Previously we have spent some time discussing how particle size can be described in terms of different standards. It’s now time to look at how the distribution of particles is important for people in the pump/dredge world.

“Particle Distribution” is simply a term used to describe how much of a specific slurry or deposit is of any given particle size. It is the key to determining how the slurry will react when we pump it and, in the case of a deposit, how it will re-suspend.

Let us first look at how distribution affects production when we are using a pump to re-suspend a solid.  Maximum production is obtained when a pump is continually in contact with the subject material. This is obtained when the material comes or flows to the pump.  If the operator continually has to move the pump to keep it supplied with solids, it is far harder to maintain a consistently high rate of production. Furthermore, if operated in a sweeping motion the pump only accesses material from one side, again reducing solids production.

In short, we want the material to move or flow to the pump. Sand at the tide line on a beach is an example of a material that flows well. As the person in the photo below found out, it is impossible to dig a hole with her feet, as the sand just flows into the hole she is trying to create.

wet sand

In terms of dredging , wet beach sand  is said to have an angle of repose of close to zero. Deposits  with  low or slight angles of repose result in a constant feed of material to the pump and high production rates . In contrast materials with  high angles of repose such as virgin deposits of aggregate , require the pump operator to constantly move the pump resulting in down time and lack of efficiency.

When dredging or relocating any deposit, we hope the material deposit has a low angle of repose, but which materials yield slight angles? Well, any deposit that has fairly uniform particle size that is not ultra-fine is a good candidate. Having all the particles of a fairly consistent size means there are no smaller particles that could fill the gaps within the deposit. Gaps result in a deposit that is difficult to compact and allow the particles to roll over each other as they flow toward the pump.

Geological deposit of blue clay

You will notice that above, I stated a qualifier of “uniform particle size that is not ultra-fine”. Materials that have a uniform ultrafine particle distribution have virtually no gaps between the particles and, as such, form a very consolidated deposit that will not flow. Blue Georgia clay, or for that matter any consolidated clay, will resist movement of any type.

Deposits with a uniform particle distribution that is ultra-fine also result in an issue when we look at re-suspending or “slurrying“ a deposit.   Let’s look at clay again.  You can shoot water at it, and it just deforms, or you can agitate it, and the surface particles wash off, but it stays fairly uniform.  Clay and other similar deposits are said to be “plastic” and virtually impossible to sufficiently agitate the solids to form a slurry.  (Note: Cutter head dredges can pump plastic materials, but they are, in effect, pumping chunks of material, not a slurry).


Particle size distribution not only effects things before the material enters the pump system it also effects the situation within the pump in two key areas.

The first area is wear. Deposits that have all large particles wear pumps far more quickly than deposits that have a mixture of particle sizes.  If there is a mixture of particle size, the smaller particles can effectively get in the way of the larger particles and stop them from directly impacting on the wear components in the pump. The same holds true for the discharge pipe.

The second concerns settling velocity.  Although the subject of settling velocities is a discussion in itself, particle distribution does affect settling velocity.  In short, the heavier or higher the viscosity of the carrier liquid, the more energy it can impose on large articles to keep them moving through a pipeline.  Hence, slurries that contain large particles, as well as smaller particles, are less likely to have to plug problems than slurries that contain just the large particles when pumped at the same flow rates/velocities.

While on the subject of settling velocities, it should be noted that slurries composed entirely of ultra-fine particles will not settle out in pipes. This allows for very low flow velocities and provides the associated benefit of low friction losses within our piping system.


Now that we understand the importance of knowing particle distribution how do we express it. First data must be obtained on the makeup of the deposit. Relating  back to previous  lessons  we must identify percentages of particles that pass through standard mesh sizes. This info can then be summarized on a ‘particle distribution graph.

The sample graph below plots the total percent of the sample that will pass any given screen size.



In this sample illustration the plotted horizontal and vertical lines are there as the source for what is referred to as the D60 and D10 numbers.

The “D number”  is just  an abbreviation to represent the percentage of product that will pass a given screen size. In the example above the D60 is approx 0.6 mm, which means that 60 % of the sample is smaller that 0.6 mm. The D10 would be approx. 0.02 mm.

I personally find this representation of particle distribution somewhat difficult to visualize. The graph below is, in my opinion, easier to visualize than the graph shown in the previous slide. However, having to calculate the area under the curve makes establishing “D numbers “ far more difficult.

I hope this brief discussion helps everyone to better understand the importance of determining and accurately communicating particle distribution.  If someone refers to a “D” number and you are having questions about a deposit, gather as much info as you can and then call our very competent application engineering department at Hevvy/Toyo Pumps.



Accessing and describing the make of a deposit or a slurry is important in the selection of pumping equipment. Particle size along with particle distribution will affect pump wear, material selection, pipeline settling velocity, and how a deposit will or will not flow and therefore affect production rates. Deposits with oversized wood or trash particles will also create a problem known as “bird nesting”. Oversized rocks will create an issue known as “paving stoning”. Both subjects for another day

Although it is important to be able to accurately describe both particle size and distribution, today we will only address particle size.

To the right are examples of the types of terms used to describe rock in the sand and gravel industry.  Useful in obtaining a general feel for the makeup of a deposit but not very analytical. To be able to further clarify, and to easily describe the size of particles, some standards have been established.   Unfortunately, as with most things in our industry, there are multiple standards!!

The basis or method of the various standards is however common, and that is “ to define a particle in accordance with the size of the particle that will pass through a specific size mesh”.  Unfortunately, the sizes of mesh and the units of measure used to describe the mesh is not common. The four particle size standards  I have worked with are; “Tyler”,  “US Bureau of Standard Screens”, “British Standard Screens”, “I.M.M. Screens” and  Wentworth.

Diameter (mm) Diameter (phi) Wentworth Size Class
4096 256 -12 -8 Gravel Boulder
64 -6 Cobble
4 -2 Pebble
2 -1 Granule
1 0 Sand Very Coarse Sand
0.5 1 Coarse Sand
0.25 2 Medium Sand
0.125 3 Fine Sand
0.0625 4 Very Fine Sand
0.0313 5 Silt Coarse Silt
0.0156 6 Medium Silt
0.0078 7 Fine Silt
0.0039 8 Very Fine Silt
0.00006 14 Mud Clay

The Wentworth standard is rarely used but does contain a useful table that provides a clear cross-reference between common terms used to describe materials and the relative particle size in mm for that material.

U.S. Bureau of Standard Screens Tyler Screens British Standard Screens I.M.M. Screens
Mesh Aperture Mesh Aperture Mesh
Mesh Aperture Mesh Aperture
Inches mm Inches mm Inches mm Inches mm
2.5 .321 7.925
3 .265 6.73 3 .263 6.680
3.5 .223 5.66 .221 5..613 3.5
4 .187 4.76 4 .185 4.699
5 .157 4.00 .156 3.962 5
6 .132 3.36 6 .131 3.327 5 .1320 3.34
7 .111 2.83 .110 2.794 7 6 .1107 2.81
8 .0937 2.38 8 .093 2.362 7 .0949 2.41 5 .100 2.51
10 .0787 2.00 .087 1.981 9 8 .0810 2.05
12 .0661 1.68 10 .065 1.651 10 .0660 1.67
8 .062 1.574
14 .0555 1.14 .055 1.397 12 12 .0553 1.40
10 .050 1.270
16 .0469 1.19 14 .046 1.168 14 .0474 1.20
12 .0416 1.056
18 .0394 1.00 .039 .991 16 16 .0395 1.00
20 .0331 .84 20 .0328 .883 18 .0336 .85
16 .0312 .792
25 .0280 .71 .0276 .701 24 22 .0275 .70
20 .025 .635
30 .0232 .59 28 .0232 .586 25 .0236 .60
35 .0197 .50 .0195 .495 32 30 .0197 .50 25 .020 .508
40 .0165 .42 35 .0164 .417 36 .0166 .421 30 .0166 .421
45 .0138 .35 .0138 .351 42 44 .0139 .353 35 .0142 .361
40 .0125 .317
50 .0117 .297 48 .0116 .295 52 .0166 .295
60 .008 .250 .0097 .246 60 60 .0099 .252 50 .01 .254
70 .008 .210 65 .0082 .208 72 .0083 .211 60 .0083 .211
80 .0070 .177 .0069 .175 80 85 .0070 .177 70 .0071 .180
100 .0059 .149 100 .0059 .147 100 .0060 .152 80 .0065 .157
90 .0055 .139
120 .0049 .125 .0049 .124 115 120 .0049 .125 100 .0050 .127
140 .0041 .105 150 .0041 .104 150 .0041 .105 120 .0042 .107
170 .0035 .088 .0035 .088 170 170 .0035 .088 150 .0033 .084
200 .0029 .074 200 .0029 .074 200 .0030 .076 170 .0029 .074
230 .0024 .062 .0024 .061 250 240 .0026 .065 200 .0025 .063
270 .0021 .053 270 .0021 .053 300 .0021 .053
325 .0017 .044 .0017 .043 325
400 .0015 .037

This was a very basic short discussion today but next time we will use the terminology and the information on particle size discussed today to delve into “Particle Distribution” and how it affects deposit flow and therefore production .



You will often hear the term “Specific Gravity” or “SG” associated with slurry. It is indeed a useful method of describing the slurry but it can also be very misleading if you don’t know the specific gravity of the solid and fully understand how it can affect the consistency of the slurry. SG also has a significant impact on centrifugal pump horsepower consumption, making it a critical part of pump motor sizing calculations.

Specific Gravity is quite simply the ratio of the weight of a substance to the weight of water. To establish a numerical value for any substance you take the weight of a known volume of the substance and divide it by the weight of water in that same volume. For example, the weight of one cubic foot of solid granite is 168 lbs, and the weight of water per cubic foot is 62.4 lbs. Granite’s SG is therefore 168/62.4 or 2.7. The graphic below illustrates the same calculation done using metric units with a sample size of one cubic meter.

As you can see the units of measure are irrelevant as long as they remain consistent and the sample volume for the substance in question is consistent with the sample size used for the baseline (water).

The phrase “solid granite” is underlined in an earlier paragraph. The key word here is solid. Materials that have air gaps come with a slight complication. If we look at the illustration below using sand or gravel made from granite the issue becomes clear.

Using simple math, dry sand made of granite appears to have a SG of 1.9, far different than the SG of 2.7 for solid granite.  The difference of course is the air gaps between the grains of sand. These gaps must be allowed for in any calculation and we use a term referred to as “bulk ratio” or “bulk density” to compensate.  (See a previous article for details on bulk ratio or bulk density.)

Fortunately, most slurry pumping applications do not contain entrained air. As such it is easy to measure the SG and use the value to help describe the type/consistency of the slurry. For example, if we were to take a plastic bucket filled with a sand slurry weighing in at 55 lbs and then fill the same bucket with water and find its weight is 41 lbs, then the SG of the sand slurry is 55/41 or 1.34.

In any given volume, the total weight of the suspended solid is the factor that yields the numerical SG.  Since this total weight is based on both the SG of the solid and the concentration of the solid, caution must be used when relying solely on the slurry SG to visualize the consistency of a given slurry.

For example; a heavy concentration of spruce wood pulp slurry would have an SG of less than one but could be thick enough to walk on. A light slurry of steel mill scale may have an SG of 1.3 but look like dirty water.

It is clear that the SG of a slurry is based on two factors, those being the SG of the solid and the volume or concentration of solids. As the example above illustrates it would be important to quote at least one of these two factors in conjunction with the SG value to provide a clear picture of the consistency of the slurry.

In addition to helping to describe a specific slurry, knowing the SG of a slurry is critical when sizing motors for centrifugal pumps. Fortunately, the math is super simple.  To establish the HP required on slurry just multiply the HP required on water by the SG of the slurry.

Enough for today.  Be sure to look for an upcoming article on the concentration of solids by weight or volume and how that relates to SG.



Previously we have discussed net positive suction pressure but avoided the role vapour pressure may have on NPSHA. Today, we will shed some light on this subject.

The study of vapour pressure starts with developing an understanding of evaporation. Evaporation of a liquid is a concept most of us generally understand, but do we really know the intricacies of the process?

Let’s start by examining a beaker of material in a liquid state. If that material is water, the beaker at a molecular level would be full of H2O water molecules. If the water was cold, say 1 degree C, then these molecules would be moving slowly and would not possess much kinetic energy. The molecular attraction between the molecules would keep almost all the surface molecules contained.

Periodically, due to random collisions of molecules, one surface molecule may accumulate enough kinetic energy to break the molecular attraction and leave the water’s surface. This molecule is now in a gaseous state and has “evaporated”.

As the temperature of a liquid increases so does the kinetic energy of the individual molecules.  As the drawing below illustrates, the increase in energy within the molecules at the elevated temperature translates into more molecules obtaining enough energy to transition into a gaseous state.

If the beaker is open to the atmosphere, air currents carry these water vapour molecules away, and the level in the beaker slowly drops. However, if the beaker is sealed as in the illustration above, the water vapour is contained and pressure starts to build.

While evaporation is occurring within the sealed beaker, a few gaseous molecules lose kinetic energy due to collisions with other molecules and return to the liquid state. A process we know as condensation.

If the liquid within the sealed container is held at a stable temperature for a long enough period of time, an equilibrium is reached. That is to say, water vapour is being formed by the process of evaporation and at the same rate vapour is condensing back to a liquid state. The pressure at which a state of equilibrium is reached at any given temperature is known as the liquid’s vapour pressure at that temperature.

Some people crudely describe the vapour pressure as the push the liquid has to jump from a liquid to a gaseous state.  When the liquid is in a container that is open to the atmosphere, the force exerted by the vapour pressure is pushing against atmospheric pressure.

When water is heated to 100 degrees C (212 F) it has a vapour pressure of approx 14.7 psi.  The pressure exerted by the water, in an effort to jump to a gaseous state, is now equal to the downward pressure on the water exerted by atmospheric pressure at sea level.  Atmospheric pressure at this point is incapable of stopping the rapid formation of water vapour throughout the entire volume of the liquid, and the water boils.

As discussed in earlier blogs, pumps often rely on atmospheric pressure to push liquid into the eye of the impeller. If the vapour pressure is pushing up against the downward force of atmospheric pressure, then it is detracting from the pressure available to push liquid into the pump.  In terms of earlier blogs, it is reducing the net positive suction head available. (NPSHA)

Looking at the formula used to calculate the net positive suction head available,

NPSHA = Ha +/- Hs – Hf + Hv – Hvp

we see vapour pressure is shown as negative or subtractive in nature. This coincides with today’s discussions of how vapour pressure acts against atmospheric pressure and detracts from the pressure available to feed liquid to a pump.

Fortunately, the majority of liquids pumped are water-based and not particularly hot but if you have a more volatile liquid or is at an elevated temperature best consider its vapour pressure if you have low NPSHA or you may be in for pumping problems.

If you need help or advice, the Hevvy/Toyo’s team of application engineers are well versed on this subject and are always willing to help.



Industrial pump manufacturers always have available pump curves for fixed-speed pumps with full-size impellers, but what if you want to run the pump at a different RPM or use a trimmed impeller?  Well, the manufacturer should be able to provide you with a modified curve but if that is not easily obtainable just use affinity laws to modify the manufacturer’s standard curve.

Affinity laws are not courtroom jargon, just some simple formulas that can be applied to operating points on a pump curve to predict a new point when pump RPM or impeller diameter is modified.

Changing the Pump RPM: When the impeller diameter of a centrifugal pump is held constant the effect of changing the speed (RPM) is in accordance with the following formulas, where  N= RPM   Q = Capacity   H = Head, and  BHP = brake horsepower.

Q1/Q2 = N1/N2

H1/H2 = (N1/N2)2

 BHP1/BHP2 = (N1/N2)3

Changing the Impeller Diameter: The effect of trimming the impeller without changing the pump speed is virtually identical to what happens when you alter only the pump speed. Therefore the formulas are also very similar, as illustrated below, where  D= impeller diameter   Q = Capacity   H = Head and  BHP = brake horsepower.

Capacity: Q1/Q2 = D1/D2

Head: H1/H2 = (D1/D2)2

BHP: BHP1/BHP2 = (D1/D2)3

Note; Before  I show you how to use these formulas to generate a special curve there are a couple of cautionary notes.  Firstly, the accuracy of affinity laws or formulas diminishes as the percent change increases.  Generally changes of approximately 15 % or less still yield acceptable accuracy.  Secondly, when trimming an impeller you are modifying the effective length of the impeller vanes and that is what results in a modified performance.  Since the impeller vane on most pumps does not start at the center of the impeller the percent change in impeller diameter does not accurately reflect the percent change in vane length. Since impeller diameter is easier to work with than vane length, affinity laws substitute it. On high flow- low head impellers this substitution can introduce significant error and actual test tank data should be used to create modified curves instead of affinity laws.

Creating a curve for a special pump speed or impeller trim is basically the same procedure. I will therefore only demonstrate one today and that is the creation of a  pump “Head–Flow” curve.

Below is a pump curve for a 1750 rpm pump.  If we needed to create a curve for 1650 rpm we would start by calculating the shut-off point. The current shut-off point is zero flow at 125 meters as indicated by the green star. The new shut-off point at the reduced rpm would be calculated using the following two calculations

The new flow point at shut-off is calculated using the flow affinity formula  N1/N2=Q1/Q2


1750/1650 = 0/Q2     therefore  Q2 is zero M3/hr

The new head point is calculated using the head affinity formula  (N1/N2)2=H1/H2


 (1750/1650)2 = 125/H2    therefore H2 is 111 meters

Plotting zero M3/hr at 111 meters on the curve below, we establish the calculated shut-off point for the pump at an rpm of 1650.  (indicated by the blue star below)

pump curve

We would next calculate the new run-out point. The current run-out point is 17.2 M3/hr at a head of 85 meters, as indicated by the red star. The run-out point at the reduced rpm would be calculated using the same two formulas as used with the shut-off point



1750/1650 = 17.2/Q2     therefore  Q2 is 16.2 M3/Hr

The new head point is calculated using the flow affinity formula  (N1/N2)2=H1/H2


 (1750/1650)2 = 85/H2    therefore H2 is 75.6 meters

Plotting 16.2 M3/hr at 75.6 meters on the curve above, we establish the calculated run-out point for the pump at an rpm of 1650.  (indicated by the orange star above)

Additional points can be calculated to fill in the points between shut off and run out using the same procedure as used above, thereby filling in some points on the 1650 rpm curve as shown in the illustration below. (black stars)

pump curve

Finally, as shown below, connect the points, and you have a 1650 pump curve.

pump curve

In closing, this is how you can create a curve by yourself, however, if it is a Toyo pump curve, you may forget this blog altogether and just call 604-298-1213 and have our application team e-mail you what you need!



In a previous blog, I quoted the number 2.31 when discussing the relationship between water pressure  & water head.  Since then I have had a request to elaborate on the number 2.31 and where it comes from.  Frankly, with many gauges still calibrated in PSI  I still relate to the US system of measure. The quoted factor therefore only applies to pressure in units of PSI and water head expressed in feet. The basis of the factor 2.31 is as follows:

  • One cubic ft of water weighs 62.4 lbs and contains 1728 cubic inches. Therefore one cubic inch weighs 62.4/1728 or 0.0361 lbs.
  • A column of water with a cross-section of one square inch, that is 12 inches tall (one foot), would weigh 0.0361 x 12 or 0.433 lbs. Exerting a pressure of .433 lbs per square inch.
  • Based on this, a pressure of 1 lb per square inch would require a one square inch column of water equal to 1/.433 or 2.31 ft tall

PS —  In the metric system the conversion factor is 1 Kilopascal [kPa] = 0.101 974  Meters of water column [mH2O]  My apologies to my fellow Canadians for not providing the metric Stay tuned, and bye for now.


A couple of months ago I used the term “Bulk Density Ratio” and said that it was a subject for another day. So today is the day.

Pea gravel

The term bulk density ratio describes the relationship of a material or substance mass as found in a particular sample vs its theoretical mass, assuming “complete compactness” and no voids of any type.

For example, Pea gravel is made up of small particles of stone, and a solid block of stone usually weighs approximately 168 lbs per cubic foot  ( 2.7 times the weight of water, an SG of 2.7).

When we measured a particular sample of pea gravel we found that it weighed  108 lbs per cubic foot (1.76 times the weight of water). But stone is stone!! Why the difference?

Clearly, the difference is the pea gravel sample is full of voids between the stones. The ratio of solid stone to the sample of pea gravel is 168 lbs / 108 lbs or 1.55

Said differently,  pea gravel occupies 1.55 times more space than solid rock for the same mass of product. This ratio is referred to as the “Bulk Density Ratio”.

If we take the weight of the pea gravel sample and put it over the weight of a sample of the same volume of solid stone and then convert it to a percentage (108/168  X 100 =  64%) we find that stone occupies 64 % of the space. Assuming it is a dry sample, air must occupy the balance of 36% of the sample’s volume.

Great information but how is it useful in the pump world? Well, the answer to that question is that Bulk Density Ratio is the key to calculating production rates when moving products like sand or gravel that are often measured by volume.  The easiest way to explain this is by using an example.

You are a contractor and your job is to remove a sand bar made up of coarse sand that is 100 meters wide by 100 meters long by 5 meters thick, or 50,000 m3 of material.  If you can pump at a rate of 800 m3/hr of slurry with a density of 1.23,  how many hours must you operate?

You can assume the coarse sand has a dry Sg of 2.7. But the pumpage is not all sand, it is a slurry with water occupying the space between the solids.  The formula below can help you determine the percent solids, by volume.

Cv =( Sm-Sl)/(Ss- Sl) = (1.23 – 1.0)/(2.7 – 1.0) = .135 or 13.5 %

  • Cw = Percent solids by weight
  • Cv = Percent solids by Volume
  • Ss = Sg of the dry solid
  • Sm = Sg of the slurry
  • Sl = Sg of the liquid

At a slurry flow rate of 800 m3/hr this equates to .135 x 800 m3 or 108 m3 of “solid” rock per hour, but the end product is not solid!!  This is where you need to know a Bulk Density Ratio to properly estimate production by volume.

Applying the bulk density ratio of 1.55 for coarse sand would mean the contactor is removing the sand bar at a rate of 108 x 1.55 or, 168 m3/hr. Based on this rate of production, the project of moving 50,000 cubic meters being pumped at a rate of  168 cubic meters per hour will take 297hrs to complete.

Hopefully, this short blog helps to clear up some of the confusion around this subject, but if you still have some questions or need help on a specific project please feel free to contact our very competent applications team.

Bye for now


It is very important when working with centrifugal pumps that there is a clear understanding of how specific gravity, commonly abbreviated to just  SG, affects or doesn’t affect pump output in terms of head and or pressure.

Let’s first define SG. The dictionary definition is simply “ the ratio of the density of a substance to that of water”.   As shown below one cubic meter of water weighs 1000 kg, while an equal volume of granite weighs 2700 Kg. ,  2.7 times as much. Hence granite is said to have an SG of 2.7.

A word of caution! Whatever the sample size, there must be no voids. Materials such as sand or gravel must be adjusted by a factor to allow for the air space between the particles. This factor is referred to as the “bulk ratio”.  A subject for another day!

Turning attention to the head. It is sometimes defined as “the vertical distance (in feet or meters) from the elevation of the energy grade line on the suction side of the pump to the elevation of the energy grade line on the discharge side of the pump.”

When applied to a centrifugal pump’s output, it is more simply described as “the height a pump can lift a liquid to”.  For example, if a centrifugal pump tries to pump straight up a 300 ft pipe, but flow stops as the water in the pipe reaches 231 ft, the pump is said to have a shut-off head of 231 ft.

You may ask, why not express the output of a centrifugal pump in PSI? Well, the answer to that question is based on the operating physics of a centrifugal pump.

Centrifugal pumps use centrifugal force generated as a liquid is spun down the vanes of the impeller to create pressure within the pump’s casing/volute. The heavier the liquid, the more force or pressure is created.  Some pump manufacturers rate performance at a stated output pressure based on water, usually in terms of PSI, but the majority of pump manufacturers use the term head, as it does not vary with liquid SG.

The basis of the statement in bold above is illustrated to the right.  The pumps are identical, but the blue column represents water being lifted while the orange column represents a liquid with an SG of 1.5 being lifted.  Notice the output pressures vary in proportion to the SG while the rating for lift remains constant.

The centrifugal pump handling a liquid that has an SG of 1.5 is accelerating a liquid weighing 1.5 times that of water down its impeller vanes, creating a force (pressure) that is 1.5 times that of an identical pump pumping water.  The column of liquid, however, also weighs 1.5 times more than water. Hence it needs a pressure equal to the pressure on water times 1.5 to lift the heavier liquid to the same height.

The obvious question then becomes, if pressure gauges measure in PSI, how do we relate that to a pump’s output in ft or meters of the head?  The full explanation of that is again a subject for another day, but the equation below and the diagram to the right will get you by for now.




I hope today’s blog helps explain the relationship between SG, Head, and Pressure. As mentioned in today’s blog, I will in future blogs address the “Bulk Ratio” and fully explain the 2.31 ratio of head to PSI.

Stay tuned, and bye for now.


Periodically I am asked if there are self-priming slurry pumps. To answer this question accurately we must address the 3 main types of pumps separately.

Rotary positive displacement pumps utilize close-tolerance parts to prevent fluid from returning from the discharge to the suction side. This close tolerance is generally tight enough to also be effective on air, making it quite capable of evacuating air from the suction line. Although in principle pumps such as rotary gear pumps, lobe pumps, and vane pumps are self-primers, they are not typically designed to run dry for extended periods. As such the priming time must be minimized to avoid overheating and pump damage.

Reciprocating positive displacement pumps that have appropriate valves will displace air just as easily as a liquid however there may be a problem. PD pumps that are classified as contact pumps (the reciprocating component is in direct contact with the product) often rely on the product for lubrication. This makes them susceptible to heat and wears problems when running dry, so like the rotating PD pumps priming time must be minimized to avoid overheating and pump damage.

On the other hand, if you have a non-contact PD pumps, such as diaphragm pumps, these can usually run dry for quite some time but be sure that the valves are designed for high suction lift self-priming applications or there will be trouble. I know this from personal experience. Years ago I installed an electric shut-off valve between the tank and the fuel pump in my old car as an anti-theft device. One day I forgot to switch it to open. As I started the car and drove up my driveway one of the valves on the fuel pump quietly came to pieces. No fixing it, a completely new pump had to be installed.

Centrifugal pumps are the most common pumps and are used worldwide for millions of applications. Not surprisingly quite a few are in applications that are deemed as self-priming, but is the pump itself really self-priming?  To answer this question we must look closely at how a centrifugal pump works.

The operational principle of this type of pump, as the name indicates, uses centrifugal force to accelerate a mass and create pressure. Liquids generally weigh over 50 lbs per cubic foot (800kg/m³) and as such, they have enough mass that when accelerated by centrifugal force they do develop a pressure, hopefully, enough to move the liquid out of the impeller and into the discharge line. As the liquid leaves the impeller a low-pressure zone is created inside the pump and atmospheric pressure pushes the liquid up the already flooded inlet pipe into the low-pressure zone, hence keeping the pump flooded with liquid.

When a pump sits void of liquid the impeller only has air to accelerate with centrifugal force. With a weight of 0.0765 lb/cu ft (1.225 kg/m³), there is close to no pressure available to effectively create a low-pressure zone and therefore no differential pressure to flood the pump.

A person could attempt to overcome this by filling the pump with water and thereby giving the impeller a mass to accelerate, but after a few rotations, the impeller would again be trying to accelerate air.  Unless the suction line was almost already flooded the pump would have no chance of pumping. Although a standard centrifugal pump clearly can not be self-priming, engineers have worked around this issue by adding auxiliary items to the standard pump.

Priming tanks are one such item.  When these tanks are added to the pump they can in effect create a self-priming unit.  The priming tank when charged with liquid prior to start up provides sufficient liquid to flood some of the impeller allowing it to create a low-pressure zone in the suction eye. As the liquid leaves the impeller it carries with it some air as it enters a second priming tank better known as the separation tank. There, through gravity, the liquid leaves the expelled air and returns to the main priming tank. This allows the cycle to continue using the same initial charge of liquid as the suction line is progressively emptied of air. When manufacturers incorporated the priming and separation tanks into the pump they created a self-priming unit but simply called it a self-priming centrifugal pump.  See the illustration below.

Compressed Air Self-Priming Pumps like the pumps described above rely on an auxiliary piece of equipment to create a pumping unit that is self-priming.  In this case, however, it is a compressed air-powered venturi system that evacuates the air from the suction line and pump casing. Manufacturers that use this method of self-priming have an auxiliary air compressor installed on the pump that is driven off the main pump shaft. It continually supplies a jet of air into a tapered tube that is attached to a high spot in the pump suction. This, in conjunction with a special air-tight valve on the discharge side, creates a low-pressure zone inside the pump head. This in turn allows atmospheric pressure to push the liquid up the suction piping, thereby flooding the pump.

So in answer to the title question… When it comes to PD pumps there is clearly no myth, self priming pumps do exist. The standard centrifugal pump however cannot prime without the addition of auxiliary equipment. As a result, do we then say standard centrifugal pumps are not self-priming pumps?  Or do we include the priming and separation tanks as part of the pump and say self-priming pumps do exist?

I’ll let you decide.

Bye for now!


I  was recently in a discussion with an engineer regarding the use of a cutter fan to supply agitation. I was very surprised to find that the engineer, although having some pump experience did not understand the difference between the title objects. Not only could this lead to backward rotation and subsequent damage to a pump but the lack of understanding would also lead to the implementation of ineffective dredging techniques. Therefore today’s blog is on Cutter Fan agitators and shaft-connected Inducers.

The Cutter Fan also known as an Agitator is connected to the main shaft of the pump. The first manufacturer to employ this type of agitation was Hevvy/Toyo Pumps on their heavy-duty submersible dredge pumps. Their patented design utilized a curved three-blade stirring attachment that was threaded onto the pump shaft just below the suction inlet.

The cutter fan or “agitator” as it is sometimes called is typically protected by a stand attached to the bottom of the pump. For added protection on the larger pumps, Toyo places a stub shaft in between the pump main shaft and the cutter fan. Operationally, all cutter fans redirect a portion of the fluid heading toward the pump suction and push or “fan” fluid away from the pump. The agitation provided by the cutter fan dislodges solids and re-suspends them into a slurry. As these solids are drawn toward the pump inlet some of the slurries is redirected by the cutter fan back down into the solids deposit providing a more effective form of agitation than a jet of purified water. This redirecting of slurry/solids continues in a cyclic fashion forming a “pocket” of high solids content slurry directly in front of the pump suction inlet. This of course maximizes the number of solids being pumped, an important feature for any dredge pump.

The Inducer, like the cutter fan, is attached to the main shaft. It can be located anywhere in front of the impeller. The optimum location is entirely application-dependent. Technically said, its normal function is to raise the inlet head by an amount sufficient to provide the NPSHR, thereby preventing significant cavitation in the pump. In short, it can help in applications where initial priming is difficult or the fluid just refuses to flow well into the pump.  Below is a picture of a style of inducer.

In Summary, cutter fans push product away from the pump to aid in agitation while inducers help draw product into the pump. If your new pump arrives with an item that looks like an inducer or maybe a cutter fan, do not guess as to which it is and use it to confirm correct rotation. Since they are both normally attached to the shaft by some form of thread, reverse rotation may result in components unscrewing during operation and some very expensive repairs. Pumps always have a rotational arrow. The cutter fan,  and sometimes the inducer, can be excellent items to watch when bumping for rotation as they are easily visible, but when wiring the pump observe the marked arrow to obtain correct rotation!!  Once you have set rotation you can now look at the blade/vane angles and it will be easy to determine whether the mystery item is a cutter fan or an inducer.

Bye for now!