viking-water spray system design

DESIGN OF WATER SPRAY SYSTEMS FORFIRE PROTECTION

I GENERAL CONSIDERATIONS

CAUTIONThese design procedures are provided to you only for the generalguidance of our water spray system designers. They containbroad outlines of the types of considerations, which enter into thedesign of water spray systems. Because of the many differenttypes of equipment and applications encountered in practice, wecannot provide general system design, which will satisfy all thevarying needs of our customers. Therefore you must rely on theexpertise of your system designer and encourage him to use allavailable information from the owner, insurance authorities andlocal governmental units. Viking does not warrant or guaranteethat following these procedures will result in satisfactory systemdesign for your particular project.

Deluge systems are used in applications where there is anextreme hazard. The principal purpose of a deluge system maybe either extinguishment or cooling or both. In the case ofvolatile liquids, particularly outdoors, the principal purpose ofthe deluge system is to cool the equipment so that it is notsignificantly damaged by the fire. In many cases it ispermissible to lose the material being stored or processed aslong as the equipment used to store or process it can be putback into service in a relatively short time when the fire isextinguished. Actual extinguishment is often accomplished byeither letting the fire burn itself out or by other means.

Cooling must accomplish two important functions. It mustkeep the structure below the temperature at which deformationor physical weakening occurs and it must limit the heat inputinto the liquid or gas contained in the equipment in order to keep the pressure in the equipment within tolerable limits.

Equipment can be expected to be exposed to heat frombasically two sources, either a spill fire where burning liquid orgas is present over, under or around the equipment or anexposure fire adjacent to the equipment but not actuallyinvolving it.

A vessel filled with liquid has a considerable capacity toabsorb heat without raising its temperature significantly. Theliquid acts as a heat sink and because of the good heat transfercharacteristics between the vessel shell and the liquid, the shell is kept relatively cool. It should be noted, however, that theinside of a tank is seldom clean and deposits tend to build uponthe lower walls and the bottom. These deposits act as aninsulator and can greatly reduce the heat transfer to the liquid.When the vessel is empty or filled with a gas, its heat absorbingcapacities are dramatically reduced, therefore, an empty vessel is much more susceptible to fire damage than a full one.Similarly, if a vessel is partially full the top portion is much moresusceptible to damage than the bottom portion.

When a gas or volatile liquid is heated, the result is arapidly increasing pressure that must either be reduced orrelieved or there is danger that the equipment will rupture. Ifportions of the equipment have been either weakened oroverstressed by the fire, these provide specific points wherefailure can more easily occur. Venting is usually employed tokeep the pressure within safe limits, however, if the fire issevere enough, the venting capacity may not be sufficient tokeep the pressure at safe levels. Cooling of the equipment maywell insure that the venting capacity is adequate.

Structural supports not encased in concrete or sufficientlyfireproofed must be protected since their failure could wellresult in either damage to or collapse of the equipment.

Adequate cooling must be provided to protect theequipment from excessive heat transfer applied by either directimpingement of flame or radiation.

In most cases it is necessary to protect all portions of theequipment. (Often very high tanks such as cracking towers areonly protected up to about 30 feet or 9M). Ideally one wouldwish to apply a uniform density of water over every bit of theexposed surface. This is an impossibility since spray patterns of sprinklers rarely, if ever, exactly match the contours of theequipment. The effects of wind and gravity further compoundthe problem. In an outdoor installation, because of windproblems, spray nozzles should be placed within two feet (.6M)of the surface being protected unless this surface is relativelyshielded from the wind.

Normally water applied at the top of the equipment will rundown the vertical sides, however, the actual amount of rundown is not entirely predictable due to wind conditions, the fact thatthe equipment may not be exactly vertical, protrusions from theequipment which may “roof off” certain areas and perhaps most important, loss of water through vaporization due to intenseheat of the fire. In addition, the equipment may be or will shortlybecome dirty This dirt is usually water repellent to some degreeand this will cause the rundown to “channel” rather than beevenly distributed. If the equipment is elevated from the ground, there will be virtually no rundown on the bottom. Therefore,rundown must be considered but cannot be taken for granted.

The designer attempting to fit a spray nozzle layout to apiece of equipment will encounter theoretical dry spots. Heshould give consideration to whether or not these will turn out to be actual dry spots and attempt to adjust his designaccordingly. A dry spot, particularly on the top seam of a tank,can be quite dangerous. If the dry spot is exposed to severeradiation, a carbon deposit can form. This deposit greatlyincreases the radiation heat transfer potential and, furthermore, because of the nature of the carbon, the spot becomes highlywater repellent. Theoretical dry spots are much more importantin the top portion of a tank where there is little chance ofrundown than they are on the tank shell or the bottom wherethere usually would be some liquid.

In designing a deluge system primarily for coolingpurposes, it is suggested that the designer follow these steps:

1. Determine equipment dimensions and water densityrequirements

2. Establish individual design areas and total design area.3. Determine individual and total design area water demand.4. Determine water supply conditions and probable available

pressure for each design area.5. Determine number and type of nozzles required to give

adequate coverage and water delivery (by trial and error).

In selecting a nozzle arrangement to give suitable coverage,the designer has a choice of a number of coverage angles anda number of capacities. It will be necessary to select a properarrangement to provide adequate density and coverage. Oftenthere will be a number of choices.

DESIGN DATA WATER SPRAY DESIGN

Form No. F_042602

April 18, 2003 Design 208 a

Revised page replaces pages 209a-o dated July 15,2002. (Renumbered page.)

II SPECIFIC DESIGN PROCEDURES

A. DESIGN PROCEDURE WATER SPRAY PROTECTIONFOR HORIZ0NTAL TANKS

CAUTIONThese design procedures are provided to you only for the generalguidance of your water spray system designers. They containbroad outlines of the types of considerations, which enter into thedesign of water spray systems. Because of the many differenttypes of equipment and applications encountered in practice, wecannot provide general system design, which will satisfy all thevarying needs of our customers. Therefore, you must rely on theexpertise of your system designer and encourage him to use allavailable information from the owner, insurance authorities andlocal governmental units. Viking does not warrant or guaranteethat following these procedures will result in satisfactory systemdesign for your particular project.

Make a detailed survey of the tank and its surroundings. Note diameter, length, height of ends, locations anddimensions of any irregularities which may impede distributionof water such as ladders, manholes, piping connections and the like. Note size, location and material of supporting structure fortank. Note proximity of other hazard producing equipment. Note the presence of dikes, walls and barriers. Determine contentsof tank and establish water density requirements for protection.

Tank Diameter = DTank End Height = hTank Shell Length = LDensity Required = d

1. Find the area of the cylindrical shell (As)See Figure 1. Shell area = Ag =TT DL

2. Find the water required for the shell (Qs)Water required is the shell area times the required density

Qs = As(d)3. Find the area of the ends (Ae)

See Figure 1. Use the proper formula for the tank end. Ifboth ends are not the same, use appropriate formula foreach end. Neglect the presence of tank fittings.

Flat End

Spherical End

Hemispherical End

4. Find the water required for theends (Qe)Water required is the end area times the required densityfor each end.

Qe1 = Ae1 (d)Qe2 = Ae2 (d)

5. Determine the water required for appendages (Qa)If there are appendages to the tank, which increase thebasic area of coverage, estimate the water required toprotect these appendages using the required shell density.

Qa1=Aa1(d)Qa =Qa1+Qa2 . . . .

6. Determine the design area for legs or supports (Ae)Concrete or fireproof steel supports generally require noprotection. A short unprotected steel support up to aboutone foot C.3M) generally requires no protection as long asit receives some rundown or splash. Longer unprotectedsupports require spray protection over the design areaindicated in Figure 2.

7. Determine water required for each leg or support (Ql)The water demand equals the design area times therequired density. (Table A)

Ql1=Al1(d)8. Determine the total amount of water required (Qtot)

The total amount of water required is the sum of the waterfor the shell, both ends, the appendages and the supports

Qtot = Qs+Qe1+Qe2+Qa+Ql1+Ql2 + . . . .9. Assume a nozzle pressure

Based upon known water supply conditions and/orassumed pump conditions and supply piping conditions,calculate pressure available to the tank. Note that in smalldiameter tanks the static pressure differential between thehighest and lowest nozzle will not be very great.

10. Select the most probable nozzle arrangementA wide range of nozzle capacities and angles are available. The objective is to obtain adequate coverage and flow witha minimum number of nozzles with no overspray.Overspray will occur on the shell when a large angle nozzle is used on a small diameter tank. For nozzles located 2 feet (.6M) from the tank surface, overspray will occur on tankssmaller than those listed:Nozzle Angle Used Minimum Tank Diameter

ft. M30 1.4 .4360 4.0 1.290 10.0 3.0120 26.0 8.0140 62.0 19.4

Nozzles can be used on tanks smaller than listed bymounting closer to tank. Eliminate those nozzles which are suitable only for larger tanks.Lay out the tank to scale and place nozzles in accordancewith the following:a. Nozzle direction

Nozzles protecting the tank shell should be pointeddirectly at the tank. Nozzles protecting the tank endshould be pointed directly at the ends except in the case of flat surfaces. In the case of vertical flat surfaces, thenozzle should be pointed down approximately 10°.Nozzles protecting supports should be located at thepoint where the tank protection will not cover thesupport and should point down the length of thesupport.

b. Distance of nozzle from surfaceUnless tanks are located indoors, where there are nowind considerations, the face of the nozzle should belocated no more than two feet (.6M) from the surface ofthe tank. Small angle nozzles or window sprinklersprotecting columns should be located close to surfaceand arranged to spray down the columns.

c. Number of nozzlesThe distance between nozzles covering the shell andthe end of the tank depends upon a number of factors.When the water spray hits the tank, there will be a

Ae = TT (D2 + h2) 4

Ae = TT D2

4

Ae = TT D2

2

DESIGN DATAWATER SPRAY

DESIGN

April 18, 2003Design 208 b

certain amount of tangential movement on the tanksurface. The amount of this tangential movementdepends upon the pressure and the angle at which thespray strikes the surface. Also it should be noted thatthe coverage area of the pattern on the shell increasesaround the circumference as opposed to the axis.Similarly the density decreases. Generally the numberof nozzles indicated in the below listed tables can bearranged to produce adequate coverage.

1. Tank End-See Table B2. Tank Shell-See Table Cd. Overspray

If nozzles are located too far from the tank, or if the tankis of a too small diameter, there will be overspray, thatis, water coming from the nozzle will not impinge on thetank and will be wasted. In order to avoid overspray.locate the nozzle closer to tank or use smaller angle.

e. Rundown considerationsRundown will occur over the top half of the tank shell,over the top half of a convex spherical end and over anentire flat end. There will be little or no rundown over the bottom half of the shell, the bottom half of a convex endand a concave end. There also may be projections from the tank, which will “roof off” certain areas, which wouldnormally be covered by rundown. These “roofed off” areas often require specific nozzle coverage. Inhorizontal tanks the designer generally should protectthe “roofed off” areas first and apply the remainder ofthe required water uniformly over the entire outsidesurface.

f. Support NozzlesConcrete or adequately fireproofed steel supportsrequire no protection. Short, unprotected steel supports up to about one foot (.3M) long generally require noprotection as long as they receive some rundown orsplash. Longer, steel supports require protection.Water should be applied on the inside of the channel orH column at or above the point where rundown orsplash becomes ineffective. Often a small angle spraynozzle or window sprinkler can be effective.

11. Find water required from each nozzle (Q nozzle)For each design area divide the water required by thenumber of nozzles discharging into the design area.

Q areaQ nozzle = Number of nozzles

12. Select the proper nozzle capacityConsult the K tables and determine the nozzle which willproduce the discharge closest to the required at theassumed pressure.

13. Determine the required nozzle pressureConsult the K tables and determine the pressure which willgive the required flow for the nozzle selected.

14. Adjust the designHydraulically calculate the piping system to produce therequired nozzle flow. For each individual design arearequired water demand must be discharged into the design area. Discharge from each nozzle must be as uniform aspossible. If total discharge of water into design area isbelow calculated requirements, add additional nozzles.

B. DESIGN PROCEDURE WATER SPRAY PROTECTIONFOR VERTICAL TANKS


Make a detailed survey of the tank and its surroundings. Note diameter, length, height of top and bottom, locations anddimensions of any irregularities which may impede distributionof water such as ladders, manholes, piping connections and the like. Note size, location and material of supporting structure fortank. Note proximity of other hazard producing equipment. Note the presence of dikes, walls and barriers. Determine contentsof tank and establish water density requirements for protection.

Tanks are generally protected using nozzles on the topand rings of nozzles around the shell at various levels. Thesenozzles must provide the proper quantity of water into a “Design Area”.

Tank Diameter = DTop Height = hShell Height = HShell Density Required = dLeg Density Required = l

1. Find the Height (L) and Number (N] of the shell design areasa. Smooth spherical top tanks (See Figure 3)

The total shell height is divided into two or more areassuch that the height of the upper area is one-third theheight of the lower areas. The height of the lower areasmust not exceed 12 feet (3.7M). (The upper area isincluded in the top design area - See Step 2-A).

L = H (Solve by trial and error)N +1/3

b. Smooth, flat or conical top tanks (See Figure 3)The shell is divided into areas of equal height notexceeding 12 feet (3.7M).

L= H (Solve by trial and error)N

c. Other considerationsIf there are appendages to the tank which “roof off”areas so that there is no coverage either by rundown ortangential flow, then these areas may constituteadditional design areas. When a flange is encounteredon the circumference of a tank, the design area beginsimmediately under the flange and extends downwarduntil the next flange or the bottom. Often the designarea of very tall tanks is limited to the lower 30 feet(9.2M) from the ground because the danger of fireexposure above this level is small. In this case topprotection may not be required. In no case should theheight of any shell design area exceed 12 feet (3.7M).


DESIGN

April 18, 2003 Design 208 c

2. Find the design area of the top (At)a. Spherical top tanks (See Figure 3)The design area equals the area of the top plus the shell area between the upper tank edge and the upper shell design area. (See Step 1-A).

Spherical top

Hemispherical top

b. Flat or conical top tanks (SeeFigure 3)The design area equals thearea of the top.

Flat top

Conical top

3. Find the total waterrequired for the top (Qt)The total water required for the top equals the design areafor the top times the required density (Table A)

Qt = At (d)4. Find the design area for the shell rings (As)

See Figure 3. For smooth tanks the design area will be thesame for each ring. For tanks with appendages orobstructions, design areas may differ.

As = TT DL5. Find the water required for the shell rings (Qs)

Water required equals design area times required density(Table A) for each ring. In the case of smooth tanks, waterdemand will be the same for each ring.Qs1 = As1(d)

6. Find the design area for the bottom (Ab)See Figure 3.

Flat bottom

Spherical bottom

Hemispherical bottom

Conical bottom

7. Find water demand for bottom (Qb)Water demand equals design area of bottom timesrequired density

Qb=Ab(d)8. Find water required for appendages (Qa)

If there are appendages to the tank which increase thebasic area of coverage, estimate this area and multiply bythe required density (Table A) to obtain the additionalwater required.

Qa1=Aal(d)Qa = Qal+Qa2+••

9. Find the design area for legs or supports (Al)Concrete or fireproof steel supports generally require noprotection. A short, unprotected steel leg up to about onefoot (.3M) generally requires no protection as long as itreceives some rundown or splash. Longer, unprotectedlegs require spray protection over the design areaindicated in Figure 2.

10. Find the water required for each leg (Ql)The water required for each leg equals the design area ofthe leg times the required density

Ql1 = Al1(dl)Ql2 = Al2(dl)

11. Find total water required (Qtot)Total water required equals the sum of the water demandfor the top, bottom, all rings, all legs, and all appendages.

12. Determine probable nozzle pressureBased upon known water supply conditions and/orassumed pump conditions and supply piping conditionsand water demand, calculate pressure available to thevarious areas of the tank. Note that in high tanks there willbe a significant static pressure difference between the topand bottom nozzles.

13. Select the most probable nozzle arrangementA wide range of nozzle capacities and angles are available. The objective is to obtain adequate coverage and flow witha minimum number of nozzles with no oversprayOverspray will occur on the shell when a large angle nozzle is used on a small diameter tank. For nozzles located 2 feet (.6M) from the tank, surface, overspray will occur on tankssmaller than those listed:Nozzle Angle Used Minimum Tank Diameter

Ft. M30 1.4 .4360 4.0 1.290 10.0 3.0120 26.0 8.0140 62.0 19.4

Nozzles can be used on tanks smaller than listed bymoving nozzle closer to tank. Eliminate those nozzleswhich are suitable only for larger tanks.Lay out the tank to scale and place nozzles in accordancewith the following:a. Distance of nozzle from surface

Unless tanks are located indoors where there are nowind considerations, the surface of the nozzle shouldbe located no more than 2 feet (.6M) from the surface of the tank. Small angle nozzles or window sprinklersprotecting columns should be located close to surfaceand arranged to spray down the columns.

b. Distance between nozzlesThe distance between nozzles covering the shell, thetop and the bottom of the tank depends upon a numberof factors. When the water spray hits the tank, there willbe a certain amount of tangential movement on the tank surface. The amount of this tangential movementdepends upon the pressure and the angle at which thespray strikes the surface.

c. Location of top nozzlesIn designing tank top protection, a balance must beachieved between a few large nozzles delivering a large

Ab = TT D2

4

At = TT (D2 + DL)

4 3

Ab = TT (D2 + h2

) 4

Ab = TT D2

2

Ab = TT D (D2 + h2

) 1/2

2 4

At = TT D2

4At = TT D (D

2 + h2 )

1/2

2 4


DESIGN

April 18, 2003Design 208 d

At = TT (D2 + h2

+ DL) 4 3

quantity of water and many small nozzles delivering asmaller quantity- Because of the rundown andtangential movement of the water, it is not necessary toeliminate all theoretical dry spots, however, theseshould be kept to a minimum as stated above. On largediameter tanks it may be possible to use nozzles ofincreasing nozzle angle moving from the center line ofthe tank to the outside edge. With flat and conical tanks, particular attention should be paid to the top edge of the tank since there could well be no rundown. The topedge of a flat or conical top tank should be protectedwith a ring of spray nozzles. Generally the number ofnozzles indicated in Table A can be arranged toproduce adequate coverage. Different angle nozzlesmay be used to achieve coverage.

d. Location of shell nozzlesThe shell rings should be located so that the spray willimpinge at the top boundary of each shell design area. It may be advantageous to stagger the nozzles in eachsuccessive ring. Generally the number of nozzlesindicated in Table B can be arranged to produceadequate coverage.

e. Location of bottom nozzlesIf the tank is mounted on the ground, no bottomprotection is required. If the tank is skirted with the skirtextending to the ground and generally enclosing thetank bottom, sufficient protection is usually provided bya conventional sprinkler or a pendent sprinkler in theupright position delivering .1 GPM/Sq. Ft. (4.9 MM/Min)over the design area. If the bottom is essentially open,however, it should be protected using somewhat thesame system as the top. The difference is that nogravity run-off can be expected. If the tank contains aliquid, the heat absorbing capacity of the bottom will beconsiderably greater than that of the top. Generally thenumber of nozzles indicated in Table A can be arranged to produce adequate coverage.

f. Location of leg nozzlesConcrete or adequately fireproofed steel legs requireno protection. A short unprotected steel leg up to about1 foot (.3M) long generally requires no protection aslong as it receives some rundown or splash. Watershould be applied on the inside of the channel or Hcolumn or as uniformly as possible around the circularcolumn at or above the point where rundown or splashbecomes ineffective. Use a small angle spray nozzle orwindow sprinkler. It may be possible to place aconventional sprinkler or a pendent sprinkler in theupright position as high as possible inside a hollowcircular column. In this case use .1 CPM/Sq. Ft.(5MM/Min.) over the leg design area.

g. Rundown considerationsRundown will occur over the top of a spherical top tankand over the vertical shell. There will be little or norundown over the bottom. There also may beprojections from the tank which will “roof off” certainareas which would normally be covered by rundown.These “roofed off” areas often require specific nozzlecoverage. In vertical tanks “roofed off” areas constituteseparate design areas and require nozzles whichprotect directly under the roof.

h. Nozzle directionNozzles protecting the tank top and bottom should bepointed directly at the tank. Nozzles protecting the tankshell should be pointed down 10". Nozzles protectinglegs should point down the length of the leg.

i. Overspray

If nozzles are located too far from the tank shell, or if the tank is of too small diameter there will be overspray,that is, water coming from the nozzle will not impinge on the tank and will be wasted. In order to avoid overspray.locate the nozzle closer to tank or use smaller angle.

14. Find water required from each nozzle (Q nozzle)For each design area divide the water required by thenumber of nozzles discharging into the design area.Q nozzle = Q area

Number of nozzles15. Select the proper nozzle capacity

Consult the K tables and determine the nozzle which willproduce the discharge closest to the required at theassumed pressure.

16. Determine the required nozzle pressureConsult the K tables and determine the pressure which willgive the required flow for the nozzle selected.

17. Adjust the designHydraulically calculate the piping system to produce therequired nozzle flow. For each individual design arearequired water demand must be discharged into the design area. Discharge from each nozzle must be as uniform aspossible. If total discharge of water into design area isbelow calculated requirements, add additional nozzles.

C. DESIGN PROCEDURE WATER SPRAY PROTECTIONFOR TRANSFORMERS


Transformers come in many sizes and configurations. Beforeattempting to design the protection it is essential to have thefollowing information:

1. Length2. Width3. Height of transformer4. Location and height of bushings5. Height and location of lightning, if any6. Size and location of oil expansion tank, if any7. Location of any switch boxes and any equipment that may

interfere with water distribution8. Size of transformer, i.e., high and low voltage9. Phase of transformer, either single or three phase

10. Direction of incoming high voltage and low voltage wire orbus bars to the transformer

11. Setting of transformer, whether surrounded by concrete orcrushed rock

12. Elevation of bottom of transformer above grade13. Location of radiators and distance between radiators.

When space between radiators exceeds 12 inches (.3M) itmust be covered

14. Size and location of, if any


DESIGN

April 18, 2003 Design 208 e

15. Estimate of possible effects of wind, and size and locationof any wind protection.

If the transformer is not existing, it will be necessary toobtain a manufacturer’s dimensional drawing of the proposedtransformer.

The drawings for the transformer should be made to alarge scale, e.g., 3/8" to 1’-0" or 1/2" to 1’-0" (1/30 or 1/25), andthere should be three views: top, side and bottom. If more thanone ring is necessary an additional plan view may benecessary.

In addition to the transformer, a detail plan should bedrawn to show the general view, such as fire walls betweentransformers, water supply location, valve location, electricalpoles and any other obstructions that may interfere with thesprinkler piping.

Transformers present particular design problems for waterspray protection, primarily because of their irregular shape andthe necessary clearances to be provided from high voltagewiring. Generally speaking, there is much more interferencewith the water flow on the surface of the transformer than thereis on a tank. For this reason protection systems fortransformers generally involve a large number of small capacity nozzles. Often it will be necessary to put more water on thetransformer than is actually required simply to achievecoverage. It is most useful to use a large scale drawing of thetransformer and project theoretical nozzle discharge patternson it to get an idea of the type of coverage to be expected.

Transformers are generally protected using rings ofnozzles around the transformer with the top ring being locatednear the top of the transformer and subsequent rings beinglocated every 12 feet (3.6M) from the top to the bottom of thetransformer or beneath each continuous obstruction. Nozzlesare also employed to spray water on the bottom of thetransformer in the event it is more than 12 inches (.3M) abovethe ground. In addition, if the ground is covered with solidmaterial such as concrete or asphalt, nozzles must be locatedto wash flammable liquid away from the transformer. Nozzlesmust be located so as to spray the proper amount of water intothe “design area”.

To determine the various design areas of the transformer,consider that the elements of the transformer are a collection ofsimple geometric figures (cylinders, cubes, etc.). Make a planand elevation view of the simplified transformer concept. If thetransformer is located 12 inches (.3M) or more above grade,also make a bottom view. Neglect small protrusions or increasesize of figure slightly to compensate. Radiators should beconsidered as a single unit unless there is more than a 12 inch(.3M) space between them. In this case, they must beconsidered as multiple units.

Required Density = dRequired Grade Density = dg

1. Determine the design area for the top and sides of thetransformer (Ats)Referring to the simplified view of the transformer, thedesign area is the total exposed outside area of thesimplified transformer concept less the area of the bottom.

2. Determine the water required for the transformer topand sides (Qts)Water required equals the design area of the top and sidestimes the required density. (Table A)

Qts = Ats(d)3. Determine the design area of the bottom (Ab)

The design area of the bottom is the area of the bottom ofthe transformer which is 12 inches (.3M) or more abovegrade.

4. Determine the water required for the bottom (Qb)Water required for the bottom equals the bottom designarea times the required density. (Table A)

Qb = Ab(d)5. Determine the design area for the grade (if any) (Ag)

The design area for the grade is the area which appears on the simplified bottom view of the transformer plus an areaextending 3 feet (.9M) on all sides of the view. Gradeprotection is required only when a nonabsorbing surfacesuch as concrete or asphalt paving is employed. Gradesurfaces such as gravel or crushed rock. do not normallyrequire nozzle protection. Grade protection is notrequired directly under the transformer unless it is locatedat least 12 inches (.3M) above grade.

6. Find the water required for the grade (Qg)Water required for the grade is equal to design area for thegrade times the required density (Table A)

(Qg) = Ag(dg)7. Find the total water required for the entire transformer

(Qtot)Total water required equals the sum of the water demandfor the top and sides, the bottom and the grade.

Qtot = Qts+Qb+Qg8. Assume a nozzle pressure

Nozzle pressures under 30 psi (2 bars) generally do notproduce adequate water throw. Based upon known watersupply conditions and/or assumed pump conditions andsupply piping conditions, calculate the pressure availableat the transformer. Note that in high assemblies there willbe a significant static pressure difference between the topand the bottom nozzles.

9. Select the probable nozzle arrangement A wide range of nozzle capacities and angles are available. The objective is to obtain adequate coverage and flow witha minimum number of nozzles with no overspray. Lay outthe transformer to scale and place nozzles in accordancewith the following:a. Minimum electrical clearances

One of the most important considerations in locating the piping around the transformer is the distance of the pipe from the electrical components or energized parts,such as bare cables, bus ducts, and the low voltage and high voltage bushings. The clearance between anyportion of the water spray equipment and theunenclosed or uninsulated electrical components, atother than ground potential, should not be less thangiven in the following table. These clearances are forthe altitude of 3,300 feet (1000M) or less. The distanceshould be increased at the rate of one percent for each330 feet (100,M) increase of altitude above 3,300 feet(1000M).


DESIGN

April 18, 2003Design 208 f

COMMONLY ACCEPTED CLEARANCES FROMWATER SPRAY

EQUIPMENT TO LIVE ELECTRICAL COMPONENTS

NominalLine

Voltage(KV)

NominalVoltage to

Ground (KV)

DesignBIL (KV)

Minimum Clearance

(Inches) (mm)

To 15 To 9 110 7 178

23 13 150 10 254

34.5 20 200 13 330

46 27 250 17 432

69 40 350 25 635

115 66 550 37 939

138 80 650 44 1117

161 93 750 52 1320

196-230 114-132 9001050

6376

16001930

287-380 166-220

1175130014251550

8798109120

2209248927683048

500 290 16751800

131142

33273606

500-700 290-400192521002300

153168184

388642674673

NOTE: When the design BIL is not available, and when nominal voltage isused for the design criteria, the highest minimum clearance listed for thisgroup shall be used.

There are design variations in the clearance required athigher voltages as shown in the table where a range ofvoltages is indicated opposite the various BIL testvalues in the high voltage portion of the table. Up tosystem voltages of 161 kv, the design BIL kv andcorresponding minimum clearances, phase to ground,have been established through long usage. At thehigher voltages, the relationship between design BIL kvand the various system voltages has not beenestablished in practice and is dependent upon severalvariables, so that the required clearance to groundshould be based upon the design BIL used, rather thanon the nominal line voltage or voltage to ground. Checklatest rules of authority having jurisdiction.

b. Distance of nozzle from surfaceUnless the transformers are located indoors wherethere are no wind conditions, the surface of the nozzleshould be located no more than 2 feet (.6M) from thevertical surface to the transformer.

c. Coverage for transformer topIt is generally not satisfactory to run piping directlyacross the top of a transformer; therefore, most of thetop coverage is obtained by spray nozzles throwing infrom the outside. It is, however, often acceptable to runa line between the transformer body and radiatorsabove the transformer. Necessary electrical clearances must be maintained. Generally 30, 60 or 90 degreespray nozzles are installed in this top loop with thenozzles located approximately 1 to 2 feet (.3 to .6m)above the transformer top and pointed so that the waterwill impinge upon the transformer. Water should not bedirected at the high voltage bushings. The above

nozzles have a maximum effective horizontal throw of 6feet at 30 psi (1.8m at 2.0 bars). It may be desirable tolocate nozzles at the corner to achieve increasedcoverage.

d. Horizontal distance between nozzlesThe horizontal distance between nozzles should besuch that their patterns intersect along the horizontalline. For nozzles located 2 feet (.6m) from the surface,the following horizontal distances should be used:

30 degrees 13 inches .33m60 degrees 28 inches .77m90 degrees 48 inches 1.22m

e. Two sets of nozzles from the same ring Because, generally speaking, small capacity nozzleswill be used, it will often be possible to extend thenozzles above and below the loop by means of a nipple. Nipples longer than 2 feet (.6m) generally requireadditional support.

f. Bottom nozzlesIf the transformer or radiator is located more than 12inches (.3m) off the ground, it is necessary to protectthe bottom. This is generally done with wide angle spray nozzles pointing upward.

g. Between radiator protection If the radiators are more than 12 inches (.3m) apart, nozzles must be arranged to spray into this space. Anozzle angle should be selected so that the conediameter at the entrance is equal to or slightly largerthan the space between the radiators.

h. Rundown considerationsRundown will occur on smooth, vertical surfaces. Projections from the surfaces, however, will “roof off”certain areas which would normally be covered byrundown. These “roofed off” areas usually requirespecific nozzle coverage.

i. Vertical distance between nozzlesFor unobstructed vertical surfaces with no “roofed off” areaand unobstructed rundown, a maximum vertical distancebetween nozzles is 12 feet (3.6m). In practice, however,unobstructed areas of this size are rarely encountered.

j. Nozzle directionNozzles protecting the transformer top should be aimed slightly down so that all of the water impinges upon thetransformer with either all on the top or some on top and some on vertical sides. Nozzles protecting verticalsides and bottom should point directly at the surface tobe protected. Nozzles covering irregular areas shouldbe located for best coverage generally spraying intocorners. Overspray must be avoided. Nozzles coveringspace between radiators should be arranged to spraydirectly into the open space.

k. OversprayIf nozzles are located too far from the surface, or if theangle is too large, there will be overspray, that is, watercoming from the nozzle will not impinge on thetransformer and will be wasted. In order to avoidoverspray, locate nozzle closer to the transformer oruse a smaller angle. Always observe electricalclearances.

l. Grade protectionIf the transformer is located 12 inches (.3m) or moreabove a non-absorbing surface such as asphalt orconcrete, nozzles must be located under thetransformer pointing down and outside the transformercovering an area 3 feet (.9m) around the transformerpointing generally outward. The purpose of the gradeprotection nozzles is to wash flammable liquid away


DESIGN

April 18, 2003 Design 208 g

from the transformer (consider grade slope). It is oftenpossible to feed both bottom protection nozzles andgrade protection nozzles from the same pipe. In somecases (usually small transformers) it may be acceptable to use an open pendent sprinkler in the upright position. No grade protection is required when readily absorbsthe flammable liquid.

m. WindOften, because of transformer configuration andelectrical clearances, it will not be possible to locatespray nozzles close to the areas that they are expectedto protect. When the installation is outside, the effect ofthe wind must be seriously considered. Small spraynozzles operating at high pressure produce small drops that are particularly susceptible to being blown away bythe wind. It may be necessary to increase water densityin questionable conditions.

10. Find the total amount of water delivered into eachdesign area by the probable nozzle arrangementUsing the assumed nozzle pressure and the smallestavailable capacity nozzles, determine the amount of waterdischarged by each nozzle and the total amount of water ineach design area.

11. Adjust the designCompare the water delivered to the design area by theprobable nozzle arrangement with that required. Increaseor reduce nozzle sizes and pressures as necessary. (Notethat because of the irregularities of transformers, manynozzles are required to provide coverage. In addition,because of electrical clearances, some of these nozzlesare required to throw at maximum distances. For thisreason it may not be possible to reduce the number ofnozzles or the operating pressure far enough to approachthe minimum. In practice it may be possible to reduce therequired pressure below 30 psi (2.0 bars); however, thisshould not be done in the design stage.)

III - TABLES & FORMULASTABLE A, COMMONLY ACCEPTED DENSITIES


GPM/Ft. Sq. MM/Min.Transformers

Top & Sides .25 10.18Bottom .25 10.18Grade .15 6.11

Pipe RacksPipe Surface .10 4.07Maximum over plan area projected on grade .50 20.35Legs .10 4.07

TanksShells .25 10.18Supports .25 10.18

ConcavedEnd

For Round

Design Area - AL = TT DlLength = lDesign Area - AL = (w + h) l

Fig. 1a - Design Area of Horizontal Tank Fig. 1b

Area of Shell = As = TT DL

Area of Ends - Flat Ae = TT D24

Area of Ends - Spherical, Convex or Concaved Ae = Π (D2 + h2)4

Area of Ends - Hemispherical Ae = TT D22

Total Area = At = As + Ae1 + Ae2Fig. 2a - Design Area of Leg Fig. 2b

For I-Beam or Channel

Fig. 2c


DESIGN

April 18, 2003Design 208 h

L = HN + 1/3

N = number of rings - must be a whole number.

L must not exceed 12 ft. or 3.7 M.

Top Design Area = At = TT ( D2 + h2 + DL)

4 3

Design Area for each Ring = A s1, As2, etc., = TT DL

Bottom Design Area = Ab = TT (D2 + h2)

4

Fig. 3b - Cylindrical Spherical Top

Fig. 3c - Spherical Design Area = At = Ab = TT D2Top and bottom 2

Fig. 3a - Design Area for Vertical Tanks

Cylindrical Flat or Conical Top

L = H N = number of rings - N must be a whole number.

L must not exceed 12 ft. or 3.7 M.

Top & Bottom Design Areas:Flat Atf = Abf TT D2

4Conical = att = Abc = TT (D

2 + h2)

4

Design Area for each Ring = A s1, As2, etc., = TT DL


DESIGN

April 18, 2003 Design 208 i

TABLE BWater Spray Protection

Vertical Tanks: Top and BottomHorizontal Tanks: Ends

Commonly accepted maximum tank diameters for effective coverage by generally uniformally spaced spray nozzles located 2 ft.(.6 M) from the surface of vertical or horizontal tank ends of Flat, Concaved or Convex form.

Ft. M Ft. M Ft. M Ft. M Ft. M1 1.4 ,43 3.0 ,86 5.0 1,52 8.5 2,6 11.0 3,42 1.8 ,52 4.0 1,22 6.5 2,0 9.5 2,9 12.5 3,83 2.4 ,74 5.0 1,52 10.5 3,2 11.0 3,3 14.0 4,24 3.0 ,86 6.0 1,84 12.0 3,6 18.0 5,5 22.5 6,85 4.0 1,22 8.5 2,6 15.0 4,6 25.0 7,6 32.0 9,76 4.7 1,43 9.7 2,9 17.5 5,4 29.0 8,8 43.0 13,17 6.4 1,95 11.0 3,3 20.0 6,1 34.0 10,4 48.0 14,88 7.1 2,2 14.0 4,3 23.0 7,0 43.0 13,1 53.0 16,19 7.9 2,4 15.5 4,7 27.5 8,4 47.0 14,2 59.0 17,810 8.5 2,6 17.0 5,2 30.0 9,2 51.0 15,5 64.0 19,411 9.2 2,8 18.0 5,5 32.0 9,7 55.0 16,6 68.0 21,512 9.8 3,0 19.0 5,8 34.0 10,4 58.0 17,5 73.0 22,2

140°

MAXIMUM TANK DIAMETER FOR VARIOUS ANGLE NOZZLESNumberof

NozzlesUsed

30° 60° 90° 120°


DESIGN

April 18, 2003Design 208 j

TABLE CWater Spray Protection

Vertical and Horizontal Tank Shells

Commonly accepted maximum tank diameters for effective coverage by equal radially spaced spray nozzles located 2 ft. (.6 M)from tank surface of vertical or horizontal tanks.

Ft. M Ft. M Ft. M Ft. M Ft. M1 -- .8* ,24 1.5* ,46 2* ,61 * * * *2 180 1.5* ,46 3* ,92 5* 1,5 * * * *3 120 2.3 ,70 4.6 1,4 8* 2,4 * * * *4 90 2.8 ,85 5.6 1,7 10.5 3,2 17* 5,2 * *5 72 3.4 1,0 6.8 2,1 12.5 3,8 20* 6,1 * *6 60 4.0 1,2 8.0 2,4 14.8 4,5 24* 7,3 * *7 53.5 4.5 1,4 9.2 2,8 16.7 5,1 26.7 8,1 * *8 45 5.2 1,6 10.4 3,2 19.5 5,9 30.8 9,3 * *9 40 5.8 1,8 11.7 3,6 21.9 6,6 35.1 10,6 * *10 36 6.5 2,0 12.9 3,9 24.5 7,4 38.8 11,7 * *11 37.7 7.1 2,2 14.2 4,3 27.7 8,2 42.6 12,9 * *12 30 7.7 2,4 15.5 4,7 29.0 8,9 45.0 13,6 58.0* 17,6

*Indicates Overspray at 2 Ft. (,6 M)

140°

MAXIMUM TANK DIAMETER FOR VARIOUS ANGLE NOZZLESNumberof

NozzlesUsed

30° 60° 90° 120°

AngleBetweenNozzles

LOCATION OF TANK RINGS - HORIZONTAL TANKS ONLY

Ft. M Ft. M30 1 ,3 2 ,660 2 ,6 4 1,290 3.5 1,1 7 2,1

120 6 1,8 12 3,7140 7.5 2,3 15 4,6

Max. DistanceNozzle to EndSeam of Tank

Max. DistanceBetween Nozzles

NozzleAngle


DESIGN

April 18, 2003 Design 208 k


DESIGN

April 18, 2003Design 208 l

IV -- WORK SHEETS

Tank Diameter = D = Tank Shell Length = L =

Tank End Height = h = Density Required = d =

STEP

1 Find Shell Area Shell Area = As = TT DL As =

2 Find Water Reqdfor Shell

3 Find Area Flat End Ae = TT D2

of Ends 4

Spherical End Ae = TT D2 = h2

4

Hemispherical End Ae = TT D2 Ae1 = Ae2 =2

4 Find Water Qe1 = Ae1 x Density = Ae1(d)Required for Qe2 = Ae2 x Density = Ae2(d)Ends Qe1 = Ae1 =

5 Determine Qa1 = Aa1 x Density = Aa1(d)Water Required Qa = Qa1 + Qa2- - -For Appendages Qa =

6 Determine Area Al1 = See Figure 2of Legs andSupports

7 Determine Ql1 = Al1 x Density = Al1(d)Water RequiredFor Each Legor Support Ql1 = Ql2 = Ql3 =

8 Determine Qtot = Qs + Qe1 + Qe2 + Qa

Total WaterRequired Qtot =

SHELL END 1 END 2 APPENDAGES SUPPORTS9 Assume Nozzle Calculate Based on Water Supply

Pressure10 Select Most Lay Out to Scale -- List Number

Probable Nozzle and Angle of Nozzles ChosenArrangement

11 Find WaterReqd. from Q Nozzle =Each Nozzle

12 Select the Consult K Tables --Proper Nozzle List Nozzle SizeCapacity

13 Determine Consult K Tables --Reqd. Nozzle List Pressure Reqd.Pressure For Reqd. Flow

14 Adjust the Calculate System and AdjustDesign as Necessary

WORK SHEET A

Number of Nozzles

Q (area)

WATER SPRAYPROTECTION FOR

HORIZONTAL TANKS

Qs - As x Density = As(d)

INSTRUCTIONS CALCULATIONS

+ Ql1 + Ql2 - - - -

Qs =

)(

Top Diameter = D = Shell Height = H =

Top Height = h = Density Required = d =

Type

STEP

1 Find Height L For Spherical Top L = Hand Number N N + 1/3of Shell Design For Flat & Conical Top L = HAreas N

L must not exceed 12 Ft. or 3.7MSolve by Trial & Error L = N =

2 Find Design Flat Top At = TT D2

Area of Top 4Portion Cylindrical Spherical At = TT D2 + h2 + DL

4 3

Conical Top - At = TT D D2 + h2 1/2

2 4

Spherical At = TT D2 At =2

3 Find Water Qt = At x DensityDemand for = At x dTop Portion Qt =

4 Find Design As - TT D LArea of ShellRings As1 = As2 =

5 Find Water Qs1 = As1 x Density

Demand for Qs2 = As2 x Density

Shell Rings Qs1 = Qs2 =

6 Find Design Flat Bottom Ab = TT D2

Area forBottom Spherical Bottom Ab = TT D2 + h2

4

Conical Bottom Ab = TT D D2 + h2 1/2

2 4

Hemispherical Bottom Ab = TT D2

2 Ab =

WORK SHEET B, Page 1


4

WATER SPRAYPROTECTION FORVERTICAL TANKS

)(

( )

)(

)(


DESIGN

April 18, 2003 Design 208 m

Top Diameter = D = Shell Height = H =

Top Height = h = Density Required = d =

Type

STEP

7 Find Water Qb = Ab x DensityDemand for = Ab x dBottom Qb =

8 Find Water Required Qa1 = Aa1 x dfor Appendages Qa = Qa + Qa2 ----

Qa =

9 Find DesignArea of Legs Al =

10 Find WaterDemand forEach Leg Ql1

11 Find Total Qtot = Qt + Qs1 + Qs2 ---- + Qb +Ql1

Water Demand Ql2 + - - - - + Qa

Qtot

TOP RING RING BOTTOM LEG12 Determine Hydraulically Calculate to

Probable Each Group of NozzlesNozzle Pressure

13 Select the Layout to Scale - List NumberMost Probable and Angle of Nozzles ChosenNozzleArrangement

14 Find Water Q Nozzle = Q (Area)Reqd. FromEach Nozzle

15 Select the Consult K Tables -Proper Nozzle List Nozzle SizeCapacity

16 Determine Consult K Tables -The Reqd. List Press. Reqd. for Reqd. FlowNozzle Pressure

17 Adjust the Calculate and Adjust asDesign Necessary

WORK SHEET B, Page 2

A l = See Fig. 4

Ql1 = Al1 x dl

Number of Nozzles


WATER SPRAYPROTECTION FORVERTICAL TANKS


DESIGN

April 18, 2003Design 208 n

Nominal Line Voltage Density Reqd. - Transformer

Nominal Voltage to Ground

Design BIL

STEP

1 Find Design Ats = Total Outside Surface AreaArea of Less Area of Plan View of BottomTop & Sides Ats =

2 Find Water Qts - Ats x DensityDemand forTop & Sides Qts =

3 Find Design Ab = Area of Plan View of BottomArea of Bottom (If Transformer is less than 12" (3,6M)

Above Grade, Ab = O) Ab =

4 Find Water Qb = Ab x DensityDemand for (If Transformer is less than 12" (3.6M)Bottom Above Grade, Ab = O) Qb =

5 Find Design Ag - AB + A Strip 3 ft. (.9M)Area for Completely Around AbGrade (If the Grade Surface is

Non Absorbing Ag = O) Ag =

6 Find Water Qg - Ag x DensityDemand for (If Grade Surface isGrade Non Absorbing Qg = O) Qg =

7 Find TotalWater Demand Qtot = Qts + Qb + Qg Qtot

Bottom

8 Assume Calculate Based onNozzle Pressure Water Supply

9 Select the Layout to Scale - List NumberProbable Nozzle and Angle of Nozzles ChosenArrangement

10 Find the Total Consult K Tables - Find WaterWater to Each Delivered by Each Nozzle atDesign Area Assumed Pressure - Total All Water

11 Adjust the Compare delivered WaterDesign with Required Water and

Adjust as Necessary

WATER SPRAYPROTECTION FORTRANSFORMERS

WORK SHEET C

Grade

Top & Sides Grade



DESIGN

April 18, 2003 Design 208 o

Form No. F_042602 Revised page replaces pages 209a-o dated July 15,2002. (Renumbered page.)

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