Loss Due to Low Grid Frequency

We've already flogged this horse once, why go into it again.

There are 10,000 web site that will tell you how to do this and EVEN DO IT FOR YOU.

Go find one and use it.

Run the motor at any speed you like. You seem bound to do this anyway, so go ahead.

You've noticed a slight reduction in power consumption. Did you compare this to the amount of work performed or any other plausible mitigating factor?

2 - 2(49.5/50)

If your company has reached the point it has to start turning down the electrical systems operating frequency to try and save money the odds are its too late anyway.

So is this experiment an effort to save money or for other things?

do power plants save money by running at a lower ower frequency? it seems to be a trend. is there any affect to the consumer?

sorry tcmtech for repeating your post.

What is the drawback in cashflow and operating profit in terms of running the motor at 1% lower speed, Bro?

You clock runs slower so you spend more time at work that you need to.

In theory if you drop the frequency by 1%, you drop the motor speed by 1%, and the output by 1%, but only in a mill that operates at 100% efficiency. As no mill can operate at 100% the calculations become a little more complicated. The mill stones will be optimised for grinding most efficiently at a certain speed. If your existing motor speed matches this optimum then you will experience a reduction in quality not throughput. You have not changed the in-feed rate to the stones so the same quantity will be milled, it just won't be milled as finely. If the 50Hz motor speed and stone efficiency are not a prefect match then you stand an equal chance of improving the flour quality as you do of reducing the flour quality.

I am amazed that any multi generator grid system could vary in frequency for any length of time by as much as 0.5Hz. All the generators a synchronised and any generator that comes onto line slightly out of sync will be motored into the common frequency. It is the easiest part of bulk generating AC electricity to get right. Voltage is allowed to fluctuate by up to 10% but frequency is normally held to +/-0.2Hz short term. It will also spend as much time over frequency as under frequency so it balances out over a longer period. All clocks and a big chink of industry depend on this.

Dear Mr.satendrakumart

It is true in the case of centrifugal pumps, centrifugal fans for industrial applications like application in boilers. The out put speed will be 49.5/50 = 0.99 i.e., 99% and speed reduction is 1%.

For the Pump or Fans, the capacity will vary the as the speed, Pressure or head will vary as the square of the speed and the power demand will vary as the cube of the speed.

Accordingly for a Frequency of 49.5 c/s, the capacity will be (0.99)C, the Head or Pressure will vary as (0.99)x(0.99) P or H = 0.9801 or 98.01% and the Power drawn will be (0.99)x(0.99)x(0.99) P = 0.9702 or 97.02% thus a fall in power demand by 2.9701%. Here the synchronous speed is considered.

For 2 tons/hr rated output, the fall will be 0.596 Tonne/Hr., if the Frequency falls to 49.5 from 50.

Thanks,

DHAYANANDHAN.S

It seems odd that a 1% speed reduction would cause a nearly 25% throughput drop. Perhaps you meant 0.0596.

If I repeat the working of Dhayanandhan sir, for 1% reduction in frequency

Fall in speed =1-0.99 = 1%

Fall in head (through put) = 1-0.98=2% ?

Fall in power consumption = 1-0.97=3% ?

so the production loss is 2% ie fall in mill out put is 0.04 ton?

Those figures are for a centrifugal pump, rather than for a flour mill.

Is it a flour or a floor mill? I still would like to know how the slight reduction in power consumption was measured in comparing one power grid frequency to another.

It sounds like good news. Why stop there, though? Why not drop the frequency to, oh, say 20Hz?

Do I here 15Hz, 15,15,15. Now I'm at 15. Do I hear 10,10,10,10 and I have a 10! Now I'm at 5. Can I get 0,0,0,0,0. Sold to the lowest bidder at 0 Hz. The expected power savings will be 100% since no work will get done using induction motors.

How to calculate the energy lost in the form of heat, due to low frequency in inductive loads?

And what about V/F ratio and its effect?

You calculate the loss the same way one calculates any real energy loss. Energy into the system minus energy applied to the load. The difference is frequently mislabeled as the energy lost. Energy is never lost, it can be unusable, the price for doing a task, delivered to the intended load but it is never lost.

All of that energy difference will not be realized in the form of heat depending on the type of inductive load you are driving. For an electric motor some of the non-thermal diversions will be moving coolant, the rotor assembly and free space radiation. The root of the thermal dissipations will be resistive losses in stator and rotor windings, bearing friction, magnetic core losses of eddy current and hysteresis.

To accurately discuss these effects requires more education than you've demonstrated here so far. Can you tell me how many watts a solenoid with 100 mH of inductance and 4 Ω of winding resistance produce if driven by a 120 VAC (RMS) 60 Hz source? Ignore hookup wire resistance. If you can, then repeat this exercise for 59.4 Hz.