How to Select a Motor

A fan is sized by its fan curve. The curve shows cfms at a particular static pressure and rpm. Horsepower required is also charted.

thanks for d commet.. unfortunately i did not have the fan chart. i got only the informations on the lable.(it's an old one)

Schaturanga,

There exist a diagram which you should look through and in the crossing of the pressure head, delivering volume, RPM of your new motor, you will find wether it works or not. Also, add approximately 20% power and that's it. In the country where I live, we call it topographic diagram for it looks like topographic chart. If you don't have it, ask one from the fabricant.

Marijan

thanks

Thanks... i thought that only static pressure is reduced when the power reduces.. because of maintaning the same speed!

If all you do is reduce the motor kw you will burn out the motor. The fan blade configuration and the rpm determine the power needed. As mentioned, you have to reduce the rpm which will reduce the cfm.

The rates of the power, RPM, capacity and pressure proportionality is ruined by friction of the fluid. The only way to be correct is to check with the diagram. Be sure of this, otherwise you would never see the diagram, they would supply the formula.

Best regards

Marijan

The answer given by JMueller is correct. The affinity laws are as he describes them. Let me try to give you a little more on that subject. There is only one way to increase or decrease either the pressure that this fan will deliver or the volume that it will deliver. That way is to increase or decrease the tip speed of the fan blade. That can be accomplished by reducing or increasing the rotative speed of the fan blade or increasing or decreasing the diameter of the fan blade. PERIOD. The reason I mention this is to let you know that you can use the smaller powered motor but you must reduce the tip speed of the fan blade. And, I assume the motor is a constant speed motor so the only other way (assuming the blade mounts on the motor shaft and therefore cannot have its speed changed) is to reduce the outside diameter of the blade itself. If you have no curves or other information from the manufacturer that would tell you what this speed should be, you can put an amp meter on the motor leads and meaure the power draw of the motor with the fan running. Note the full load amps (FLA) of the motor from the nameplate. You must be under that number. Then reduce the diameter of the fan by carefully cutting the blades off a little at a time until the amp draw is within the nameplate FLA. This is a tedious exercise and not the best way to do it. Reducing the rotative speed is the best way through a series of pulleys and belts if that is possible. Of course, depending on how important using that smaller motor is to you, you can always use a variable frequency drive to reduce the speed of the rotating fan. Rather expensive, but workable. Finally, see if you can buy a new blade that is smaller or a new motor that is bigger.

It is not a good idea to use smaller motor on the same blower, to small motor will most likely not have enough power to rotate the blower and propably blow a fuse. The lable on the blower tells you how much power you need (11kw at 3540rpm ) to get 4500cfm. But this also depends on the situation, if you are blowing in a long tunnel or somehow the airflow is partly closed at either suction or pressure side- the load is increased on the motor and therefore reducing your chance to use smaller motor than labeled. It may be possible to use the 7,5kw without blowing the fuses if you have frequency converter to start and run the motor, with frequency converter you can speed up slowly reducing the start load, and then you can run the motor at more speed than rated to get even higer CFM, and also increse the torque, but this will shorten the lifetime of the bearings in the motor. BIGGER IS BETTER.

Increase in static will decrease motor amps. Cfm will also be reduced. Study JMuellers answer. I personally have fan curves available for all my hvac equipment so I tend to get lazy. Thanks for the exellant information Jmueller.

The answers all seem to have the same basic error.If you reduce the throughput of the fan by restricting the output,(e.g.with an orifice plate),the load on the motor will REDUCE.At the limit of no output flow the load on the motor will be just the windage of the fan,as it is not accelerating the air.The same factors apply to centrifugal liquid pumps.

Most large fans will seriously overload if the output is unrestricted but whenever I have been involved in big fans the test electricians have always to be persuaded to start up with outlet dampers shut and slowly open up to full load after abortive attempts and overload trips with everything wide open! I've won a few tenners on the result.

Run your fan and see what the phase current and voltage is at present.Calculate the power from 1.73 x V X I X cos phi,if you dont know the power factor (cos phi) 0.85 will do.. Often designers oversize the motor for various reasons,and you may find that the motor is not fully loaded,possibly actually only consuming 7.5kW if you're lucky,and you could fit a smaller motor without any changes.

If the original fan motor is loaded to more than 7.5kW then if you fit a smaller motor you will need to restrict the output to reduce the volume by experiment (shutter in duct) until the new motor is consuming its full load current.This volume may not be enough for your requirements but as others have explained the exact results can only be predicted with a fan curve of your particular fan. The fan will be at a new operating point with increased static pressure but reduced volume.Some of our previous contributors could maybe guide us to the likely results from a typical fan curve.

As there is an increased static pressure you will not get the best volume from the fan as some of the power will go into the extra pressure only to lose it at the restriction. As others have rightly said to get the best you would need to alter the fan speed by the pulleys if used,but you need the fan curve to establish the best speed.

Have a good old experiment and the best of luck.

My response to guest is not incrrect. We are saying the same thing. Increase static, cfm is reduced and amp draw is reduced. Guest is incorrect!

Open, please, website < http://mysite.verizon.net/restkgfn/airpolcontrol >

I have been following the comments on this thread, and note some apparent contradictions.

The Commoner makes a good case with the wise additions of checking actual motor load with an ammeter and the possibility of trimming the diameter of the fan to reduce its load. All too often the motors are poorly matched to the fan's actual draw (typical and understandable bias for engineers wanting to leave room for error). Practically, however, trimming the fan diameter is tricky because of problems with balance (vibration, increased bearing wear, etc.), not to mention also that a backward (or forward) curved blower is quite different from a bladed fan.

The chief drawback to the answer I posted earlier is contained in the phrase: "These assume that the efficiency of the fan is constant, which is a fair assumption for only modest changes in speed." Changing the static pressure the fan is working against directly changes the efficiency of the fan--a higher static pressure on the same fan at the same speed reduces the flow because the efficiency is reduced.

The post referencing the "London Falls" equipment has a good comparison of two fans (20" and 12" diameter) which achieve the same air flow but at significantly different static pressures and RPM's. If that post were to also note the power requirements for the two fans being compared, you would see that the 12" fan at the much higher speed would have about twice the power consumption as the 20" fan at the lower speed. When you consider that to move the same volume through a tube sized at less than 40% the area of another tube, its velocity is 2.5 times higher--this means a much higher loss of energy to friction.

This is why good engineering design of air handling systems will use increased duct sizes for lower static pressures and friction, and also use variable speed drives instead of dampers to control air volume. The very good reminders to look at the manufacturer's fan curves should be heeded. Only by doing this can we have a good match of equipment with energy efficiency.

The original post was not asking to achieve this good match, only to see if a smaller motor could be used with a particular fan that was present. Therefore, if the fan is overloading its newer and smaller motor at the same speed, its speed can be reduced or flow restricting dampers can be installed (which reduces the fan efficiency as well as its load).

Good discussion--John M.

OK,

This discussion goes from one to the other end. The question which was never put would be in my opinion:

Why Schaturanga changes the motor. Difference in the price is not substantial. If you have enough power in spare, the motor will deliver exact power for the case, and won't spend any substantial surplus. In addition you are on the safe side.

Best regards

Marijan

• I WAS CHANGE MY BLOWER MOTOR AS 7,5KW , THERFORE I HAVE TO CHANGE MY IMPEELER DESIGN, REALLY , I WANT TO KNOW ABOUT HOW TO CHANGE IMPELLER OUTER DIAMETER ACCORDING TO THE BLOWER MOTER PIOWER, ALSO I WANT TO KNOW HOW TO MEASURE STATIC PRESSURE IN THE BLOWER?

Guest,

As far as I know, a backward curved or forward curved blower is a wheel with many little blades on it mounted in parallel. Unless it is a small one out of molded material, it typically is made out of separate pieces of sheet metal and has a hub at one end of each of the blades and a ring at the other end (to hold them in place). Unlike a fan, which is with separate blades that extend radially out of a hub.

It is difficult to change the "diameter" of a blower by trimming it. However, you can change the effective length of the blades or vanes in it by wrapping a portion of it with a durable material that reduces the working length of all the vanes. Since we don't know the actual current draw on your motor, we don't know if the 11kW one was oversized to the load.

Another thing not mentioned in this thread is that when you change the motor, you need to ensure that the new one has proper running overload protection. Some motors have this built-in, but most do not. If yours is with an IEC style starter or MCP then its dial range may include the new motor's Full Load Amps (FLA). If it doesn't, you will need a new O/L unit. If it is a NEMA style starter you will need new heaters. If your new motor is overloaded and you don't get the protection set for it, I'll guarantee that you will burn-up the new motor (but it may take a few hours to a few weeks to do so).

--JMM

Marigan Topic,

Yes, the larger motor will always succeed in operating the equipment. However, the efficiency of energy used will be less. This oversizing of the motor results in a poorer power factor and extra costs for the electricity used. Some utilities meter the KVARS as well as the KW, so a poorer power factor can directly increase your operating costs.

How do we define "substantial surplus" or "enough power to spare". In a world with rapidly increasing demands for energy in the developing countries, many concerns about air pollution from electric generation plants, and the uncertainties of the effects of CO₂ on us all; I think a prudent approach would be to always minimize energy consumption whenever possible. Most users don't even realize that the extra costs for a "premium" efficiency electric motor (with only 2-6% better efficiency than a standard one) can easily be repaid many times during the life of the motor, via savings in electric power charges. Even today, some "professional" motor re-winders do not use burn-out ovens with temperature controls suitable to prevent heat damage to the stator laminations' magnetic properties. Therefore many re-winds are to a lower efficiency than the original motor.

--JMM