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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
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[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,
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[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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[250]
= 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
Length of Pipe (l) = 13.1234 ft
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
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[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
Length of Pipe (l) = 4.4291 ft
No Fitting of pipes used.
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
Length of Pipe (l) = 5.4133 ft
1 L Section (d =32) = 1 x 3 ft = 3 ft --- (Table 2)
Total Equivalent Length (L) = 8.4133 ft
For Diameter = 40mm
Total Length of Pipe (l) = 0.2296 ft
No Fitting of pipes used.
Total Equivalent Length (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
Length of Pipe (l) = 17.6837 ft
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,
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[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
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
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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[253]
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
2) Assuming pressure at ⑦I Point sprinkler
P = 16 psi
Area Covered by ⑦I Sprinkler =96.87 ft2(8.4 m2)
P = (Q /K)2
16 = [(Q₇) I /5.6]²
(Q₇) I = 22.4 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 ⑦II,
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 = 2.163 psi
Pressure required at ⑦,
P = Pressure at ⑦I Point sprinkler + Frictional loss through Pipe ⑦I to ⑦
P = 16 + 2.163
P = 18.163 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
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
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
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
Nominal Length of Pipe ⑦ to ⑧,
l (d = 25mm) = 10.8268 ft
l (d = 50mm) = 0.6561ft
1T Section (d = 25mm) = 5 ft ---- (Table 2)
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[254]
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 + 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
3) Assuming pressure at ⑦I Point sprinkler
P = 16.686 psi
Area Covered by ⑦I Sprinkler =96.87 ft2(8.4 m2)
P = (Q /K)2
16.68 = [(Q₇) I /5.6]²
(Q₇) I = 22.8753 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 = 2.249 psi
Pressure required at ⑦,
P = Pressure at ⑦I Point sprinkler + Frictional loss through Pipe ⑦I to ⑦
P = 16.686 + 2.249
P = 18.9353 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
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of ⑦II to ⑦,
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[255]
L = 7.0853 ft
P at ⑦ = Pressure at Sprinkler ⑦II + Frictional loss through Pipe ⑦II to ⑦
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
Nominal Length of Pipe ⑦ to ⑧,
l (d = 25mm) = 10.8268 ft
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 + 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)
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[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
No Fitting of pipes (d= 50mm) used.
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
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[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)
No Fitting of pipes (d= 40mm) used.
Total equivalent length of pipe ⑨ to ⑨IV,
L (d= 25) = 9.9068 ft
L (d= 40) = 0.5577 ft
Total pressure at ⑨,
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)
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[258]
1T Section (d= 32mm) =6 ft x 1= 6 ft --(Table 2)
No Fitting of pipes (d= 50mm) used.
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
Total pressure at ⑨,
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)
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[259]
No Fitting of pipes (d= 40mm) used.
Total equivalent length of pipe ⑨ to ⑨IV,
L (d= 25) = 9.9068 ft
L (d= 40) = 0.5577 ft
Total pressure at ⑨,
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,
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[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
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[261]
Fig. 2
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[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
(See the above 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,
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[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
(See the above 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)
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[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
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[265]
(See the above Fig. 5)
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.
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[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