Modifications, Edits and Additions to the CD

Corrected and Edited papers

A009. Affnity laws for centrifugal pumps

C012-1 Carbonic acid

D035. Dynamic balance

I011 Inducer

M004 Magnetite

M029 Motor selection

N001 NPSHA in USCS units

O003-1 Orfice

P033. Pressure head

P015 Piping recommendations

P042 Some common misconceptions about centrifugal pumps

S039 Shaft bending, the L3D4 formula

S072 Specific speed

SA005-1 Temperature limits for elastomers

V026 Vortexing Lliquid

W007 Wear ring clearance

Additional Papers

F040 Seal FAQ

C075 Iron corrosion

C076 Corrosion problems with submersible pumps

S116 Surging pump


Change the eighth paragraph from the botom to:

" NPSHA (net positive suction head available) = 34 + 5 - 22.7 - 0.62 - 2.34 = 13.9 feet. This is enough to stop cavitation also.

P033. PRESSURE HEAD - The last formula in the article shows a (x 2) It should read


Under the paragraph CALCULATING SHAFT DEFLECTION Change the first formula to:



There are occasions when you might want to vary the amount of fluid you are pumping or change the discharge head of a centrifugal pump. There are four ways you could do this:

Of the four methods the first two are the only sensible ones unless you are prepared to buy a new pump. In the following paragraphs we will learn what happens when we change either the pump speed or impeller diameter, and as you would guess, other characteristics of the pump are going to change along with the speed or diameter.

To determine what is going to happen, you begin by taking the new speed or impeller diameter and divide it by the old speed or impeller diameter. Since changing either one will have approximately the same affect,

I will be referring only to changing the speed in this part of the discussion.

As an example:

The capacity or amount of fluid you are pumping will vary directly with this number.

100 Gallons per minute x 2.0 = 200 Gallons per minute

50 Cubic meters per hour x 0,5 = 25 Cubic meters per hour

The head produced by the pump varies by the square of the number.

A 50 foot head x 4 (22) = 200 foot head

A 20 meter head x 0,25 (0,52) = 5 meter head

The horsepower required changes by the cube of the number

A 9 Horsepower motor was required to drive the pump at 1750 rpm. How many horsepower is required now that you are going to 3500 rpm?
9 x 8 (23) = 72 Horsepower is now required.

Likewise if a 12-kilowatt motor was required at 3000 rpm. and you decreased the speed to 1500 the new kilowatts required would be:

12 x 0,125 (0,53) = 1,5 kilowatts required for the lower rpm.

The following relationships are not exact, but they give you an idea of how speed and impeller diameter affects other pump functions.

The net positive suction head required by the pump manufacturer (NPSHR) varies by the square of the number.

A 3 meter NPSHR x 4 (22) = 12 meter NPSHR

10 foot NPSHR x 0,25 (0,52) = 2.5 foot NPSHR

The amount of shaft run out (deflection) varies by the square of the number

As an example: If you put a dial indicator on the shaft and noticed that the total run out at 1750 rpm. was 0.005 inches, then at 3500 rpm the run out would be 0.005" x 4 or 0.020 inches.

Likewise if you had 0,07 mm. run out at 2900 rpm. and you slowed that shaft down to 1450 rpm the run out would decrease to 0,07 mm x 0,25 or 0,018 mm.

The amount of friction loss in the piping varies by 90% of the square of the number. Fittings and accessories vary by the square of the number.

As an example: If the system head loss was calculated or measured at 65 meters, at 1450 rpm. the loss at 2900 rpm. would be:

65 meters x 4 = 260 x 0.9 = 234 Meters

If you had a 195 foot loss at 3500 rpm. the loss at 1750 rpm. would be: 195 x 0.25 = 48.75 / 90% = 54.17 feet of head loss.

 The wear rate of the components varies by the cube of the number.

At 1750 rpm. the impeller material is wearing at the rate of 0.020 inches per month. At 3500 rpm the rate would increase to:

0.020 " x 8 or 0.160 inches per month. Likewise a decrease in speed would decrease the wear rate eight times as much.

I started this discussion by stating that a change in impeller speed or a change in impeller diameter has approximately the same affect. This is true only if you decrease the impeller diameter to a maximum of 10%. As you cut down the impeller diameter, the housing is not coming down in size so the affinity laws do not remain accurate below this 10% maximum number.

The affinity laws remain accurate for speed changes and this is important to remember when we convert from stuffing box packing to a balanced mechanical seal. After the conversion to a mechanical seal we sometimes experience an increase in motor speed rather than a drop in amperage. The affinity laws will help you to predict the final outcome of the change.

The affinity laws also explain the effect on capacity and head when you change motor speed with a variable frequency motor (variable speed driver).

You can use the following formulas to supplement the Affinity Laws. Please keep in mind that these numbers are based on the fluid flowing through the correct size clean pipe.

Product build-up and pipe roughness are variables that will affect the final figures so consider the following "ball park" rather than exact numbers.

Please use these keys when you read the following ratios:

If you are not familiar with raising a number to some power, please look at the following examples:

32 means 3 x 3 = 9

35 means 3 x 3 x 3 x 3 x 3 = 2433

32.5 means to multiply the square of 3 (9) by the square root of 3 (1.732) = 15.6.

The piping friction loss will vary as the square of the capacity ratio

Example: assume you looked at the friction loss charts and learned that 300 gpm. flowing through a pipeline will suffer 20 feet of friction head loss. Then 500 gpm through the same line will lose:

= 56 feet of head loss.

The pump's capacity varies as the square root of the head on the liquid

Example: if a 160 foot head would deliver 300 gpm. through a specified pipeline, a 100 foot head would deliver:

= 237 gpm

The friction loss in the piping is inversely proportional to the fifth power of the pipe diameter ratio

Example: Assume a 3 inch diameter pipe can handle 300 gpm with a 20 foot friction loss. The same flow rate through a 2 inch diameter pipe would create:

= 152 foot loss

The same flow through a 4 inch line would create:

= 5 foot loss

The capacity of a pipe would vary as the 2.5 power of the diameter ratio

Example: assume that a 3 inch diameter discharge pipe delivers 300 gpm. under a specified head. Under the same head, a 2 inch pipe will deliver:

= 109 gpm.


This acid forms when carbon dioxide combines with water to form carbonic acid,

CO2 + H2O = H2CO3

The affect is to lower the pH of boiler water causing the operator to increase the amount of water treatment to maintain the proper pH.

The carbon dioxide enters the condensate system through pump packing, the stem of valves above the water line and gasketed flanges.


Everyone agrees that balancing the rotating components of a centrifugal pump is a good idea, but it is seldom done. Evidently it doesn't appear to be too important or it would be receiving some sort of priority when the pump is being overhauled or rebuilt.

To understand the importance of dynamic balance visualize yourself going down the highway in your automobile at sixty miles an hour, and you throw off a small, lead wheel weight; suddenly you notice a severe vibration in the steering wheel that makes you feel very uncomfortable. Do you have any idea how many rpms the wheels were making at sixty miles an hour or 100 kilometer per hour? Do you think it was slower or faster than the rpm of your centrifugal pump? Let's figure it out in the inch size and then we will do it in metric:

A typical fourteen inch automobile wheel has a tire that is approximately twenty five inches in diameter. This means that the circumference of that tire is 25 inches times 3.14 (pi) or 78.5 inches. Divide the 78.5 inches by 12 and you get 6.5 feet for the circumference of the tire.

At sixty miles an hour you car is going a mile a minute or 5280 feet a minute. Since the 6.5 feet represents one revolution of the wheel we divide that into the 5280 feet and we get 812 rpm at sixty miles an hour.

A common metric tire would have a diameter of 635 mm. Multiply that by 3.14 and you would get just a little bit less than two meters for the circumference. At 100 Km/hr you would be going 1.7 Km or 1700 meters/ minute.

1700 divided by two meters for one revolution of the wheel is 850 rpm.

This means that if a small lead weight can become that significant at 812 rpm or 850 rpm what does an out of balance shaft in your pump do at electric motor speeds?

Several things:

What cause a pump shaft to go out of dynamic balance assuming it was balanced at pump assembly?

Balancing the impeller

You can purchase two plane balancing equipment to do in house dynamic balancing or you can send the assembly to an outside vendor. The newer electronic vibration analysis equipment has a balancing program available for an additional charge over the vibration package. Check with your vendor for its availability.If you send the rotating assembly to an outside vendor for balancing be aware of several potential problems:


A small low head, axial flow impeller that attaches to the conventional impeller of a centrifugal pump to increase the pump's suction head (pressure), and prevent cavitation problems.

The main purpose of the inducer is not to generate an appreciable portion of the total pump head, but to increase the suction pressure of a conventional impeller. They reduce NPSH required and permit the pump to run at a higher speed, with a given NPSH available.

An inducer should be specified any time the calculated suction specific speed number is above 18,000 in USCS units or 11,000 in SI units. The inducer will reduce the net positive suction head required (NPSHR) of the pump or permit the pump to run at a higher speed.

The inducer flow angle is some where between five and ten degrees with typically two vanes and no more than four. Inducers have been used successfully with suction specific speed numbers of approximately 24,000 (14,700 metric).

Although the efficiency of the inducer is low, it will not reduce the pump overall efficiency significantly.

Not all pump manufacturers have this feature available


Magnetite is the common names for black ferric oxide (Fe3O4 ), a protective coating the forms on the inside of iron pipe to reduce further rapid corrosion.

Magnetite is sold commercially as a polishing compound and is known as "black rouge".

Once the magnetite forms on the mechanical seal sliding parts several events can occur that will fail the mechanical seal prematurely:

A red form of magnetite, hematite, (Fe2O3) is also found on the inside of the iron piping we often find in hot water systems. Like the black version (Fe3O4 ) it is also very abrasive.


Electric motors operate at their best power factor and efficiency when fully loaded so you do not want to purchase a motor that is too big, and common sense dictates that one that is too small is even worse.

In the following paragraphs we are going to learn how to select the correct motor for your centrifugal pump application.

Let's assume we will be selecting the motor for the pump described by the pump curve shown below.

The first thing we must do is decide what diameter impeller we will be using. The above curve shows impeller diameters from "A " to "E".

I have selected letters rather than numbers so that we can work the examples in either metric or inch units.

For our example we will use impeller size "A". You will want to look to the right hand side of the curve to select the last efficiency line. In this case it is the 50% line. This will give you the maximum capacity for that size impeller.

Note the capacity at this point (400) and then transfer this capacity and impeller size to a second graph (many times this information is part of the pump curve or located very close to the pump curve) that is supplied by the pump manufacturer.

The second graph will look something like the one illustrated below.

The numbers on the left side of the graph show either the brake horsepower or the kilowatts being consumed. You can select the appropriate units for your application. According to this graph we will be using about 20 (brake horsepower or kilowatts) at the last efficiency line (400).

One assumption we made during this selection process was that the specific gravity of the fluid we were pumping was one (like fresh water). If the fluid has a higher or lower specific gravity we must multiply the number on the left-hand side of the graph by the specific gravity of the fluid to get the correct horsepower or kilowatts for your application.

If the pump were sized correctly for the application it would run within ten percent of its best efficiency point. For impeller size "A" that would be approximately 325 (as shown on the first graph) so we are going to take advantage of the pump service factor (I'll explain that in a few minutes) to give us the needed horse power if we should occasionally run at this higher capacity (400) or get into any other temporary overload condition such as starting a pump that is rotating backwards.

The service factor rating is supplied by the motor manufacturer and is usually available in three ranges:

Motors are available in a variety of horsepower and kilowatt ratings. Typical horsepower ratings would be: 0.5, 1.0, 1.5, 2.0, 3.0, 5.0, 7.0, 10, 15, 20, 30, etc.

Our graph showed that we needed a 17 horse power motor, but a 15 horse power motor will work in this application because of the service factor (15 x 1.15 = 17.25 horsepower available). Keep in mind that any heat generation computations made by the motor manufacturer were made for the motor when it was running at its rated horsepower and not at the service factor rating. All this means is that the motor will run hotter than anticipated, but still within acceptable limits.

Oil refinery applications use a second factor recommended by the American Petroleum Institute (API.). This organization specifies that the factor should be used as an additional safety margin. These factors are:

If we take the same example as noted above, and insert the API (American Petroleum Institute) additional requirement, we would come up with:

There are instances where you can combine the two service factors and come up with a compromise. As an example, suppose that the horse power requirement was 8.7 instead of the 20:

According to the A.P.I. (American Petroleum Institute) you would need 8.7 x 1.25 = 10.8 horsepower, so you would have to go to a 15 horse power motor because there is nothing in between 10 and 15 horsepower. According to the above information a 10 horse power motor has a service factor rating of 1.15 so, 10 x 1.15 = 11.5 horsepower or more than enough to satisfy the API (American Petroleum Institute) recommendation.

Electric motors are sized considering the specific gravity of the liquid being pumped. If a low specific gravity pump is tested with water, or any higher specific gravity fluid, the increase in motor amperage could burn out the motor.


Net positive suction head required (NPSHR). This is the minimum head required to stop the pump from caviating when it is pumping cold water.

The pump curve that came with your pump shows the NPSH required for any given impeller size and capacity.

This number was determined by pumping cold water through the pump while reducing the suction head, until the pump showed a reduction in discharge head of three percent (3%) due to the low suction head and any formation of bubbles within the pump.

This point is called "the point of incipient cavitation".

ROTARY PUMPS NPIP (Net Positive Inlet Pressure)

Positive displacement or rotary pump people do not use the term "head", they use the term "pressure" instead so NPSHR would not be an appropriate term.

Rotary pumps are often selected to move liquids with a low vapor pressure point, or fluids with a lot of entrained bubbles.

This means that NPIP required (NPSH in centrifugal pumps) is difficult to test. The Hydraulic Institute establishes the point at the first indication of any of the following.


There are several reasons you might want to install a restrictive device or orifice in a piping system.

The equation for flow through an orifice is a simple one to understand. Only the units are somewhat awkward.

Q = AV

Experience shows that the actual flow is quite different than calculated because of the different shapes of the various orifices. Look at the diagram on the following page and you will see some of these popular shapes. Each has been assigned a "K" value.

We will enter that "K" value into our equation and the new equation becomes:


To make the equation easier to handle we can express the velocity "V" as:

If you do not know how to convert pressue to head, use this formula:

It would also make sense to convert some of the terms in our equation to terms that are more convenient to use. As an example:

Putting all of this together gives us a new formula that looks like this:

Let's plug in some numbers and calculate a flow through a typical orifice.


Q = 25 x 0.049 x 0.62 x 4.47

Q = 3.40 gallons per minute

If we want to solve for the orifice area:

If you are uncomfortable working with the orifice area in square inches you can use the diameter instead. Use the following equation:

Inserting the 0.049 square inches we calculated from the prior formula we get

We made our formula more user friendly by substituting some conversions and now we can make our calculations in gallons per minute and square inches, but the formulas would be better if we could measure the orifice diameter rather than the orifice area

I took you through this exercise to show you how the formulas we use in these papers are derived. We will re-write the flow and orifice diameter formulas again and maybe this time they will be simple enough for anybody to use. We will start with the flow formula and then fix the orifice formula:

The formula for calculating the orifice diameter becomes:

Let's see if the formulas still work. Here are the numbers:

We will begin by solving for flow (Q)

Well that worked, now let's try for orifice size:

All of these above numbers were generated assuming that you were moving water through the orifice. If you are making calculations for a liquid other than water you will have to factor in the viscosity of that liquid compared to water.

We also made an assumption that the orifice diameter is not greater than 30% of the pipe diameter. There is another formula we use for a less restrictive orifice.

Any time the ratio of the orifice diameter to the pipe diameter is greater than 30% (0.30) you should modify the formula. The modifier (M) looks like this:

When you are using the modifier, the formulas look like this:

Now we will see what happens when a 0.250 inch (1/4) orifice is put into a smaller cross section 0.500 inch (1/2) pipe, assuming the other numbers stay the same:

This means that you would have to multiply by 1.03, so the 3.46 gpm we got in the last calculation would become 3.56 gpm.

How accurate are these predicted numbers? Anytime you make a calculation using flow as a as part of the equation, you will run into some variables that will affect your results:


There are entire books written on this subject so in the next few paragraphs I will try to give you some of the highlights of what you should know about piping as it relates to pumps and seals.

The optimum pipe size will consider the installed cost of the pipe (the cost increases with size) and the pump power requirements (the power required increases with pipe friction)

In multiple pump arrangements we would prefer to have the suction bells in separate bays so that one pump suction will not interfere with another. If this is not practical, a number of units can be installed in a single large sump provided that:

The metric numbers are:


Specific speed is a term used to describe the geometry (shape) of a pump impeller. People responsible for the selection of the correct size pump can use this specific speed information to:

Specific speed is defined as, "the speed of an ideal pump geometrically similar to the actual pump, which when running at this speed will raise a unit of volume, in a unit of time through a unit of head".

The performance of a centrifugal pump is expressed in terms of pump speed, total head, efficiency and required flow. This information is available from the pump manufacturer's published curves. Specific speed is calculated from the following formula, using data from these published pump curves at the pump's best efficiency point (BEP):

The following chart gives you a graphic picture of the impeller shape represented by this number:

The major use of the specific speed number is to help you specify pumps that are more efficient.

Pumps are traditionally divided into three types: radial flow, mixed flow, and axial flow. When you look at the above chart you can see there is a gradual change from the radial flow impeller, which develops pressure principally by the action of centrifugal force, to the axial flow impeller, which develops most of its head by the propelling or lifting action of the vanes on the liquid.

In the specific speed range of approximately 1000 to 6000, double suction impeller are used as frequently as the single suction impellers.

If you substitute other units for flow and head the numerical value of Ns will vary. The speed is always given in revolutions per minute (rpm.). Here is how to alter the Specific Speed number (Ns) if you use other units for capacity and head:

United States ....... Q = gpm, and H = feet, divide the NS by 1.63

British ...................Q = Imp. Gpm, and H = feet, divide the NS by 1.9

Metric ...................Q = m3/hour and H = meters, divide the NS by 1.5

As an example we will make a calculation of NS in both metric and U.S. units:

If the above results were describing an actual application, we would notice that it was a low specific speed, radial flow pump, meaning it would be a large pump with a low efficiency.

Going to 2900 rpm. or higher would increase the Ns to 1000 or more, meaning a smaller pump with a much higher efficiency but this higher rpm would have other possible consequences :

The following diagram illustrates the relationship between specific speed (Ns) and pump efficiency. In general, the efficiency increases as Ns increases.

Specific speed also relates to the shape of the individual pump curve as it describes head, capacity, power consumption and efficiency.

In the above diagram you will note that

Here is another curve to show you the relationship between specific speed, capacity and horsepower requirements:

Keep in mind that efficiency and power consumption were calculated at the best efficiency point (BEP). In practice most pumps operate in a throttled condition because the pump was oversized at the time it was purchased. Lower specific speed pumps may have lower efficiency at the best efficiency point, but at the same time will have lower power consumption at reduced flow than many of the higher specific speed designs.

The result is that it might prove to be more economical to select a lower specific speed design if the pump had to operate over a broad range of capacity.

The clearance between the impeller and the tongue of the volute has a bearing on efficiency, pressure pulsations and cavitation. For high efficiency you would want a small clearance, but this produces larger pressure pulsations and the increased flow in this area can reduce the fluid pressure enough to cause flashing of the product and a type of cavitation known as The vane passing syndrome.

For impellers up to fourteen inches in diameter (355 mm) this clearance should be a minimum of four percent of the impeller diameter. If you are using greater than fourteen-inch diameter impellers the clearance should be at least six percent of the impeller diameter. Also remember that as this clearance increases the impeller experiences some slippage. That is the major reason that we do not like to remove more than ten percent of the impeller diameter when trimming is called for.

If you work in both metric and imperial units, as I do, the subject of specific speed becomes very confusing because both systems use the same specific speed numbers to describe the impeller shape. They do this even though they use a different set of units to arrive at the same number.


Fluorocarbon (Viton®)
-15 +400°
-25 +205°
Ethylene propylene
-70 +300°
-55 +150°
-20 +450°
-30 +230°
0 +500°
-20 +260°
-45 +300°
45 +150°
Buna N
-65 +225°
-55 +105°
Buna S
-75 +250°
-60 +120°
-23 +400
-30.5 +205
-50 +250
-45 +120
-80 +375
-62 +190
-75 +450
-59 +232


Vortexing of the fluid in a suction sump or pit sounds a lot like cavitation problems and will cause excessive shaft deflection that is harmful to:

One way to tell if you have a cavitation or vortexing problem is to remember that vortexing problems are intermittent as the vortices form. Cavitation once started tends to stay with you. Proper pit or sump design can eliminate this vortexing problem, but what do you do if the installation is not new and the problem exists? There could be several things that could have caused the vortexing problem:

Maybe the original design was bad and that is causing the problem. Although this is a very large subject there are a few guide lines you might check out:

The hydraulic Institute ANSI recommendation: ANSI/HI 9.8-1998 Section 9.8.7 recommends the following formula to determine the minimum submergence of a submersible pump::

Now we will take a look at what you can do with an existing installation. Remember that a low velocity and straight line flow to all pumps is always desired. If you are getting vortexing problems you might be able to:

In the next few illustrations I will show you the recommended sump dimensions to prevent vortexing and eddy flows.

The first chart shows the recommended dimensions:


The next two charts show where the dimensions came from:


About the screens:


Wear rings should be replaced when their clearance doubles. This additional clearance will increase the pump power requirements with the amount of power varying according to the specific speed (NS ) of the impeller

If the wear-ring clearance is too large the pump will take on excessive vibration caused by internal recirculation. This can cause seal and bearing component damage. Another problem is that the pump will not meet its designed capacity because of the internal recirculation.

When replacing wear rings, it is not uncommon to bore out the stationary ring and machine the rotating ring oversize to get the correct clearance. If you do this on a double-ended pump, be sure to do both sides of the impeller to prevent upsetting the balanced hydraulic forces and thrusting the impeller to the end that was not bored oversize.

Also be aware that many wear rings supplied by OEM manufacturers are out of round and must be machined to get the proper clearance inside the stationary ring


Eliminate the paragraph: You should never throttle the suction of a centrifugal pump

ans. If the product you are pumping is explosive and close to its vapor point, suction throttling my be your only option. Discharge throtteling would produce additional heat that might be dangeous.

F040 The most asked questions about mechanical seals

What is considered good life for a mechanical seal?

Why do most seals fail prematurely?

What are the most common causes of component damage?

What are the most common causes for the lapped seal faces to open?

Do seal faces have to be lubricated? Can they run dry?

Do seal faces have to be kept cool?

When should you use two hard faces?

Why not standardize on two hard faces?

Do seals have to leak.?

Why do most original equipment seal designs frett and damage the shaft under the dynamic elastomer or spring loaded Teflon.?

Do you have to flush most slurry applications?

I am looking for a simple solution to a difficult problem. Do discharge recirculation filters or cyclone separators installed between the pump discharge and the stuffing box make sense in slurry applications?

If I put a higher fluid pressure barrier fluid between dual seals, shouldn't that keep the faces clean?

Do you need a higher pressure barrier fluid between dual seals?

How does seal hydraulic balance work?

Is it O.K. to have a third party rebuild my mechanical seals?

Should I be using split mechanical seals?

If I touch the lapped faces, are they ruined?

Why should you not use stainless steel springs or stainless steel bellows in mechanical seals?

Why not standardize on Teflon as the preferred rubber part in a mechanical seal?

Why not mount the seal outside the stuffing box and then dirt and solids will not get into the springs and sliding parts of the mechanical seal?

What is a cartridge seal?

Do I need the new gas seals if I want to seal fugitive emissions?

Why does my outside mounted seal make a whistling sound?

Every time I remove a rubber bellows seal from my pump it is stuck to the shaft. Why?

When my metal bellows seal fails because of breakage at the plates, the break is always near the end fittings and never in the middle of the bellows. How is that explained?

® DuPont Dow elastomer


Iron, exposed to moist air, will react with oxygen in the air to form iron oxide. This oxidation process is called rusting.

iron + oxygen iron(III) oxide. 4Fe(s) + 3O2(g) 2Fe2O3(s)
Types of corrosion
Combined effect of mechanical factors and corrosion.

C076 Corrosion problems with submersible pumps

S116 Surging pump

Some thoughts on pump surging
System Stiffness

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