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Which one is more energy efficient: 1

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nbucska

Electrical
Jun 1, 2000
2,191
Which one is more energy efficient:
1.) Continuous PWM control OR
2.) Bang-bang control ?

Any way to estimate the difference ?
<nbucska@pcperipherals.com>
 
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It depends.
Assuming that you are referring to AC motor control, and by &quot;bang-bang&quot; you mean an electro-mechanical starter, and by &quot;PWM&quot; you mean an AC variable frequency drive.

If you are running at full speed. the &quot;bang-bang&quot; control is more efficient. The PWM drives will be from 95-97% efficient at full speed. The starter will be 99.99% efficient.

If however you are using a process control system that chokes off the output of your machine, a VF drive can possibly save energy. For example, think of a throttling valve on a pump. The motor runs at full speed and the valve is partially closed to control flow. The pump will consume slightly less energy, but not by much. If instead you modulate the pump speed with the drive, the pump will use even less energy at the same flow rate. The VF drive will save energy by virtue of the &quot;affinity law&quot; which states that in a variable torque load such as a centrifugal pump, the power consumed is proportional to the cube of the speed of the motor. So, if you can turn your motor speed down to 70% of full speed, the energy used by the pump is .7 x .7 x .7, or 34.3% of the energy used at full speed. Even though the VF drive is typically less than 95% efficient at lower speeds, the extreme savings more than offset it.
Quando Omni Flunkus Moritati

 
rbucska, jraef:

Wow!!! I was afraid to ask about “Bang-bang.”

For what it’s worth I have heard numbers from 1% up to 5% loss for PWM drives due to heat dissipation and harmonics. One thing that offsets this is if you are being charged a PF correction fee, the input of a PWM will appear as unity. You may get a reduced power cost by using a drive.

David
 
d23, if you use a pwm drive, the true power factor is not unity, but typically in the order of 0.7 or less. The angle between the voltage and current is close to unity, but the current is very distorted so the harmonics cause a poor power factor. This will show up on KVA metering. (Power factor is the ratio betwen KW and KVA)

nbucska. I assume you options are using a variable speed drive at part speed, or switching the motor On and Off on demand.
This is very dependant on the application. In some cases, the machine operating at full speed, is below optimum efficiency and operating at reduced speed can improve the machine efficiency. This is true for pumps that are partially throttled, but not for pumps that are pumping with no head.
If you use a pwm control to vary the speed and optimise the operating efficiency of the machine, and as part of it's operating cycle you spend time at rated speed, you can actually pay more for the operation of the machine due to KVA Maximum demand charges.
To give a more definative answer, we need more detail of the application, operating cycle and what your tarrifs / concerns are.
Best regards, Mark Empson
 
Suggestion: Among other factors, some of them mentioned, there are:
1. Stability concerns, namely, the bang-bang controls can introduce oscillations and even resonances.
2. PWM can introduce common-mode currents and common-mode voltages.
3. Active Front End (AFE) regenerative controls could be considered.
4. Various stresses and material fatigue due to bang-bang controls need to be considered since they reduce the life-cycle.
5. Etc.
 
To add further to marke's comments, the input power factor to a VSD is proportional to load. The PF of the VSD is constant, but as the input current increases (due to motor current) the waveform becomes increasingly distorted (harmonics).
It is therfore impossible for a VSD manufacturer to quote true PF.
 
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