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
Double
Tyler
Series
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 .

Cheers,

RJ

We have previously talked about “slurry specific gravity” with a view to helping everyone understand its numerical value, but how does it really help pump owners? Today, we will look at how the knowledge of specific gravity can help us determine production rates when the game’s name is moving solids.

Dealing with plain liquids is all about how much liquid you can move in a certain period of time. It is always measured in units of volume per time (e.g. USGPM or litres per second); however, when using a liquid to move a solid, we need to be able to quantify the number of solids moved or, in other words, the amount of production.

When dealing with a slurry, the production rate becomes a function of both volume (flow rate) and concentration of solids (density of the slurry). You will need both values to determine the rate of production. Flow rate is easily established with flow meters, and the density of the slurry can be established by weighing a known volume of slurry and using it to determine the specific gravity (Sg) of the slurry.

Production can be measured in terms of volume of product/solid moved or in terms of weight of product/solid moved. To use a sample to determine concentration by volume directly requires the drying /complete removal of the carrier liquid from a known volume of slurry. This can be a time-consuming process and, as such, is normally avoided. Instead, we usually solve for the concentration of solids by weight first, and if required, use a formula and the solids by weight to establish the concentration by volume.

To determine a concentration by weight, we must first establish the Sg of the slurry. To do this, we simply obtain a sample of the slurry and use the measured weight in relation to that of water, thereby giving you the Sg of the slurry (You can refer to last month’s lesson for more detail). With the slurry Sg, the known Sg of the dry solid, and the Sg of the carrier liquid, we can use the following formula to calculate the concentration of solids by weight.

Sgm =   1 / {(Cw / Sgs ) + [(1 – Cw) / Sgl]}

Sgm = specific gravity of slurry   |   Cw  = concentration of solids by weight

Sgs =  specific gravity of the solids    |   Sgl =  specific gravity of liquid without solids

Example: Given that the slurry Sg is 1.23, the Sg of the solids is 2.7, and the Sg of the liquid is 1.0, the concentration of solids by weight would be as follows:

Sgm =   1 / {(Cw / Sgs ) + [(1 – Cw) / Sgl]} 

1.23 = 1 /  {(Cw / 2.7)  + [(1 – Cw) / 1.0]}

Solving for Cw, we obtain 0.30 or 30%  

If you are like me and have trouble with the mathematics of solving the equation above, first solve for the concentration of solid by volume, then use it to convert to solids by weight. This is done as follows.

If we know the slurry Sg, the Sg of the dry solid, and the Sg of the carrier liquid, we can use the following formula to calculate the concentration of solids by volume.

Cv=(Sgm –  Sgl)/(Sgs-Sgl)

Sgm = specific gravity of slurry   |   Cv  = concentration of solids by Volume

Sgs =  specific gravity of the solids   |   Sgl =  specific gravity of liquid without solids

Example: Given that the slurry Sg is 1.23, the Sg of the solids is 2.7, and the Sg of the liquid is 1.0, the concentration of solids by volume would be as follows:

Cv =  (1.23 – 1.0) / (2.7 – 1.0) = 0 .23/1.7 = .135

Or 13.5% by volume

It is important to note that the mathematical concentration of solids by volume does not take into account another important factor, the bulk density ratio.  As such, volumetric concentration on its own can not be directly used for real-world volumetric calculations.

Please see a previous lesson titled “Bulk Density” for more info on that subject.

We can now apply the formula below to convert from a concentration by volume to a concentration by weight.

Cw=Cv x Sgs / Sgm

Sgm = specific gravity of slurry    |    Cv  = concentration of solids by Volume

Sgs =  specific gravity of the solids    |    Cw  = concentration of solids by weight

Example:  Given that the slurry Sg is 1.23, the Sg of the solids is 2.7, and the concentration of solids by Volume  is 13.5%, then the concentration of solids by weight would be as follows:

Cw =   13.5 x  2.7 / 1.23   =   30% by weight 

As you can see from the examples used above, a slurry with an Sg of 1.23 may have a concentration by weight of 30% while also having a concentration by volume of 13.5 %. This difference illustrates the need to clearly identify whether the percentage of solid is being stated by weight or volume when discussing production figures.

In closing, remember to factor in “bulk density” if any real-world calculations of solids by volume are required.

 

Cheers

RJ

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.

Cheers,

RJ

What is a slurry?  The dictionary definition of slurry is a thin mixture of an insoluble substance, such as cement, clay, or coal, with a liquid, such as water or oil. The word thin when referring to the mixture is important as it is used to separate a slurry from a paste. A paste is a thick mixture of an insoluble solid with a liquid.

So when is a slurry too thick to be called a slurry and would better be referred to as a paste?  The general rule is if it flows such as the coal slurry shown on the image below does, it is a slurry.

Pipeline slurry transport

If the product does not flow under its own weight, such as the photo of the soldering paste below, it is referred to as a paste.

A slurry can be described in many ways, but when selecting a pump, it is first important to define why you are pumping a solid. Here, there are generally  three options:

  1. You are trying to relocate a liquid that may be contaminated with unwanted solids.
  2. You are trying to move a slurry of a fixed percent solids content.
  3. You are trying to use a liquid to relocate a solid.

A good example of the unwanted solid slurry would be the dewatering of a mine. Pumps in this type of application are normally made of cast iron, and every effort is made to minimize solids content. When the solids content is very low, generally less than 2% slurry, pump manufacturers refer to this as a dirty water application (not a true slurry application).

Dirty water Stock Photos, Royalty Free Dirty water Images | Depositphotos

The second category, where you are trying to move a slurry with a fixed percent solids content, is most often associated with an application that is part of an ongoing process.  Often, these are non-settling solids or ones transported at such a rate as to prevent settling.  A good example of this would be the transportation of pulp stock in the pulp and paper industry.

The third category is the toughest. Here, you are trying to move as much solids as possible, generally with as little liquid as you can. An example of this type of application would be the relocation of mine tailings.  Here, every effort is made to maximize solids content so as not to waste money/energy moving a liquid that is not required other than as a carrier for the tailings. Depending on the specific gravity of the dry solid slurries of 70% solids (by weight) are not uncommon and still flow well.

Hevvy/Toyo Pumps, while making all types of slurry pumps, concentrates mostly on the last category discussed today.  Within this category, there are a host of parameters that are used to help describe a slurry. Things like particle distribution, particle shape, particle hardness, SG of the dry solid, bulk density, percent solids content and PH of the liquid just to name a few. These are all subjects for another day, but I will get to each of them soon.

Cheers RJ

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.

Cheers

RJ

In the realm of heavy centrifugal pumps and industrial rotary equipment, ensuring the longevity and optimal performance of your assets is paramount. When it comes to repairs, there’s an often-underestimated gem in the world of maintenance: true OEM repair.

  1. Authenticity translates to reliability

When your industrial equipment undergoes an OEM repair, you are essentially breathing new life into it with components designed and crafted by the very same experts who created the equipment originally.

This ensures a seamless fit, impeccable compatibility, and adherence to the exacting standards that the manufacturer upholds.

  1. Cost Savings

By leveraging the expertise only a manufacturer can have of its products, OEM repair services conduct a thorough diagnosis to identify and replace only the damaged or worn-out parts, sparing you from the expense of an entirely new pump.

This financial advantage not only safeguards your bottom line but also liberates valuable capital that can be allocated to other critical areas of your business operations.

  1. Reduced Downtime

Downtime can result in significant losses. An OEM dedicated repair team, well-versed in the intricacies of your equipment, can swiftly pinpoint the issue; sourcing of the necessary components isn’t an issue as they often have the most-used parts in stock.

This efficiency can be particularly crucial if you operate around the clock or have tight production deadlines.

 

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

OR

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

OR

 (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

N1/N2=Q1/Q2

OR

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

OR

 (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!

Cheers

RJ

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.

RJ

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

RJ

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.

RJ