Hydraulics for Fire Protection International Fire Sprinkler Association

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 Review of Computer Calculations 2

Study of Water • Hydraulics – The 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. • Hydrokinetics – The study of water in motion. 3

Hydraulic Focus • Pressure • Flow 4

Pressure Types • Atmospheric Pressure – Caused by the weight of air, varies with altitude • Lower at high altitudes, higher at low altitudes • 14. 7 psi at sea level • Gage Pressure – The actual reading on a gage, does not account for atmospheric pressure. (psig) • Absolute Pressure – The sum of atmospheric pressure and gage pressure. (psia) 5

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. 6

Elevation Pressure 62. 4 lbs/ft 2 1 ft 0. 433 lbs/in 2 1 ft • A cubic foot of water results in a static pressure at its base of 62. 4 lbs/ft 2 • Converted to square inches a column of water 1 -foot high exerts a pressure of 0. 433 lbs/in 2 7

Elevation Pressure (continued) • Pressure (psi) = 0. 433 X Elevation (ft) 15 ft What is the pressure difference? 8

Elevation Pressure Example (Continued) • What is the pressure at the hydrant? Pressure (psi) = 0. 433 x Elevation (ft) 200 ft P=? 6 ft 9

Elevation Pressure 2 nd Example How high is the water? • ? ft P=47 psi 6 ft 10

Pressure Types (continued) • Residual Pressure (PR) – The pressure at a given point in a conduit or appliance with a specific volume of water flowing. 11

Pressure Types (continued) • Normal Pressure (PN) – The pressure created on the walls of pipe or tanks holding water. – This is the pressure read by most gages. PN • Velocity Pressure (PV) – The pressure associated with the PV flow of water measured in the same direction as the flow. 12

Calculating Velocity Pressures Pn = P t – P v Where: Pn = normal pressure (psi) Pt = total pressure (psi) Pv = velocity pressure (psi) Velocity pressure can be found as follows: 13

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. 14

Flow (Q) • The quantity (of water) which passes by a given point in a given period of time • Generally measured in gallons per minute (gpm) or cubic feet per second (ft 3/sec) • Uses the term “Q” in most equations 15

Flow Equation Q=Ax. V • Q = flow in ft 3/sec • A = cross sectional area of pipe in ft 2 • V = water velocity in ft/sec • Q is a constant for any given closed system. 16

Flow Equation (continued) Q = A x V = constant flow X gpm When the pipe size changes flow remains constant: Q = A 1 x V 1 = A 2 x V 2 17

Flow Example 1 6 -inch 5. 7 ft/s ? 3 -inch • 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

Flow Example 1 Solution 6 -inch 5. 7 ft/s ? 3 -inch How fast is it flowing when the pipe size is reduced to 2 inch? A 1 = r 2 = (3 in)2 = 28. 3 in 2 A 2 = r 2 = (1. 5 in)2 = 7. 1 in 2 19

Flow from an Outlet • Dependent upon a number of factors – Size of the orifice – Construction of the device – Material used in the device – Other components near the device (e. g. screens) • For a sprinkler, that ability is determined experimentally in a laboratory 20

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 21

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. 22

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? 23

Flow from a Sprinkler (continued) • The flow equation can be rearranged to solve for pressure or k-factor: 24

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? 25

K-factor Calculation Example • What is the K-factor for an outlet that is flowing 65 gpm at 30 psi? • 26

Friction Loss (PL) • Occurs when water flows in pipes, hoses, or other system devices • Caused by water in contact with walls • Used to account for losses in energy from water making turns or traveling difficult paths 27

Formulas for Calculating Friction Loss • Hazen-Williams formula – Fire sprinkler systems – Water-spray systems • Darcy-Weisbach formula – Anti-freeze systems – Water mist systems – Foam-water systems • Fanning formula 28

Hazen-Williams Formula • Most common for sprinkler calculations • Assumes water is at room temperature but is still accurate with temperature variations • Based on C-factor, flow, and pipe size • Calculates the amount of friction loss in ONE FOOT of pipe 29

Hazen-Williams Formula Where: PL = friction loss (psi/ft) Q = flow (gpm) C = roughness coefficient (based on pipe material) • di = interior pipe diameter (inches) • • 30

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) Galvanized (all) Plastic (listed, all) Cement-lined cast or ductile iron Copper tube or stainless steel Asbestos cement Concrete 120 150 140 140 31

Inside Diameters (di) List for steel and copper in Table A. 6. 3. 2 and Table A. 6. 3. 5 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. 423 3 -inch 3. 068 3. 260 2. 907 2. 95 4 -inch 4. 026 4. 260 3. 857 N/A 32

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? ? 40 2 -inch schedule 40 pipe 32 ft PL= 0. 112 psi/ft 33

Hazen-Williams Example (continued) • What will a gage at the other end read? ? 40 2 -inch schedule 40 pipe 32 ft • PL = 0. 112 psi/ft • Friction Loss = 0. 112 psi/ft x 32 ft = 3. 6 psi • Gage Pressure = 40 psi – 3. 6 psi 36 psi 34

Fittings • Energy losses through fittings are caused by turbulence in the water • To determine losses through fittings “equivalent length” is used • NFPA has a table to provide equivalent pipe lengths • Table is based on schedule 40 steel in a wet pipe system with C Values of 120. 35

Equivalent Length Chart Fittings & Valves Expressed in Equivalent Feet of Pipe ¾ in 1 ¼ in 1 ½ in 2 ½ 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 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 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 2 Swing Check - 5 7 9 11 14 16 19 22 27 32 45 55 65 8 2 7 9 11 12 9 13 16 18 3 4 5 6 36

Adjusting Equivalent Lengths • NFPA 13 table is based on schedule 40 steel pipe for a wet system • All others need to be adjusted for: – Change in pipe material • C-factor other than 120 – Change in interior diameter • Other than those for schedule 40 steel 37

Adjusting for C-Factor Table 22. 4. 3. 2. 1 C Value Multiplier Value of C 100 130 140 Multiplying 0. 713 1. 16 1. 33 Factor 150 1. 51 • Begin with the equivalent length value from the table • Multiply the length by the factor above for the appropriate C-factor 38

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 39

Fittings (continued) • All fittings must be accounted for in the calculations – Including tees, elbows, valves, etc. – Some may have pressure loss or equivalent length values from manufacturer’s listing information • Special provisions: – Fittings connected directly to sprinklers – Fittings where water flows straight through without changing direction 40

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? 41

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 pipe • Adjustments are needed for: – Type K Copper – Interior diameter 42

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 150 – Per Table 22. 4. 3. 2. 1: Multiplier = 1. 51 • Adjustment for inside diameter – 3 -inch copper has an inside diameter of 2. 907 -inch 43

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 ft – Adjustment for C-factor = 1. 51 – Adjustment for diameter = 0. 77 • The equivalent length for a 3 -inch Type K Copper standard turn elbow is: 7 ft x 1. 51 x 0. 77 = 8. 14 ft 44

Hydraulic Calculation Principles 45

The Layout Process 1. Define the Hazard 2. Analyze the Structure 3. Analyze the Water Supply 4. Select the Type of System 5. Select the Sprinkler Type(s) and Locate Them 6. Arrange the Piping 7. Arrange Hangers and Bracing (where needed) 8. Include System Attachments 9. Hydraulic Calculations 10. Notes and Details for Plans 11. As-Built Drawings 46

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 supply 47

Hydraulics Affected By… • Piping Configuration – Tree Systems – Loop Systems – Gridded Systems – Multi-purpose Systems • Can be any of the configurations above with at least one domestic fixture tied into the piping. • Pipe Size and Material • Pipe Fittings 48

Tree System ch n ra B es n Li Cross Main Riser Branch lines and sprinklers are fed from only one direction 49

Loop System Allows smaller cross mains because each branch line is fed from two directions. 50

Grid System Allows smaller cross mains and branch lines since each sprinkler is fed by at least two paths. 51

Hydraulics Affected By… (continued) • Type of Sprinkler – Standard Spray Sprinklers – Extended Coverage Sprinklers – Control Mode Specific Application (CMSA) Sprinklers – Early Suppression Fast Response (ESFR) Sprinklers – Residential Sprinklers – In-rack Sprinklers – Specially Listed Sprinklers 52

Hydraulics Affected By… (continued) • Design Method – Density/Area Method – Room Design Method – Special Design- NFPA special arrangements for residential, stairs, chutes, etc. 53

Hydraulic Calculation Principles • Provide enough water from each sprinkler to control or extinguish fire • Provide water for all sprinklers which are likely to open • Minimize pipe size for material cost, but not create large pressure loss due to friction 54

Hydraulic Calculation Approaches • Density/Area Method • Room Design Method • Special Design Approaches – Residential Sprinklers – ESFR Sprinklers – Specially Listed Sprinklers – Water Curtains – Other 55

Density/Area Method • Density is the flow of water that lands in a single square foot under the sprinkler • Measured in flow divided by unit area – English units: gpm/ft 2 • Flow required from a sprinkler is calculated by multiplying selected density by the coverage area 56

Density/Area Curves 57

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 58

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. 59

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 line – This 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 60

Density/Area Curves • Total Number of Sprinklers to Calculate – Design Area ÷ Area Per Sprinkler • Number of Sprinklers per Branch Line Where: S is the distance between sprinklers on the branch line 61

Density/Area Example 2 • The sprinkler system in an OH 2 occupancy has a discharge density of 0. 2 gpm/ft 2 over 1500 ft 2 (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? 62

Density/Area Example 2 (continued) Which sprinklers on the 3 rd line should be added? E D C B A 10 9 8 7 6 5 4 3 2 1 63

Area Adjustments • Dry-Pipe Systems – Increase area 30% (Section 11. 2. 3. 2. 5) • Double Interlock Preaction Systems – Increase Area 30% (Section 11. 2. 3. 2. 5) • Extra Hazard Occupancy with High Temperature Sprinklers – Decrease Area 25%, but minimum of 2000 ft 2 (Section 11. 2. 3. 2. 6) 64

Area Adjustments (continued) • Quick Response Sprinklers (11. 2. 3) – Area of operation can be reduced 25 to 40% depending on ceiling height when: • • • Wet pipe system only Light or ordinary hazards 20 ft maximum ceiling height No unprotected ceiling pockets No less than 5 sprinklers in design area – Area may be less than 1500 ft 2 65

Quick Response Area Adjustment • Ceiling Height <10 ft Ceiling Height (ft) v. % Reduction – Reduction is 40% 40 • Between 10 and 20 ft 35 – Y = (-3 x/2)+55 30 • Ceiling Height is 20 ft 25 20 – Reduction is 25% 15 10 • Over 20 ft Ceiling Height 5 – No reduction allowed 0 0 5 10 15 20 66

Area Adjustments (continued) • Sloped Ceilings – Area of operation is increased by 30% if pitch exceeds 2 in 12 (rise in run). This is an angle of 9. 46° rise run 67

Area Adjustments (continued) • Unsprinklered concealed spaces minimum 3, 000 ft 2 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 68

Area Adjustments (continued) • Unsprinklered concealed spaces minimum 3, 000 ft 2 applied after all other adjustments unless: – Flame spread rating 25 or less – Spaces 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 69

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 system • 30% increase for slope • Using 1500 ft 2 as the selected operation area – 1500 ft 2 x 1. 3 = 2535 ft 2 design area • There is no change in the density. 70

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 ft • 25% decrease for ceiling height • 30% increase for slope • Using 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. 71

Room Design Method • Density based on Figure 11. 2. 3. 1. 1 • Calculate all of the sprinklers in the most demanding room (usually the largest) • Walls must have fire resistance rating equal to the required water supply duration Table 11. 2. 3. 1. 2 Light Hazard 30 minutes Ordinary Hazard 60 – 90 minutes Extra Hazard 90 – 120 minutes 72

Room Design Method (continued) • Light Hazard – Doors must have automatic or self closers, or – Calculations must include two sprinklers from each adjoining space • Ordinary and Extra Hazard – Doors 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 73

Room Design Method Example Which room is the most demanding? Light Hazard, no door closers 2 1 3 4 6 7 5 11 8 9 15 10 12 13 14 74

RDM Example Solution Room # 1 2 Sprinklers in Room 6 2 Sprinklers Calculated 10 6 3 4 5 6 7 8 8 4 4 7 1 1 12 7 11 5 5 2 Room Sprinklers # in Room 9 1 10 3 11 12 13 14 15 3 1 1 1 6 Sprinklers Calculated 6 10 12 3 3 3 11 75

Special Design Approaches • Specially Listed Sprinklers – Minimum flow and/or pressure included in the listing of the sprinkler – Uses the area of calculation from the density area method • ESFR Sprinklers – 12 sprinklers calculated – 4 sprinklers over 3 branch lines • Other – Provides the number of sprinklers to be calculated and minimum pressure or flow necessary 76

Hydraulically Most Remote Sprinklers • Location of Open Sprinklers – Usually highest and farthest from system riser – Can 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 supply • Several sets of calculations may need to be done to find the most demanding values 77

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 78

City Water Mains • Information from the local water authority • Flow testing near the site • Need the following information: – Static Pressure – Residual Flow 79

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 system • If 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 pump – use larger pipe to reduce friction loss – maintain higher water level in an elevated tank – install tank at higher elevation 80

Step-by-Step Calculations 1. 2. 3. 4. 5. Identify hazard category Determine sprinkler spacing Determine piping arrangement Calculate amount of water needed per sprinkler Calculate number and location of open sprinklers in the hydraulically most demanding area 6. Start at most remote sprinkler and work towards the water supply calculating flows and pressures 7. Compare demand with supply 81

Example • Ordinary Hazard Group 2 occupancy • 12 ft ceiling height • Quick Response standard spray sprinklers with a k-factor of 5. 6 • Wet pipe sprinkler system • Sprinklers on 10 ft x 12. 5 ft spacing 82

Example: Plan View 123 ft 5 ft 80 ft 38 ft 10 ft 5. 5 ft 12. 5 ft 83

Example (continued) 1. 2. 3. 4. Select hazard category: OH 2 Determine sprinkler spacing: 10 ft x 12. 5 ft Determine piping arrangement: Done Calculate amount of water per sprinkler a) Select Density/Area Method b) Pick point from density/area curve: 0. 2 gpm/ft 2 over 1500 ft 2 c) 0. 2 gpm/ft 2 x 125 ft 2 = 25 gpm/sprinkler 84

Example (continued) 1. 2. 3. 4. Select hazard category: OH 2 Determine sprinkler spacing: 10 ft x 12. 5 ft Determine piping arrangement: Done Calculate amount of water per sprinkler: 25 gpm 5. Calculate number & location of open sprinklers a) Area Adjustment(s): QR Reduction: % = (-3 x/2) + 55 = [-3(12)/2] + 55 = 37% 1500 ft 2 x 0. 63 = 945 ft 2 b) 945 ft 2 ÷ 125 ft 2 per sprinkler = 7. 56 = 8 sprinklers c) 1. 2(945)0. 5/10 = 3. 7 = 4 sprinklers per branch line 85

Example: Hydraulic Remote Area 123 ft 5 ft 80 ft 38 ft 10 ft 5. 5 ft 12. 5 ft 86

Determine the Starting Pressure • Most remote sprinkler needs 25 gpm • Sprinkler k = 5. 6 • Starting information for the first sprinkler: – 25 gpm at 19. 9 psi • Next work back to the water supply adding pressure losses and flows throughout the system 87

Information Needed for Calculations • Select initial pipe sizes • Locate nodes on all places where: a. Flow (Q) takes place, b. Type of pipe or system changes (C), and c. Diameter (di) changes. • Layout calculation paths starting with primary path then attachment paths • Fill in hydraulic calculation sheets 88

Hydraulic Calculation Paths Locate the system nodes: 5 ft 38 ft 10 ft BL 2 4 8 3 7 2 1 6 5 TOR 12. 5 ft 5. 5 ft BL 1 89

Main Calculation Path • Start at the most remote sprinkler (#1) • Path: 1 2 3 4 BL 1 BL 2 TOR Balancing Point 90

Auxiliary Calculation Path • Section that connects into the “balancing point” • Auxiliary Path A: 5 6 7 8 BL 2 K-Factor needed to balance 91

Balancing Flows • Only one pressure is inside the pipe – Use the higher pressure 25 gpm @ 15 psi 35 psi 125 gpm @ 35 psi Q = 125 + 38. 2 Q = 163. 2 gpm • Calculate an equivalent K-factor for the portion of the pipe with the lower pressure • Calculate the actual flow using the K-factor and the new pressure. 92

Hydraulic Calculation Work Sheets • Hand calculations or computer calculations must present certain information • Computer generated sheets have been standardized by NFPA 13 (since January 2008) • Traditional hand calculation sheets have minor variations from the computer standard • The path created is the order the calculations will be completed 93

Calculation Work Sheets • Column Headings: Node 2 Elev. 1 K(ft) factor Elev. 2 (ft) Flow – this step (q) Total Flow (Q) Nom. ID Fittings – amount & Actual length ID L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) Total (PT) Pressure Node 1 Elevation (Pe) Notes Friction (Pf) 94

Work Sheets: Nodes Node 1 Elev. 1 (ft) Node 2 Elev. 2 (ft) 1 17. 5 2 17. 5 • 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

Work Sheets: Elevation Elev. 1 Node 1 (ft) Kfactor Elev. 2 Node 2 (ft) 1 17. 5 2 17. 5 8. 0 • 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

Work Sheets: K-Factor Elev. 1 (ft) Kfactor Elev. 2 (ft) 17. 5 8. 0 • The “K-Factor” column relates to the sprinklers Flow – or other flowing orifices this step used in the system. (q) • For devices, such as sprinklers, the K-factor Total is found in the Flow (Q) manufacturer’s cut sheets. • At balancing points the 26. 0 K-factor would be the calculated value. 26. 0 97

Work Sheets: Flow – Kthis step factor (q) Nom. ID Total Actual Flow (Q) ID 8. 0 26. 0 1 ¼-inch 26. 0 1. 380 • 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

Work Sheets: Pipe Size • The pipe size column is split into two rows – Flow – nominal and actual. Nom. Fittings this step • The “Nominal ID” is ID – (q) noted for the diameter amount Total Actual & length of the pipe in that step. • The “Actual ID” is the Flow (Q) ID real inside diameter 26. 0 1 ¼-inch used for the friction loss calculation. 26. 0 1. 380 99

Work Sheets: Fittings Nom. ID Fittings L (ft) – Actual amount F (ft) & length T (ft) ID 1 ¼-inch 1. 380 10 -- • 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. 10 100

Work Sheets: Lengths Fittings – amount & length L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 10 C = 120 -10 0. 056 • 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

Work Sheets: Friction Loss F (ft) Friction Loss T (ft) (psi/ft) 10 -10 C = 120 0. 056 Pressure L (ft) C-Factor • The friction loss column contains the CTotal (PT) factor for the pipe in Elevation that step and the amount of friction loss (Pe) per foot of pipe. Friction • The Hazen-Williams (Pf) formula is used to determine the friction 10. 6 loss and 3 decimal -places are used. 0. 5 102

Work Sheets: Pressure Friction Loss (psi/ft) C = 120 0. 056 Pressure C-Factor • “PT” is the total pressure at the start of that step Total (PT) • “Pe” is the pressure from Elevation the elevation change in Notes (Pe) the step Friction • “Pf” is the pressure in the (Pf) step caused by friction 10. 6 Q = 0. 2 (130) • The pressure is totaled = 26 gpm on the next line to start -P = (26/8. 0)2 = 10. 6 psi that step. 0. 5 103

Pressure Work Sheets: Notes Total (PT) • The “Notes” column is Elevation used to list additional Notes (Pe) information such as equations for flows, Friction pressures, elevation (Pf) pressure and equivalent Q = 0. 2 (130) = 10. 6 K-factors. 26 gpm -P = (26/8. 0)2 = 0. 5 10. 6 psi 104

Calculation Exercise Node 2 Elev. 1 K(ft) factor Total Flow (Q) Elev. 2 (ft) S 4 16. 5 BL 1 15. 5 BL 2 15. 5 Flow – this step (q) 5. 6 -- Nom. ID Fittings – amount & Actual length ID 25. 0 1 ½inch 128. 6 1. 610 -- 2 -inch 2 T = 16 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 10 C = 120 13 C = 120 Total (PT) Pressure Node 1 Elevation (Pe) Notes Friction (Pf) 20. 0 2. 067 • Determine the values for the highlighted boxes. 10 5

Calculation Exercise Solution Node 2 S 4 BL 1 BL 2 Elev. 1 K(ft) factor Total Flow (Q) Elev. 2 (ft) 16. 5 5. 6 15. 5 Flow – this step (q) 25. 0 128. 6 -- -128. 6 Nom. ID Fittings – amount & Actual length ID 1 ½inch 1. 610 2 -inch 2. 067 2 T = 16 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 10 C = 120 16 26 13 -13 0. 505 C = 120 0. 150 Total (PT) Pressure Node 1 Elevation (Pe) Notes Friction (Pf) 20. 0 0. 4 13. 1 33. 5 0 1. 9 10 6

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 psi – Residual pressure = 60 psi – Flow at 60 psi = 1000 gpm 107

The Layout Process 1. Define the Hazard 2. Analyze the Structure 3. Analyze the Water Supply 4. Select the Type of System 5. Select the Sprinkler Type(s) and Locate Them 6. Arrange the Piping 7. Arrange Hangers and Bracing (where needed) 8. Include System Attachments 9. Hydraulic Calculations 10. Notes and Details for Plans 11. As-Built Drawings 108

Identify the Hazard • In accordance with NFPA 13 hazard classifications, an electronics factory is classified as an Ordinary Hazard Group I occupancy. 109

The Layout Process 1. Define the Hazard 2. Analyze the Structure completed 3. Analyze the Water Supply 4. Select the Type of System 5. Select the Sprinkler Type(s) and Locate Them 6. Arrange the Piping 7. Arrange Hangers and Bracing (where needed) 8. Include System Attachments 9. Hydraulic Calculations 10. Notes and Details for Plans 11. As-Built Drawings 110

Available Water Supply 120 Flow Test Summary Sheet 110 100 90 psi static pressure 90 Pressure (psi) 80 60 psi residual pressure at 1000 gpm 70 60 50 40 30 20 100 200 300 400 500 600 700 800 Flow (gpm) 900 1000 111

The Layout Process 1. Define the Hazard 2. Analyze the Structure 3. Analyze the Water Supply 4. Select the Type of System 5. Select the Sprinkler Type(s) and Locate Them 6. Arrange the Piping 7. Arrange Hangers and Bracing (where needed) 8. Include System Attachments 9. Hydraulic Calculations 10. Notes and Details for Plans 11. As-Built Drawings 112

System Details • Type of System: – Wet pipe system • Type of Sprinkler: TY 3121 – 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 113

Electronics Factory Plan View N Mains are Schedule 10 Branch lines are Schedule 40 53 ft 12. 5 ft 100 ft 10 ft 5 ft from north wall and 6 ft from west wall 5 ft from south wall and 6. 5 ft from east wall 200 ft 114

Electronics Factory Elevation View Alarm Check Valve – Viking J-1 N Gate Valve 5 ft 12. 5 ft between branch lines All branch lines are on a 1 ft riser nipple 4 -inch PVC (ID – 4. 240 inches) Long Turn Elbow 7 ft 3 -inch Schedule 10 15 ft 18 ft Riser is 1 ft away from the east wall. 42 ft 115

Electronics Factory Isometric View 1 -inch 1 ¼-inch 1 ½-inch 1 ft riser nipple 1 ½-inch 15 ft main and riser 3 -inch N Underground 4 -inch PVC 116

The Layout Process 1. Define the Hazard 2. Analyze the Structure 3. Analyze the Water Supply 4. Select the Type of System 5. Select the Sprinkler Type(s) and Locate Them 6. Arrange the Piping 7. Arrange Hangers and Bracing (where needed) 8. Include System Attachments 9. Hydraulic Calculations 10. Notes and Details for Plans 11. As-Built Drawings 117

Select a Design Approach • Use the density/area method • A point from the density/area curves need to be selected 118

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 ft 2. – Wet pipe system, ordinary hazard, and a ceiling height of 18 ft 119

Area Reduction For QR Sprinklers – y = % reduction in area – x = ceiling height Ceiling Height (ft) v. % Reduction 40 35 30 25 20 15 10 5 0 0 5 10 15 20 120

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 2 121

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 122

Forming the Design Area • Continue to add branch lines until 9 sprinklers are included 123

12. 5 ft 9 Mains are Schedule 10 N 53 ft 5 6 7 8 100 ft 1 2 3 4 10 ft Remote Area Branch lines are Schedule 40 200 ft 124

Information Needed for Calculations 1. Select initial pipe sizes completed 2. Locate nodes on all places where: a) Flow (Q) takes place, b) Type of pipe or system changes (C), and c) Diameter (di) changes. 3. Layout calculation paths starting with primary path then attachment paths 4. Fill in hydraulic calculation sheets 125

Node Locations - Isometric 15 26 37 89 1 ft riser nipple 1 ½-inch 1 L 2 L 3 L B B B TOR 15 ft main and riser 3 -inch N FF M Underground 4 -inch PVC CW 1 -inch 1 ¼-inch 4 1 ½-inch 126

Information Needed for Calculations • Select initial pipe sizes • Locate nodes on all places where: – Flow (Q) takes place, – Type of pipe or system changes (C), and – Diameter (di) changes. • Layout calculation paths starting with primary path then attachment paths • Fill in hydraulic calculation sheets 127

Calculation Paths • Main Path: 1 2 3 BL 3 • Auxiliary Paths: 5 6 9 BL 3 7 BL 1 4 BL 2 TOR 8 BL 2 128

Information Needed for Calculations • Select initial pipe sizes • Locate nodes on all places where: – Flow (Q) takes place, – Type of pipe or system changes (C), and – Diameter (di) changes. • Layout calculation paths starting with primary path then attachment paths • Fill in hydraulic calculation sheets 129

Starting Sprinkler Values • Each sprinkler covers an area of 125 ft 2 • Each sprinkler is required to deliver a density of 0. 15 gpm/ft 2 • Minimum flow per sprinkler: 125 ft 2 x 0. 15 gpm/ft 2 = 18. 8 gpm • Minimum pressure for 18. 8 gpm: 130

Data Summary… Water Supply Information: Static Pressure Residual Pressure Hazard Classification System Type Ceiling Height Density/Area Adjusted to Sprinkler Type K – Factor Area Per Sprinkler Minimum Sprinkler Flow Minimum Pressure 90 psi @ 0 gpm 60 psi @ 1000 gpm OH - 1 Wet 18 feet 0. 15 gpm/ft 2 / 1500 ft 2 0. 15 gpm/ft 2 / 1080 ft 2 Quick Response – Standard Spray K =5. 6 125 ft 2 18. 8 gpm 11. 3 psi 131

Start With the Most Remote Sprinkler 1 2 3 4 1 ½-inch BL 1 1 -inch 1 ¼-inch 5 6 7 8 3 -inch 9 BL 2 BL 3 132

Starting the Calculation Sheet Elev. 2 (ft) 1 17. 0 2 17. 0 5. 6 Flow – this step (q) Total Flow (Q) Nom. ID Fittings – amount & Actual length ID 18. 8 1 -inch 18. 8 1. 049 L (ft) C-Factor F (ft) Total (PT) Pressure Node 2 Elev. 1 (ft) K-factor Node 1 Elevation (Pe) T (ft) Friction Loss (psi/ft) 10 C = 120 11. 3 -- 0. 116 -- 10 Notes Friction (Pf) 1. 2 12. 5 Now determine the friction loss 133

Second Sprinkler Calculation 1 2 3 4 1 ½-inch 18. 8 gpm @ 11. 3 psi 1 ¼-inch 1 ½-inch 134

Node 2 1 2 Elev. 1 (ft) Elev. 2 (ft) 17. 0 5. 6 17. 0 2 17. 0 3 17. 0 5. 6 Flow – this step (q) Total Flow (Q) 18. 8 Nom. ID Fittings – amount & Actual length ID 1 -inch L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 10 C = 120 -- 0. 116 Total (PT) Pressure Node 1 K-factor Second Sprinkler - Calculation Sheet Elevation (Pe) Friction (Pf) 11. 3 -- 18. 8 1. 049 10 19. 8 38. 6 1 ¼inch 10 C = 120 12. 5 ---- 0. 116 -- 1. 380 What is the total flow? Notes 1. 2 13. 7 What is the friction loss? 10 135

Third Sprinkler Calculation 18. 8 gpm @ 11. 3 psi 19. 8 gpm @ 12. 5 psi 1 2 3 4 1 ½-inch 1 ¼-inch 1 ½-inch 136

Node 2 2 3 Elev. 1 (ft) Elev. 2 (ft) 17. 0 5. 6 17. 0 3 17. 0 4 17. 0 5. 6 Flow – this step (q) Total Flow (Q) Nom. ID Fittings – amount & Actual length ID L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 10 C = 120 1 ¼inch ---- 38. 6 1. 380 10 20. 7 59. 3 1 ¼inch 10 19. 8 1. 380 What is the total flow? -- 0. 116 Total (PT) Pressure Node 1 K-factor Third Sprinkler - Calculation Sheet Elevation (Pe) Notes Friction (Pf) 12. 5 -1. 2 C = 120 13. 7 0. 256 -- 2. 6 16. 3 What is the friction loss? 10 137

Fourth Sprinkler Calculation 18. 8 gpm @ 11. 3 psi 2 19. 8 gpm @ 12. 5 psi 20. 7 gpm @ 13. 7 psi 1 3 4 1 ½-inch 138

Node 2 Elev. 1 (ft) Elev. 2 (ft) 3 17. 0 4 17. 0 5. 6 Flow – this step (q) Total Flow (Q) Nom. ID Fittings – amount & Actual length ID L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 20. 7 1 ¼inch 10 C = 120 59. 3 1. 380 10 22. 6 81. 9 1 ½inch -- 0. 256 2 T 19 C = 120 16 ft 16 0. 219 35 Total (PT) Pressure Node 1 K-factor Fourth Sprinkler - Calculation Sheet Elevation (Pe) Notes Friction (Pf) 13. 7 -2. 6 16. 3 0. 4 Pe = BL 1 16. 0 1. 610 7. 7 1 * 0. 433 24. 4 What is the total flow? What is the friction loss? Are there any fittings? 139

Branch Line 1 BL 2 is identical to BL 1 therefore an equivalent K-factor can be used to balance the flow. 2 81. 9 gpm @ 3 24. 4 psi 4 1 BL 1 3 -inch BL 2 140

Node 2 4 BL 1 Elev. 1 (ft) Elev. 2 (ft) 17. 0 16. 0 5. 6 Flow – this step (q) Total Flow (Q) 22. 6 81. 9 -81. 9 Nom. ID Fittings – amount & Actual length ID 1 ½inch 1. 610 3 -inch 2 T = 16 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 19 C = 120 16 35 13 -- 0. 219 C = 120 0. 007 Total (PT) Pressure Node 1 K-factor Branch Line 1 - Calculation Sheet Elevation (Pe) Notes Friction (Pf) 16. 3 0. 4 7. 7 Pe = 1 * 0. 433 24. 4 KBL 1 = 16. 58 0. 1 24. 5 What is the friction loss? What is the new flow? Are there any fittings? What is the total flow? BL 2 16. 0 3. 260 13 141

Branch Line 2 BL 2 is identical to BL 1 therefore the K-factor from BL 1 can be used to calculate the flow. 5 6 81. 9 gpm @ 24. 4 psi 7 8 BL 1 BL 2 3 -inch BL 3 142

Elev. 2 (ft) Branch Flow – Line 2 - Calculation Sheet Nom. this step (q) Total Flow (Q) Fittings – amount & Actual length ID ID BL 1 16. 0 -- 3 -inch BL 2 16. 0 81. 9 3. 260 BL 2 BL 3 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 12. 5 C = 120 -12. 5 0. 007 C = 120 Total (PT) Pressure Node 2 Elev. 1 (ft) K-factor Node 1 Elevation (Pe) Friction (Pf) 24. 4 -0. 1 3 -inch KBL 1 = 16. 58 24. 5 82. 1 --0. 026 0. 3 16. 0 3. 260 12. 5 164. 0 24. 8 What is the friction loss? What is the flow? Aretotal there any fittings? 16. 0 Notes 143

Branch Line 3 81. 9 gpm @ 24. 4 psi 82. 1 gpm @ 24. 5 psi BL 1 9 BL 2 BL 3 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. 144

Elev. 1 (ft) Node 2 Elev. 2 (ft) 9 17. 0 BL 3 16. 0 BL 3 5. 6 Flow – this step (q) Total Flow (Q) Nom. ID Fittings – amount & Actual length ID 18. 8 1 ½inch 18. 8 1. 610 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 19 C = 120 2 T 16 0. 014 16 ft 35 Total (PT) Pressure Node 1 K-factor Branch Line 3 - Calculation Sheet Elevation (Pe) Notes Friction (Pf) 11. 3 0. 4 0. 5 12. 2 Pe = 1 * 0. 433 Now determine the friction loss Fittings? Starting flow value is used: q = 0. 15 gpm/ft 2 * 125 ft 2 145

Branch Line 3 81. 9 gpm @ 24. 4 psi 82. 1 gpm @ 24. 5 psi BL 1 9 BL 2 18. 8 gpm @ 12. 2 psi BL 3 146

BL 2 BL 3 Elev. 2 (ft) 16. 0 FF 1. 0 Nom. ID Fittings – amount & Total Actual length Flow (Q) ID 82. 1 16. 0 BL 3 Flow – this step (q) 164. 0 26. 8 190. 8 3 -inch 3. 260 Fittings L (ft) F (ft) C-Factor T (ft) Friction Loss (psi/ft) 12. 5 C = 120 -12. 5 182. 8 0. 026 C = 120 Total (PT) Pressure Node 2 Elev. 1 (ft) 5. 38 Node 1 K-factor Branch Line 3 – Balancing Point Elevation (Pe) Notes Friction (Pf) 24. 5 -0. 3 24. 8 q = 16. 58 (24. 5)0. 5 q = 5. 38 (24. 8)0. 5 147

Branch Line 3 - Fittings BL 3 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. 148

Node 2 BL 2 Elev. 1 (ft) Elev. 2 (ft) 16. 0 Flow – this step (q) Nom. ID Fittings – amount & Total Actual length Flow (Q) ID 82. 1 3 -inch BL 3 16. 0 164. 0 3. 260 BL 3 16. 0 26. 8 3 -inch FF 1. 0 190. 8 3. 260 L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) 12. 5 C = 120 -12. 5 ACV=10 182. 8 El=9. 4 GV=1. 3 0. 026 C = 120 Total (PT) Pressure Node 1 K-factor Branch Line 3 – Balancing Point (continued) Elevation (Pe) Friction (Pf) 24. 5 -0. 3 24. 8 6. 5 6. 9 38. 2 Friction Loss? 20. 7 0. 034 203. 5 Notes q = 16. 58 (24. 5)0. 5 q = 5. 38 (24. 8)0. 5 Pe = 15 * 0. 433 149

System Underground N Underground pipe is PVC. M 12 ft 42 ft CW Long Turn Elbow 5 ft FF C-factor is 150 per Table 4 -inch PVC 22. 4. 4. 7. (ID – 4. 240 inches) Long turn elbows are typically used underground and cause less turbulence to the flow in the pipe. main and riser 3 -inch 150

Node 2 BL 3 FF Elev. 1 (ft) Elev. 2 (ft) 16. 0 1. 0 Flow – this step (q) Nom. ID L (ft) C-Factor F (ft) T (ft) Friction Loss (psi/ft) ACV=10 182. 8 C = 120 Fittings – amount & Total Actual length Flow (Q) ID 26. 8 190. 8 3 -inch 3. 260 FF 1. 0 -- 4 -inch CWM -4. 0 190. 8 4. 240 El=9. 4 20. 7 GV=1. 3 203. 5 El 54 Total (PT) Pressure Node 1 K-factor Underground – Calculation Sheet 0. 034 C = 150 Elevation (Pe) Notes Friction (Pf) 24. 8 6. 5 6. 9 q = 5. 38 (24. 8)0. 5 Pe=15 * 0. 433 38. 2 GV Equivalent length of fittings? 151

System Underground - Fittings per Table 22. 4. 3. 1. 1: Long turn elbow – 6 feet Gate valve – 2 feet main and riser 3 -inch N M 42 ft CW Long Turn Elbow 12 ft 4 -inch PVC (ID – 4. 240 inches) 5 ft FF 152

Node 2 BL 3 FF Elev. 1 (ft) Elev. 2 (ft) 16. 0 1. 0 Flow – this step (q) Nom. ID Fittings – amount & Total Actual length Flow (Q) ID 26. 8 190. 8 3 -inch 3. 260 FF 1. 0 -- 4 -inch CWM -4. 0 190. 8 4. 240 L (ft) F (ft) C-Factor T (ft) Friction Loss (psi/ft) ACV=10 182. 8 C = 120 El=9. 4 20. 7 GV=1. 3 203. 5 El 54 GV 15. 5 69. 5 0. 034 C = 150 0. 009 Total (PT) Pressure Node 1 K-factor Underground – Calculation Sheet Elevation (Pe) Notes Friction (Pf) 24. 8 6. 5 6. 9 q = 5. 38 (24. 8)0. 5 Pe=15 * 0. 433 38. 2 2. 2 0. 6 41. 0 Friction loss? Pe = 5 * 0. 433 153

Node 2 FF CWM Elev. 1 (ft) Elev. 2 (ft) 1. 0 -4. 0 Hose -4. 0 CWM -4. 0 Flow – this step (q) Nom. ID Fittings – amount & Total Actual length Flow (Q) ID -190. 8 250 440. 8 4 -inch 4. 240 L (ft) F (ft) C-Factor T (ft) Friction Loss (psi/ft) El 54 C = 150 GV 15. 5 69. 5 0. 009 Total (PT) Pressure Node 1 K-factor Hose Stream – Calculation Sheet Elevation (Pe) Notes Friction (Pf) 38. 2 2. 2 0. 6 Pe = 5 * 0. 433 41. 0 System demand is 441 gpm @ 41 psi. Hose stream for OH 1? 154

Water Supply 120 Flow Test Summary Sheet 110 100 90 psi static pressure 90 Pressure (psi) 80 60 psi residual pressure at 1000 gpm 70 60 50 System Demand 190. 8 gpm @ 41. 0 psi 40 With Hose Stream 441 gpm @ 41. 0 psi 30 20 100 200 300 400 500 600 700 800 Flow (gpm) 900 1000 1100 155

Another Example • Comparison of various sprinkler options for protecting a Storage Occupancy – Commodity: Computers in corrugated cardboard boxes with appreciable plastic trim (Class IV) • What are the options? 156

Comparison Example (continued) • For comparison purposes all systems will be: – Palletized Storage – 20 -foot Maximum Storage Height – Installed per NFPA 13 – Wet Type System – Class IV Commodity 157

Sprinkler System Comparison (General Storage, Class IV, 20 -ft high) ESFR CMSA High Ordinary Temperature Sprinkler Type K-Factor 5. 6 8. 0 11. 2 14. 0 16. 8 25. 2 Design Area 2000 2000 1950 2600 1200 # Sprinklers 20 20 15 20 12 12 12 Flow/ Sprinkler 39 39 39 44. 27 30 30 34. 8 44. 3 78 55 99 99 178 Pressure Total Flow 48. 5 23. 8 12. 9 10 28. 7 14 10 10 50 25 50 35 50 900 900 1018 690 800 1019 1345. 5 1265 1366 158 2456

Reviewing Computer Calculations • Checking the input data • Checking the output data 159

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? 160

Reviewing Inputs (continued) • Check number of sprinklers in design area • Check 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 161

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 162

Reviewing Outputs • Does the flow into each node equal the flow out of the node? 75 gpm 50 gpm 25 gpm Correct 163

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 necessary • Make sure elevation changes are recorded in the pipes when applicable 164

Reviewing Outputs (continued) • Check that the minimum density has been met – The 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. 165

Thank you for attending! National Fire Sprinkler Association 40 Jon Barrett Road Patterson, NY 12563 (845) 878 -4200 www. nfsa. org 16 6