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| Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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#1
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Another sparkies question about generators
If a motor is used as a generator will it put out about the same
wattage as it consumes when used as a motor? Thanks, Eric |
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#2
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Another sparkies question about generators
Yes, about the same probably ... but ....
Be aware using a motor as a generator is only really practical when considering DC machines. The common induction motor is not easily adapted to generation and is almost never done in a home-shop environment. Bob Swinney "Eric R Snow" wrote in message ... If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric |
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#3
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Another sparkies question about generators
On Tue, 26 Aug 2003 15:21:30 GMT, "Bob Swinney"
wrote: Yes, about the same probably ... but .... Be aware using a motor as a generator is only really practical when considering DC machines. The common induction motor is not easily adapted to generation and is almost never done in a home-shop environment. Bob Swinney Not true! Look he http://www.fortunecity.com/greenfiel.../inductge.html -Carl |
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#4
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Another sparkies question about generators
"Eric R Snow" wrote in message
... If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Well, no. The power it consumes is the mechanical power it puts out, plus electrical and mechanical losses. If you put the same mechanical power in, using it as a generator, you have losses going the *other* way, too. So the difference in power in versus power out, for a fixed value of mechanical power in and out, is the sum of those two sets of losses -- the losses it has as a motor, plus the losses it has as a generator. The efficiency of the motor depends on its design and size. A small DC motor used as a generator starts off in the hole, because its efficiency as a *motor* isn't all that great (maybe 70%). And, at the very best, you have equal losses working the other way. Big, quality motors are much more efficient -- into the 90%+ range. To make an efficient generator out of a brush-type permanent-magnet DC motor, you may have to shift the phase angle of the brushes slightly. I have here in the mess I call my shop an old DC motor that was modified by someone to allow fine adjustments of the brushes for that express purpose. Unfortunately, it's a series/parallel-wound electromagnet motor, which requires a lot of knowledge to get it properly excited. My understanding is that small, modern, permanent magnet DC motors run at something close to their motor efficiency when run as generators, but not quite. Ed Huntress |
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Another sparkies question about generators
On Tue, 26 Aug 2003 16:21:26 GMT, "Ed Huntress"
wrote: Unfortunately, it's a series/parallel-wound electromagnet motor, which requires a lot of knowledge to get it properly excited. I knew a woman like that once, only it was cash, not knowledge. -Carl |
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Another sparkies question about generators
It will be good for the same amount of current and therefore the same
amount of wattage if the voltage is the same. What are you trying to do and what kind of motor are you thinking of using? Induction motors will work well especially if you use the grid to excite them. Dan Eric R Snow wrote in message . .. If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric |
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#7
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Another sparkies question about generators
"Carl Byrns" wrote in message
... On Tue, 26 Aug 2003 16:21:26 GMT, "Ed Huntress" wrote: Unfortunately, it's a series/parallel-wound electromagnet motor, which requires a lot of knowledge to get it properly excited. I knew a woman like that once, only it was cash, not knowledge. -Carl If you can just figure out where to hook the wires, and how much voltage to use... Ed Huntress |
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Another sparkies question about generators
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#9
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Another sparkies question about generators
On Tue, 26 Aug 2003 10:38:51 -0700, Eric R Snow
wrote: On 26 Aug 2003 09:41:36 -0700, (Dan Caster) wrote: It will be good for the same amount of current and therefore the same amount of wattage if the voltage is the same. What are you trying to do and what kind of motor are you thinking of using? Induction motors will work well especially if you use the grid to excite them. Dan Eric R Snow wrote in message . .. If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Well. I was thinking about using a brushless DC motor to act as a generator for boat lights. The ones that have magnets in the center and the rotating field is produced by the drive electronics. If I spin the magnets then I'm hoping to get usable power out of the thing. 100 watts should be plenty. Eric Those old DC servo motors (like old DASDI drive motors) make good generators. |
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Another sparkies question about generators
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Another sparkies question about generators
On Tue, 26 Aug 2003 07:12:47 -0700, Eric R Snow
wrote: If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power. Jim |
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#13
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Another sparkies question about generators
Gary sez:
"Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney "Gary Coffman" wrote in message ... On Tue, 26 Aug 2003 23:57:15 +0100, wrote: On Tue, 26 Aug 2003 07:12:47 -0700, Eric R Snow wrote: If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power. Gary |
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Another sparkies question about generators
"Bob Swinney" wrote in message news:BBR2b.205710$Oz4.54331@rwcrnsc54... Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney Ah, I read it carefully, and I read it the same way that Gary does. It isn't entirely clear why "Operated at the same speed as a generator, 80% of the original motor input power is available as generated output," unless he's talking about armature resistance losses. There are other losses but that seems to be the subject. Then he counts armature resistance losses again, no? Ed Huntress "Gary Coffman" wrote in message ... On Tue, 26 Aug 2003 23:57:15 +0100, wrote: On Tue, 26 Aug 2003 07:12:47 -0700, Eric R Snow wrote: If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power. Gary |
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Another sparkies question about generators
On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney"
wrote: Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney "Gary Coffman" wrote in message .. . On Tue, 26 Aug 2003 23:57:15 +0100, wrote: On Tue, 26 Aug 2003 07:12:47 -0700, Eric R Snow wrote: If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power. Gary Greetings Bob, Maybe you can tell me what is going on with a stepper I just measured. It is rated at 5 volts, 5 ohms, and 7.5 degrees. When spun by hand the resistance to turning is quite high. Spun faster by hand it of course gets way easier to turn. Spinning the shaft in the lathe, with no load, gets voltage readings from 3 volts @ 330 rpm to 14 volts @ 1800 rpm. Those voltage readings are no load. Using a lamp as the load the current measured @ 1800 rpm is .5 amps. At 550 rpm it is about.3 amps. The lamp is a twelve volt lamp. It doesn't glow at all. When the voltage is measured with the lamp connected it drops to .2 volts. Id s it because there are so many poles on the rotor that it behaves this way? If all the coils were wired in parallel and the rotor replaced with one with just 4 poles would this improve the situation? Thanks, Eric |
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#16
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Another sparkies question about generators
On Tue, 26 Aug 2003 14:47:48 -0700, Eric R Snow
wrote: I thought about using a big stepper I have but it saturates at the rpm that it would be running at. At least I think that's what happens. If all the wires are twisted together and the shaft spun the resistance increases and then at a certain rpm drops dramatically. I suppose if there were less poles in the rotor this wouldn't happen. ERS You are getting AC out of the servo - and poly phase at that. You need some diodes to make them work properly. Tying wires together makes the phases work against each other. |
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Another sparkies question about generators
Eric,
Sorry, but any thing I could offer would only be a guess. Steppers aren't my cup of tea but I do know they are classed as synchronous motors. Here's a guess though - your stepper rated 5 volts, 5 ohms; this implies the motor is maybe rated at only 5 watts, if I'm "guessing" correctly. Bob Swinney "Eric R Snow" wrote in message ... On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney" wrote: Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney "Gary Coffman" wrote in message .. . On Tue, 26 Aug 2003 23:57:15 +0100, wrote: On Tue, 26 Aug 2003 07:12:47 -0700, Eric R Snow wrote: If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Only if it's 100% efficient. The simplest case to look at is a low speed permanent magnet DC motor where the armature resistance is the major loss component. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. For the same internal heating, an 80% efficient machine used at the same speed as a generator will deliver about 0.8 x 0.8 = 64% of its rated motor input power. Gary Greetings Bob, Maybe you can tell me what is going on with a stepper I just measured. It is rated at 5 volts, 5 ohms, and 7.5 degrees. When spun by hand the resistance to turning is quite high. Spun faster by hand it of course gets way easier to turn. Spinning the shaft in the lathe, with no load, gets voltage readings from 3 volts @ 330 rpm to 14 volts @ 1800 rpm. Those voltage readings are no load. Using a lamp as the load the current measured @ 1800 rpm is .5 amps. At 550 rpm it is about.3 amps. The lamp is a twelve volt lamp. It doesn't glow at all. When the voltage is measured with the lamp connected it drops to .2 volts. Id s it because there are so many poles on the rotor that it behaves this way? If all the coils were wired in parallel and the rotor replaced with one with just 4 poles would this improve the situation? Thanks, Eric |
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#18
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Another sparkies question about generators
On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney" wrote:
Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney In this case I'm not ready to grant Jim deity status. Perhaps you'd better go back and read it twice, Bob. Here it is. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Now this is true. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. This is at best misleading. There is no "original motor input power" here. The motor isn't being used as a motor. It is being used as a generator. There's *external* mechanical power driving the motor shaft. This could be equal to 50% of the "original motor input power", or it could be 80% of the "original motor input power", or it could be 100% of the "original motor input power", or it could be 200% of the "original motor input power". In other words, it is *totally independent* of whatever power the motor may have drawn when being used as a motor. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. *Now* he can account for armature losses. If they are 20%, then the generated output will be 20% less than the mechanical input power applied to the generator. Gary |
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Another sparkies question about generators
"Eric R Snow" wrote in message ... If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Check out this site. http://www.qsl.net/ns8o/Induction_Generator.html Richard W. |
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Another sparkies question about generators
On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney"
wrote: Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. I respect both contributors, agree with Gary here. Loss is I^2 R regardless of which way the current is going. Power out is power in minus I^2 R, whether power "in" is electrical or mechanical. I think the reason the stepper motor wimps out at higher speeds has to do with coil inductance. Reactance grows with frequency which is proportional to speed. The same issue limits how fast they can be driven with given available voltage. I wonder if a brushless DC motor would commutate without external excitation. Something has to supply the electronics that makes it brushless. What are you going to spin it with, Eric? Most electric-start engines (including outboards) already have alternators capable of a good deal more than 100 watts. My local surplus store sometimes has DCPM brush-type 12-volt motors. I tested a couple in my collection today. One has resistance of 0.121 ohms, so at 8.33 amps (100 watts at 12 volts out) the loss would be 8.4 watts and the efficiency would be 100/108.4 or 92%. The price sticker on it says $5.95 but I've had it a while. Ya never know what the Ax-Man will have on any given day, but if you'd like I'd be glad to make the selfless sacrifice (snort!) of visiting the Ax-Man to see what he might have for you. This particular motor weighs 3 lb. I'm near Minneapolis. |
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Another sparkies question about generators
Don Foreman wrote: (clip) I respect both contributors, agree with Gary here (clip) ^^^^^^^^^^^^^^ I respect everybody, but I agree with Gary also. Look at it this way. Say, you have this 80% efficient motor running, and drawing 1 kw. Then 800 watts of power will appear at the output shaft, and 200 watts will be lost as heat in the electrical and maehanical losses of the motor. It is the motor's ability to dissipate heat that determines the motor rating. If you load it more, it will develop more power, but overheat. Now, let's say we couple this motor to the shaft of an identical motor, hooked up as a generator. It is receiving 800 watts of power as input on the shaft, and, being 80% efficient, it will put out 640 watts of power, and throwing off 160 watts as heat. In this cnfiguration, you to count the losses twice, because you are running two machines in series. But that wasn't the question. The question was, can the motor, run as a generator, develop about the same power? Note that the second unit is only dissipating 160 watts of heat, so it is not being run at capacity. If you had the power available from some other source (windmill, waterwheel, gas engine...whatever) you could boost the load until the heat losses reached the design limit of 200 watts, and you would be running the motor as a generator at the same power it is rated for as a motor. So, you may count the losses twice only if you are running two units. And you CAN load to the same power level, regardless of which way the power is flowing. |
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Another sparkies question about generators
Seems we are having basic math problems again. Look at it this way (which
is presumably what Jim Pentagrid meant). You have a power rating for a motor. The question was, how much of the *motor's power rating* is available if it is run as a generator? Assume the motor is 80% efficient and that it is exactly as capable of being a generator as it was a motor; i.e., also 80%. Take the input power rating for the motor X 80%. This is the input power rating of that motor ran as a generator. Incidentally that amount of power is input as mechanical motion. Now, the motor-as-generator, is 80% efficient. The mechanical input power is converted to electrical power at 80%. That would be 80 X 80 or aprox. 64% of the original input power rating for the motor. Bob Swinney "Gary Coffman" wrote in message ... On Tue, 26 Aug 2003 22:57:37 GMT, "Bob Swinney" wrote: Gary sez: "Why do you think you can count the same resistance twice? The current only goes through it once, regardless of whether it is being used as a motor or a generator." He is not counting anything twice. Go back and read the post carefully. You can rely on anything that Jim says as gospel. Bob Swinney In this case I'm not ready to grant Jim deity status. Perhaps you'd better go back and read it twice, Bob. Here it is. With an 80% efficient motor, 20% of the input power is lost in armature resistance. Now this is true. Operated at the same speed as a generator, 80% of the original motor input power is available as generated output. This is at best misleading. There is no "original motor input power" here. The motor isn't being used as a motor. It is being used as a generator. There's *external* mechanical power driving the motor shaft. This could be equal to 50% of the "original motor input power", or it could be 80% of the "original motor input power", or it could be 100% of the "original motor input power", or it could be 200% of the "original motor input power". In other words, it is *totally independent* of whatever power the motor may have drawn when being used as a motor. However it now has to travel through the armature resistance before it reaches the output terminals and this loses a further 20% of the available power. *Now* he can account for armature losses. If they are 20%, then the generated output will be 20% less than the mechanical input power applied to the generator. Gary |
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#23
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Another sparkies question about generators
On Wed, 27 Aug 2003 06:52:42 -0700, Eric R Snow wrote:
On Wed, 27 Aug 2003 00:39:31 GMT, clare @ snyder.on .ca wrote: On Tue, 26 Aug 2003 14:47:48 -0700, Eric R Snow wrote: I thought about using a big stepper I have but it saturates at the rpm that it would be running at. At least I think that's what happens. If all the wires are twisted together and the shaft spun the resistance increases and then at a certain rpm drops dramatically. I suppose if there were less poles in the rotor this wouldn't happen. ERS You are getting AC out of the servo - and poly phase at that. You need some diodes to make them work properly. Tying wires together makes the phases work against each other. If I tie the wires in pairs, so there are 4 pairs in an 8 wire motor the effect is the same. I know AC is coming out. That makes no difference when measuring the output when used as agenerator. The load was resistive and the meter set to AC. ERS How are you combining the outputs of the pairs? Since each pair produces a different phase, you can't just combine them across a resistive load. You'll have phase cancellation if you try that. But if you diode rectify each phase, you can then combine them. For maximum output, you need to use a bridge rectifier for each phase (each winding), then combine all the DC + and all the DC - leads from all the bridges to form the net output to feed to the resistive load. (If you're sure you know the winding senses, you could star connect the windings and use individual rectifiers, but the bridge is insensitive to phasing sense, and produces 1.41 times as much output voltage too.) Alternatively, if you don't want to rectify to DC, you can use a separate resistive load for each pair. Each load only gets a fraction of the total output, but at least you won't have phase cancellation. This would require 4 bulbs for your lighting application, but that's not necessarily bad. It gives you redundancy. Note that saturation isn't what would limit current at high RPM (you can't saturate a PM alternator). What could is the increase in reactance of each winding as the frequency (RPM) is increased. Inductive reactance increases proportionate to frequency. This reactance is effectively in series with the load, and will limit the maximum current the load can draw. But that shouldn't be an issue for any reasonable RPM. Note that conventional auto alternators are 6 phase (2 more than you have), and they work up to 10,000 RPM. So I don't think that the number of poles is an issue. Gary |
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#24
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Another sparkies question about generators
On Wed, 27 Aug 2003 14:56:12 GMT, "Bob Swinney" wrote:
Seems we are having basic math problems again. Look at it this way (which is presumably what Jim Pentagrid meant). You have a power rating for a motor. The question was, how much of the *motor's power rating* is available if it is run as a generator? Assume the motor is 80% efficient and that it is exactly as capable of being a generator as it was a motor; i.e., also 80%. Take the input power rating for the motor X 80%. This is the input power rating of that motor ran as a generator. Incidentally that amount of power is input as mechanical motion. Now, the motor-as-generator, is 80% efficient. The mechanical input power is converted to electrical power at 80%. That would be 80 X 80 or aprox. 64% of the original input power rating for the motor. No. You can't count the efficiency twice. The motor is 80% efficient, so it can convert 80% of the electrical power fed to it to mechanical motion. Conversely, when used as a generator, it can convert 80% of the mechanical power fed to it to electrical power. In both cases, the missing 20% goes to heating the armature resistance (ignoring bearing and windage losses). You can't *count that twice* because it can't be used as a motor and a generator *simultaneously*. It is used as one or the other, and in each case, efficiency is 80%, *not* 64%. Lets run some numbers. Suppose we have a motor rated at 1 hp electrical (746 watts) input. If it is 80% efficient, then it will produce 0.8 hp (596.8 watts) on its output shaft. Now if we decide to use it as a generator, we feed it 1 hp (746 watts) mechanical on its *input* shaft, and then we can extract 0.8 hp (596.8 watts) electrical on its *output* terminals. All we've done is show that, as an electrical machine, the motor is reversible. Efficiency is the same either way. Now I'll grant that if you were silly enough to only feed the generator with 0.8 hp on its input shaft, you'd only get 477.4 watts out of it. But that doesn't mean its efficiency has been reduced to 64%. You just arbitrarily decided not to feed in enough power to get 596.8 watts out at the generator's 80% conversion efficiency. Gary |
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#25
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Another sparkies question about generators
Bob Swinney wrote:
.... Assume the motor is 80% efficient and that it is exactly as capable of being a generator as it was a motor; i.e., also 80%. Take the input power rating for the motor X 80%. This is the input power rating of that motor ran as a generator. ... no no no, this is the _output_ power rating used as a generator!! So the input to get the rated output would be 100/80 = 125% of rating. A 20% loss in the windings gives 100% of rating as output. Bob |
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#26
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Another sparkies question about generators
Nope. The first figure arrived at was the power input rating of the motor.
Other calculations are based on this. Bob Swinneuy "Gary Coffman" wrote in message ... On Wed, 27 Aug 2003 14:56:12 GMT, "Bob Swinney" wrote: Seems we are having basic math problems again. Look at it this way (which is presumably what Jim Pentagrid meant). You have a power rating for a motor. The question was, how much of the *motor's power rating* is available if it is run as a generator? Assume the motor is 80% efficient and that it is exactly as capable of being a generator as it was a motor; i.e., also 80%. Take the input power rating for the motor X 80%. This is the input power rating of that motor ran as a generator. Incidentally that amount of power is input as mechanical motion. Now, the motor-as-generator, is 80% efficient. The mechanical input power is converted to electrical power at 80%. That would be 80 X 80 or aprox. 64% of the original input power rating for the motor. No. You can't count the efficiency twice. The motor is 80% efficient, so it can convert 80% of the electrical power fed to it to mechanical motion. Conversely, when used as a generator, it can convert 80% of the mechanical power fed to it to electrical power. In both cases, the missing 20% goes to heating the armature resistance (ignoring bearing and windage losses). You can't *count that twice* because it can't be used as a motor and a generator *simultaneously*. It is used as one or the other, and in each case, efficiency is 80%, *not* 64%. Lets run some numbers. Suppose we have a motor rated at 1 hp electrical (746 watts) input. If it is 80% efficient, then it will produce 0.8 hp (596.8 watts) on its output shaft. Now if we decide to use it as a generator, we feed it 1 hp (746 watts) mechanical on its *input* shaft, and then we can extract 0.8 hp (596.8 watts) electrical on its *output* terminals. All we've done is show that, as an electrical machine, the motor is reversible. Efficiency is the same either way. Now I'll grant that if you were silly enough to only feed the generator with 0.8 hp on its input shaft, you'd only get 477.4 watts out of it. But that doesn't mean its efficiency has been reduced to 64%. You just arbitrarily decided not to feed in enough power to get 596.8 watts out at the generator's 80% conversion efficiency. Gary |
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#27
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Another sparkies question about generators
On Wed, 27 Aug 2003 11:15:57 -0400, Gary Coffman
wrote: On Wed, 27 Aug 2003 06:52:42 -0700, Eric R Snow wrote: On Wed, 27 Aug 2003 00:39:31 GMT, clare @ snyder.on .ca wrote: On Tue, 26 Aug 2003 14:47:48 -0700, Eric R Snow wrote: I thought about using a big stepper I have but it saturates at the rpm that it would be running at. At least I think that's what happens. If all the wires are twisted together and the shaft spun the resistance increases and then at a certain rpm drops dramatically. I suppose if there were less poles in the rotor this wouldn't happen. ERS You are getting AC out of the servo - and poly phase at that. You need some diodes to make them work properly. Tying wires together makes the phases work against each other. If I tie the wires in pairs, so there are 4 pairs in an 8 wire motor the effect is the same. I know AC is coming out. That makes no difference when measuring the output when used as agenerator. The load was resistive and the meter set to AC. ERS How are you combining the outputs of the pairs? Since each pair produces a different phase, you can't just combine them across a resistive load. You'll have phase cancellation if you try that. But if you diode rectify each phase, you can then combine them. For maximum output, you need to use a bridge rectifier for each phase (each winding), then combine all the DC + and all the DC - leads from all the bridges to form the net output to feed to the resistive load. (If you're sure you know the winding senses, you could star connect the windings and use individual rectifiers, but the bridge is insensitive to phasing sense, and produces 1.41 times as much output voltage too.) Alternatively, if you don't want to rectify to DC, you can use a separate resistive load for each pair. Each load only gets a fraction of the total output, but at least you won't have phase cancellation. This would require 4 bulbs for your lighting application, but that's not necessarily bad. It gives you redundancy. Note that saturation isn't what would limit current at high RPM (you can't saturate a PM alternator). What could is the increase in reactance of each winding as the frequency (RPM) is increased. Inductive reactance increases proportionate to frequency. This reactance is effectively in series with the load, and will limit the maximum current the load can draw. But that shouldn't be an issue for any reasonable RPM. Note that conventional auto alternators are 6 phase (2 more than you have), and they work up to 10,000 RPM. So I don't think that the number of poles is an issue. Gary Well, I wanted to use a stepper but couldn't get one to put out enough juice. But, a drill motor from HF draws 11 amps when loaded. It spins a gearbox with a 36 t0 1 ratio with 500 rpm out. So the motor spins at 18,000 rpm. Spinning this same motor at 10,000 rpm gives 9 volts @ 1.5 amps for 13.5 watts. Since I was looking for 12 watts that will be O.K.. If LEDs are used with a six volt battery,( the one from the drill), then that should be plenty of light. I need to figure out how to mount the motor with a belt drive to get the proper rpm and seal the motor so it lasts more than a week exposed to salt air. ERS |
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Another sparkies question about generators
I have a string trimmer permanent magnet motor in my junk waiting for
a possibel project. You can have it if you want it. I will dig it out and let you know what I can. It is several times larger than a bike generator and could be what you want. Dan Eric R Snow wrote in message news The outboard is tiny. 2 or 3 hp I think. Bicycle generators put out about 6 watts. 20 watts is my goal. This motor is a pull start. Thanks for the offer of finding a small motor. I am still looking through my stuff for something that will work. ERS |
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Another sparkies question about generators
I did not recall correctly. The string trimmer motor is a series
wound motor and does not work well as a generator. Dan Eric R Snow wrote in message The outboard is tiny. 2 or 3 hp I think. Bicycle generators put out about 6 watts. 20 watts is my goal. This motor is a pull start. Thanks for the offer of finding a small motor. I am still looking through my stuff for something that will work. ERS |
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Another sparkies question about generators
"Richard W." wrote in message ...
"Eric R Snow" wrote in message ... If a motor is used as a generator will it put out about the same wattage as it consumes when used as a motor? Thanks, Eric Check out this site. http://www.qsl.net/ns8o/Induction_Generator.html Richard W. String trimmer motors make pretty decent cieling fans powered off a PV solar panel. |
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Another sparkies question about generators
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Another sparkies question about generators
On Wed, 27 Aug 2003 06:52:42 -0700, Eric R Snow
wrote: On Wed, 27 Aug 2003 00:39:31 GMT, clare @ snyder.on .ca wrote: On Tue, 26 Aug 2003 14:47:48 -0700, Eric R Snow wrote: I thought about using a big stepper I have but it saturates at the rpm that it would be running at. At least I think that's what happens. If all the wires are twisted together and the shaft spun the resistance increases and then at a certain rpm drops dramatically. I suppose if there were less poles in the rotor this wouldn't happen. ERS You are getting AC out of the servo - and poly phase at that. You need some diodes to make them work properly. Tying wires together makes the phases work against each other. If I tie the wires in pairs, so there are 4 pairs in an 8 wire motor the effect is the same. I know AC is coming out. That makes no difference when measuring the output when used as agenerator. The load was resistive and the meter set to AC. ERS Just wanted to be sure you understood the wiring of a polyphase device. If you get out of phase windings in parallel, you get low output and lots of heat. |
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Another sparkies question about generators
On Wed, 27 Aug 2003 23:13:24 -0400, Ned Simmons
wrote: In article , says... I wonder if a brushless DC motor would commutate without external excitation. Something has to supply the electronics that makes it brushless. Every brushless motor I've worked with has had a PM armature--no excitation required. Else why would it be called brushless? Separate issues. No field excitation is required because it's PM. It's brushless because electronics accomplish commutation, but electronics don't do anything without voltage to work with. Though I've never tried it, I'd expect one would make a very good 3-phase generator. They certainly do a good job of pumping up the bus voltage on the servo amp when delivering negative torque to an overhauling load. I agree, but it may be necessary to substitute diodes for the internal elex. For 12 watts a battery would be a lot simpler if perhaps not nearly as much fun. |
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Another sparkies question about generators
SNIP
For 12 watts a battery would be a lot simpler if perhaps not nearly as much fun. Not as much fun and not guaranteed to have power for the lights as long as the motor is running. Besides, I don't want a big battery pack. Now, don't tell me about reliability and how more things to supply power just means more thing to go wrong  .ERS |
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Another sparkies question about generators
Eric R Snow wrote:
For 12 watts a battery would be a lot simpler if perhaps not nearly as much fun. Not as much fun and not guaranteed to have power for the lights as long as the motor is running. Besides, I don't want a big battery pack. Now, don't tell me about reliability and how more things to supply power just means more thing to go wrong .How about a small 12V battery, say 8 to 12 Ah, and two generators or alternators, one belt driven off the auxiliary, if it doesn't already have one, and one on the mast wind driven. Ted |
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Another sparkies question about generators
On Thu, 28 Aug 2003 06:59:19 -0700, Eric R Snow
wrote: SNIP For 12 watts a battery would be a lot simpler if perhaps not nearly as much fun. Not as much fun and not guaranteed to have power for the lights as long as the motor is running. Besides, I don't want a big battery pack. Now, don't tell me about reliability and how more things to supply power just means more thing to go wrong .ERS I would much rather help you succeed than be a failure forecaster! Have you thought about if you'll need voltage regulation, and if so how you'll accomplish that? You may want a motor with rated voltage higher than 12 volts. Most outboards max out at about 5000 to 5500 RPM but you may want lights at less than wide-open throttle. Small DCPM motors typically spin pretty fast at rated voltage so they'd need to spin similarly fast to generate rated voltage. However, a higher-voltage motor will produce more volts per RPM because they have more turns on the windings. 90 volts is a common DCPM voltage, I don't think a 1-amp 90VDC motor would be very big and it would make 12 volts at fairly low RPM. I'm going to Ax-Man today to see what they have in motors. |
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Another sparkies question about generators
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Another sparkies question about generators
On Thu, 28 Aug 2003 12:02:27 -0500, Don Foreman
wrote: On Thu, 28 Aug 2003 06:59:19 -0700, Eric R Snow wrote: SNIP For 12 watts a battery would be a lot simpler if perhaps not nearly as much fun. Not as much fun and not guaranteed to have power for the lights as long as the motor is running. Besides, I don't want a big battery pack. Now, don't tell me about reliability and how more things to supply power just means more thing to go wrong .ERS I would much rather help you succeed than be a failure forecaster! Have you thought about if you'll need voltage regulation, and if so how you'll accomplish that? You may want a motor with rated voltage higher than 12 volts. Most outboards max out at about 5000 to 5500 RPM but you may want lights at less than wide-open throttle. Small DCPM motors typically spin pretty fast at rated voltage so they'd need to spin similarly fast to generate rated voltage. However, a higher-voltage motor will produce more volts per RPM because they have more turns on the windings. 90 volts is a common DCPM voltage, I don't think a 1-amp 90VDC motor would be very big and it would make 12 volts at fairly low RPM. I'm going to Ax-Man today to see what they have in motors. The motor I tested that comes closest to what I want is 2.25 long and 1.5 dia.. At 10,000 rpm it was putting out 9 volts @ 1.5 amps. It was $10.00 at H.F.. It came complete with a drill chuck, batteries, charger and a drill motor case wrapped around it. For 8 bucks they have one with a keyed chuck. If two were connected in series then spun to 15,000 rpm max then with a cheap regulator the batteries could be charged and the lights lit. Now, if a similar sized motor came along with a higher voltage output so only one would be needed then that would be great. I'm curious to see what Ax-Man has. ERS |
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Another sparkies question about generators
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Another sparkies question about generators
On Thu, 28 Aug 2003 16:56:10 -0400, Ned Simmons
wrote: Every brushless motor I've worked with has had a PM armature--no excitation required. Else why would it be called brushless? Separate issues. No field excitation is required because it's PM. I thought that's what I said? I only meant that not requiring field excitation is not why it's called brushless. Some PM motors do have brushes. Though I've never tried it, I'd expect one would make a very good 3-phase generator. They certainly do a good job of pumping up the bus voltage on the servo amp when delivering negative torque to an overhauling load. I agree, but it may be necessary to substitute diodes for the internal elex. I assume you're talking about substituting diodes for the output stage of the amp here. Obviously if you're using the motor as a generator, there's no need to actively commutate the motor windings, but you do need to rectify the motor's output if you want DC. If the brushless motor has internal commutation electronics they may need to be circumvented with diodes. Maybe some use external amps? |
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