 psi

Total Pressure Required at ⑰,

P = Pressure loss in pipe between ⑯ & ⑰ + Total pressure required at ⑯

= 5.3719 + 141.6065 = 146.9784 psi

P. For Nodal Point at ⑱

From Basement 1 to Basement 2

Nominal pipe length (L) = 5000mm = 16.4042 ft

Total equivalent length of pipe = 16.4042 ft

So,

Total pressure required at Nodal Point ⑱,

P = Frictional loss through pipe ⑰ to ⑱ + Elevation loss through pipe ⑰ to ⑱ +Total Required pressure at ⑰

= 9.9144 x 10-8 x (Q₁₃)1.85 x 16.4042 + 0.433 x 16.4042 + 146.9784

= 0.0435 + 7.1030

P = 154.0814 psi

Total pressure required at ⑱= 154.0814 psi

Fig. 5

Total pressure required at ⑱= 154.0814 psi





[265]


Q. At Pump Outlet ⑲,

Elevation pressure loss from ⑱ to ⑲,

Height of 2nd basement = 7000mm = 22.9658 ft,

Elevation loss = 0.4333 X 22.9658 = 9.9442 psi

Nominal pipe length from ⑱ to ⑲,

l = 650+800+3300+2800+5000+12000(Extra inside pump R/m)

= 24550mm = 80.5446 ft

10L Section (d=150mm) =14 ft x10 -----(Table 2)

=140 ft

2 Butter Fly Valves at pump Outlet (d=150mm)

= 10ft x 2 = 20 ft ----(Table 2)

Total equivalent length of pipe ⑱ to ⑲

L = 240.5446 ft

Pressure loss between ⑱ to ⑲,

p = Frictional loss through Pipe ⑱ to ⑲ + Elevation loss through pipe ⑱ to ⑲

= 9.9144 X 10-8 X (Q₁₃)1.85 X 240.5446 + 9.9442

= 0.638 + 9.9442

= 10.5822 psi

So, Total pressure required at pump outlet ⑲,

P = Pressure loss between ⑱ to ⑲ + Total Required pressure at ⑱

= 10.5822 + 154.0814 psi

= 164.6636 psi

III. RESULT

Minimum pressure required at Pump Outlet,

P = 164.6636 psi = 165 psi = 12 Bar

Minimum Flow Rate Through Pump,

Q = 248 gpm = 938.783 lpm

AMAO Considered = 904 ft² = 84 m²

(As per IS15105) [3]

Minimum Pressure required at each sprinkler = 10.152 psi = 0.701 bar [3]

ABBREVIATIONS

AMAO, Assume Maximum Area of Operation; SFES, Sprinkler Fire Extinguishing system;

IV. CONCLUSION

The obtained result is matched with the Software generated calculation.

From the above example, we can easily perform Automatic Fire Sprinkler system calculation. Moreover, we can

say that, bigger diameter pipe would have lower frictional loss than smaller diameter pipe. Research paper

would help the Fire Fighting Project Engineers to analyze the flow of water through Fire system pipe. This

calculation would make the system to work precisely in fire hazard condition rather than thumb rule

calculations. It has many benefits from building the efficient and cheapest Fire system to the reliable systems.

ACKNOWLEDGEMENT

I would like to thank my father Mr. Gurunath Shirsat for sharing his past 15 years of experience of this field.

Special thanks to Mr. Siddesh Ghugare & Team for providing me the possible infrastructure to complete this

work. Also, I would like to give sincere thanks to Mr. Aanish sir for guiding me throughout the period.





[266]

V. REFERENCES [1] https://en.wikipedia.org/wiki/Hydraulic_calculat ion

[2] http://arco-hvac.ir/wp- content/ uploads/2015 /05/James_D._Lake_Nfpa _13_Automatic

_Sprinkler_SystemBookZZ.org_.pdf

[3] https://archive.org/details/gov.in.is.15105.2002

[4] https://www.hdfire.com/pdf/alarm-valve/HD_247_Alarm_Valve_Model-H.pdf

[5] http://www.kirloskarpumps.com/product-pump-end-suction-pumps.aspx

[6] https://archive.org/details/nationalbuilding01

[7] https://archive.org/details/nationalbuilding02

[8] https://www.tyco-fire.com /index.php? P=tdsect &S=S4

[9] http://www.newagefireprotection.com/product-detail.php?productId=112

[10] https://fdocuments.in/document/automatic-sprinkler-system-calculations.html

https://www.tyco-fire.com/

https://fdocuments.in/document/automatic-sprinkler-system-calculations.html