I have read many of the posts regarding so called "magic boxes and energy savers" and it is apparent most of them are selling snake oil. I have a friend that is considering a sales job with this company http://www.powerdexems.com/smartplug.aspx selling yet another energy saving device, any thoughts?
Energy Saver?
- Thread starter CooperToo
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Arround here we use these;http://www.cabelas.com/catalog/prod...8891&type=product&rid=10&WTz_l=YMAL;IK-229563
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110405-2355 EDT
CooperToo:
The description is sort of hogwash. But if this reduces a high input voltage, or at least the volt-time integral over one half cycle, then it might provide some power saving.
Get one of the devices. Get a Kill-A-Watt power meter, in the right places about $25 or so. Make sure it works down to 85 V, some don't. Get a 15 A Variac, and a half horsepower water pump that can be run dry.
Under no load adjust the input voltage to the Kill-A-Watt from 95 V to 135 V in 5 V steps. Monitor the power, watts, and not VA to the pump motor at each voltage and plot. Next put the pump in a tank of water with a hose of 10 or more feet vertically to provide a load on the motor. Make the hose output elevation high enough to nearly fully load the motor. Repeat the test.
Then put the power saving device after the Kill-A-Watt and rerun both tests.
Very important to report back the results of these tests.
.
CooperToo:
The description is sort of hogwash. But if this reduces a high input voltage, or at least the volt-time integral over one half cycle, then it might provide some power saving.
Get one of the devices. Get a Kill-A-Watt power meter, in the right places about $25 or so. Make sure it works down to 85 V, some don't. Get a 15 A Variac, and a half horsepower water pump that can be run dry.
Under no load adjust the input voltage to the Kill-A-Watt from 95 V to 135 V in 5 V steps. Monitor the power, watts, and not VA to the pump motor at each voltage and plot. Next put the pump in a tank of water with a hose of 10 or more feet vertically to provide a load on the motor. Make the hose output elevation high enough to nearly fully load the motor. Repeat the test.
Then put the power saving device after the Kill-A-Watt and rerun both tests.
Very important to report back the results of these tests.
.
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
While this device appears to be a soft start unit with a current transformer that gives a feed back as to the amount of current the motor is pulling, which kind of doesn't make sense as this would work backwards from the description they claim.
Here is a PDF on it
http://www.powerdexems.com/docs/productdocs_smartplug.pdf
while a soft start will reduce the power use during start up, the savings would only reflect the amount of starting up the motor, and the graph shows that the unit at full load is the same with or without the unit in-line?
While soft start technology has been around for years and predates VFDs, which offer much greater savings if properly implemented into a very controlled system, like that that has been done with constant velocity well pumps.
There reference to saving on a peak demand charge would not even apply to a dwelling.
I can see some savings, but from the fact they are claiming that motors are running under loaded in factory design systems like AC units, refrigerators, well pumps and such I don't see it.
Here is a PDF on it
http://www.powerdexems.com/docs/productdocs_smartplug.pdf
while a soft start will reduce the power use during start up, the savings would only reflect the amount of starting up the motor, and the graph shows that the unit at full load is the same with or without the unit in-line?
While soft start technology has been around for years and predates VFDs, which offer much greater savings if properly implemented into a very controlled system, like that that has been done with constant velocity well pumps.
There reference to saving on a peak demand charge would not even apply to a dwelling.
I can see some savings, but from the fact they are claiming that motors are running under loaded in factory design systems like AC units, refrigerators, well pumps and such I don't see it.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110406-1353 EDT
Data on a cheap import drill press. Mechanical load near constant consisting of spindle not drilling, belt, bearings, and windage.
Volts -- Watts
100 ---- 302
110 ---- 319
120 ---- 348
130 ---- 389
With a heavier mechanical load there should be less variation with voltage.
.
Data on a cheap import drill press. Mechanical load near constant consisting of spindle not drilling, belt, bearings, and windage.
Volts -- Watts
100 ---- 302
110 ---- 319
120 ---- 348
130 ---- 389
With a heavier mechanical load there should be less variation with voltage.
.
Electric-Light
Senior Member
It's often required by the PoCo when the starts/hour and/or the hp size of motor is disagreeable with PoCo policies for across the line starting.
Even wye-delta starting is a form of soft starting. Starting a 480v motor in wye and switching to delta with a centrifugal switch or a timer would reduce starting current as if the motor was started on 277v service.
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
For refrigerator's, AC compressors, disposals, dishwashers, clothes dryers? in a dwelling?
see the PDF: http://www.powerdexems.com/docs/productdocs_smartplug.pdf
And look at the PS-1B on the left
Last edited:
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110407-1047 EDT
If the mechanical load on the output of a motor is unchanged, then how do you reduce power consumption on the input side of the motor? The obvious answer is to increase the motor efficiency.
Thus, how do you do that? You lower the losses in the motor. How do you do that from outside the motor?
If frequency and waveform are unchanged, then I think the only possibility is to adjust the input voltage to whatever value optimizes efficiency. At full load this may be a rather flat curve. I don't know. At zero or light load I believe this is achieved by operating at a low but reasonable voltage.
If one distorts the waveform, then I do not know if efficiency can be improved. But my guess is that a sine wave input at some optimum voltage for the load will provide the greatest efficiency.
What is the basic theory of the device manufactured by PowerDEX? It is not clear, but my guess is that in some fashion they are adjusting the volt-time integral to the motor. This may have some merit.
The kind of experiment I described in an earlier post needs to be performed. An old Sears pump I have shows a large variation in input power under no load as voltage is adjusted. Similar to the drill press motor I presented in a previous post. I am not ready to run the water pumping experiment yet. CooperToo --- you and your friend should be running this of experiment. You are most directly involved with the question.
A residential refrigerator of the on-off type is unlikely to see much increase in efficiency unless the supply voltage is rather high. This motor will be running at a design maximum load.
.
If the mechanical load on the output of a motor is unchanged, then how do you reduce power consumption on the input side of the motor? The obvious answer is to increase the motor efficiency.
Thus, how do you do that? You lower the losses in the motor. How do you do that from outside the motor?
If frequency and waveform are unchanged, then I think the only possibility is to adjust the input voltage to whatever value optimizes efficiency. At full load this may be a rather flat curve. I don't know. At zero or light load I believe this is achieved by operating at a low but reasonable voltage.
If one distorts the waveform, then I do not know if efficiency can be improved. But my guess is that a sine wave input at some optimum voltage for the load will provide the greatest efficiency.
What is the basic theory of the device manufactured by PowerDEX? It is not clear, but my guess is that in some fashion they are adjusting the volt-time integral to the motor. This may have some merit.
The kind of experiment I described in an earlier post needs to be performed. An old Sears pump I have shows a large variation in input power under no load as voltage is adjusted. Similar to the drill press motor I presented in a previous post. I am not ready to run the water pumping experiment yet. CooperToo --- you and your friend should be running this of experiment. You are most directly involved with the question.
A residential refrigerator of the on-off type is unlikely to see much increase in efficiency unless the supply voltage is rather high. This motor will be running at a design maximum load.
.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110411-1247 EDT
First, some data from one of my freezers:
Just after the beginning of the of the cycle
Volts .... Watts
105 ....... 309
110 ....... 317
120 ....... 325
125 ....... 334
130 ....... 345
A few minutes later, but not yet at steady-state
Volts .... Watts
100 ....... 293
105 ....... 295
110 ....... 298
120 ....... 313
125 ....... 322
Note: 295/322 = 0.916, an 8% reduction.
This freezer runs about a 50% duty cycle. Thus, the energy consumption is about 0.150*8760 = 1314 KWH/year. My cost about 1314*0.132 = $174/year. If I reduced this by 10%, then the saving is $17/year.
Based upon what I believe the Smart Plug does and somewhat confirmed by talking to them I believe the device will save some energy. I would want their small device to be priced in the $25 to $50 range, but guess what its price is? $400. Knowing roughly what would be in the box it is way over priced.
On to the technical aspects. I would not call it a scam. The limited description is probably not too inaccurate. The description was not written by a electrical engineer in the United States. It might result by translation from a description by a foreign engineer. The PowerDEX company in Nevada is not likely where this was designed and being manufactured. Note: it is a west coast company so you can not call too early. The person I talked with has some knowledge about the product, is not an electrical engineer, and has some experimental experience with the product.
Their estimated saving is possible in some cases, but this needs to be considered in terms of the product cost and its relationship to the load to which it is connected.
They say it should not or can not be used with things that include electronics. It should be made perfectly clear that this device is designed to be used with a single induction motor. They describe it as using the peak of the sine wave. More likely it is a Triac or SCR with phase shift control to reduce the output voltage.
Clearly if applied to my freezer there would be no reasonable payback period.
They gave me an illustration of where there is a good payback time frame. A 1 HP pump motor for a pool that is run continuously. They also admit that they need to find those applications where the load device runs continuously, and the rating of the load is near the rating the Smart Plug. The load also probably needs to run near its rating, and there is a high supply voltage. They have real world data from the pump application that indicates a payback within about 3 years.
Without the device to play with I can not judge how the improvement varies with input voltage. My guess is that if the nominal voltage is around 110 V there won't be much improvement, whereas, at 125 V their percentages might work out. No way would I speed $400 to experiment with this device. At $50 I might.
I would not call this a scam, but rather a product with limited application at its current price.
.
First, some data from one of my freezers:
Just after the beginning of the of the cycle
Volts .... Watts
105 ....... 309
110 ....... 317
120 ....... 325
125 ....... 334
130 ....... 345
A few minutes later, but not yet at steady-state
Volts .... Watts
100 ....... 293
105 ....... 295
110 ....... 298
120 ....... 313
125 ....... 322
Note: 295/322 = 0.916, an 8% reduction.
This freezer runs about a 50% duty cycle. Thus, the energy consumption is about 0.150*8760 = 1314 KWH/year. My cost about 1314*0.132 = $174/year. If I reduced this by 10%, then the saving is $17/year.
Based upon what I believe the Smart Plug does and somewhat confirmed by talking to them I believe the device will save some energy. I would want their small device to be priced in the $25 to $50 range, but guess what its price is? $400. Knowing roughly what would be in the box it is way over priced.
On to the technical aspects. I would not call it a scam. The limited description is probably not too inaccurate. The description was not written by a electrical engineer in the United States. It might result by translation from a description by a foreign engineer. The PowerDEX company in Nevada is not likely where this was designed and being manufactured. Note: it is a west coast company so you can not call too early. The person I talked with has some knowledge about the product, is not an electrical engineer, and has some experimental experience with the product.
Their estimated saving is possible in some cases, but this needs to be considered in terms of the product cost and its relationship to the load to which it is connected.
They say it should not or can not be used with things that include electronics. It should be made perfectly clear that this device is designed to be used with a single induction motor. They describe it as using the peak of the sine wave. More likely it is a Triac or SCR with phase shift control to reduce the output voltage.
Clearly if applied to my freezer there would be no reasonable payback period.
They gave me an illustration of where there is a good payback time frame. A 1 HP pump motor for a pool that is run continuously. They also admit that they need to find those applications where the load device runs continuously, and the rating of the load is near the rating the Smart Plug. The load also probably needs to run near its rating, and there is a high supply voltage. They have real world data from the pump application that indicates a payback within about 3 years.
Without the device to play with I can not judge how the improvement varies with input voltage. My guess is that if the nominal voltage is around 110 V there won't be much improvement, whereas, at 125 V their percentages might work out. No way would I speed $400 to experiment with this device. At $50 I might.
I would not call this a scam, but rather a product with limited application at its current price.
.
Electric-Light
Senior Member
Well, the wattage maybe going down, but do you know if the efficiency is increasing?
It's fairly possible that the rotational speed is decreasing with increase in slip caused by reduced voltage which in turn reduces the refrigeration capacity and increasing the duty cycle.
Many refrigeration equipment are dual rated for 50/60 Hz. It spins slower at 50Hz and not surprisingly, the rated refrigeration capacity is lower at 50Hz.
It's fairly possible that the rotational speed is decreasing with increase in slip caused by reduced voltage which in turn reduces the refrigeration capacity and increasing the duty cycle.
Many refrigeration equipment are dual rated for 50/60 Hz. It spins slower at 50Hz and not surprisingly, the rated refrigeration capacity is lower at 50Hz.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110412-0706 EST
Electric-Light:
I can be quite sure that the lower wattage was a function of improved motor efficiency. Although a better test would be to measure the total energy used over exactly two or three refrigeration cycles.
Note that in going from 110 to 105 to 100 that there was little change in power vs that from 110 to 125 in my second batch of measurements.
.
Electric-Light:
I can be quite sure that the lower wattage was a function of improved motor efficiency. Although a better test would be to measure the total energy used over exactly two or three refrigeration cycles.
Note that in going from 110 to 105 to 100 that there was little change in power vs that from 110 to 125 in my second batch of measurements.
.
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
I can see this since pole slippage would be proportional to the load/voltage.
I wonder if it can sense the current going back up when pole slippage starts, this would maximize the efficiency by lowering the voltage just to the point of pole slippage but not allowing the motor to go below this point which would increase current, DC motors operated on AC like drill and saw motors would have a problem as the RMP is based upon the voltage and current, so I don't think this device would work for them, many kitchen type hand tools, IE; mixers have DC type motors in them.
This circuit is kind of a reverse current regulator, so instead of dropping the voltage as current increases, it increases voltage if current increases.
I think I just answered my question in red, with the above statement:roll:
Last edited:
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
If the load is reduced on a AC motor, then voltage can be reduced also without slippage, if slippage does occur it will increase the current, which will increase the wattage, the point is with this devise is to balance the voltage at the point the slippage starts, RPM is a function of the frequency of the AC not voltage.
A very good example is try to pull the armature out of a solenoid at the first it is easy to move it, but as your pulling it out it gets harder, this is because the current is increasing, put an amp meter on it, and you will see the current climb very fast as you pull it out. this is why relays have very little hold in current, but much higher pull in current.
gar
Senior Member
- Location
- Ann Arbor, Michigan
- Occupation
- EE
110415-0852 EDT
hurk27:
In an induction motor slip does not just start, it always exists even at no load because there is actually some load. That load is windage and rotating friction. For small values of slip the magnitude of slip is approximately proportional to torque load. See p 184 of "Alternating-Current Machinery", by Bailey and Gault, 1951. The entire chapter 12 is on induction motors. In the range from 0 to 5% slip it is similar to a DC PM motor relative to slip vs torque, approximately linear.
The induced frequency in the rotor is proportional to slip. Thus, at small values of slip the frequency is very low and the inductive reactance in the rotor can be ignored compared to the resistive component. Induced voltage in the rotor is proportional to slip frequency. Thus, current in the rotor for small values of slip is defined by the rotor resistance and rotor induced voltage. Since R is approximately constant then rotor I is proportional to slip.
A synchronous motor is quite different relative to slip and torque. There is no slip until synchronization is lost. There is just a mechanical phase shift of rotor position relative to the rotating magnetic vector from the stator. This phase shift is approximately proportional to torque load.
.
hurk27:
In an induction motor slip does not just start, it always exists even at no load because there is actually some load. That load is windage and rotating friction. For small values of slip the magnitude of slip is approximately proportional to torque load. See p 184 of "Alternating-Current Machinery", by Bailey and Gault, 1951. The entire chapter 12 is on induction motors. In the range from 0 to 5% slip it is similar to a DC PM motor relative to slip vs torque, approximately linear.
The induced frequency in the rotor is proportional to slip. Thus, at small values of slip the frequency is very low and the inductive reactance in the rotor can be ignored compared to the resistive component. Induced voltage in the rotor is proportional to slip frequency. Thus, current in the rotor for small values of slip is defined by the rotor resistance and rotor induced voltage. Since R is approximately constant then rotor I is proportional to slip.
A synchronous motor is quite different relative to slip and torque. There is no slip until synchronization is lost. There is just a mechanical phase shift of rotor position relative to the rotating magnetic vector from the stator. This phase shift is approximately proportional to torque load.
.
Electric-Light
Senior Member
Ideally yes, but all except synchronous motors slip. Some substantially so.
Mechanical clocks and timers that do not contain any time base are very accurate and they use a small synchronous motor which obeys frequency strictly.
Good examples are multi-speed HVAC fans, desk fans etc. When you switch speed, the fan turns at a different speed, because of intentional slippage and as such they're extremely inefficient. If you over-volt them, they'll spin faster, but no faster than field speed (1800 or 3600)
Motor is 3450RPM in North America 2840RPM in Europe. (and their dividends)
The absolute fastest a line frequency induction motor can turn is 3,600 RPM on 60Hz.
A very good example is try to pull the armature out of a solenoid at the first it is easy to move it, but as your pulling it out it gets harder, this is because the current is increasing, put an amp meter on it, and you will see the current climb very fast as you pull it out. this is why relays have very little hold in current, but much higher pull in current.[/QUOTE]
Electric-Light
Senior Member
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