energy saving using drives
- Thread starter panthripu
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- San Francisco Bay Area, CA, USA
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- Electrical Engineer
100A, plus whatever losses you introduced with the VFD.
In your statement, BOTH conditions say there is not load on the motor. If that situation does not change, then if the No Load Current is 100A at full speed, it is still 100A at 1/3 speed, at least as far as what is being drawn from the utility side.
You are somehow getting caught up in the marketing BS that is rampant in the VFD reseller's world (and I am a memeber of that guild).
Motor current is irrelevant anyway other than with regards to overload protection, so why do you care? Motor kW is important in terms of operating cost, so that is going to be related to LOAD on the motor. No change in load, no change in operating costs.
petersonra
Senior Member
- Location
- Northern illinois
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- engineer
1500 RPM? Must be in some place with 50 Hz power.
Presuming the no load current @1500 RPM you are asking about is the motor current, chances are at 500 RPM it will be less, but there is no way to directly calculate it.
It might also be that the PF will drop as the frequency drops and the current may not change as much as you might expect.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
120416-2145 EDT
panthripu:
Your title was labeled energy saving, but then you asked about current. If energy saving is your interest, then you want to talk about input power.
Following are results of a small Bodine brushless motor. All data measured at the controller input. No external load. This motor has a built in gear box with maximum output RPM of 42. The motor is 22B2BEBL-D3. On the output of the gear box is a cogged belt driving a shaft with ball bearing support. The belt ratio produces an output shaft speed somewhat above 30 RPM. At the moment I do not want to check actual output RPM.
Amps Watts VoltAmp Speed dial position
0.18 -- 16 ---- 23 --------- 100
0.16 -- 9.5 --- 21 ---------- 50
0.12 -- 5.4 --- 15 ---------- 20
Unloaded I would expect your motor at 1/3 speed would have a somewhat lower input power than at full speed.
.
panthripu:
Your title was labeled energy saving, but then you asked about current. If energy saving is your interest, then you want to talk about input power.
Following are results of a small Bodine brushless motor. All data measured at the controller input. No external load. This motor has a built in gear box with maximum output RPM of 42. The motor is 22B2BEBL-D3. On the output of the gear box is a cogged belt driving a shaft with ball bearing support. The belt ratio produces an output shaft speed somewhat above 30 RPM. At the moment I do not want to check actual output RPM.
Amps Watts VoltAmp Speed dial position
0.18 -- 16 ---- 23 --------- 100
0.16 -- 9.5 --- 21 ---------- 50
0.12 -- 5.4 --- 15 ---------- 20
Unloaded I would expect your motor at 1/3 speed would have a somewhat lower input power than at full speed.
.
If this about energy saving and the motor is on no load, why not just turn it off?
energy saving
energy saving
Thanks guys for your quick reply
What ever values i mentioned are just imagenary. You are right , my intention is to save energy but i wish to know what if i keep the motor running no load. Or if i used drive to run the motor at reduced speed in no load condition how much energy am i going to save.
In other words , you are right...i want to know the KW in no load conditoin with rated speed and 50% of rated speed or 10% of rated speed.
Other option is quite very common which is to switch off the motor but as this kind of situation (no loading ) comes quite frequently per day , i dont want to damage the motor by frequently switching ON and OFF.
By the way its three phase 460V induction motor.
energy saving
Thanks guys for your quick reply
What ever values i mentioned are just imagenary. You are right , my intention is to save energy but i wish to know what if i keep the motor running no load. Or if i used drive to run the motor at reduced speed in no load condition how much energy am i going to save.
In other words , you are right...i want to know the KW in no load conditoin with rated speed and 50% of rated speed or 10% of rated speed.
Other option is quite very common which is to switch off the motor but as this kind of situation (no loading ) comes quite frequently per day , i dont want to damage the motor by frequently switching ON and OFF.
By the way its three phase 460V induction motor.
With a variable frequency inverter the motor is "soft started" so is not subject to the thermal and mechanical stress that direct on line starting would cause.
Unless there were compelling reasons to keep it running, I'd turn it off rather that let it idle.
If you don't need to vary the speed, a soft starter would be a cheaper option. The is a reduced voltage type starter which ramps the motor voltage over a period of time.
- Location
- Bremerton, Washington
The motor is dumb and responds to the load. If there is too much load, the motor just keeps pulling current until it burns up.
But with a variable torque load such as a water pump, it responds to the affinty laws, and that says if you reduce the speed 50% the power drops by 1/8.
But with a variable torque load such as a water pump, it responds to the affinty laws, and that says if you reduce the speed 50% the power drops by 1/8.
- Location
- San Francisco Bay Area, CA, USA
- Occupation
- Electrical Engineer
The Pump Afinity Law you refer to applies to flow, not just speed. If you reduce the flow mechanically, i.e. a valve, you get the same effect for the most part. But a valve adds losses to the system in the form of turbulence, heat and pressure drop. So if you remove the valve and reduce the flow by reducing the speed with the VFD, you get the reduced energy WITHOUT the losses associated with the valve. That is the energy savings realized by a VFD. It's still significant, I just like to make sure people understand that it isn't everything the marketing people hype it up to be.
It's a very common misconception that is heavily promoted by the VFD industry, but the reality is that you don't need the VFD in order to realize most of the effects of the pump afinity laws. The VFD basically just IMPROVES on them in many cases.
But back to the original question and modification. Just slowing down an UNLOADED motor with a VFD will not change the unloaded energy consumption by anything more than a very slight reduction in losses inside of the motor that are based on magnetizing current (I wouldn't expect more than 1 or 2%), because it is STILL the same motor load. So as Besoeker said, the only REAL energy saver in that instance is the Off switch. Then also like he said, if you don't need the variable speed, use a Soft Starter.
It's a very common misconception that is heavily promoted by the VFD industry, but the reality is that you don't need the VFD in order to realize most of the effects of the pump afinity laws. The VFD basically just IMPROVES on them in many cases.
But back to the original question and modification. Just slowing down an UNLOADED motor with a VFD will not change the unloaded energy consumption by anything more than a very slight reduction in losses inside of the motor that are based on magnetizing current (I wouldn't expect more than 1 or 2%), because it is STILL the same motor load. So as Besoeker said, the only REAL energy saver in that instance is the Off switch. Then also like he said, if you don't need the variable speed, use a Soft Starter.
mike_kilroy
Senior Member
- Location
- United States
another perspective..... if truly not loaded, a motor with 100amp 460v 3ph no load (magnetizing current) would be rated likely around 200 amps full load, so probably around a 175hp. That size motor probably requires around 2 amps REAL power to turn (overcome friction) at 500 or 1500rpm speed. So if truly run no load, yes, turning it down to 500rpm when not in use will save electricity since it is power in vs power out for the vfd: @ 1500rpm it will pull real power of 460v,3ph,2amps or 1591 watts. At 500rpm it will pull (500/1500)*460v again times 1.73* 2 amps or 530 watts. vfd will add 2-3% in losses in either case, so for this, ignore that.
so power in is 1591 watts @ 1500rpm from POCO, or 530 watts @ 500rpm from POCO.
As you said, this is for short periods or time during the day, so why bother? shut it off!
so power in is 1591 watts @ 1500rpm from POCO, or 530 watts @ 500rpm from POCO.
As you said, this is for short periods or time during the day, so why bother? shut it off!
MSU
Member
- Location
- Hattiesburg, MS
There are two basic sources of KW flow to a motor operating at 'no load'. One is electrical losses expressed as load or copper losses, core losses, parasitic losses including stray flux. The mechanical losses consist of windage and friction. Windage losses follow the fan curves for the style of fan employed. Friction losses depend on the bearing system employed and RPM. Total losses at NL are typically less than half of FL losses. Electrical losses are the majority of the FL Loss number. If the motor is 94.3% Eff, the total losses are 100% - 94.3% = 5.7% . 93.5-95.5% is a typical efficiency for motors in the NLA range which you suggest. 100NLA or so implies 200-250HP @460V. 460V systems are typically 60Hz, implying 1800RPM. The same motor typically will operate on a 380V/50Hz system at 1500RPM, with a HP reduction proportionate to the reduction in RPM or applied Hz. This assumes adequate core for 50 HZ operation. Some deration is required if core losses are elevated due to Hz reduction.
In general terms, windage is reduced according to the fan curves, friction roughly proportionately to speed, and electrical losses proportionately to current reduction.
On a typical V/Hz drive, the output voltage is reduced proportionately to Hz, so at say half speed, half voltage is applied. Half voltage at constant current is roughly half power, assuming PF remains constant. One can simply read the NLA off of the drive at various frequencies and estimate KW flow. Most drives will also display KW, so one could adjust the speed as desired and directly read KW flow from the drive display at the desired speeds.
In general terms, windage is reduced according to the fan curves, friction roughly proportionately to speed, and electrical losses proportionately to current reduction.
On a typical V/Hz drive, the output voltage is reduced proportionately to Hz, so at say half speed, half voltage is applied. Half voltage at constant current is roughly half power, assuming PF remains constant. One can simply read the NLA off of the drive at various frequencies and estimate KW flow. Most drives will also display KW, so one could adjust the speed as desired and directly read KW flow from the drive display at the desired speeds.
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