International Fire Sprinkler Association wwwfiresprinklerglobal 1 Course Outline Definitions and Equations Hydraulic Calculation Principles Hydraulic Calculation Process Example Calculation ID: 688860 Download Presentation
The Profession of Fire Protection Engineering A career path choice for University of St Thomas Engineering Students Minnesota Chapter Society of Fire Protection Engineers Presentation to UST students
in . Europe and Poland. Every year, devastating forest fires take place in Europe and around the world, destroying thousands of hectares of forests and affecting population. When the scale of a fire exceeds the capacity of a country to extinguish it, the EU Civil Protection Mechanism can be activated to provide a rapid and effective response..
Current Status. Victoria. mandatory fire sprinkler protection. Queensland. . mandatory to upgrade fire protection . sprinklers are one option. New South Wales. . retrofitting is common across many types of buildings.
SUNY . PPAA. Winter 2012 Conference. Barbara Boyle, SUNY. Bill Held, SUCF. 1. Deutsche . Bank. 2. 3. . 4. August 2007 Fatal Fire. Robert . Beddia. , 53. Joseph . Graffagnino. , 33. 5. 3 contractor supervisors criminally indicted.
within the SRA. Inspections: The inspection authority shall inspect for compliance with these regulations. When inspections are conducted, they should occur prior to: the issuance of the use permit
Four Components for a Fire. Fuel. (any combustible material: solid, liquid, or gas). Oxygen. (in the air we breath, can ignite fire). Heat. (the energy needed for the fuel to give off enough vapours for a fire to ignite).
Hydraulics of Semi Circular Weirs. Q=CLH. t. 3/2. L = Effective Length of Weir. H. t. = Total Head (Still Pool). = H + V. 2. /2g = Energy Grade Line. . H & V measured 3H upstream from weir.
Hydraulics of Semi Circular Weirs. Q=CLH. t. 3/2. L = Effective Length of Weir. H. t. = Total Head (Still Pool). = H V. 2. /2g = Energy Grade Line. . H & V measured 3H upstream from weir.
Hydraulics of Semi Circular Weirs. Q=CLH. t. 3/2. L = Effective Length of Weir. H. t. = Total Head (Still Pool). = H + V. 2. /2g = Energy Grade Line. . H & V measured 3H upstream from weir.
11 April 2017 .  . Orlando, FL USA .  Casey . Grant PE, Executive Director. Research Update on. 2017 Metro Fire Chiefs Conference. © Fire Protection Research Foundation and National Fire Protection Association. All Rights Reserved.
International Fire Sprinkler Association. www.firesprinkler.global. 1. Course Outline. Definitions and Equations. Hydraulic Calculation Principles. Hydraulic Calculation Process. Example Calculation.
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Slide1
Hydraulics for Fire Protection
International Fire Sprinkler Associationwww.firesprinkler.global
1Slide2
Course Outline
Definitions and EquationsHydraulic Calculation PrinciplesHydraulic Calculation Process
Example CalculationReview of Computer Calculations
2Slide3
Study of Water
HydraulicsThe science which defines the mechanical principles of water at rest or in motion.Hydrostatics
The scientific laws that define the principles of water at rest.HydrokineticsThe study of water in motion.
3Slide4
Hydraulic Focus
Pressure
4
FlowSlide5
Pressure Types
Atmospheric PressureCaused by the weight of air, varies with altitudeLower at high altitudes, higher at low altitudes
14.7 psi at sea levelGage PressureThe actual reading on a gage, does not account for atmospheric pressure. (psig)Absolute PressureThe sum of atmospheric pressure and gage pressure. (psia)
5Slide6
Pressure Types
(continued)Static Pressure (Ps
)The potential energy available within a system when no water is flowing. Pressure is created by elevating water above a source, or it can be created mechanically with pumps or pressure tanks.
6Slide7
Elevation Pressure
A cubic foot of water results in a static pressure at its base of 62.4 lbs
/ft2 Converted to square inches a column of water 1foot high exerts a pressure of 0.433 lbs/in2
7
1 ft
62.4 lbs/ft
2
1 ft
0.433 lbs/in
2Slide8
Elevation Pressure
(continued)Pressure (psi) = 0.433 X Elevation (ft)
8
5 ft
15 ft
What is the pressure difference?Slide9
Elevation Pressure Example
(Continued)What is the pressure at the hydrant?
9
Pressure (psi) = 0.433 x Elevation (ft)
P=?
200 ft
6 ftSlide10
Elevation Pressure 2nd Example
How high is the water?
10
P=47 psi
? ft
6 ftSlide11
Pressure Types
(continued)Residual Pressure (PR)
The pressure at a given point in a conduit or appliance with a specific volume of water flowing.
11Slide12
Pressure Types
(continued)Normal Pressure (P
N)The pressure created on the walls of pipe or tanks holding water.This is the pressure read by most gages.Velocity Pressure (P
V
)
The pressure associated with the flow of water measured in the same direction as the flow.
12
P
N
P
VSlide13
Calculating Velocity Pressures
P
n = Pt – Pv
Where:
P
n
= normal pressure (psi)
P
t
= total pressure (psi)
P
v
= velocity pressure (psi)
Velocity pressure can be found as follows:
13Slide14
Using Velocity Pressure
When velocities are high in a closed system the pressure needs to be accounted for in the calculations
It can reduce the flows and pressures needed in a system 510 percent
In most sprinkler systems velocities are low and their pressures create a minor effect, therefore velocity pressures can be ignored.
It should be used at points where large flows take a 90degree turn in the piping.
14Slide15
Flow (Q)
The quantity (of water) which passes by a given point in a given period of timeGenerally measured in gallons per minute (gpm) or cubic feet per second (ft
3/sec)Uses the term “Q” in most equations
15Slide16
Flow Equation
Q = A x V
Q = flow in ft3/secA = cross sectional area of pipe in
ft
2
V = water velocity in
ft/sec
Q is a constant for any given closed system.
16Slide17
Flow Equation
(continued)Q = A x V = constant flow
17
When the pipe size changes flow remains constant:
Q = A
1
x V
1
= A
2
x V
2
A
1
x V
1
= A
2
x V
2
X gpmSlide18
Flow Example 1
If water is flowing at 5.7 ft
/sec in 6inch pipe, how fast is it flowing when the pipe size is reduced to 3inch?
18
?
6inch
3inch
5.7
ft/sSlide19
Flow Example 1 Solution
How fast is it flowing when the pipe size is reduced to 2inch?
19
A
1
=
r
2
= (3 in)
2
= 28.3 in
2
A
2
= r
2
= (1.5 in)
2
= 7.1 in
2
?
6inch
3inch
5.7
ft/sSlide20
Flow from an Outlet
Dependent upon a number of factorsSize of the orificeConstruction of the device
Material used in the deviceOther components near the device (e.g. screens)For a sprinkler, that ability is determined experimentally in a laboratory
20Slide21
Flow from an Outlet
(continued)Where:Q is the flow (gpm)
di is the diameter of opening (inches)Pv is the measured velocity pressure (psi)
CD is the discharge coefficient of the device
This is used when testing water supplies to determine the amount of flow
21Slide22
Flow from a Sprinkler
Where:
Q is flow (gpm)k is kfactor determined in the sprinkler listing (gpm/psi½)
P
is the pressure (psi)
The diameter of the opening and discharge coefficient are incorporated into the empirical determination of kfactor.
22Slide23
Sprinkler Flow Example
A sprinkler is being installed with a kfactor of 5.6. If the pressure at the sprinkler is 20 psi, how much water will exit the sprinkler?
23Slide24
Flow from a Sprinkler
(continued)The flow equation can be rearranged to solve for pressure or kfactor:
24Slide25
Pressure Calculation Example
What is the pressure for a sprinkler that has a kfactor of 17.6 and the expected flow is 83 gpm?
25Slide26
Kfactor Calculation Example
What is the Kfactor for an outlet that is flowing 65 gpm at 30 psi?
26Slide27
Friction Loss (P
L)Occurs when water flows in pipes, hoses, or other system devices
Caused by water in contact with wallsUsed to account for losses in energy from water making turns or traveling difficult paths
27Slide28
Formulas for Calculating Friction Loss
HazenWilliams formulaFire sprinkler systems
Waterspray systemsDarcyWeisbach formulaAntifreeze systemsWater mist systemsFoamwater systemsFanning formula
28Slide29
HazenWilliams Formula
Most common for sprinkler calculationsAssumes water is at room temperature but is still accurate with temperature variations
Based on Cfactor, flow, and pipe sizeCalculates the amount of friction loss in ONE FOOT of pipe
29Slide30
HazenWilliams Formula
Where:PL = friction loss (psi/
ft)Q = flow (gpm)C = roughness coefficient (based on pipe material)di = interior pipe diameter (inches)
30Slide31
Roughness Coefficient
Table 22.4.4.7 HazenWilliams C Values
Pipe or Tube
C Value
Unlined case or ductile
iron
100
Black
steel (dry systems including
preaction
)
100
Black steel
(wet systems including deluge)
120
Galvanized (all)
120
Plastic (listed,
all)
150
Cementlined
cast or ductile iron
140
Copper
tube or stainless steel
150
Asbestos cement
140
Concrete
140
31Slide32
Inside Diameters (d
i)List for steel and copper in Table A.6.3.2 and Table A.6.3.5
32
Nominal Pipe Size
Schedule
40
Schedule 10
Type K Copper
CPVC*
1inch
1.049
1.097
0.995
1.101
1 ¼inch
1.380
1.442
1.245
1.394
1 ½inch
1.610
1.682
1.481
1.598
2inch
2.067
2.157
1.959
2.003
2 ½inch
2.469
2.635
2.435
2.423
3inch
3.068
3.2602.907
2.95
4inch
4.026
4.260
3.857
N/ASlide33
HazenWilliams Example
If a pressure gage is reading 40 psi at one end of a 32foot section of 2inch schedule 40 pipe (C = 120) flowing at 110 gpm, what will a gage at the other end read?
33
40
?
2inch schedule 40 pipe
32 ft
P
L
= 0.112 psi/ftSlide34
HazenWilliams Example
(continued)What will a gage at the other end read?
PL = 0.112 psi/ftFriction Loss = 0.112 psi/ft x 32
ft
= 3.6 psi
Gage Pressure = 40 psi – 3.6 psi
36 psi
34
40
?
2inch schedule 40 pipe
32 ftSlide35
Fittings
Energy losses through fittings are caused by turbulence in the waterTo determine losses through fittings “equivalent length” is used
NFPA has a table to provide equivalent pipe lengthsTable is based on schedule 40 steel in a wet pipe system with C Values of 120.
35Slide36
Equivalent Length Chart
36
Fittings & Valves
Fittings & Valves Expressed in Equivalent Feet of Pipe
¾ in
1 in
1 ¼ in
1 ½ in
2 in
2 ½ in
3 in
3 ½ in
4 in
5 in
6 in
8 in
10 in
12 in
45° Elbow
1
1
1
2
2
3
3
3
4
5
7
9
11
12
90° Standard Elbow
2
2
3
4
5
6
7
8
10
12
14
18
22
27
90° Long Turn Elbow
1
2
2
2
3
4
5
5
6
8
9
13
16
18
Tee/Cross
3
5
6
8
10
12
15
17
20
25
30
25
50
60
Butterfly Valve




6
7
10

12
9
10
12
19
21
Gate Valve




1
1
1
1
2
2
3
4
5
6
Swing Check

5
7
9
11
14
16
19
22
27
32
45
55
65Slide37
Adjusting Equivalent Lengths
NFPA 13 table is based on schedule 40 steel pipe for a wet systemAll others need to be adjusted for:
Change in pipe materialCfactor other than 120Change in interior diameterOther than those for schedule 40 steel
37Slide38
Adjusting for CFactor
Table 22.4.3.2.1 C Value Multiplier
Value
of C
100
130
140
150
Multiplying Factor
0.713
1.16
1.33
1.51
38
Begin with the equivalent length value from the table
Multiply the length by the factor above for the appropriate CfactorSlide39
Adjusting for Inside Diameter
Begin with the equivalent length value from the table
Multiply the length by the factor above calculated for the inside diameter of the pipe being used
39Slide40
Fittings
(continued)All fittings must be accounted for in the calculationsIncluding tees, elbows, valves, etc.
Some may have pressure loss or equivalent length values from manufacturer’s listing informationSpecial provisions: Fittings connected directly to sprinklersFittings where water flows straight through without changing direction
40Slide41
Equivalent Length Exercise
What is the equivalent pipe length of Type K copper tube which used for a 3inch standard turn 90degree elbow?
41Slide42
Equivalent Length Exercise Solution
What is the equivalent pipe length of Type K copper tube which used for a 3inch standard turn 90degree elbow?
NFPA 13 Table 22.4.3.1.1 :3inch 90degree elbow = 7 ft of pipeAdjustments are needed for:Type K Copper
Interior diameter
42Slide43
Equivalent Length Exercise Solution
(continued)What is the equivalent pipe length of Type K copper tube which used for a 3inch standard turn 90degree elbow?
Adjustment for material (Cfactor)Copper has a CFactor of 150Per Table 22.4.3.2.1: Multiplier = 1.51Adjustment for inside diameter
3inch copper has an inside diameter of 2.907inch
43Slide44
Equivalent Length Exercise Solution
(continued)What is the equivalent pipe length of Type K copper tube which used for a 3inch standard turn 90degree elbow?
Apply the factors:Equivalent pipe length per Table 22.4.3.1.1 = 7 ftAdjustment for Cfactor = 1.51Adjustment for diameter = 0.77The equivalent length for a 3inch Type K Copper standard turn elbow is:
7 ft x 1.51 x 0.77 = 8.14 ft
44Slide45
Hydraulic Calculation Principles
45Slide46
The Layout Process
Define the Hazard
Analyze the StructureAnalyze the Water Supply
Select the Type of System
Select the Sprinkler Type(s) and Locate Them
Arrange the Piping
Arrange Hangers and Bracing (where needed)
Include System Attachments
Hydraulic Calculations
Notes and Details for Plans
AsBuilt Drawings
46Slide47
Hydraulic Calculations
Verify that the amount of water specified by the design approach can be delivered to control or extinguish a fire Confirm and adjust pipe sizing to accomplish control or extinguishment
Determine size and adequacy of water supply47Slide48
Hydraulics Affected By…
Piping ConfigurationTree SystemsLoop Systems
Gridded SystemsMultipurpose SystemsCan be any of the configurations above with at least one domestic fixture tied into the piping.Pipe Size and MaterialPipe Fittings
48Slide49
Tree System
49
Branch Lines
Riser
Cross Main
Branch lines and sprinklers are fed from only one directionSlide50
Loop System
Allows smaller cross mains because each branch line is fed from two directions.
50Slide51
Grid System
Allows smaller cross mains and branch lines since each sprinkler is fed by at least two paths.
51Slide52
Hydraulics Affected By…
(continued)Type of Sprinkler
Standard Spray SprinklersExtended Coverage SprinklersControl Mode Specific Application (CMSA) SprinklersEarly Suppression Fast Response (ESFR) SprinklersResidential Sprinklers
Inrack Sprinklers
Specially Listed Sprinklers
52Slide53
Hydraulics Affected By…
(continued)Design Method
Density/Area MethodRoom Design MethodSpecial Design NFPA special arrangements for residential, stairs, chutes, etc.
53Slide54
Hydraulic Calculation Principles
Provide enough water from each sprinkler to control or extinguish fireProvide water for all sprinklers which are likely to openMinimize pipe size for material cost, but not create large pressure loss due to friction
54Slide55
Hydraulic Calculation Approaches
Density/Area MethodRoom Design MethodSpecial Design ApproachesResidential Sprinklers
ESFR SprinklersSpecially Listed SprinklersWater CurtainsOther
55Slide56
Density/Area Method
Density is the flow of water that lands in a single square foot under the sprinklerMeasured in flow divided by unit area
English units: gpm/ft2Flow required from a sprinkler is calculated by multiplying selected density by the coverage area
56Slide57
Density/Area Curves
57Slide58
Density/Area Example 1
A sprinkler system has been installed with standard spray sprinklers spaced 10 feet by 11 feet 6 inches apart. If this is an Ordinary Hazard Group 2 occupancy and the discharge density is 0.2 gpm/ft
2, what is the minimum required flow from a sprinkler?Coverage Area:A = 10
ft
x 11.5
ft
= 115 ft
2
Density times area equals flow:
0.2 gpm/ft
2
x 115 ft
2
= 23 gpm
58Slide59
Density/Area Method
(continued)
Fire Rectangle: “…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 used…”Different remote area geometry may be required by other authorities.
59Slide60
Fire Rectangle
When sprinklers are evenly spaced, the long leg of the rectangle can be divided by the distance between the sprinklers on a branch lineThis determines the number of sprinklers per line
If the last line of sprinklers being added to the design area does not need the same number of sprinklers the most demanding ones are added to the calculations
60Slide61
Density/Area Curves
Total Number of Sprinklers to Calculate Design Area ÷ Area Per SprinklerNumber of Sprinklers per Branch Line
61
Where:
S is the distance between sprinklers on the branch lineSlide62
Density/Area Example 2
The sprinkler system in an OH2 occupancy has a discharge density of 0.2
gpm/ft2 over 1500 ft2
(selected from Figure 11.2.3.1.1)
, each sprinkler covers 115 ft
2
, how many sprinklers will be in the design area?
1500 ft
2
÷
115 ft
2
= 13.04
14 sprinklers
If sprinklers along the branch line are 10 ft apart, how many sprinklers/line are calculated?
62Slide63
Density/Area Example 2 (continued)
Which sprinklers on the 3rd line should be added?
63
1
7
6
5
4
3
2
10
9
8
A
E
D
C
BSlide64
Area Adjustments
DryPipe SystemsIncrease area 30% (Section 11.2.3.2.5)Double Interlock
Preaction SystemsIncrease Area 30% (Section 11.2.3.2.5)Extra Hazard Occupancy with High Temperature SprinklersDecrease Area 25%, but minimum of 2000 ft
2
(Section 11.2.3.2.6)
64Slide65
Area Adjustments
(continued)Quick Response Sprinklers (11.2.3.2.3)Area of operation can be reduced 25 to 40% depending on ceiling height when:
Wet pipe system onlyLight or ordinary hazards20 ft maximum ceiling height
No unprotected ceiling pockets
No less than 5 sprinklers in design area
Area may be less than 1500 ft
2
65Slide66
Quick Response Area Adjustment
Ceiling Height <10
ft
Reduction is 40%
Between 10 and 20
ft
Y = (3x/2)+55
Ceiling Height is 20
ft
Reduction is 25%
Over 20
ft
Ceiling Height
No reduction allowed
66Slide67
Area Adjustments
(continued)Sloped CeilingsArea of operation is increased by 30% if pitch exceeds 2 in 12 (rise in run). This is an angle of 9.46°
67
rise
runSlide68
Area Adjustments
(continued)Unsprinklered
concealed spaces minimum 3,000 ft2 applied after all other adjustments unless:Noncombustible or limitedcombustible space with minimal combustible loading and No access
Limited access an no occupancy or storage
Filled with noncombustible insulation
Light/Ordinary hazard occupancies with wood joists or solid noncombustible or limitedcombustible construction subdivided into 160 ft
3
areas, including areas under insulation directly on joists
68Slide69
Area Adjustments (continued)
Unsprinklered
concealed spaces minimum 3,000 ft2 applied after all other adjustments unless:Flame spread rating 25 or lessSpaces constructed of fire retardant materials defined by NFPA 703
Spaces over isolated small rooms < 55 ft
2
Vertical pipe chases under 10 ft
2
meeting 8.15.1.2.14
Exterior columns under 10 ft
2
supporting
sprinklered
canopies
Light/Ordinary hazard occupancies with noncombustible or limitedcombustible ceilings attached to composite wood joists directly or with 1inch metal channels subdivided into 160 ft
3
areas
69Slide70
Multiple Adjustments Example 1
Compound adjustments based on original area of operation selected from Figure 11.2.3.1.1.Drypipe system installed under slope of 4 in 12
30% increase for dry system30% increase for slopeUsing 1500 ft2 as the selected operation area
1500 ft
2
x 1.3 x 1.3 = 2535 ft
2
design area
There is no change in the density.
70Slide71
Multiple Adjustments Example 2
Compound adjustments based on original area of operation selected from Figure 11.2.3.1.1.
QR sprinklers under 3 in 12 slope, ceiling height is 20 ft25% decrease for ceiling height30% increase for slopeUsing 1500 ft
2
as the operation area
1500 ft
2
x 0.75 x 1.3 = 1463 ft
2
design area
There is no change in density.
71Slide72
Room Design Method
Density based on Figure 11.2.3.1.1Calculate all of the sprinklers in the most demanding room (usually the largest)
Walls must have fire resistance rating equal to the required water supply duration72
Table 11.2.3.1.2
Light Hazard
30 minutes
Ordinary Hazard
60 – 90 minutes
Extra
Hazard
90 – 120 minutesSlide73
Room Design Method
(continued)Light Hazard
Doors must have automatic or self closers, or Calculations must include two sprinklers from each adjoining spaceOrdinary and Extra HazardDoors must have automatic or self closers with appropriate fire resistance ratings.
Corridors/Narrow Rooms
When protected with a singlerow of sprinklers, calculate maximum of 5 sprinklers or 75 feet
73Slide74
Room Design Method Example
Which room is the most demanding? Light Hazard, no door closers
74
1
9
8
7
6
5
4
3
2
10
11
12
14
13
15Slide75
RDM Example Solution
Room #
Sprinklers in Room
Sprinklers Calculated
Room #
Sprinklers in Room
Sprinklers Calculated
1
6
10
9
1
6
2
2
6
10
3
10
3
8
12
11
3
12
4
4
7
12
1
3
5
4
11
13
1
3
6
7
5
14
1
3
7
1
5
15
6
11
8
1
2
75Slide76
Special Design Approaches
Specially Listed SprinklersMinimum flow and/or pressure included in the listing of the sprinkler
Uses the area of calculation from the density area methodESFR Sprinklers12 sprinklers calculated4 sprinklers over 3 branch lines
Other
Provides the number of sprinklers to be calculated and minimum pressure or flow necessary
76Slide77
Hydraulically Most Remote Sprinklers
Location of Open SprinklersUsually highest and farthest from system riserCan be hard to locate in gridded systems
Special sprinklers may be more demanding due to flow characteristics instead of those farther away from the water supplySeveral sets of calculations may need to be done to find the most demanding values
77Slide78
Types of Water Supplies
City Water Mains (public supply)Reservoir, Lake, Pond, River, etc.Private Water Mains (NFPA 24)Water Tanks (NFPA 22)
Fire Pumps (NFPA 20)Could be used as part of any of the water supplies
78Slide79
City Water Mains
Information from the local water authorityFlow testing near the siteNeed the following information:
Static PressureResidual PressureResidual Flow
79Slide80
Water Supply Summary
If the system demand is NOT within the capacity of the water supply, alterations are need to the supply or to the systemIf the supply is too low on flow:
arrange a secondary water source (e.g. tank, lake, pond, etc.)If the supply is too low on pressure:install a fire pumpuse larger pipe to reduce friction loss
maintain higher water level in an elevated tank
install tank at higher elevation
80Slide81
StepbyStep Calculations
Identify hazard category
Determine sprinkler spacingDetermine piping arrangement
Calculate amount of water needed per sprinkler
Calculate number and location of open sprinklers in the hydraulically most demanding area
Start at most remote sprinkler and work towards the water supply calculating flows and pressures
Compare demand with supply
81Slide82
Example
Ordinary Hazard Group 2 occupancy12 ft ceiling heightQuick Response standard spray sprinklers with a kfactor of 5.6
Wet pipe sprinkler systemSprinklers on 10 ft x 12.5 ft spacing
82Slide83
Example: Plan View
83
80 ft
123 ft
38 ft
10 ft
12.5 ft
5 ft
5 ft
5 ft
5.5 ftSlide84
Example
(continued)Select hazard category:
OH2Determine sprinkler spacing: 10 ft x 12.5 ft
Determine piping arrangement:
Done
Calculate amount of water per sprinkler
Select
Density/Area Method
Pick point from density/area curve:
0.2 gpm/ft
2
over 1500 ft
2
0.2 gpm/ft
2
x 125 ft
2
=
25 gpm/sprinkler
84Slide85
Example
(continued)
Select hazard category: OH2Determine sprinkler spacing: 10 ft x 12.5 ft
Determine piping arrangement: Done
Calculate amount of water per sprinkler:
25 gpm
Calculate number & location of open sprinklers
Area Adjustment(s):
QR Reduction
: % = (3x/2) + 55 = [3(12)/2] + 55 = 37%
1500 ft
2
x 0.63
= 945 ft
2
945 ft
2
÷ 125 ft
2
per sprinkler = 7.56 =
8 sprinklers
1.2(945)
0.5
/10 = 3.7 =
4 sprinklers per branch line
85Slide86
Example: Hydraulic Remote Area
86
80 ft
123 ft
38 ft
10 ft
12.5 ft
5 ft
5 ft
5 ft
5.5 ftSlide87
Determine the Starting Pressure
Most remote sprinkler needs 25 gpmSprinkler k = 5.6
Starting information for the first sprinkler: 25 gpm at 19.9 psiNext work back to the water supply adding pressure losses and flows throughout the system
87Slide88
Information Needed for Calculations
Select initial pipe sizesLocate nodes on all places where:
Flow (Q) takes place, Type of pipe or system changes (C), andDiameter (di) changes.
Layout calculation paths starting with primary path then attachment paths
Fill in hydraulic calculation sheets
88Slide89
Hydraulic Calculation Paths
89
38 ft
10 ft
12.5 ft
5 ft
5 ft
5 ft
5.5 ft
Locate the system nodes:
1
2
3
4
5
6
7
8
BL1
BL2
TORSlide90
Main Calculation Path
Start at the most remote sprinkler (#1)Path:
90
1
2
3
4
BL1
TOR
BL2
Balancing PointSlide91
Auxiliary Calculation Path
Section that connects into the “balancing point”Auxiliary Path A:
91
5
6
7
8
BL2
KFactor needed to balanceSlide92
Balancing Flows
Only one pressure is inside the pipe
Use the higher pressureCalculate an equivalent Kfactor for the portion of the pipe with the lower pressureCalculate the actual flow using the Kfactor and the new pressure.
92
25 gpm @ 15 psi
125 gpm @ 35 psi
35 psi
Q = 125 + 38.2
Q = 163.2 gpmSlide93
Hydraulic Calculation Work Sheets
Hand calculations or computer calculations must present certain informationComputer generated sheets have been standardized by NFPA 13 (since January 2008)
Traditional hand calculation sheets have minor variations from the computer standardThe path created is the order the calculations will be completed
93Slide94
Calculation Work Sheets
Column Headings:
94
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)Slide95
Work Sheets: Nodes
The “Node” column is used to coordinate the hydraulic calculation with the sprinkler plans
Node 1 is the starting location of the calculation step and Node 2 is where that step ends and the next will begin.
95
Node
1
Elev.
1 (ft)
Node 2
Elev. 2 (ft)
1
17.5
2
17.5Slide96
Work Sheets: Elevation
The “Elevation” column is coordinated to the node location in the system.
For calculation purposes the centerline height is used.Elevation is used for pressure calculations further in the step.
96
Node
1
Elev.
1 (ft)
Kfactor
Node 2
Elev. 2 (ft)
1
17.5
8.0
2
17.5Slide97
Work Sheets: KFactor
The “KFactor” column relates to the sprinklers or other flowing orifices used in the system.
For devices, such as sprinklers, the Kfactor is found in the manufacturer’s cut sheets.At balancing points the Kfactor would be the calculated value.
97
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Elev. 2 (ft)
Total Flow (Q)
17.5
8.0
26.0
17.5
26.0Slide98
Work Sheets: Flow
The “Flow” column is used for both adding the new flow for the step and finding the total flow at that point in the system.
The top line is “q” where the new flow is added.
The bottom line is “Q” where the flow is totaled.
At the first step, the lines will be equal.
98
Kfactor
Flow – this step (q)
Nom. ID
Total Flow (Q)
Actual ID
8.0
26.0
1 ¼inch
26.0
1.380Slide99
Work Sheets: Pipe Size
The pipe size column is split into two rows – nominal and actual.
The “Nominal ID” is noted for the diameter of the pipe in that step.
The “Actual ID” is the real inside diameter used for the friction loss calculation.
99
Flow – this step (q)
Nom. ID
Fittings – amount & length
Total Flow (Q)
Actual ID
26.0
1 ¼inch
26.0
1.380Slide100
Work Sheets: Fittings
The “Fittings” column is used for listing the equivalent lengths of any fittings between Node 1 and Node 2 for that step.
Many branch lines do NOT have any fittings that need to be accounted for in the step.
100
Nom. ID
Fittings – amount & length
L (ft)
Actual ID
F
(ft)
T (ft)
1 ¼inch
10
1.380

10Slide101
Work Sheets: Lengths
The length column sums the physical lengths (centertocenter) of the step with the equivalent lengths.
“L” is the physical length“F” is the equivalent length from the fittings“T” is the total length for that step
101
Fittings – amount & length
L (ft)
CFactor
F
(ft)
Friction Loss (psi/ft)
T (ft)
10
C
= 120

0.056
10Slide102
Work Sheets: Friction Loss
The friction loss column contains the Cfactor for the pipe in that step and the amount of friction loss per foot of pipe.
The HazenWilliams formula is used to determine the friction loss and 3 decimal places are used.
102
L (ft)
CFactor
Pressure
Total (P
T
)
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
10
C
= 120
10.6

0.056

10
0.5Slide103
Work Sheets: Pressure
“P
T” is the total pressure at the start of that step“Pe” is the pressure from the elevation change in the step“P
f
” is the pressure in the step caused by friction
The pressure is totaled on the next line to start that step.
103
CFactor
Pressure
Total (P
T
)
Notes
Friction Loss (psi/ft)
Elevation
(
P
e
)
Friction (P
f
)
C
= 120
10.6
Q
= 0.2 (130) = 26 gpm
P = (26/8.0)
2
= 10.6 psi
0.056

0.5Slide104
Work Sheets: Notes
The “Notes” column is used to list additional information such as equations for flows, pressures, elevation pressure and equivalent Kfactors.
104
Pressure
Total (P
T
)
Notes
Elevation
(
P
e
)
Friction (P
f
)
10.6
Q
= 0.2 (130) = 26 gpm
P = (26/8.0)
2
= 10.6 psi

0.5Slide105
Calculation Exercise
Determine the values for the highlighted boxes.
105
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
S4
16.5
5.6
25.0
1 ½inch
2T = 16
10
C
= 120
20.0
BL1
15.5
128.6
1.610
BL1
15.5


2inch
13
C
= 120
BL2
15.5
2.067Slide106
Calculation Exercise Solution
106
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
S4
16.5
5.6
25.0
1 ½inch
2T = 16
10
C
= 120
20.0
16
0.505
0.4
BL1
15.5
128.6
1.610
26
13.1
BL1
15.5


2inch
13
C
= 120
33.5

0.150
0
BL2
15.5
128.6
2.067
13
1.9Slide107
Full System Hydraulic Calculation
An electronics factory is being built.Water supply tests were done near the site and produced the following information:
Static pressure = 90 psiResidual pressure = 60 psiFlow at 60 psi = 1000 gpm
107Slide108
The Layout Process
Define the Hazard
Analyze the StructureAnalyze the Water Supply
Select the Type of System
Select the Sprinkler Type(s) and Locate Them
Arrange the Piping
Arrange Hangers and Bracing (where needed)
Include System Attachments
Hydraulic Calculations
Notes and Details for Plans
AsBuilt Drawings
108Slide109
Identify the Hazard
In accordance with NFPA 13 hazard classifications, an electronics factory is classified as an
Ordinary Hazard Group I occupancy.
109Slide110
The Layout Process
Define the Hazard
Analyze the StructureAnalyze the Water Supply
Select the Type of System
Select the Sprinkler Type(s) and Locate Them
Arrange the Piping
Arrange Hangers and Bracing (where needed)
Include System Attachments
Hydraulic Calculations
Notes and Details for Plans
AsBuilt Drawings
110
completedSlide111
Flow (gpm)
Available Water Supply
Water Supply
111
Pressure (psi)
100
Flow Test Summary Sheet
0
60 psi residual pressure at 1000 gpm
90 psi static pressure
20
80
120
110
100
90
70
60
50
40
30
10
200
300
400
500
600
700
800
900
1000
1100Slide112
The Layout Process
Define the Hazard
Analyze the StructureAnalyze the Water Supply
Select the Type of System
Select the Sprinkler Type(s) and Locate Them
Arrange the Piping
Arrange Hangers and Bracing (where needed)
Include System Attachments
Hydraulic Calculations
Notes and Details for Plans
AsBuilt Drawings
112Slide113
System Details
Type of System:
Wet pipe system
Type of Sprinkler: TY3121
Standard spray quick response upright sprinkler with a Kfactor of 5.6
Typical Sprinkler Spacing:
Sprinklers are 10
ft
apart on the branch lines, and 12.5
ft
between branch lines
113Slide114
200 ft
Electronics Factory Plan View
114
100 ft
53 ft
5 ft from north wall and 6 ft from west wall
12.5 ft
10 ft
Branch lines are Schedule 40
Mains are Schedule 10
N
5 ft from south wall and 6.5 ft from east wallSlide115
All branch lines are on a 1 ft riser nipple
Electronics Factory Elevation View
115
N
15 ft
5 ft
Gate Valve
Alarm Check Valve – Viking J1
Riser is 1 ft away from the east wall.
Long Turn Elbow
12.5 ft between branch lines
18 ft
4inch PVC
(ID – 4.240 inches)
42 ft
7 ft
3inch
Schedule 10Slide116
Electronics Factory Isometric View
116
N
15 ft
1 ft riser nipple
1 ½inch
1 ½inch
1inch
1 ¼inch
1 ¼inch
1 ½inch
main and riser
3inch
Underground
4inch PVCSlide117
The Layout Process
Define the Hazard
Analyze the StructureAnalyze the Water Supply
Select the Type of System
Select the Sprinkler Type(s) and Locate Them
Arrange the Piping
Arrange Hangers and Bracing (where needed)
Include System Attachments
Hydraulic Calculations
Notes and Details for Plans
AsBuilt Drawings
117Slide118
Select a Design Approach
Use the density/area methodA point from the density/area curves need to be selected
118Slide119
Check for Area Adjustments
Quick response sprinklers (in light hazard or ordinary hazard with wet pipe system, reduce design area based on maximum ceiling height, where it is less than 20 ft
)Original design area, from the area/density curve, is 1500 ft2.Wet pipe system, ordinary hazard, and a ceiling height of 18
ft
119Slide120
Area Reduction For QR Sprinklers
y = % reduction in area
x = ceiling height
120Slide121
Design
Area
(
continued
)
Starting with 1500 ft
2
design area
Applying the 28% reduction in area:
100%  28% = 72%
1500 ft
2
* 0.72 = 1080 ft
2
New design area is
1080 ft
2
Density remains at
0.15
gpm
/ft
2
121Slide122
Design Area
(continued)
Design area is 1080 ft
2
Each sprinkler, spaced 10
ft
x 12.5
ft
, is covering 125 ft
2
How many sprinklers are in the design area?
1080 / 125 = 8.64 sprinklers =
9 sprinklers
122Slide123
Forming the Design Area
Continue to add branch lines until 9 sprinklers are included
123Slide124
200 ft
Remote Area
124
100 ft
53 ft
12.5 ft
10 ft
Branch lines are Schedule 40
Mains are Schedule 10
N
1
2
3
4
5
6
7
8
9Slide125
Information
Needed for Calculations
Select initial pipe sizesLocate nodes on all places where:Flow (Q) takes place,
Type of pipe or system changes (C), and
Diameter (
d
i
) changes.
Layout calculation paths starting with primary path then attachment paths
Fill in hydraulic calculation sheets
125
completedSlide126
N
15 ft
1 ft riser nipple
1 ½inch
1 ½inch
1inch
1 ¼inch
1 ¼inch
1 ½inch
main and riser
3inch
Underground
4inch PVC
1
2
3
4
5
6
7
8
9
BL1
BL2
BL3
TOR
FF
Node Locations  Isometric
126
CWMSlide127
Information Needed for Calculations
Select initial pipe sizesLocate nodes on all places where:Flow (Q) takes place,
Type of pipe or system changes (C), andDiameter (di) changes.Layout calculation paths starting with primary path then attachment pathsFill in hydraulic calculation sheets
127Slide128
Calculation Paths
Main Path:
Auxiliary Paths:128
1
2
3
4
BL1
TOR
BL2
BL3
5
6
7
8
BL2
9
BL3Slide129
Information Needed for Calculations
Select initial pipe sizesLocate nodes on all places where:Flow (Q) takes place,
Type of pipe or system changes (C), andDiameter (di) changes.Layout calculation paths starting with primary path then attachment pathsFill in hydraulic calculation sheets
129Slide130
Starting Sprinkler Values
Each sprinkler covers an area of 125 ft2Each sprinkler is required to deliver a density of 0.15
gpm/ft2Minimum flow per sprinkler:125 ft2
x 0.15 gpm/ft
2
=
18.8 gpm
Minimum pressure for 18.8 gpm:
130Slide131
Data Summary…
Water Supply Information:
Static Pressure 90 psi @ 0 gpm Residual Pressure 60 psi @ 1000 gpmHazard Classification OH  1System Type Wet
Ceiling Height 18 feet
Density/Area 0.15 gpm/ft
2
/ 1500 ft
2
Adjusted to 0.15 gpm/ft
2
/ 1080 ft
2
Sprinkler Type Quick Response – Standard Spray
K – Factor K =5.6
Area Per Sprinkler 125 ft
2
Minimum Sprinkler Flow 18.8 gpm
Minimum Pressure 11.3 psi
131Slide132
1inch
3inch
Start With the Most Remote Sprinkler
132
1 ¼inch
1 ¼inch
1 ½inch
1 ½inch
1
2
3
4
5
6
7
8
9
BL1
BL2
BL3Slide133
Starting the Calculation Sheet
133
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
1
17.0
5.6
18.8
1inch
10
C = 120
11.3


2
17.0
18.8
1.049
10
Now determine the friction loss
0.116
1.2
12.5Slide134
Second Sprinkler Calculation
134
1 ¼inch
1 ¼inch
1 ½inch
1 ½inch
1
2
3
4
18.8 gpm @ 11.3 psiSlide135
Second Sprinkler  Calculation Sheet
135
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
1
17.0
5.6
18.8
1inch
10
C = 120
11.3

0.116

2
17.0
18.8
1.049
10
1.2
2
17.0
5.6
1 ¼inch
10
C
= 120
12.5


3
17.0
1.380
10
19.8
What is the total flow?
38.6
What is the friction loss?
0.116
1.2
13.7Slide136
Third Sprinkler Calculation
136
1 ¼inch
1 ½inch
1 ½inch
1
2
3
4
18.8 gpm @ 11.3 psi
19.8 gpm @ 12.5 psiSlide137
Third Sprinkler  Calculation Sheet
137
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
2
17.0
5.6
19.8
1 ¼inch
10
C
= 120
12.5

0.116

3
17.0
38.6
1.380
10
1.2
3
17.0
5.6
1 ¼inch
10
C = 120
13.7


4
17.0
1.380
10
20.7
What is the total flow?
59.3
What is the friction loss?
0.256
2.6
16.3Slide138
Fourth Sprinkler Calculation
138
1 ½inch
1 ½inch
1
2
3
4
18.8 gpm @ 11.3 psi
19.8 gpm @ 12.5 psi
20.7 gpm @ 13.7 psiSlide139
Fourth Sprinkler  Calculation Sheet
139
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
3
17.0
5.6
20.7
1 ¼inch
10
C = 120
13.7

0.256

4
17.0
59.3
1.380
10
2.6
4
17.0
5.6
1 ½inch
19
C = 120
16.3
BL1
16.0
1.610
22.6
What is the total flow?
81.9
What is the friction loss?
0.219
7.7
24.4
Are there any fittings?
2T
16 ft
16
35
P
e
=
1 * 0.433
0.4Slide140
Branch Line 1
140
3inch
1
2
3
4
BL1
BL2
81.9 gpm @ 24.4 psi
BL2 is identical to BL1 therefore an equivalent Kfactor can be used to balance the flow.Slide141
Branch Line 1  Calculation Sheet
141
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
4
17.0
5.6
22.6
1 ½inch
2T = 16
19
C = 120
16.3
P
e
= 1 * 0.433
16
0.219
0.4
BL1
16.0
81.9
1.610
35
7.7
BL1
16.0
3inch
13
C = 120
24.4

BL2
16.0
3.260
13

What is the total flow?
81.9
What is the friction loss?
0.007
0.1
24.5
Are there any fittings?
K
BL1
=
16.58
What is the new flow?Slide142
Branch Line 2
142
3inch
5
6
7
8
BL1
BL2
BL3
81.9 gpm @ 24.4 psi
BL2 is identical to BL1 therefore the Kfactor from BL1 can be used to calculate the flow.Slide143
Branch Line 2  Calculation Sheet
143
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
BL1
16.0

3inch
12.5
C = 120
24.4
K
BL1
= 16.58

0.007

BL2
16.0
81.9
3.260
12.5
0.1
BL2
16.0
3inch
12.5
C = 120
24.5


BL3
16.0
3.260
12.5
82.1
What is the total flow?
164.0
What is the friction loss?
0.026
0.3
24.8
Are there any fittings?Slide144
Branch Line 3
144
9
BL1
BL2
BL3
81.9 gpm @ 24.4 psi
82.1 gpm @ 24.5 psi
BL3 is different from the other two calculated for this system. It will have to be calculated from the farthest point working toward the branch line and then balanced.Slide145
Branch Line 3  Calculation Sheet
145
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
9
17.0
5.6
18.8
1 ½inch
19
C = 120
11.3
BL3
16.0
18.8
1.610
BL3
Now determine the friction loss
0.014
0.5
12.2
Starting flow value is used:
q = 0.15
gpm
/ft
2
* 125 ft
2
Fittings?
2T
16 ft
P
e
=
1 * 0.433
0.4
16
35Slide146
Branch Line 3
146
9
BL1
BL2
BL3
81.9 gpm @ 24.4 psi
82.1 gpm @ 24.5 psi
18.8 gpm @ 12.2 psiSlide147
Branch Line 3 – Balancing Point
147
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
BL2
16.0
82.1
3inch
12.5
C = 120
24.5
q =
16.58 (24.5)
0.5

0.026

BL3
16.0
164.0
3.260
12.5
0.3
BL3
16.0
5.38
3inch
182.8
C = 120
24.8
q =
5.38 (24.8)
0.5
FF
1.0
3.260
26.8
Fittings
190.8Slide148
Branch Line 3  Fittings
148
BL3 to FF has an elbow, alarm check valve, and gate valve.
3inch El – std 90 per Table 22.4.3.1.1:
7 feet
3inch ACV – Viking Model J1 per cut sheets:
10 feet
3inch GV per Table 22.4.3.1.1:
1 foot
El and GV need to be adjusted for Schedule 10 pipe sizes.Slide149
Branch Line 3 – Balancing Point
(continued)
149
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
BL2
16.0
82.1
3inch
12.5
C = 120
24.5
q =
16.58 (24.5)
0.5

0.026

BL3
16.0
164.0
3.260
12.5
0.3
BL3
16.0
26.8
3inch
ACV=10
182.8
C = 120
24.8
q =
5.38 (24.8)
0.5
El=9.4
20.7
FF
1.0
190.8
3.260
GV=1.3
203.5
0.034
Friction Loss?
6.9
P
e
= 15 * 0.433
6.5
38.2Slide150
System Underground
150
N
main and riser
3inch
FF
CWM
5 ft
Long Turn Elbow
4inch PVC
(ID – 4.240 inches)
42 ft
12 ft
Underground pipe is
PVC.
Cfactor
is
150
per Table 22.4.4.7.
Long turn elbows are typically used underground and cause less turbulence to the flow in the pipe.Slide151
Underground – Calculation Sheet
151
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
BL3
16.0
26.8
3inch
ACV=10
182.8
C = 120
24.8
q =
5.38 (24.8)
0.5
P
e
=15 * 0.433
El=9.4
20.7
0.034
6.5
FF
1.0
190.8
3.260
GV=1.3
203.5
6.9
FF
1.0

4inch
El
54
C = 150
38.2
GV
CWM
4.0
190.8
4.240
Equivalent length of fittings?Slide152
System Underground  Fittings
152
N
main and riser
3inch
FF
CWM
5 ft
Long Turn Elbow
4inch PVC
(ID – 4.240 inches)
42 ft
12 ft
Fittings per Table 22.4.3.1.1:
Long turn elbow – 6 feet
Gate valve – 2 feetSlide153
Underground – Calculation Sheet
153
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
BL3
16.0
26.8
3inch
ACV=10
182.8
C = 120
24.8
q =
5.38 (24.8)
0.5
P
e
=15 * 0.433
El=9.4
20.7
0.034
6.5
FF
1.0
190.8
3.260
GV=1.3
203.5
6.9
FF
1.0

4inch
El
54
C = 150
38.2
GV
15.5
CWM
4.0
190.8
4.240
69.5
0.009
Friction loss?
0.6
P
e
= 5 * 0.433
2.2
41.0Slide154
Hose Stream – Calculation Sheet
154
Node
1
Elev.
1 (ft)
Kfactor
Flow – this step (q)
Nom. ID
Fittings – amount & length
L (ft)
CFactor
Pressure
Total (P
T
)
Notes
Node 2
Elev. 2 (ft)
Total Flow (Q)
Actual ID
F
(ft)
Friction Loss (psi/ft)
Elevation
(
P
e
)
T (ft)
Friction (P
f
)
FF
1.0

4inch
El
54
C = 150
38.2
P
e
= 5 * 0.433
GV
15.5
0.009
2.2
CWM
4.0
190.8
4.240
69.5
0.6
Hose
4.0
41.0
CWM
4.0
250
Hose stream for OH1?
440.8
System demand is 441 gpm @ 41 psi.Slide155
Flow (gpm)
Water Supply
155
Pressure (psi)
100
Flow Test Summary Sheet
0
60 psi residual pressure at 1000 gpm
90 psi static pressure
20
80
120
110
100
90
70
60
50
40
30
10
200
300
400
500
600
700
800
900
1000
1100
System Demand
190.8 gpm @ 41.0 psi
With Hose Stream
441 gpm @ 41.0 psiSlide156
Another Example
Comparison of various sprinkler options for protecting a Storage OccupancyCommodity: Computers in corrugated cardboard boxes with appreciable plastic trim (Class IV)
What are the options?156Slide157
Comparison Example
(continued)For comparison purposes all systems will be:Palletized Storage
20foot Maximum Storage HeightInstalled per NFPA 13Wet Type System Class IV Commodity
157Slide158
Sprinkler System Comparison
(General Storage, Class IV, 20ft high)
Sprinkler Type
KFactor
Design Area
# Sprinklers
Flow/ Sprinkler
Pressure
Total Flow
Ordinary Temperature
5.6
2000
20
39
48.5
900
8.0
2000
20
39
23.8
900
11.2
2000
20
39
12.9
900
14.0
2000
20
44.27
10
1018
High Temperature
5.6
2000
20
30
28.7
690
8.0
2000
20
30
14
690
11.2
2000
20
34.8
10
800
14.0
2000
20
44.3
10
1019
CMSA
11.2
1950
15
78
50
1345.5
11.2
2600
20
55
25
1265
ESFR
14.0
1200
12
99
50
1366
16.8
1200
12
99
35
1366
25.2
1200
12
178
50
2456
158Slide159
Reviewing Computer Calculations
Checking the input dataChecking the output data
159Slide160
Reviewing Inputs
Are the hydraulic nodes the same on the calculation and the plans?Are the sprinklers used in the calculation the same as on the plans?
Are the pipes the same type, schedule and size as on the plans?Is the water supply information the same as the flow test or design basis?
160Slide161
Reviewing Inputs
(continued)Check number of sprinklers in design areaCheck location of sprinklers to verify that the most demanding area is being calculated
Check the number of sprinklers in design area on each branch line
161Slide162
Reviewing Fire Pump Inputs
How does the software program treat fire pumps?When the program only inputs one data point (rated flow and pressure of the fire pump), there will be variance from the actual fire pump curve.
When the program inputs at least 3 data points, it can produce the correct performance curve for the fire pump
162Slide163
Reviewing Outputs
Does the flow into each node equal the flow out of the node?
163
Correct
75 gpm
25 gpm
50 gpmSlide164
Reviewing Outputs
(continued)Check friction loss between hydraulic nodes with HazenWilliams formula
Check equivalent lengths of fittings to make sure that the Cfactor and the inside diameter adjustments have been made where necessaryMake sure elevation changes are recorded in the pipes when applicable
164Slide165
Reviewing Outputs (continued)
Check that the minimum density has been metThe design area used should be multiplied by the density.
The value in the hydraulic calculations should be higher than the minimum density times area due to pressure losses in the system.Typical systems run 10 to 20 percent above the minimum.
165Slide166
Thank you for attending!
National Fire Sprinkler Association40 Jon Barrett RoadPatterson, NY 12563
(845) 8784200www.nfsa.org
166
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