PUMP MAGAZINE: Questions and Answers (51-60)


Question #51 Dear Doctor: 


I want to know the parameters and the equations for the calculation of the thickness of the walls of the impellers and volutes of the centrifugal pumps. 


Also, could you comment on the viscosity effects for the selection of centrifugal pumps, for example for the pumping of pulps and amoniacales. 


Cordial greetings, 


Zoilo Suárez Pérez, EMNI


Answer: Dear Zoilo,


Structural analysis of the pump pressure components, such as casings, is done by the pump designers using pressure vessel equations, such as ASME Code, or similar. Impellers are not considered pressure containing vessels, and their mechanical designs are based on keyway stresses, blades bending rigidity, and so on. If you like, we can perform a design, or analysis of the existing pump design, for your conditions.


With regard to viscosity, Hydraulic Institute published standards, showing a procedure to correct for the viscosity, if different then water. Typically, centrifugal pumps are not used for viscosities above 200-500 cP, as flow, head, and efficiency degrade significantly. Please note that corrections for pulp not as straightforward as for other viscous products. Instead of the viscosity, pulp applications often use percent paper stock, and apply viscous drag correction in that way. Paper stock in water behaves a bit different then simply a viscous fluid.


For more details, apply SEARCH function for the related Articles and Q&A questions within the Pump Magazine. Also, we cover these subjects in greater depth in our regular Pump School training sessions. The next one is February 27-28 in New York, then in March in Houston. Refer to section PUMP SCHOOL (click) of Pump Magazine.



Dr. Lev Nelik, P.E., Apics

Pumping Machinery



Question #52 Dear Doctor: 


My question is about a system, with centrifugal pump pumping water from an open tank to another open tank. Water goes up about 15 m, then a short horizontal section, and then goes down 9 m vertically into the tank, so that the net difference in water levels is 6 m. My questions are:


a) For sizing the pump, do I have to consider the difference in water levels or the highest point of the system? I was told to consider only the difference in water levels, as long as the shut-off head of the pump is greater than the static head of the highest point of the system to guarantee the correct pump start-up. Is it right?


b) I am afraid of a gravitational flow to happen in the vertical pipe just before the entrance to the tank. Is it possible? Is there any air entrance to the pipe in that vertical section (which may have negative pressure)?


Thank you very much,

Carlos Gonzalez

Bilbao, Spain


Answer: Dear Carlos,


This is what it sounds like you have:




1) When the pump starts, it has to lift water 15 m – this is what the original static head is. Once the water reaches the top and flows over, the pump would be working only against the 6 m elevation. You were advised correctly in principle, but the problem could with actual numbers (as they say, the devil is in the details):




What they tell you is your pump would normally be working at about BEP (best efficiency point), which is your 6 m, but during the initial filling-of-the-pipe operation, it would be at the Shut-off condition, which actually (as the discharge pipe gets filled and level in it rises) also increases, until reaches the maximum of 15 m.


Unfortunately, most centrifugal pump curves have approximately 20% rise from the BEP to shut-off, depending on the design. The 30% is realistic maximum (axial flow turbine pumps are exception, so if this is what you have, let us know). This means that 6 m BEP head may rise to about 8 m, or so, - but certainly not as high as 15 m.


You need to select the pump that produces at least 15 at shut-off, and actually some more to be safe – perhaps 20 m. Then, your BEP head would be about 12 m or so, i.e. more then 6 m you will need when pump gets going. You would need to install the discharge to throttle the excess flow, unless you can live with more flow, i.e. a bigger pump.


2) I am not sure what you mean in your second point. However, your concern with air is a good one. The discharge pipe should terminate UNDER the water level, - and the deeper the better, to prevent entrained air. The air can airlock your pipe and also cause undesirable turbulence (“foaming”) at the discharge tank. This is a classical problem at paper mills, where tanks feed another pump, and the agitated air/water/paper mixture causes problem with the follow-up pump and equipment.


Also, a check valve may be a good idea, otherwise the water can get siphoned out from the discharge tank back to suction pump through the pump when it is stopped.


I hope this helps,



Dr. Lev Nelik, P.E., Apics

Editor, Pump Magazine


            Input from our readers:

For practical purpose it is always beneficial to have pumps having 10-15% more head and flow capacity within operating range. The driver may be designed for end of the curve operation to avoid overloading problem due to efficiency loss in long run leading to repeated stoppage hampering production process. 


Here I reaffirm the suggestions by Dr. Nelik to select pump of 20 M head or above to have operational flexibility. To avoid siphoning of fluid from discharge tank to pump suction when pump is idle, few small holes may be drilled on the elbow of discharge pipe right above tank, which would equalize the pressure difference causing siphon action (please refer the figure) .Installation of check valve is also a solution, but many times it mal-functions, thus not 100% reliable.                                                                                                                                                                                                                


Sourav Kumar Chatterjee,

Manager, Rotary Equipment

HPCL Mumbai, India



Question #53 Dr. Pump,


I have been posed with a question.  One paper coating Progressing Cavity pump causes pipe vibration. I have been told that running two pumps in parallel at slightly different speeds could reduce or eliminate this vibration. Is that possible? How is that done?


Your help is very much needed.


Brian Huza

Sales Engineer

Van Bergen & Markson


            Answer: Dear Brian,


The first thing you need to do is understand the cause of a problem. Pipe vibration can happen for many reasons, to name a few:

a)     Acoustic resonance

b)     Structural resonance

c)     Poor support

d)     Poor foundation


A pump acts as a forcing function, but there must be something in the system to be forced! In other words, you must look at both: pump itself, as well as a system it is reacting with. Progressing Cavity pumps are known to run rough, by the nature of their rotor design, as well as connection of the rotor to the drive shaft. Such “nutating” rotors can create problems, which is why PC pumps usually run at slow speeds (there are also other reasons for that, such as rotor/stator interference fit). If the problem is more with piping, and not as much with a pump, then two pumps simply exaggerate the problem, and force the piping twice is violently. Reducing a speed of one pump could, in theory, result in combining frequencies, with resultant harmonics being at new frequency. If the pipe resonance happened to hit at a pump running frequency, then “shifting away” into a new harmonics can “tune the system out”. There is good mathematics behind it, such as Fourier series, but we will not go into that.


In practice, however, such approach is a bit like scratching a left ear with a right foot. It may work, but looks kind of strange. How is this done? - You can change gearbox ratios, or VFD drive speed, which is a simple thing to try, - and if it works – you solved it. Unfortunately, VFD drives are not often used with PC pumps, so getting one is an added expense at this point. Changing gearbox ratio is easier, but still takes some doing.


Look at the root cause. They are findable, although not always straightforward. You may have someone knowledgeable join you to help troubleshooting, as this may save you time.



Dr. Lev Nelik, P.E., Apics

Pumping Machinery



Question #54 Dr. Pump,


Can you tell me how leakage varies with wear ring clearance for a given p/p. How does the calculated values compare with actual values, for clearance starting at the recommended value to twice and then on to three times the recommended value? Impeller speed appears to have a small influence on leakage, but this is not supported by the theory.





            Answer: Dear Al,


It was found that a wear ring leakage can be closely estimated by treating the clearance thorough the sharp edge orifice of the equal area, and then divide the result in half. This works fairly well for plain rings. For grooved rings, the dynamics are more complex, as secondary flows make things less straightforward, as well as depends on the way the grooves are cut.


This will get you in a ballpark. As wear ring clearance opens up, the area changes, so calculate it the same way. If you want to get more theoretical, a good source is Schlichting book on turbulent flow mechanics.


By the way, as wear rings clearance opens up, other things happen, - not just reduced efficiency due to leakage. In multi-stage boiler feed pumps, this is a big issue, and effects a critical speed - very significantly. API-610 recognized this a long time ago, and looks at this aspect with great attention.



Dr. Lev Nelik, P.E., Apics

Pumping Machinery




Question #55 Hello, Dr.Pump, -


I am very impressed with your articles, and they really help me understand more about pumps and pumping systems. I want to ask you about the Pulsation Dampener requirement for our Reciprocating Pump. It is used for feed water system in steam generator. My pulsation dampener is Hydril product that using Nitrogen as gas cushion. We have experienced many leaks in our Dampener, and our operators had an initiative to modify current pulsation dampener. The modification is they inject the feed water (180 psig) into the dampener throughout the point for filling the nitrogen. We don't use nitrogen anymore inside dampener. Currently, the pump doesn't have high vibration, but the suction line has vibration higher than another suction line that still using original dampener system. The suction pressure is pulsating between 100 to 300 psig (is that too high?)


Is that OK to modify our current Dampener system? My suction pressure is about 180 psig, - is it possible to have cavitation in our pump? I need your information before I decide whether this modification can or cannot be used in our system.


Thank you and regards,


Patria Legawa

Facility Engineer, Caltex

FOMT - Duri Operation


            Answer: Dear Patricia,


We have asked Andy Shelton of Pentair Pump company (also known as Aplex company) to comment. Aplex manufactures reciprocating pumps, and are familiar with pulsation dampeners, which are almost always applied for the reciprocating units. This is what Andy advises:

It is not advisable to modify your dampener they way you have described.  Without the nitrogen volume in the dampener, you have effectively removed the dampener from the system.  The suction line vibration indicates that the pump is not being fed properly.  It is a common mistake to think that high suction pressure means the pump is getting plenty of liquid.  There must be something in the piping system to accommodate the pumps uneven demand for liquid.  In short, the dampener needs the nitrogen to function properly and the pump needs the dampener.

Andy Shelton

By the way, Patricia, - what did Hydril say? I would think they would likewise advise against using liquid, and use originally intended Nitrogen. Also, why not use Nitrogen?


If any of our readers are from Hydril, you may want to assist Patricia with this application, locally.


Dr. Lev Nelik, P.E., Apics

Editor, Pump Magazine




Question #56 Dear sir,


Please clarify the following:


When two pumps are available (from different vendors) with same capacity

and with 10% and 19% higher shut off head, which one is to be chosen?


I think that a very steeply rising curve should not be used when a single pump is used for

transfer purpose. Flat curve should not be used when controlled is by valve.


(Higher shut off head means, higher work will be done on the fluid, i.e. the fluid temperature

will increase and may even flash in the pump casing, which may be detrimental to the pump).


Thank you,



            Answer: Dear Mr. Anbazhagan, -


It is always better to have a continuously rising head-capacity curve. This prevents instabilities when several pumps operate in parallel. It really does not matter if it rises 10 or 19%. However, for marginal hydraulic designs, pump manufacturers may have somewhat a tendency to “flatten” a curve, when in fact the actual test would show a slight “droop” near shut-off. To prevent such marginal or questionable situations, it is better to choose a higher rising design. API-610, in fact, specifies that5 pumps must have continuously rising curves, especially when operating in parallel.


Regarding the energy imparted to fluid near shut-off – it doubtfully makes practical difference either way. The issue is not to operate the pump near shut-off to begin with. If pumps stay away from the shut-off area, and operate close to BEP, the whole issue is academic. Of course, it is not possible to operate at a very limited region near BEP, and manufacturers define the MCSF (minimum continuous stable flow), which, depending on Specific Speed, and Suction Specific Speed, is around 30-40% of BEP. If pumps run into this “bad” region, other problems happen – high radial thrust, seal leaks, bearing problems, couplings failures, - to name a few.


There are several other articles on this subject, and curve stability at PUMP MAGAZINE, and you can locate them via SEARCH function.



Dr. Lev Nelik, P.E.

Pump Magazine




Question #57 How to determine or calculate up to what liquid level of a closed stock drum (with Pentane liquid) can I pump out liquid without cavitation.


Pump: Centrifugal

          Normal suction pressure: 1.05 barg

          Discharge Pressure: 4.05 barg

          Normal Capacity: 6.2 m3/hr

          NPSH Available: 10m

          specific gravity of Pentane @ pumping temp:  .612

          pumping temp. : 32 deg. C

          Vapor Pressure at pumping temp. : .94 barg





            Answer: Dear Ronnie,


First of all, it is a good idea to draw a simple system sketch, to better visualize what you got:

Is this correct? I am not sure where you measure 1.05 bar(g). I assume it is on top of the tank, since you say that it is a closed tank? If the picture is true – we can continue, otherwise we need stop and clarify. Pay attention to details, and ask questions until you are sure of exactly what is going on. Only then proceed with analysis.


Then, let’s think about the basics. The higher the level of liquid in the tank – the higher is pressure at the pump inlet. That’s good. Because if the pressure at the pump inlet is too low, and especially when it gets too close to vapor pressure – the liquid begins to vaporize (essentially boil), and cavitation begins. Since it is suction situation we are interested in, the discharge side information is not really needed at this point. All we need to do is determine what the available NPSHA is and to make sure it is greater then the required NPSHR (with some safety margin). The NPSHR, however, can not be determined from the sketch – instead, you need to get it from the pump manufacturer curve – it will show the NPSHR plot versus flow.


Let’s try to see what is level in the tank “H”, for the conditions shown. Let’s remember the definitions: the NPSHA is essentially the “excess of total suction head over the vapor pressure head”. Suction head can be calculated based on the liquid level in the tank, and then subtracting the losses between the tank and the pump. But – we don’t really need to bother with that either – because, fortunately, there is a suction gage installed – and we can simply read it (it is always better to read the data then guess or calculate from another datum).


So, 1.05 bar(gage) is (x14.5) = 15.2 psi(gage). I will use US units, and will get everything to “feet”. But you can do the same in metric. To get from “gage” to “absolute” units, we add 14.7 psi, i.e. 15.2+14.7 = 19.9 psi(a). Then convert to feet: (x2.31/SG), i.e. 19.9x2.31/.612 = 75.1 feet. To get the suction head at the pump suction flange, we need to add liquid level, and subtract hydraulic losses. Then subtract vapor pressure (.94 bar(g) = 13.6 psi(g) = 28.3 psi(a) = 106.9 feet). NPSHA=10 m = 32.8 feet.


So, 32.8 = 75.1 + H – hlosses – 106.9


Solving: (H – hlosses) = 32.8+106.9-75.1 = 64.6 feet


If losses were zero, the liquid level would be 64.6 feet. But there are losses, so the tank level is probably higher then 64.6 feet.


You next need to find out what is the required NPSHR at that flow. Let’s say, just for example, that NPSHR = 6m (19.7 feet), i.e. you have 10-6 = 4 m margin. Usually, the recommended margin is about 1.5 m (5 feet), i.e. your NPSHA can not be less then 6+1.5 = 7.5 m = 24.6 feet. For the NPSHA=10 m (= 32.8 feet), you would have 32.8 – 24.6 = 8.2 feet (2.5 m) for the level to drop before you get to the margin level.


Obviously, I made some assumptions, just to get the process going for you. Now that you know what information you need to get (NPSHR from the curve, and estimate the losses in the suction line), - and just follow the same logic, - the answer will come out easily.


By the way, I somehow suspect the suction gage is not where I interpreted it on the sketch, because the level in the tank (64.6 feet) sounds high (?), unless it is a really big tank.


I hope this helps, - also, use SEARCH function within the web site we have, and look up other articles and questions on cavitation, NPSHA, NPSHR, etc. – you may find it helpful.


Last, but not least, join in with others at the next session of Pump School! – the information is in Pump School section.




Dr. Lev Nelik, P.E.

Pumping Machinery


Dear Dr. Lev Nelik,


A clarification sketch on our system:


                                                       1 barg     


















No suction pressure gauge installed by the pump; only gauge at discharge.


We are maintaining/controlling the horizontal stock drum pressure to 1 barg using nitrogen. We maintain that pressure during pumping. Our target is to zero out the content. Starting level in the tank is at 450 mm from the tank bottom. The distance or height of tank bottom to pump suction centerline is about 1.5m.


My question is still: can I pump this out without cavitation to zero level in the drum?





Ronnie, - I am still trying to understand the numbers, and to confirm how 10 m is arrived at.


The pressure in the tank, on the liquid surface is 1.05 barg, which is equivalent to 17.5 meters (in "gage" units). This, adding to 1.5m liquid column from the pump to the bottom of a tank, plus another 0.45m level in the tank itself, gives 17.5 + 1.5 + 0.45 = 19.5 m


The vapor pressure of 0.94 barg, when also expressed in meters, is 15.7 meters. Thus, the "excess" pressure (suction head over vapor pressure) - which is what NPSHA is, would be: 19.5 - 15.7 = 3.8 meters


Thus, where did NPSHA=10m come from?


I might be mistaken in interpreting something, but let's get this step understood first, before we go on.


Lev Nelik

Pumping Machinery


Dear Lev,


The 10m NPSH available is based on the pump Data Sheet from manufacturer. I guess we need to find out where did that come from. Probably from the older spec, and we now need to go over the numbers again, to make sure NPSHA is sufficient.






Question #58

Dear Pump Magazine,


I came across your interesting site on the internet whilst searching for fluid flow information to assist me with my dissertation prepared in connection with an Institute membership application.  I have kept the following free of detail and calculation and have concentrated the philosophy. My paper concerned the support systems required to allow Steam Turbo Generator Trials to be conducted on a ship whilst the vessel was in Dry Dock.


Under normal circumstances (Fig. 1) its Sea Water cooling systems suction and discharge from the Condensers are at the same relative heights - statically the system being in equilibrium with no net flow.  The systems centrifugal pump when running in slow speed created a flow of 480cubic metres/hour in the system.


In dry dock (Fig. 2) the requirement was to ascertain if a support system designed to connect to the vessels inlet and discharge points was able to support trials on the Turbo Generation Machinery with the ships pump operating in slow speed. i.e. efficient Condenser operations.


As part of the heat transfer calculation, I calculated the volumetric flow of cooling water through the ships circulating water system when fed from temporary supplies connected to the ships hull valves which in turn were fed from the Dry Dock Water Flood main.  For the purposes of this initial calculation I took into account all losses in the temporary supply and discharge system but ignored the relatively short ships system and arrived at a figure in the order of 250 cubic metres/hour.  I then mistakenly added this figure to the known ships system/pump performance figure of 480 cubic metres/hour to give total flow of about 730 cubic metres/hour.

I say mistakenly as I have been informed this is not sound.


My question is: by adding an external supply to a vessels cooling system which has a known system performance, what is the criteria for assessment of whether the temporary supplies are able to support the suction requirements of the ships/systems pump.  Is this the Net Pump Suction Head and in reality is this all I needed to prove that I could satisfy? and what then would be the best method of assessing the probable flow in the system?


The arrangements are shown on Fig. 1 and 2 below, and I would be grateful for an explanation.



Fig. 1 - 

1m Approx









SW SYSTEM PUMP @ 480m³/hr







Cooling system, Vessel afloat at sea






























= 12.8 METRES



















































Roger Sandell



Answer: Dear Roger, -


Let me see if I understand correctly: in dry dock, the boat is essentially hooked up to a piping network in front of pumps, and also a piping network after the pumps. Thus, friction losses are added in front of, and after the pump. Let’s assume the pump runs at the same speed.


Pump operating point is an intersection of a pump curve and a system curve. In this sense, regarding performance, the system is what is AFTER the pump. By adding more pipes means adding more friction losses, i.e. a system curve is still a parabola, but a steeper one. Since the pump curve is the same (same running speed), the intersection point is closer to low flow. To know exactly where - do what you started to do - calculate additional friction losses and get a steeper system curve - plot it and find intersection with the pump curve - you get new flow (lower then when operating at sea - where discharge pipe is much shorter.


The next step is suction side. Once you determine the new flow (in step above), look at the pump curve (manufacturer should have supplied), and see what is NPSHR (required) at that flow. You next need to calculate new NPSHA (available) to make sure it is greater than required NPSHR. Things are tougher now - because new long pipe added losses in suction. But calculation is still straightforward. First, find the water level above the pump centerline (I am not sure where it is from the sketch, but assume it is at the same level where the main feed line is - and if so, it is 12.8 - 1.4 = 11.4 meters above the pump centerline. Next calculate the losses in the line for the flow (which was determined in step above). Say the flow was determined to be about 300 m/hr. Then calculate friction losses in suction piping system in front of a pump. Let’s say, for example (I will guess just to make a point - but you should calculate exactly), that suction line losses add to about 4 meters. Then: 11.4 - 4 = 7.4 meters. Next, subtract vapor pressure (for cold water it is approximately 0.5 m) – so you end up with NPSHA = 11.4 - 4 - .5 = 6.9 m


Now compare this NPSHA with NPSHR (using pump data curve), and if NPSHA>NPSHR - the pump will work, and not cavitate. Usually, some safety margin is added, say about 1m or so, to make sure you do not run the pump right at the edge of the NPSHA.


Does this help a little? Let me know.


By the way, Sims Pump and Valve company supplies pumps for the Navy, as well as Marine industry. They make special structured engineered composite impellers, casings, wear rings, bushings, etc. This engineered composites are 80% lighter then metal, but resist salt ocean water very much better then bronze, or even stainless steel, - which is why it is becoming a choice material for US Navy pumps.


I would be interested to get in contact with the people that you are working with on these applications, to explore this subject further. You can see some information on that on our site, section Editorial, as well as Useful Links, section Sims Pumps.




Dr. Lev Nelik, P.E.

Pump Magazine


Follow-up: Lev,


A great help, thank you. I'll keep in touch through your site re possible applications for Sims.


One omission from my previous data was that the prime mover is an Axial Flow pump.  I calculated the suction head addition from the Dry Dock system to be a few metres larger than that in normal service (allowing for losses), and once the additional frictional losses on the discharge side (due to the added Dry Dock system) were accounted for (as you indicated) there remained a head advantage which overall translated in a small degree of additional flow to the pump flow of normal service.  I guess this is not an unreasonable expectation (within the limits of the pumps design capability)?



Roger Sandell


Answer to the follow-up:

Roger, - it is always best to make a very basic sketch (even by hand) of a system, - you should try to get in a habit of doing that when handling pump issues. Stick with the basics, and keep it simple. First, draw a pump H-Q curve. The draw a system curve – it will be a parabola (H versus Q), but off-set along Head axis by the amount of static head. Static head is a difference between the discharge static pressure and suction static pressure. For open tanks it is simply a difference between the levels of liquids in tanks.


What you are telling me is that you suction level is higher in a dry dock, but the discharge level is about the same. That means the difference in levels is less – as compared to what it is at sea. In fact, you probably have a negative static head, which is possible, i.e. water wants to flow from a higher level in a suction tank into a discharge tank – even without help from the pump.


Now think what this means – the system parabola gets “lowered” down the Head axis. For example: say in sea suction level is at 5 feet (referenced to some reference plane), and discharge level is 7 feet – i.e. static head is 7-2=2 feet. Then, the rising parabola is a function of flow, and represents losses when pump is running (the static head is there regardless if pump works or idle).


Say then in a dry dock the level at the dock wall that hold the inlet water is higher-  say at 30 feet level. And say it discharges to some other vessel the level of water in which is at 8 feet. The difference (static head) is –(negative) 8-30=-22 feet – i.e. the water would flow even without the pump. Now visualize the new parabola going lower, i.e. it will intersect a pump curve (which is the same) at higher flow.


But losses in discharge are, as you estimated, higher (longer pipes). This would make a new parabola steeper, tending to get to lower flow. The net effect is whichever “wins” – if lowered static head is more significant then the added friction – then you right, the parabola, even though steeper – is sitting lower – and may intersect the pump curve at higher flow. So – it depends on numbers.




Lev Nelik

Pump Magazine




Your explanations have been clear and helpful.  As a result of contacting you I shall certainly be much more confident of any assessment I carry out in future.


Kind regards


Roger Sandell



Roger, - thanks. I am glad it was helpful.


Lev Nelik




Question # 59 Dear sir,


How to calculate the MTBF (mean time between failures) of a complete pump, as well as individual components, such seal, bearing and coupling? Are stand-by pumps also to be taken into consideration or only running pumps?


Also please clarify the difference between MTBF and MTBR. Do you have the data of world class MTBF for pumps, seals, bearings & couplings?


D.R. Patil,

DY Manager, Rotary Group



 Everyone calculates Mean Time Between Failures/Repairs (MTBF/R) different.  What is worse is that everyone defines what constitutes a “failure” differently.  In my “other life” at Exxon some affiliates decided to exempt field repairs form the calculation.  That is if the pump failure could be corrected in the field - it would not be captured as a failure. Some companies who use stand-by spares divide the population number by 2 to account for the active population only.  At Exxon we calculated the number as: (Summation of failures/month) /(# of pumps in population/2) in a rolling 12 month period.  This gives you the mean time to failure for the population.  I am guessing at the formula a little since I do not remember it exactly.

The number is usually calculated by the larger chemical and petrochemical companies as a means of trying to improve the life of the equipment.  It requires that you document the population of pumps/motors/compressors/turbines.  It also requires creating a detailed history and completing a failure analysis after each failure.  At Exxon there were 6 to 8 engineers and 3 to 4 technicians doing this all the time.  We had accurate numbers and kept track of all the failures in paper folders, like dossiers, for each piece of equipment.   But we never got to the point of calculating MTBR/F for components.  I am sure it can be done, especially today with the greater availability of data base, and computers.  It still represents a significant commitment of time and a good team of dedicated people.

Luis Rizo

Pump Magazine




Question # 60 Dear sir,


Could you please tell me how to calculate BEP?


Answer:  Best Efficiency Point is the pump flow at which the sum of all pump losses are at their minimum. To the right of the BEP the losses are mainly internal hydraulic friction within the impeller passages, diffuser vanes or volute, as well as incidence losses. To the left of the BEP are mainly losses associated with fluid recirculation and other instabilities.  Even at BEP there are losses, - but they are the less then to the right or to the left of the BEP. Detailed calculation is the area of hydraulic designers, and Pumping Machinery can perform such calculations when a pump needs to be re-rated. For example, if the originally selected pump has 1000 gpm, but operates at 200 gpm, we can redesign the impeller, to make it a better fit (i.e. minimize losses) at the 200 gpm, - in other words, we would move the BEP point to the left. Such method is an effective method to upgrade the pump design, without major changes to the piping or casing, - an economical approach to pump upgrade and retrofit work.


On a pump curve, supplied by the manufactures in their catalogs, the BEP point is where the efficiency is the highest. At each impeller diameter cut, there is its own BEP point, although efficiency is higher at the BEP for the larger impeller diameter, as compared to efficiency at the BEP point at the smaller impeller diameter cut.


Lev Nelik

Pump Magazine




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