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8/11/2019 Cavitation Orifices and Valves ML030700448.pdf http://slidepdf.com/reader/full/cavitation-orifices-and-valves-ml030700448pdf 1/22 4 - i. -. ______ -; -. - -.- - WATER HAM ME R/CAVITATIO N Assessment & Prevention "* WHAT IS WATERHAMMER " WATERHAMMER CATEGORIES "* WATERHAMMER PHENOMENON "* WATERHAMMER ASSESSMENT "* WATERHAMMER PREVENTION "* INDUSTRY EXPERIENCES 0 0 S 0 0 WHAT IS CAVITATION WHERE IT OCCURS HOW TO IDENTIFY EFFECTS OF CAVITATION CAVITATION DAMAGE DESIGN CONSIDERATIONS SUBHASH C. KHURANA FLORIDA POWER & LIGHT L FPL

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4 - i. -. ______ -;

-. - -.- -

WATER HAMMER/CAVITATIO N

Assessment & Prevention

"* WHAT IS WATERHAMMER" WATERHAMMER CATEGORIES

"* WATERHAMMER PHENOMENON"* WATERHAMMER ASSESSMENT"* WATERHAMMER PREVENTION"* INDUSTRY EXPERIENCES

0

0S

0

0

WHAT IS CAVITATIONWHERE IT OCCURS

HOW TO IDENTIFYEFFECTS OF CAVITATION

CAVITATION DAMAGEDESIGN CONSIDERATIONS

SUBHASH C. KHURANAFLORIDA POWER & LIGHT

LFPL

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Cavitaiton Trng.Page 1 of 6

CAVITATION

1.0 What is Cavitation

In a flowing liquid if the static pressure is dropped below thevapor pressure corresponding to the liquid temperature (or thetemperature is increased above the saturation temperaturecorresponding to the liquid pressure), vapor bubbles (cavities)will be generated. If these bubbles are carried downstreams bythe liquid to a region of higher static pressure (or lowertemperature) then the bubbles will collapse. the region of theflow where bubbles exist is the cavitating region, whereas the

observed damage is at the location of the bubble collapse.

Cavitation is distinguished from flashing or boi-in- flow in thatno pressure recovery occurs downstream in the latter type offlow, and therefore no bubble collapse occurs.

2.0 Where Cavitation is likely to occur in Systems

a Orifices (Pressure profile across an Orifice, Figure-l)

* Valves

* Venturies

* Pump Suction

0 Pipe Fittings (Elbows, TEEs, etc.), (Figures 2 and 3)

0 Bumps

3.0 Identification of Cavitation

This process of bubble formation, growth, and collapse producesoscillation in the flow regime that can be identified by distinct

sound. The sound may range from occasional popping (at the onsetof cavitaion) to a sound like frying bacon as cavitation levelincreases. Extensive cavitation sounds like gravel flowingthrough the system. Because the flow oscillation associated withcavitation is broad band, pipe vibtation will also be associatedwith cavitation.

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Cavitaiton Trng.Page 2 of 6

4.0 Effects of Cavitation/Damage

Cavitation produces undesirable effects on piping system andcomponents. It generates undesirable noise, may cause erosion,restrict flow capacity of the system, and may cause pipe failuredue to pipe failure.

The erosion observed in cavitating fluid is due to bubblecollapse near the pipe wall. Studies have shown that as a resultof bubble collapse the liquid jets with velocities of 300 to 2000feet per second will impact the pipe wall. At this high impactvelocity, the local pressure at the wall could be as high as10,000 atm. The continuous bomardment of the metal surface willresult in the pitted configuration observed in most damagedcomponents.

5.0 Design Considerations

For system susceptible to cavitation, the designer should ensurethat under operating conditions, the system is either free fromcavitation or has acceptable level of cavitation. Levels ofcavitation are defined below.

Incipient Cavitation

Critical Cavitation

Incipient Damage

Choking Cavitation -

The onset of cavitation. Noobjectionable noise and no damage.

Continuous light cavitation usuallyadopted as the design criterion. Noiseand vibtation are acceptable and onlyminor damage after long periods.

Onset of surface pitting after shortperiod of operation.

Flow is not increased if downstreampressure is decreased. Maximum vibrationand noise.

Cavitation at pump suction is well taken care by satisfyingVendor's NPSH requirements. Valves and Orifices are genarally thecommon source of cavitation problems encountered in power plants.The following sections provide basic guidance on how to avoidcavitation of orifices and valves.

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Cavitaiton Trng.Page 3 of 6

ORIFICES

(See Figure-1 for a pressure profile across an Orifice)

Cavitaiton Index

a = (Pd - PV)/(P- - Pd)

- (Pd - Pv)/AP (Reference 1)

Where: a = Orifice Cavitation Index

Pu = Upstream Pressure, psia

Pd = Downstream Pressure, psia

Pv = Vapor Pressure, psia

The above calculated "cavitation Index" is compared toaccepatable "Cavitation Level" (see Figure 5) which isexperimently determined . If the cavitaion index is equal to, orhigher than the cavitaion level, the orifice size is acceptable.If a single orifice fails to satisfy cavitation criterion,multiple orifices are required. The spacings between orificesmust allow complete pressure recovery, which is achieved byplacing the orifices 6-8 pipe diameters apart. The "f," ratio(orifice diameter/pipe inside diameter) of orifices shall be inincreasing order along the flow. Sample Problem-I illustratesselection of orifice(s) to have cavitation free operation.When enough space is not available, multi hole type orifice plate(perforated plate), or a cone type orifice is 'recommended.

Other definitions of Cavitation Parameter

(hu-h.)a = and

U 2 /2g

(hu-hv)

a = (Reference 2)(hu-hd)

Where: a = Cavitation parameterhu = Upstream headhv = Vapor headhv = Downstream headU = Upstream velocityg = Gravitational constant

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Cavitaiton Trng.Page 4 of 6

Reference-2 utilizes reference velocities for a component sizeand upstream head at which a type of cavitation commences. Thesevelocities are then corrected to other component sizes andupstream head conditions using empirical relationships withexperimentally derived coefficients.

VALVES

Colorado State University tested some valves for Hydraulic andCavitation characteristics. The test data indicate that thecritical cavitation index does not only depend on the valveopening (area ratio) but is also dependent on VALVE SIZE,UPSTREAM PRESSURE, and VALVE TYPE. Figure-9 through-12 showresults of such tests. Figure-12 also includes a curve for thecritical cavitation index for a thin orifice plate for comparisonto that of a valve.

For a valve to operate completely cavitation free, operatingconditions should never approach those corresponding to thecritical cavitation index. Operation beyond critical cavitationis never recommended.

Analytically, it may not always be possible to accurately predictwhen a valve will start to cavitate. Therefore, monitoring fornoise and vibration levels at the required operating point isrecommended.

Reference 2 provides guidelines for determining cavitationparameter for valves. Sample Problem-2 is used as an illustrationfor a butterfly valve.

6.0 References

1. Technical Paper, "Eliminating cavitation from pressurereducing orifices", CHEMICAL ENGINEERING DECEMBER 12,1983

2. INTERNAL FLOW SYSTEMS, by Donald S. Miller, Volume 5 in theBHRA ( British Hydromechanics Research Association )FluidEngineering Series

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Cavitaiton Trng.Page 5 of 6

SAMPLE PROBLEM-I -/- / &'-- - I0.

Pump discharges water to atmosphere (See Figure-4).

Pipe size: 14-inch, Schedul

Flow: 4500 gpm

Pressure drop across orifice(s):

Fluid temperature:

e 40

120 psid

1600 F

Determine the size of orifice(s) for cavitation free operation.Neglect piping friction loss. Use Figure 5 for comparingcalculated cavitation index (a).

SAMPLE PROBLEM-2

Water flows from a reservoir at higher elevation to the reservoirat lower elevation, and flow is to be controlled by a butterfyvalve (See Figure-6). Determine the cavitation free range ofcontrol for the A-type butterfly valve from fully closed tofully open position. Loss coefficients for butterfly valve typesare shown in Figure-7. Accepatance criteria is criticalcavitation velocity..

Pipe Size:Valve Size:Water Temperature:

K-for Piping between "A" to "B" :K-for Piping between "C" to "D" :K-for butterfly valves types:

30-inch, Schedule-4030-inch diameter900 F

177

Shown in Figure-7

Based on Reference-2, Cavitation velocities for a butterfly valveare determined as follows:

Ui or U, = Cl. (Ur or Ucr).[(hu-hv)/50] 0 .39

and

Uch = Urch [ (hu-hb) /50] 0.5

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Cavitaiton Trng.Page 6 of 6

Where:

U1 , U,, and Uch are incipient, critical or choking cavitationvelocity (to be determined) for a given butterfly valve.C1 is correction factor for the valve size used, Figure-8

Ui,. U,,, and Uch are the reference cavitation velocities(incipient, critical, or choking as applicable) for the base

valve, shown in Figure-B.

h = upstream head of the valve (meters)

h, = vapor head (meters)

"50" = "h,-hv" value (in meters) for the base valve (size0.3m diameter)

CAUTION: USE DIMENSIONAL UNITS APPROPRIATELY

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ORIFICE

II

Pu Pd

PvP

PRESSURE PROFILE OF A SINGLE-ORIFICE SYSTEM

Pu = Pressure upstream of the orifice

Pc = Pressure at the vena-contracta A473

Pd = Pressure downstream of the orifice

Pv = Vaper pressure of the flowing fluid

FIGURE - 1

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Brim

I

FIGURE - 3

---------

, !I,

ti

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INCORRECT SOLUTION:(1 stage orifice)

CORRECT SOLUTION:(2 oriface stages)

2nd 1st

_ 7 Ii0-2P -2

A471

FIGURE - 4

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Critical Cavitation Index

for

Thin Plate Orifice

FIGURE - S

- . - 1* 9 - I- -

---. 1L-J - + - 1 IiT -i

I.

- I--t-I-t-4-I-I-I t-t-1-I I

NO CAVITATION /

CAVITATION

- ..- . ..-.

L_~- - -- - ---

-.. -...- -•"-. ... - .. ..-

0.0 0.1 0.2 0.3 0.4 0.5 0.6

DIAMETER RATIO (d/D) -

0.7 0.8 0.9

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70 FT.

RESERVOIRD

RESERVOIR

B

REFERENCEVALVE -0 FT.

FIGURE - 6

A477

200 FT.

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1000

C

100

50

" A

10

1.00.1

0 10 20 30 40 50 60 70 80 go

Vahv owniN (deg,)

Loss Coefficients for Butterfly Valves(from Reference 2, Figure 14.19)

FIGURE - 7

lI

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L=s coeffi~ent K

Cavitation Velocities for Butterfly Valves(from Reference 2, Figure 6.22)

U

IA.CC

aU

10.2

1.4 - 7 --2m---"

1.0 - --ma oO." ........_. , , . , .., - - -. , - - - - - " - -'- - - - "- - - - - " - - --

0.1 101

LOUcoff @McKV

Correction Factors for Valve Size(from Reference 2, Figure 6.23)

FIGURE - 8iIi

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I ....... 11. FPiGURE - 9

12-in. Rjuttes4L.VVv?! with24-in. expansion downstream

Upstream pressure 00 psi

0

1

0 10

Normal tests

Air injection into stem- -

- -J

S.... rnRf 70 tz0 iu "IU

"Test Valve Opening'- 9o

116

5

4

3

2

I

(1 '0•,.Iv

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"FIGURE - 10

to 20 30 40 50 60 70 80

Test Valve Opening - 7

6

5

4

0

u

u;

u

3

z

i

00

, 4 .

A',C•, 'z-

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FIGURE- 12

3.0 - -

8-in. Nqg1j&.y e•

1.,xpansion P u

o 14-in. 153 psi* 14-in. 65 psi -

& 12-in. 153 psio 12-In. 65 psi* 14- in. 153 psi (swOeep tests)

S2.0A 12-in. 153 psi (sweep tests) -

0 8-in. 65 psi

>0

u

"u . ."o.

F//

0 20 40 60 80 103

Test Valve Opening - I

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FIGURE - 12

10 20 30 40 50 60 70

'rest Valve Opening - I

Orifice Area Ratio (d/D) - 0/0

6

5

t

ULa4

u 2

I

00 80

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Cavitation Trng.AttachmentPage 1 of 3

Eliminating cavitation frompressure-reducing orices

Multiple orifices can avert the cavitation, and resulting noiseand vibration, caused by single orifices in hydraulic systems

P.ztiik C. Tuisg .. d Miltd MQiuie, Ouarw Hydro

o Fig I illustrates flow through a sing orifice in apipe: P. is the pressure upstream of the orifice; P, thevena-contracta pressure; and P,. the pressure dow'n

stream of the orifice. The last pressure. Pj. must be

higher than the vapor pressure. P,. of the flowing fluid.If PA falls belou P,. bubbles form. the first stage of

c,,vitation. As the fluid moves downstream, the bubblescollapse under the higher pressure. Pg. in the final stageof ca-itation. The implosions generate noise and vibration. and accelerate the erosion of the piping []].

If P, is kept below P,. t'o-phase flow continues downstream as a flashing mixture without cavitation.

Cavitation criterionTo avoid orifice camitation. the orifice size selected for

the pressure-drop requirement must be checked againstan orifice cavitation index, which is then compared to a

cavitationlevel If the index is equal to. or higher than.

the cavitation level. the orifice size is acceptable.The index was defined b. Tullis and Govindarajan 12]:

0,- (P 4 - PY(.W. - Pd) - (pd - P.-AP (1)They also experimentally determined acceptable cavitation levels. These. ',. vs. orifice-to-pipe-dia. raitios. P. for%anouspipe sizes are shou n in Fig. 2. At these cavitationLevels,noise is io, and stead% and no damage will result.

When a single orifice fails to satisf' the cas'itationcriterion.two or more orifices in series are required. In multipke-orifice systems. the spacings between orifices mustallow complete pressure recover)', which is achievedwhen'the circumference of the orifice jet expanding toward the pipe ^-allequals that of the inside pipe. Fig. 3relates these distances to P [3].

Design of single-orifice systemsTo design a single-orifice system free from cavitation:1. Size the orifice for the required flowrate and pres

sure drop by means o7 !-. (2):Ah- - X(V412)p (2)

Determine the resistance factor. K. for the orifice byinserting the floeate and orifice pressure drop into Eq.(2). Having the K value, find the orifice size from Fig.1LX"

2. Check that the orifice sizeobtabwd from Step I sat-

is'ies the cavitation criterion. Calculate the cavitationindex. o'. for the orifice %ia Eq (1), and get the corresponding acceptable cavitation level, a,,. from Fig.Compare a' to a',: If a ý c',. the size of the single orificeacceptable; if o < a,. multiple orifices are required

Design of multiple-orifice systemsEach orifice in a multiple system (illustrated in Fig.

is sized to meet the c•vitation criterionI. Assume a P value for the fast orifice. OR 1 (the o

furthest downstream). With this 9 value, determine tcorresponding resistance factor. K. from Fig 5.

2. Substitute the value for K obtained in Step I in

Eq. (2) to find the AP across OR,.3. Substitute this calculated AP into Eq. (1) to detemine the ca2itation index. a,.

4. With the Pvalue assumed in Step 1. find the accepable caiLtation leveL o in Fig. 2.

5. Compare oto o',. If ao:q the assumed 0 i satfactory. (To have the minimum number of orifices forspecific flourate and pressure-drop requirement. tvalue of a should be kept as dose as possible to Or.)0r< 0o, choose a larger p value and repeat Stepsthrough 5 until a'o'.r

6. The downstream pressure. PIM. of the second o

CMUKAL LXWALLIUWG ZNWEMM U. Mo

Pressure profile of a single-orifice system Fig.

I~P. pP.

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Acceptable cavitation level related -to orifice-to-pipe-dia. ratio 121 , . Fig. 2

- •. ... - .• ,

, .. --: . 2•.cL

I..:

0.2 -

1 -0 2~

vistarne downstfeano

for compltce pressure ri

I"\ I ' I--------- P11

1 I IOR3 OR2 oR,

A Pipe flow W=. VX Resistance factor!n Man, Gowra. igh

p~d 3akpreuuur. L~a

P. OrirKr upavam prm=a. kPMP. Vapor prmur. M•

A, Pipe internial radlus,-V Tho.wVktiz). =Vsa Dinanc for orie jet drcuzmfu to fm pipe &

¢mlference. =I Du-ner rato o oroif/"tpipe

•-•). hem4r Cavitation budesw, AmzpLabk tavisaogkevel

fKce. OR-,. is set equal to the uptream pressure. PI..the first orifice. OR,. etc. (as indicated in Fig. 4)

7. The last orifice (the one funhest uPTreim) is sizas was described for a single-orifice system

8. After all the orifices have been sized. determine tneccssarý distances between the orifices for cavitatio

free operation b%means of Fig. 3. For example. find t

proper distance bet%een ORI and OR2 (Fig 4j. enter t

value of J for OR 2 into Fig. 3 and obtain the correspond

ing value of x!Rr Having the pipe radius. R,. determinthe proper distance;-x, between OR, and OR.

Example illustrates the procedureDesign an orifice system for a 186 0-kg s floTate

water through a NPS12 Sch. 80 pipe. The upstreapressure and temperature are 10.340 kPa and 138respec-ve.. and the orifice backpressure is 690 IPi Tvapor pressure and density of water at I38'C: are 340and 928.0 Ig'm'. respectivel.. The pipe's inside radius

0.145 m. and its flou' area is 0.066 mi.First determine the flow velocity through the pipe

V - m'pA = 186.0:(928 00X0.066, - 3.0 m s

1. Trý a single orifice, sizing it via Eq. (21.(10.340 - 690() I.SM = A(3.0"2] 92F 0

SWith K - 2.300. P = 0.186. from Fig. 5.

2. Check whether the orifice size P= 0.186 sati

the cavitation criterion. Via Eq. (1):arn (69 - 340)'00.340 - 690. = 0.036

With P - 0.186. Fig 2 gives: a, 1I.I.

As (a -- 0.036) < (ct- 1.1). one onficr systemnot meet the criterion, and a multiple s%stem musdesigned:

l. Assume that 1, = 0.4 for the first orifice. OR,one furthest downstream). With this 191.A - 90

2. Determine the pressure drop via Eq (21.AP - 90 1(3.0)2f2]92.0 - 375.8 LRa

3. Substitute this AP value into Eq. (1):

a, - (690 - 340)f375.8 - 0.93

4. With ,0.41 ,A ,-.3, from Fig. 5.

5. Because (0, - 0.9) < (q, - 1.3). the cavitationriOn has no( been met. Tr" a larger orifice size--e

P-Ps - 0.45. Repeating the "foregoing procedure" -,1.62-and a, = 1.6. which satisf% thecrflterion.

-6The sizes of the remaining orifices are determi

SPr:or'usl described-i.e. the downstream press

i

(Ca Tr1 "I

afa multiple-orifice system Fig 4

I

Cavitation Trng.AttachmentPage 2 of 3

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10 0.1O-OWAI

RetationshiP of rnistame factor to cwific&-to-pipe-dia. ratio for "rp-edged orificn 141

CKLUSCAL gw.&ýP*G DICU"" $2. IM3

cavitation Trng.AttachmentPage 3 of 3

DIM a 0

fig. S

Ibe, siýravick C lr;ýj'ý des;r eftw-M wWOrwww Hvdro (700 L ftwrmn.-ACTorama Cku-MSG Dfi-CjMWA,,:,I plimot 416-W--=W HOVý7.-ý - ImpmuWam wchý V-d

Md Lr dnwn of PTUM %-k-p- -and mWed eq-Prwww2HOkI" OfAwwd'W&Cdearrwrmw from shi-ý

t;nrmm *(%%w=wn wWw- USA,; In= the Vwwrwn Of M-wt- Ow

a '70"'d CArrw" in Omano. a VwvIkw of the Anwn= Sm 0( M-M--JJ'r'w' WWthe Othm O(wrrAImuck, in wdwwW Pubb

WICU MAUAWOM S & wneac deW",pm-1 ý,h OnL&rw14-ldfuH*map--b-b- widudedrWn ww.. "Mm of Vwjw-2' "Pup-"

,d d..---J P-- Wj'(' IS " wwC"r-Kw CDC--q

imm aw U-V'w- -( bar'IdeTwjadrAL jw a a =,ed

Lw=Onmb" of Lhe . S:

Im rni.-r-w b'r- a'd Owa`St a Undboa affhad r-,T.Imbýwl MWO(bM--'

7

p2d. of the d OrifiCC- OR2. is Wt equal to dW UPSET'CaMpressure. PI.. of the first orifikx. Mi. CM

7. Size the final oTifice. as his been dcscrR)ed.8. Lasd). find the dis=ccs beturctm the orifices.

J. Adadry. EAw

References1. Jaime. Ir-C&wwxanin eancral Vamjawww- ant Ca-d5rWars. VaL34.

No.vmbrr 1961

2 TW6 j ?..NW GOVWAMYULR. Cavaum WWSOW-5-kVISM SMOr6CCL:'- j dfPiv"wfir Dn. Anwn= Sm. Ci Crei Lngwwcm MwCh 1273

T"Wd=. a G- MW%-*m U P. -1-M ri, ei L'w Ve" GWWN=7AwwrIMASm. of MechanciJ Engvwo-LP*v he 73-%AITM-9

4. XaAwwkv- U- mid Twq. T C- Ssmg of TW*mbag Onfa=L.Saw%4PYOW Ar GPft&bowwC.D"mub" IM LwAvr=mn.Wj-wd 'Aabs.L W- nuid TlwwTbrwACIýTwoorifiawis soon.Toom Awný Sm 4(XFAwmwdLnmwm jounn 1157

OLIcnpffwL M 'Wýand MMn. j G- The Fkp- of Sowrxwd VAw TbrONOThrombM CwVcm TvoL Amomm SK #f AdmArmait LegwAm JWY M 1

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