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### Transcript of ANALYTICAL CALCULATION OF AUTOMATIC SPRINKLER FIRE

( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[247]
EXTINGUISHING SYSTEM (SFES)
*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
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[248]
(See the above Fig. 1)
A. At point 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,
Q = 5.6 x 10.1526 = 17.842 gpm Area Covered by Sprinkler at = 96.87 ft2 (9m2)
From Nodal point to Nodal point , frictional pressure loss in pipe line shall be calculated from Hazen -
Williams empirical formula,
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[249]
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)
d = 1.380 inch (32mm)
d = 1.610 inch (40mm)
d = 2.067 inch (50mm)
d = 2.469 inch (65mm)
d = 3.068 inch (80mm)
d = 4.026 inch (100mm)
d = 6.065 inch (150mm)
For to Nodal point,
Nominal length of pipe (l)= 0.9842 ft (300mm)
For finding equivalent length of fitting refer
Table 2
So total equivalent Length
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[250]
p = 5.0989 X 10-4 X (17.842)1.85 X 5.9842
= 0.6304 psi
B. For to Nodal Point
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 node,
(P) = 10.7824 psi
Area Covered by Sprinkler at = 96.87 ft2 (9m2)
10.7824 = Pressure at Sprinkler + frictional loss through pipe between to
10.7824= (Q/K)² + 5.0989 x 10-4 x Q1.85 x L
10.7824= (Q/5.6)² + 5.0989 x 10-4 x Q1.85 x 2.164
Q= 18.1859 gpm
C. For Nodal Point to Nodal Point
Total Flow Rate through Pipe (Q) = Q +Q
= 17.842 + 18.1859
= 36.0279 gpm
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,
No Fitting of pipes used.
Total Equivalent Length (L) = 0.5249 ft
Pressure loss through Pipe between to ,
p = Pressure loss due to (d = 25mm) + Pressure loss due to (d = 32mm)
p = 5.0989 x 10-4 x Q1.85 x L + 1.3410 x 10-4 x Q1.85 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
(P) = 8.6061 + Pressure required at Nodal Point
= 8.6061 + 10.7824
= 19.3885 psi
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[251]
So,
(P) = 19.3885 psi
Area Covered by Sprinkler= 72.65 ft2 (6.74 m2)
19.3885 = Pressure at Sprinkler + Pressure loss through pipe between to
19.3885 = (Q/K)² + 5.0989 x 10-4 x Q1.85 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= (Q/5.6)² + 5.098 x 10-4 x Q1.85 x 4.4291
Q = 24.1337 gpm
Total Flow Rate (Q) = Q + Q
= 36.0279 + 24.1337
= 60.1616 gpm
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
No Fitting of pipes used.
Total Equivalent Length (L) = 0.2296 ft
Pressure loss through Pipe between Nodal Point to Point ,
(p) = Pressure loss due to (d = 32mm) + Pressure loss due to (d = 40mm)
P = 1.3410 x 10-4 x Q1.85 x L + 6.3301 x 10-5 x Q1.85 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
P = 21.6255 psi
Area Covered by Sprinkler at = 90.4 ft2 (8.4 m2)
21.6255 = [(Q)I / 5.6]2
(Q)I = 26.041 gpm
Total Flow Rate required between point to Nodal Point ,
Q= (Q)I + Q = 86.2034 gpm
G. For point to Nodal Point
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,
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
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[252]
p = 6.3301 x 10-5 x (86.2034)1.85 x 29.6837
p = 7.1556 psi
p = 7.1556 + 21.6255
p = 28.7811 psi
So,
(P) = 28.7811 psi
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 I Point sprinkler
P = 15 psi
P = (Q /K)2
(Q) I = 21.688 gpm
Nominal Length of Pipe I to = 11.4829 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of pipe I to ,
L =13.4829 ft
p = 2.038 psi
Pressure required at ,
P = Pressure at I Point sprinkler + Frictional loss through Pipe I to
P = 15 + 2.038
P = 17.038 psi
Area Covered by II Sprinkler =96.87 ft2 (8.4 m2)
Nominal Length of Pipe II to = 5.0853 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of pipe II to
L = 7.0853 ft
P at = Pressure at Sprinkler II + Frictional loss through Pipe II to
17.038 = 5.0989 x 10-4 x [(Q) II]1.85 x 7.0853 + [(Q) II /5.6]²
(Q) II = 22.334gpm
(Q) = 44.022 gpm
l (d = 25mm) = 10.8268 ft
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[253]
No Fitting of pipes (d= 50mm) used.
Total equivalent length of pipe to ,
L (d = 25mm) = 15.8268 ft
L (d = 50mm) = 0.6561ft
Pressure required at Nodal point ,
P = Frictional Loss through to (d =25) + Frictional loss through to (d =50) + Pressure at
P = 5.0989 x 10-4 x (Q)1.85 x 15.8268 + 1.8747 x 10-5 x (Q)1.85 x 0.6561 + 17.038
P = 25.9161 psi
Total pressure at junction = 25.9161 psi (need pressure equal to 28.7811 psi)
Neglected
P = 16 psi
P = (Q /K)2
(Q) I = 22.4 gpm
Nominal Length of Pipe I to = 11.4829 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of pipe I to II,
L =13.4829 ft
p = 2.163 psi
Pressure required at ,
P = Pressure at I Point sprinkler + Frictional loss through Pipe I to
P = 16 + 2.163
P = 18.163 psi
Area Covered by II Sprinkler = 96.87 ft2(8.4 m2)
Nominal Length of Pipe II to = 5.0853 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of pipe II to ,
L = 7.0853 ft
P at = Pressure at Sprinkler II + Frictional loss through Pipe II to
18.163 = 5.0989 x 10-4 x [(Q) I I]1.85 x 7.0853 + [(Q) II /5.6]²
(Q) II = 23.064gpm
(Q) = 45.464 gpm
l (d = 25mm) = 10.8268 ft
l (d = 50mm) = 0.6561ft
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[254]
Total equivalent length of pipe to ,
L (d = 25mm) = 15.8268 ft
L (d = 50mm) = 0.6561ft
Pressure required at Nodal point ,
P = Frictional Loss through to (d =25) + Frictional loss through to (d =50) + Pressure at
P = 5.0989 x 10-4 x (Q)1.85 x 15.8268 + 1.8747 x 10-5 x (Q)1.85 x 0.6561 + 18.163
P = 27.5865 psi
Total pressure at junction = 27.5865 psi (need pressure equal to 28.7811 psi)
Neglected
* Note: - if pressure at sprinkler I = 15 psi, then total pressure at junction = 25.9161 psi
& If pressure at sprinkler I = 16 psi, then total pressure at junction = 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 – pressure due to (p = 16 psi)
= 28.7811 – 27.5865
= 1.1946 psi
So, pressure at sprinkler I = 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
(Q) I = 22.8753 gpm
Nominal Length of Pipe I to = 11.4829 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of pipe I to ,
L =13.4829 ft
p = 2.249 psi
Pressure required at ,
P = Pressure at I Point sprinkler + Frictional loss through Pipe I to
P = 16.686 + 2.249
P = 18.9353 psi
Area Covered by II Sprinkler = 96.87 ft2(8.4 m2)
Nominal Length of Pipe II to = 5.0853 ft
1 L Section (d = 25mm) = 2 ft ---- (Table 2)
Total equivalent length of II to ,
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[255]
L = 7.0853 ft
P at = Pressure at Sprinkler II + Frictional loss through Pipe II to
18.9353 = 5.0989 x 10-4 x [(Q) I I]1.85 x 7.0853 + [(Q) II /5.6]²
(Q) II = 23.5517 gpm
(Q) = (Q) I + (Q) II
(Q) = 46.427 gpm
l (d = 25mm) = 10.8268 ft
l (d = 50mm) = 0.6561ft
No Fitting of pipes (d= 50mm) used.
Total equivalent length of pipe to ,
L (d = 25mm) = 15.8268 ft
L (d = 50mm) = 0.6561ft
Pressure required at Nodal point ,
P = Frictional Loss through to (d =25) +Frictional loss through to (d =50) + Pressure at
P = 5.0989 x 10-4 x (Q)1.85 x 15.8268 + 1.8747 x 10-5 x (Q)1.85 x 0.6561 + 18.9353
P = 28.7314 psi
Total pressure at junction = 28.7314 psi (need pressure 28.7811 psi)
Accepted
Q = Q + Q
= 86.2034 + 46.427
1T Section (d = 50) = 10 ft ---- (Table 2)
Total equivalent length of pipe to ,
L = 15.9055 ft
p = 1.8747 x 10-5 x Q1.85 x L
= 1.8747 x 10-5 x (132.6304)1.85 x 15.9055
p = 2.5198 psi
Total required pressure at junction ,
P = Pressure at + Frictional loss through Pipe to
= 28.7314 + 2.5198
Q = (Q)I + (Q)III + (Q)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 I Node sprinkler,
P = 20 psi
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[256]
(Q)I = 25 gpm
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 I to III,
L (d = 25mm) = 8.2649 ft
L (d = 32mm) = 9.9369 ft
L (d = 40mm) = 0.0656 ft
Frictional loss through pipe I to III,
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 [(Q)I]1.85 x 8.2649 +1.3410 x 10-4 x [(Q)I]1.85 x 9.9369 + 6.3301 x 10-5 x [(Q)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 (Q)III,
P = Pressure at sprinkler (Q)I + Frictional loss through pipe I to III
= 20 + 2.1406
Area Covered by Sprinkler II= 85.57 ft2 (7.97m2)
P = (Q /K)2
(Q)II = 26.35 gpm
= 25 + 26.35
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 to III
L= 32.9369 ft
= 3.0450 psi
Total pressure at ,
P = Pressure loss through pipe to III + Pressure at Nodal point III
= 3.0450 + 22.1406
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
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[257]
Area Covered by Sprinkler IV= 96.87 ft2 (9 m2)
Nominal Length of Pipe to IV,
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 to IV,
L (d= 25) = 9.9068 ft
L (d= 40) = 0.5577 ft
Total pressure at ,
P = pressure at IV Sprinkler + Frictional loss through (d=25mm) + Frictional loss through (d=40mm)
25.1856 = [(Q)IV /5.6]² + 5.0989 x 10-4 x [(Q)IV]1.85 x 9.9068 + 6.3301 x 10-5 x [(Q)IV]1.85 x 0.5577
(Q)IV = 26.8276 gpm
Flow Rate at ,
= 51.35 + 26.8276
Nominal Length of Pipe to ,
l = 2000mm = 6.5616 ft
1T Section (d = 40mm) =8 ft x 1= 8 ft --(Table 2)
Total equivalent length of pipe to
L = 14.5616 ft
p = 6.3301 x 10-5 x (Q)1.85 x 14.5616
= 6.3301 x 10-5 x (78.1776)1.85 x 14.5616
p = 2.9296 psi
Neglected
P = 22.27 psi
P = (Q /K)2
(Q)I = 26.427 gpm
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)
e-ISSN: 2582-5208 International Research Journal of Modernization in Engineering Technology and Science
( Peer-Reviewed, Open Access, Fully Refereed International Journal )
Volume:03/Issue:09/September-2021 Impact Factor- 6.752 www.irjmets.com
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[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 I to III,
L (d= 25mm) = 8.2649 ft
L (d= 32mm) = 9.9369 ft
L (d= 40mm) = 0.0656 ft
Frictional loss through pipe I to III,
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 [(Q)I]1.85 x 8.2649 +1.3410 x 10-4…