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Contents
  1. 1Purpose and history
  2. 2Design density and design area
  3. 3From the most remote sprinkler to the source
    1. 3.1Worked example
  4. 4Pipe friction and the Hazen–Williams formula
  5. 5Looped and gridded systems
  6. 6Comparing demand with the water supply
  7. 7Use in fire investigation
  8. 8Limits and common misconceptions
  9. 9See also
  10. 10Notes
  11. 11References

Hydraulic calculation is the method used to show that the pipes and water supply of a sprinkler system can deliver the required flow and pressure to every sprinkler expected to open in a fire.[1] The designer starts from a prescribed water density over a design area and determines the flow and pressure needed at the most remote sprinkler.[1][2] The calculation then works back towards the source, adding the pressure lost to pipe friction, fittings and differences in height.[2] The result, called the system demand, is a specific flow at a specific pressure.[3] It must lie below the curve that describes the available water supply.[4]

In the United States, the sprinkler standard NFPA 13 of the National Fire Protection Association (NFPA) requires the empirical Hazen–Williams formula for pipe friction.[3] Hydraulically calculated designs have been in common use since the early 1970s, as an alternative to the older pipe schedule method.[5] Hydraulic calculation is the only part of sprinkler design that is widely calculated. The ability of a water spray to suppress a fire can only be estimated with very rudimentary methods.[6]

Purpose and history

The pipe schedule method uses predefined tables of pipe sizes for light, ordinary and extra hazard occupancies.[5] Each table lists the maximum number of sprinklers that a pipe of a given size may supply, which made design quick but subjective.[5] In the 1970s the capabilities of pipe schedule systems, proven over a hundred years, were codified into curves of water density against design area.[6] This step introduced hydraulic calculation into what had become a cookbook method of design.[6]

Hydraulic calculation brought two practical gains. Designers could take advantage of strong water supplies to build more economical systems, and they could determine the flows and pressures available at points in the piping.[6] The second gain opened the way to special sprinklers, which are approved for specific protection goals.[6] The calculation only covers the flow of water through the pipes. Most protection criteria in NFPA 13 result from experience and judgement, and only some, such as those for early suppression fast response sprinklers, rest on full-scale tests.[6]

Fleming distinguishes four basic types of system: wet pipe, dry pipe, deluge and preaction.[7] A wet pipe sprinkler system, by far the most common type, keeps water under pressure in its pipes.[7] Heat at one sprinkler then makes only that sprinkler discharge water over its own area.[7] A dry pipe system holds air or nitrogen in the pipes until a sprinkler opens, and is used where water in the pipes could freeze.[7]

The first and most critical design step is the occupancy hazard classification.[6] If the hazard is underestimated, a fire may overpower the sprinklers, and no calculation method exists for this step.[6] Sprinkler hazard classification in NFPA 13 distinguishes light hazard, ordinary hazard groups 1 and 2, and extra hazard groups 1 and 2.[8] Offices are examples of light hazard occupancies.[8]

Design density and design area

The hazard class determines the minimum water density and the minimum area of sprinkler operation, which are the inputs for the hydraulic calculation.[8] The density is a water flow per unit floor area, expressed in mm/min or L/(min·m²), or in gallons per minute per square foot (gpm/ft²).[5][1] The designer reads the sprinkler design density and design area from an area and density curve in NFPA 13.[1] Any point on or to the right of the curve is acceptable, so a high density over a small area and a low density over a large area are both possible.[1] In either case the fire is expected to be controlled by the sprinklers within the design area.[1]

Fleming gives an example curve that allows anything from 0.1 gpm/ft² over 5,000 ft² to 0.17 gpm/ft² (6.9 mm/min) over 1,500 ft² (139 m²).[1] A designer might select 0.15 gpm/ft² (6.1 mm/min) over 2,400 ft², which is about 223 m².[1] Water is provided only for the sprinklers within this area, because the others are not expected to open.[1] NFPA 13 divides the design area by the area per sprinkler and rounds up.[1] With sprinklers spaced at 12 ft by 15 ft (3.66 m by 4.57 m), each covers 180 ft², and the design area holds 2,400 / 180 = 13.3, so 14 sprinklers.[1]

Chart of design area against water density: a falling straight line from 465 m2 at 4.1 mm/min to 139 m2 at 6.9 mm/min, region to its right shaded.
Sample area and density curve converted to SI units. Any point on or to the right of the line is acceptable; the marked point is the selection of 6.1 mm/min over 223 m². Based on Fleming (2002), p. 4-74.

The remote design area must be a rectangle with its long side along the branch lines.[1] Its length is 1.2 times the square root of the design area, here 1.2 × 49 = 59 ft, or five sprinklers spaced 12 ft apart along a branch line.[1] The 1999 edition of NFPA 13 modifies the design area for dry systems, for quick response sprinklers under flat smooth ceilings and for rooms in highly compartmented buildings.[1] Since 1985 it has also allowed a design area of four sprinklers for dwelling units with residential sprinklers.[1]

Plan of a sprinkler system: a riser and cross main feed six branch lines; a shaded rectangle of 14 sprinklers lies at the far end.
Plan of a tree sprinkler system with the remote design area of 14 sprinklers at the far end of the branch lines. The calculation starts at the most remote sprinkler and runs back to the supply. Based on Fleming (2002), pp. 4-74 to 4-75.

For light hazard occupancies, the 2002 edition of NFPA 13 requires 0.1 gpm/ft² over 1,500 ft², about 4.1 mm/min over 139 m².[9] With sprinklers spaced 14 ft by 14 ft, each sprinkler covers 196 ft², and the design area needs eight sprinklers.[9] Each of them must then discharge at least 19.6 gpm, about 74 L/min.[9]

From the most remote sprinkler to the source

The calculation starts at the most remote sprinkler in the design area.[1] In a tree system, where each sprinkler is fed from one direction only, this is generally the end sprinkler on the branch line farthest from the riser.[1] The area that this sprinkler protects is taken as S × L. Here S is twice the larger distance to the next sprinkler on the line, and L is twice the larger distance to the next branch line.[2] The minimum flow from the sprinkler is this area multiplied by the required density.[2]

The flow through a sprinkler orifice rises with the square root of the water pressure in the pipe. All other properties of the orifice are lumped into one experimental constant, the K-factor, which testing laboratories determine when they list a sprinkler model.[2] NFPA 13 sets a minimum pressure of 7 psi (0.48 bar) at the end sprinkler, to ensure a proper spray pattern.[2]

Q=KP⇔P=(QK)2

Worked example

Fleming considers a standard spray sprinkler that protects 12 ft by 14 ft, or 15.6 m², at a density of 6.1 mm/min (0.15 gpm/ft²).[2] The minimum flow is 15.6 × 6.11 ≈ 95.4 L/min (25.2 gpm). A standard sprinkler with a nominal orifice of half an inch has a K-factor of 5.6 gpm/psi1/2, which is 80.7 L/(min·bar1/2).[2] The required pressure is then (95.4 / 80.7)² = 1.40 bar, or 20.2 psi, well above the minimum of 0.48 bar.[2]

From the end sprinkler the calculation proceeds backwards towards the source of supply.[2] At each point it adds the pressure needed to overcome friction and differences in height, while keeping the minimum pressure at the remote sprinkler.[2] A water column produces a pressure of 0.433 psi per foot of height, about 0.098 bar per metre.[10] This follows from the specific weight of water, 9.81 kN/m³.[11] Beyond the design area no further sprinkler flow is added, only an allowance for hose streams.[2]

Sprinklers closer to the supply receive a higher pressure, so each discharges more water than the minimum.[2] This effect is called hydraulic increase.[3] It explains why the total demand of a system is not simply the density multiplied by the design area.[3] At the end of the calculation, the demand is known as a specific flow at a specific pressure.[3]

Pipe friction and the Hazen–Williams formula

Water flowing through a pipe loses pressure through friction with the pipe wall and turbulence in the flow.[12] Flows in fire protection piping are almost always turbulent.[12] NFPA 13 prescribes the empirical Hazen–Williams formula to estimate this friction loss.[3] In words, the loss per metre of pipe grows with the flow to the power 1.85 and falls steeply with the internal diameter. A roughness coefficient C represents the condition of the pipe wall; smoother pipes have higher values. In SI units the formula reads:[13]

p=6.05×105Q1.85C1.85D4.87

Symbol Description Unit
p friction loss per metre of pipe bar/m
Q flow rate through the pipe or sprinkler L/min
C Hazen–Williams roughness coefficient none
D internal pipe diameter mm
K K-factor of a sprinkler L/(min·bar1/2)
P water pressure at the sprinkler bar

The US form used in NFPA 13 is p=4.52Q1.85/(C1.85d4.87), with the loss in psi per foot, the flow in gpm and the diameter in inches.[3] NFPA 13 stipulates the C values, which are meant to represent the inside roughness of aged pipe.[3] Steel pipe in dry and preaction systems takes C = 100, and steel pipe in wet and deluge systems takes C = 120.[3] Cement-lined cast or ductile iron takes 140, and copper tube and listed plastic pipe take 150.[3]

Fittings and valves cause extra losses through turbulence.[3] Designers usually express them as equivalent lengths of straight pipe, which are added to the real pipe length.[3] A standard elbow in 1 in. Schedule 40 steel pipe, for example, counts as 2 ft (0.61 m) of pipe.[3] The table of equivalent lengths assumes C = 120 and is multiplied by factors for other pipe materials.[3]

In the continuation of the worked example, the second sprinkler is fed through 3.66 m (12 ft) of 1 in. Schedule 40 steel pipe with a bore of 26.6 mm and C = 120.[3] The formula gives 0.045 bar/m (0.20 psi/ft), so the pipe loses 0.17 bar (2.4 psi).[3] The second sprinkler therefore needs 1.56 bar (22.6 psi) and discharges about 101 L/min (26.6 gpm).[3] If four elbows were added to avoid a column, the pressure would rise to 24.2 psi and the flow to 27.5 gpm.[14] NFPA 13 allows designers to ignore velocity pressure, because the effect is usually small and ignoring it gives a more conservative design.[14]

Side view of a pipe with two sprinklers 3.66 m apart: the end sprinkler gets 1.40 bar and 95 L/min, the second 1.56 bar and 101 L/min.
Two sprinklers at the end of a branch line. Friction in 3.66 m of pipe adds 0.17 bar, so the second sprinkler discharges more than the end sprinkler. Based on Fleming (2002), pp. 4-75 to 4-76.

Looped and gridded systems

In a tree system each sprinkler is fed along a single path, so the calculation can follow the pipes in one direction.[1] In looped and gridded systems, water reaches a point along several paths, and the split of flow between them is unknown at the start.[15] For grids, computers are generally used, because the problem becomes a set of non-linear equations.[15] By hand, the Hardy Cross method assumes a flow distribution and corrects it step by step until the pressure losses along different routes agree.[15]

NFPA 13 requires the calculated pressures to balance within 0.5 psi (0.03 bar) at each hydraulic junction.[15] In practice this means repeating the corrections until the loss from the design area back to the supply is almost the same along any path.[15] Pipes in series or in parallel can first be replaced by one equivalent pipe, which reduces the work.[15]

Comparing demand with the water supply

The strength of a fire water supply is tested by measuring two pressures at a hydrant.[4] The static pressure is read with no extra flow, and the residual pressure is read while a known flow is taken from another hydrant.[4] The fall in pressure between the two readings reflects friction in the supply network, and it is proportional to the flow to the power 1.85.[4] The available pressure at any other flow can therefore be estimated from the two measurements, with the formula below; the formula follows from this proportionality but is not printed in the handbook. On graph paper with a flow scale to the power 1.85, the water supply curve becomes a straight line.[4]

Pavail(Q)=Ps−(Ps−Pr)(QQr)1.85

Symbol Description Unit
Pavail pressure available from the supply at flow Q bar
Ps static pressure (no extra flow) bar
Pr residual pressure during the test flow bar
Qr flow during the test L/min

Fleming describes a test with a static pressure of 6.9 bar (100 psi) and a residual pressure of 5.5 bar (80 psi) at 3,785 L/min (1,000 gpm).[4] Reading a sketch graph, the handbook gives about 90 psi (6.2 bar) available at 1,703 L/min (450 gpm).[4] This article's own calculation with the formula above gives 100 − 20 × 0.451.85 = 95.4 psi, or 6.58 bar, for the same flow. The system demand, including the hose stream allowance, should lie below the supply curve.[4] In a demand calculation the pressure and flow at the end sprinkler are related back to the supply and compared with this curve.[16] If the demand lies above the curve, a pump, a tank or other provisions must be added.[16]

Line chart: available supply pressure falls from 6.9 bar at zero flow to 5.5 bar at 3785 L/min; a margin line lies 0.69 bar lower.
Water supply curve from a flow test with 6.9 bar static and 5.5 bar residual pressure at 3,785 L/min, with a margin of 10 percent of the static pressure. Based on Fleming (2002), p. 4-81, and Hopkins et al. (2005), p. 88.

Good practice leaves a margin of safety between demand and supply.[16] Hopkins et al. describe it as a fixed percentage of the static pressure, such as 10 or 20 percent, or a fixed pressure, such as 10 or 20 psi.[16] The margin covers fluctuations in the supply and later changes to the system, such as extra fittings and pipe added during tenant fit-outs.[16]

Where the supply lacks pressure, a fire pump can make up the difference.[17] In the United States, NFPA 20 requires a horizontal fire pump to deliver at least its rated pressure at its rated flow.[17] At 150 percent of the rated flow it must still deliver at least 65 percent of the rated pressure.[17] By interpolating between such points, Fleming shows that a pump rated at 500 gpm and 100 psi can be expected to give at least 86 psi at 600 gpm.[18] Tanks must hold enough water for the minimum duration in NFPA 13, at the required pressure even when nearly empty.[18]

Use in fire investigation

Hydraulic calculation also works in the opposite direction. In a supply calculation, the water supply and the piping are given, and the flow is followed forward to find the largest density that the system can deliver.[16] The National Institute of Standards and Technology used this approach to assess the sprinkler water supplies of the World Trade Center buildings.[19] Representative floors were analysed with 4, 8 to 15, and 18 to 25 open sprinklers.[19]

For the 99th and 107th floors of the North Tower (WTC 1), up to 24 sprinklers could operate while keeping an average density of at least 0.1 gpm/ft².[20] That is three times the number of sprinklers in the design area.[20] On the 98th floor the system could supply 17 sprinklers at 7 psi and 0.1 gpm/ft², over twice the design area.[20] In the mid-level zone, the flow to one supply riser was limited to about 440 gpm, which limited the systems on the 87th and 32nd floors to 19 and 17 sprinklers.[20]

Limits and common misconceptions

The Hazen–Williams formula has no sound theoretical basis and contains no properties of the fluid.[13] It applies only to water at about 15.6 °C and gives acceptable results only with a careful choice of C, which depends on pipe material, age and water quality.[13] It is not suitable for C values well below 100.[21] In rough pipes the loss rises with the flow to the power 2 rather than 1.85, which makes high flows in old pipes uncertain.[21]

Titus compares both methods for 500 gpm (1,893 L/min) through 100 ft (30.5 m) of 4 in. (102 mm) steel pipe.[21][22] The more theoretical Darcy–Weisbach equation gives 5.98 psi, while Hazen–Williams gives 10.06 psi with C = 100 and 5.40 psi with C = 140.[22] The result therefore depends strongly on the chosen coefficient, whereas the Darcy–Weisbach result is far less sensitive to the assumed roughness.[22]

A common misconception is that a hydraulically adequate system will suppress any fire. The calculation shows only that water is delivered; how well that water suppresses a fire is judged empirically.[6] Another misconception is that every sprinkler opens in a fire. In a wet pipe system, heat opens individual sprinklers, and incident data show that more than half of all fires are controlled or extinguished by one or two sprinklers.[7][23] In Australia, the data sheets of the Australian Building Codes Board base their effectiveness estimates on wet pipe systems designed and installed to the Australian Standard AS 2118 and serviced to AS 1851. The edition that applies is the one prescribed by the National Construction Code at the time of installation.[24]

See also

Notes

  1. ↑ 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 Fleming (2002), p. 4-74.
  2. ↑ 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 2.11 2.12 Fleming (2002), p. 4-75.
  3. ↑ 3.00 3.01 3.02 3.03 3.04 3.05 3.06 3.07 3.08 3.09 3.10 3.11 3.12 3.13 3.14 3.15 3.16 Fleming (2002), p. 4-76.
  4. ↑ 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 Fleming (2002), p. 4-81.
  5. ↑ 5.0 5.1 5.2 5.3 Hopkins et al. (2005), p. 17.
  6. ↑ 6.0 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 Fleming (2002), p. 4-73.
  7. ↑ 7.0 7.1 7.2 7.3 7.4 Fleming (2002), p. 4-72.
  8. ↑ 8.0 8.1 8.2 Hopkins et al. (2005), p. 16.
  9. ↑ 9.0 9.1 9.2 NFPA (2002), as cited in Hopkins et al. (2005), p. 87.
  10. ↑ Fleming (2002), p. 4-78.
  11. ↑ Titus (2002), p. 4-44.
  12. ↑ 12.0 12.1 Titus (2002), p. 4-50.
  13. ↑ 13.0 13.1 13.2 Titus (2002), p. 4-52.
  14. ↑ 14.0 14.1 Fleming (2002), p. 4-77.
  15. ↑ 15.0 15.1 15.2 15.3 15.4 15.5 Fleming (2002), p. 4-79.
  16. ↑ 16.0 16.1 16.2 16.3 16.4 16.5 Hopkins et al. (2005), p. 88.
  17. ↑ 17.0 17.1 17.2 Fleming (2002), p. 4-82.
  18. ↑ 18.0 18.1 Fleming (2002), p. 4-83.
  19. ↑ 19.0 19.1 Hopkins et al. (2005), p. 89.
  20. ↑ 20.0 20.1 20.2 20.3 Hopkins et al. (2005), p. 138.
  21. ↑ 21.0 21.1 21.2 Titus (2002), p. 4-54.
  22. ↑ 22.0 22.1 22.2 Titus (2002), p. 4-56.
  23. ↑ Hopkins et al. (2005), p. 18.
  24. ↑ Australian Building Codes Board (2020a), Data Sheet C2, p. 70.

References

  • Australian Building Codes Board. (2020a). Data sheets: Fire Safety Verification Method (Handbook annex, Print version 1.1).
  • Fleming, R. P. (2002). Automatic sprinkler system calculations. In P. J. DiNenno, D. Drysdale, C. L. Beyler, W. D. Walton, R. L. P. Custer, J. R. Hall Jr., & J. M. Watts Jr. (Eds.), SFPE handbook of fire protection engineering (3rd ed., pp. 4-72–4-87). National Fire Protection Association; Society of Fire Protection Engineers.
  • Hopkins, M., Schoenrock, J., & Budnick, E. (2005). Fire suppression systems (NIST NCSTAR 1-4B). National Institute of Standards and Technology.
  • Titus, J. J. (2002). Hydraulics. In P. J. DiNenno, D. Drysdale, C. L. Beyler, W. D. Walton, R. L. P. Custer, J. R. Hall Jr., & J. M. Watts Jr. (Eds.), SFPE handbook of fire protection engineering (3rd ed., pp. 4-44–4-71). National Fire Protection Association; Society of Fire Protection Engineers.