Centrifugal Air Blowers

Here are the three fan laws:

http://www.delhi-industries.com/Engineering/Tips/FanLaws.asp

To directly answer your question - it is Law #1: Speed is directly proportional to output volume.

The fan law is still applicable at a high altitude? How to correct the speed for the altitude compensation?

Fan law is still applicable at various altitudes, but mass flow is proportional to air density at non-standard conditions. Power is also proportional to air density.

CFM1=CFM2 (RPM1/RPM2)

PRESSURE1 = PRESSURE2 ((RPM1/RPM2)^2) (DENSITY1/DENSITY2)

HORSEPOWER1 = HORSEPOWER2 ((RPM1/RPM2)^3) (DENSITY1/DENSITY2)

PENDING ALTITUDE, YOU CAN FIND YOUR AIR DENSITY WHICH IS DEPENDENT ON DRY BULB AND WET BULB TEMPERATURES AND NOT JUST ON ALTITUDE PRESSURE

IF VELOCITY IS GREATER THAN 100 FT/SEC I WOULD ALSO USE COMPRESSIBILITY FACTOR - ALL CAN BE FOUND IN ANSI/AMCA 210-99

Hi, nzChinese!

Further to the above reiteration of the obvious in basic physics for you by Guest:

I'm concerned that you are getting too much information to understand how it applies to your original question.

Don't be confused. There is no change in pressure by pushing air through a centrifugal pump. A centrifugal pump is only a fan that takes the air in through a larger passage and pushes it out through a smaller passage so that it can be sent a longer distance in a more effective manner so it can do work. In fact, it is rarely used to pump air, being more widely recognized as a water or (less often) a fuel pump.

The only change in pressure that any pump could make in air pressure could take place if you pushed more air and more of it and more of it and more of it into a closed space that it could not escape from; and a centrifugal pump could not do that very well as it's not designed for that purpose. If you tried to do that with a centrifugal pump, the air would simply stop flowing through the pump.

^{CFM1=CFM2 (RPM1/RPM2)} This means that no matter how much air you run through your blower, you can't increase it to more air magically coming out than went in. Want more air? Increase the RPMs. You still only get out as much as you put in.

^{PRESSURE1 = PRESSURE2 ((RPM1/RPM2)^2) (DENSITY1/DENSITY2)} This means that since the density of the air (measured e.g. in grams/cc or kg/m³) never changed magically by going through the blower either, it's still the same old air with the same old density, and therefore the pressure (e.g. measured in mass per unit area...probably atmospheric) going in is the same pressure that comes out. The person who figured this out for us was Bernoulli.

According to Wikipedia, "Bernoulli's equation can be summarized in the following memorable word equation:

static pressure + dynamic pressure = total pressure^[10]

Every point in a steadily flowing fluid, regardless of the fluid speed at that point, has its own unique static pressure p, dynamic pressure q, and total pressure p₀.

The significance of Bernoulli's principle can...be summarized as "total pressure is constant along a streamline." Furthermore, if the fluid flow [air is a fluid for this purpose] originated in a reservoir, the total pressure on every streamline is the same and Bernoulli's principle can be summarized as "total pressure is constant everywhere in the fluid flow."

^{HORSEPOWER1 = HORSEPOWER2 ((RPM1/RPM2)^3) (DENSITY1/DENSITY2)} This is just going to be confusing for you, because it means that different densities of air, which you are not dealing with, require a change in horsepower application. No point in confusing you with advanced concepts to answer your simple inquiry. Your air density is the same going in and out of your pump because it is the same air.

Another point of interest, which has no bearing on your centrifugal pump, is "^{PENDING ALTITUDE, YOU CAN FIND YOUR AIR DENSITY WHICH IS DEPENDENT ON DRY BULB AND WET BULB TEMPERATURES AND NOT JUST ON ALTITUDE PRESSURE.}" While this is an interesting thing to notice, keep in mind that if you want to know what the density of your ambient air is for any reason, keep this rather complicated method for another time. Just check a thermometer and a barometer and check the gauge pressure on a distance-above-or-below-sea-level scale. If the temperature and pressure is in the normal range, the density of air will be somewhere around 1.2 kg/m^3. Do you really need that information? No. And there's no point in getting out your sling hygrometer to calculate the humidity/altitude components of the air either.

Guest has also included this proposition: ^{IF VELOCITY IS GREATER THAN 100 FT/SEC I WOULD ALSO USE COMPRESSIBILITY FACTOR - ALL CAN BE FOUND IN ANSI/AMCA 210-99}. This amount of additional information might confuse you too, until it comes to calculating the amount of work your pump can do with the air flow it produces. However, it is important to realize that air is not compressed through a centrifugal pump, as the amount of air coming out is still the same as the amount being put in. Just gets speeded up through narrower outlets. A "compressibility factor" does not mean "it is compressed". It is a mathematical concept called a "factor" used as a shortcut to

• determine how the velocity of the air might be useful in compression.
• how compressible a gas is, and
• when testing blowers to help determine the work that can be done by air at greater or lesser velocities by using the greater amount of air at speed to create a difference in the amount of air at any given moment (pressure) on a surface between the high velocity air side and the low-to-zero velocity side of that surface.

In other words, when the high velocity air is used to do work by changing the pressure on a surface as though the air were pressurized by changing the amount of air on that surface at a given moment, there is an accumulation of more air on that surface. And this is where your confusion arises. The pump only changes the velocity. The velocity is converted into an air pressure differential only during its application to do work.

In another example of the application of change of air velocity through a narrower or wider aperture, air traveling at a high velocity excites its molecules and causes heat generation (HOT PIPE). Air traveling at a very slow velocity can give rise to heat absorption (COLD PIPE). (A really neat application of this effect is called a "Venturi Tube"), and further to the idea of employing a "factor", a Venturi Tube might make use of a 'temperature conversion factor' that takes into account the size and type of material used for the aperture and velocity of the air through it.

WHERE DOES THE CHANGE OF AIR PRESSURE FIGURE INTO ALL OF THIS?

Well, not from your pump.

If you wanted to really compress the air, a centrifugal pump would not be powerful enough to force it into closed space like a tank under pressure. For that purpose, you would probably employ a piston-style pump instead.

You can speed the air up, but unless you compress it in a closed chamber to change its ambient pressure, which a centrifugal pump does not do, you can't change its total or overall pressure. You can, however, and a centrifugal pump does, restrict its outflow through a narrow orifice, which in order to compensate for the restriction, speeds it up...this is called increased velocity. And the increased velocity can be applied to create pressure differences.

In your OP, you seem to have "increased velocity", since it can exchange 'speed' back into 'original speed' for energy conversion , confused with "compression". They are not the same, and no, the air does not get compressed or more dense by being forced through the smaller aperture of the centrifugal pump.

It will, however, use the extra energy imparted by its increased velocity to do things like travel longer distances through a line before petering out, and performing tasks like pushing buttons. Again, the build-up or concentration of high velocity air on a small surface like a button can move the surface more easily than low velocity air at the same pressure because the difference in amount of air per second loading the high velocity side (MORE) of the button exposed to the air flow, and the low velocity side (LESS) not exposed to the air flow causes a difference in the ambient air pressure affecting the surface of the button on both sides, and sucks the button down towards the low velocity side.

But if the button gets stuck or doesn't move, then the air on the upstream side might compress a _{very small amount} (centrifugal pumps being what they are) in the narrower passageway until the pressure differential builds up enough to suck the button toward the lower pressure side.

In many pneumatic systems (systems that use air to conduct messages that activate task mechanisms), mostly compressed --rather than pumped-- air is used. The compressed air comes from a pressurized tank that has been filled by a piston-style pump and is admitted to the control lines through valves that ensure the pressure remains constant during signalling. It is thought that using pumped air is ineffective because of the difficulty in maintaining constant flows of air for signal efficiency.

Mark

Hi,

How do you explain e.g. a "turbo" in a car comprised of an exhaust gas turbine and a air 'compressor' or centrifugal pump sitting oin the same shaft and charging the cylinders with more air.

You need a "higher pressure" in front of the intake valve in order to feed more air into the cylinder(s). The turbine running at high rpm provides that pressure of as much as 0.5 bar or so.

I could give you other exaples in indutry, where a centrifugal blower builds up a fluidized bed. The air exiting the nozzles is above ambient pressure. Below the nozzles the pressure is higher than that by the pressure loss of the nozzles.

What say you? Mark

Hi, Guest!

Same thing. The air exiting the nozzles is at ambient pressure, but at a greater velocity. The velocity as applied against a resistance transforms into pressure. In starting up a heavy duty diesel, compressed air is used because it can move the cylinders. Keeping it going with a blower ensures a greater mass of air movement per unit time than scavenging.

Mark

errata:

"Keeping it going with a blower ensures a greater mass of air movement per unit time than scavenging"

should read

Keeping it going with a blower ensures a greater volume of air movement per unit time than scavenging.

When I wrote "mass", I was thinking of a more "massive" amount of air (i.e. volume), not in terms of 'mass/volume/density/etc. etc.'. I know that a greater mass per unit time makes no sense whatever and anyone reading it would guess my intent, but I thought I'd better put in the correction just the same.

Mark

Hi, nzchinese!

The velocity of the driver and size of the flow-through aperature are directly proportional to the velocity of the air.

In addition to motor speed, the diameter of the driver is directly proportional to the velocity of the driver.

The mass of the air and changes in air pressure are not affected by the blower.

Volume of air affected by the blower is calculated as flow-through multiplied by time.

Mark

Centrifugal blower is obbeying Fan laws.

As per fan laws

Speed (N-RPM) is directly proportional to Air volume (V-m³/Sec)

Speed (N-RPM) is directly Squared proportional to [Static Pressure ]²(P-Pa)

Speed (N-RPM) is directly Cubical proportional to [Power]³ (P-kw)

You may confirm pl

Regards,

Prasad

Hi, nzchinese!

I don't know if you're still looking at the replies to your question. But if you are not, here is a short answer:

the speed (rpm) is proportional to its output pressure. NOT CORRECT

the speed (rpm) is proportional to its output mass air flow. NOT CORRECT

the speed (rpm) is proportional to its output volume air flow. CORRECT

I hope you got a good mark on your test. You can look up Centrifugal Pump in Wikipedia to learn more about it. The idea of a volute casing is what makes it unique and a genius of an idea for the movement of fluids.

Mark

Thank you Mark for your useful information.

My project is to make an air pressure source that should keep pressure constant at any altitude. the requirement is: Pressure output is 18cmH2O at flow rate of 150LPM.

I'd like to know how to correct the blower speed to compensate the altitude without a pressure gauge.

In your point of view, what type of pump I should choos: Centrifugal or Axial?

Hi, nzchinese!

Since this is a project and its intent is to give you a chance to show what you can do, I think I will leave you alone now, and let you perform on your own.

Good luck with your project. Post back again to tell us how you did.

All the information you are looking for is available on the web or in text books.

Mark

Dear Mr nzchinese,

1st Statement is wrong. 2nd and 3rd statement is correct.

DHAYANANDHAN.S