Blog Purpose

The CTG Technical Blog is intended as a source of information on subjects related to industrial and precision cleaning technology. The writer of the blog, John Fuchs, has 40+ years of experience covering the entire gamut of cleaning. Mr. Fuchs has extensive knowledge of ultrasonic cleaning technology having been employed by Blackstone-Ney ultrasonics and its predecessors since 1968. The blog is also intended to serve as a forum for discussion of subjects related to cleaning technology. Questions directed to the blog will be responded to either in the blog (if the topic has general interest) or directly by email. Emails with questions about the current blog should be entered in the comments section below. Off-topic questions related to cleaning may be sent to

Laminar Flow

September 11th, 2015

When a liquid or a gas flows uniformly and without turbulation, the result is called “laminar flow.” The most visual example of laminar flow for most of us are those arching water displays where perfectly shaped slugs of crystal clear water flys gracefully through the air from one point to another.

Paragon Pools

Laminar flow streams of water    (

When I first saw this effect at Disney World many years ago, I was confident that they had added something to the water (maybe glycerin?) to make it act that way.  I now know that it’s all done by creating a laminar flow.  Almost more amazing than the water flying through the air is the fact that there is no splash when the water re-connects with a pool of water at the other end.  In cleaning applications, the lack of splash may be the most significant property of laminar flow but it’s not always in a liquid.  Let’s dig a little deeper.

By the way, just for fun, try duplicating the above with your garden hose some time – – good luck!

Laminar flow in a tube or pipe means that all of the liquid is moving as a uniform mass at the same speed and in the same direction.

The top illustration above depicts laminar flow.  Notice that the flow is straight and uniform both in the tube and after exit.  In the lower illustration, turbulent flow results in non-uniform discharge.

The top illustration above depicts laminar flow through a tube or pipe. Notice that the flow is straight and uniform both in the tube and after exit. In the lower illustration, turbulent flow results in non-uniform discharge.

Think flow, not pressure, think low velocity, think smooth, think straight!

Although the above focused on laminar flow in a stream of water, laminar flow is also possible in nearly any confined space as well – tanks, ducts, ovens, etc.  Laminar flow is an important design parameter for several applications in industrial cleaning.  For example, it is important in rinsing applications where the most efficient exchange of water in a tank is important to minimize water use.  In cleaning, laminar flow can help assure that contaminants are  removed from the cleaning site so that they can be collected by filtration or disposed of efficiently.  In the case of clean rooms, laminar flow of air in laminar flow benches and, in most cases, the entire clean room helps assure that contaminants are not re-distributed into the clean environment. Laminar flow is easier to grasp in concept than it is to produce.Pressure and velocity are enemies  to creating laminar flow in both liquids and gasses.  Even minor irregularities in the surfaces containing the flow can lead to turbulence fast-moving flows.  Smooth walls are therefore important.  Ducting or piping should be large with flow rates as low as possible for minimum turbulation and maximum linearity.  Nozzles and fans immediately upstream of a space requiring laminar flow are totally contra-indicated.  Linearizing flow usually requires a device to straighten the flow.  In the case of a gas, laminar flow may be produced by flowing the gas for some distance through a series of parallel small tubes.  A pile of drinking straws is often used as an example in laboratory and academic situations.  Porous metals, fins/louvers and layers of open cell expanded foam are also practical solutions.  Equally important to generating a laminar flow is making sure that it is properly received at the end of its flow.  Any obstruction to flow will create turbulation.

In the case of liquids there are a number of schemes to produce laminar flow.  In cleaning, laminar flow is usually required in a tank of some description.  The use of nozzles (no matter how many) won’t do the job nor will a large number of small holes drilled in a plate with pressurized liquid behind it (which is how a lot of people approach laminar flow).  Successful laminar flow can be achieved using porous metal or thicknesses of metal or plastic sponge in line with the flow.  In any case, the entire side wall must constitute the source, not just a small window.  And again, liquid must be received and removed from the other side of the tank in the same volume as it is introduced and without creating turbulation.

In the design of cleaning machines, laminar flow is something not to be taken lightly or approached without a good understanding of the principles and mechanics involved.  It is, as I said above, a little more complex than it sounds and totally counterintuitive but, when used properly, can be a very effective tool.

–  FJF  –

Ultrasonic “Double Boiler”

September 8th, 2015

Maximum ultrasonic performance requires the most efficient transfer of ultrasonic vibrations from the ultrasonic transducers to the liquid in the process tank.  This is generally accomplished by applying ultrasonic transducers directly to the exterior surfaces of a tank containing the process liquid.  Transducer attachment techniques favor attachment to metals like stainless steel.  What if the chemistry to be used in the tank is not compatible with stainless steel?  One alternative that inevitably comes to mind is to use a metal tank with an internal coating that will be compatible with the chemistry.  Unfortunately, ultrasonic vibrations have been shown to quickly de-bond such coatings.

An option which has been proven practical in a number of applications is the use of a “double-boiler” approach where a secondary tank is positioned or floated in a primary tank made of stainless steel or another suitable metal.  Water is used to couple ultrasonic energy from the transducers mounted on the exterior of the primary tank to the walls of the secondary tank as shown below.

Double Boiler

Setting up a tank in tank system is, admittedly, a bit cumbersome and maximizing ultrasonic penetration into the inner tank may require some experimentation and tweeking.  The following are a few tips that may be helpful – –

Glass is one material known to transmit ultrasonic energy well.  In cases where it is possible, pyrex beakers make excellent secondary tanks using one or more in a single primary tank.  In most cases, beakers are either suspended by their rims using a fixture or simply “floated” in the primary tank by filling each beaker to a level that will give it some stability.  Glass beakers are seamless and, although it may be my imagination, the rounded bottom configuration seems to focus sound waves in the beaker.  In no case should the beaker be set on the bottom of the primary tank as contact between the beaker and the tank bottom will cause erosion caused by surface cavitation.  The use of beakers as described here can also be employed to evaluate various cleaning chemicals without mixing up a batch large enough to fill a standard cleaning tank.  The clear beaker also facilitates viewing the cleaning in progress.

In larger applications, quartz or plastic insert tanks are usually used.  Designing such a system requires that several factors be considered including the tank material, the shape and thickness of the material, fabrication techniques, clearances and placement of the secondary tank in the primary tank as well as other considerations that are, unfortunately, not well defined.  For example, the laws of physics say that maximum transmission will be achieved if the thickness of the secondary tank bottom is equal to 1/2 of the wavelength of sound in the material.  Although this may be practical at higher ultrasonic and megasonic frequencies, it would indicate a bottom thickness of up to several inches at the more common ultrasonic frequencies below 100kHz – a design parameter that would be difficult to meet and has not been met (to my knowledge) in any existing equipment.  In general, it has been assumed that a thinner internal tank construction will result in better ultrasonic transmission.

It can not be argued that any obstruction to the transmission of ultrasonic energy will be of consequence.  Yet, as of this time, a “double boiler” construction is the only answer we have to cavitating liquids that are not compatible with the materials of construction of commercially available ultrasonic tanks.

–  FJF  –

More Isn’t Always Better

August 14th, 2015

Earlier blogs have shown how difficult and deceptive it can be to relate energy consumed to work delivered.  In this blog I would like to offer an example of this difficulty as it relates directly to ultrasonic transducers and cleaning.  First of all, most (if not all) ultrasonic devices used for cleaning utilize a number of individual vibrators attached to a plate or diaphragm which, in turn, is vibrated by the individual vibrators to produce sound waves in a liquid media contacting the plate on the side opposite the individual vibrators.  Sometimes we call the entire assembly consisting of the vibrators and the plate the transducer, but in other cases we refer to the individual vibrators as transducers as well.  We have also discussed in earlier blogs that these vibrators (or transducers) depend on resonance at a particular frequency or frequencies (in the case of multiple frequency transducers) to deliver the desired high amplitude of vibration required to produce cavitation and implosion in the liquid.  Each of the individual vibrators is capable of converting approximately 40 watts of electrical energy into mechanical energy as vibration.  The actual power depends on the design of the vibrator which varies from manufacturer to manufacturer.  The power of each vibrator is confined to some degree by the fact that only so much energy can be transmitted per given area of the output face of the transducer into the liquid in contact with it.  Increasing the power to the vibrators is deleterious to their function and efficiency of energy transmission.

Note – An analogy to the above can be found in the incandescent light bulb.  Increasing the output of the bulb entails increasing its size accordingly to accommodate a larger filament and provide the necessary cooling, etc. to provide a practical life.  The ultrasonic vibrator, however, because of the need for resonance is limited in size and, therefore, capacity.

Our discussions regarding ultrasonic transducers to this point have focused on the vibration produced along the axis of the transducer (push and pull if you will).  There are, however, many other modes of vibration as well.  For example, the following illustration shows that as the vibrator attached to a plate vibrates up and down, a resulting wave moves through the plate lateral to the axial vibration.  This is called a shear wave.

A vibration at one point of a plate initiates shear waves in the plate which radiate like ripples on the surface of a pond from where a pebble has been dropped.

A vibration at one point of a plate initiates shear waves in the plate which radiate like ripples on the surface of a pond from where a pebble has been dropped.

The next illustration shows an ideal situation where vibrators are spaced in such a way that each adds to the lateral motion of the shear wave initiated by the adjacent vibrator.

Ideally spaced transducers will complement one another in the propagation of the shear wave in the plate.

Ideally spaced transducers will complement one another in the propagation of the shear wave in the plate.

At times, there is an irresistible temptation to increase ultrasonic power in the interest of, supposedly, providing more ultrasonic energy.  This can be driven by actual need but, in many cases, is just for the sake of an assumed more powerful and, hence, effective ultrasonic transducer.  Since the transducer plate size is often limited, and since each vibrator is capable of only providing a limited amount of power, the inevitable result is that more vibrators are added.  This, of course, requires the vibrators to be placed closer and closer together which, in turn, may lead to a situation where some of the vibrators act as the wheel rotating in the opposite direction discussed in an earlier blog.

Improper placement of ultrasonic vibrators can result in adjacent vibrators working in opposition with regard to the shear wave.  The result is a reduction in useful vibrational energy transmitted into the liquid contacting the plate.

Improper placement of ultrasonic vibrators can result in adjacent vibrators working in opposition with regard to the shear wave. The result is a reduction in useful vibrational energy transmitted into the liquid contacting the plate.

In the above case, the red and black vibrators are working in opposition to each other.  Each, however, is consuming its allotted electrical energy.  So, although the power consumed has been increased, the resulting ultrasonic effect has not benefited linearly with the added power and, in some cases, may be reduced.

The above is an example of just one of many considerations in designing an effective ultrasonic cleaning system – – and just one example of a case where more may not be better.

–  FJF  –

Cleaning Sintered or Porous Parts

August 11th, 2015

Most surfaces that we encounter in industrial cleaning are relatively smooth and contiguous.  We have talked earlier about the difficulties of cleaning blind holes, threads, capillary spaces and other challenging configurations. The one surface we haven’t yet explored is that of a material that is itself porous.  Sintered materials including metal, ceramic, glass, plastic and others immediately come to mind.

Note – Sintering is a process in which small particles of metal or any of many other materials either isolated or in combination with others are compressed, heated or in some other way caused to fuse together to a greater or lesser degree depending on the application to form a solid shape.  For some parts, the fusion is complete rendering a part that is essentially solid and without porosity.  This process is an economical way to make odd-shaped metal parts that would be difficult to produce using other means (such as machining) as well as to produce a stronger part overall.  Partial sintering yields parts that will absorb or actually allow liquids or particles below a particular size  pass through them.  Sintered parts impregnated with lubricants are used in some bearings to provide a self-lubricating feature as the lubricant leaches out of the sintered material over the life of the bearing.  Porous sintered material is also used as a filter in some applications.

Filters of many varieties including those that are intended to be cleaned and reused, as well as those that are not, offer a cleaning challenge because of their porous nature.  As discussed below, some filters are, indeed, able to be cleaned while others are difficult to impossible to clean depending on the material and features of construction.  For a little more about cleaning filters, check out this previous blog.

Ultrasonic cleaning has proven effective in cleaning sintered parts and is recommended almost universally, especially for cleaning sintered metal and ceramic filters.  Successful cleaning, however, may require a little more than simply plunking the part into an ultrasonic cleaner for 30 minutes or so.  Cleaning a porous part provides something of a dual challenge.  Even if the part only needs  to be “superficially clean,” there is always a potential for contamination seeping out from the interior and re-contaminating the surface over time.

When cleaning a porous part, the ideal situation is to use a method that will cause cleaning liquid to penetrate and pass through the porous piece.  Without liquid flow, contaminants interior to the piece will not be removed easily if at all.  Sprays, turbulation, agitation and even ultrasonics frequently fail to reach the interior of a porous part.

When the contaminant consists either partially or wholly of particles, as in the case of filters, the challenge is further magnified as particles become trapped or “wedged” in the porosity.  Here, the solution is to provide flow in a direction opposite to that which caused the particles to become trapped in the first place.  In the case of filters, this is not difficult.  Simply flow through the filter in the direction opposite to that normally seen. If the direction of flow is indeterminate a flow originating from many directions (as in rotating the part) will probably give the best results.

In cases where it is not possible to force liquid to flow through the porous material, similar results can be achieved by multiple dipping of the part allowing the absorbed liquid to drain out between dips.  This can also be assisted by the application of a vacuum (using a vacuum hose or other direct means of connecting to the part) between immersions.

Under no conditions should cleaning a porous part be considered a “no brainer.”  And, for the same reasons that these parts are difficult to clean, cleanliness testing is similarly difficult.  Remember, in order to find contaminants, one generally needs a cleaning method that is better than that used for production cleaning.  In the case of porous parts, there may not be a better means available.  Extraction using a vacuum or long periods of immersion offer the most likely candidates.

–  FJF  –

Electricity Behind the Walls

August 4th, 2015

Electricity is not something we give considerable attention to in our daily lives.  At home or at work (or wherever) there are electrical outlets.  We simply plug devices into the outlets and the devices light up, heat, rotate, vibrate, charge or do whatever they’re supposed to do.  What’s behind the outlet, generally, is of little concern to us.  However, how this all happens is interesting and, sometimes, important.  Those of us who have done some world travel know that electricity in various countries and regions varies in voltage, but what we may not know, is that there are some major differences in the connections behind the walls that may, in some cases, be of significant consequence.

15A OutletLet’s start with the United States where electricity is available at an outlet that looks like the one shown at the right.  Used in most of North America, this is a standard 120 volt outlet found in homes and offices.  Most of us know that the two “slots” are the actual electrical conductors and that the other hole which is sort of a “D” shape turned on its side is the ground.  Of the two slots, the wider (taller) is intended to be the “ground” while other is the “hot” side of the circuit.  The “ground” slot and the “D” are both at ground potential but are connected to the distribution box by separate wires.  This information is usually plenty for the casual user of electrical outlets but it is interesting what goes on behind the walls.

The electricity starts out at the power plant.  Before it reaches the distribution panel in your house, there is a transformer which reduces one phase of a three phase source from many thousands of volts down to that appropriate to power various devices to be connected.  In general this is 120/240 volts in a “split phase” configuration.

Note – 120 Volt power may also be derived directly from a three phase source with each phase providing 208 Volts.  In that case, wiring is totally different and will be the subject for another blog.

"Split Phase" delivery of 120 and 240 volt power in the United States.

“Split Phase” delivery of 120 and 240 volt power in the United States.

A transformer reduces the voltage from the primary source down to 240 volts RMS in the secondary coil.  The secondary coil has a center tap which is connected to ground (ground rod, water pipe or other “solid” ground).  As the diagram above shows, this configuration provides two 120 volt RMS sources.  Devices operating at 120 volts RMS (outlets, lamps, appliances, etc.) are connected to one of these two sources in a somewhat arbitrary manner with 1/2 of the devices operating with alternating current that is 180 degrees out of phase with the others.  These two phases are the reason for the term split phase.

In the case of devices that use large amounts of electricity (stove, heat, air conditioning, etc.) the differential between the two phases is used to provide a 240 volt RMS source as shown below.

Split phase power source provides 240 volts by the differential of the two 120 volt sources that are 180 degrees out of phase as shown above.

Split phase power source provides 240 volts by the differential of the two 120 volt sources that are 180 degrees out of phase as shown above.

The important thing to note here is that NEITHER of the “hot” wires is ground although devices operating from 240 volts RMS are required to have a “ground” wire connected to the earth ground for the purpose of safety!  There are hybrid devices that utilize both 120 and 240 volts RMS (some stoves and air conditioners for example).  These devices must have 4 wires.  One connected to each side of the 240 volts RMS source, one connected to the centertap/ground of the transformer (to act as the neutral for the 120 volt RMS source(s) ) and another connected to the earth ground for safety.

The above is a convention for the United States.  Other countries and areas (sometimes even areas within a single country) may derive various voltages using totally different schemes.  This may make things a little tricky, especially when it comes to grounding.

–  FJF  –

Three Phase Power – Why and How

July 15th, 2015

Three phase power is commonly used where large amounts of electrical power are required.  Examples are power transmission lines and large industrial machines which require considerable power.  What is three phase power anyway?

Let’s start by thinking of a two wire transmission line as shown below connecting the source of alternating current power to the location where it will be used.

Two Wire Transmission

The amount of power that can be delivered to the destination is limited by the size of the wires and the distance.  Now let’s say that there is a need to double the amount of power transmitted from the source to the destination.  There are a few simple ways to do this.  The first, and maybe most obvious, would be to add two more wires thereby duplicating the capacity of the single transmission line.  Or, the size of the wire could be increased to accommodate more amperage without an increased loss of voltage at the destination.  Or a transformer could be used at the source to increase the voltage (and lower the current in the transmission line) and a second transformer at the destination to reduce the voltage to that required.  This option, of course, would require insulators with increased capacity along the transmission line to prevent additional losses.  Larger wire and/or transformers and insulators are costly but feasible in some instances.

Note – A previous blog discussed the fact that wire is limited in its capacity to conduct amps but that increased voltage doesn’t matter as long as the wire is adequately insulated.

But there is another practical option which is why the following illustration looks much more familiar to us than the one above.

Three Wire Transmission

When one looks at the current in a single phase circuit, it  can be seen that the full current carrying capacity of the wires is not utilized at all times.  In fact, at some points in time the wire is not carrying any current (or voltage) at all.

Absolute Current Single Phase

With only two wires, there is no way to more fully utilize their current carrying capacity for more of the time.  Adding a third wire, however, offers the opportunity to capture this unused capacity as shown below.

Using three wires instead of two allows more efficient utilization of the wires' current carrying capacity.

Using three wires instead of two allows a greater utilization of the wires’ current carrying capacity for more of the time.  In the diagram at the right, more current is flowing for more of the time.

With 3 phase power, there are 3 alternating current sources (one between any 2 of the 3 wires).  Each is out of sync (phase) with the other two by 1/3 of the 60 cycle frequency.  This asynchronous scheme allows each wire to do “multiple duty” contributing to each of the phases during different parts of the cycle.  The added efficiency is such that by increasing the number of wires by 50 percent (from 2 to 3) provides a 73 percent increase in power transmission capability using wires of the same capacity and with the same voltage delivered on each phase compared to that of a single phase.  Because of this increased utilization of the wires’ capacity and the fact that three phase power is usually available at higher voltages than single phase power, three phase power is favored wherever large amounts of electricity are to be used or transmitted.  Virtually all power transmission lines utilize the benefits of a three phase configuration and high voltage to transmit power over long distances with maximum efficiency which is why we see groups of three wires littering the countryside.

–  FJF  –


A Little About RMS (Root Mean Squared)

July 10th, 2015

Most of us are aware that the power we get from the power line coming into our house or factory comes as alternating current.  Basically, the voltage on the “hot” wire as referenced to ground varies, going negative then positive 60 times each second (60 Hz).

AC vs DC

The benefit and compelling argument for using alternating current to transmit electrical energy is that it is relatively easy to change the voltage of the alternating current using transformers (which can not be done with direct current).  The ability to convert voltage makes it possible to distribute large amounts of electrical power over great distances using conductors of practical size.

Note – The amount of current (amps) that a wire or other conductor can carry without a significant loss of power is determined by its material of construction, size (gauge) and length.  Long distance power transmission lines operate at very high voltage to transmit the same amount of power with smaller gauge wires than would be required at a lower voltage.

The RMS voltage of any waveform will produce the same heating effect on a resistive load as if the same direct current voltage were applied to the same resistive load.  RMS stands for Root, Mean, Squared.  To understand this better, let’s start with the portion of the sinusoidal wave where the voltage is positive.  To find the RMS voltage, we would take the voltage at every point and, first, calculate their square roots.  Having done this for every point, we would add all of these square roots together and then divide by the total number of points.  Finally, we would take the average of the square roots and square it to get the RMS voltage.  Of course, this can be done using some simple calculus but I’ll leave that for the math geeks – concept is what we’re going for here.  The result of all this effort will give us the RMS voltage.  One should not confuse RMS with average voltage which, except for a few special cases, is somewhat skewed from taking the simple average of the voltage over time by the fact that we are going for power, not really voltage, and power includes a squared factor (P=I²R).  In the diagram below, the blue area (representing the sinusoidal voltage) will produce the same heating effect in a resistive load as the yellow area which represents the RMS voltage.

RMS Positive Cycle

Now let’s look at what happens when the sinusoidal voltage goes negative – this is a bit more tricky.  The square root of a negative number is, well, difficult.  As we all know, if you multiply a negative number by a negative number, the result is always positive – – almost always.  Through the magic of mathematics, there are imaginary numbers which are represented by the addition of an i to a number so that, for example, the square root of -2 is 1.414i .  Squaring 1.414i gives you 2, not -2.  As a result, the RMS value remains positive despite the fact that the voltage is negative for 1/2 of each cycle.  This makes sense as the heating effect on a resistor is the same no matter which way the current is flowing.

RMS Both Cycles

Finding the RMS voltage of a sinusoidal wave signal is easy because some pretty simple rules apply.  The RMS voltage of a sinusoidal is always the peak voltage divided by the square root of 2 (1.414).  Yes, this means that the peak voltage of 120 volt RMS power is around 170 volts and that the peak to peak (maximum positive to maximum negative) is 340 volts.  And guess what, 340 divided by the square root of 2 is 240!  But that’s another blog.

Most of us will never encounter the need to determine the RMS voltage of a waveform other than one that is sinusoidal.  But it is important to remember that electrical energy can come in many wave shapes.  The simple rules above ONLY apply if the voltage is sinusoidal.  In order to accurately determine the RMS voltage of shapes other than those that are sinusoidal (square waves, sawtooth waves and ramp functions, for example) requires different math.  Most common meters accurately measure RMS voltage ONLY if the signal is sinusoidal!

–  FJF  –


Ultrasonics – The Trouble With Watts – Part 2

June 26th, 2015

As discussed in a previous blog, one problem with watts is that watts do not equal energy.  Let’s carry that thought a little further again using the vehicle analogy I put forth in the initial blog of this series.

In the ultrasonic world it is common to associate the effectiveness of an ultrasonic cleaning tank its rate of energy consumption (Watts)  either from the power source or in the delivery line to the transducer.  Preceding blogs illustrate (in part) why this association is basically flawed from a strictly physical standpoint.  But the problem goes even deeper.  Imagine a vehicle traveling from point A to point B.  How do you determine how much gasoline it will take to accomplish this?  The “goal” is to move the vehicle from point A to point B.  Using the laws of physics, we can easily determine how much energy this will take.  The amount of energy required to accomplish this task is a constant and never changes.  To move the vehicle from A to B requires applying a force over distance.  This is the actual amount of energy (gasoline) needed and requires adding the energy required to overcoming inertia to start the motion (acceleration), the energy required to overcome the inertia of the moving object once the destination is reached (braking) as well as static and moving friction as the vehicle is moved (surface roughness, bearings, etc.) all in addition to that constant amount of energy that is required to move from point A to point B.  All of these things are pretty straight-forward and stand to reason.

Just for fun, let’s throw in some other grantedly rather absurd but possible variables which could have a significant effect on the amount of energy required to move the vehicle from point A to point B.  Let’s say, for example, that this is a four-wheeled vehicle with four wheel drive but that one of the wheels is geared to rotate opposite to the direction of the other 3.  It is not difficult to imagine, in this case, that more energy would be required to move from point A to point B because energy delivered to the counter-rotating wheel is totally lost and, in fact, results in the other three wheels expending more energy than they would otherwise.  In another scenario, let’s imagine that the road surface is icy.  Now the vehicle may not move at all no matter how much energy is expended.  In summary, it becomes obvious that the amount of energy required to move a vehicle from point A to point B can vary widely depending on a number of other conditions.  Under ideal conditions, all of the energy delivered should contribute to the desired result but that is seldom the case.

And yet in the ultrasonic world, we seem hung up on watts as a way to express energy in an ultrasonic cleaning tank no matter how inappropriate that might be.  In fact, the only energy that really counts is that released as cavitation bubbles implode near the surface to be cleaned.  So, what happens to all that energy represented (as it is) in watts that doesn’t find its way to the site where cleaning is actually taking place?  We’ll take that up in an upcoming blog.

–  FJF  –

Ultrasonic “Shadowing”

June 22nd, 2015

We in the ultrasonics industry have long been aware of an effect which is sometimes called “ultrasonic shadowing.”  In general, this is what happens when parts being cleaned are positioned in such a way that parts cast an “ultrasonic shadow” which prevents parts in the shadow from being effectively cleaned.  This phenomenon, although we know it exists is quite difficult to characterize and predict.

Ultrasonic shadowing does not produce a distinct shadow as the case when an opaque object casts a shadow from a distinct light source like the sun or a bare light bulb.  The following illustration will help clarify this point.

An object will cast a more distinct shadow on a bright, sunny day.

An object will cast a more distinct shadow on a bright, sunny day.  A larger object will produce a darker shadow on a cloudy day as more of the diffused light is blocked.

On a bright, sunny day, an opaque object held between the sun and a surface such as a sidewalk will cast a distinctly outlined shadow of the object.  On a cloudy day, however, where the sunlight is distributed and diffused by a layer of clouds (for example), the outline of the shadow becomes less distinct as the light from a distributed source is able to illuminate what would be the shadow from a more concentrated light source such as the sun.  Even in the cloudy day scenario, however, if the object casting the shadow is large enough, there will be a dark area where even the diffused light can not reach.  The ultrasonic shadow is more like that produced on a cloudy day.  This is because the ultrasonic energy source is distributed like the light on a cloudy day and not a point source like the sun.  Even if the ultrasonic energy was delivered from a single point, there would likely be a distribution due to reflections and general diffusion of the energy produced by the transmitting liquid.

The other effect that tends to “temper” the sharpness of ultrasonic shadowing is the fact that, for the most part, items being cleaned are at least somewhat transparent to ultrasonic energy.  It can be easily demonstrated that a thin plate of stainless steel held between an ultrasonic source and an object being ultrasonically cleaned has very little effect on the transmission of ultrasonic waves.  That is not to say that this effect will go on indefinitely.  As more thin plates are added, more and more ultrasonic energy will be lost to the slight attenuation in each plate and also to reflections between the plates.  A similar effect can be seen if a light source is obscured by several sheets of glass that are separated from one another.  Eventually, all of the light is reflected away.

The “take away” here is that although ultrasonic shadowing is real and does have an effect on the strength of the ultrasonic field, it is not a sharply defined phenomenon.  Racking of multiple parts for ultrasonic cleaning is possible even if one part is partially “hidden” from the ultrasonic source.  At some point, however, there will be a reduction of the ultrasonic field and subsequent cleaning effect as part density becomes more than the ultrasonic energy can overcome.  The best way to determine if cleaning problems are related to ultrasonic shadowing either by the number of parts or as an effect of the fixture is to try cleaning a single part (hung on a wire, for example) in the tank.  If it is possible to clean a single part but not a full load of racked parts then there is a possibility that ultrasonic shadowing may be the cause.

– FJF  –

The Trouble With Watts

June 8th, 2015

Some time ago I wrote a paper entitled “What is a Watt.”  Although this paper seems lost in history (I can not find a copy of it), I can remember the point that I was trying to make when I wrote it.  In essence, it made the case that a watt is only an instantaneous measure of the rate of energy development or consumption (power), not a measurement of overall energy.  Part of the problem is the fact that “Watt” is a somewhat unique unit of measure.  A watt equals a rate of energy consumption or generation of 1 joule of energy per second.  If you substitute joules per second for watts, it is easy to see that watts does not equal energy (joules) but, rather, a rate of generation or consumption of energy (joules).  You need to integrate the rate of energy consumption or generation over the amount of time the energy is applied to achieve a measure of accumulated energy.  I say that a watt is unique because in most cases the rate of achieving a total quantity is defined as the amount of that quantity per unit of time  eg. miles per hour.  An example? – There is no unit that I know of that is the equivalent of miles per hour.  The closest is the knot which is one nautical mile per hour.  Using a reverse analogy, it might be simpler if we adopted the term Andretti as the equivalent of one mile per hour. Our speedometers would be marked in Andrettis, not miles per hour. In simple terms, Andretti’s would be to miles as watts are to joules.

Watts can be related to miles per hour as the rate of travel but not the total distance covered.  Miles per hour, for example, is not an appropriate response to the question, “How far is it from Detroit to Cleveland.”  Similarly, watts, by itself, is not an appropriate answer to, “How much ultrasonic energy is there in this ultrasonic tank.”  The dimension of time is missing.  Think of it this way – – If you were on your way from Detroit to Cleveland you could drive at a constant speed of 65 miles per hour or, you could drive at a speed of 80 miles per hour from Detroit to Toledo (60 miles) and then at a speed of 45 miles per hour from Toledo to Cleveland (118 miles) on the snow-covered Ohio Turnpike.  You could boast that you had traveled at 80 miles per hour in the course of your trip but, in truth, you had really only travelled at a speed of 52.8 miles per hour overall for an accumulated distance of 178 miles in 3.37 hours.  If you had travelled at a continuous 80 miles per hour for the entire trip, you would have been at your destination in 2.23 hours.  In summary, miles per hour does not describe the distance travelled any more than watts describes the amount of energy in an ultrasonic cleaning tank.

Sometimes speed is important while at others, it is distance that counts.  Although both move by moving their legs, a sprinter wants speed while a hiker wants distance.  In the ultrasonic world, any power in a liquid above that required to produce cavitation at a given frequency will provide cleaning.  A higher rate of delivering power (watts) to the liquid will result in larger numbers of more energetic cavitation bubble collapses and could result in faster and, perhaps better, cleaning but not always as we will see in upcoming blogs. 

–  FJF  –