A BRIEF INTRODUCTION TO CENTR IFUGA L P U MP S Joe Evans, Ph.D This publication is based upon an introductory, half day class that I presented many years ago. It is designed to provid e the new comer with an entry level knowledge of centrifugal pump theory and operation. Of equal importance, it w il l make him aware of those areas that will require additional study if he is to become truly proficient. You will notice references to our “Puzzler”. These brain teasing discussions investigate the specifics of pumps, motors, and their controls and can be downloaded from the ed ucation section of ou r w eb site. This introduction, on the other hand, is much more structured a nd sticks to the basics. Some of the figures you will see have been redu ced in size substantially. If you have difficulty reading them, just increase their size using the Acrobat scaler. If you have comments or suggestions for improving our educational material, please email me at j eva ns @ p acific liqu id .com. INTRODUCTION Definition & Description B y d efi nition, a centrifugal pu m p is a machine. More specif ically, it is a machine that imparts energy to a fluid. This energy infusion can cause a liquid to flow, rise to a higher level, or both. The centrifugal pump is an extremely simple machine. It is a member of a f amily known as rotary machines and consists of two basic parts: 1) the rotary element or impeller and 2) the stationary elemen t or casing (volute). The figure below is a cross section of a centrifugal pu mp and shows the two basic parts. VOLUTE IMPELLER Figure 1 The centrifugal pump’s function is as simple as i ts design. It is fi ll ed w ith liquid and the impeller is rotated. Rotation impar ts energy to the liquid causing it to exit the impeller’s vanes at a greater velocity than it possessed w hen i t entered. This outw ard flow redu ces the pressure at the impeller eye, allowing more liquid to enter. The l iquid that exits the impeller is collected in the casing (volute) where its velocity is converted to pressure before it leaves the pu mp ’ s d isc harge. A Very Brief H istory The centrifugal pump was developed in Europe in the late 1600’s and was seen in the United States in the early 1 800’ s. Its w ide spread use, however, has occurred only i n the last seventy -f ive years. Prior to that time, the vast majority of pumping applications involved positi ve displacement pumps. The increased p opulari ty of centrif ugal p um ps is due largely to the comparatively recent development of high speed electric motors, steam turbines, and internal combustion engines. The centrifugal pum p is a relati vely high speed machine and the development of high speed drivers has made possible the developm ent of compact, ef fi cient pumps.
This publication is based upon anintroductory, half day class that I presentedman y years ago. It is designed to provid ethe new comer with an entry levelknowledge of centrifugal pump theory andoperation. Of equal imp ortance, it will makehim aware of those areas that will requireadditional study if he is to become trulyproficient.
You will notice references to our “Puzzler”.These brain teasing discussions investigatethe specifics of pumps, motors, and theircontrols and can be downloaded from theeducation section of our web site. Thisintroduction, on the other hand, is muchmore stru ctured and sticks to the basics.
Some of the figures you will see have beenredu ced in size substantially. If you havedifficulty reading them, just increase their sizeusing the Acrobat scaler. If you have
comments or suggestions for improving oureducational material, please email me at [email protected].
INTRODUCTION
Definition & Description
By definition, a centrifugal pu mp is a
machine. More specifically, it is a machinethat imparts energy to a fluid . This energyinfusion can cause a liquid to flow, rise to ahigher level, or both.
The centrifugal pump is an extremely simplemachine. It is a member of a family knownas rotary machines and consists of two basicparts: 1) the rotary element or impeller and 2)the stationary element or casing (volute). The
figure below is a cross section of a centrifugalpu mp and show s the two basic parts.
VOLUTE
IMPELLER
Figure 1
The centrifugal pump’s function is as simpleas its design. It is filled with liquid and theimpeller is rotated. Rotation impar ts energyto the liquid causing it to exit the impeller’svanes at a greater velocity than it possessedwhen it entered. This outw ard flow redu cesthe pressure at the impeller eye, allowingmore liquid to enter. The liquid that exits theimpeller is collected in the casing (volute)where its velocity is converted to pressure
before it leaves the pu mp ’s d ischarge.
A Very Brief History
The centrifugal pump was developed inEurope in the late 1600’s and was seen in theUnited States in the early 1800’s. Its widespread use, however, has occurred on ly in thelast seventy-five years. Prior to that time, thevast majority of pumping applicationsinvolved positive displacement pum ps.
The increased p opu larity of centrifugal pum psis due largely to the comparatively recentdevelopment of high speed electric motors,steam turbines, and internal combustionengines. The centrifugal pum p is a relativelyhigh speed machine and the development of high speed drivers has made possible thedevelopm ent of compact, efficient p um ps.
Since the 1940's, the centrifugal pump hasbecome the pump of choice for manyapp lications. Research and development hasresulted in both improved performance andnew materials of construction that havegreatly expand ed it's field of applicability. Itis not uncommon today to find efficiencies of
93%+ for large pumps and better than 50%for small fractional horsepow er u nits.
Modern centrifugal pumps have been built tomeet conditions far beyond what wasthought possible fifty to sixty years ago.Pumps capable of delivering over 1,000,000gallons per minute at heads of more than 300feet are common in the nuclear powerindu stry. And , boiler feed pump s have beendeveloped that d eliver 300 gallons per m inuteat m ore than 1800 feet of head .
THEORY
In operation, a centrifugal pump “slings”liquid out of the impeller via centrifugal force.Now centrifugal force, itself, is a topic of debate. Although I will not go into detailhere, it is considered by many, includingmyself, to be a false force. For our purposeshere, we will assume that it is a real force.Refer to the “False Force Puzzler” for moreinformation.
Centrifugal Force
A classic example of the action of centrifugalforce is shown below. Here, we see a pail of water swinging in a circle. The swinging pail
Figure 2
generates a centrifugal force that holds thewater in the pail. Now, if a hole is bored inthe bottom of the pail, water will be thrownout. The distance the stream carries (tangent
to the circle) and the volume that flows out(per un it time) depends up on the velocity ( inft/ sec) of the rotating pail. The faster the pailrotates the greater the centrifugal force andtherefore the greater the volume of waterdischarged and the distance it carries.
The description above could be consideredthat of a crude centrifugal pump (sans voluteof course). It dem onstrates that the flow andhead (pressure) developed by a centrifugalpu mp d epends upon the rotational speed and,more precisely, the peripheral velocity of itsimpeller (pail).
Periph eral Velocity & Head
Gravity is one of the more important forces
that a centrifugal pu mp m ust overcome. Youwill find that the relationship between finalvelocity, due to gravity, and initial velocity,du e to impeller speed, is a very useful one .
If a stone is dropped from the top of abuilding it's velocity will increase at a rate of 32.2 feet per second for each second that itfalls. This increase in velocity is known asacceleration due to gravity. Therefore if weignore the effect of air resistance on the fallingstone, we can predict the velocity at which itwill strike the ground based upon its initialheight and the effect of acceleration due togravity.
The equation that describes the relationship of velocity, height, and gravity as it applies to afalling bod y is:
For example if a stone is dropped from abuilding 100 feet high:
v2 = 2 x 32.2 ft/ sec2 x 100 ft
v2 = 6440 ft2 / sec2
v = 80.3 ft/ sec
The stone, therefore, will strike the ground ata velocity of 80.3 feet per second .
This same equation allows us to determinethe initial velocity required to th row the stoneto a heigh t of 100 feet. This is the casebecause the final velocity of a falling bodyhappens to be equal to the initial velocityrequired to launch it to height from which it
fell. In the example above, the initial velocityrequired to throw the stone to a height of 100feet is 80.3 feet per second, the same as itsfinal velocity.
The same equation applies when pumpingwater with a centrifugal pu mp . The velocityof the water as it leaves the impellerdeterm ines the head developed . In otherwords the water is “thrown” to a certainheight. To reach this height it mu st start withthe sam e velocity it would at tain if it fell from
that height.
If we rearrange the falling body equation weget:
h = v2 / 2g
Now we can determine the height to wh ich abody (or water) will rise given a particularinitial velocity. For examp le, at 10 Ft per Sec:
h = 10 ft/ sec x 10 ft/ sec / 2 x 32.2 ft/ sec2
h = 100 ft2 / sec2 / 64.4 ft/ sec2
h = 1.55 ft
If you were to try this with several differentinitial velocities, you would find out that thereis an interesting relationship between theheight achieved by a body and its initial
velocity. This relationship is one of thefundamental laws of centrifugal pumps andwe will review it in detail a little later. As afinale to this section, let's apply w hat w e havelearned to a practical app lication.
Problem: For an 1800 RPM pu mp , find the
impeller diameter necessary to develop ahead of 200 feet.
First we must find the initial velocity requiredto d evelop a head of 100 feet:
v2 = 2 gh v2 = 2 x 32.2 ft/ sec2 x 200 ft
V2 = 12880 ft2 / sec2
v = 113 ft/ sec
We also need to know the number of rotations the impeller und ergoes each second :
1800 RPM / 60 sec = 30 RPS
Now we can compute the number of feet apoint on the impellers rim travels in a singlerotation:
Once a pump has been d esigned an d is readyfor production, it is given a complete andthorough test. Calibrated instrum ents areused to gather accurate data on flow, head,
horsepower, and net positive suction headrequired. Dur ing the test, data is recorded atshu t off (no flow), full flow, and 5 to 10 pointsbetween. These points are plotted on a graphwith flow along th e X axis (abscissa) and headalong the Y axis (ord inate). The efficiency,brake horsepower, and Net Positive SuctionHead Required (NPSHR) scales are alsoplotted as ordinates. Therefore, all values of head, efficiency, brake horsepower, andNPSHR are plotted versus capacity andsmooth curves are draw n throu gh the points.
This curve is called the Characteristic Curvebecause it shows all of the operatingcharacteristics of a given pu mp . The curvebelow is an example.
Figure 3
For publication purposes, it is much moreconvenient to d raw several curves on a singlegraph. This presentation method shows a
number of Head-Capacity curves for onespeed and several impeller diameters or oneimpeller diameter and several differentspeeds for the same pu mp . This type of curve is called an Iso-Efficiency or Comp ositeCharacteristic Curve. You will note that theefficiency and horsepower curves arerepresented as contour lines. Also theNPSHR curve applies only to the Head-Capacity curve for the full size impeller.
NPSHR will increase somewhat for smallerdiameters. The curve below is a typicalComp osite Curve.
Figure 4
The method of reading a performance curveis the same as for any other graph . Forexample; in Figure 3, to find the head,horsepower and efficiency at 170 GPM, youlocate 170 GPM on the abscissa and read th ecorresponding data on the ordinate. Thepoint on the head / capacity curve that alignswith the h ighest point on the efficiency curveis known as the Best Efficiency Point or BEP.In Figure 3, it occurs at approximately 170GPM @ 125’TDH .
The composite curve shown in Figure 4 isread in much the same manner. Head is readexactly the same way; however, efficiencymu st be interpolated since the Head-Capacitycurve seldom falls exactly on an efficiencycontour line. Horsepower can also beinterpolated as long as it falls below ahorsepow er contour line. If the Head-Capacity curve intersects or is slightly abovethe horsepower contour line, brake
horsepower may need to be calculated.NPSHR is found in the same manner as head.We will discuss N PSHR in d etail a little later.
Calculating Brake Horsepower (BHP)
Usually, if a specific Head-Capacity poin t fallsbetween two horsepower contour lines, thehigher horsepower motor is selected.Sometimes, however, we may need to know
the exact horsepower requirement for thatpoint of operation. If so Brake BHP for acentrifugal pump can be calculated as follows:
BHP = GPM X Head / 3960 X Efficiency
For example, in Figure 4, the BHP required at
170 GPM for the 5 1/ 2" impeller is:
BHP = 170 X 90 / 3960 X .74
BHP = 15300 / 2860.4
BHP = 5.22
Many operating points will fall between thevarious curves, so it is important that youunderstand a composite curve well enough tointerpolate and find the approximate values.
A p um p is typically d esigned for one specificcondition but its efficiency is usually highenough on either side of the design point toaccommodate a considerable capacity range.Often, the middle one third of the curve issuitable for application use.
Different pumps, although designed forsimilar head and capacity can vary widely inthe shape of their Characteristic Curves. Forinstance, if two pumps are designed for 200GPM at 100' TDH, one may develop a shut off head of 110' while the other may develop ashu t off head of 135'. The first pump is said tohave a flat curve while the second is said tohave a steep curve. The steepness of thecurve is judged by the ratio of the head atshut off to that at the best efficiency point.Each type of curve has certain applications forwhich it is best suited.
Op eration in Series (Booster Service)
When a centrifugal pump is operated with apositive suction pressure, the resultingdischarge pressure will be the sum of thesuction pressure and the pressure normallydeveloped by the pump when operating atzero suction pressure. It is this quality of acentrifugal pump that makes it ideally suitedfor use as a booster pu mp . This quality alsomakes it practical to build multi-stage
(multiple impeller) pu mp s. A booster pu mptakes existing p ressure, whether it be from anelevated tank or the discharge of anotherpu mp , and boosts it to some higher pressure.
Two or more pumps can be used in series toachieve the same effect. The figure below
shows the curves for two identical pumpsoperated in series. Both the head andhorsepower at any given point on thecapacity curve are add itive. Capacity,however, remains the same as that of eitherone of the pu mp s.
Figure 5
Parallel Operation
Two or more pu mp s may also be operated inparallel. The curves developed d uring paralleloperation are illustrated below.
Pum ps op erating in parallel take their suctionfrom a common header or supply anddischarge into a common discharge. AsFigure 6 illustrates, the flows and horsepow erare add itive. You w ill notice that, while headdoes not change, flow is almost doubled atany given point.
THE AFFINITY LAWS
The Centrifugal Pump is a very capable andflexible machine. Because of th is it isunnecessary to design a separate pump foreach job. The per forman ce of a centrifugalpump can be varied by changing the impellerdiameter or its rotational speed. Eitherchange produces approximately the sameresults. Reducing impeller diameter is
probably the most common change and isusu ally the most economical. The speed canbe altered by changing pulley diameters orby changing the speed of the dr iver. In somecases both speed and impeller diameter arechanged to obtain the d esired results.
When the driven speed or impeller diameterof a centrifugal pump changes, operation of the pump changes in accordance with threefund amental laws. These laws are known asthe "Laws of Affinity". They state that:
1) Capacity varies directly as the change inspeed or imp eller diameter
2) Head varies as the square of the change inspeed or imp eller diameter
3) Brake horsepower varies as the cube of the change in speed or imp eller diameter
If, for example, the pump speed or impellerdiameter were doubled:
1) Capacity will dou ble
2) Head will increase by a factor of 4 (2 tothe second p ower)
3) Brake horsepow er will increase by a factorof 8 (2 to the third pow er)
These principles apply regardless of thedirection (up or down) of the speed orchange in diameter.
Consider the following example. A pu mpoperating at 1750 RPM, delivers 210 GPM at75' TDH, and requires 5.2 brake horsepower.
What will happen if the speed is increased to2000 RPM? First we find th e speed ra tio.
Speed Ratio = 2000/ 1750 = 1.14
From the laws of Affinity:
1) Capacity varies directly or:
1.14 X 210 GPM = 240 GPM
2) Head varies as the square or:
1.14 X 1.14 X 75 = 97.5' TDH
3) BHP var ies as the cube or:
1.14 X 1.14 X 1.14 X 5.2 = 7.72 BHP
Theoretically the efficiently is the same forboth conditions. By calculating several pointsa new curve can be draw n.
Whether it be a speed change or change inimpeller diameter, the Laws of Affinity giveresults that are approximate. The discrepancybetween the calculated values and the actualvalues obtained in test are due to hydraulicefficiency changes that result from themod ification. The Laws of Affinity givereasonably close results w hen the changes arenot more than 50% of the original speed or15% of the original d iameter.
Here, we have only described the affinitylaws and applied them to a few examples.For an in depth discussion and proof of theirvalidity, see the “Affinity Pu zzler”.
The Specific Gravity of a substance is the ratioof the weight of a given volume of thesubstance to that of an equal volume of waterat standard temperature and pressure (STP).Assuming the viscosity of a liquid is similar tothat of water the following statements willalways be true regardless of the specificgravity:
1) A Centrifugal pum p will always developthe same head in feet regardless of a liquid’sspecific gravity.
2) Pressure will increase or decrease in directpropor tion to a liquid ’s specific gravity.
3) Brake HP required will vary d irectly with aliquid’s specific gravity.
The Figure below illustrates the relationshipbetween pressure (in psi) and head (in ft) forthree liquids of differing sp ecific gravity.
Figure 7
We can see that the level in each of the threetanks is 100 feet. The resulting pressu re at the
bottom of each varies substantially as a resultof the varying specific gravity. If, on theother hand we keep pressure constant asmeasured at the bottom of each tank, thefluid levels will vary similarly.
A centrifugal pump can also develop 100' of
head when pumping water, brine, andkerosene. The resulting pressures, however,will vary just as those seen in Figure 7. If thatsame pump requires 10 HP when pumpingwater, it will require 12 HP when pumpingbrine and only 8 HP when pumpingkerosene.
The preceding discussion of Specific Gravityillustrates why centrifugal pump head (orpressure) is expressed in feet. Since pu mpspecialists work w ith many liquid s of varying
specific gravity, head in feet is the mostconvenient system of designating head.When selecting a pump, always rememberthat factory tests and curves are based onwater at STP. If you are working with otherliquid s always correct the HP required for thespecific gravity of the liquid being pu mp ed.
The Effect of Viscosity
Viscosity is a fluid property that isindepen dent of specific grav ity. Just asresistivity is the inherent resistance of aparticular conductor, viscosity is the internalfriction of a fluid. The coefficient of viscosityof a fluid is the measure of its resistance toflow. Fluid s hav ing a high viscosity aresluggish in flow. Examples include molassesand heavy oil. Viscosity usually varies greatlywith temperatu re with viscosity decreasing astemperatu re rises.
The instrument used to measure viscosity is
the viscometer. Although there are many, theSaybolt Universal is the most common. Itmeasures the time in seconds required for agiven quantity of fluid to pass through astandard orifice under STP. The un it of measurement is the SSU or Seconds SayboltUniversal.
High viscosity can gum up (pun intend ed) theinternals of a centrifugal pu mp . Viscous
liquids tend to reduce capacity, head, andefficiency while increasing the brake HP. Ineffect this tends to steepen the head - capacityand HP curves while lowering the efficiencycurve.
The performance of a small Centrifugal
Pum p, han dling liquids of various viscosities,is shown graph ically in the figure below.
Figure 8
Normally, small and medium sizedcentrifugal pumps can be used to handleliquids with viscosities up to 2000 SSU. Below50 SSU the Characteristic Curves remainabout the same as those of water; however,there is an imm ediate decrease in efficiency
when viscosity increases over that of water.Viscosities over 2000 SSU are usually bettersuited for positive displacement pum ps.
FRICTION AND FRICTION HEAD
Friction occurs when a fluid flows in oraround a stationary object or when an object
moves throu gh a fluid . An automobile andan aircraft are subjected to the effects of friction as they move through ouratmosph ere. Boats create friction as theymove throu gh the water. Finally liquid screate friction as they move through a closedpipe.
A great deal of money has been spent on thedesign and redesign of boats, aircraft, andautos to reduce friction (often referred to asdrag). Why? Because friction produces heatand where there is heat there is energy,wasted energy that is. Making these vehicles"slippery" reduces friction therefore reducingthe energy required to get them from point Ato point B.
As water moves through a pipe its contact
with the pipe wall creates friction. As flow (ormore correctly velocity) increases, so doesfriction. The more water you try to cramthrough a given pipe size the greater thefriction and thus the greater the energyrequired to push it through. It is because of this energy that friction is an extremelyimportant component of a pum ping system.
Laminar flow describes the flow of a liquid ina smooth pipe. Und er conditions of laminarflow, the fluid nearest the pipe wall movesmore slowly than that in the center. Actuallythere are many gradients between the pipewall and the center. The smaller the pipediameter, the greater the contact between theliquid and the wall thus the greater thefriction. The figure below shows laminar flowthrou gh two cross sections of a pipe. Thevector lengths in the right hand drawing areproportional to the velocity of the flowingliquid.
Friction or Friction Head is defined as theequivalent head in feet of liquid necessary toovercome the friction caused by flow th rougha pipe and its associated fittings.
Friction tables are universally available for awide range of pipe sizes and materials. They
are also available for various pipe fittings andvalves. . These tables show the friction lossper 100 feet of a specific pipe size at variousflow rates. In the case of fittings, friction isstated as an equivalent length of pipe of thesame size. An example is show n below.
Figure 10
It is evident from the tables, that frictionincreases with flow. It is also eviden t thatthere is an optimum flow for each pipe size,after which friction can use up adisproportionate amoun t of the pum p outpu t.For example, 100 GPM for short lengths of 2"pipe may be acceptable; however, over
several hundred feet the friction loss will beun acceptable. 2 1/ 2" pipe will redu ce thefriction by 2/ 3 per 100 feet and is a mu chbetter choice for a 200 foot pipeline. Forsystems that operate continuously 3" pipemay be the ap prop riate choice.
Many friction tables show both friction lossand fluid velocity for a given flow rate.Generally it is wise to keep fluid velocityun der 10 feet per second . If this ru le isfollowed , friction w ill be minimized.
The friction losses for valves and fittings canalso add up . 90 degree turns and restrictivevalves add the most friction. If at all possiblestraight through valves and gentle turnsshould be used.
Consider this problem: What head must apump develop if it is to pump 200 GPMthrough a 2.5" pipe, 200 feet long, and to anelevation of 75'? Use the table in Figure 10.
Elevation = 75'
Friction 2.5” Pipe = 43' / 100ft or 86'
Total head = 75' + 86' = 161' TDH
Use of 3" pipe in the above problem willreduce friction head by 30% and although itwill cost more initially, it will pay for itself inenergy savings over a fairly short period of time.
The “Hot and Cold” Puzzler delves deeperinto fluid friction and its consequences.
SUCTION CONDITIONS
Suction conditions are some of the mostimportant factors affecting centrifugal pumpoperation. If they are ignored during the
design or installation stages of an application,they will probably come back to haun t you.
Suction Lift
A pump cannot pull or "suck" a liquid up itssuction pipe because liquids do not exhibit
tensile strength. Therefore, they cannottransmit tension or be pulled. When a pum pcreates a suction, it is simply reducing localpressure by creating a partial vacuum.Atmospheric or some other external pressureacting on the surface of the liquid pushes theliquid u p the suction pipe into the pum p.
Atmospheric pressure at sea level is calledabsolute pressure (PSIA) because it is ameasurement using absolute zero (a perfectvacuum ) as a base. If pressure is measured
using atmospheric pressure as a base it iscalled gauge pressure (PSIG or simp ly PSI).
Atmospheric pressure, as measured at sealevel, is 14.7 PSIA. In feet of head it is:
Head = PSI X 2.31 / Specific Gravity
For Water it is:
Head = 14.7 X 2.31 / 1.0 = 34 Ft
Thus 34 feet is the theoretical maximumsuction lift for a pump pu mp ing cold water atsea level. No p um p can attain a suction lift of 34 ft; however, well designed ones can reach25 ft quite easily.
You will note, from the equation above, thatspecific gravity can have a major effect onsuction lift. For examp le, the theoret icalmaximum lift for br ine (Specific Gravity = 1.2)at sea level is 28 ft.. The realistic maximu m is
arou nd 20ft. Remember to always factor inspecific gravity if the liquid being pumped isanyth ing bu t clear, cold (68 degrees F) water.
In addition to pump design and suctionpiping, there are two physical properties of the liquid being pumped that affect suctionlift.
1) Maximum suction lift is dependent upon
the pressure applied to the surface of theliquid at the suction source. Maximumsuction lift decreases as pressure d ecreases.2) Maximum suction lift is dependent uponthe vapor pressure of the liquid beingpumped. The vapor pressure of a liquid is thepressure necessary to keep the liquid from
vaporizing (boiling) at a given temperature.Vapor pressure increases as liquidtemperature increases. Maximu m suction liftdecreases as vapor p ressure rises.
It follows then, that the m aximu m suction liftof a centrifugal pump varies inversely withaltitud e. Conversely, maximum suction liftwill increase as the external pressure on itssource increases (for example: a closedpressure vessel). The figure below shows therelationship of altitude and atmospheric
pressure.
Figure 11
A pu mp ing app lication located at an elevationof 5000 feet will experience a reduction inatmospheric pressure of approximately sixfeet. This will result in a reduction in NPSHA(discussed in the next section) by the sameamou nt. Elevation mu st be factored into apu mp ing app lication if the installation is morethan a few hun dr ed feet above sea level
The maximum suction lift of a liquid varies
inversely with the temperature of the liquid.The higher the temperature, the higher thevapor pressure and thus suction lift isdecreased. If a centrifugal pum p is used topump a liquid that is too hot the liquid willboil or vaporize in the pum p suction. Thiscondition is called cavitation and will bediscussed in more detail later.
The figure below shows the relationshipbetween vapor pressure and temperature forclear,cold water.
Figure 12
At a temperature of 70 degrees F, a pressureof only one foot is required to keep water inthe liquid state. As its temperature rises,however, more and more pressure isrequired. At about 210 degrees, a pressure of 34 feet or, sea level atmospheric pressure, isrequired. As it rises to 212 degrees the waterwill boil unless some additional pressure isapp lied. When pu mp ing liquid s at elevatedtemperatures, the liquid’s vapor pressure atthat temperature must be included in theNPSHA calculation.
Capacity and Suction Lift
The suction lift of a centrifugal pump alsovaries inversely with pu mp capacity. This is
illustrated in the figure at the top of theadjoining column . Figure 13 shows how thehead - capacity curve falls off quickly atvariou s suction lifts. You will notice thatmaximum suction lift increases as pumpcapacity decreases. For this reason pu mp sused in high suction lift applications areselected to operate in a range considerably tothe left of their peak efficiency.
Figure 13
Net Positive Suction Head (NPSH)
Net Positive Suction Head Required (NPSHR)is a fun ction of a specific pum p design. Insimple terms it is the pressure, measured atthe centerline of the pump suction, necessaryfor the pump to function satisfactorily at agiven flow. Although NPSHR varies withflow, temp erature and altitud e have no effect.
Net Positive Suction H ead Available (NPSHA)is a characteristic of the system in which thepu mp operates. It depend s up on theelevation or pressure of the suction supply,friction in the suction line, altitude of the
installation, and the vapor pressure of theliquid being pumped.
Both available and required NPSH vary withthe capacity of a given pump and suctionsystem. NPSHA is d ecreased as the capacity isincreased due to the increased friction lossesin the suction piping. NPSHR increasesapp roximately as th e square of capacity sinceit is a function of the velocities and friction inthe pum p inlet. NPSHA can be calculated asfollows:
NPSHA = H a + Hs - Hvp
Where:Ha = Atmosp heric pressure in feetHs = Total suction head or lift in feetHvp = Vapor pressure in feet
For example a pu mp installed at an altitude of 2500 ft and has a suction lift of 13 ft whilepu mp ing 50 degree water. What is NPSHA?
NPSHA = Ha + Hs - Hvp
NPSHA = 31 - 13 - .41
NPSHA = 17.59 ft
Often a two foot safety margin is subtractedfrom NPSHA to cover unforeseencircumstances. When selecting a pu mp forthe conditions above, the NPSHR as shownon the pump's characteristic curve should be15.59 ft or less (17.59 - 2).
Working in the opposite direction, we have apump that requires 8 ft of NPSH at 120GPM.
If the p um p is installed at an altitude of 5000 ftand is pumping cold water at 60 degrees,what is the m aximu m suction lift it can attain?
NPSH = Ha + Hs - Hvp
8 + 2 = 28.2 - Hs - .59
Hs = 28.2 - 8 - 2 - .59
Hs = 17.61 ft (Includ ing the 2 ft margin of safety)
The preceding has dealt only with water. Thesame general p rinciples apply to other liquids;however, vapor pressure must be factoredinto the equations. For roughapproximations where the vapor pressure isunknown, a pump will usually operatesatisfactorily if the NPSHA is equal to orgreater than that required for water undersimilar cond itions. This method m ay be usedonly when the viscosity of the liquid is
app roximately the same as water.
Some of the illustrations used in thisintroduction were adapted from CentrifugalPump Manual by F.E. Myers, a member of the Pentair Pum p Group .
7 6 1 A h u a H o n o lu l u , H I 9 6 8 1 9
Ph o n e 8 0 8 . 5 3 6 . 7 6 9 9 Fa x 8 0 8 . 5 3 6 . 8 7 6 1
