Finding Suitable Size of The Motor By Comparing No-Load and Load Currents

Shaft power = torque x angular velocity.

Bear in mind that the torque can go up depending on what is happening inside the kiln.

Motor power = shaft power / motor efficiency.

Supply current = motor power / voltage / power factor.

What is the supply voltage?

What is the business objective in fitting a smaller motor?

Hello jcl

The other thing to remember is: How old is your existing motor?

If you motor is many years old, it will be a large framed one, with a large diameter shaft, run quite cool, just be not working hard (giving long life, no bearing worries), and be able to cope with sudden load changes.

If you replace with a more modern motor, it will be much smaller, small diameter shaft, run hot enough to fry an egg on the casing, working fully all the time, (will need repairs and bearings more often), and sudden load changes may give the motor a "heart attack".

Kind Regards....

Hi jcl,

First when you want to design some machine or some system that was drived by motor, You have to calculate the load of that machine or system, then the load of shaft of motor, and finally must to choose correct motor.

Please see carefully on motor's name plate 42A current with what the voltage is and what the voltage is using. And your machine was running with maxi load and maxi speed or not, what the current start is with maxi load? These informations are found easily in operation manual of your machine.

10KW CONT.duty is fine!

Dear JReaf,

Thank you very much for your very good explaination.

Acctually I am trying to reduce the electrical consumption of the factory. So I thought I could reduce the electricity bill by replacing small motors to our 5 kilns (soon it will be 10 kilns)

The motor specificatins are as follows,

Output : 22kW

Current : 42.5 A

Voltage : 380 V ( but our supply voltage is 400V, I think it is also another problem)

Speed : 1470 r/min 50 Hz Ins Cl. B

Rating : S1

Spec of Magnetic Clutch,

Excite voltage : 70 V

Excite current : 2.88 A

Ins. : B

Rating : S1

So according to the infomation I have given, do you think I can't save much electricity by changing motors for 5 kilns.

Please advice.

"Acctually I am trying to reduce the electrical consumption of the factory. So I thought I could reduce the electricity bill by replacing small motors to our 5 kilns (soon it will be 10 kilns)"

Reducing energy consumption and reducing the energy bill are not always the same thing. Replacing the motors will not likely do much to reduce the energy consumption; if you run the kilns 24/7/365 then a small incremental improvement in efficiency may add up over time, but that must be weighed against the cost of replacing the motors. Include the labor, downtime and risk of further downtime if there happened to be a valid reason for having 22kW motors in the first place that you may be overlooking. Don't forget that you are second-guessing the original engineering decision, something that comes with risks if you don't have all the facts as to why the original engineer decided on those motors.

But if your energy BILL includes penalties for having poor power factor, then you may be able to reduce the bill even if the energy stays the same, because it is likely that you have poor power factor with such a light load. So if you choose not to replace the motors, at least look into Power Factor Correction capacitors.

As I said earlier, I would also evaluate the need for the clutch drive system. They are notoriously inefficient but may have been put in when VFDs didn't yet exist or were too expensive. Replacing those with VFDs would be an immediate way to save energy and may not require the replacement of the motors (if done right) or the addition of PFC capacitors, they will take care of both issues as well.

Acctually I am trying to reduce the electrical consumption of the factory. So I thought I could reduce the electricity bill by replacing small motors to our 5 kilns (soon it will be 10 kilns)

The solutions of Mr.JRaef are perfect, you should do it.

I want tell more about your opinion. The power of motor is 22kw, this mean the max. load of it is 22kw it depents on your load, if your load is P~400V*18.5A~7.4kw, mean your motor power is output also 7.4kw,.... Just image that after you install new motor with power 400V*25A~10kW, and you measure current again the result will be 18.5A too. If your kilns runed with max load this means you can not add any workpieces into your kilns or your kilns contained max designed weight,...( I don't belive that) and your current is 18.5A too, please tell this to kiln manufacturer for improving 5 kilns that will be installed in the near future and you will save some money: machine cost, electricity bill, place cost(smaller machine is cheaper than and smaller dim).

You can use the rate of electricity bill and sale (or production quantity) to improve your project instead of only electricity bill. In our factory, I had rearranged lightings system by turn on some needed ereas while producing or improved productivity,....

Good luck and success

ha ha, just seem that there are some unneccessary parts in your factory. reducing lighting is just suitable only for your factory, man

Hello JRaef

Your answer is consistently good, so I gave you a Good Answer tick too.

Cheers and Kind Regards....

Dear friend,

You can measure directly KW consumed with the help of a LOAD MANAGER and then divide it by eff. of motor say 0.9 which will give you theoratical value of KW required. You can select nearby standard rating in such a way that motor is loaded to 80%.

With regards

D.D.Desai

Dear friend.

There is an almost linear relation between the motor output power and the speedreduction of a motorshaft.

Take a look at the motors nameplate, it gives the RPM at full load.

I expect it to be 2935 RPM (3522 for 60Hz) at a load of 22 kW.

At no load the speed will be 3000 RPM (3600 for 60 Hz).

This means that the motor speed decrease with (3000-2935) / 22 RPM/kW

(ita est 2.95 RPM / kW)

Now let run you motor at the usual applicable load and measure the shaft speed with a tacho meter or stroboscoop.

If you measure 2950 RPM (3540 for 60 Hz) then the difference between 3000 and 2950 gives 50 RPM

So the motor gives 50/2.95 equals to 16.9 kW.

I suppose that you motor is a 2 pole motor. If it has an other pole number, the numbers will change.

dear sir,

15kw motor will fulfill your requirement but it will take 21Amps when loaded.your power factor with 22kw motor is 0.6 but this will improve your power factor to 0.8

I will add a single point may it had been forgetten by others.

very important to descide when you select the motor there is another factor to be considered not only the motor KW , you need it for continuous duty, short duty,intermitten duty.......etc

for example you can't use 20 KW motor of a pump or fan to replace a lift or escalator burned motor with the same KW. it will not work for a long time and will burn

I developed a spreadsheet to estimate power draw based on motor amp draw vs full load amp draw. Using typical full load power factor and efficiency numbers I come up with a power draw estimate of 4.3 kW at 18.5 amps. Actual power being drawn by the mechanical load is more like 2.9 kW (a lightly loaded motor is not very efficient).

If it is a 3 phase Induction Motor,the thumb Rule is....kW of the motor = 1/2 the maximum load current.In your case,10 kW should be quite OK.

Injineri.RC

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