Energy Saving for Constant Torque Loads

What is the nature of the electrical tariff:

• per kWh?
• Maximum demand?
• other?

Is there a power factor penalty imposed by the utility supply company?

what is ur application? if it is hoist then i will suggest don't go for saving.

The original poster has already stated that it is a mixer. What type, and what it is mixing, how much of it and at what temperature and pressure have not yet been stated.

You have to answer Pw's question. If you are billed kwh's and/or kw demand then it does not matter. You are billed for what you use.

I know that this is not what we were asked to answer but I cannot help myself from pointing out that there is no such thing as a constant torque motor. If there was a constant torque motor then there could be a motor with an infinite angular velocity.

At redfred : Perhaps it is a DC motor with a constant current flowing in the armature ;-)

I still have a problem, at the moment there seems to be no speed regulation in your application. And you want a constant torque, how is it possible that the load is going from 35 to 15 kW? Mechanical speed regulation ?

Power still equals torque times angular velocity

Based on your figures, your average power is 21kW, compared to a minimum of 15kW. So you could save 6kW , assuming 24hrs/day, 365 days/year, 0.12$/kWh that costs about ~$6,000.

So the answer is yes, you could save money.

Most modern VSD's have a function that lets you run at (approximately) constant power and depending on your circumstances that would seem worth looking at.

Also check if you actually need to mix as fast as you do, a gearbox change might also lower the required power (cost).

It still depends upon the electrical tariff charged. Oh, well. Try another thread. _{<unsubscribes>}

You need to re-read the question. He did not say that he reduced the load from 21kw to 15 kw. He said it varies between those figures. So there is no apparent savings.

Just a thought (need to be studied in details): Let there be two motors ,one 15 kw for 70% of duration and other 20 kw for peak load.This 20kw motor to get coupled and synchronize to take care of peak load of 35kw (15+20=35).Friends,this is no solution ,just a point for discussion.

First you need to find out the power factor. If the power factor is very low such as .7 below than there is slight saving. Having variable power is directly to do with the load and quality of power.

My guess is that your motor is constant speed and that the Kilowatts vary depending on the load in the mixer. The motor will only use the amount of energy it needs to maintain the speed. If this is the case, using a variable speed drive will not save any money except for starting, and only if you exceed your demand factor. You will actually have a slight loss dependant upon how efficient the drive is.

Other than a POSSIBLE small savings from a VERY small improvement in power factor- assuming that the site is paying a PF penalty, there is virtually NO value in adding anything to this drive.

In the future- if the motor needs replacement- and if the "tracked" loads are typical and not lighter than what might be expected, there is an argument for installing a replacement motor closer to the 35 kW peak power load observed.

Other than that- look elsewhere for savings- you are paying for what you are using, whether that is demand or kWh.

ONE possible option- depending on operations and the electric rate charged- would be to schedule the mixing operation so that the "high" loads occur at night and the lighter loads occur during peak electrical demand hours. That MIGHT generate a small savings due to the electric rates charged- if they are "time of day" based.

Elroy hit the nail on the head. A motor uses what it has to use. You cannot dictate to a motor how many kW it will consume for a load task, it is what it is.