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pat young
Unregistered guest
Posted From: 71.130.76.20
Rating: N/A
Votes: 0 (Vote!) | | Posted on Tuesday, July 11, 2006 - 9:26 am: | |
Howdy Busnuts
I have a whole bunch of inverters, and have had the old tube type (well, maybe transistor, but it was large, heavy and sucked power like a industrial chupacabra.)
So here's my questions:
Do the new inverters only drain the battery as much as the load is? I think my old one would pull 500 watts as soon as you turned it on, or put even a small load to it. I know a 100 watt light bulb drained the battery in an hour, and it was a good battery, in my little diesel Nissan pickup.
Has anyone used a 1000 watt alternator, for example, and put about a 750 watt load on it (a 7 inch grinder) and had the power come out sorta wimpy? This 1000 watt I bought from Pep Boys, a Vector, didn't put out adequate current to spin the grinder fast. Is that typical? Are most inverters overrated, or only the cheapies? I think now I should always get an inverter about 30-40 per cent bigger than the load.
Thanks folks
Patrick in Fresburg
(559) 251-3814
WheelchairBusProject@yahoo.com |
TomC
Unregistered guest
Posted From: 66.218.33.156
Rating: N/A
Votes: 0 (Vote!) | | Posted on Tuesday, July 11, 2006 - 11:27 am: | |
Certain modified sine wave inverters will not spin up some kind of motors as fast. My fan on the stove hood is slower on inverter, but my bathroom heater is the same!? The only way is to get a true sine wave inverter that is close to real electricity generation as possible. More expensive, but worth it in the long run. Xantrax makes a 2500 watt true sine wave inverter with load sharing, 130 amp battery charger, etc. Check with Dick Wright at Wrico International in Eugene, Or. Good Luck, TomC |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 952
Registered: 5-2001
Posted From: 65.74.70.16
Rating: N/A
Votes: 0 (Vote!) | | Posted on Tuesday, July 11, 2006 - 8:52 pm: | |
Actually, Pat, you should go for much more than 30 to 40% more than your load, where it involves a motor. The reason is that most inverters are rated in volt-amps, not watts.
If a motor has a 60% power factor, this means that true power will only be 60% of the volt-amps. A 2012 inverter will often be rated 2000 volt-amps, but at 60% power factor, you will only get 1200 watts before you overload the inverter.
To answer your first question, most of the modern inverters will be over 80% efficient for most of their range. Some will break 90%, except with very light or very heavy loads.
Most inverters without a search function will take over an amp when idling; given some time, that gets expensive. The search function will cut power consumption down to a fraction of that, so you don't need to turn them off when you are not using the 120 volt power.
To get a feel for the size of a load on the battery when using a 12 volt inverter, multiply the AC line current of the load by 11. A 10 amp grinder could take 110 amps.
Your good battery in your Nissan was never meant to put out that kind of power for any length of time. Four golf cart batteries could do a pretty decent job.
For what it's worth.
Tom Caffrey PD4106-2576
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1609
Registered: 1-2001
Posted From: 70.105.143.139
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Votes: 0 (Vote!) | | Posted on Wednesday, July 12, 2006 - 9:18 am: | |
I was under the impression that power factor does not create an kw load on the source since it is only out of phase current. As long as you do not exceed the amperage rating of the inverter it should not create any problem
However, this would only be true for a true sine wave inverter. The modified sine wave inverters do not like motor loads, period.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
Rating: N/A
Votes: 0 (Vote!) | | Posted on Wednesday, July 12, 2006 - 4:50 pm: | |
I myself have never heard of rating inverters by load power factor. The control electronics will/should go to great lengths to make the current and voltage waveforms stay in phase. If you could graph the duty cycle of the controller's PWM output to the switching transistors, you'd see a close correspondence to the electrical waveform for a resistive load, and a radical departure therefrom as the load becomes more reactive. The effective limits are a function of the beefiness of the power handling components, which translates into a certain ampacity, and the transformer ratio, which determines the peak voltage that can be delivered and thus the ability to oppose a load's back EMF.
Aside from that, doesn't motor power factor approach unity as the motor is loaded, anyhow?
Power factor isn't nearly so much of an issue as starting surge. You can get a cheap inverter, in which case you need massive overcapacity, because it's cheap and can't surge worth a damn, or you can get a good inverter and not have to derate it.
Expect any inverter to consume 10-20 watts at idle; true sine wave inverters will use more than modwave units. Search mode will consume less, but may not keep clocks happy.
And, most notably: inverters such as we're speaking of here aren't designed to run directly from an alternator, which you seem to say you're wanting to do. Regardless, unless you know exactly what conditions (RPM, temperature) are required for your alternator to produce 1KW, how do you know that you're getting it? |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1610
Registered: 1-2001
Posted From: 66.168.162.246
Rating: N/A
Votes: 0 (Vote!) | | Posted on Wednesday, July 12, 2006 - 5:13 pm: | |
High quality induction motors typically have a power factor of 0.90 lagging or higher, at full load although many are rated at 0.80 pf.
During the start cycle the pf may drop as low as 0.40 or lower for a few cycles. This pf current is out of phase with the kw current and is generally what causes problems.
On the other hand synchronous motors typically operate at 1.0 pf and pf is adjustable by changing the excitation current, which is a separate DC power supply.
Richard |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 953
Registered: 5-2001
Posted From: 65.74.70.16
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Votes: 0 (Vote!) | | Posted on Wednesday, July 12, 2006 - 10:56 pm: | |
Richard, I tested a buffing machine on a Trace modified sine wave inverter, using a Kill-A-Watt. It reports volts, amps and power factor, so I thought that it would be a good test.
When the buffer was unloaded, the meter reported 1645 VA and 29% power factor, for a power consumption of 440 watts.
When I stood the macine up on it's brush, the VA dropped to 1545 and gave a 60% power factor for a power consumption of 840 watts.
Now, I realize that the meter could be reading wrong, but I have nothing that says it is. The Trace RV 2012 did a fine job of powering the buffer, but the power consumption that it reported when the buffer was unloaded looked awfully high.
The Trace control panel reported 140 amps unloaded and 175 loaded. If any numbers were off, I would think that they would be the ones reported by the Trace.
For what it's worth.
Tom Caffrey
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1611
Registered: 1-2001
Posted From: 70.105.143.139
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 9:26 am: | |
That is very good information Tom and sounds generally as to what I would expect.
Calculating it out, 440 watts at 0.29 pf equals 1517 VA, as opposed to the 1645 VA indicated.
840 watts at 0.6 pf equals 1400 VA as opposed to the 1545 VA indicated.
So there is some error in the Kill-A-Watt unit but only about 10% which is pretty accurate for that type device.
BTW, most single phase electrical power equipment is rated for 1.0 pf duty and I suspect the inverters may be rated this way. Unfortunately I do not have one available to check.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
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Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 11:43 am: | |
Tom, the general behavior of true power ramping up with load, via the power factor approaching unity, is about what I'd have thought to be the case. The question then becomes: what does the inverter do with the apparent power component of the total power waveform? Does it waste it as heat, buffer it in its capacitors, or recycle it to the batteries?
It'd be nice to see that load test performed on a true sine wave unit, for the following reason: from a design-of-experiments perspective, we really don't know how the Kill-A-Watt meter performs its sampling and derives its measurements, so we thus don't know how it's going to respond to non-sinusoidal waveforms.
Taking a measurement can be an amazingly involved process. Indeed, about all we can gather, from its published spec of "0.2% accuracy", whatever exactly that means, is that such a figure implies a 10 bit A/D converter (1 sign bit, and 9 bits yielding a 1-in-512 parts, or 0.2% resolution). However, one might also make that claim via an 8 bit converter that is oversampled by 4x or more.
Yet another question is whether the unit has a 2 channel A/D, or at least a 2 channel sample and hold that feeds a single channel A/D, so that it can gather simultaneous voltage and current samples, or whether it alternates channel sampling and thus suffers some error.
And, the definition of a base, i.e. "adequate", sampling rate is also open to question. One would like to think that the unit took "many" samples of each half of the waveform, but who knows - e.g. it might only sample positive half cycles, or every other cycle, etc. It might be doing only peak detection, and inferring power factor via the time difference between voltage and current peaks - truly obtuse things are possible, and have likely been done at on at least one occasion.
Etc.
Richard, I seem to recall at least one inverter specification for an operating range of power factors; I think that it was +/- 45 degrees. |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1613
Registered: 1-2001
Posted From: 70.105.143.139
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 1:09 pm: | |
The apparent power caused by a lagging power factor actually utilizes no real power (watts). It is, as it name implies, apparent power.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 1:33 pm: | |
Richard, apparent power still causes I2R heating losses in all the wiring and circuitry that it passes through. While in wiring that may be relatively trivial, in the inverter it could conceivably be more significant, as the currents themselves are still quite real (they are made "apparent" only by virtue of being out of phase with the voltage waveform), and must therefore be processed by the unit. |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1614
Registered: 1-2001
Posted From: 70.105.143.139
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 1:48 pm: | |
Herman, you are correct and I am not disputing that at all.
However, I was taught that I2R losses, where ever they occur, are identified as real power, not apparent power.
Please note that I have been out of school for over 50 years and some of this basic electronic fundamentals training was long in the past with generally no need to go back and refresh. Take pity on an old man.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
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Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 2:56 pm: | |
Richard, no intent at all to beat you over the head or such; it's apparent that apologies are in order. Really! |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1615
Registered: 1-2001
Posted From: 66.168.162.246
Rating: N/A
Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 4:35 pm: | |
No apology needed but accepted nevertheless. I think we are all trying to learn something on these boards and discussions like this helps us all.
Richard |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 954
Registered: 5-2001
Posted From: 65.74.70.16
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Votes: 0 (Vote!) | | Posted on Thursday, July 13, 2006 - 9:08 pm: | |
Richard and Herman, thanks for the comments. What kind of caught my attention was the fact that VA FELL when the buffer was loaded.
I've seen this before when using a motor on straight sine wave power. The motor I was using drew full line current with no load, which did not increase with a load.
I finally drew the conclusion that a motor could well draw full current with no load and that an applied load would mostly just shift the power factor.
I didn't mention it before, but the inverter I was using for the tests gave some conflicting reports for the shunt current and the inverter current. The Trace reports them both on the remote.
At the time, I thought the problem came from the shunt value being an average and the inverter value being instantaneous. For now, I am just regarding that part of the information as suspect.
I do think that the I2R losses are the reason that the inverter carries a VA rating instead of one in watts. Trace also specifies that power factor can be from +1 to -1.
I have noticed that the Magnum brand has SOME of it's published ratings in watts, but that's the only case I can recall.
I've been monkeying around with these things because I have gotten involved in some alternative energy projects in central Washington with some neighbors of property that we own.
We're 2 miles from the grid, and the utilities are thinking that they will be able to get $100k or so per mile of line. I see it as a good excuse to avoid the utilities.
I've got the neighbors saving something like $300 per month in fuel on an investment of $4,000, including labor. They like the payback on that real well.
Thanks for all the help.
Tom Caffrey PD4106-2576
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1617
Registered: 1-2001
Posted From: 70.105.143.139
Rating: N/A
Votes: 0 (Vote!) | | Posted on Friday, July 14, 2006 - 10:28 am: | |
Tom, although I never paid much attention to it, I seem to remember that no load amps was only slightly less than full load amps. The only thing that changed was power factor as apparent power changed to real power. Unless you were monitoring with a power factor instrument there would be no way to tell.
You would not believe the many discussions I have had with customers who complained about high inefficiency because there units were drawing so many amps at low or no load, and thinking they were paying for this.
One thing that many may not be aware of is that household electric meters measure only true power and not apparent power.
Industrial meters may or may not read apparent power and some utilities actually come into the plant and measure apparent power and charge extra if the pf is below a certain number. 0.8 I seem to remember.
Richard |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 956
Registered: 5-2001
Posted From: 65.74.70.16
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Votes: 0 (Vote!) | | Posted on Friday, July 14, 2006 - 10:16 pm: | |
Richard, I was aware thet house meters were supposed to be true power. However, I have seen some real monkeybusiness at our local utility, so I wouldn't trust what they have to say very far.
For example, some of our meters are not connected to the neutral, so I have to believe that whatever current is being monitored is being multiplied by 240 volts.
Does a meter of this kind average the two legs or does it just read one leg?
We one recieved a bill for one month's electricity at a small barn here in town that was for something like $13,000. When I took it to the utility, I was told that the meter had turned over to zero and started over.
I've thought about that story, and I've decided that I wasn't told the whole story, because if what they said was true, they would have sent us a check for a bunch of money!
For what it's worth.
Tom Caffrey PD4106-2576
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1620
Registered: 1-2001
Posted From: 66.168.162.246
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Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 7:51 am: | |
I believe all current is measured before the neutral is developed. As long as both hot legs are monitored, it really does not matter what the neutral amps are. At least that is my belief.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
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Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 11:07 am: | |
Apparently (i.e., it seems that), it's easier and cheaper to measure watts instead of VA, and thus meters, residential ones at least, have thus been wattmeters, particularly since, traditionally, residential loads have been largely resistive in nature and thus exhibit good power factor. So, whatever reactive power you use has basically been free.
As for whether neutral is involved: for 120/240V split phase service, of course it is, as otherwise they'd only be measuring one leg; the meter sums the leg currents.
Industrial loads typically have a higher reactive component, and thus the things Richard speaks of, i.e. about onsite inspections by the utility, placement onsite of power factor correction capacitors, and/or surcharges for low power factor loads, come into play.
And as houses incorporate more complex loads, e.g. electronics with their switching power supplies (e.g. microwave ovens, fluorescent lights, etc.), they become less resistive, and in response utilities are at least contemplating substitution of wattmeters with VA meters.
Customers should pay for the apparent component of their power usage, because a utility's costs to deliver a current flow are irrespective of its phase angle with respect to voltage, e.g. the cost of the metal needed to comprise the required wire gauges throughout their network, and the network's I2R losses. And yes, it is a very real, err, actual, form of inefficiency on the customer's part to use loads that don't themselves perform power factor correction. |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1622
Registered: 1-2001
Posted From: 66.168.162.246
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Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 12:54 pm: | |
Herman, I need a little help understanding this statement:
"As for whether neutral is involved: for 120/240V split phase service, of course it is, as otherwise they'd only be measuring one leg; the meter sums the leg currents"
My impression would be that since all the amps flow in either one or the other of the two hot legs and they are 180 degrees out of phase, and only unbalanced current flows in the neutral, it should be a simple matter to sum these two currents. Please help.
Richard |
herman
Unregistered guest
Posted From: 69.181.162.81
Rating: N/A
Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 2:07 pm: | |
Sorry, Richard, you're right, you will indeed get a correct value if you do measure at *both* hots, which satisfies the N-1 criteria.
However, the neutral is 'developed' upstream of the meter, and passes through the meter box, where it may or may not be used in some way. |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1623
Registered: 1-2001
Posted From: 66.168.162.246
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Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 2:17 pm: | |
OK thanks Herman. I really did not remember whether the neutral was brought in from the pole or not. Guess I could have walked outside and looked!
Richard
(Message edited by drivingmisslazy on July 15, 2006) |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 958
Registered: 5-2001
Posted From: 65.74.70.16
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Votes: 0 (Vote!) | | Posted on Saturday, July 15, 2006 - 9:36 pm: | |
Richard, in the case of no neutral at the meter, the neutral has no connnection to the meter, even though it passes within a few inches.
Herman, are you saying that the L1 to L2 voltage is applied to the coil at the disk, and the field from the current on each leg adds to the disk movement?
If that's the case, then the voltage is just being averaged for each leg, isn't it?
While that wouldn't be very significant where the voltage stays in the normal range, I would think that some brownout conditions might have some pretty large errors, particularly if one leg was dropped by an outage.
Thinking on it, I can see that the utility would lose as easily as the customer if there was an error, so maybe it's not a big deal.
Thanks for the discussion.
Tom Caffrey PD4106-2576
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1624
Registered: 1-2001
Posted From: 66.168.162.246
Rating: N/A
Votes: 0 (Vote!) | | Posted on Sunday, July 16, 2006 - 8:33 am: | |
I really do not think the meter is a voltage device. It is a current (AMP) device. It really should not matter what the voltage is. At least that is what I believe.
I strongly suspect that there are current shunts in there and all the amps flow thru the shunts. The voltage drop across the shunt would be proportional to the amperage flow. Maybe 100 amps would create 1 volt or something like that.
The voltage differential between the two shunts would be immaterial, whether 100 volts or 1000 volts.
Richard
(Message edited by drivingmisslazy on July 16, 2006) |
Tom Caffrey (Pvcces)
Registered Member
Username: Pvcces
Post Number: 960
Registered: 5-2001
Posted From: 65.74.70.16
Rating: N/A
Votes: 0 (Vote!) | | Posted on Sunday, July 16, 2006 - 7:53 pm: | |
Richard, the meter can't read true power if it does not sample voltage, so I believe that it has to use what it can. In this case, the voltage between L1 and L2 is all it's got.
For what it's worth.
Tom Caffrey PD4106-2576
Suncatcher
Ketchikan, Alaska |
Richard Bowyer (Drivingmisslazy)
Registered Member
Username: Drivingmisslazy
Post Number: 1625
Registered: 1-2001
Posted From: 66.168.162.246
Rating: N/A
Votes: 0 (Vote!) | | Posted on Sunday, July 16, 2006 - 8:50 pm: | |
In the case of the home meter, it knows what the nominal voltage is. Maybe they do measure voltage also. I really do not know.
Richard |
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