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Hydraulics for Fire Protection

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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Hydraulics for Fire Protection






Presentation on theme: "Hydraulics for Fire Protection"— Presentation transcript:

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 1-foot 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 5-10 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 90-degree 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 6-inch pipe, how fast is it flowing when the pipe size is reduced to 3-inch?

18

?

6-inch

3-inch

5.7

ft/sSlide19

Flow Example 1 Solution

How fast is it flowing when the pipe size is reduced to 2-inch?

19

A

1

=

 r

2

=  (3 in)

2

= 28.3 in

2

A

2

=  r

2

=  (1.5 in)

2

= 7.1 in

2

?

6-inch

3-inch

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 k-factor 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 k-factor.

22Slide23

Sprinkler Flow Example

A sprinkler is being installed with a k-factor 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 k-factor:

24Slide25

Pressure Calculation Example

What is the pressure for a sprinkler that has a k-factor of 17.6 and the expected flow is 83 gpm?

25Slide26

K-factor Calculation Example

What is the K-factor 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

Hazen-Williams formulaFire sprinkler systems

Water-spray systemsDarcy-Weisbach formulaAnti-freeze systemsWater mist systemsFoam-water systemsFanning formula

28Slide29

Hazen-Williams Formula

Most common for sprinkler calculationsAssumes water is at room temperature but is still accurate with temperature variations

Based on C-factor, flow, and pipe sizeCalculates the amount of friction loss in ONE FOOT of pipe

29Slide30

Hazen-Williams 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 Hazen-Williams 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

Cement-lined

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*

1-inch

1.049

1.097

0.995

1.101

1 ¼-inch

1.380

1.442

1.245

1.394

1 ½-inch

1.610

1.682

1.481

1.598

2-inch

2.067

2.157

1.959

2.003

2 ½-inch

2.469

2.635

2.435

2.423

3-inch

3.068

3.2602.907

2.95

4-inch

4.026

4.260

3.857

N/ASlide33

Hazen-Williams Example

If a pressure gage is reading 40 psi at one end of a 32-foot section of 2-inch schedule 40 pipe (C = 120) flowing at 110 gpm, what will a gage at the other end read?

33

40

?

2-inch schedule 40 pipe

32 ft

P

L

= 0.112 psi/ftSlide34

Hazen-Williams 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

?

2-inch 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 materialC-factor other than 120Change in interior diameterOther than those for schedule 40 steel

37Slide38

Adjusting for C-Factor

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 C-factorSlide39

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 3-inch standard turn 90-degree elbow?

41Slide42

Equivalent Length Exercise Solution

What is the equivalent pipe length of Type K copper tube which used for a 3-inch standard turn 90-degree elbow?

NFPA 13 Table 22.4.3.1.1 :3-inch 90-degree 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 3-inch standard turn 90-degree elbow?

Adjustment for material (C-factor)Copper has a C-Factor of 150Per Table 22.4.3.2.1: Multiplier = 1.51Adjustment for inside diameter

3-inch copper has an inside diameter of 2.907-inch

43Slide44

Equivalent Length Exercise Solution

(continued)What is the equivalent pipe length of Type K copper tube which used for a 3-inch standard turn 90-degree elbow?

Apply the factors:Equivalent pipe length per Table 22.4.3.1.1 = 7 ftAdjustment for C-factor = 1.51Adjustment for diameter = 0.77The equivalent length for a 3-inch 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

As-Built 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 SystemsMulti-purpose 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

In-rack 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

Dry-Pipe 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 limited-combustible 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 limited-combustible 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 limited-combustible ceilings attached to composite wood joists directly or with 1-inch 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.Dry-pipe 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 single-row 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

Step-by-Step 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 k-factor 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

K-Factor needed to balanceSlide92

Balancing Flows

Only one pressure is inside the pipe

Use the higher pressureCalculate an equivalent K-factor for the portion of the pipe with the lower pressureCalculate the actual flow using the K-factor 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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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)

K-factor

Node 2

Elev. 2 (ft)

1

17.5

8.0

2

17.5Slide97

Work Sheets: K-Factor

The “K-Factor” column relates to the sprinklers or other flowing orifices used in the system.

For devices, such as sprinklers, the K-factor is found in the manufacturer’s cut sheets.At balancing points the K-factor would be the calculated value.

97

Elev.

1 (ft)

K-factor

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

K-factor

Flow – this step (q)

Nom. ID

Total Flow (Q)

Actual ID

8.0

26.0

1 ¼-inch

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 ¼-inch

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 ¼-inch

10

1.380

--

10Slide101

Work Sheets: Lengths

The length column sums the physical lengths (center-to-center) 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)

C-Factor

F

(ft)

Friction Loss (psi/ft)

T (ft)

10

C

= 120

--

0.056

10Slide102

Work Sheets: Friction Loss

The friction loss column contains the C-factor for the pipe in that step and the amount of friction loss per foot of pipe.

The Hazen-Williams formula is used to determine the friction loss and 3 decimal places are used.

102

L (ft)

C-Factor

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

C-Factor

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 K-factors.

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ½-inch

2T = 16

10

C

= 120

20.0

BL1

15.5

128.6

1.610

BL1

15.5

--

--

2-inch

13

C

= 120

BL2

15.5

2.067Slide106

Calculation Exercise Solution

106

Node

1

Elev.

1 (ft)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ½-inch

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

--

--

2-inch

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

As-Built 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

As-Built 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

As-Built Drawings

112Slide113

System Details

Type of System:

Wet pipe system

Type of Sprinkler: TY3121

Standard spray quick response upright sprinkler with a K-factor 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 J-1

Riser is 1 ft away from the east wall.

Long Turn Elbow

12.5 ft between branch lines

18 ft

4-inch PVC

(ID – 4.240 inches)

42 ft

7 ft

3-inch

Schedule 10Slide116

Electronics Factory Isometric View

116

N

15 ft

1 ft riser nipple

1 ½-inch

1 ½-inch

1-inch

1 ¼-inch

1 ¼-inch

1 ½-inch

main and riser

3-inch

Underground

4-inch 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

As-Built 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 ½-inch

1 ½-inch

1-inch

1 ¼-inch

1 ¼-inch

1 ½-inch

main and riser

3-inch

Underground

4-inch 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

1-inch

3-inch

Start With the Most Remote Sprinkler

132

1 ¼-inch

1 ¼-inch

1 ½-inch

1 ½-inch

1

2

3

4

5

6

7

8

9

BL1

BL2

BL3Slide133

Starting the Calculation Sheet

133

Node

1

Elev.

1 (ft)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

1-inch

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 ¼-inch

1 ¼-inch

1 ½-inch

1 ½-inch

1

2

3

4

18.8 gpm @ 11.3 psiSlide135

Second Sprinkler - Calculation Sheet

135

Node

1

Elev.

1 (ft)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

1-inch

10

C = 120

11.3

--

0.116

--

2

17.0

18.8

1.049

10

1.2

2

17.0

5.6

1 ¼-inch

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 ¼-inch

1 ½-inch

1 ½-inch

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ¼-inch

10

C

= 120

12.5

----

0.116

--

3

17.0

38.6

1.380

10

1.2

3

17.0

5.6

1 ¼-inch

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 ½-inch

1 ½-inch

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ¼-inch

10

C = 120

13.7

--

0.256

--

4

17.0

59.3

1.380

10

2.6

4

17.0

5.6

1 ½-inch

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

3-inch

1

2

3

4

BL1

BL2

81.9 gpm @ 24.4 psi

BL2 is identical to BL1 therefore an equivalent K-factor can be used to balance the flow.Slide141

Branch Line 1 - Calculation Sheet

141

Node

1

Elev.

1 (ft)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ½-inch

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

3-inch

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

3-inch

5

6

7

8

BL1

BL2

BL3

81.9 gpm @ 24.4 psi

BL2 is identical to BL1 therefore the K-factor from BL1 can be used to calculate the flow.Slide143

Branch Line 2 - Calculation Sheet

143

Node

1

Elev.

1 (ft)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

--

3-inch

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

3-inch

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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 ½-inch

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

3-inch

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

3-inch

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.

3-inch El – std 90 per Table 22.4.3.1.1:

7 feet

3-inch ACV – Viking Model J-1 per cut sheets:

10 feet

3-inch 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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

3-inch

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

3-inch

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

3-inch

FF

CWM

5 ft

Long Turn Elbow

4-inch PVC

(ID – 4.240 inches)

42 ft

12 ft

Underground pipe is

PVC.

C-factor

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

3-inch

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

--

4-inch

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

3-inch

FF

CWM

5 ft

Long Turn Elbow

4-inch 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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

3-inch

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

--

4-inch

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)

K-factor

Flow – this step (q)

Nom. ID

Fittings – amount & length

L (ft)

C-Factor

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

--

4-inch

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

20-foot Maximum Storage HeightInstalled per NFPA 13Wet Type System Class IV Commodity

157Slide158

Sprinkler System Comparison

(General Storage, Class IV, 20-ft high)

Sprinkler Type

K-Factor

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 Hazen-Williams formula

Check equivalent lengths of fittings to make sure that the C-factor 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) 878-4200www.nfsa.org

166