Introduction to Pump Curves Gould Pumps

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Introduction to Pump Curves

A performance curve is nor-mally a curved line drawn ona grid or graph of vertical andhorizontal lines. This curvedline represents the perfor-mance of a specific pump. Thevertical and horizontal linesrepresent units of measurethat illustrate that perfor-mance.

In our application there is atank or well full of water. Wewant to use the water for a

Introduction to Pump Curves

specific process or in a home.Often the water is at a lowerlevel and gravity will notallow it to flow uphill, so apump is used. A pump is amachine used to transfer ormove a volume of water (orfluid) a given distance. Thisvolume is measured over aperiod of time expressed ingallons per minute (gpm) orgallons per hour (gph). Thisvolume is also referred to ascapacity or flow.

The pump develops energycalled discharge pressure ortotal dynamic head (tdh). Thispressure is expressed in unitsof measure called pounds persquare inch (psi) or feet ofhead(ft.). NOTE: 1 psi willpush a column of water up apipe a distance of 2.31’. Aperformance curve is used todetermine which pump bestmeets the system require-ments.

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Example 1Example 1

The graph below would beused to illustrate a pump’sperformance. It is importantto determine the value of eachgrid line or square. Onthe lefthand side of the graph TotalDynamic Head (TDH) isshown with the unit of mea-sure in feet. The numbers startat the bottom, left hand cor-ner with 0 and go up thevertical axis. This is the abilityof the pump to producepressure expressed in feet of

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head, which is a term used bymany engineers. Sometimesthe measure will say totaldynamic head -- feet of water.This is another term whichmeans a gauge was installed atthe discharge of the pump anda reading was taken in psi.This reading is converted tofeet (1 psi = 2.31 feet) andwater was the liquid beingpumped.

The other unit of measure is

gallons per minute and isshown across the bottom(horizontal axis). Start at thebottom, left corner with 0 andgo to the right. These numbersrelate to the ability of thepump to produce flow ofwater expressed as capacity ingallons per minute (GPM).

Also shown are the metricmeasures in meters for theTDH and cubic meters perhour for the capacity.

4

To produce a performancecurve the pump is operatedusing a pressure gauge, throt-tling valve and flowmeterinstalled in the discharge pipe.The pump is run with thethrottling valve completelyclosed so there is no flow anda reading is taken from thepressure gauge. This is re-

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Example 2aExample 2a

ferred to as the maximumshutoff pressure. Covert thispsi reading (1 psi = 2.31 ft.)to feet of head and this is thepump’s maximum Total Dy-namic Head (TDH) at 0capacity. A mark is placed onthe graph to indicate thisperformance (Point 1). Thevalve is opened until the

flowmeter reads 5 gpm andanother pressure reading istaken from the gauge. Asecond point (2) is marked onthe graph to indicate thisperformance. This process iscontinued at 5 gpm incre-ments across the graph.

5

We now connect all thepoints. This curved line iscalled a head/capacity curve.

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Example 2bExample 2b

Head (H) is expressed in feetand capacity (Q) is expressedin gallons per minute (GPM).

The pump will always runsomewhere on the curve.

6

Example 3

There are many different typecurves shown in our catalog.Here is a composite perfor-mance curve (more than onepump) for the 18GS perfor-mance submersible. There is aseparate curve for each horse-power size.

Example 3

Let’s compare two sizes:

1. First look at the 18GS07,3/4 HP, 6 impellers. At 15GPM capacity this modelwill make 158 feet.

2. Now look at the 18GS20,2 HP, 14 impellers. At 15GPM capacity this modelwill make 360 feet.

When you add impellers, thepump makes more pressure(expressed in feet). Thisallows the pump to go deeperin a well, but also takes morehorsepower.

COMPOSITE PERFORMANCE CURVES

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20Ft.

1GPM

RECOMMENDED RANGE6 – 28 GPM

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Example 4

Here is a different kind ofcurve you will find used onsome grids called “efficiencycurve”. Efficiency as a per-centage is shown on the scaleat the right side of the grid.

1. Find 425 GPM at 500’.From this point read verti-cally UP until you touch theefficiency curve line. Thengo horizontally to the rightto find the percent. In thisexample it would be 72%.

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Example 4

2. Find 425 GPM at 800’.From this point read verti-cally DOWN until youtouch the efficiency curveline. Then go horizontallyto the right to find thepercent. In this example itwould be 72%.

The highest point of theefficiency curve line is calledthe “B.E.P.” or “Best Effi-ciency Point”.

Pump manufacturers describepumps in terms of the flowrate at the best efficiencypoint. In Example 4, thepump would be classified as a425 gallon per minute pump.

We have included a sheetcalled (4A) “How to figurehorsepower and operatingcost”. You may want to makecopies and pass this out. Thissheet shows why, with largerpumps, you want to select theone that has the best effi-ciency.

EFFICIENCY CURVE LINE SHOWN BY ITSELF

8

How to Figure Horsepower andOperating Cost

BHP = Brake Horsepower 3960 = ConstantSp. Gravity = Specific Gravity of Water is (1) WATTS = Volts x Amps1 HP = 746 Watts

Note: No motor is 100% efficient, so we say 1 HP = 1000 Watts which is (1) kilowatt or KW

The formula for figuring brake horsepower (BHP) is as follows:

GPM x Total Dynamic Head (TDH) x Sp. Gravity 3960 x Efficiency (Decimal)

Select a submersible pump for the following application:100 GPM at 375 ft. TDH 460V

There are several which will meet this application:90L15-11 65% eff. 15 HP motor (14.8 HP)100H15-8 67% eff. 15 HP motor (16.4 HP)150H15-6 67% eff. 15 HP motor (14.1 HP)5CLC015-9 74% eff. 15 HP motor (12.8 HP) Note: Impeller trim to 375’ TDH

90L15 = 14.8 HP 5CLC015 Cost = $4,6005CLC015 = 12.8 HP 90L15 Cost = 4,280Difference = 2.0 HP Cost difference = $ 320

We assume — 1 kilowatt hour (KWH) cost $ .10/KWH. We multiply the HP difference 2.0 HP x $.10 and the Cost Savings per hour $ .20. If the pump runs 10 hours per day then the cost savings is$2.00 per day. In 30 days, the savings is $60. Take the cost difference and divide by the cost savingsper day ($320 / $2.00) and the difference in price is paid back in 160 days.

How to Figure Horsepower andOperating Cost

9

Find the point 350 GPM at300 feet. What is the effi-ciency? 70% is correct. Whenyour capacity/head that youplan to use falls between twohorsepower sizes, the curveline you always use is the oneabove or the larger horse-power.

We have decided to use the 40HP pump and our head re-quirements are 300 feet. Wehave to install a throttlingvalve, and throttle the pumpcapacity to 350 GPM. If wedon’t, the pump will run witha capacity of 420 GPM.

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10 HP/1 STG.

COMPOSITE PERFORMANCE CURVES

Example 5Example 5

To determine this find 350GPM at 300’ again. From thispoint go horizontally to theright until you touch the 40HP, H-Q curve line. A pumpwill always run somewhere onits curve. At 420 GPM what isthe pump’s efficiency? 72%.

There is one other consider-ation called upthrust, whichmay occur when running thepump at open discharge. Notethat each curve line stops justpast 575 GPM. Under normalconditions the thrust in asubmersible pump is down-wards, toward the motor. At

abnormally high flows, whichresults when there is insuffi-cient head on the pump, thisthrust is upset and goes theopposite way because theimpeller is not developingsufficient head. This is calledupthrust. To prevent this,install a throttling valve andthrottle the pump to 575GPM when running the pumpopen discharge.

10

Here is a curve that showsHead/Capacity and also showssuction lift.

The upper left part of thecurve is divided into twoparts. Find 4 GPM and readvertically up. If your suctionlift is 20’ or 25’ the pump willmake 114 feet of head. If yoursuction lift is 5’, 10’ or 15’,the pump will make a littlemore — 119 feet.

We always show what thepump will do at all points onthe curve. We would not

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Example 6Example 6

recommend or select thispump to run below 16 gallonsper minute.

The vertical curve lines on theright also relate to suction liftin 5’ increments. Find 45GPM at 70’. This pump willwork if your suction lift is 22’or less. Remember as yourcapacity increases, your suc-tion lift ability decreases.

When considering capacity, ingallons per minute, and thesuction lift you are trying toovercome, you must stay to

the left of the vertical suctioncurve line.

Find 45 GPM at 70’ again. Ifyou had a 20’ lift to over-come, this pump would workfine.

Find 50 GPM at 70’. If youhad a 20’ lift to overcome, thispump will not work. You areto the right of the 20’ suctioncurve line. The next higherhorsepower model may be thepump to meet this require-ment.

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Example 7

Here is a new curve lineshown called BHP (BrakeHorsepower). BHP is chartedto the right just past efficiencyscale.

Find 110 GPM at 71’. Fromthis point read vertically downuntil you touch the BHP curve

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LIFT 20’ 15’ 10’ 5’6.1m 4.6 m 3.1 m 1.5 m

H - Q

Model 3656 SP3HP ODP5 1/8 dia. Imp.

Example 7

COMPOSITE PERFORMANCE CURVESMODEL 3656 SP 3500 RPM

line. Then go horizontally tothe right to find the horse-power required. In this case, itis 3.1 BHP. The motor used is3 HP with 1.15 service factor(3 x 1.15 = 3.45). We reallyhave 3.45 HP to use.

Find 110 GPM at 71’ again.What is the efficiency? 66% iscorrect.

12

Example 8

Pictured is a single line curvefor a centrifugal pump.

Find 70 GPM at 56 ft. Sincethe point we need is below thehead/capacity curve line, thepump will exceed our require-ments.

We can have the pump meetour exact requirements byproper trimming of the impel-ler. This will establish a newcurve for the pump. The curvewill run approximately paral-lel to the curve shown andthrough the point required;i.e., 70 GPM at 56 ft.

What happens if we don’t trimthe impeller to meet our exactrequirements? The pump willrun somewhere on its curve.Our head requirement is 56 ft.Follow 56 ft. horizontally tothe right until you meet thecurve line. Your capacity is 80

Example 8

GPM. Can this installationhandle the additional capac-ity? Notice also that efficiency(EFF), brake horsepower(BHP), and net positive suc-tion head required (NPSHR)also change, with the capacityof 80 GPM.

If we use a throttling valve onthe discharge side of the pumpwe can regulate the capacity ofthe pump to 70 GPM usingthe standard impeller. Readvertically up from 70 GPMuntil you meet the curve line.The pump will make 581⁄2 ft.Can this installation handlethe extra head?

Remember, trimming theimpeller will make the pumpmeet your exact requirements,70 GPM at 56 ft. (The head/capacity requirements shouldalways be furnished when thepump is ordered.)

What other information isshown? Efficiency, brakehorsepower, and net positivesuction head required. All ofthese values are measured onthe left hand side of the grid.

Find 70 GPM at 56 ft. again.

What is the efficiency? 70.5%is correct.

What is the brake horsepowerrequired? 1.5 BHP is correct.

What is the net positive suc-tion head required?

NPSHR = 2.2 ft. You will haveto determine the NPSHA.

NPSHA = Net Positive SuctionHead available.

NPSHA has to be greater thanNPSHR for the pump to work.

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Example 8Example 8continued

NET POSITIVE SUCTION HEAD (NPSH) AND CAVITATION

Centrifugal Pump Fundamentals

The Hydraulic Institute defines NPSHas the total suction head in feetabsolute, determined at the suctionnozzle and corrected to datum, lessthe vapor pressure of the liquid in feetabsolute. Simply stated, it is ananalysis of energy conditions on thesuction side of a pump to determine ifthe liquid will vaporize at the lowestpressure point in the pump.The pressure which a liquid exerts onits surroundings is dependent upon itstemperature. This pressure, calledvapor pressure, is a unique character-istic of every fluid and increases withincreasing temperature. When thevapor pressure within the fluidreaches the pressure of the surround-ing medium, the fluid begins tovaporize or boil. The temperature atwhich this vaporization occurs willdecrease as the pressure of thesurrounding medium decreases.A liquid increases greatly in volumewhen it vaporizes. One cubic foot ofwater at room temperature becomes1700 cu. ft. of vapor at the sametemperature.It is obvious from the above that if weare to pump a fluid effectively, wemust keep it in liquid form. NPSH issimply a measure of the amount ofsuction head present to prevent thisvaporization at the lowest pressurepoint in the pump.NPSH Required is a function of thepump design. As the liquid passesfrom the pump suction to the eye ofthe impeller, the velocity increasesand the pressure decreases. There arealso pressure losses due to shock andturbulence as the liquid strikes the

impeller. The centrifugal force of theimpeller vanes further increases thevelocity and decreases the pressure ofthe liquid. The NPSH Required is thepositive head in feet absolute requiredat the pump suction to overcomethese pressure drops in the pump andmaintain the liquid above its vaporpressure. The NPSH Required varieswith speed and capacity within anyparticular pump. Pumpmanufacturer’s curves normallyprovide this information.NPSH Available is a function of thesystem in which the pump operates. Itis the excess pressure of the liquid infeet absolute over its vapor pressureas it arrives at the pump suction. Fig.4 shows four typical suction systemswith the NPSH Available formulasapplicable to each. It is important tocorrect for the specific gravity of theliquid and to convert all terms tounits of “feet absolute” in using theformulas.In an existing system, the NPSHAvailable can be determined by a gagereading on the pump suction. Thefollowing formula applies:

NPSHA = PB - VP ± Gr + hV

WhereGr = Gage reading at the

pump suction expressed infeet (plus if above atmo-spheric, minus if belowatmospheric) corrected tothe pump centerline.

hv= Velocity head in the suctionpipe at the gage connection,expressed in feet.

Cavitation is a term used to describethe phenomenon which occurs in apump when there is insufficientNPSH Available. The pressure of theliquid is reduced to a value equal toor below its vapor pressure and smallvapor bubbles or pockets begin toform. As these vapor bubbles movealong the impeller vanes to a higherpressure area, they rapidly collapse.The collapse, or “implosion” is sorapid that it may be heard as arumbling noise, as if you werepumping gravel. The forces duringthe collapse are generally high enoughto cause minute pockets of fatiguefailure on the impeller vane surfaces.This action may be progressive, andunder severe conditions can causeserious pitting damage to the impel-ler.The accompanying noise is the easiestway to recognize cavitation. Besidesimpeller damage, cavitation normallyresults in reduced capacity due to thevapor present in the pump. Also, thehead may be reduced and unstableand the power consumption may beerratic. Vibration and mechanicaldamage such as bearing failure canalso occur as a result of operating incavitation.The only way to prevent the undesir-able effects of cavitation is to insurethat the NPSH Available in the systemis greater than the NPSH Required bythe pump.

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NET POSITIVE SUCTION HEAD (NPSH) AND CAVITATION

4a SUCTION SUPPLY OPEN TO ATMOSPHERE– with Suction Lift

4b SUCTION SUPPLY OPEN TO ATMOSPHERE– with Suction Head

4c CLOSED SUCTION SUPPLY– with Suction Lift

4d CLOSED SUCTION SUPPLY– with Suction Head

NPSHA = PB – (VP + LS + hf)

NPSHA = PB + LH – (VP + hf)

NPSHA = p – (LS + VP + hf)

NPSHA = p + LH – (VP + hf)

p

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CL

PB = Barometric pressure, in feet absolute.VP = Vapor pressure of the liquid at maximum pumping temperature, in feet absolute (see page 16).p = Pressure on surface of liquid in closed suction tank, in feet absolute.LS = Maximum static suction lift in feet.LH = Minimum static suction head in feet.hf = Friction loss in feet in suction pipe at required capacity.Note: See Vapor Pressure Chart in Technical Manual.

p

Centrifugal Pump Fundamentals

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Centrifugal Pump FundamentalsVAPOR PRESSURE OF WATER

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Here is a centrifugal pumpcurve showing informationthat we have already covered,but in a different format.

This is a Model 3656, 11⁄2 x 2-6 ODP.

11⁄2 = Discharge size in inches2 = Suction size in inches6 = Basic diameter of the

impeller in inches.

The top head/capacity curveline shows an actual impellerdiameter of 515⁄16 inches.

Our requirement is 140 GPMat 95 ft. We can properly trimthe impeller to meet ourrequirement, since the pump

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MODELS 3656/3756 S GROUP PERFORMANCE CURVES

Example 9Example 9

with the 5-15/16 inches diam-eter will more than meet ourneeds.

What happens if we install thepump and don’t trim theimpeller? Our head require-ment is 95 ft. Follow 95 ft.horizontally to the right untilyou meet the curve line. Yourcapacity is 157 GPM.

If we use a throttling valve onthe discharge side of the pumpwe can regulate the capacity ofthe pump to 140 GPM usingthe standard impeller. Readvertically up from 140 GPMuntil you meet the curve line.The pump will make 109 ft.

Can the installation handlethis extra head?

Remember, trimming theimpeller to the proper diam-eter will make the pump meetyour exact requirements, 140GPM at 95 ft. (The head/capacity requirements shouldalways be furnished when thepump is ordered.)

Consider our original require-ment again: 140 GPM at 95ft. What is the brake horse-power required? A little lessthan 5 HP. What is the effi-ciency? 71% is correct. Whatis the NPSHR? 13 ft. is correct.

17

Example 10

To better understand brakehorsepower as shown in thesetwo curves, consider thefollowing: Brake horsepower(BHP) is the actual horse-power delivered to the pumpshaft. The formula for figuringbrake horsepower is:

Brake Horsepower =

GPM x Feet Head x SP.GR.3960 x Pump Efficiency

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Example 10

Compare the two curves at150 gallons per minute.Horsepower is given so wedon’t have to calculate what itis. The grid shows an ODPmotor (Open Drip Proof)which has a 1.15 servicefactor.

5 HP x 1.15 SF =

5.75 available horsepower

The 515⁄16” diameter impellerat 150 GPM uses more than 5HP, but does not exceed 5.75HP.

BHP =

150 x 100 x 1.0 = 5.37 BHP3960 x 70.5%


18

Example 11

Now look at the grid forTEFC motor (totally enclosed— fan cooled).

A TEFC motor has a 1.0service factor.

5 HP x 1.0 SF =

5 available horsepower

Example 11

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The 55/8” diameter impellerat 150 GPM does not exceedthe 5 HP line.

BHP =

150 x 82 x 1.0 = 4.64 BHP3960 x 67%

19

Example 12Example 12

A System Head Curve is theeasiest and most accurate wayto decide which pump is bestsuited for an application. Thefactors used to create a SystemHead Curve are gallons perminute, friction loss data andpipe size, total head (lift ordepth to water) and pressuredesired (expressed in feet).

To be able to pick the bestpump for any given job wemust be given some data.

Let’s look at a 50 gpm irriga-tion system.

Pumping Level - 250’ at 50gpmPump Set - 280’Well Depth - 300’Need 50 psi (115’) to operatethe sprinkler heads.Distance to 1st lateral is1000’.The only variable or control-lable item is the pipe frictionloss which varies with pipesize. We cannot change theflow, the pumping level, thepressure or the length of pipe.We will look at the differencebetween using 11⁄2”, 2” and 3”pipe.

Pipe GPM Friction Loss/ Times Plus 365’Size 100’ of 11⁄2” 12.8 Total Head11⁄2” 30 6.26 80 44511⁄2” 50 16.45 210 575 TDH11⁄2” 70 31.73 406 771

Pipe GPM Friction Loss/ Times Plus 365’Size 100’ of 2” 12.8 Total Head2” 30 1.81 23 3882” 50 4.67 60 425 TDH2” 70 8.83 113 478

Pipe GPM Friction Loss/ Times Plus 365’Size 100’ of 3” 12.8 Total Head3” 30 .26 4 3693” 50 .66 9 374 TDH3” 70 1.24 16 381

PUMP SYSTEM HEAD CURVEWe have 1000’ of offset pipeand 280’ of drop pipe for atotal pipe length of 1280’.Divide the 1280 by 100,because F.L. tables show theloss per 100’ of pipe, to get amultiplier of 12.8.

We add the Total Head to theF.L. to find the TDH. Plot theFlow/TDH on any curve toshow the System Head Curvefor that system.

Pumping Level 250’PSI in feet 115’Total Head 365’Friction Loss +

= TDH

You can see that there is a bigdifference in TDH due to thepipe used. The 11⁄2” pipewould require a 15 hp pump,the 2” and 3” can use a 7.5hp. The 2” pipe will give us46 gpm and the 3” will yield

53 gpm. The difference of 7gpm is 420 gph or 10,080gpd. If you do not need theextra water use 2” and savemoney on the pipe. If youneed all the water you can getthe 3” pipe will be much

cheaper over the life of thepipe than using a larger pump.There has never been a cus-tomer who wanted less waternext year than this year. Largepipe leaves room to grow thesystem if necessary.

20

Example 12continued

Example 12

GPM GPH3⁄8” 1⁄2” 3⁄4” 1” 11⁄4” 11⁄2” 2” 21⁄2” 3”ft. ft. ft. ft. ft. ft. ft. ft. ft.

25 1,500 38.41 9.71 4.44 1.29 .54 .1930 1,800 13.62 6.26 1.81 .75 .2635 2,100 18.17 8.37 2.42 1.00 .3540 2,400 23.55 10.70 3.11 1.28 .4445 2,700 29.44 13.46 3.84 1.54 .5550 3,000 16.45 4.67 1.93 .6660 3,600 23.48 6.60 2.71 .9370 4,200 31.73 8.83 3.66 1.2480 4,800 11.43 4.67 1.58

Plastic Pipe: Friction Loss (in feet of head) per 100 Ft.

Size of fittings, Inches 1⁄2” 3⁄4” 1” 11⁄4” 11⁄2” 2” 21⁄2” 3” 4” 5” 6” 8” 10”90° Ell 1.5 2.0 2.7 3.5 4.3 5.5 6.5 8.0 10.0 14.0 15 20 2545° Ell 0.8 1.0 1.3 1.7 2.0 2.5 3.0 3.8 5.0 6.3 7.1 9.4 12Long Sweep Ell 1.0 1.4 1.7 2.3 2.7 3.5 4.2 5.2 7.0 9.0 11.0 14.0Close Return Bend 3.6 5.0 6.0 8.3 10.0 13.0 15.0 18.0 24.0 31.0 37.0 39.0Tee-Straight Run 1 2 2 3 3 4 5Tee-Side Inlet or Outlet 3.3 4.5 5.7 7.6 9.0 12.0 14.0 17.0 22.0 27.0 31.0 40.0or Pitless AdapterBall or Globe Valve Open 17.0 22.0 27.0 36.0 43.0 55.0 67.0 82.0 110.0 140.0 160.0 220.0Angle Valve Open 8.4 12.0 15.0 18.0 22.0 28.0 33.0 42.0 58.0 70.0 83.0 110.0Gate Valve-Fully Open 0.4 0.5 0.6 0.8 1.0 1.2 1.4 1.7 2.3 2.9 3.5 4.5Check Valve (Swing) 4 5 7 9 11 13 16 20 26 33 39 52 65In Line Check Valve(Spring) 4 6 8 12 14 19 23 32 43 58or Foot Valve

Equivalent Number of Feet Straight Pipe for Different Fittings

21


0 5

20 40 60 80 10000

FLOW RATE

0

50

75

100

200

300

400

100

500150

700

TOTA

L D

YNA

MIC

HEA

D

METERS FEET

10

GPM0

10

20

30

40

50

60% EFF

175

25

125

600

200

15

55

45

35

25

15

5

m3/hr20

MODEL 55GSSIZE COMPOSITERPM 3450

225

250800

55GS15 – 5 STAGES

55GS20 – 7 STAGES

55GS30 – 9 STAGES

55GS75 – 22 STAGES



11/2” Pipe

2” Pipe

3” Pipe

22

Example 13Example 13

The System Head Curve canbe very steep when the frictionloss makes up a larger part ofthe Total Dynamic Head. Wesee this clearly on a sewagejob.

Total Lift 20’200’ of 2” Pipe3 Baths2” Solids Handling Pump

GPM Friction Loss x 200 = +Dis. Head (Lift) = TDHat GPM 100 (Total Dynamic Head)

20 .86 2 1.72 20’ 22’30 1.81 2 3.62 20’ 24’40 3.11 2 6.22 20’ 26’50 4.67 2 9.34 20’ 29’60 6.6 2 13.2 20’ 33’80 11.43 2 22.86 20’ 43’100 17.0 2 34 20’ 54’120 24.6 2 49.2 20’ 69’

As you can see from theplotted system curve it crosses5 pump curves. You can nowaccurately pick the best pumpfor the job or tell exactly howmuch a certain pump willpump in a given situation. It isa good way to show customersthat there is always more thanone pump for every applica-tion. It also shows the wisdom

of properly sizing the pipe tothe flow to save horsepowerand electricity.

2” PVC Pipe x 200’ long

SEWAGE SYSTEM HEAD CURVE

0 20 40 60 80 240 U.S. GPM0

10

20

30

40

FEET

TOTA

L D

YNA

MIC

HEA

D

0

5

10

15

METERS

FLOW RATE

SERIES: 3887BHF2" SOLIDSRPM: 3500Enclosed Impeller

100 120 140

50

0 3010 20 m3/h50

100

60

70

80

90

20

25

30

160 180 200 220

40

WS20BHF

WS15BHF

WS10BHF

WS07BHF

WS03BHF

5 FT

10 GPM

WS05BHF

20’ Discharge Head (VerticalLift)We need to pump at least 30gpm, 2” solids.

This curve showsthat we could use 5different pumps onthis job and pumpbetween 33 gpm at25’ TDH and 100gpm at 55’ TDH.

The pumps willoperate where thepump curve andsystem curve cross.The longer thedischarge pipe thesteeper the curve.

23

NotesNotes

24

©2001 Goulds PumpsPrinted in U.S.A.CINTRO3

SENECA FALLS, NEW YORK 13148(315) 568-2811

Goulds Pumps, and the ITT Engineered BlocksSymbol are registered trademarks andtradenames of ITT Industries.

www.goulds.com

Documents

Introduction to Pump Curves Gould Pumps