3-00 Hydraulic Principles

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September 2006Page 1 of 59

HYDRAULICS OF FIRE PROTECTION SYSTEMS

Table of Contents PageTHEORY OF WATER FLOW IN PIPES ........................................................................................................ 4

Bernoullis Theorem and Applications .................................................................................................... 4Hazen-Williams Formula ......................................................................................................................... 6Pressure Loss at Fittings ........................................................................................................................ 7Discharge from Nozzles ......................................................................................................................... 7Discharge Coefficient .............................................................................................................................. 8Theoretical Discharge ........................................................................................................................... 13Nozzle K Factor .................................................................................................................................... 13Combined Sprinkler Discharge and Pipe Flow .................................................................................... 14

TESTING WATER SUPPLIES ..................................................................................................................... 15Purposes ............................................................................................................................................... 15Hazards ................................................................................................................................................. 15Flow and Pressure Measurements ....................................................................................................... 15Test Planning ........................................................................................................................................ 17Private Water Systems ......................................................................................................................... 18Public Water Systems .......................................................................................................................... 19Test Arrangement ................................................................................................................................. 21Test Procedure ..................................................................................................................................... 22Determining Hydraulic Gradients .......................................................................................................... 23Special Water Test Techniques and Considerations ............................................................................ 26

PRESENTING WATER SUPPLY DATA ...................................................................................................... 28Supply Curves (Fig. 16) ........................................................................................................................ 28Friction-Loss Curve for a Single Pipeline (Fig. 17) .............................................................................. 28Curve Combination - Methods and Meaning ....................................................................................... 30Calculation of Yield from a Single Supply ............................................................................................ 31Calculation of Yield from Supplies in Parallel (Fig. 21) ........................................................................ 35Calculation of Yield from Supplies in Series (Fig. 22) ......................................................................... 36

HYDRAULICS OF SPRINKLER SYSTEMS ............................................................................................... 37Sprinkler System Demand Specifications ............................................................................................ 37Use of Velocity Pressure ...................................................................................................................... 39Sprinkler System Flow and Pressure Adjustments .............................................................................. 46Multilevel Systems ................................................................................................................................ 46Looped and Gridded Systems .............................................................................................................. 47Overlay Systems ................................................................................................................................... 57

EXISTING SYSTEMS - CONVERTING TO NEW SPECIFICATIONS ....................................................... 57

List of FiguresFig. 1. Relationships between various hydraulic factors in typical piping ..................................................... 5

Fig. 2. Typical hydrant butts ........................................................................................................................... 8Fig. 3. Water flow devices ............................................................................................................................ 16Fig. 4. Typical pressure gauge locations ..................................................................................................... 16Fig. 5. Pitot tube with gauge and air chamber ............................................................................................ 16Fig. 6. Taking pitot readings at hose nozzle ................................................................................................. 17Fig. 7. Simplified, freehand line diagram of water supplies ......................................................................... 17Fig. 8. Single-source water supply ............................................................................................................... 18Fig. 9. Single source with underground loop ................................................................................................ 18

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Fig. 10. Two sources with underground loop ............................................................................................... 19Fig. 11. Typical public water system near a plant site ................................................................................. 20Fig. 12. Two dead-end pipes from a public water system near a facility site .............................................. 21Fig. 13. Single source with partial loop ........................................................................................................ 22Fig. 14. Relations between water pressure and elevation in pipe having uniform flow .............................. 23

Fig. 15. Pipeline for which hydraulic gradient is computed (upper) and plot of profile andhydraulic gradient to the pipe line (lower) ...................................................................................... 25

Fig. 16. Supply curves .................................................................................................................................. 29Fig. 17. Friction loss curve for a single pipeline .......................................................................................... 30Fig. 18. Combination of supply and friction loss curves to determine yield ................................................ 31Fig. 19. Typical loop in distribution system .................................................................................................. 32Fig. 20. Friction loss and supply curves in parallel pipe (loop) systems ..................................................... 33Fig. 21. Calculation of yield from supplies in parallel. Note: Curves (e), (g), and (h) do not show more

than 150% of rated pump flow because that flow is the maximum guaranteed for allpumps and all installations. ............................................................................................................ 34

Fig. 22. Calculation of yield from supplies in series .................................................................................... 35Fig. 23. Sprinkler layout used in waterflow calculation examples ................................................................ 38Fig. 24. Flow and pressure adjustments for specific array of operating sprinklers ..................................... 47Fig. 25. Demand lines for a multilevel sprinkler system .............................................................................. 48Fig. 26. Typical looped feed sprinkler system ............................................................................................. 49Fig. 27. Graphical determination of flow split in loop .................................................................................. 51

List of TablesTable 1. Hazen-Williams Pipe Coefficients For Underground Use ................................................................. 7Table 2. Hazen-Williams Pipe Coefficients For Sprinkler System Use .......................................................... 7Table 3. Typical Discharge Coefficients .......................................................................................................... 8Table 4. Theoretical Discharge Through Circular Orifices, U.S. Gallons Per Minute (l/min). Computed

For Discharge Coefficient of 1.00 (Seldom Reached In Practice). Reduce the Given DischargeBy Multiplying By A Coefficient Suited to the Particular Opening Used. See Table 3. .................... 9

Table 5. Values of K For Various Discharge Orifices. .................................................................................. 14Table 6. Data For Example of Gradient Test. .............................................................................................. 24Table 7. Pump Discharge Pressures ............................................................................................................ 36

Table 8. Sprinkler System Calculation For Example. ................................................................................... 41Table 8. Sprinkler System Calculation (cont d.) ........................................................................................... 42Table 9. Velocity Pressure Factors ............................................................................................................... 43Table 10. Sprinkler System Calculation For Example .................................................................................. 44Table 10. Sprinkler System Calculation For Example (cont d) ..................................................................... 45Table 11. Comparison Of Results For Different Calculation Methods .......................................................... 46Table 12. Sprinkler System Calculation For Example 1 ............................................................................... 52Table 12. Sprinkler System Calculation For Example 1. {cont d) ................................................................ 53Table 13. Sprinkler System Calculation For Example 2 .............................................................................. 55Table 13. Sprinkler System Calculation For Example 2 (cont d) ................................................................. 56Table 14. Comparison of Results From Examples 1 and 2 ......................................................................... 57Table 15. Conversion Factors and Formulas ............................................................................................... 57

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Fig. 9. Single source with underground loop ................................................................................................ 17Fig. 10. Two sources with underground loop ............................................................................................... 18Fig. 11. Typical public water system near a plant site ................................................................................. 19Fig. 12. Two dead-end pipes from a public water system near a facility site .............................................. 20Fig. 13. Single source with partial loop ........................................................................................................ 21

Fig. 14. Relations between water pressure and elevation in pipe having uniform flow .............................. 22Fig. 15. Pipeline for which hydraulic gradient is computed (upper) and plot of profile and

hydraulic gradient to the pipe line (lower) ...................................................................................... 24Fig. C-1. 1 1 212 in. diameter drain ............................................................................................................... 28Fig. C-2. 2 in. diameter drain ....................................................................................................................... 29Fig. C-3. 3 in. diameter drain. ...................................................................................................................... 30Fig. 16. Supply curves .................................................................................................................................. 31Fig. 17. Friction loss curve for a single pipeline .......................................................................................... 32Fig. 18. Combination of supply and friction loss curves to determine yield ................................................ 33Fig. 19. Typical loop in distribution system .................................................................................................. 34Fig. 20. Friction loss and supply curves in parallel pipe (loop) systems ..................................................... 35Fig. 21. Calculation of yield from supplies in parallel. Note: Curves (e), (g), and (h) do not show more

than 150% of rated pump flow because that flow is the maximum guaranteed for allpumps and all installations. ............................................................................................................ 36

Fig. 22. Calculation of yield from supplies in series .................................................................................... 37Fig. 23. Sprinkler layout used in waterflow calculation examples ................................................................ 40Fig. 24. Flow and pressure adjustments for specific array of operating sprinklers ..................................... 49Fig. 25. Demand lines for a multilevel sprinkler system .............................................................................. 50Fig. 26. Typical looped feed sprinkler system ............................................................................................. 51Fig. 27. Graphical determination of flow split in loop .................................................................................. 53

List of TablesTable 1. Hazen-Williams Pipe Coefficients For Underground Use ................................................................. 6Table 2. Hazen-Williams Pipe Coefficients For Sprinkler System Use .......................................................... 6Table 3. Typical Discharge Coefficients .......................................................................................................... 7Table 4. Theoretical Discharge Through Circular Orifices, U.S. Gallons Per Minute (l/min). Computed

For Discharge Coefficient of 1.00 (Seldom Reached In Practice). Reduce the Given

Discharge By Multiplying By A Coefficient Suited to the Particular Opening Used. See Table 3. ... 8Table 5. Values of K For Various Discharge Orifices. .................................................................................. 13Table 6. Data For Example of Gradient Test. .............................................................................................. 23Table C-1. Equivalent Lengths of Fittings (ft) .............................................................................................. 27Table 7. Pump Discharge Pressures ............................................................................................................ 39Table 8. Sprinkler System Calculation For Example. ................................................................................... 43Table 8. Sprinkler System Calculation (contd.) ........................................................................................... 44Table 9. Velocity Pressure Factors ............................................................................................................... 45Table 10. Sprinkler System Calculation For Example .................................................................................. 46Table 10. Sprinkler System Calculation For Example (contd) ..................................................................... 47Table 11. Comparison Of Results For Different Calculation Methods .......................................................... 48Table 12. Sprinkler System Calculation For Example 1 ............................................................................... 54Table 12. Sprinkler System Calculation For Example 1. {contd) ................................................................ 55Table 13. Sprinkler System Calculation For Example 2 .............................................................................. 57

Table 13. Sprinkler System Calculation For Example 2 (contd) ................................................................. 58Table 14. Comparison of Results From Examples 1 and 2 ......................................................................... 59Table 15. Conversion Factors and Formulas ............................................................................................... 59

Hydraulics of Fire Protection Systems 3-0FM Global Property Loss Prevention Data Sheets Page 3



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Automatic sprinklers with adequate water supplies are the best protection for most fire hazards. To obtainthe most economical system for any location, fire protection systems must be designed individually. To designthese systems, knowledge of water flow in pipes is needed. This data sheet covers basic theory of waterflow in pipes and applications, including water supply testing and sprinkler system hydraulics.

September 2006. Minor editorial changes were made for this revision.May 2006. Data Sheet 3-0 was updated to remove any reference to Data Sheet 2-76, which was madeobsolete.

THEORY OF WATER FLOW IN PIPES

Bernoullis Theorem and Applications

Problems of water flow in pipes are usually solved by procedures based on Bernoulli s theorem, which statesthat in steady flow, without friction, the sum of velocity head, pressure head, and elevation head is constantfor a particle throughout its course. This theorem can be expressed by the following equation:

(v2 /2g) + (p/w) + z = H (equation 1)

where v = velocity, ft/s (m/s)g = acceleration of gravity = 32.2 ft/s 2 (9.81 m/s 2 )p = pressure, lb/ft 2 (Pa)w = weight of water per unit volume = 62.4 lb/ft 3 (9810 N/m 3 )z = elevation head (or potential head), distance above an assumed datum, ft (m)H = total head of water, ft (m)

In equation 1 the terms v 2 /2g and p/w express velocity head and pressure head, respectively.

Velocity head = h v = v2 /2g or v = 2gh v (equation 2)

Pressure head = h p = p/w or p = wh p (equation 3)

Because total head (H) is constant, a change in velocity results in the conversion of velocity head to pressurehead, or vice versa.

For a pipeline flowing full between points a and b, Bernoulli s theorem can be modified to include friction,as follows:

(va 2 /2g) + (p a /w) + za = (vb 2 /2g) + (p b /w) + zb +hab (equation 4)

where h ab is the total dynamic head lost between points a and b.

Figure 1 illustrates the relationships between various factors in typical piping. Heads are indicated by heightsto which water rises in the vertical tubing. Velocity magnitudes and directions are indicated by arrows. Ateach location, B, C, D, E, and F, pressure head, h pF , hpB , etc., is a measure of the potential energy of waterin the pipe; velocity head, h vF , etc., is a measure of the kinetic energy of the water; and the sum, h pF + hvF ,etc., total head, is a measure of the total energy of the water.

Flow rate in a pipeline or discharge through an orifice can be expressed in terms of velocity and cross-sectional area of the stream:

Q = Av or v = Q/A (equation 5)

where Q = flow rate, ft 3 /s (m 3 /s)A = cross-sectional stream area, ft 2 (m 2 )v = average water velocity, ft/s (m/s)

In Figure 1(a), Q F = QB because no water flows out of the pipe (once stabilized in the vertical tubing).Therefore, from equation 5, A F vF = AB vB , and since the pipe does not change size, A F = AB , so that v F =vB . This means kinetic energy is the same at F and B, and h vF = hvB . Pressure head decreases linearlybetween F and B. The difference, h pF - hpB , is friction loss or the total dynamic head lost between F and B.(See equation 4.) The loss rate per unit length is (h pF - hpB )/L, where L is length between F and B.

In Figure 1(b), water is flowing out of the pipe at B, and the fitting between D and E reduces cross-sectionarea. Flow at F is the sum of flows at B and C, Q F = Q B + Q C . Since Q B > 0, Q C must be less than Q F , (AC vC< AF vF ). Cross section areas at F and C are the same, so v C < vF . This means kinetic energy at C is less




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than at F or h vC < hvF . Pressure head, h pC , is shown slightly less than h pF because of friction loss straightthrough the fitting. Flow out of the pipe at B has substantially increased velocity, because all potential energyis given up to friction or converted to kinetic energy. Thus, total head at F, h pF + h vF , equals h FB + h vB , frictionloss between F and B plus velocity head at B. From experimental observation, friction loss between F andB for discharging water is approximately equal to velocity head at F, h FB hvF .

Between C and D, Q C = QD , vC = vD , hvC = hvD , and h pC drops slightly because of friction, to h pD .

Fig. 1. Relationships between various hydraulic factors in typical piping




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Table 1. Hazen-Williams Pipe Coefficients For Underground Use

Kind of Pipe Water Corrosiveness Cast-iron, unlined Mild Moderate Severe 10 years old 105 90 7515 years old 100 75 6020 years old 95 65 5530 years old 85 55 4550 years old 75 50 40Cast-iron, unlined, new 120Cast-iron, cement-lined 140Cast-iron, bitumastic-enamel-lined 140Cement-asbestos 140Approved plastic-lined steel 145*Approved glass-fiber-reinforced plastic 160*Approved PVC 150**If using the Hazen-Williams formula, use these coefficients. If using nominal pipe size and Table 1 from Data Sheet 2-89, usethe following artificial C coefficients to compensate for differences in internal diameter between these pipes and Schedule 40steel pipe:FM Approved plastic-lined steel 165; FM Approved glass-fiber-reinforced plastic 180; FM Approved PVC 165.

Fire protection pipe, normally without flow, is expected to deteriorate less rapidly than pipe subject tocontinuous or intermittent draft. Cement-lined, bitumastic-enamel-lined, or cement asbestos pipe is relativelysmooth with little or no reduction in carrying capacity over a reasonable period of time.

Unlined steel pipe exposed to water for various time periods has a wide range of C values. To allow for quickearly deterioration under field conditions, therefore, use approximate C values from Table 2 for sprinklersystems.

Table 2. Hazen-Williams Pipe Coefficients For Sprinkler System Use

C Value* System Type Kind of Pipe Wet Dry

Hydraulic Design Steel 120 100Copper ** 150 150

Pipe ScheduleSteel 120 100

Copper ** 150 150* Use indicated values unless measurements indicate other values should be used.

**Use indicated values with actual pipe size in Hazen-Williams formula. Use C = 130 with nominal pipe size tocompensate for differences in internal diameter between copper tube and Schedule 40 steel pipe.

Pressure Loss at Fittings

In typical public water or fire protection pipe, losses arising from changes in flow direction and changes invelocity are called loss due to fittings. Such losses are proportional to velocity head (v 2 /2g) and can beequated to losses in a length of straight pipe. Data Sheet 2-89 gives equivalent pipe lengths for various fittings.In underground pipe computations, fitting loss is generally a small portion of total loss and is not normallyconsidered. In sprinkler system computations, fitting loss is normally considered.

Experiments have shown that pressure loss due to fittings occurs primarily downstream from the fittings andinvolves cavitation and turbulence. Thus, tees have larger losses than short radius elbows, which have largerlosses than long radius elbows, etc.

Discharge from Nozzles

From equations 2 and 5, it follows that

Q = A2gh v (equation 7)

In a jet discharging from a nozzle, all the available head (velocity head plus pressure head) is converted tovelocity head, which can be measured with a Pitot gauge. When velocity head and nozzle diameter areknown, theoretical discharge can be computed from equation 7.




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Discharge Coefficient

Actual discharge is less than computed discharge because velocity is not uniform over the cross section ofthe stream. Therefore, a correction factor, or discharge coefficient, is needed in order to use equation 7.Thus

Q = cA2gh v (equation 8)

Table 3 gives typical discharge coefficients.

Table 3. Typical Discharge Coefficients

Type of Orifice c Hydrant butt, smooth, well-rounded outlet* 0.80Hydrant butt, square outlet* 0.70Hydrant butt, inset outlet* 0.60Smooth Underwriter nozzles 0.97Deluge nozzles 0.99Open pipe, smooth and well-rounded, at least 10 diameters long 0.90Open pipe, burred opening or less than 10 diameters long 0.80

*See Fig 2.

Discharge coefficients used in calculations of flow from hydrants depend upon the character of the hydrant

outlet. Figure 2 shows three general types of hydrant outlets.

Fig. 2. Typical hydrant butts




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Table 4. Theoretical Discharge Through Circular Orifices, U.S. Gallons Per Minute (l/min). Computed For Discharge Coefficient of 1.00 (Seldom Reached In Practice). Reduce the Given Discharge By Multiplying By A Coefficient Suited to

the Particular Opening Used. See Table 3.




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*For pressure in bars, multiply by 0.01.




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Hydrants with orifice plates or valved butts are not in any of the above categories. Use hoses with nozzlesto determine flow.

In convenient fire protection units, equation 8 may be rewritten

Q = acd 2 P v (equation 9)

where Q = rate of flow, gpm (l/min)a = constant = 29.8 (0.666) (0.0666)

(Use 0.666 for P v in bars and 0.0666 for P v in kPa.)c = discharge coefficientd = orifice diameter, in. (mm)P v = velocity pressure or Pitot pressure, psi (bar) (kPa)

Theoretical Discharge

Table 4 gives theoretical discharges in gpm (l/min) for various orifice diameters and velocity (Pitot) pressureswhen c = 1.00. To obtain actual discharge, find theoretical discharge corresponding to known P v and d, andmultiply by appropriate coefficient from Table 3.

Nozzle K Factor

Equation 8 can also be written in the form

Q = KP v (equation 10)

where K = acd 2

K is theoretically constant for a given orifice. If nozzle K is known and P v is measured, discharge in gpm(l/min) is found from equation 10. Nominal values of K for various nozzles are given in Table 5.

Sprinkler K is calculated from a variation of equation 10:

Q = KP n (equation 11)

where P n is measured upstream from the sprinkler in a pipe or reservoir.

When P n is measured in a reservoir, P v is essentially zero, so we use P n = P t and write:

Q = KP t (equation12)




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Although technically this method applies only when the sprinklers are at exactly the same elevation, it issatisfactory in single-story sections of sprinkler systems unless the ceiling is steeply pitched.

TESTING WATER SUPPLIES

The performance of public and private water supplies for fire protection service is best determined fromperiodic flow tests. Such tests are particularly important for the design of new fire protection systems. Siteselection and timing must be carefully chosen to reveal conditions that might exist at the time of the fire.Testing procedures are based on principles of water flow in pipes and discharge from orifices. Available fireflow must be determined even if known to be less than demand or even if water supply is less than reliable.

Purposes

There are several reasons for flow testing water supplies:

1. Fire flow available to a given area can be compared with demands specified in existing standards.

2. Fire pumps and drivers are subject to numerous malfunctions that can be discovered and correctedpromptly.

3. Closed valves and other obstructions become apparent.

4. Water supply deterioration can be detected.

5. The ability of plant personnel to use the fire protection system can be assessed.

6. Weakened underground pipe sometimes ruptures under fire pump or fire department pressures. It is farbetter that this condition be discovered during a test than during a fire.

Hazards

Ensure qualified personnel direct flow tests because personal injury and property damage can result fromimproperly conducted tests. Some of the more common problems that arise are:

1. Improperly secured nozzles can work loose and injure personnel.

2. Excessive flows may draw vacuums in high buildings and in hilly country, possibly contaminating watersupplies, damaging boilers, and interrupting industrial processes. Avoid reducing pressure in public mainsbelow 20 psi (1.38 bar) (138 kPa).

3. Local flooding at low spots, such as truck docks, basements, pits, tunnels, etc., is possible.

4. Water damage can occur to storage if it is located outdoors or in low-lying buildings where water canflow through doorways.

5. High voltage electrical equipment can be shorted by solid hose streams, with danger to personnel. Notethat spray from fire protection equipment avoids this problem.

6. Rapid valve operation results in water hammer, which can rupture piping and damage sprinkler equipment.

Flow and Pressure Measurements

Flow tests are usually made by discharging water through one or more hydrant or nozzle outlets. At the sametime readings are taken with pressure gauges at sprinkler risers, nonflowing hydrants, or other direct

connections to supply pipes (Figs. 3 and 4).Flow rate is determined by measuring the velocity pressure of each stream with a Pitot gauge (Fig. 5). ThePitot orifice is held firmly in the center of the jet. The knife edge is held directly against the nozzle end withthe blade at right angles to the nozzle axis. Velocity pressure is read directly on the gauge in psi (bar) (kPa)(Fig. 6). The corresponding discharge in gpm (l/min) is then obtained from tables or equations 9 or 10. Notedischarge coefficients and exact diameter for all nozzles and use reliable pressure gauges.

Pitot readings under 10 psi (0.69bar) (69 kPa) are questionable. Avoid them whenever practical by reducing the size or number of orifices.

Pressures observed at sprinkler risers, nonflowing hydrants, or other direct connections to the supply pipeare called static if there is no test flow. Pressures observed with test streams flowing are called residual. Whentesting, always read both static and residual pressures. Handle gauges carefully and calibrate them yearly




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Test Planning

Flow tests must indicate the ability of water supplies to meet water demands in the demand area. Beforetesting, plan a course of action, indicating where to make flows, where to read pressures, and which valveoperating sequence to use.

Occasionally, physical conditions require changes in plans. Such changes are usually simple. For example,a given hydrant can be dangerous to use without hoses because of land slope or high voltage electricalequipment, while an adjacent hydrant is relatively safe.

When accuracy is needed, plan to use hose streams rather than hydrant butts. Accuracy might be needed

if costly recommendations are being considered.Check for local regulations that may affect planning. For example, if may be necessary to notify the publicwater service before flow testing private systems.

For public water systems and large plants, condense water supply information onto a single small sheet ofpaper in diagram form (Fig. 7). Show all sources and connections as well as major loops and valves.

Fig. 6. Taking pitot readings at hose nozzle

Fig. 7. Simplified, freehand line diagram of water supplies




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When planning flow tests and when testing, keep the following fundamentals in mind:

1. Make a flow equal to or greater than demand near the demand area.

2. Keep fire protection in service at all times.

3. Flow each major loop and each source or connection (except pressure tanks).4. Reopen all valves and flow all sources together.

Private Water Systems

Water supplies range from simple to complex, and planning varies accordingly. Figure 8 illustrates a simplesupply. Flow the hydrant and read static and residual pressures at the riser below the alarm check valve.Flow sufficient water to meet the combined sprinkler and hose stream demand.

If there is a single source with an underground loop (Fig. 9), plan as follows:

1. Select hydrant and riser nearer the demand area and read static pressure at the riser.

2. Close valve A and flow hydrant (south loop), reading residual pressure at the riser.

3. Open valve A.

4. Close valve B and flow hydrant (north loop), reading residual pressure at the riser.

5. Open valve B and flow hydrant (both loops) with flow rate sufficient to prove that water supply satisfieswater demand. Read residual pressure at the riser.

6. Read static pressure.

If there are two sources or two connections and an underground loop (Fig. 10), plan as follows:

1. Select hydrant and riser nearest the demand area.

2. Close valves B and C and read static pressure at the riser.

3. Flow hydrant (gravity tank, south loop), reading residual pressure at the riser.

4. Open valve B.

Fig. 8. Single-source water supply

Fig. 9. Single source with underground loop




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5. Close valve A and flow hydrant (gravity tank, north loop), reading residual pressure at the riser.

6. Read static pressure at the riser and open valve C.

7. Close valve B and flow hydrant (public), reading static and residual pressures at the riser.

8. Open valves A and B and flow hydrant (both sources, both loops) with flow rate sufficient to prove thatwater supply satisfies water demand. Read residual pressure at the riser.

9. Read static pressure at the riser.

Tests of more complex supplies are planned similarly, keeping in mind the fundamentals.

Not only is it necessary to select a flow to prove the ability of the supply to meet the demand, but total flowmust draw from all sources. A test that does not draw from all sources of a multi-source system, whenfeasible, lacks completeness. For example, typical gravity tank pressure is 50 psi (3.45 bar) (345 kPa) andtypical public pressure is 80 psi (5.52 bar) (552 kPa). If the test result with all valves open is reported as80 psi (5.52 bar) (552 kPa) static, 1000 gpm (3790 l/min), 51 psi (3.52 bar) (352 kPa) residual, the water supplyis seriously misrepresented because there is no flow from the gravity tank. Gravity tanks can increase flowdramatically with small pressure losses, depending on relative locations of gravity tank, public connections,and flowing hydrant.

Public Water Systems

To test public water supplies at proposed facility sites, visit the local water service to diagram appropriatesections of the water map and to arrange for tests. Previous test results are sometimes on file and may beuseful for planning tests. Plan a course of action before testing.

Generally, flow from a single hydrant indicates whether or not the supply is adequate for the demand. Incase of high demands, flow two or more hydrants. Select hydrants to give significant results. Ensure pressurereadings and flows are as close to plant connection points as practical. Figure 11 shows a typical situation.

1. Example (Fig. 11)

If all pipes are the same size, flow either hydrant, with pressure readings at the other, since most of the frictionloss occurs in the part of pipe A that is out of the picture.

2 . Example (Fig. 11)

If pipes B, C, and D are small and A is large, test flow at either hydrant has little effect on pressure in A.With test flow at either hydrant, the nonflowing hydrant is about halfway from A to the flowing hydrant via pipethat is carrying part of the test flow. Therefore, pressure at the nonflowing hydrant reflects about half thechange near the flowing hydrant. Such pressure readings may be acceptable, but it is preferable to flow thewest hydrant and read pressures north of it.


If pipe C is very small, pressure at the nonflowing hydrant is much closer to pressure in pipeA than to pressureat the flowing hydrant. Read pressures at nearby hose bibs or flow the west hydrant and read pressuresnorth of it. Make sure hose bibs are supplied from the pipe under test and that pressure is unaffected bydomestic draft or pressure regulator during measurements.

Fig. 10. Two sources with underground loop




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If pipes C and D are small and A and B are large, test flow at west hydrant with pressure readings at easthydrant represents available supply at the west side of the facility site adequately.

Test flow at east hydrant with pressure readings at a nearby hose bib represents available supply at theeast side of the facility site. Make sure hose bibs are supplied from the pipe under test and that pressureis unaffected by domestic draft or pressure regulator during measurements.

Test flow at the east hydrant with pressure readings at the west hydrant is not representative, because testflow must pass through small pipes C and D where pressure loss per unit length is relatively high. Westhydrant pressures are thus much higher than pressures near the flowing hydrant.

Similarly, high flows available at the west hydrant do not represent water supply at the east hydrant.


Another situation that occurs sufficiently often to merit special attention is illustrated in Figure 12. Since thetwo dead-end pipes are part of a single water system, flow at A reduces flow available at B and vice-versa.Therefore, to determine total flow available to the facility site, flow hydrants A and B simultaneously and readresidual pressures as near as possible to both A and B.

Then, in case the facility is supplied from only one pipe, flow A and B separately and read residual pressuresupstream from each flowing hydrant.

Fig. 11. Typical public water system near a plant site




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Public water systems often require special treatment. Some items needing attention follow:

1. If systems are direct or intermittent pumping, determine facilities in operation during tests.

2. If there is elevated storage, test with pumps shut down, if possible.

3. If special fire pressure is available, test if possible, and determine (a) notification required, (b) time requiredto obtain service, and (c) facilities in operation.

4. Check for pressure regulating valves.

5. Check for variations in operating procedures from day to night or from summer to winter.

Test Arrangement

When using hose streams lash nozzles to substantial supports such as posts, trees, or deeply driven stakes.Do not hold nozzles by hand.

When hoses are not available and waterflow from hydrants or pump headers must be redirected to avoiddamage, use elbows and piping with adapters from hose thread to pipe thread. Ensure pipe downstream fromelbows is at least ten diameters long to obtain smooth flow. If Pitot pressures are erratic or higher at orificeedges than at centers, pipes must be longer than ten diameters to smooth flows.

Read static and residual pressures at locations where they respond to the normal pressure existing in pipesthrough which test water is flowing. Ensure there is no check valve or shut valve between the column offlowing water and the point where residual pressure is read. This precaution reduces the probability of readingtrapped overnight high pressures or of finding no pressure drop when flowing from a typical public watersupply.

Use FM Global valve supervision procedures.

Make sure tanks are full and that valves controlling tank fill lines are shut before starting test.

Ensure nonflowing butts on the flowing hydrant are tightly capped.

Ensure all valves closed during water tests are shut tightly.

Rsidual pressure readings measure normal pressure where the column of still water starting in the gaugemeets the column of flowing water, adjusted for elevation difference. This location should be near the demandarea. Figure 13 shows how test flow and residual pressure can be taken near the demand area withmeaningless results because the junction of the two water columns is much further away. Read residualpressure at B with test flow at C so the column of still water starting in the gauge meets the column of flowingwater a short distance from the demand area. If residual pressure is read at A with test flow at C, the columnof still water starting in the gauge meets the column of flowing water at D, indicating available water supplyat the city connection and defeating the purpose of making a flow test near the point of highest demand. Ifboth sprinkler systems are hydraulically designed, read residual pressures at both A and B to obtain the bestwater supply information for calculating both systems.

Fig. 12. Two dead-end pipes from a public water system near a facility site




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Test Procedure

The last step before testing is to make sure any central station, fire service, or proprietary supervisory serviceshave been notified that a flow test is being conducted. Then proceed with the test as follows:

1. Close valves to isolate the desired source and underground loop. Use FM Global valve supervisionprocedures.

2. Read static pressure and bleed off any trapped high pressure.

3. Inspect area for possible damage from discharge.

4. If flowing an open hydrant butt, determine orifice diameter and coefficient. (See Table 3.)

5. Open hydrant valve wide or open header valves to create desired flow.

6. Make sure discharge is doing no damage.

7. Allow ample time for residual pressure to stabilize. It often takes several minutes for pressure to stopchanging.

8. Read pressures at flowing orifices with Pitot gauge and note residual pressure.

9. Close hydrant or header valves slowly.10. Read static pressure.

11. Convert Pitot readings to flows using Table 4 or equation 9 and plot, as illustrated in Figure 16.

12. Compare with anticipated results to determine if test setup is correct or if problems exist.

13. Correct problems and replan test sequence if necessary.

14. Proceed with test sequence and record results. When testing pumps, achieve desired flow by addingor removing nozzle tips to change orifice size, changing number of streams, or throttling valves. To alter flowfrom hydrants, change orifice size or number of streams. Do not throttle hydrant valves.

Fig. 13. Single source with partial loop




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For most tests, readings at two flow rates with all valves open are recommended to provide a check on results,and if public water is a supply, to help assess the effect of normal public (non-test) flow or automatic startingpumps.

15. Compare with previous tests or with anticipated results to ensure that no further testing is needed.

Dismantle test setup.16. Make sure all valves are locked or sealed open and follow FM Global valve supervision procedures.

17. Refill gravity and pump suction tanks and make sure fire pumps are left on automatic.

18. Complete FM Global valve supervision procedures.

19. Notify supervisory service that test is complete and ascertain that valve tamper signals and flow signalswere transmitted properly.

Determining Hydraulic Gradients

If fire flow tests show that yield from a water supply is unaccountably low, identify the causes by determiningthe hydraulic gradient. Low yields can be caused by partly closed valves, excessively tuberculated pipes,or obstructions. Pipe size errors on plans can make yields appear abnormal.

Hydraulic gradients isolate locations of abnormally high losses, which cause low yields. The hydraulic gradientis a profile of residual pressure. It assumes a uniform rate of flow and simultaneous residual pressurereadings at various points along the pipe. In practice, moving one gauge progressively from test point to testpoint while test flow is maintained yields maximum consistency in pressure readings.

If there are tees, crosses, bends, valves, or meters in pipe being tested, obtain the pressure loss in thesedevices from Data Sheet 2-89 and deduct it from the observed drop in pressure before calculating coefficientsfor given pipe. Relations between pressure and elevation in pipe having uniform flow are illustrated in Figure14.

Fig. 14. Relations between water pressure and elevation in pipe having uniform flow




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Explanation of Table 6:

Columns 1 to 5, inclusive, contain data from Figure 15.

Column 6. Static - residual pressure = total loss

Column 7. Difference in total loss, station to station

Fig. 15. Pipeline for which hydraulic gradient is computed (upper) and plot of profile and hydraulic gradient to the pipe line (lower)




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Column 8. Loss per foot (per meter):

10/910 = 0.011 psi (A to B) (0.69/277 = 0.00249 bar [0.249 kPa])

20/500 = 0.04 psi (B to C) (1.38/152 = 0.00909 bar [0.909 kPa])

Column 9. f c =Column 8

Loss/ft (m) (from Data Sheet 2-89, Table 9)

For 8-in. (203-mm) pipe (0.011 psi/ft [2.49 mbar/m] [0.249 kPa/m])

0.011 psi per ft/0.011 psi per ft = 1(A to B)

(2.49 mbar per m/2.49 mbar per m = 1)

0.040 psi per ft/0.011 psi per ft = 3.6 (B to C)

(9.09 mbar per m/2.49 mbar per m = 3.6)

For 6-in. (152-mm) pipe (0.046 psi/ft [10.37 mbar/m] [1.037 kPa/m])

0.046 psi per ft/0.046 psi per ft = 1(C to D)

(10.37 mbar per m/10.37 mbar per m = 1)

Column 10. Values of C from Data Sheet 2-89, Table 14, corresponding to f c

Column 11. Assume zero elevation at station with the maximum static pressure, in this case Station D, wherestatic = 85 psi (5.86 bar) (586 kPa). Elevations of other stations are found by subtracting respective staticreadings from 85 psi (5.86 bar) (586 kPa). Thus:

A = 85 - 81 = 4 psi (5.86 - 5.59 = 0.28 bar [28 kPa]).

B = 85 - 75 = 10 psi (5.86 - 5.17 = 0.69 bar [69kPa]) etc.

Column 12. Gradient elevations are obtained by adding residual pressure to gauge elevations. Thus:

A = 71 + 4 = 75 psi (4.90 + 0.27 = 5.17 bar [517 kPa])

B = 55 + 10 = 65 psi (3.79 + 0.69 = 4.48 bar [448 kPa]) etc.

Differences in gradient elevation equal losses between corresponding stations.

Recommendations:

1. Clean and line 8-in. (203-mm) pipe BC, restoring C to 100.

2. Replace 6-in. (152-mm) pipe CD with 8-in. (203-mm) cement-lined pipe.

Show expected gradient with recommendations completed (Fig. 15).

Loss B to C: 1 500 0.011 = 5.5 psi (1 152 0.00249 = 0.38 bar [38 kPa])

Gradient elevation at C = 65 - 5.5 = 59.5 psi (4.48 - 0.38 = 4.1 bar [410 kPa])

Loss C to D: 0.537 435 0.011 = 2.8 psi (0.537 133 0.00249 = 0.18 bar [18 kPa])

Gradient elevation at D = 59.5 - 2.8 = 56.7 psi (4.1 - 0.18 = 3.92 bar [392 kPa])

Gradient elevation at E = 56.7 - 10 = 46.7 psi (3.92 - 0.69 = 3.23 bar [323 kPa])

Special Water Test Techniques and Considerations

Water testing is often less straightforward than has been indicated thus far. This section presents someproblems that arise occasionally, and their possible solutions.

Problem: There are no pressure gauge mounting fittings below check valves on sprinkler risers.

Solution: Mount the gauge on the fitting above the check valve and open the 2-in. (51-mm) drain just enoughto allow the indicated pressure to drop and stabilize. At flows low enough to avoid adding significantly tothe test flow, essentially no pressure is lost in the check valve.




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Problem: There are no hydrants, sprinkler risers, or hose bibs within 1000 ft (305 m) of the flowing hydrant.

Solution: Mount the gauge on the second butt of the flowing hydrant. Indicated pressure differs from nearbyresidual pressures by friction loss in hydrant and supply pipe. Losses in hydrants vary considerably, butaverage from 2 psi (0.14 bar) (14 kPa) at 500 gpm (1890 l/min) to 5 psi (0.34 bar) (34 kPa) at 1000 gpm

(3790 l/min).




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Problem: Pitot tube is unavailable.

Solution: Mount the gauge on the second butt of the flowing hydrant. Indicated pressure equals velocitypressure at flowing butt plus friction loss in butt minus velocity pressure in hydrant barrel. Friction loss in buttapproximately equals velocity pressure in hydrant barrel, so indicated pressure approximately equals Pitot

pressure.If a hose and nozzle are used on the hydrant, subtract friction loss in the hose from indicated pressure toobtain the approximate Pitot reading.

PRESENTING WATER SUPPLY DATA

Investigation of water supplies for fire protection may be expedited by using graphs that provide means forthe effective presentation of hydraulic conditions and indicate what changes or improvements are needed.

There are two kinds of flow curves supply curves and loss curves - that appear as curved lines on linearcross-section paper. The same curves appear as straight lines on semi-exponential paper having linearsubdivisions for pressure on the vertical axis and exponential (N 1.85 ) subdivisions for flow on the horizontalaxis. This semi-exponential (N 1.85 ) paper fits the Hazen-Williams pressure-drop-flow relation, given inequation 6. Numerical values on the vertical and/or horizontal scales of N 1.85 paper may be multiplied ordivided by any constant as may be necessary to fit problems onto sheets or to magnify portions of curvesto facilitate study.

N1.85 paper is particularly useful for the following purposes:

1. Showing supply curves (Fig. 16)

2. Calculating friction-loss curve in a single pipeline (Fig. 17)

3. Combining friction-loss and supply curves to determine yield (Fig. 18)

4. Solving parallel-pipe (loop) systems (Fig. 19 and 20)

5. Calculating yield from combinations of supplies (Fig. 21 and 22)

Supply Curves (Fig. 16)

Supply curves show the rate of flow available at any residual pressure. They are drawn by plotting flow againstresidual pressure obtained from water test data or from data based on characteristics of specific supplies.Typical supply curves shown in Figure 16 are obtained by plotting the supply available at discharge outletsor at typical points in a public water system.

When plotting a fire-flow curve with supply from a public water system, three or more test points, includingstatic, are desirable to check on accuracy and to avoid errors that might be introduced by extrapolation beyondlimits of the test data. In addition, extra pumps may come on line automatically with higher flows. If threetest points are used, a curve instead of the usual straight line may appear because the water system, inaddition to test flow, is also supplying a normal demand of unknown quantity.

True static pressure may be higher than observed static pressure because of friction loss to the test locationdeveloped by normal flow. Under such circumstances, the curve drawn between points representingobserved conditions will give a reasonably accurate indication of the available fire flow.

Friction-Loss Curve for a Single Pipeline (Fig. 17)

To plot a friction loss curve for a pipeline of uniform size throughout, determine the losses through the pipeat two different rates of flow; plot these points and connect them with a straight line. Because the loss iszero with no flow, it is customary to use the 0-0 coordinate as one of the two points. Select the other pointnear the top or right-hand edge of the graph.

Example: Plot the friction loss curve for 1,120 ft (342 m) of 8-in. (203-mm) pipe having Hazen-Williams Cof 90. Friction loss per foot (per meter) from equation 6 is p = cQ 1.85 /C1.85 d4.87 . Data Sheet 2-89 tabulatesp and multipliers, f c = (C 2 /C1 )

1.85 (Table 14, DS 2-89), to convert from C 1 to C 2 . Total friction loss is

P f = pfc L (equation 16)

where L is pipe length.




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1. Assume 1,500 gpm (5680 l/min) flow.

2. From Data Sheet 2-89, Tables 14 and 9,fc = 1.22 for converting C 1 = 100 to C 2 = 90p = 0.024 psi/ft (5.4 mbar/m) (0.54 kPa/m) for 8-in. (203-mm) pipe, for assumed flow, and for C = 100.

3. From equation 16, P f = 0.024 1.22 1120 = 32.8 psi Use 33 psi. (0.0054 1.22 342 = 2.25 bar [225kPa]. Use 2.25 bar [225 kPa].)

4. Plot a point at 33 psi (2.25 bar) (225 kPa) and 1500 gpm (5680 l/min). Connect it to the 0-0 point with astraight line (Fig. 17). This is the required friction loss curve. It illustrates loss in the given pipe at any flow,and conversely, flow through the pipe at any loss.

Fig. 16. Supply curves




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To plot the friction loss curve for a single pipeline made up of various lengths and diameters of pipe, proceedas follows:

1. Determine friction loss for each element of the pipeline as described above, using the same rate of flowfor each element.

2. Add friction losses for individual elements.

3. Plot the sum of friction losses at the selected flow.4. Draw a line connecting this point with the 0-0 point. This is the friction loss curve for the entire pipeline.

Curve Combination - Methods and Meaning

The following statements summarize relationships between operations with curves and actual situations:

1. Two curves can be combined vertically or horizontally.

2. Vertical combination is accomplished by selecting a flow and adding or subtracting pressures.

3. Horizontal combination is accomplished by selecting a pressure and adding or subtracting flows.

4. Vertical combination or flow selection implies a series (end-to-end) arrangement of pipes, sources, andelevations, since inflow equals outflow.

a. Elevation is normally interpreted as fixed pressure (horizontal line) in series with the source (or in serieswith the pipe system) and is independent of flow. It is subtracted from the source curve if higher thanthe source, and added if lower.

b. Pump curves are not straight lines. Generally, combined source curves are not straight lines. Pipecurves are straight lines. Elevation curves combined with pipe curves are straight lines.

5. Horizontal combination or pressure selection implies a parallel (side-by-side) arrangement of pipes,sources, and hoses, since pressure difference across the system is the same by any route.

a. Hose demand is normally interpreted as a fixed flow (vertical line) in parallel with the source andindependent of pressure. Hose demand is then subtracted from the source curve at several selectedpressures. Combined source curve and hose demand is not a straight line.

Fig. 17. Friction loss curve for a single pipeline




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Calculation of Yield from a Single Supply

Combination of friction loss and supply curves in pipe systems (Fig. 18).

In Figure 18, supply and friction loss curves are combined vertically to indicate yield at a point some distancefrom the source, as at B in the example shown.

A test made at point A on the public main indicates that flow of 950 gpm (3600 l/min) reduces static pressureof 100 psi (6.89 bar) (689 kPa) to residual pressure of 90 psi (6.21 bar) (621 kPa). Static pressure at PointB is 80 psi (5.52 bar) (552 kPa). The difference in static pressure is due to elevation. A is connected to Bby 767 ft (234 m) of 6-in. (152-mm) pipe with a known coefficient of 110.

Curve (a) is the supply curve of water available at point A, developed from test data as follows:

Fig. 18. Combination of supply and friction loss curves to determine yield




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1. Plot one point at 100 psi (6.89 bar) (689 kPa) and no flow.

2. Plot one point at 950 gpm (3600 l/min) and 90 psi (6.21 bar) (621 kPa).

3. Connect the two points with a straight line, curve (a).

Curve (b) is the friction loss curve in the pipe between A and B, which is developed as for Figure 17. Flowof 1,500 gpm (5680 l/min) is assumed, for which friction loss from equation 16 using Data Sheet 2-89, Tables9 and 14, is

P f = 0.098 0.838 767 = 63 psi(0.0222 0.838 234 = 4.35 bar [435 kPa])

Curve (c) is the supply curve at B before correction for elevation. It is obtained by subtracting friction losscurve (b) from supply curve (a) as follows:

1. At no flow, static pressure is 100 psi (6.89 bar) (689 kPa) and there is no friction loss. Plot a point at thispressure and flow.

2. At 1,500 gpm (5860 l/min), residual pressure is 77 psi (5.31 bar) (531 kPa) and friction loss is 63 psi (4.35bar) (435 kPa), a difference of 14 psi (0.97 bar) (97 kPa). Plot a point at 1,500 gpm (5680 l/min) and 14psi (0.97 bar) (97 kPa).

3. Connect these two points by a straight line, curve (c). Where, as in discharge from a centrifugal pump,supply curve (a) plots as a curve rather than a straight line, friction loss at a number of flow rates must bededucted from the supply curve to develop a true curve representation of yield at B.

4. Draw curve (d) parallel to, and 20 psi (1.38 bar) (138 kPa) below, curve (c). Curve (d) represents supply,

or net yield, at B and is obtained by correcting supply curve (c) for elevation difference.Combination of friction loss and supply curves in parallel-pipe (loop) systems (Fig. 19 and 20)

Solutions for parallel-pipe (loop) problems are based on two principles:

1. Pressure difference between any two points of a network is the same by any route through the network.

2. Flow toward any point equals flow away from the point.

Example:

In the example of Figures 19 and 20, it is not possible to make a flow test at the proposed facility site, buta test at point A is available. The problem involves the water supply to expect at point C when pipe B isinstalled. The test at point A indicates that flow of 1,400 gpm (5,300 l/min) reduces pressure from 70 psi

Fig. 19. Typical loop in distribution system




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(4.83 bar) (483 kPa) static to 60 psi (4.14 bar) (414 kPa) residual. Static pressure at point C is 95 psi (6.55bar) (655 kPa). The difference in static pressures is due to elevation difference.

To obtain the water supply curve at point C:

1. Plot supply curve (d) (water available at point A).

2. Plot friction loss curves (a) and (b) for each side of the loop. This applies the second principle above.Flow into one end of a pipe equals flow out the other end, regardless of size changes or bends.

a. Assume 1,500 gpm (5680 l/min) in each side.

b. From Data Sheet 2-89, Table 3, P B = 0.059 psi/ft (13.3 mbar/m) (1.33 kPa/m).

c. Since Table 3 in Data Sheet 2-89 is generated for C = 140, and proposed pipe has C = 140, f c = 1and equation 16 yields:

P fB = 0.059 1320 = 77.9 psi. (0.0133 403 = 5.36 bar [536 kPa]).

d. From Data Sheet 2-89, Table 9,

P C = 0.098 psi/ft (22.2 mbar/m) (2.22 kPa/m), also f c = 0.838 for converting C 1 = 100 to C 2 = 110.

Fig. 20. Friction loss and supply curves in parallel pipe (loop) systems




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e. Equation 16 yields:

P fC = 0.098 0.838 1260 = 103.5 psi (0.0222 0.838 384 = 7.15 bar [715 kPa])

3. Plot friction loss curve (c) for the entire loop by adding the flow available through each side at someconvenient pressure. This applies both principles. Friction loss is the same across each side. Flow at A equalsthe sum of flows at B and D, and equals flow at C.

a. To add curves, it is easiest to select the highest round-number pressure at which the sum of flows ison the graph paper. At 25 psi (1.72 bar) (172 kPa), 710 gpm (2690 l/min) is available through pipe C,and 810 gpm (3060 l/min) through pipe B.

b. Add these quantities and plot a point at 25 psi (1.72 bar) (172 kPa) and 1520 gpm (5750 dm).

c. Connect this point to the 0-0 coordinate by a straight line. This is the friction loss curve (c) for the entireloop.

Fig. 21. Calculation of yield from supplies in parallel. Note: Curves (e), (g), and (h) do not show more than 150% of rated pump flow because that flow is the maximum guaranteed for all pumps and all installations.




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4. Plot supply curve (e) at point C, before elevation correction, by subtracting friction loss through the loop(curve (c)) from the available supply at point A (curve (d)).

At 1,400 gpm (5300 l/min),60 - 21 = 39 psi(4.14 - 1.45 = 2.69 bar [269 kPa].)

At 0 gpm (0 l/min),70 - 0 = 70 psi(4.83 - 0 = 4.83 bar [483 kPa].)

5. Plot supply curve (f) 25 psi (1.72 bar) (172 kPa) above curve (e). This corrects for elevation as indicatedby static pressures. Curve (f) is the expected water supply (yield) at point C.

Calculation of Yield from Supplies in Parallel (Fig. 21)

In the example shown, public water tests at point A indicate that flow of 600 gpm (2270 l/min) reduces staticpressure from 105 psi (7.24 bar) (724 kPa) to 75 psi (5.17 bar) (517 kPa) residual. The centrifugal fire pumpat B is rated 750 gpm (2840 l/min) at 100 psi (6.89 bar) (6.89 kPa). The connection between A and B is 1,640ft (500 m) of 8-in. (203-mm) cast iron pipe with C of 70. Points B and C are 35 ft (10.7 m) higher than pointA.

Fig. 22. Calculation of yield from supplies in series




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Supplies from fire pump and public water connection are additive at any point C between the two sourcesafter proper allowances are made for elevation difference and friction loss between individual sources andpoint of demand.

To plot water supply curve (yield) at point C from fire pump and public water connection in simultaneous

operation:1. Plot supply curve (a), public water available at point A, by the method given for Figure 16.

2. Plot friction loss curve (b), for the pipe between points A and C, by the method given for Figure 17. Usingequation 16

P f = 0.011 1.94 655 = 14 psi(0.0025 1.94 200 = 0.97 bar [97 kPa]) at 1,000 gpm (3790 l/min)

3. Plot supply curve (c) at point C from public water at point A before elevation correction by subtractingfriction loss indicated by curve (b) from curve (a). This is vertical combination of a series arrangement.

4. Correct curve (c) for elevation difference by drawing curve (d) 15 psi (1.03 bar) (103 kPa) below curve(c) to represent net yield at point C from point A (vertical combination).

5. Plot pump characteristic curve (e).

6. Plot friction loss curve (f) for pipe between points B and C. Using equation 16

P f = 0.011 1.94 985 = 21 psi(0.0025 1.94 300 = 1.45 bar [145 kPa]) at 1,000 gpm (3790 l/min).

7. Plot supply curve (g), showing yield at point C from the fire pump at point B, by subtracting friction lossindicated by curve (f) from curve (e) (vertical combination).

8. Add curves (d) and (g) to form curve (h) (horizontal combination of a parallel arrangement). This is therequired water supply curve (yield) at point C from fire pump and public water connection in simultaneousoperation. To illustrate: At 80 psi (5.52 bar) (552 kPa), curve (d) (300 gpm [1140 l/min]) + curve (g) (825 gpm[3120 l/min]) = 1,125 gpm (4260 l/min).

Calculation of Yield from Supplies in Series (Fig. 22)

In the example shown, a test of the public water connection at point A indicates that flow of 1,000 gpm (3790l/min) reduces pressure from 34 psi (2.34 bar) (234 kPa) static to 30 psi (2.07 bar) (207 kPa) residual. A1,000-gpm (3790-l/min), 60 psi (4.14 bar) (414 kPa) fire pump takes suction from the public main through211 ft (65 m) of 8-in. (203-mm) pipe with C = 80 and discharges through 163 ft (50 m) of 6-in. (152-mm) pipewith C = 120 to point C. Points A, B, and C are at the same elevation.

To plot the water supply curve (yield) at point C from the booster pump:

1. Plot water supply curve (a) at point A available from public water.

2. Plot friction loss curve (b) for the pump suction pipe between points A and B. Using equation 16,

P f = 0.024 1.52 211 = 7.7 psi(0.0054 1.52 65 = 0.53 bar [53 kPa]) at 1,500 gpm (5680 l/min)

3. Plot suction-supply curve (c) to the pump at point B by subtracting curve (b) from curve (a) (vertical

combination).4. Plot characteristic curve (d) of the pump at zero suction pressure.

5. Plot pump discharge curve (e) by adding curves (c) and (d) at several flows as given in Table 7 (verticalcombination). Note this means booster pump suction pressure equals public residual pressure at all flowrates.

Table 7. Pump Discharge Pressures

Flow, gpm (l/min)Net head of pump, psi (bar)(kPa) Curve (d) +

Residual suction pressure, psi (bar) (kPa)

Curve (c) =

Pump discharge pressure, psi (bar) (kPa)

Curve (e)0 72 35 107




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(0) (4.96)(496) (2.41)(241) (7.38)(738)500 69 33 102

(1890) (4.76)(476) (2.28)(228) (7.03)(703)750 65 30 95

(2840) (4.48)(448) (2.07)(207) (6.55)(655)1000 60 26 86

(3790) (4.14)(414) (1.79)(179) (5.93)(593)1250 52 22 74

(4730) (3.59)(359) (1.52)(152) (5.1 )(510)1500 39 17 56

(5680) (2.69)(269) (1.17)(117) (3.86)(386)

6. Plot friction loss curve (f) for pipe between points B and C. Using equation 16

P f = 0.046 0.72 163 = 5.4 psi(0.0104 0.72 50 = 0.37 bar [37 kPa]) at 1,000 gpm (3790 l/min).

7. Plot supply curve (g), showing yield at point C, by subtracting curve (f) from curve (e) (vertical combination).This is the yield available at point C from the booster pump.

HYDRAULICS OF SPRINKLER SYSTEMS

Pressures and discharges of sprinklers in simultaneous operation are not readily calculated by any exactformula. The most practical methods for calculating water flow in sprinkler piping are to use manual calculationmethods for either tree-type of looped sprinkler systems as outlined in the data sheet, or a computer softwareprogram. Computations by the methods explained in this section agree closely with test work.

Sprinkler calculations are based on hydraulic principles of discharge from circular orifices and water flow inpipes. Discharge depends on total pressure and friction loss. Figure 1 and equations 12 to 15, withassociated text, illustrate relationships between flows, discharges, and pressures.

While sprinkler system computations can be made precisely, final results can vary 5% or more from actualflows measured experimentally. The reasons are:

1. Pipe coefficients vary widely with size and age and from location to location.

2. FM Approved sprinkler K values vary 5%.3. Water supplies can easily be different from when tested.

In this data sheet, assumed values are nominal and indicated accuracy in examples is for illustration andcomparison only.

Sprinkler System Demand Specifications

Demand specifications for sprinkler protection often are expressed as rate of water application (density) fora given area (area of demand). Sprinkler systems so designed must be able to discharge a specified densityover the hydraulically most remote area (defined in Data Sheet 2-8N, Chapter 7) as well as over every othersimilar area covered by the systems. Typically, the remote area is calculated and the pattern of pipe sizesestablished therein is extended to the rest of the system.




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Calculation Techniques

Sprinkler system calculations rely heavily on equations 12 and 13. Since square root is convenient, equation12, Q = K P t is often used solely. However, N

1.85 paper is also convenient, so entire sprinkler systems areoften represented by equation 13, Q = KP t

0.54 . (Note that Q = Kp 0.54 approximates p = (Q/K) 1.85 closely.)

For most sprinkler systems and demand areas, use Q = Kp0.54

or N1.85

paper, but for individual sprinklersor small groups of sprinklers, use Q = Kp 0.5 .

Calculate branch line pressure and discharge with some or all sprinklers operating by assuming a pressureat the end sprinkler in the demand area and working back to the cross main. Pressure losses in sprinklerfittings need not be considered. Proceed as follows:

1. At the end sprinkler (pointA1 of Figure 23), assume a total pressure and calculate discharge using equation12 and sprinkler K from Table 5. Accuracy to 1 gpm at 0.1 psi is ample (approximately 4 l/min and 7 mbar[0.7 kPa]).

2. At the second sprinkler, point A2, add friction loss in the end pipe to total pressure at the end sprinklerto obtain total pressure.

3. Calculate discharge from the second sprinkler, using total pressure at point A2, equation 12, and sprinklerK from Table 5.

4. Calculate discharges and pressures for third and subsequent sprinklers on the same branch line similarly.

To obtain total pressure at the cross main for a single branch line, add the following:

1. Total pressure at sprinkler nearest to cross main

2. Friction loss in pipe between nearest sprinkler and cross main

3. Pressure loss in elbow (or tee) at top of nipple connecting branch line to cross main (obtained from DataSheet 2-89)

4. Friction loss in nipple if over 6 in. (152 mm) long (ignore elevation)

5. Pressure loss in tee at bottom of nipple

Using flow and pressure thus obtained, calculate K from equation 12 to represent the entire branch line.

Fig. 23. Sprinkler layout used in waterflow calculation examples




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If the first cross main nipple supplies two branch lines, the top fitting is a tee instead of an elbow (Point 7of Fig. 23), and nipple and branch line sizes may all be different.

If two branch lines connected to a single tee are operating, calculate discharges (Q i and Q j) and cross mainpressures (P ti and P tj) from each as shown in the example. If P ti does not equal P tj , they must be equalized

because two different pressures at a single point are impossible. (This condition is equivalent to therequirement that pressure difference between any two points of a network must be the same by any routethrough the network.) To equalize, select the lower (say P tj) and adjust it up to P ti.

Using equation 14, adjust Q j to Q jP ti /P tj . For single branch lines or rows, equation 14 is more realistic thanequation 15. It is unnecessary to recalculate other flows and pressures. Using flow and pressure thusobtained, calculate K from Equation (12) to represent both branch lines.

When several branch lines are operating and all are alike, K is the same for each. Thus at point 9, branchrow K is the same as at point 8 and total pressure is the sum of friction loss in the end length of cross mainand total pressure at point 8. Flow to branch row CD is then P t (at point 9) times K. Calculate remainingrows similarly.

Often branch rows alternate in length, e.g., 15 sprinklers, 16 sprinklers, 15 sprinklers, etc. Find one K for15-sprinkler rows and another for 16-sprinkler rows and proceed as above using appropriate K for each row.

When flow and pressure are obtained for any point supplying all operating sprinklers, add elevation pressureand friction loss in pipe and fittings back to a selected part of the supply system (riser, fire pump, etc.).

The following is common nomenclature on calculation forms:

P t = P n + P v = total pressureP n = P t - P v = normal (pipe, gauge, or net) pressureP v = velocity pressureP f = friction loss (Equation 16)P e = elevation pressureQ = flow rateC = Hazen-Williams coefficientK = 5.6 (81 bar)(8.1 kPa) for 1 2-in. (13 mm) sprinklers (Table 5).

Set up calculations in tabular form starting with no. 1 sprinkler on branch line A. The note column is used

for preliminary calculations.Example (Fig. 23)

Given: A wet pipe sprinkler system with indicated pipe sizes and sprinklers 8 ft (2.44 m) apart, on branchlines 10 ft (3.05 m) apart.

Required: Find pressure needed at point 18 to maintain flow of 960 gpm (3630 l/min) with indicated 40sprinklers operating.

Result: 70.3 psi (4.85 bar) (485 kPa) at top of riser. (The computation is given in Table 8.)

In reporting results of calculations, round off to the nearest 10 gpm and 1 psi. (If calculations are in metricunits, equivalent values are about 40 l/min and 70 mbar or 7 kPa).

Use of Velocity Pressure

This section is included to acquaint the reader with a technique commonly used in the industry. There isno need to use velocity pressure in sprinkler system calculations. In order to compensate for different methodsof calibrating sprinkler K, velocity pressure is sometimes deducted from total pressure to obtain normalpressure, and sprinkler discharge is calculated from equation 11. This technique provides more nearlyaccurate results when sprinkler K is determined for normal pressure in the pipe or fitting, or for total pressurein the sprinkler. No compensation is needed and velocity pressure should not be used when sprinkler K isdetermined for total pressure in the pipe as it is in Table 5.

When actual internal pipe diameter is known, equation 17, a variation of equation 9, gives values of velocitypressure.

P v = Q2 /cd 4 (17)

where:




40/59

P v = velocity pressure, psi (bar) (kPa)Q = Flow, gpm (l/min)c = constant = 888 (0.443 bar) (0.00443 kPa)

(Use 0.443 for P v in bars and 0.00443 for P v in kPa).d = actual internal pipe diameter, in. (mm).

Table 9 gives values of cd 4 for schedule 40 steel pipe. Use equation 17 and actual internal pipe diameterfor all other pipe.

Calculation Techniques With Velocity Pressure

When velocity pressure is considered, it is done as follows:

1. At the end sprinkler (pointA1 of Figure 23) assume a total pressure and calculate discharge using equation12. Velocity pressure is not considered at the end sprinkler.

2. At the second sprinkler (point A2 of Figure 23) add friction loss in the end pipe to total pressure at theend sprinkler to obtain total pressure at Point A2.




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T a b l e 8 . S p r i n k l e r S y s t e m C a l c u l a t i o n ( c o n t d

. )

N o w c a r r y f l o w d o w n t o c r o s s m a i n

, c o m p u t e b r a n c h l i n e K

, a n d f i n i s h c a l c u l a t i o n

.

S p k r

. o r

N o z z l e

I d e n t . &

L o c a t i o n

F l o w

, g p m

( l / m i n )

P i p e S i z e i n

.

( m m )

P i p e

F i t t i n g s

a n d

D e v i c e s

E q u i v

. P i p e

L e n g t h

, f t ( m )

F r i c t i o n

L o s s

, p s i / f t

( m b a r / m )

( k P a / m )

P r e s s u r e S u m m a r y , p s i

( b a r ) ( k P a )

N o r m

a l

P r e s s u r e

,

p s i ( b a r )

( k P a )

N o t e s

7

q Q 2 1 4 ( 8 1 0 )

2 ( 5 1 )

T

L g t h 1 ( 0

. 3 0 5 )

F i t 1 0 ( 3

. 0 5 )

T o t 1 1 ( 3

. 3 5 )

0 . 3 8 4

( 8 7 ) ( 8

. 7 )

P t

2 2 . 1

( 1 . 5

2 ) ( 1 5 2 )

P e

P f

4 . 2 ( 0

. 2 9 ) ( 2 9 )

P t

P v

P n

R i s e r n i p p l e =

1 f t ( 0

. 3 0 5 m ) . I n c l u d e l e n g t h a n d

i g n o r e e l e v a t i o n . I n c l u d e t e e a t b o t t o m o f n i p p l e

.

8

q Q 2 1 4 ( 8 1 0 )

2 ( 5 1 )

L g t h 1 0 ( 3

. 0 5 )

F i t

T o t 1 0 ( 3

. 0 5 )

0 . 3 8 4

( 8 7 ) ( 8

. 7 )

P t

2 6 . 3

( 1 . 8

1 ) ( 1 8 1 )

P e

P f

3 . 8 ( 0

. 2 6 ) ( 2 6 )

P t

P v

P n

K =

2 1 4 /

2 6

. 3 =

4 1 . 7

( 8 1 0 /

1 . 8 1 =

6 0 2 [ 6 0 . 2 ] )

( U s e 6 0 2 f o r p i n b a r s a n d 6 0 . 2

f o r p i n k P a . )

9

q 2 2 9 ( 8 6 7 )

Q 4 4 3 ( 1 6 8 0 )

2 1 2 ( 6

3 )

L g t h 1 0 ( 3

. 0 5 )

F i t

T o t 1 0 ( 3

. 0 5 )

0 . 6 2 2

( 1 4 1 ) ( 1 4 . 1 )

P t

3 0 . 1

( 2 . 0

7 ) ( 2 0 7 )

P e

P f

3-00 Hydraulic Principles

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