ChapChap-3
Fire Fighting
Design Recommendations & Hydraulic –Calculation By Dr. Ali Hammoud BAU-2005
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DEVELOPING A SPRINKLER PLAN FOR APPROVAL To prepare a sprinkler plan for approval, it is necessary
to: to: 1. Collect and review data. 2. Trace or scan building outline. 3. Draw necessary building details. 4. Determine branch line logic. 5. Determine system type and configuration. 6. Determine the hazard class of the occupancy. occupancy. 7. Determine the area protected by each sprinkler. 8. Determine the number of branch lines. 9. Determine the distance between branch lines. 10.Determine 10.Determine the maximum allowable distance between sprinklers. RefRef-3
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11- Pipe sizing tables. 12- Friction Loss Formula ” HazenHazen-Williams” Williams” 13- Hydraulic Calculation for sprinkler systems 1414-exanple problem
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Installation Requirements Based on NFPA-13 System Protection Area Limitations. 1. Light hazard — 52,000 ft2 (4831 m2) 2. Ordinary hazard — 52,000 ft2 (4831 m2) 3. Extra hazard —Hydraulically calculated 40,000 ft2 (3716 m2) 4. Storage — High-piled storage — 40,000 ft2 (3716 m2) RefRef-1
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7-Area of Coverage •
Determination of the Protection Area of Coverage
1. Along Branch Lines. Is the distance between sprinklers, defined as “S” 2. Between Branch Lines. Is the perpendicular distance to the sprinkler on the adjacent branch line, defined as “L” As = S X L •
The maximum allowable protection area of coverage for a sprinkler (As) shall be in accordance with the value indicated in each type or style of sprinkler. The maximum area of coverage of any sprinkler shall not exceed 400 ft2 (36 m2). 5
L Area protected by one sprinkler
S
AS
As = S X L
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Area of Coverage (for Light Hazard) Protection Areas and Maximum Spacing (Standard Spray Upright/Standard Spray Pendent) for Light Hazard
•
Construction type
Max Area Max Spacing As (m2) S (m)
Noncombustible
20.9
4.6
Combustible
15.6
4.6
Dr. Ali Hammoud BAUBAU- 2005
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Area of Coverage ( for Ordinary Hazard) Hazard •
Protection Areas and Maximum Spacing (Standard Spray Upright/Standard Spray Pendent) for Ordinary Hazard
Construction type All
Max Area Max Spacing S (m) As (m2) 12.1 4.6
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Area of Coverage (for High Hazard) •
Protection Areas and Maximum Spacing (Standard Spray Upright/Standard Spray Pendent) for High Hazard
Construction type All, ρ >= 0.25
Max Area As (m2) 9.3
Max Spacing “S” (m) 3.7
All, ρ < 0.25
12.1
4.6
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Practical Spacing “S” of up-right and pendent spray sprinkler The maximum distance between sprinklers, either on branch lines or between branch lines, shall be as follows:(as per NFPANFPA-13.item 44-4) : Light Hazard 15ft (4.5 m) Ordinary Hazard 15ft Extra – Hazard 12ft (3.6 m) High – Piled storage 12ft The distance from sprinklers to wall shall exceed oneone-half of allowable distance between sprinklers. 10
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8- Determine the Number of branch Lines. The number of branch lines per bay is determined by dividing the width of the bay, or the distance between column lines in feet, by the maximum allowable distance between sprinklers, (Lmax) . If the result is not a whole number, round up to the nearest whole number. The formula below can be used for this calculation:
Number of branch lines = (Width of bay)/ (Lmax) For example, an ordinary hazard system installed in a bay that is is 35 ''-0" wide would require three branch lines, since NFPA 13 mandates a maximum of 15 ''-0" between branch lines:
35' /15’ /15’ = 2.3 rounded to 3 branch lines
The space between the beams along the column lines in exposed construction is 11 called a bay.
Number of branch Lines 35' /15’ /15’ = 2.3 rounded to 3 branch lines
RefRef-3
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9- Determine the distance between branch Lines (L). The acceptable distance “L” between the lines in the bay is obtained by dividing the width of the bay by the number of branch lines
L=
Total Width of bay Number of branch lines
For example, example, continuing with the building in the previous example and calculate calculate the distance L:
L= 35' /3 branches = 11.667 or 11’ 11’-8” .
However the distance from the end lines to the walls is L/2 = 5’ 5’-10” 10” 13
10- Determine the maximum allowable distance between sprinklers (S).
NFPA 13 allows Smax to be 15 ft for light hazard and ordinary and 12 ft for extra hazard ,S is also obtained by the same formula ;
S = AS / L For example, for an ordinary hazard the NFPA 13 would require As =130 ft2 ( 12.1 m2 ) per sprinkler . However for the previous example L =11’ =11’-8” the maximum spacing “S “, becomes: S= 130 /11’ /11’-8” = 11.14 ft ≅ 11’ 11’ ft.
( 1ft =12” =12”)
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Ordinary hazard spacing example in one bay of building . Refer to drawing and check to be certain that sprinkler spacing “S “ is acceptable : As= Sx L 11’ 11’ x 11’ 11’-8” =128.3 ft2
From NFPANFPA-13
Amax =130 ft2. Layout is Acceptable RefRef-3
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Maximum Distance from Walls & ceiling Spacing (Standard Spray Upright/Standard Spray Pendent)
• The distance from sprinklers to walls shall not exceed one-half of the allowable distance ( L/2). • Sprinklers shall be located a minimum of 4 in. (102 mm) from a wall.
RefRef-1
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Deflector Position Standard Pendent and Upright Spray Sprinklers
• Distance Below Ceilings: deflector and the ceiling shall be a minimum of 1 in. (25.4 mm) and a maximum of 12 in. (305 mm) • Under obstructed construction, construction the sprinkler deflector shall be located within the horizontal planes of 1 in. in to 6 in. in (25.4 mm to 152 mm) below the structural members and a maximum distance of 22 in. in (559 mm) The deflector distance is the dimension from of the ceiling to the the top of the sprinkler deflector. 17
11- Pipe sizing tables • Schedule size for light Hazard Occupancies systems, • Schedule size for Ordinary Hazard Occupancies systems. • Schedule size for Extra Hazard occupancies
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Schedule size for light Hazard Occupancies systems Light Hazard Pipe Schedules (as per NFPANFPA-13.table613.table6-5.2.2) Steel 1 in.…………… in.…………… 2 sprinklers 1¼ in.………… in.………… 3 sprinklers 1½ in.………… in.………… 5 sprinklers 2 in.…………… in.…………… 10 sprinklers 2½ in.……… in.……… 30 sprinklers 3 in.………… in.………… 60 sprinklers 3½ in.……… in.……… 100 sprinklers
Copper 1 in.…………… in.…………… 2 sprinklers 1¼ 1¼ in.…………… in.…………… 3 sprinklers 1½ 1½ in.…………… in.…………… 5 sprinklers 2 in.……………… in.……………… 12 sprinklers 2½ 2½ in.…………… in.…………… 40 sprinklers 3 in.…………… in.…………… 65 sprinklers 3½ 3½ in.…………… in.…………… 115 sprinklers
RefRef-1
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Schedule size for Ordinary Hazard Occupancies systems Ordinary Hazard Pipe Schedule (as per NFPANFPA-13.table613.table6-5.3.2(a)) Steel Copper 1 in. …………… 2 sprinklers 1 in. ……………… 1¼ in.…………… 1¼ in.…………… 3 sprinklers 1¼ in. …………… 1½ in.…………… 1½ in.…………… 5 sprinklers 1½ in. …………… 2 in.……………… 2 in. ……………… in.……………… 10 sprinklers 2½ in.…………… 2½ in.…………… 20 sprinklers 2½ in. …………… 3 in. ………………40 3 in. ……………… ………………40 sprinklers 3½ in. ……………65 3½ ……………65 sprinklers 3½ in. …………… 4 in. ………………100 4 in. ……………… ………………100 sprinklers 5 in. ………………160 5 in. ……………… ………………160 sprinklers 6 in. ………………275 6 in. ……………… ………………275 sprinklers
2 sprinklers 3 sprinklers 5 sprinklers 12 sprinklers 25 sprinklers 45 sprinklers 75 sprinklers 115 sprinklers 180 sprinklers 300 sprinklers
RefRef-1
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Schedule size for Extra Hazard occupancies
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Remark “ Sprinklers outlet ” •Minimum 11-in. (25.4(25.4-mm) outlets shall be provided The branch lines and the sprinklers spacing and distance are now be dimensioned on the drawing with respect to the bays . Pipe sizing is than be completed from the pipe Schedule
RefRef-1
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11
H.WH.W-1
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12-Friction Loss Formula ” Hazen-Williams”
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Friction Loss Formula ” HazenHazen-Williams” Williams” Pipe friction losses shall be determined on the basis of the HazenHazen-Williams formula, as follows: 1.85
Pf =
4.52 × Q C 1.85 × d 4.87
psi / foot
Where p = frictional loss factor in psi per foot of pipe Q = flow in gpm C = friction loss coefficient for new black steel pipe C=120 d = actual internal diameter of pipe in inches For SI units, the following equation shall be used: used:
Pf = 6.05(
Q 1.85 ) × 10 5 1.85 4.87 C ×d
Where pm = frictional resistance in bar per meter of pipe Qm = flow in L/min C = friction loss coefficient dm = actual internal diameter in mm
Bar / m
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Actual internal diameter
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N.B. Choosing the Hydraulically most demanding area The hydraulically most demanding area may be the geographically most remote area, which is the area whose linear distance from the sprinkler system riser is the longest ,in feet . Sometimes , the hydraulically most demanding area is not the most remote area. However “ When in doubt , calculate it out “ which means that if you are unsure which is the hydraulically most demanding area , perform calculations for other potential hydraulically most demanding areas until all uncertainty is removed.
RefRef-3
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13-Hydraulic -Calculation Procedures The procedure consists of the following steps: 1. Selecting occupancy, 2. Selecting a hydraulic density, 3. Determining the length of the hydraulically most demanding area, 4. Determining the number of sprinklers flowing along the length of the design area, 5. Determining the configuration of sprinklers in the hydraulically most demanding area, 6. Determining the minimum flow at the hydraulically most demanding sprinkler, 7. Determining the minimum pressure at the hydraulically most demanding sprinkler, 8. Determining friction loss in each pipe segment. 29
Hydraulic calculations con’t 1- Selection of occupancy ( refer to Chp.1) The correct selection of the occupancy classification of a building or a portion of a building is the foundation for meaningful and reliable hydraulic calculations. Careful selection of the occupancy is the most critical decision that a fire protection systems designer makes during the hydraulic calculation process. A list of occupancies are summarized below; 30
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Hydraulic calculations con’t 2- Selecting a Hydraulic density The area density curves shown in the accompanied figure 1 may be used as the basis for your calculation in selecting the density. These curves are a function of a design density and the total design area of sprinkler operation. The design density is the quantity of water per square foot of remote area that has been found by experience to be effective in controlling a fire of given occupancy. occupancy. The design area (A ) is an area whose size is related to the occupancy, where in all sprinklers sprinklers in the area is expected to actuate at the same time. time.
RefRef-3
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Area-Density Method The area under which the maximum number of sprinklers shall operate among one riser grid, is selected from the area-density curve Example: From the extra hazard occupancy group 1 (curve 4) we choose the point having the maximum density and the lower protected area that is: (A A = 2500ft², 0.3gpm/ft²) Fig. 1
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Area/density curves. “Remarks” • If the areas of sprinkler operation is less than 1500 ft2 (139 m2) for light and ordinary hazard occupancies, the density for 1500 ft2 (139 m2) shall be selected that is ρ=0.1 and 0.15 respectively. • However for extra hazard occupancies if the calculated areas of sprinkler operation is less than 2500 ft2 (232 m2), the density for 2500 ft2 (232 m2) shall be used. RefRef-1
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Area-Density Method • The area under which the maximum number of sprinklers shall operate among one riser grid, is selected from the areadensity curve • Example for ordinary hazard group 1 (curve 2) we choose the point having the maximum density and the lower protected area (1500ft², 0.15gpm/ft²) Fig. 1
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Example of hydraulically most demanding area 1
2
3 5
4 RefRef-1
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Hydraulic calculations Con’t
3- Determining the Length of the Design Area NFPA 13 provides a formula with which the hydraulically most demanding area is sized: The minimum length of the rectangle is 1.2 times the square root of the design area (A (A):
Lmin = 1.2
A = 1.2 1500 = 46.5 ft
A designer has selected a design area of 1.500 feet2 from the area/density curves in Figure 1. Determine the minimum length of the design area. “L” shown in Figure 2. This “ L =46.5 ft“ ft“ length is a minimum, because the length of the design area must be increased so that the edge of the design falls at the point between two sprinklers. We will make this adjustment in the next step. 36
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Figure 2.
Lmin = 1.2
A RefRef-3
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Hydraulic calculations Con’t 4- Number of sprinklers flowing along the length of the design area of the Design Area
The number of sprinkler flowing along the length of the design area area is determined by dividing Lmin by the spacing between sprinklers ( S).
Ns =
Lmin S
For example ; From previous we get Lmin= 46.5 ft and it is known from NFPANFPA- 13 the sprinkler spacing for ExtraExtra- hazard group I , S= 12’ 12’. Therefore:
Ns =
46 .5 feet = 3.88 that is 4 sp. 12 feet / sprinkler
The actual length of the design area Lact, is determined by multiplying the actual number of sprinklers , Ns , by the spacing between sprinklers ( S).
Lact = Ns × S = 4 × 12 ' = 48 ft
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Hydraulic calculations Con’t 4- Configuration of sprinklers in the design Area Once we have determined the actual length of the design area [most remote area] , we must determine its width. As follows:
W =
A Lact
1500 ft 2 A W = = = 31.25 ft Lact 48 ft From Fig 3 the area A = 48' x 30' = 1440 square feet, containing 12 sprinklers. This falls slightly short of the required 1.500 square foot design area. Since each sprinkler covers a 12' x 10' area or 120 square feet we know that adding one sprinkler to 1440 sq ft results in an area of 1 440 sq ft plus 120 sq ft, or 1560 sq ft. which meets the desired minimum criteria. The question is where is this 39 thirteenth sprinkler added? This fina1 sprinkler is always added closest to the crossmain to assure the hydraulically most demanding area.
Given : A wet pipe sprinkler system , Occupancy ordinary hazard group I Area of operation =1500 Sq. Ft , Density = 0.15 gpm per sq. foot Lmin= 46.5 feet , Ns = 4 sprinklers along the length of the design area Lact = 48 feet , W= 31.25 feet , Width covered by 3 branch lines is 30’ 48’ x 30’ = 1440 sq.ft. [ 12 sprinklers] 14440 sq.ft is less 1500 sq.ft. must add one more sprinkler [13]
Fig.3
closest to the crossmain
40
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In short ,the Number of sprinklers in each branch line is determine as follows: Where the design is based on area/density method, the design area shall be a rectangular area having a dimension parallel to the branch lines at least 1.2 times the square root of the area of sprinkler operation (A) used, which shall permit the inclusion of sprinklers on both sides of the cross main. Number of Sprinklers on one branch line 1.2 × A S Where S = Spacing between sprinklers (ft) N=
Light Hazard Ordinary Hazard Extra – Hazard High – Piled storage
15ft (4.5 m) 15ft 12ft (3.6 m) 12ft
A = Remonte area from Area − Density curve (ft 2 )
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Example of estimating the total number of sprinklers operating at the same time & the number of sprinklers in each branches
RefRef-1 42
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Hydraulic calculations Con’t 5- Minimum flow at the hydraulically most Demanding sprinkler ( one sprinkler ) The minimum required flow at the hydraulically most demanding sprinkler sprinkler is obtained by multiplying the design density by the area covered by by one sprinkler .
Q = d × ( As ) Q = d × ( As ) = 0.15 gpm / sq. ft × 120 sq. ft . = 18 gpm This is the minimum required flow at the hydraulically most demanding demanding sprinkler. For ordinary Hazard group I “As” As” is taken as 130 ft2. 43
Hydraulic calculations
6- Determining the minimum pressure at the hydraulically most demanding sprinkler
The flow at a sprinkler head is determined by the formula:
Q=K P The flow at a sprinkler (Q) is equal to the sprinkler discharge coefficient (K) times the square root of the pressure (P). K is commonly referred to as the K-factor ( coefficient of discharge) . Each sprinkler that is tested and listed for use on a sprinkler system has a unique KK-factor, factor or orifice coefficient, that is calculated for that sprinkler. Sprinklers having nominal orifices of 1/2" generally have K-factors ranging from 5.3 to 5.8.
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Flow Characteristic & Water spreading S.I Units
Q=K √P Where Q= flow rate in L/min K= K factor P= pressure in Bar 45
Sprinkler Discharge Characteristics “K” Percent of Nominal l½ in. Discharge
Thread Type
Nominal Orifice Size Marked On Frame
1.3-1.5
25
½ in. NPT
Yes
1.8-2.0
33.3
½ in. NPT
Yes
9.5
2.6-2.9
50
½ in. NPT
Yes
7/16
11.0
4.0-4.4
75
½ in. NPT
Yes
5.3-5.8
100
½ in. NPT
NO
140
¾ in. NPT or
NO
200
½ in. NPT or
Yes
¾ in. NPT
Yes
¾ in NPT
Yes
Nominal Orifice Size (in)
(mm)
¼
6.4
5/16
8.0
3/8
K Factor1
½
12.7
17/32
13.5
5/8
15.9
7.4-8.2
11.0-11.5 ¾
19.0
13.5-14.5
K= 5.6
250
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7- Friction Loss Formula “Hazen-Williams formula” HazenHazen-Williams formula is one of the most popular friction loss formula, recognized by NFPA 13 and considered as a standard formula for the calculation of the pressure drop.
1)- Calculations shall begin at the hydraulically most remote sprinkler. Discharge at each sprinkler shall be based on the calculated pressure at that sprinkler. 2)- Pipe friction loss shall be calculated in accordance with the HazenHazen-Williams formula.
4.52 × Q1.85 Pf = 1.85 C × d 4.87
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14-Example Problem The next figure present a plan of wet pipe system layout Assuming the following data: Density, Density,0.15 gpm/ft2 calculated over the most remote 1500 sq.ft. sq.ft. In this example take [As= 120 ft2],S=12 ft & L= 10ft and the total number of Sp. is 13 sprinklers in most remote area. 1. The minimum flow at the most remote sprinkler =18 =18 gpm 2. The minimum required pressure at the most remote sprinkler=10.33 Psi. Psi. Size the piping system & determine the pump duty. Knowing that , the branch lines are schedule 40 black steel. (C=120 ) Crossmain lines are schedule 10 black steel. 48
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Sectional view of the building
RefRef-3. 50
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Solution in steps 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.
Number of sprinklers riser needed Most remote area Area of coverage and sprinklers spacing Expected total number of sprinklers per floor Pipe sizing Determination of the most remote area Number of sprinklers in the most remote area Number of sprinklers along one branch Flowrate of the most remote sprinkler Residual pressure of the most remote sprinkler Total effective length between two adjacent sprinklers Friction loss formula Calculating pump total pressure Calculating pump total flowrate 51
Number of sprinklers riser needed Based on NFPA 1313-4-2.1
Since the parking area is less than 4831m2, one sprinkler riser is sufficient 52
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Area-Density Method
Ad is the most remote area and is equal to 1500 ft2 for ordinary Hazard 1 53
Area of Coverage/sprinkler spacing Protection Areas and Maximum Spacing (Standard Spray Upright/Standard Spray Pendent) for Ordinary Hazard. NFPANFPA-13 4.4.1.1 Protection area (As) Construction System ft2 m2 type type All All 130 12.1
Maximum spacing (s) ft m 15
4.6
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4. Expected total number of sprinklers per floor
The expected number of sprinklers is determined as follows : Number of sprinklers =Total area of the car parking / As Area of the car parking 13200 ft2 ( 110 ft x 120 ft) ,divided by the sprinkler coverage area As =120 ft2 we get 110 sprinklers.
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Pipe sizing 1 in. …………… 2 sprinklers 1¼ in.…………… in.…………… 3 sprinklers 1½ in.…………… in.…………… 5 sprinklers 2 in.……………… 10 sprinklers in.………………10 2½ in.…………… in.…………… 20 sprinklers
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Determining the flowrate of the most remote sprinkler
q = density × As
(gpm) (gpm
Step 5- Determining the residual pressure of the
most remote sprinkler
q p= k
2
(psi)
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Determining the total effective length between two adjacent sprinklers
T = L+F
T =total effective length L= Pipe length F= Equivalent pipe length
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Friction Loss Formula Pipe friction losses shall be determined on the basis of the HazenHazen-Williams formula, as follows:
P = 4.52
Q1.85 C 1.85 × d 4.87
Where, p = frictional resistance in psi per foot of pipe Q = flow in gpm C = friction loss coefficient black steel C=120 d = actual internal diameter of pipe in inches
Dr. Ali Hammoud
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T = L+F = 12 + 0 = 12
P
f
=T×P
Where, Pf =friction loss in psi between two adjacent sprinklers
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Calculate Total pressure drop PT
PT = Pf + PT previous + Pe Where, PT = total pressure in psi Pe = elevation pressure in psi
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Calculate flow q of the second sprinkler “2” The flow of both sprinklers
q = K PT
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12+6 =18’
Remember ! 1st Branch
4 Sprinklers flow q= 79.56 gpm Whereas 4 x 18 gpm =72 gpm
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18 gpm 19.08 gpm 37.08 gpm 20.16 gpm 57.24 gpm 22.32 gpm 79.56 gpm 79.56 gpm
Flow distribution in Branch 1
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At = Top of riser nipple & AB = Bottom of riser nipple Pressure drop due to elevation 1.5 ‘ Pe= Z x 0.433 Psi/ft =1.5 x0.433 =0.65 Psi AT
F = Pressure drop due Tee = 10 ft ‘
BB
AB
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At = Top of riser nipple & AB = Bottom of riser nipple Pressure drop due to elevation 1.5 ‘ Pe= Z x 0.433 Psi/ft =1.5 x0.433 = 0.65 Psi Pressure drop due to [ 2” Tee ] = 10 ft Friction loss Psi/100 ft from Hazen William equation with Q= 79.56 gpm & D=2” we get Pf=0.064 Psi. The total equivalent length T = (1.5 +10) = 11.5 ft The pressure drop Pf= 11.5x 0.064= 0.74 ft The total pressure drop Pt = 21.35+ 0.65+ 0.74 =22.74 ft.
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AB = Bottom of riser nipple
Pressure drop in line A-B =10 x 0.007 = 0.07 where D= 3 , q=79.56 gpm and ,L= 10 ft
Branch Line “K” calculation: Imagine a huge sprinkler installed at nod BB having K= 16.68.This K represents the total flow of sprinklers 5 thru 8 @ node BB with an orifice capable of discharging the total flow for 4 sprinklers at nodes 5 thru 8.
Q=K x√P K= 79.56/√22.74 =16.68
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BB = Bottom of riser nipple & CB For Pt = 22.81 Psi & K= 16.68 the flow in branch BB-C q= 79.99 gpm
79.66 68gpm
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Sprinkler 13 flow rate Node DB To adjust the pressure for the pressure differential between the demand of sprinkler 13 at DB and sprinklers 1-12 at DB we use the balancing equation.
QAdjusted = Qlow ×
P high P low
=18×
23.05 11.37
= 25.62 gpm
It is important to note that at node 13 ,we use the same minimum flow that we calculated for the most remote sprinkler (1 ) i.e 18 gpm as low flow for sprinkler 13. 69
Dr. Ali Hammoud
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Flow distribution in the main pipes 264.3 gpm
25 gpm
239.3 gpm
80.08 gpm 159.22 gpm
79.66 gpm 79.56 gpm
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Node DB Pressure drop From D-E L=5+10+10+10+10+10=55 ft. F= 15 ft one Tee 3” Leffective =70 ft Pressure drop = 70x 0.062 ft= =4.32 ft.
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Sectional view of the building
E
90 Elbow =10 ft
Pressure drop From E-F L=58’-6”. F=10’ [4” Elbow ],Leffective =68.5 ft Pressure drop = 68.5x 0.018 = 1.32 ft. Total Pressure is the summation of all Pt = 37.94 ft 73
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Fire Hose streams: •Add 100 gpm for light hazard occupancies •250 gpm for ordinary hazard occupancies •500 gpm for extra hazard occupancies 75
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Equivalent Pipe Lengths of Valves and Fittings (shall be used with a HazenHazen-Williams C factor of 120 only)
Dr. Ali Hammoud BAUBAU- 2005
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Example problem II
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Example of determining the number of sprinklers to be calculated
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Example from the NFPA 13
Car parking
9 4
6
5
7
8
81
1 2 3 4
7 9
4 DN
5
6 L=70 ft
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Hydraulic calculation example
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Obstructions to sprinkler discharge pattern development
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Obstruction Sprinklers Location C e ilin g O bstruction
W all
A f
O p e n w o o d
[D
w e b s te e l o r tru s s
− 8 ]+ B
A f 3C or 3D Ventilation ducts
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Sprinklers under pitched roofs with sprinkler directly under peak; branch lines run up the slope.
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Sprinklers at pitched roofs; branch lines run up the slope
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Drainage • All sprinkler pipe and fittings shall be so installed that the system can be drained Riser or Main Size 1. Up to 2 in. 2. 2/21in., 3 in., 31/2 in. 3. 4in. and larger
Size 3/4 in. 1 ¼ in. 2 in.
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System test connection on wet pipe system
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Aboveground Pipe and Tube • Ferrous Piping (Welded and Seamless) Specification for Black and Hot-Dipped ZincCoated (Galvanized) Welded and Seamless Steel Pipe for Fire Protection Use ASTM A 795 • Specification for Welded and Seamless Steel Pipe ANSI/ASTM A 53 • Copper Tube (Drawn, Seamless) Specification for Seamless Copper Tube ASTM B 75 Specification for Seamless Copper Water Tube ASTM B 88 92
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Underground Pipe • Piping shall be listed for fire protection service and comply with the AWWA standards • Cement Mortar Lining for Ductile Iron Pipe and Fittings for Water • Polyethylene Encasement for Ductile Iron Pipe Systems • Ductile Iron and Gray Iron Fittings, 3-in. Through 48-in., for Water and Other Liquids • Rubber-Gasket Joints for Ductile Iron Pressure Pipe and Fittings • Flanged Ductile Iron Pipe with Ductile Iron or Gray Iron Threaded Flanges • Ductile Iron Pipe, Centrifugally Cast for Water • Steel Water Pipe 6 in. and Larger • Coal-Tar Protective Coatings and Linings for Steel Water Pipelines Enamel and Tape — Hot Applied • Cement-Mortar Protective Lining and Coating for Steel Water Pipe 4 in. and Larger — Shop Applied 93 • Field Welding of Steel Water Pipe
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References 1. NFPA-2004 2. F.Fall “ Building services & equipment” Vol 1-2-3 3. R. gagnon Fire protection systems. 4. The Design Project & AutoCAD drawing by Upland –engineering “Dr. Hammoud” 5. Photos From Webs…. 95
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