Hydraulics for Fire Protection Section references throughout the presentation are for NFPA 13, 2010 Edition unless otherwise noted. Revision 13 May 2011 - vbv International Fire Sprinkler Association www.firesprinkler.global

Course Outline Definitions and Equations Hydraulic Calculation Principles Hydraulic Calculation Process Example Calculation Review of Computer Calculations Most of the sections have clear breaks between them.

Study of Water Hydraulics Hydrostatics Hydrokinetics 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. Definition of hydraulics. Examining the principles of water at rest as well as water flowing.

Hydraulic Focus Pressure Flow This is the test apparatus for determining a fire sprinkler k-factor. It is run at a known flow and pressure. With these two items as the focus of this section of class, definitions and equations will be presented in order to determine these values in the hydraulic calculations later on.

Pressure Types Atmospheric Pressure Gage Pressure Absolute 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) Atmospheric pressure is the weight of air in the atmosphere over an area. 14.7 lbs is the weight of a one inch square column of air as tall as the atmosphere measured at sea level. The change in atmospheric pressure may be important when installing a system at high elevations. In many cases it is not necessary to use atmospheric pressure in calculations done for sprinkler systems. However, there are times when the absolute pressure will need to be used and therefore the atmospheric pressure must be accounted for in the equation. Gage pressure is the pressure read off of a gage. It does not include atmospheric pressure as the atmospheric pressure is also acting on the gage. Absolute pressure is the sum of atmospheric pressure and gage pressure. In a few equations, it is necessary to use the absolute pressure in order to arrive at a correct answer.

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. Static pressure is the amount of pressure available when normal resting conditions exist. The amount of static pressure can increase with elevation of an object or in our case the water.

Elevation Pressure 62.4 lbs/ft2 0.433 lbs/in2 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 The value of water pressure in psi can be used in order to determine pressure changes due to elevation differences.

Elevation Pressure (continued) Pressure (psi) = 0.433 X Elevation (ft) 5 ft 15 ft This is the formula for pressure created through elevation changes with water. This formula is only appropriate for use when pressure is expressed in psi and elevation in feet. The 0.433 is a constant based on the weight of water (shown on the previous slide). Give the class a minute to determine what the pressure difference will be. P = 0.433 x 10 ft = 4.3 psi What is the pressure difference?

Elevation Pressure Example (Continued) What is the pressure at the hydrant? Pressure (psi) = 0.433 x Elevation (ft) 200 ft P=? The amount of pressure available from a water tank is important to know. The measurement is to the water line because we are trying to determine what the gage will read. When used for calculation purposes, the lowest part of the tank (just above where the feed line connects to the tank) would be used for determining pressure availability for a system because the number will change as the water level decreases. It should be noted that when gages are involved, rounding to the whole number is close enough. There is much fluctuation from gage needles or interpretation between marked values that a whole number is the appropriate level of accuracy. 6 ft

Elevation Pressure 2nd Example How high is the water? ? ft P=47 psi This is the equation we use to calculate pressure due to elevated water. Now, the equation is being used in a different direction, the elevation is being determined. Again, rounding is acceptable. 6 ft

Pressure Types (continued) Residual Pressure (PR) The pressure at a given point in a conduit or appliance with a specific volume of water flowing. Residual pressure is the pressure of water while flowing. Discussion of water flow tests will occur later on in this presentation.

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. Velocity Pressure (PV) The pressure associated with the flow of water measured in the same direction as the flow. PN A pressure gage in the wall of the pipe would read normal pressure available from the water inside the system. The velocity pressure comes from the movement of the water (or fluid) in the piping. This pressure acts in the same direction as the water. Pn PV

Calculating Velocity Pressures Pn = Pt – Pv Where: Pn = normal pressure (psi) Pt = total pressure (psi) Pv = velocity pressure (psi) Velocity pressure can be found as follows: The equation on the bottom comes from Section 22.4.2.2. It is not required for the velocity pressure to be calculated in NFPA 13, but other water-based fire protection systems such as water spray (NFPA 15) or water mist (NFPA 750) may require it to be used.

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. For most NFPA 13 systems, the calculation is conservative without accounting for the velocity pressure. Most scenarios where velocity pressure would become a concern with an NFPA 13 system are self-regulating because of the high friction loss values that typically accompany the high velocities. It is permitted to use the velocity pressures though for any system and may prove beneficial if the calculation is close to what is available from the water supply.

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 (ft3/sec) Uses the term “Q” in most equations Definition of flow.

Q = A x V Flow Equation Q = flow in ft3/sec A = cross sectional area of pipe in ft2 V = water velocity in ft/sec Q is a constant for any given closed system. This is one of the basic equations utilized in the industry. Most flow meters work on the concept of measuring the velocity of water flowing past a point and knowing the diameter of the opening. From this equation, the meter would then calculates the flow. Units are important when working with this equation, as flow is more common in the sprinkler industry in gpm and not ft3/sec.

Flow Equation (continued) Q = A x V = constant flow X gpm When the pipe size changes flow remains constant: Q = A1 x V1 = A2 x V2 A1 x V1 = A2 x V2 Since the flow in any closed system is constant, then the velocity must change each time the diameter (or area) of the piping changes. The last line on this slide shows the relationship between the area and velocity at one point in a system, and the area and velocity at a second point in the system.

Flow Example 1 ? 3-inch 6-inch 5.7 ft/s 6-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? Example: How changes in diameter effect velocity in the pipe

Flow Example 1 Solution ? 6-inch 3-inch 5.7 ft/s How fast is it flowing when the pipe size is reduced to 2-inch? A1 =  r2 =  (3 in)2 = 28.3 in2 A2 =  r2 =  (1.5 in)2 = 7.1 in2 It is okay to use in2 in this equation, unlike the previous slide, because the areas cancel out in the equation. This means it is only important that the area for each section of pipe is in the same units and not what the units are. The general rule of thumb is that when the pipe diameter is cut in half the velocity will quadruple in order to keep the flow constant.

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

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

k is k-factor determined in the sprinkler listing (gpm/psi½) 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.

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?

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

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? NICET uses one decimal place on pressure values for their calculations. Discuss: The minimum pressure requirement of 7 psi from NFPA 13 or minimum values from manufacturer cut sheets.

K-factor Calculation Example What is the K-factor for an outlet that is flowing 65 gpm at 30 psi? Sprinklers are listed with a nominal k-factor, with values taken to one decimal place, for use in calculations. However, when a k-factors are calculated for use within the calculation, two decimal places should be used.

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 The loss in energy within the system due to water flow is commonly called friction loss. When the water molecules contact the inside surface of walls it slows down. The water in the center of the pipe is usually flowing the fastest. When the water is laminar flow then the friction loss has less of an effect than when it is turbulent flow.

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 The Fanning formula has similarities to the Darcy-Weisbach formula. It was used as the predecessor before calculators and computers allowed for easier calculations for more complet formulas such as the Darcy-Weisbach.

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 In water-based fire protection systems, it is common to use the Hazen-Williams formula for friction loss calculations. It was developed over 100 years ago by two gentlemen (Mr. Hazen and Mr. Williams) who basically measured the flow of water in different pipes and the difference in pressures at both ends. They then developed a mathematical formula which fit the data they observed. This is called an empirically derived formula and should be accepted as long as the data collection is valid. This formula is valid for velocities less than 30 ft/s. Above this point the equation may not be accurate. Other industries, specifically the civil engineering community, have other methods and formulas for calculating friction loss. These are acceptable under NFPA 13, as long as they are also accepted practices and applicable to water-based closed systems.

Hazen-Williams Formula Where: PL = friction loss (psi/ft) Q = flow (gpm) C = roughness coefficient (based on pipe material) di = interior pipe diameter (inches) The Hazen Williams formula must be in the units specified for the formula to work correctly, as the constant 4.52 relates to these units.

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) Black steel (wet systems including deluge) 120 Galvanized (all) Plastic (listed, all) 150 Cement-lined cast or ductile iron 140 Copper tube or stainless steel Asbestos cement Concrete The table is from the 2010 Edition of NFPA 13. The roughness coefficient is one of the safety factors built into the Hazen-Williams calculations. The typical C-factor value for new steel pipe is between 140 and 150, but NFPA 13 recognizes that the pipe will corrode over time (as it ages) due to its contact with water, air, and anything carried by the water into the system. Therefore, C-factors of 100 or 120 depending on the type of system are used. The lower C-factors represent rougher material piping and the higher C-factors are smoother.

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 2.435 2.423 3-inch 3.068 3.260 2.907 2.95 4-inch 4.026 4.260 3.857 N/A CPVC pipe is made by a few manufacturers, data for ID should be taken from the design and installation information provided by the manufacturer(s). This is just a section of information available in NFPA 13 as well as other resources for inside pipe diameters. Discuss why the focus is on the inside diameter. Note that the tables also include outside diameters for spacing in the building, sizing hangers, etc.

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 Example of how to use the Hazen-Williams formula. Basically we want to know how much pressure will be lost as the water flows through this pipe. PL= 0.112 psi/ft

Hazen-Williams Example (continued) What will a gage at the other end read? 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 40 ? 2-inch schedule 40 pipe 32 ft Calculate the lose over the 32-foot section of pipe.

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.

Equivalent Length Chart 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 2 3 4 5 7 9 11 12 90° Standard Elbow 6 8 10 14 18 22 27 90° Long Turn Elbow 13 16 Tee/Cross 15 17 20 25 30 50 60 Butterfly Valve - 19 21 Gate Valve Swing Check 32 45 55 65

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 There is a handout in the workbook that shows the calculation for a 2-inch fitting in three different materials. The point is to discuss that the fittings offer corrections for the friction loss that is caused by the turbulence of the water. A 2-inch elbow will lose approximately 0.5 psi no matter the material due to the water turning. The change in equivalent length is so that the value comes out to be 0.5 psi. This is demonstrated in the handout.

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 Begin with the equivalent length value from the table Multiply the length by the factor above for the appropriate C-factor

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

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 The straight through flow applies to steel and copper. This does not apply to CPVC pipe. Other specially listed piping should be checked by review of the manufacturer’s installation instructions.

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?

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

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

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

Hydraulic Calculation Principles This is a section break to switch gears. It takes the focus from familiarity with the equations to the process of the hydraulic calculation.

The Layout Process Define the Hazard Analyze the Structure Analyze 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 The intention of this slide is to point out where hydraulics fit in the overall picture. This is helpful for those who actually plan the layout of the system. It is also relevant to those who review the submittal package, so that hydraulics are noted as an important piece that needs to be looked at but that it is clearly noted there are also other important features that are looked at that can alter the hydraulics such as spacing.

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 2

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

Branch lines and sprinklers are fed from only one direction Tree System Branch Lines Cross Main Riser Branch lines and sprinklers are fed from only one direction

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

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

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 Different starting pressures, different coverage areas and different k-factors can change the entire calculation.

Hydraulics Affected By… (continued) Design Method Density/Area Method Room Design Method Special Design- NFPA special arrangements for residential, stairs, chutes, etc. This can change the number of sprinklers that need to be calculated.

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

Hydraulic Calculation Approaches Density/Area Method Room Design Method Special Design Approaches Residential Sprinklers ESFR Sprinklers Specially Listed Sprinklers Water Curtains Other Note that the residential calculations for sprinklers will be different where NFPA 13D or NFPA 13R are followed compared with NFPA 13.

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/ft2 Flow required from a sprinkler is calculated by multiplying selected density by the coverage area

Density/Area Curves Emphasize that the lowest point is the most common to select due to the smallest amount of calculation area. Note that sliding up the curve may have benefits, such as insurance requirements for more calculation area, combustible concealed space requiring 3000 ft2 minimum, etc. 6

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/ft2, what is the minimum required flow from a sprinkler? Coverage Area: A = 10 ft x 11.5 ft = 115 ft2 Density times area equals flow: 0.2 gpm/ft2 x 115 ft2 = 23 gpm

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. Discuss that it may not always be a perfect rectangle. The long side of the rectangle needs to be parallel to the branch line. Fire has been empirically studied to show that the general shape of the effected area is rectangular. Since there is no way to tell where a fire would occur over the life of the building, the longer of that rectangular needs to be parallel to the branch line as that forces more water inside a single pipe and is therefore more demanding on the system.

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 Discuss that the fire rectangle calculation from the previous slide is to determine the minimum length of the design area along the dimension of the branch lines. When this is divided by the distance between the sprinklers (where a typical length is appropriate) the calculation then determines the number of sprinklers to be calculated on a line. There can never be a partial value and no matter what the decimal value is, the number of sprinklers is round up to the next whole number. Verification should be done at the end of laying out which sprinklers will be calculated to make sure the total area needed is actually covered by the selected sprinklers.

Density/Area Curves Total Number of Sprinklers to Calculate Design Area ÷ Area Per Sprinkler Number of Sprinklers per Branch Line Discuss that we never add a partial sprinkler. When dividing by area per sprinkler or distance between sprinklers (S), caution is needed because this only applies where there is uniform spacing. With non-uniform spacing the length in feet should be laid along the branch line in order to determine the number of sprinklers needed in the calculation. When the length is laid out, be careful to make sure that a partial area hasn’t been entered and another sprinkler needed along the line. Where: S is the distance between sprinklers on the branch line 12

Density/Area Example 2 1500 ft2 ÷ 115 ft2 = 13.04 14 sprinklers 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 ft2, how many sprinklers will be in the design area? 1500 ft2 ÷ 115 ft2 = 13.04 14 sprinklers If sprinklers along the branch line are 10 ft apart, how many sprinklers/line are calculated? This is continued from Example 1.

Density/Area Example 2 (continued) Which sprinklers on the 3rd line should be added? E 10 5 Assume this is the plan that goes with the previous problem. Discuss selecting sprinklers when it is not a perfect rectangle. D 9 4 C 8 3 B 7 2 A 6 1

Area Adjustments Dry-Pipe Systems Double Interlock Preaction 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 ft2 (Section 11.2.3.2.6) Discuss that these apply to the density/area method. Some can be large benefits to the calculation while others are a necessity but cause significantly larger demands. 7

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 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 ft2

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 This can be used whenever the criteria list is met. This includes concealed spaces.

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 There are no units for the slope. Sometimes it is written 1 in 6. run

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 ft3 areas, including areas under insulation directly on joists If this needs to be used, it may be beneficial to slide up the density/area curves to a larger starting area that corresponds to a lower density, since the large area will have to be applied anyways.

Area Adjustments (continued) Unsprinklered concealed spaces minimum 3,000 ft2 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 ft2 Vertical pipe chases under 10 ft2 meeting 8.15.1.2.14 Exterior columns under 10 ft2 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 ft3 areas This is continuation of the list from the previous slide.

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 ft2 as the selected operation area 1500 ft2 x 1.3 x 1.3 = 2535 ft2 design area There is no change in the density. Discuss that 1.3 as a multiplier is the same as 130% or calculating 30% and then adding it to the 1500 ft2.

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 ft2 as the operation area 1500 ft2 x 0.75 x 1.3 = 1463 ft2 design area There is no change in density. Discuss that after area adjustments the layout of the remote area is the same process for determining the number of sprinklers in the design area and the number of sprinklers calculated per branch line.

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 Not always the largest room, especially with mixed hazards. In order to use the lower values for ordinary and extra hazard, the system must be electrically supervised for waterflow alarms and supervisory devices.

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 Discuss that the fire rating on doors is not identical to that of the walls but it is appropriate for the rating of the compartment.

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

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

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

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

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

City Water Mains Information from the local water authority Flow testing near the site Need the following information: Static Pressure Residual Pressure Residual Flow To determine the adequacy of the available water supply, a flow test must be conducted. Some water authorities have projection programs (i.e. City of LA). These predict years ahead what condition the water supply will be in and the amounts of flow and pressure that will be available for the system. Some water authorities have additional cushions that need to be maintained on the water supply. Some require 10 psi margin others use values that the public main pressure is not lowered beyond certain points such as 15 psi minimum.

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 Velocity pressure calculations may help where demands are close to available amounts.

Step-by-Step Calculations 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 Start at most remote sprinkler and work towards the water supply calculating flows and pressures Compare demand with supply The box is the focus of this lesson. Step #5 includes locating where the hydraulically most demanding area is laid in the system.

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

Example: Plan View 123 ft 5 ft 10 ft 38 ft 80 ft 12.5 ft 5.5 ft 5 ft 16

Example (continued) Select hazard category: OH2 Determine 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/ft2 over 1500 ft2 0.2 gpm/ft2 x 125 ft2 = 25 gpm/sprinkler 4

Example (continued) Calculate number & location of open sprinklers Select hazard category: OH2 Determine 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 ft2 x 0.63 = 945 ft2 945 ft2 ÷ 125 ft2 per sprinkler = 7.56 = 8 sprinklers 1.2(945)0.5/10 = 3.7 = 4 sprinklers per branch line 4

Example: Hydraulic Remote Area 123 ft 5 ft 10 ft 38 ft 80 ft Discuss that it is located on the bottom side because of the main location and distance the water travels to the most remote sprinkler. 12.5 ft 5.5 ft 5 ft 5 ft 16

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

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 Discuss differences in selecting extra nodes for computer calculations. In hand calculations the extra steps are undesirable due to the extra effort, but in computer calculations it allows for easy access to change pipe sizes and optimize the system.

Hydraulic Calculation Paths Locate the system nodes: 5 ft 10 ft 38 ft BL1 BL2 TOR 4 8 12.5 ft 3 7 2 6 5.5 ft 5 ft 1 5 5 ft

Main Calculation Path 1 2 3 4 BL1 BL2 Start at the most remote sprinkler (#1) Path: BL1 BL2 1 2 3 4 TOR Balancing Point

Auxiliary Calculation Path Section that connects into the “balancing point” Auxiliary Path A: BL2 5 6 7 8 K-Factor needed to balance

Balancing Flows Only one pressure is inside the pipe Use the higher pressure 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. 25 gpm @ 15 psi 35 psi 125 gpm @ 35 psi Q = 125 + 38.2 Q = 163.2 gpm

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 Discuss that the velocity pressure column is not required on the computer generated standard because it is not used all the time or required to be used by NFPA 13. Also, there is no step number or reference number columns, although this is not necessary with computers it is a convenience when performing hand calculations.

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

Work Sheets: Nodes Node 1 Elev. 1 (ft) Node 2 Elev. 2 (ft) 1 17.5 2 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. Node 1 Elev. 1 (ft) Node 2 Elev. 2 (ft) 1 17.5 2 Emphasize that the plans need to indicate the hydraulic node points on them.

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. Node 1 Elev. 1 (ft) K-factor Node 2 Elev. 2 (ft) 1 17.5 8.0 2

Work Sheets: K-Factor Elev. 1 (ft) K-factor Flow – this step (q) 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. Elev. 1 (ft) K-factor Flow – this step (q) Elev. 2 (ft) Total Flow (Q) 17.5 8.0 26.0

Work Sheets: Flow K-factor Flow – this step (q) Nom. ID Total Flow (Q) 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. K-factor Flow – this step (q) Nom. ID Total Flow (Q) Actual ID 8.0 26.0 1 ¼-inch 1.380

Fittings – amount & length 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. Flow – this step (q) Nom. ID Fittings – amount & length Total Flow (Q) Actual ID 26.0 1 ¼-inch 1.380 Remind the students that the actual inside diameters for steel and copper are in the annex of the standard. Also add that others can be found from manufacturers.

Fittings – amount & length 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. Nom. ID Fittings – amount & length L (ft) Actual ID F (ft) T (ft) 1 ¼-inch 10 1.380 -- Reminder that sprinklers directly attached to a fitting on the branch line do not get listed in the fitting column or need an equivalent length for the calculation.

Fittings – amount & length Friction Loss (psi/ft) 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 Fittings – amount & length L (ft) C-Factor F (ft) Friction Loss (psi/ft) T (ft) 10 C = 120 -- 0.056

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. L (ft) C-Factor Pressure Total (PT) F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 10 C = 120 10.6 -- 0.056 0.5 Emphasize that NICET uses 3 decimal places for their values.

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

Work Sheets: Notes Pressure Total (PT) Notes Elevation (Pe) Friction (Pf) 10.6 Q = 0.2 (130) = 26 gpm P = (26/8.0)2 = 10.6 psi -- 0.5 The “Notes” column is used to list additional information such as equations for flows, pressures, elevation pressure and equivalent K-factors.

Fittings – amount & length Friction Loss (psi/ft) Calculation Exercise Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) S4 16.5 5.6 25.0 1 ½-inch 2T = 16 10 C = 120 20.0 BL1 15.5 128.6 1.610 -- 2-inch 13 BL2 2.067 Determine the values for the highlighted boxes.

Calculation Exercise Solution Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 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 -- 2-inch 13 33.5 0.150 BL2 2.067 1.9 Discuss the difference in friction loss from the pipe size since there was no added flow. Note that if the system doesn’t work because of the available water the first step on the screen is a good place to look at increasing the pipe diameter and saving a significant amount of pressure from friction, which only accumulates throughout the system as the calculation goes on. It is good to take a break here because the next section begins the full walk-through for a hydraulic calculation.

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

The Layout Process Define the Hazard Analyze the Structure Analyze 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 Go through the process with the example. The first item is always to define the hazard.

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

The Layout Process completed Define the Hazard Analyze the Structure Analyze 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 completed The second step is to analyze the structure for the purpose of where the sprinklers can possibly be located as well as how the system will be supported. For this exercise, it is assumed to be done for us. Step 3 is to analyze the water supply to learn what is available for the system. This also can help, with experience, determine the type of sprinkler that may be most efficient for the job.

Available Water Supply 100 Flow Test Summary Sheet 60 psi residual pressure at 1000 gpm 90 psi static pressure 20 80 120 110 90 70 60 50 40 30 10 200 300 400 500 600 700 800 900 1000 1100 Pressure (psi) Flow (gpm)

The Layout Process Define the Hazard Analyze the Structure Analyze 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 These items can be grouped because the focus of today’s lesson is the hydraulic calculations so many of the other decisions have been made so time can be spent on the calculation portion of the process.

System Details Type of System: Type of Sprinkler: TY3121 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 The cut sheet for the TY3121 is included in the student workbook.

Electronics Factory Plan View 5 ft from north wall and 6 ft from west wall 10 ft 12.5 ft 53 ft Mains are Schedule 10 100 ft Branch lines are Schedule 40 5 ft from south wall and 6.5 ft from east wall 200 ft

Electronics Factory Elevation View Riser is 1 ft away from the east wall. 3-inch Schedule 10 All branch lines are on a 1 ft riser nipple Alarm Check Valve – Viking J-1 18 ft 15 ft 12.5 ft between branch lines N Gate Valve 5 ft The cut sheets for the alarm check valve are in the student workbook. 4-inch PVC (ID – 4.240 inches) 7 ft Long Turn Elbow 42 ft

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

The Layout Process Define the Hazard Analyze the Structure Analyze 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 This is where the calculations actually begin.

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

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

Area Reduction For QR Sprinklers y = % reduction in area x = ceiling height

Design Area (continued) Starting with 1500 ft2 design area Applying the 28% reduction in area: 100% - 28% = 72% 1500 ft2 * 0.72 = 1080 ft2 New design area is 1080 ft2 Density remains at 0.15 gpm/ft2

Design Area (continued) Design area is 1080 ft2 Each sprinkler, spaced 10 ft x 12.5 ft, is covering 125 ft2 How many sprinklers are in the design area? 1080 / 125 = 8.64 sprinklers = 9 sprinklers

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

Branch lines are Schedule 40 Remote Area N 1 5 2 6 3 7 53 ft 4 8 9 Mains are Schedule 10 100 ft 10 ft 12.5 ft Branch lines are Schedule 40 200 ft

Information Needed for Calculations completed 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 The pipe sizes have been done for this exercise. The nodes now need to be located on the drawing

Node Locations - Isometric 1 5 1-inch 2 6 1 ¼-inch 3 7 1 ¼-inch 4 8 9 1 ½-inch 1 ½-inch 1 ft riser nipple 1 ½-inch BL1 BL2 BL3 TOR main and riser 3-inch 15 ft N FF Underground 4-inch PVC CWM

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 The paths are the plan for the calculation.

Calculation Paths 1 2 3 4 5 6 7 8 9 BL1 BL2 BL3 BL2 BL3 Main Path: Auxiliary Paths: BL1 BL2 1 2 3 4 BL3 TOR BL2 5 6 7 8 BL3 9

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 Once the “plan” (from the calculation path) has been laid out, the calculation sheets get filled in with the data.

Starting Sprinkler Values Each sprinkler covers an area of 125 ft2 Each sprinkler is required to deliver a density of 0.15 gpm/ft2 Minimum flow per sprinkler: 125 ft2 x 0.15 gpm/ft2 = 18.8 gpm Minimum pressure for 18.8 gpm:

Data Summary… Water Supply Information: Static Pressure 90 psi @ 0 gpm Residual Pressure 60 psi @ 1000 gpm Hazard Classification OH - 1 System Type Wet Ceiling Height 18 feet Density/Area 0.15 gpm/ft2 / 1500 ft2 Adjusted to 0.15 gpm/ft2 / 1080 ft2 Sprinkler Type Quick Response – Standard Spray K – Factor K =5.6 Area Per Sprinkler 125 ft2 Minimum Sprinkler Flow 18.8 gpm Minimum Pressure 11.3 psi

Start With the Most Remote Sprinkler 1 2 1-inch 5 3 1 ¼-inch 6 4 1 ¼-inch 7 1 ½-inch 1 ½-inch 8 BL1 3-inch 9 BL2 BL3

Starting the Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 1 17.0 5.6 18.8 1-inch 10 C = 120 11.3 -- 2 1.049 0.116 1.2 12.5 Now determine the friction loss

Second Sprinkler Calculation 18.8 gpm @ 11.3 psi 1 2 3 1 ¼-inch 4 1 ¼-inch 1 ½-inch 1 ½-inch

Second Sprinkler - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 1 17.0 5.6 18.8 1-inch 10 C = 120 11.3 -- 0.116 2 1.049 1.2 1 ¼-inch 12.5 ---- 3 1.380 19.8 0.116 38.6 1.2 13.7 What is the friction loss? What is the total flow?

Third Sprinkler Calculation 18.8 gpm @ 11.3 psi 1 2 19.8 gpm @ 12.5 psi 3 4 1 ¼-inch 1 ½-inch 1 ½-inch

Third Sprinkler - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 2 17.0 5.6 19.8 1 ¼-inch 10 C = 120 12.5 ---- 0.116 -- 3 38.6 1.380 1.2 13.7 4 20.7 0.256 59.3 2.6 16.3 What is the friction loss? What is the total flow?

Fourth Sprinkler Calculation 18.8 gpm @ 11.3 psi 1 2 19.8 gpm @ 12.5 psi 3 20.7 gpm @ 13.7 psi 4 1 ½-inch 1 ½-inch

Fourth Sprinkler - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 3 17.0 5.6 20.7 1 ¼-inch 10 C = 120 13.7 -- 0.256 4 59.3 1.380 2.6 1 ½-inch 19 16.3 BL1 16.0 1.610 2T 22.6 16 35 16 ft 0.4 Pe = 1 * 0.433 0.219 81.9 7.7 24.4 What is the total flow? What is the friction loss? Are there any fittings?

Branch Line 1 BL2 is identical to BL1 therefore an equivalent K-factor can be used to balance the flow. 1 2 81.9 gpm @ 24.4 psi 3 4 BL1 3-inch BL2

Branch Line 1 - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 4 17.0 5.6 22.6 1 ½-inch 2T = 16 19 C = 120 16.3 Pe = 1 * 0.433 16 0.219 0.4 BL1 16.0 81.9 1.610 35 7.7 3-inch 13 24.4 -- BL2 3.260 -- KBL1 = 16.58 0.007 81.9 0.1 What is the new flow? 24.5 What is the friction loss? Are there any fittings? What is the total flow?

Branch Line 2 BL2 is identical to BL1 therefore the K-factor from BL1 can be used to calculate the flow. 5 6 81.9 gpm @ 24.4 psi 7 8 BL1 BL2 3-inch BL3

Branch Line 2 - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) BL1 16.0 -- 3-inch 12.5 C = 120 24.4 KBL1 = 16.58 0.007 BL2 81.9 3.260 0.1 24.5 BL3 82.1 0.026 0.3 164.0 24.8 What is the friction loss? What is the total flow? Are there any fittings?

Branch Line 3 81.9 gpm @ 24.4 psi 82.1 gpm @ 24.5 psi BL1 9 BL2 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. BL3

Branch Line 3 - Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) 9 17.0 5.6 18.8 1 ½-inch 19 C = 120 11.3 BL3 16.0 1.610 2T 16 35 0.4 0.014 Pe = 1 * 0.433 16 ft 0.5 12.2 Now determine the friction loss Fittings? Starting flow value is used: q = 0.15 gpm/ft2 * 125 ft2

Branch Line 3 BL1 9 BL2 BL3 81.9 gpm @ 24.4 psi 82.1 gpm @ 24.5 psi

Branch Line 3 – Balancing Point Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) BL2 16.0 82.1 3-inch 12.5 C = 120 24.5 q = 16.58 (24.5)0.5 -- 0.026 BL3 164.0 3.260 0.3 5.38 182.8 24.8 q = 5.38 (24.8)0.5 FF 1.0 Discuss that the TOR may be a good spot to locate a node. However, it is not required based on the current setup of this system. The end value would be the same even if this was broken into 2 or 3 steps. 26.8 190.8 Fittings

BL3 to FF has an elbow, alarm check valve, and gate valve. Branch Line 3 - Fittings 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.

Branch Line 3 – Balancing Point (continued) Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) BL2 16.0 82.1 3-inch 12.5 C = 120 24.5 q = 16.58 (24.5)0.5 -- 0.026 BL3 164.0 3.260 0.3 26.8 ACV=10 182.8 24.8 q = 5.38 (24.8)0.5 El=9.4 20.7 FF 1.0 190.8 GV=1.3 203.5 6.5 0.034 6.9 Pe = 15 * 0.433 38.2 Friction Loss?

Underground pipe is PVC. System Underground main and riser 3-inch N 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. FF 5 ft 12 ft 4-inch PVC (ID – 4.240 inches) Long Turn Elbow CWM 42 ft

Underground – Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) BL3 16.0 26.8 3-inch ACV=10 182.8 C = 120 24.8 q = 5.38 (24.8)0.5 Pe=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 -- 4-inch El 54 C = 150 38.2 GV CWM -4.0 4.240 Equivalent length of fittings?

System Underground - Fittings main and riser 3-inch Fittings per Table 22.4.3.1.1: Long turn elbow – 6 feet Gate valve – 2 feet N FF 5 ft 12 ft 4-inch PVC (ID – 4.240 inches) Long Turn Elbow CWM 42 ft

Underground – Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) BL3 16.0 26.8 3-inch ACV=10 182.8 C = 120 24.8 q = 5.38 (24.8)0.5 Pe=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 -- 4-inch El 54 C = 150 38.2 GV 15.5 CWM -4.0 4.240 69.5 2.2 Pe = 5 * 0.433 0.009 0.6 41.0 Friction loss?

Hose Stream – Calculation Sheet Node 1 Elev. 1 (ft) K-factor Flow – this step (q) Nom. ID Fittings – amount & length L (ft) C-Factor Pressure Total (PT) Notes Node 2 Elev. 2 (ft) Total Flow (Q) Actual ID F (ft) Friction Loss (psi/ft) Elevation (Pe) T (ft) Friction (Pf) FF 1.0 -- 4-inch El 54 C = 150 38.2 Pe = 5 * 0.433 GV 15.5 0.009 2.2 CWM -4.0 190.8 4.240 69.5 0.6 Hose 41.0 250 440.8 System demand is 441 gpm @ 41 psi. Hose stream for OH1?

Flow Test Summary Sheet 60 psi residual pressure at 1000 gpm Water Supply 100 Flow Test Summary Sheet 60 psi residual pressure at 1000 gpm 90 psi static pressure 20 80 120 110 90 70 60 50 40 30 10 200 300 400 500 600 700 800 900 1000 1100 Pressure (psi) System Demand 190.8 gpm @ 41.0 psi With Hose Stream 441 gpm @ 41.0 psi Flow (gpm)

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? This example is summarized to show possible variations.

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

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 23.8 11.2 12.9 14.0 44.27 10 1018 High Temperature 30 28.7 690 14 34.8 800 44.3 1019 CMSA 1950 15 78 50 1345.5 2600 55 25 1265 ESFR 1200 12 99 1366 16.8 35 25.2 178 2456

Reviewing Computer Calculations Checking the input data Checking the output data

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?

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

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

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

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

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.

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